标签归档:computer-vision

如何消除数独方块中的凸度缺陷?

问题:如何消除数独方块中的凸度缺陷?

我当时在做一个有趣的项目:使用OpenCV(如Google护目镜等)从输入图像中解决数独问题。我已经完成了任务,但是最后我遇到了一个小问题。

我使用OpenCV 2.3.1的Python API进行了编程。

以下是我所做的:

  1. 读取图像
  2. 找到轮廓
  3. 选择面积最大的那个(也有些等同于正方形的那个)。
  4. 找到拐角点。

    例如下面给出:

    在此处输入图片说明

    请注意,绿线正确地与数独的真实边界重合,因此数独可以正确变形。请检查下一张图片)

  5. 使图像变形为完美的正方形

    例如图片:

    在此处输入图片说明

  6. 执行OCR(为此我使用了我在OpenCV-Python的简单数字识别OCR中给出的方法)

而且该方法效果很好。

问题:

退房 这张图片。

在此图像上执行步骤4会得到以下结果:

在此处输入图片说明

画出的红线是原始轮廓,是数独边界的真实轮廓。

画出的绿线是近似轮廓,它将是变形图像的轮廓。

数独顶部的绿线和红线之间当然有区别。因此,在扭曲时,我并没有获得数独的原始边界。

我的问题 :

如何在数独的正确边界(即红线)上扭曲图像,或者如何消除红线和绿线之间的差异?OpenCV中有什么方法吗?

I was doing a fun project: Solving a Sudoku from an input image using OpenCV (as in Google goggles etc). And I have completed the task, but at the end I found a little problem for which I came here.

I did the programming using Python API of OpenCV 2.3.1.

Below is what I did :

  1. Read the image
  2. Find the contours
  3. Select the one with maximum area, ( and also somewhat equivalent to square).
  4. Find the corner points.

    e.g. given below:

    enter image description here

    (Notice here that the green line correctly coincides with the true boundary of the Sudoku, so the Sudoku can be correctly warped. Check next image)

  5. warp the image to a perfect square

    eg image:

    enter image description here

  6. Perform OCR ( for which I used the method I have given in Simple Digit Recognition OCR in OpenCV-Python )

And the method worked well.

Problem:

Check out this image.

Performing the step 4 on this image gives the result below:

enter image description here

The red line drawn is the original contour which is the true outline of sudoku boundary.

The green line drawn is approximated contour which will be the outline of warped image.

Which of course, there is difference between green line and red line at the top edge of sudoku. So while warping, I am not getting the original boundary of the Sudoku.

My Question :

How can I warp the image on the correct boundary of the Sudoku, i.e. the red line OR how can I remove the difference between red line and green line? Is there any method for this in OpenCV?


回答 0

我有一个可行的解决方案,但是您必须自己将其转换为OpenCV。它用Mathematica编写。

第一步是通过将每个像素除以关闭操作的结果来调整图像的亮度:

src = ColorConvert[Import["http://davemark.com/images/sudoku.jpg"], "Grayscale"];
white = Closing[src, DiskMatrix[5]];
srcAdjusted = Image[ImageData[src]/ImageData[white]]

在此处输入图片说明

下一步是找到数独区域,因此我可以忽略(遮盖)背景。为此,我使用连接组件分析,然后选择凸面面积最大的组件:

components = 
  ComponentMeasurements[
    ColorNegate@Binarize[srcAdjusted], {"ConvexArea", "Mask"}][[All, 
    2]];
largestComponent = Image[SortBy[components, First][[-1, 2]]]

在此处输入图片说明

通过填充此图像,我得到了数独网格的蒙版:

mask = FillingTransform[largestComponent]

在此处输入图片说明

现在,我可以使用二阶导数滤波器在两个单独的图像中查找垂直线和水平线:

lY = ImageMultiply[MorphologicalBinarize[GaussianFilter[srcAdjusted, 3, {2, 0}], {0.02, 0.05}], mask];
lX = ImageMultiply[MorphologicalBinarize[GaussianFilter[srcAdjusted, 3, {0, 2}], {0.02, 0.05}], mask];

在此处输入图片说明

我再次使用连接的分量分析从这些图像中提取网格线。网格线比数字长得多,因此我可以使用卡尺长度来仅选择与网格线相连的组件。按位置对它们进行排序,对于图像中的每个垂直/水平网格线,我得到2×10的蒙版图像:

verticalGridLineMasks = 
  SortBy[ComponentMeasurements[
      lX, {"CaliperLength", "Centroid", "Mask"}, # > 100 &][[All, 
      2]], #[[2, 1]] &][[All, 3]];
horizontalGridLineMasks = 
  SortBy[ComponentMeasurements[
      lY, {"CaliperLength", "Centroid", "Mask"}, # > 100 &][[All, 
      2]], #[[2, 2]] &][[All, 3]];

在此处输入图片说明

接下来,我将每对垂直/水平网格线进行放大,将它们放大,计算出像素间的交点,并计算结果的中心。这些点是网格线的交点:

centerOfGravity[l_] := 
 ComponentMeasurements[Image[l], "Centroid"][[1, 2]]
gridCenters = 
  Table[centerOfGravity[
    ImageData[Dilation[Image[h], DiskMatrix[2]]]*
     ImageData[Dilation[Image[v], DiskMatrix[2]]]], {h, 
    horizontalGridLineMasks}, {v, verticalGridLineMasks}];

在此处输入图片说明

最后一步是为通过这些点的X / Y映射定义两个插值函数,并使用这些函数变换图像:

fnX = ListInterpolation[gridCenters[[All, All, 1]]];
fnY = ListInterpolation[gridCenters[[All, All, 2]]];
transformed = 
 ImageTransformation[
  srcAdjusted, {fnX @@ Reverse[#], fnY @@ Reverse[#]} &, {9*50, 9*50},
   PlotRange -> {{1, 10}, {1, 10}}, DataRange -> Full]

在此处输入图片说明

所有操作都是基本的图像处理功能,因此在OpenCV中也应该可行。基于样条的图像转换可能会更困难,但我认为您并不是真的需要它。可能使用您现在在每个单个单元格上使用的透视变换,将获得足够好的结果。

I have a solution that works, but you’ll have to translate it to OpenCV yourself. It’s written in Mathematica.

The first step is to adjust the brightness in the image, by dividing each pixel with the result of a closing operation:

src = ColorConvert[Import["http://davemark.com/images/sudoku.jpg"], "Grayscale"];
white = Closing[src, DiskMatrix[5]];
srcAdjusted = Image[ImageData[src]/ImageData[white]]

enter image description here

The next step is to find the sudoku area, so I can ignore (mask out) the background. For that, I use connected component analysis, and select the component that’s got the largest convex area:

components = 
  ComponentMeasurements[
    ColorNegate@Binarize[srcAdjusted], {"ConvexArea", "Mask"}][[All, 
    2]];
largestComponent = Image[SortBy[components, First][[-1, 2]]]

enter image description here

By filling this image, I get a mask for the sudoku grid:

mask = FillingTransform[largestComponent]

enter image description here

Now, I can use a 2nd order derivative filter to find the vertical and horizontal lines in two separate images:

lY = ImageMultiply[MorphologicalBinarize[GaussianFilter[srcAdjusted, 3, {2, 0}], {0.02, 0.05}], mask];
lX = ImageMultiply[MorphologicalBinarize[GaussianFilter[srcAdjusted, 3, {0, 2}], {0.02, 0.05}], mask];

enter image description here

I use connected component analysis again to extract the grid lines from these images. The grid lines are much longer than the digits, so I can use caliper length to select only the grid lines-connected components. Sorting them by position, I get 2×10 mask images for each of the vertical/horizontal grid lines in the image:

verticalGridLineMasks = 
  SortBy[ComponentMeasurements[
      lX, {"CaliperLength", "Centroid", "Mask"}, # > 100 &][[All, 
      2]], #[[2, 1]] &][[All, 3]];
horizontalGridLineMasks = 
  SortBy[ComponentMeasurements[
      lY, {"CaliperLength", "Centroid", "Mask"}, # > 100 &][[All, 
      2]], #[[2, 2]] &][[All, 3]];

enter image description here

Next I take each pair of vertical/horizontal grid lines, dilate them, calculate the pixel-by-pixel intersection, and calculate the center of the result. These points are the grid line intersections:

centerOfGravity[l_] := 
 ComponentMeasurements[Image[l], "Centroid"][[1, 2]]
gridCenters = 
  Table[centerOfGravity[
    ImageData[Dilation[Image[h], DiskMatrix[2]]]*
     ImageData[Dilation[Image[v], DiskMatrix[2]]]], {h, 
    horizontalGridLineMasks}, {v, verticalGridLineMasks}];

enter image description here

The last step is to define two interpolation functions for X/Y mapping through these points, and transform the image using these functions:

fnX = ListInterpolation[gridCenters[[All, All, 1]]];
fnY = ListInterpolation[gridCenters[[All, All, 2]]];
transformed = 
 ImageTransformation[
  srcAdjusted, {fnX @@ Reverse[#], fnY @@ Reverse[#]} &, {9*50, 9*50},
   PlotRange -> {{1, 10}, {1, 10}}, DataRange -> Full]

enter image description here

All of the operations are basic image processing function, so this should be possible in OpenCV, too. The spline-based image transformation might be harder, but I don’t think you really need it. Probably using the perspective transformation you use now on each individual cell will give good enough results.


回答 1

Nikie的答案解决了我的问题,但他的答案是在Mathematica中。因此,我认为我应该在这里给出其OpenCV改编版。但是在实施之后,我可以看到OpenCV代码比nikie的mathematica代码大得多。而且,我在OpenCV中找不到nikie完成的插值方法(尽管可以使用scipy完成,但是我会在时间到时告诉它。)

1.图像预处理(关闭操作)

import cv2
import numpy as np

img = cv2.imread('dave.jpg')
img = cv2.GaussianBlur(img,(5,5),0)
gray = cv2.cvtColor(img,cv2.COLOR_BGR2GRAY)
mask = np.zeros((gray.shape),np.uint8)
kernel1 = cv2.getStructuringElement(cv2.MORPH_ELLIPSE,(11,11))

close = cv2.morphologyEx(gray,cv2.MORPH_CLOSE,kernel1)
div = np.float32(gray)/(close)
res = np.uint8(cv2.normalize(div,div,0,255,cv2.NORM_MINMAX))
res2 = cv2.cvtColor(res,cv2.COLOR_GRAY2BGR)

结果:

结帐结果

2.找到数独广场并创建蒙版图像

thresh = cv2.adaptiveThreshold(res,255,0,1,19,2)
contour,hier = cv2.findContours(thresh,cv2.RETR_TREE,cv2.CHAIN_APPROX_SIMPLE)

max_area = 0
best_cnt = None
for cnt in contour:
    area = cv2.contourArea(cnt)
    if area > 1000:
        if area > max_area:
            max_area = area
            best_cnt = cnt

cv2.drawContours(mask,[best_cnt],0,255,-1)
cv2.drawContours(mask,[best_cnt],0,0,2)

res = cv2.bitwise_and(res,mask)

结果:

在此处输入图片说明

3.查找垂直线

kernelx = cv2.getStructuringElement(cv2.MORPH_RECT,(2,10))

dx = cv2.Sobel(res,cv2.CV_16S,1,0)
dx = cv2.convertScaleAbs(dx)
cv2.normalize(dx,dx,0,255,cv2.NORM_MINMAX)
ret,close = cv2.threshold(dx,0,255,cv2.THRESH_BINARY+cv2.THRESH_OTSU)
close = cv2.morphologyEx(close,cv2.MORPH_DILATE,kernelx,iterations = 1)

contour, hier = cv2.findContours(close,cv2.RETR_EXTERNAL,cv2.CHAIN_APPROX_SIMPLE)
for cnt in contour:
    x,y,w,h = cv2.boundingRect(cnt)
    if h/w > 5:
        cv2.drawContours(close,[cnt],0,255,-1)
    else:
        cv2.drawContours(close,[cnt],0,0,-1)
close = cv2.morphologyEx(close,cv2.MORPH_CLOSE,None,iterations = 2)
closex = close.copy()

结果:

在此处输入图片说明

4.查找水平线

kernely = cv2.getStructuringElement(cv2.MORPH_RECT,(10,2))
dy = cv2.Sobel(res,cv2.CV_16S,0,2)
dy = cv2.convertScaleAbs(dy)
cv2.normalize(dy,dy,0,255,cv2.NORM_MINMAX)
ret,close = cv2.threshold(dy,0,255,cv2.THRESH_BINARY+cv2.THRESH_OTSU)
close = cv2.morphologyEx(close,cv2.MORPH_DILATE,kernely)

contour, hier = cv2.findContours(close,cv2.RETR_EXTERNAL,cv2.CHAIN_APPROX_SIMPLE)
for cnt in contour:
    x,y,w,h = cv2.boundingRect(cnt)
    if w/h > 5:
        cv2.drawContours(close,[cnt],0,255,-1)
    else:
        cv2.drawContours(close,[cnt],0,0,-1)

close = cv2.morphologyEx(close,cv2.MORPH_DILATE,None,iterations = 2)
closey = close.copy()

结果:

在此处输入图片说明

当然,这不是很好。

5.查找网格点

res = cv2.bitwise_and(closex,closey)

结果:

在此处输入图片说明

6.纠正缺陷

在这里,nikie进行某种插值,对此我了解不多。而且我找不到此OpenCV的任何相应功能。(也许在那里,我不知道)。

查看此SOF,它说明了如何使用SciPy进行此操作,我不想使用它:OpenCV中的图像转换

因此,在这里,我将每个子正方形的四个角用作每个变角透视图。

为此,首先我们找到质心。

contour, hier = cv2.findContours(res,cv2.RETR_LIST,cv2.CHAIN_APPROX_SIMPLE)
centroids = []
for cnt in contour:
    mom = cv2.moments(cnt)
    (x,y) = int(mom['m10']/mom['m00']), int(mom['m01']/mom['m00'])
    cv2.circle(img,(x,y),4,(0,255,0),-1)
    centroids.append((x,y))

但是结果质心将不会被排序。查看下图以查看其顺序:

在此处输入图片说明

因此,我们从左到右,从上到下对它们进行排序。

centroids = np.array(centroids,dtype = np.float32)
c = centroids.reshape((100,2))
c2 = c[np.argsort(c[:,1])]

b = np.vstack([c2[i*10:(i+1)*10][np.argsort(c2[i*10:(i+1)*10,0])] for i in xrange(10)])
bm = b.reshape((10,10,2))

现在看下面他们的命令:

在此处输入图片说明

最后,我们应用转换并创建尺寸为450×450的新图像。

output = np.zeros((450,450,3),np.uint8)
for i,j in enumerate(b):
    ri = i/10
    ci = i%10
    if ci != 9 and ri!=9:
        src = bm[ri:ri+2, ci:ci+2 , :].reshape((4,2))
        dst = np.array( [ [ci*50,ri*50],[(ci+1)*50-1,ri*50],[ci*50,(ri+1)*50-1],[(ci+1)*50-1,(ri+1)*50-1] ], np.float32)
        retval = cv2.getPerspectiveTransform(src,dst)
        warp = cv2.warpPerspective(res2,retval,(450,450))
        output[ri*50:(ri+1)*50-1 , ci*50:(ci+1)*50-1] = warp[ri*50:(ri+1)*50-1 , ci*50:(ci+1)*50-1].copy()

结果:

在此处输入图片说明

结果几乎与nikie相同,但是代码长度很大。也许可以使用更好的方法,但是在那之前,这种方法行之有效。

关于方舟。

Nikie’s answer solved my problem, but his answer was in Mathematica. So I thought I should give its OpenCV adaptation here. But after implementing I could see that OpenCV code is much bigger than nikie’s mathematica code. And also, I couldn’t find interpolation method done by nikie in OpenCV ( although it can be done using scipy, i will tell it when time comes.)

1. Image PreProcessing ( closing operation )

import cv2
import numpy as np

img = cv2.imread('dave.jpg')
img = cv2.GaussianBlur(img,(5,5),0)
gray = cv2.cvtColor(img,cv2.COLOR_BGR2GRAY)
mask = np.zeros((gray.shape),np.uint8)
kernel1 = cv2.getStructuringElement(cv2.MORPH_ELLIPSE,(11,11))

close = cv2.morphologyEx(gray,cv2.MORPH_CLOSE,kernel1)
div = np.float32(gray)/(close)
res = np.uint8(cv2.normalize(div,div,0,255,cv2.NORM_MINMAX))
res2 = cv2.cvtColor(res,cv2.COLOR_GRAY2BGR)

Result :

Result of closing

2. Finding Sudoku Square and Creating Mask Image

thresh = cv2.adaptiveThreshold(res,255,0,1,19,2)
contour,hier = cv2.findContours(thresh,cv2.RETR_TREE,cv2.CHAIN_APPROX_SIMPLE)

max_area = 0
best_cnt = None
for cnt in contour:
    area = cv2.contourArea(cnt)
    if area > 1000:
        if area > max_area:
            max_area = area
            best_cnt = cnt

cv2.drawContours(mask,[best_cnt],0,255,-1)
cv2.drawContours(mask,[best_cnt],0,0,2)

res = cv2.bitwise_and(res,mask)

Result :

enter image description here

3. Finding Vertical lines

kernelx = cv2.getStructuringElement(cv2.MORPH_RECT,(2,10))

dx = cv2.Sobel(res,cv2.CV_16S,1,0)
dx = cv2.convertScaleAbs(dx)
cv2.normalize(dx,dx,0,255,cv2.NORM_MINMAX)
ret,close = cv2.threshold(dx,0,255,cv2.THRESH_BINARY+cv2.THRESH_OTSU)
close = cv2.morphologyEx(close,cv2.MORPH_DILATE,kernelx,iterations = 1)

contour, hier = cv2.findContours(close,cv2.RETR_EXTERNAL,cv2.CHAIN_APPROX_SIMPLE)
for cnt in contour:
    x,y,w,h = cv2.boundingRect(cnt)
    if h/w > 5:
        cv2.drawContours(close,[cnt],0,255,-1)
    else:
        cv2.drawContours(close,[cnt],0,0,-1)
close = cv2.morphologyEx(close,cv2.MORPH_CLOSE,None,iterations = 2)
closex = close.copy()

Result :

enter image description here

4. Finding Horizontal Lines

kernely = cv2.getStructuringElement(cv2.MORPH_RECT,(10,2))
dy = cv2.Sobel(res,cv2.CV_16S,0,2)
dy = cv2.convertScaleAbs(dy)
cv2.normalize(dy,dy,0,255,cv2.NORM_MINMAX)
ret,close = cv2.threshold(dy,0,255,cv2.THRESH_BINARY+cv2.THRESH_OTSU)
close = cv2.morphologyEx(close,cv2.MORPH_DILATE,kernely)

contour, hier = cv2.findContours(close,cv2.RETR_EXTERNAL,cv2.CHAIN_APPROX_SIMPLE)
for cnt in contour:
    x,y,w,h = cv2.boundingRect(cnt)
    if w/h > 5:
        cv2.drawContours(close,[cnt],0,255,-1)
    else:
        cv2.drawContours(close,[cnt],0,0,-1)

close = cv2.morphologyEx(close,cv2.MORPH_DILATE,None,iterations = 2)
closey = close.copy()

Result :

enter image description here

Of course, this one is not so good.

5. Finding Grid Points

res = cv2.bitwise_and(closex,closey)

Result :

enter image description here

6. Correcting the defects

Here, nikie does some kind of interpolation, about which I don’t have much knowledge. And i couldn’t find any corresponding function for this OpenCV. (may be it is there, i don’t know).

Check out this SOF which explains how to do this using SciPy, which I don’t want to use : Image transformation in OpenCV

So, here I took 4 corners of each sub-square and applied warp Perspective to each.

For that, first we find the centroids.

contour, hier = cv2.findContours(res,cv2.RETR_LIST,cv2.CHAIN_APPROX_SIMPLE)
centroids = []
for cnt in contour:
    mom = cv2.moments(cnt)
    (x,y) = int(mom['m10']/mom['m00']), int(mom['m01']/mom['m00'])
    cv2.circle(img,(x,y),4,(0,255,0),-1)
    centroids.append((x,y))

But resulting centroids won’t be sorted. Check out below image to see their order:

enter image description here

So we sort them from left to right, top to bottom.

centroids = np.array(centroids,dtype = np.float32)
c = centroids.reshape((100,2))
c2 = c[np.argsort(c[:,1])]

b = np.vstack([c2[i*10:(i+1)*10][np.argsort(c2[i*10:(i+1)*10,0])] for i in xrange(10)])
bm = b.reshape((10,10,2))

Now see below their order :

enter image description here

Finally we apply the transformation and create a new image of size 450×450.

output = np.zeros((450,450,3),np.uint8)
for i,j in enumerate(b):
    ri = i/10
    ci = i%10
    if ci != 9 and ri!=9:
        src = bm[ri:ri+2, ci:ci+2 , :].reshape((4,2))
        dst = np.array( [ [ci*50,ri*50],[(ci+1)*50-1,ri*50],[ci*50,(ri+1)*50-1],[(ci+1)*50-1,(ri+1)*50-1] ], np.float32)
        retval = cv2.getPerspectiveTransform(src,dst)
        warp = cv2.warpPerspective(res2,retval,(450,450))
        output[ri*50:(ri+1)*50-1 , ci*50:(ci+1)*50-1] = warp[ri*50:(ri+1)*50-1 , ci*50:(ci+1)*50-1].copy()

Result :

enter image description here

The result is almost same as nikie’s, but code length is large. May be, better methods are available out there, but until then, this works OK.

Regards ARK.


回答 2

您可以尝试对任意扭曲使用某种基于网格的建模。而且,由于数独已经是一个网格,所以它应该不会太难。

因此,您可以尝试检测每个3×3子区域的边界,然后分别对每个区域进行变形。如果检测成功,它将为您提供更好的近似值。

You could try to use some kind of grid based modeling of you arbitrary warping. And since the sudoku already is a grid, that shouldn’t be too hard.

So you could try to detect the boundaries of each 3×3 subregion and then warp each region individually. If the detection succeeds it would give you a better approximation.


回答 3

我想补充一点,上述方法仅在数独板直立时才有效,否则高度/宽度(反之亦然)的比率测试很可能会失败,并且您将无法检测数独的边缘。(我还想补充一点,如果不垂直于图像边界的线,则sobel操作(dx和dy)将仍然有效,因为线相对于两个轴仍然具有边缘。)

为了能够检测直线,您应该进行轮廓分析或逐像素分析,例如ContourArea / boundingRectArea,左上角和右下角点…

编辑:我设法通过应用线性回归并检查错误来检查一组轮廓是否形成一条线。但是,当直线的斜率太大(即> 1000)或非常接近0时,线性回归的效果较差。因此,在线性回归之前应用上述比率测试(在大多数赞成的答案中)是合乎逻辑的,对我来说确实有用。

I want to add that above method works only when sudoku board stands straight, otherwise height/width (or vice versa) ratio test will most probably fail and you will not be able to detect edges of sudoku. (I also want to add that if lines that are not perpendicular to the image borders, sobel operations (dx and dy) will still work as lines will still have edges with respect to both axes.)

To be able to detect straight lines you should work on contour or pixel-wise analysis such as contourArea/boundingRectArea, top left and bottom right points…

Edit: I managed to check whether a set of contours form a line or not by applying linear regression and checking the error. However linear regression performed poorly when slope of the line is too big (i.e. >1000) or it is very close to 0. Therefore applying the ratio test above (in most upvoted answer) before linear regression is logical and did work for me.


回答 4

为了去除未切割的角,我应用了伽玛校正,其伽玛值为0.8。

伽玛校正之前

绘制红色圆圈以显示缺少的角。

伽玛校正后

代码是:

gamma = 0.8
invGamma = 1/gamma
table = np.array([((i / 255.0) ** invGamma) * 255
                  for i in np.arange(0, 256)]).astype("uint8")
cv2.LUT(img, table, img)

如果缺少某些关键点,这是对Abid Rahman的回答的补充。

To remove undected corners I applied gamma correction with a gamma value of 0.8.

Before gamma correction

The red circle is drawn to show the missing corner.

After gamma correction

The code is:

gamma = 0.8
invGamma = 1/gamma
table = np.array([((i / 255.0) ** invGamma) * 255
                  for i in np.arange(0, 256)]).astype("uint8")
cv2.LUT(img, table, img)

This is in addition to Abid Rahman’s answer if some corner points are missing.


回答 5

我认为这是一个很棒的帖子,也是ARK的一个很好的解决方案。很好地布置和解释。

我正在研究类似的问题,并完成了整个工作。进行了一些更改(例如,xrange到range,cv2.findContours中的参数),但是应该可以立即使用(Python 3.5,Anaconda)。

这是上述元素的汇编,并添加了一些缺少的代码(即,标记点)。

'''

/programming/10196198/how-to-remove-convexity-defects-in-a-sudoku-square

'''

import cv2
import numpy as np

img = cv2.imread('test.png')

winname="raw image"
cv2.namedWindow(winname)
cv2.imshow(winname, img)
cv2.moveWindow(winname, 100,100)


img = cv2.GaussianBlur(img,(5,5),0)

winname="blurred"
cv2.namedWindow(winname)
cv2.imshow(winname, img)
cv2.moveWindow(winname, 100,150)

gray = cv2.cvtColor(img,cv2.COLOR_BGR2GRAY)
mask = np.zeros((gray.shape),np.uint8)
kernel1 = cv2.getStructuringElement(cv2.MORPH_ELLIPSE,(11,11))

winname="gray"
cv2.namedWindow(winname)
cv2.imshow(winname, gray)
cv2.moveWindow(winname, 100,200)

close = cv2.morphologyEx(gray,cv2.MORPH_CLOSE,kernel1)
div = np.float32(gray)/(close)
res = np.uint8(cv2.normalize(div,div,0,255,cv2.NORM_MINMAX))
res2 = cv2.cvtColor(res,cv2.COLOR_GRAY2BGR)

winname="res2"
cv2.namedWindow(winname)
cv2.imshow(winname, res2)
cv2.moveWindow(winname, 100,250)

 #find elements
thresh = cv2.adaptiveThreshold(res,255,0,1,19,2)
img_c, contour,hier = cv2.findContours(thresh,cv2.RETR_TREE,cv2.CHAIN_APPROX_SIMPLE)

max_area = 0
best_cnt = None
for cnt in contour:
    area = cv2.contourArea(cnt)
    if area > 1000:
        if area > max_area:
            max_area = area
            best_cnt = cnt

cv2.drawContours(mask,[best_cnt],0,255,-1)
cv2.drawContours(mask,[best_cnt],0,0,2)

res = cv2.bitwise_and(res,mask)

winname="puzzle only"
cv2.namedWindow(winname)
cv2.imshow(winname, res)
cv2.moveWindow(winname, 100,300)

# vertical lines
kernelx = cv2.getStructuringElement(cv2.MORPH_RECT,(2,10))

dx = cv2.Sobel(res,cv2.CV_16S,1,0)
dx = cv2.convertScaleAbs(dx)
cv2.normalize(dx,dx,0,255,cv2.NORM_MINMAX)
ret,close = cv2.threshold(dx,0,255,cv2.THRESH_BINARY+cv2.THRESH_OTSU)
close = cv2.morphologyEx(close,cv2.MORPH_DILATE,kernelx,iterations = 1)

img_d, contour, hier = cv2.findContours(close,cv2.RETR_EXTERNAL,cv2.CHAIN_APPROX_SIMPLE)
for cnt in contour:
    x,y,w,h = cv2.boundingRect(cnt)
    if h/w > 5:
        cv2.drawContours(close,[cnt],0,255,-1)
    else:
        cv2.drawContours(close,[cnt],0,0,-1)
close = cv2.morphologyEx(close,cv2.MORPH_CLOSE,None,iterations = 2)
closex = close.copy()

winname="vertical lines"
cv2.namedWindow(winname)
cv2.imshow(winname, img_d)
cv2.moveWindow(winname, 100,350)

# find horizontal lines
kernely = cv2.getStructuringElement(cv2.MORPH_RECT,(10,2))
dy = cv2.Sobel(res,cv2.CV_16S,0,2)
dy = cv2.convertScaleAbs(dy)
cv2.normalize(dy,dy,0,255,cv2.NORM_MINMAX)
ret,close = cv2.threshold(dy,0,255,cv2.THRESH_BINARY+cv2.THRESH_OTSU)
close = cv2.morphologyEx(close,cv2.MORPH_DILATE,kernely)

img_e, contour, hier = cv2.findContours(close,cv2.RETR_EXTERNAL,cv2.CHAIN_APPROX_SIMPLE)

for cnt in contour:
    x,y,w,h = cv2.boundingRect(cnt)
    if w/h > 5:
        cv2.drawContours(close,[cnt],0,255,-1)
    else:
        cv2.drawContours(close,[cnt],0,0,-1)

close = cv2.morphologyEx(close,cv2.MORPH_DILATE,None,iterations = 2)
closey = close.copy()

winname="horizontal lines"
cv2.namedWindow(winname)
cv2.imshow(winname, img_e)
cv2.moveWindow(winname, 100,400)


# intersection of these two gives dots
res = cv2.bitwise_and(closex,closey)

winname="intersections"
cv2.namedWindow(winname)
cv2.imshow(winname, res)
cv2.moveWindow(winname, 100,450)

# text blue
textcolor=(0,255,0)
# points green
pointcolor=(255,0,0)

# find centroids and sort
img_f, contour, hier = cv2.findContours(res,cv2.RETR_LIST,cv2.CHAIN_APPROX_SIMPLE)
centroids = []
for cnt in contour:
    mom = cv2.moments(cnt)
    (x,y) = int(mom['m10']/mom['m00']), int(mom['m01']/mom['m00'])
    cv2.circle(img,(x,y),4,(0,255,0),-1)
    centroids.append((x,y))

# sorting
centroids = np.array(centroids,dtype = np.float32)
c = centroids.reshape((100,2))
c2 = c[np.argsort(c[:,1])]

b = np.vstack([c2[i*10:(i+1)*10][np.argsort(c2[i*10:(i+1)*10,0])] for i in range(10)])
bm = b.reshape((10,10,2))

# make copy
labeled_in_order=res2.copy()

for index, pt in enumerate(b):
    cv2.putText(labeled_in_order,str(index),tuple(pt),cv2.FONT_HERSHEY_DUPLEX, 0.75, textcolor)
    cv2.circle(labeled_in_order, tuple(pt), 5, pointcolor)

winname="labeled in order"
cv2.namedWindow(winname)
cv2.imshow(winname, labeled_in_order)
cv2.moveWindow(winname, 100,500)

# create final

output = np.zeros((450,450,3),np.uint8)
for i,j in enumerate(b):
    ri = int(i/10) # row index
    ci = i%10 # column index
    if ci != 9 and ri!=9:
        src = bm[ri:ri+2, ci:ci+2 , :].reshape((4,2))
        dst = np.array( [ [ci*50,ri*50],[(ci+1)*50-1,ri*50],[ci*50,(ri+1)*50-1],[(ci+1)*50-1,(ri+1)*50-1] ], np.float32)
        retval = cv2.getPerspectiveTransform(src,dst)
        warp = cv2.warpPerspective(res2,retval,(450,450))
        output[ri*50:(ri+1)*50-1 , ci*50:(ci+1)*50-1] = warp[ri*50:(ri+1)*50-1 , ci*50:(ci+1)*50-1].copy()

winname="final"
cv2.namedWindow(winname)
cv2.imshow(winname, output)
cv2.moveWindow(winname, 600,100)

cv2.waitKey(0)
cv2.destroyAllWindows()

I thought this was a great post, and a great solution by ARK; very well laid out and explained.

I was working on a similar problem, and built the entire thing. There were some changes (i.e. xrange to range, arguments in cv2.findContours), but this should work out of the box (Python 3.5, Anaconda).

This is a compilation of the elements above, with some of the missing code added (i.e., labeling of points).

'''

https://stackoverflow.com/questions/10196198/how-to-remove-convexity-defects-in-a-sudoku-square

'''

import cv2
import numpy as np

img = cv2.imread('test.png')

winname="raw image"
cv2.namedWindow(winname)
cv2.imshow(winname, img)
cv2.moveWindow(winname, 100,100)


img = cv2.GaussianBlur(img,(5,5),0)

winname="blurred"
cv2.namedWindow(winname)
cv2.imshow(winname, img)
cv2.moveWindow(winname, 100,150)

gray = cv2.cvtColor(img,cv2.COLOR_BGR2GRAY)
mask = np.zeros((gray.shape),np.uint8)
kernel1 = cv2.getStructuringElement(cv2.MORPH_ELLIPSE,(11,11))

winname="gray"
cv2.namedWindow(winname)
cv2.imshow(winname, gray)
cv2.moveWindow(winname, 100,200)

close = cv2.morphologyEx(gray,cv2.MORPH_CLOSE,kernel1)
div = np.float32(gray)/(close)
res = np.uint8(cv2.normalize(div,div,0,255,cv2.NORM_MINMAX))
res2 = cv2.cvtColor(res,cv2.COLOR_GRAY2BGR)

winname="res2"
cv2.namedWindow(winname)
cv2.imshow(winname, res2)
cv2.moveWindow(winname, 100,250)

 #find elements
thresh = cv2.adaptiveThreshold(res,255,0,1,19,2)
img_c, contour,hier = cv2.findContours(thresh,cv2.RETR_TREE,cv2.CHAIN_APPROX_SIMPLE)

max_area = 0
best_cnt = None
for cnt in contour:
    area = cv2.contourArea(cnt)
    if area > 1000:
        if area > max_area:
            max_area = area
            best_cnt = cnt

cv2.drawContours(mask,[best_cnt],0,255,-1)
cv2.drawContours(mask,[best_cnt],0,0,2)

res = cv2.bitwise_and(res,mask)

winname="puzzle only"
cv2.namedWindow(winname)
cv2.imshow(winname, res)
cv2.moveWindow(winname, 100,300)

# vertical lines
kernelx = cv2.getStructuringElement(cv2.MORPH_RECT,(2,10))

dx = cv2.Sobel(res,cv2.CV_16S,1,0)
dx = cv2.convertScaleAbs(dx)
cv2.normalize(dx,dx,0,255,cv2.NORM_MINMAX)
ret,close = cv2.threshold(dx,0,255,cv2.THRESH_BINARY+cv2.THRESH_OTSU)
close = cv2.morphologyEx(close,cv2.MORPH_DILATE,kernelx,iterations = 1)

img_d, contour, hier = cv2.findContours(close,cv2.RETR_EXTERNAL,cv2.CHAIN_APPROX_SIMPLE)
for cnt in contour:
    x,y,w,h = cv2.boundingRect(cnt)
    if h/w > 5:
        cv2.drawContours(close,[cnt],0,255,-1)
    else:
        cv2.drawContours(close,[cnt],0,0,-1)
close = cv2.morphologyEx(close,cv2.MORPH_CLOSE,None,iterations = 2)
closex = close.copy()

winname="vertical lines"
cv2.namedWindow(winname)
cv2.imshow(winname, img_d)
cv2.moveWindow(winname, 100,350)

# find horizontal lines
kernely = cv2.getStructuringElement(cv2.MORPH_RECT,(10,2))
dy = cv2.Sobel(res,cv2.CV_16S,0,2)
dy = cv2.convertScaleAbs(dy)
cv2.normalize(dy,dy,0,255,cv2.NORM_MINMAX)
ret,close = cv2.threshold(dy,0,255,cv2.THRESH_BINARY+cv2.THRESH_OTSU)
close = cv2.morphologyEx(close,cv2.MORPH_DILATE,kernely)

img_e, contour, hier = cv2.findContours(close,cv2.RETR_EXTERNAL,cv2.CHAIN_APPROX_SIMPLE)

for cnt in contour:
    x,y,w,h = cv2.boundingRect(cnt)
    if w/h > 5:
        cv2.drawContours(close,[cnt],0,255,-1)
    else:
        cv2.drawContours(close,[cnt],0,0,-1)

close = cv2.morphologyEx(close,cv2.MORPH_DILATE,None,iterations = 2)
closey = close.copy()

winname="horizontal lines"
cv2.namedWindow(winname)
cv2.imshow(winname, img_e)
cv2.moveWindow(winname, 100,400)


# intersection of these two gives dots
res = cv2.bitwise_and(closex,closey)

winname="intersections"
cv2.namedWindow(winname)
cv2.imshow(winname, res)
cv2.moveWindow(winname, 100,450)

# text blue
textcolor=(0,255,0)
# points green
pointcolor=(255,0,0)

# find centroids and sort
img_f, contour, hier = cv2.findContours(res,cv2.RETR_LIST,cv2.CHAIN_APPROX_SIMPLE)
centroids = []
for cnt in contour:
    mom = cv2.moments(cnt)
    (x,y) = int(mom['m10']/mom['m00']), int(mom['m01']/mom['m00'])
    cv2.circle(img,(x,y),4,(0,255,0),-1)
    centroids.append((x,y))

# sorting
centroids = np.array(centroids,dtype = np.float32)
c = centroids.reshape((100,2))
c2 = c[np.argsort(c[:,1])]

b = np.vstack([c2[i*10:(i+1)*10][np.argsort(c2[i*10:(i+1)*10,0])] for i in range(10)])
bm = b.reshape((10,10,2))

# make copy
labeled_in_order=res2.copy()

for index, pt in enumerate(b):
    cv2.putText(labeled_in_order,str(index),tuple(pt),cv2.FONT_HERSHEY_DUPLEX, 0.75, textcolor)
    cv2.circle(labeled_in_order, tuple(pt), 5, pointcolor)

winname="labeled in order"
cv2.namedWindow(winname)
cv2.imshow(winname, labeled_in_order)
cv2.moveWindow(winname, 100,500)

# create final

output = np.zeros((450,450,3),np.uint8)
for i,j in enumerate(b):
    ri = int(i/10) # row index
    ci = i%10 # column index
    if ci != 9 and ri!=9:
        src = bm[ri:ri+2, ci:ci+2 , :].reshape((4,2))
        dst = np.array( [ [ci*50,ri*50],[(ci+1)*50-1,ri*50],[ci*50,(ri+1)*50-1],[(ci+1)*50-1,(ri+1)*50-1] ], np.float32)
        retval = cv2.getPerspectiveTransform(src,dst)
        warp = cv2.warpPerspective(res2,retval,(450,450))
        output[ri*50:(ri+1)*50-1 , ci*50:(ci+1)*50-1] = warp[ri*50:(ri+1)*50-1 , ci*50:(ci+1)*50-1].copy()

winname="final"
cv2.namedWindow(winname)
cv2.imshow(winname, output)
cv2.moveWindow(winname, 600,100)

cv2.waitKey(0)
cv2.destroyAllWindows()

如何检测圣诞树?[关闭]

问题:如何检测圣诞树?[关闭]

可以使用哪些图像处理技术来实现检测以下图像中显示的圣诞树的应用程序?

我正在寻找可以在所有这些图像上使用的解决方案。因此,需要训练haar级联分类器模板匹配的方法不是很有趣。

我正在寻找可以使用任何编程语言编写的东西,只要它仅使用开源技术即可。该解决方案必须使用此问题上共享的图像进行测试。有6个输入图像,答案应显示每个图像的处理结果。最后,对于每个输出图像,必须绘制红线以包围检测到的树。

您将如何以编程方式检测这些图像中的树木?

Which image processing techniques could be used to implement an application that detects the Christmas trees displayed in the following images?

I’m searching for solutions that are going to work on all these images. Therefore, approaches that require training haar cascade classifiers or template matching are not very interesting.

I’m looking for something that can be written in any programming language, as long as it uses only Open Source technologies. The solution must be tested with the images that are shared on this question. There are 6 input images and the answer should display the results of processing each of them. Finally, for each output image there must be red lines draw to surround the detected tree.

How would you go about programmatically detecting the trees in these images?


回答 0

我有一种我认为很有趣的方法,与其他方法有所不同。与其他方法相比,我的方法的主要区别在于如何执行图像分割步骤-我使用了来自python scikit-learn 的DBSCAN聚类算法;它经过优化,可找到可能不一定具有单个清晰质心的某种无定形形状。

在最高层,我的方法很简单,可以分解为大约3个步骤。首先,我应用一个阈值(或者实际上是两个单独且不同的阈值的逻辑“或”)。与其他许多答案一样,我假设圣诞树将是场景中较亮的对象之一,因此第一个阈值只是一个简单的单色亮度测试;0-255范围(黑色为0,白色为255)上的值大于220的所有像素将保存到二进制黑白图像。第二个阈值尝试寻找红光和黄光,这在六张图像的左上角和右下角的树木中尤为突出,并且在大多数照片中普遍使用的蓝绿色背景下表现出色。我将rgb图像转换为hsv空间,并要求色相在0.0-1.0范围内小于0.2(大致对应于黄色和绿色之间的边界)或大于0.95(对应于紫色与红色之间的边界),另外我还要求明亮,饱和的颜色:饱和度和值都必须高于0.7。这两个阈值过程的结果在逻辑上“或”在一起,黑白二进制图像的结果矩阵如下所示:

对HSV和单色亮度进行阈值设置后的圣诞树

您可以清楚地看到,每个图像都有一个大的像素簇,大致对应于每棵树的位置,加上一些图像还具有一些其他的小簇,它们对应于某些建筑物的窗户上的灯光,或者对应于背景场景在地平线上。下一步是使计算机识别这些是单独的群集,并使用群集成员ID号正确标记每个像素。

为此,我选择了DBSCAN。相对于其他集群算法,这里有一个很好的视觉比较,可以比较DBSCAN通常的行为。正如我之前说的,它非常适合非晶形形状。此处显示了DBSCAN的输出,其中每个集群以不同的颜色绘制:

DBSCAN集群输出

查看此结果时,需要注意一些事项。首先,DBSCAN要求用户设置一个“接近”参数以调节其行为,该参数有效地控制了一对点必须分开的程度,以便算法声明新的单独簇,而不是将测试点聚结到已经存在的集群。我将此值设置为每个图像对角线大小的0.04倍。由于图像的大小从大约VGA到大约HD 1080不等,因此这种比例相关的定义至关重要。

另一个值得注意的点是,在scikit-learn中实现的DBSCAN算法具有内存限制,对于此示例中的某些较大图像而言,这是相当大的挑战。因此,对于一些较大的图像,我实际上必须每个群集“抽取”(即,仅保留每个第3或第4像素并丢弃其他像素),以保持在此范围内。作为这种剔除处理的结果,在某些较大的图像上很难看到其余的单个稀疏像素。因此,仅出于显示目的,上述图像中的颜色编码像素已被有效地稍微“扩张”了一点,以使其更加突出。出于叙述目的,这纯粹是一种修饰操作;尽管在我的代码中有评论提到此膨胀,

识别并标记了聚类后,第三步也是最后一步很容易:我只是在每个图像中选取最大的聚类(在这种情况下,我选择根据成员像素的总数来衡量“大小”,尽管可以却很容易地使用某种类型的度量标准来衡量物理范围)并计算该集群的凸包。凸包然后成为树的边界。通过此方法计算的六个凸包在下面以红色显示:

带有计算边界的圣诞树

源代码是为Python 2.7.6编写的,它取决于numpyscipymatplotlibscikit-learn。我将其分为两部分。第一部分负责实际的图像处理:

from PIL import Image
import numpy as np
import scipy as sp
import matplotlib.colors as colors
from sklearn.cluster import DBSCAN
from math import ceil, sqrt

"""
Inputs:

    rgbimg:         [M,N,3] numpy array containing (uint, 0-255) color image

    hueleftthr:     Scalar constant to select maximum allowed hue in the
                    yellow-green region

    huerightthr:    Scalar constant to select minimum allowed hue in the
                    blue-purple region

    satthr:         Scalar constant to select minimum allowed saturation

    valthr:         Scalar constant to select minimum allowed value

    monothr:        Scalar constant to select minimum allowed monochrome
                    brightness

    maxpoints:      Scalar constant maximum number of pixels to forward to
                    the DBSCAN clustering algorithm

    proxthresh:     Proximity threshold to use for DBSCAN, as a fraction of
                    the diagonal size of the image

Outputs:

    borderseg:      [K,2,2] Nested list containing K pairs of x- and y- pixel
                    values for drawing the tree border

    X:              [P,2] List of pixels that passed the threshold step

    labels:         [Q,2] List of cluster labels for points in Xslice (see
                    below)

    Xslice:         [Q,2] Reduced list of pixels to be passed to DBSCAN

"""

def findtree(rgbimg, hueleftthr=0.2, huerightthr=0.95, satthr=0.7, 
             valthr=0.7, monothr=220, maxpoints=5000, proxthresh=0.04):

    # Convert rgb image to monochrome for
    gryimg = np.asarray(Image.fromarray(rgbimg).convert('L'))
    # Convert rgb image (uint, 0-255) to hsv (float, 0.0-1.0)
    hsvimg = colors.rgb_to_hsv(rgbimg.astype(float)/255)

    # Initialize binary thresholded image
    binimg = np.zeros((rgbimg.shape[0], rgbimg.shape[1]))
    # Find pixels with hue<0.2 or hue>0.95 (red or yellow) and saturation/value
    # both greater than 0.7 (saturated and bright)--tends to coincide with
    # ornamental lights on trees in some of the images
    boolidx = np.logical_and(
                np.logical_and(
                  np.logical_or((hsvimg[:,:,0] < hueleftthr),
                                (hsvimg[:,:,0] > huerightthr)),
                                (hsvimg[:,:,1] > satthr)),
                                (hsvimg[:,:,2] > valthr))
    # Find pixels that meet hsv criterion
    binimg[np.where(boolidx)] = 255
    # Add pixels that meet grayscale brightness criterion
    binimg[np.where(gryimg > monothr)] = 255

    # Prepare thresholded points for DBSCAN clustering algorithm
    X = np.transpose(np.where(binimg == 255))
    Xslice = X
    nsample = len(Xslice)
    if nsample > maxpoints:
        # Make sure number of points does not exceed DBSCAN maximum capacity
        Xslice = X[range(0,nsample,int(ceil(float(nsample)/maxpoints)))]

    # Translate DBSCAN proximity threshold to units of pixels and run DBSCAN
    pixproxthr = proxthresh * sqrt(binimg.shape[0]**2 + binimg.shape[1]**2)
    db = DBSCAN(eps=pixproxthr, min_samples=10).fit(Xslice)
    labels = db.labels_.astype(int)

    # Find the largest cluster (i.e., with most points) and obtain convex hull   
    unique_labels = set(labels)
    maxclustpt = 0
    for k in unique_labels:
        class_members = [index[0] for index in np.argwhere(labels == k)]
        if len(class_members) > maxclustpt:
            points = Xslice[class_members]
            hull = sp.spatial.ConvexHull(points)
            maxclustpt = len(class_members)
            borderseg = [[points[simplex,0], points[simplex,1]] for simplex
                          in hull.simplices]

    return borderseg, X, labels, Xslice

第二部分是用户级脚本,该脚本调用第一个文件并生成上面的所有图:

#!/usr/bin/env python

from PIL import Image
import numpy as np
import matplotlib.pyplot as plt
import matplotlib.cm as cm
from findtree import findtree

# Image files to process
fname = ['nmzwj.png', 'aVZhC.png', '2K9EF.png',
         'YowlH.png', '2y4o5.png', 'FWhSP.png']

# Initialize figures
fgsz = (16,7)        
figthresh = plt.figure(figsize=fgsz, facecolor='w')
figclust  = plt.figure(figsize=fgsz, facecolor='w')
figcltwo  = plt.figure(figsize=fgsz, facecolor='w')
figborder = plt.figure(figsize=fgsz, facecolor='w')
figthresh.canvas.set_window_title('Thresholded HSV and Monochrome Brightness')
figclust.canvas.set_window_title('DBSCAN Clusters (Raw Pixel Output)')
figcltwo.canvas.set_window_title('DBSCAN Clusters (Slightly Dilated for Display)')
figborder.canvas.set_window_title('Trees with Borders')

for ii, name in zip(range(len(fname)), fname):
    # Open the file and convert to rgb image
    rgbimg = np.asarray(Image.open(name))

    # Get the tree borders as well as a bunch of other intermediate values
    # that will be used to illustrate how the algorithm works
    borderseg, X, labels, Xslice = findtree(rgbimg)

    # Display thresholded images
    axthresh = figthresh.add_subplot(2,3,ii+1)
    axthresh.set_xticks([])
    axthresh.set_yticks([])
    binimg = np.zeros((rgbimg.shape[0], rgbimg.shape[1]))
    for v, h in X:
        binimg[v,h] = 255
    axthresh.imshow(binimg, interpolation='nearest', cmap='Greys')

    # Display color-coded clusters
    axclust = figclust.add_subplot(2,3,ii+1) # Raw version
    axclust.set_xticks([])
    axclust.set_yticks([])
    axcltwo = figcltwo.add_subplot(2,3,ii+1) # Dilated slightly for display only
    axcltwo.set_xticks([])
    axcltwo.set_yticks([])
    axcltwo.imshow(binimg, interpolation='nearest', cmap='Greys')
    clustimg = np.ones(rgbimg.shape)    
    unique_labels = set(labels)
    # Generate a unique color for each cluster 
    plcol = cm.rainbow_r(np.linspace(0, 1, len(unique_labels)))
    for lbl, pix in zip(labels, Xslice):
        for col, unqlbl in zip(plcol, unique_labels):
            if lbl == unqlbl:
                # Cluster label of -1 indicates no cluster membership;
                # override default color with black
                if lbl == -1:
                    col = [0.0, 0.0, 0.0, 1.0]
                # Raw version
                for ij in range(3):
                    clustimg[pix[0],pix[1],ij] = col[ij]
                # Dilated just for display
                axcltwo.plot(pix[1], pix[0], 'o', markerfacecolor=col, 
                    markersize=1, markeredgecolor=col)
    axclust.imshow(clustimg)
    axcltwo.set_xlim(0, binimg.shape[1]-1)
    axcltwo.set_ylim(binimg.shape[0], -1)

    # Plot original images with read borders around the trees
    axborder = figborder.add_subplot(2,3,ii+1)
    axborder.set_axis_off()
    axborder.imshow(rgbimg, interpolation='nearest')
    for vseg, hseg in borderseg:
        axborder.plot(hseg, vseg, 'r-', lw=3)
    axborder.set_xlim(0, binimg.shape[1]-1)
    axborder.set_ylim(binimg.shape[0], -1)

plt.show()

I have an approach which I think is interesting and a bit different from the rest. The main difference in my approach, compared to some of the others, is in how the image segmentation step is performed–I used the DBSCAN clustering algorithm from Python’s scikit-learn; it’s optimized for finding somewhat amorphous shapes that may not necessarily have a single clear centroid.

At the top level, my approach is fairly simple and can be broken down into about 3 steps. First I apply a threshold (or actually, the logical “or” of two separate and distinct thresholds). As with many of the other answers, I assumed that the Christmas tree would be one of the brighter objects in the scene, so the first threshold is just a simple monochrome brightness test; any pixels with values above 220 on a 0-255 scale (where black is 0 and white is 255) are saved to a binary black-and-white image. The second threshold tries to look for red and yellow lights, which are particularly prominent in the trees in the upper left and lower right of the six images, and stand out well against the blue-green background which is prevalent in most of the photos. I convert the rgb image to hsv space, and require that the hue is either less than 0.2 on a 0.0-1.0 scale (corresponding roughly to the border between yellow and green) or greater than 0.95 (corresponding to the border between purple and red) and additionally I require bright, saturated colors: saturation and value must both be above 0.7. The results of the two threshold procedures are logically “or”-ed together, and the resulting matrix of black-and-white binary images is shown below:

Christmas trees, after thresholding on HSV as well as monochrome brightness

You can clearly see that each image has one large cluster of pixels roughly corresponding to the location of each tree, plus a few of the images also have some other small clusters corresponding either to lights in the windows of some of the buildings, or to a background scene on the horizon. The next step is to get the computer to recognize that these are separate clusters, and label each pixel correctly with a cluster membership ID number.

For this task I chose DBSCAN. There is a pretty good visual comparison of how DBSCAN typically behaves, relative to other clustering algorithms, available here. As I said earlier, it does well with amorphous shapes. The output of DBSCAN, with each cluster plotted in a different color, is shown here:

DBSCAN clustering output

There are a few things to be aware of when looking at this result. First is that DBSCAN requires the user to set a “proximity” parameter in order to regulate its behavior, which effectively controls how separated a pair of points must be in order for the algorithm to declare a new separate cluster rather than agglomerating a test point onto an already pre-existing cluster. I set this value to be 0.04 times the size along the diagonal of each image. Since the images vary in size from roughly VGA up to about HD 1080, this type of scale-relative definition is critical.

Another point worth noting is that the DBSCAN algorithm as it is implemented in scikit-learn has memory limits which are fairly challenging for some of the larger images in this sample. Therefore, for a few of the larger images, I actually had to “decimate” (i.e., retain only every 3rd or 4th pixel and drop the others) each cluster in order to stay within this limit. As a result of this culling process, the remaining individual sparse pixels are difficult to see on some of the larger images. Therefore, for display purposes only, the color-coded pixels in the above images have been effectively “dilated” just slightly so that they stand out better. It’s purely a cosmetic operation for the sake of the narrative; although there are comments mentioning this dilation in my code, rest assured that it has nothing to do with any calculations that actually matter.

Once the clusters are identified and labeled, the third and final step is easy: I simply take the largest cluster in each image (in this case, I chose to measure “size” in terms of the total number of member pixels, although one could have just as easily instead used some type of metric that gauges physical extent) and compute the convex hull for that cluster. The convex hull then becomes the tree border. The six convex hulls computed via this method are shown below in red:

Christmas trees with their calculated borders

The source code is written for Python 2.7.6 and it depends on numpy, scipy, matplotlib and scikit-learn. I’ve divided it into two parts. The first part is responsible for the actual image processing:

from PIL import Image
import numpy as np
import scipy as sp
import matplotlib.colors as colors
from sklearn.cluster import DBSCAN
from math import ceil, sqrt

"""
Inputs:

    rgbimg:         [M,N,3] numpy array containing (uint, 0-255) color image

    hueleftthr:     Scalar constant to select maximum allowed hue in the
                    yellow-green region

    huerightthr:    Scalar constant to select minimum allowed hue in the
                    blue-purple region

    satthr:         Scalar constant to select minimum allowed saturation

    valthr:         Scalar constant to select minimum allowed value

    monothr:        Scalar constant to select minimum allowed monochrome
                    brightness

    maxpoints:      Scalar constant maximum number of pixels to forward to
                    the DBSCAN clustering algorithm

    proxthresh:     Proximity threshold to use for DBSCAN, as a fraction of
                    the diagonal size of the image

Outputs:

    borderseg:      [K,2,2] Nested list containing K pairs of x- and y- pixel
                    values for drawing the tree border

    X:              [P,2] List of pixels that passed the threshold step

    labels:         [Q,2] List of cluster labels for points in Xslice (see
                    below)

    Xslice:         [Q,2] Reduced list of pixels to be passed to DBSCAN

"""

def findtree(rgbimg, hueleftthr=0.2, huerightthr=0.95, satthr=0.7, 
             valthr=0.7, monothr=220, maxpoints=5000, proxthresh=0.04):

    # Convert rgb image to monochrome for
    gryimg = np.asarray(Image.fromarray(rgbimg).convert('L'))
    # Convert rgb image (uint, 0-255) to hsv (float, 0.0-1.0)
    hsvimg = colors.rgb_to_hsv(rgbimg.astype(float)/255)

    # Initialize binary thresholded image
    binimg = np.zeros((rgbimg.shape[0], rgbimg.shape[1]))
    # Find pixels with hue<0.2 or hue>0.95 (red or yellow) and saturation/value
    # both greater than 0.7 (saturated and bright)--tends to coincide with
    # ornamental lights on trees in some of the images
    boolidx = np.logical_and(
                np.logical_and(
                  np.logical_or((hsvimg[:,:,0] < hueleftthr),
                                (hsvimg[:,:,0] > huerightthr)),
                                (hsvimg[:,:,1] > satthr)),
                                (hsvimg[:,:,2] > valthr))
    # Find pixels that meet hsv criterion
    binimg[np.where(boolidx)] = 255
    # Add pixels that meet grayscale brightness criterion
    binimg[np.where(gryimg > monothr)] = 255

    # Prepare thresholded points for DBSCAN clustering algorithm
    X = np.transpose(np.where(binimg == 255))
    Xslice = X
    nsample = len(Xslice)
    if nsample > maxpoints:
        # Make sure number of points does not exceed DBSCAN maximum capacity
        Xslice = X[range(0,nsample,int(ceil(float(nsample)/maxpoints)))]

    # Translate DBSCAN proximity threshold to units of pixels and run DBSCAN
    pixproxthr = proxthresh * sqrt(binimg.shape[0]**2 + binimg.shape[1]**2)
    db = DBSCAN(eps=pixproxthr, min_samples=10).fit(Xslice)
    labels = db.labels_.astype(int)

    # Find the largest cluster (i.e., with most points) and obtain convex hull   
    unique_labels = set(labels)
    maxclustpt = 0
    for k in unique_labels:
        class_members = [index[0] for index in np.argwhere(labels == k)]
        if len(class_members) > maxclustpt:
            points = Xslice[class_members]
            hull = sp.spatial.ConvexHull(points)
            maxclustpt = len(class_members)
            borderseg = [[points[simplex,0], points[simplex,1]] for simplex
                          in hull.simplices]

    return borderseg, X, labels, Xslice

and the second part is a user-level script which calls the first file and generates all of the plots above:

#!/usr/bin/env python

from PIL import Image
import numpy as np
import matplotlib.pyplot as plt
import matplotlib.cm as cm
from findtree import findtree

# Image files to process
fname = ['nmzwj.png', 'aVZhC.png', '2K9EF.png',
         'YowlH.png', '2y4o5.png', 'FWhSP.png']

# Initialize figures
fgsz = (16,7)        
figthresh = plt.figure(figsize=fgsz, facecolor='w')
figclust  = plt.figure(figsize=fgsz, facecolor='w')
figcltwo  = plt.figure(figsize=fgsz, facecolor='w')
figborder = plt.figure(figsize=fgsz, facecolor='w')
figthresh.canvas.set_window_title('Thresholded HSV and Monochrome Brightness')
figclust.canvas.set_window_title('DBSCAN Clusters (Raw Pixel Output)')
figcltwo.canvas.set_window_title('DBSCAN Clusters (Slightly Dilated for Display)')
figborder.canvas.set_window_title('Trees with Borders')

for ii, name in zip(range(len(fname)), fname):
    # Open the file and convert to rgb image
    rgbimg = np.asarray(Image.open(name))

    # Get the tree borders as well as a bunch of other intermediate values
    # that will be used to illustrate how the algorithm works
    borderseg, X, labels, Xslice = findtree(rgbimg)

    # Display thresholded images
    axthresh = figthresh.add_subplot(2,3,ii+1)
    axthresh.set_xticks([])
    axthresh.set_yticks([])
    binimg = np.zeros((rgbimg.shape[0], rgbimg.shape[1]))
    for v, h in X:
        binimg[v,h] = 255
    axthresh.imshow(binimg, interpolation='nearest', cmap='Greys')

    # Display color-coded clusters
    axclust = figclust.add_subplot(2,3,ii+1) # Raw version
    axclust.set_xticks([])
    axclust.set_yticks([])
    axcltwo = figcltwo.add_subplot(2,3,ii+1) # Dilated slightly for display only
    axcltwo.set_xticks([])
    axcltwo.set_yticks([])
    axcltwo.imshow(binimg, interpolation='nearest', cmap='Greys')
    clustimg = np.ones(rgbimg.shape)    
    unique_labels = set(labels)
    # Generate a unique color for each cluster 
    plcol = cm.rainbow_r(np.linspace(0, 1, len(unique_labels)))
    for lbl, pix in zip(labels, Xslice):
        for col, unqlbl in zip(plcol, unique_labels):
            if lbl == unqlbl:
                # Cluster label of -1 indicates no cluster membership;
                # override default color with black
                if lbl == -1:
                    col = [0.0, 0.0, 0.0, 1.0]
                # Raw version
                for ij in range(3):
                    clustimg[pix[0],pix[1],ij] = col[ij]
                # Dilated just for display
                axcltwo.plot(pix[1], pix[0], 'o', markerfacecolor=col, 
                    markersize=1, markeredgecolor=col)
    axclust.imshow(clustimg)
    axcltwo.set_xlim(0, binimg.shape[1]-1)
    axcltwo.set_ylim(binimg.shape[0], -1)

    # Plot original images with read borders around the trees
    axborder = figborder.add_subplot(2,3,ii+1)
    axborder.set_axis_off()
    axborder.imshow(rgbimg, interpolation='nearest')
    for vseg, hseg in borderseg:
        axborder.plot(hseg, vseg, 'r-', lw=3)
    axborder.set_xlim(0, binimg.shape[1]-1)
    axborder.set_ylim(binimg.shape[0], -1)

plt.show()

回答 1

编辑注释:我编辑了这篇文章,以(i)根据要求单独处理每棵树图像,(ii)考虑对象的亮度和形状,以提高结果的质量。


下面介绍一种考虑物体亮度和形状的方法。换句话说,它寻找具有三角形形状且具有明显亮度的物体。它使用Marvin图像处理框架以Java实现。

第一步是颜色阈值。此处的目的是将分析重点放在亮度很高的物体上。

输出图像:

源代码:

public class ChristmasTree {

private MarvinImagePlugin fill = MarvinPluginLoader.loadImagePlugin("org.marvinproject.image.fill.boundaryFill");
private MarvinImagePlugin threshold = MarvinPluginLoader.loadImagePlugin("org.marvinproject.image.color.thresholding");
private MarvinImagePlugin invert = MarvinPluginLoader.loadImagePlugin("org.marvinproject.image.color.invert");
private MarvinImagePlugin dilation = MarvinPluginLoader.loadImagePlugin("org.marvinproject.image.morphological.dilation");

public ChristmasTree(){
    MarvinImage tree;

    // Iterate each image
    for(int i=1; i<=6; i++){
        tree = MarvinImageIO.loadImage("./res/trees/tree"+i+".png");

        // 1. Threshold
        threshold.setAttribute("threshold", 200);
        threshold.process(tree.clone(), tree);
    }
}
public static void main(String[] args) {
    new ChristmasTree();
}
}

在第二步中,将图像中最亮的点放大以形成形状。该过程的结果是具有明显亮度的物体的可能形状。应用洪水填充分割,可以检测到断开的形状。

输出图像:

源代码:

public class ChristmasTree {

private MarvinImagePlugin fill = MarvinPluginLoader.loadImagePlugin("org.marvinproject.image.fill.boundaryFill");
private MarvinImagePlugin threshold = MarvinPluginLoader.loadImagePlugin("org.marvinproject.image.color.thresholding");
private MarvinImagePlugin invert = MarvinPluginLoader.loadImagePlugin("org.marvinproject.image.color.invert");
private MarvinImagePlugin dilation = MarvinPluginLoader.loadImagePlugin("org.marvinproject.image.morphological.dilation");

public ChristmasTree(){
    MarvinImage tree;

    // Iterate each image
    for(int i=1; i<=6; i++){
        tree = MarvinImageIO.loadImage("./res/trees/tree"+i+".png");

        // 1. Threshold
        threshold.setAttribute("threshold", 200);
        threshold.process(tree.clone(), tree);

        // 2. Dilate
        invert.process(tree.clone(), tree);
        tree = MarvinColorModelConverter.rgbToBinary(tree, 127);
        MarvinImageIO.saveImage(tree, "./res/trees/new/tree_"+i+"threshold.png");
        dilation.setAttribute("matrix", MarvinMath.getTrueMatrix(50, 50));
        dilation.process(tree.clone(), tree);
        MarvinImageIO.saveImage(tree, "./res/trees/new/tree_"+1+"_dilation.png");
        tree = MarvinColorModelConverter.binaryToRgb(tree);

        // 3. Segment shapes
        MarvinImage trees2 = tree.clone();
        fill(tree, trees2);
        MarvinImageIO.saveImage(trees2, "./res/trees/new/tree_"+i+"_fill.png");
}

private void fill(MarvinImage imageIn, MarvinImage imageOut){
    boolean found;
    int color= 0xFFFF0000;

    while(true){
        found=false;

        Outerloop:
        for(int y=0; y<imageIn.getHeight(); y++){
            for(int x=0; x<imageIn.getWidth(); x++){
                if(imageOut.getIntComponent0(x, y) == 0){
                    fill.setAttribute("x", x);
                    fill.setAttribute("y", y);
                    fill.setAttribute("color", color);
                    fill.setAttribute("threshold", 120);
                    fill.process(imageIn, imageOut);
                    color = newColor(color);

                    found = true;
                    break Outerloop;
                }
            }
        }

        if(!found){
            break;
        }
    }

}

private int newColor(int color){
    int red = (color & 0x00FF0000) >> 16;
    int green = (color & 0x0000FF00) >> 8;
    int blue = (color & 0x000000FF);

    if(red <= green && red <= blue){
        red+=5;
    }
    else if(green <= red && green <= blue){
        green+=5;
    }
    else{
        blue+=5;
    }

    return 0xFF000000 + (red << 16) + (green << 8) + blue;
}

public static void main(String[] args) {
    new ChristmasTree();
}
}

如输出图像所示,检测到多种形状。在此问题中,图像中只有几个亮点。但是,实施此方法是为了处理更复杂的情况。

在下一步中,将分析每个形状。一种简单的算法可以检测形状类似于三角形的形状。该算法逐行分析对象形状。如果每个形状线的质心几乎相同(给定阈值),并且质量随着y的增加而增加,则对象具有三角形的形状。形状线的质量是该线中属于该形状的像素数。想象一下,您将对象水平切片并分析每个水平段。如果它们彼此居中,并且长度以线性模式从第一段到最后一段增加,则您可能有一个类似于三角形的对象。

源代码:

private int[] detectTrees(MarvinImage image){
    HashSet<Integer> analysed = new HashSet<Integer>();
    boolean found;
    while(true){
        found = false;
        for(int y=0; y<image.getHeight(); y++){
            for(int x=0; x<image.getWidth(); x++){
                int color = image.getIntColor(x, y);

                if(!analysed.contains(color)){
                    if(isTree(image, color)){
                        return getObjectRect(image, color);
                    }

                    analysed.add(color);
                    found=true;
                }
            }
        }

        if(!found){
            break;
        }
    }
    return null;
}

private boolean isTree(MarvinImage image, int color){

    int mass[][] = new int[image.getHeight()][2];
    int yStart=-1;
    int xStart=-1;
    for(int y=0; y<image.getHeight(); y++){
        int mc = 0;
        int xs=-1;
        int xe=-1;
        for(int x=0; x<image.getWidth(); x++){
            if(image.getIntColor(x, y) == color){
                mc++;

                if(yStart == -1){
                    yStart=y;
                    xStart=x;
                }

                if(xs == -1){
                    xs = x;
                }
                if(x > xe){
                    xe = x;
                }
            }
        }
        mass[y][0] = xs;
        mass[y][3] = xe;
        mass[y][4] = mc;    
    }

    int validLines=0;
    for(int y=0; y<image.getHeight(); y++){
        if
        ( 
            mass[y][5] > 0 &&
            Math.abs(((mass[y][0]+mass[y][6])/2)-xStart) <= 50 &&
            mass[y][7] >= (mass[yStart][8] + (y-yStart)*0.3) &&
            mass[y][9] <= (mass[yStart][10] + (y-yStart)*1.5)
        )
        {
            validLines++;
        }
    }

    if(validLines > 100){
        return true;
    }
    return false;
}

最后,如下图所示,原始图像中突出显示了每个形状类似于三角形且具有明显亮度的位置(在本例中为圣诞树)。

最终输出图像:

最终源代码:

public class ChristmasTree {

private MarvinImagePlugin fill = MarvinPluginLoader.loadImagePlugin("org.marvinproject.image.fill.boundaryFill");
private MarvinImagePlugin threshold = MarvinPluginLoader.loadImagePlugin("org.marvinproject.image.color.thresholding");
private MarvinImagePlugin invert = MarvinPluginLoader.loadImagePlugin("org.marvinproject.image.color.invert");
private MarvinImagePlugin dilation = MarvinPluginLoader.loadImagePlugin("org.marvinproject.image.morphological.dilation");

public ChristmasTree(){
    MarvinImage tree;

    // Iterate each image
    for(int i=1; i<=6; i++){
        tree = MarvinImageIO.loadImage("./res/trees/tree"+i+".png");

        // 1. Threshold
        threshold.setAttribute("threshold", 200);
        threshold.process(tree.clone(), tree);

        // 2. Dilate
        invert.process(tree.clone(), tree);
        tree = MarvinColorModelConverter.rgbToBinary(tree, 127);
        MarvinImageIO.saveImage(tree, "./res/trees/new/tree_"+i+"threshold.png");
        dilation.setAttribute("matrix", MarvinMath.getTrueMatrix(50, 50));
        dilation.process(tree.clone(), tree);
        MarvinImageIO.saveImage(tree, "./res/trees/new/tree_"+1+"_dilation.png");
        tree = MarvinColorModelConverter.binaryToRgb(tree);

        // 3. Segment shapes
        MarvinImage trees2 = tree.clone();
        fill(tree, trees2);
        MarvinImageIO.saveImage(trees2, "./res/trees/new/tree_"+i+"_fill.png");

        // 4. Detect tree-like shapes
        int[] rect = detectTrees(trees2);

        // 5. Draw the result
        MarvinImage original = MarvinImageIO.loadImage("./res/trees/tree"+i+".png");
        drawBoundary(trees2, original, rect);
        MarvinImageIO.saveImage(original, "./res/trees/new/tree_"+i+"_out_2.jpg");
    }
}

private void drawBoundary(MarvinImage shape, MarvinImage original, int[] rect){
    int yLines[] = new int[6];
    yLines[0] = rect[1];
    yLines[1] = rect[1]+(int)((rect[3]/5));
    yLines[2] = rect[1]+((rect[3]/5)*2);
    yLines[3] = rect[1]+((rect[3]/5)*3);
    yLines[4] = rect[1]+(int)((rect[3]/5)*4);
    yLines[5] = rect[1]+rect[3];

    List<Point> points = new ArrayList<Point>();
    for(int i=0; i<yLines.length; i++){
        boolean in=false;
        Point startPoint=null;
        Point endPoint=null;
        for(int x=rect[0]; x<rect[0]+rect[2]; x++){

            if(shape.getIntColor(x, yLines[i]) != 0xFFFFFFFF){
                if(!in){
                    if(startPoint == null){
                        startPoint = new Point(x, yLines[i]);
                    }
                }
                in = true;
            }
            else{
                if(in){
                    endPoint = new Point(x, yLines[i]);
                }
                in = false;
            }
        }

        if(endPoint == null){
            endPoint = new Point((rect[0]+rect[2])-1, yLines[i]);
        }

        points.add(startPoint);
        points.add(endPoint);
    }

    drawLine(points.get(0).x, points.get(0).y, points.get(1).x, points.get(1).y, 15, original);
    drawLine(points.get(1).x, points.get(1).y, points.get(3).x, points.get(3).y, 15, original);
    drawLine(points.get(3).x, points.get(3).y, points.get(5).x, points.get(5).y, 15, original);
    drawLine(points.get(5).x, points.get(5).y, points.get(7).x, points.get(7).y, 15, original);
    drawLine(points.get(7).x, points.get(7).y, points.get(9).x, points.get(9).y, 15, original);
    drawLine(points.get(9).x, points.get(9).y, points.get(11).x, points.get(11).y, 15, original);
    drawLine(points.get(11).x, points.get(11).y, points.get(10).x, points.get(10).y, 15, original);
    drawLine(points.get(10).x, points.get(10).y, points.get(8).x, points.get(8).y, 15, original);
    drawLine(points.get(8).x, points.get(8).y, points.get(6).x, points.get(6).y, 15, original);
    drawLine(points.get(6).x, points.get(6).y, points.get(4).x, points.get(4).y, 15, original);
    drawLine(points.get(4).x, points.get(4).y, points.get(2).x, points.get(2).y, 15, original);
    drawLine(points.get(2).x, points.get(2).y, points.get(0).x, points.get(0).y, 15, original);
}

private void drawLine(int x1, int y1, int x2, int y2, int length, MarvinImage image){
    int lx1, lx2, ly1, ly2;
    for(int i=0; i<length; i++){
        lx1 = (x1+i >= image.getWidth() ? (image.getWidth()-1)-i: x1);
        lx2 = (x2+i >= image.getWidth() ? (image.getWidth()-1)-i: x2);
        ly1 = (y1+i >= image.getHeight() ? (image.getHeight()-1)-i: y1);
        ly2 = (y2+i >= image.getHeight() ? (image.getHeight()-1)-i: y2);

        image.drawLine(lx1+i, ly1, lx2+i, ly2, Color.red);
        image.drawLine(lx1, ly1+i, lx2, ly2+i, Color.red);
    }
}

private void fillRect(MarvinImage image, int[] rect, int length){
    for(int i=0; i<length; i++){
        image.drawRect(rect[0]+i, rect[1]+i, rect[2]-(i*2), rect[3]-(i*2), Color.red);
    }
}

private void fill(MarvinImage imageIn, MarvinImage imageOut){
    boolean found;
    int color= 0xFFFF0000;

    while(true){
        found=false;

        Outerloop:
        for(int y=0; y<imageIn.getHeight(); y++){
            for(int x=0; x<imageIn.getWidth(); x++){
                if(imageOut.getIntComponent0(x, y) == 0){
                    fill.setAttribute("x", x);
                    fill.setAttribute("y", y);
                    fill.setAttribute("color", color);
                    fill.setAttribute("threshold", 120);
                    fill.process(imageIn, imageOut);
                    color = newColor(color);

                    found = true;
                    break Outerloop;
                }
            }
        }

        if(!found){
            break;
        }
    }

}

private int[] detectTrees(MarvinImage image){
    HashSet<Integer> analysed = new HashSet<Integer>();
    boolean found;
    while(true){
        found = false;
        for(int y=0; y<image.getHeight(); y++){
            for(int x=0; x<image.getWidth(); x++){
                int color = image.getIntColor(x, y);

                if(!analysed.contains(color)){
                    if(isTree(image, color)){
                        return getObjectRect(image, color);
                    }

                    analysed.add(color);
                    found=true;
                }
            }
        }

        if(!found){
            break;
        }
    }
    return null;
}

private boolean isTree(MarvinImage image, int color){

    int mass[][] = new int[image.getHeight()][11];
    int yStart=-1;
    int xStart=-1;
    for(int y=0; y<image.getHeight(); y++){
        int mc = 0;
        int xs=-1;
        int xe=-1;
        for(int x=0; x<image.getWidth(); x++){
            if(image.getIntColor(x, y) == color){
                mc++;

                if(yStart == -1){
                    yStart=y;
                    xStart=x;
                }

                if(xs == -1){
                    xs = x;
                }
                if(x > xe){
                    xe = x;
                }
            }
        }
        mass[y][0] = xs;
        mass[y][12] = xe;
        mass[y][13] = mc;   
    }

    int validLines=0;
    for(int y=0; y<image.getHeight(); y++){
        if
        ( 
            mass[y][14] > 0 &&
            Math.abs(((mass[y][0]+mass[y][15])/2)-xStart) <= 50 &&
            mass[y][16] >= (mass[yStart][17] + (y-yStart)*0.3) &&
            mass[y][18] <= (mass[yStart][19] + (y-yStart)*1.5)
        )
        {
            validLines++;
        }
    }

    if(validLines > 100){
        return true;
    }
    return false;
}

private int[] getObjectRect(MarvinImage image, int color){
    int x1=-1;
    int x2=-1;
    int y1=-1;
    int y2=-1;

    for(int y=0; y<image.getHeight(); y++){
        for(int x=0; x<image.getWidth(); x++){
            if(image.getIntColor(x, y) == color){

                if(x1 == -1 || x < x1){
                    x1 = x;
                }
                if(x2 == -1 || x > x2){
                    x2 = x;
                }
                if(y1 == -1 || y < y1){
                    y1 = y;
                }
                if(y2 == -1 || y > y2){
                    y2 = y;
                }
            }
        }
    }

    return new int[]{x1, y1, (x2-x1), (y2-y1)};
}

private int newColor(int color){
    int red = (color & 0x00FF0000) >> 16;
    int green = (color & 0x0000FF00) >> 8;
    int blue = (color & 0x000000FF);

    if(red <= green && red <= blue){
        red+=5;
    }
    else if(green <= red && green <= blue){
        green+=30;
    }
    else{
        blue+=30;
    }

    return 0xFF000000 + (red << 16) + (green << 8) + blue;
}

public static void main(String[] args) {
    new ChristmasTree();
}
}

这种方法的优点在于,由于它可以分析物体的形状,因此可能会与包含其他发光物体的图像一起使用。

圣诞节快乐!


编辑注2

讨论了此解决方案与其他解决方案的输出图像的相似性。实际上,它们非常相似。但是这种方法不仅可以分割对象。它还从某种意义上分析了对象的形状。它可以处理同一场景中的多个发光物体。实际上,圣诞树不必是最亮的圣诞树。我只是为了中止讨论而中止。样本中存在偏差,即仅寻找最亮的对象,便会找到树木。但是,我们真的要在这一点上停止讨论吗?在这一点上,计算机实际上能识别出类似于圣诞树的物体吗?让我们尝试缩小这一差距。

下面给出的结果只是为了阐明这一点:

输入图像

在此处输入图片说明

输出

在此处输入图片说明

EDIT NOTE: I edited this post to (i) process each tree image individually, as requested in the requirements, (ii) to consider both object brightness and shape in order to improve the quality of the result.


Below is presented an approach that takes in consideration the object brightness and shape. In other words, it seeks for objects with triangle-like shape and with significant brightness. It was implemented in Java, using Marvin image processing framework.

The first step is the color thresholding. The objective here is to focus the analysis on objects with significant brightness.

output images:

source code:

public class ChristmasTree {

private MarvinImagePlugin fill = MarvinPluginLoader.loadImagePlugin("org.marvinproject.image.fill.boundaryFill");
private MarvinImagePlugin threshold = MarvinPluginLoader.loadImagePlugin("org.marvinproject.image.color.thresholding");
private MarvinImagePlugin invert = MarvinPluginLoader.loadImagePlugin("org.marvinproject.image.color.invert");
private MarvinImagePlugin dilation = MarvinPluginLoader.loadImagePlugin("org.marvinproject.image.morphological.dilation");

public ChristmasTree(){
    MarvinImage tree;

    // Iterate each image
    for(int i=1; i<=6; i++){
        tree = MarvinImageIO.loadImage("./res/trees/tree"+i+".png");

        // 1. Threshold
        threshold.setAttribute("threshold", 200);
        threshold.process(tree.clone(), tree);
    }
}
public static void main(String[] args) {
    new ChristmasTree();
}
}

In the second step, the brightest points in the image are dilated in order to form shapes. The result of this process is the probable shape of the objects with significant brightness. Applying flood fill segmentation, disconnected shapes are detected.

output images:

source code:

public class ChristmasTree {

private MarvinImagePlugin fill = MarvinPluginLoader.loadImagePlugin("org.marvinproject.image.fill.boundaryFill");
private MarvinImagePlugin threshold = MarvinPluginLoader.loadImagePlugin("org.marvinproject.image.color.thresholding");
private MarvinImagePlugin invert = MarvinPluginLoader.loadImagePlugin("org.marvinproject.image.color.invert");
private MarvinImagePlugin dilation = MarvinPluginLoader.loadImagePlugin("org.marvinproject.image.morphological.dilation");

public ChristmasTree(){
    MarvinImage tree;

    // Iterate each image
    for(int i=1; i<=6; i++){
        tree = MarvinImageIO.loadImage("./res/trees/tree"+i+".png");

        // 1. Threshold
        threshold.setAttribute("threshold", 200);
        threshold.process(tree.clone(), tree);

        // 2. Dilate
        invert.process(tree.clone(), tree);
        tree = MarvinColorModelConverter.rgbToBinary(tree, 127);
        MarvinImageIO.saveImage(tree, "./res/trees/new/tree_"+i+"threshold.png");
        dilation.setAttribute("matrix", MarvinMath.getTrueMatrix(50, 50));
        dilation.process(tree.clone(), tree);
        MarvinImageIO.saveImage(tree, "./res/trees/new/tree_"+1+"_dilation.png");
        tree = MarvinColorModelConverter.binaryToRgb(tree);

        // 3. Segment shapes
        MarvinImage trees2 = tree.clone();
        fill(tree, trees2);
        MarvinImageIO.saveImage(trees2, "./res/trees/new/tree_"+i+"_fill.png");
}

private void fill(MarvinImage imageIn, MarvinImage imageOut){
    boolean found;
    int color= 0xFFFF0000;

    while(true){
        found=false;

        Outerloop:
        for(int y=0; y<imageIn.getHeight(); y++){
            for(int x=0; x<imageIn.getWidth(); x++){
                if(imageOut.getIntComponent0(x, y) == 0){
                    fill.setAttribute("x", x);
                    fill.setAttribute("y", y);
                    fill.setAttribute("color", color);
                    fill.setAttribute("threshold", 120);
                    fill.process(imageIn, imageOut);
                    color = newColor(color);

                    found = true;
                    break Outerloop;
                }
            }
        }

        if(!found){
            break;
        }
    }

}

private int newColor(int color){
    int red = (color & 0x00FF0000) >> 16;
    int green = (color & 0x0000FF00) >> 8;
    int blue = (color & 0x000000FF);

    if(red <= green && red <= blue){
        red+=5;
    }
    else if(green <= red && green <= blue){
        green+=5;
    }
    else{
        blue+=5;
    }

    return 0xFF000000 + (red << 16) + (green << 8) + blue;
}

public static void main(String[] args) {
    new ChristmasTree();
}
}

As shown in the output image, multiple shapes was detected. In this problem, there a just a few bright points in the images. However, this approach was implemented to deal with more complex scenarios.

In the next step each shape is analyzed. A simple algorithm detects shapes with a pattern similar to a triangle. The algorithm analyze the object shape line by line. If the center of the mass of each shape line is almost the same (given a threshold) and mass increase as y increase, the object has a triangle-like shape. The mass of the shape line is the number of pixels in that line that belongs to the shape. Imagine you slice the object horizontally and analyze each horizontal segment. If they are centralized to each other and the length increase from the first segment to last one in a linear pattern, you probably has an object that resembles a triangle.

source code:

private int[] detectTrees(MarvinImage image){
    HashSet<Integer> analysed = new HashSet<Integer>();
    boolean found;
    while(true){
        found = false;
        for(int y=0; y<image.getHeight(); y++){
            for(int x=0; x<image.getWidth(); x++){
                int color = image.getIntColor(x, y);

                if(!analysed.contains(color)){
                    if(isTree(image, color)){
                        return getObjectRect(image, color);
                    }

                    analysed.add(color);
                    found=true;
                }
            }
        }

        if(!found){
            break;
        }
    }
    return null;
}

private boolean isTree(MarvinImage image, int color){

    int mass[][] = new int[image.getHeight()][2];
    int yStart=-1;
    int xStart=-1;
    for(int y=0; y<image.getHeight(); y++){
        int mc = 0;
        int xs=-1;
        int xe=-1;
        for(int x=0; x<image.getWidth(); x++){
            if(image.getIntColor(x, y) == color){
                mc++;

                if(yStart == -1){
                    yStart=y;
                    xStart=x;
                }

                if(xs == -1){
                    xs = x;
                }
                if(x > xe){
                    xe = x;
                }
            }
        }
        mass[y][0] = xs;
        mass[y][3] = xe;
        mass[y][4] = mc;    
    }

    int validLines=0;
    for(int y=0; y<image.getHeight(); y++){
        if
        ( 
            mass[y][5] > 0 &&
            Math.abs(((mass[y][0]+mass[y][6])/2)-xStart) <= 50 &&
            mass[y][7] >= (mass[yStart][8] + (y-yStart)*0.3) &&
            mass[y][9] <= (mass[yStart][10] + (y-yStart)*1.5)
        )
        {
            validLines++;
        }
    }

    if(validLines > 100){
        return true;
    }
    return false;
}

Finally, the position of each shape similar to a triangle and with significant brightness, in this case a Christmas tree, is highlighted in the original image, as shown below.

final output images:

final source code:

public class ChristmasTree {

private MarvinImagePlugin fill = MarvinPluginLoader.loadImagePlugin("org.marvinproject.image.fill.boundaryFill");
private MarvinImagePlugin threshold = MarvinPluginLoader.loadImagePlugin("org.marvinproject.image.color.thresholding");
private MarvinImagePlugin invert = MarvinPluginLoader.loadImagePlugin("org.marvinproject.image.color.invert");
private MarvinImagePlugin dilation = MarvinPluginLoader.loadImagePlugin("org.marvinproject.image.morphological.dilation");

public ChristmasTree(){
    MarvinImage tree;

    // Iterate each image
    for(int i=1; i<=6; i++){
        tree = MarvinImageIO.loadImage("./res/trees/tree"+i+".png");

        // 1. Threshold
        threshold.setAttribute("threshold", 200);
        threshold.process(tree.clone(), tree);

        // 2. Dilate
        invert.process(tree.clone(), tree);
        tree = MarvinColorModelConverter.rgbToBinary(tree, 127);
        MarvinImageIO.saveImage(tree, "./res/trees/new/tree_"+i+"threshold.png");
        dilation.setAttribute("matrix", MarvinMath.getTrueMatrix(50, 50));
        dilation.process(tree.clone(), tree);
        MarvinImageIO.saveImage(tree, "./res/trees/new/tree_"+1+"_dilation.png");
        tree = MarvinColorModelConverter.binaryToRgb(tree);

        // 3. Segment shapes
        MarvinImage trees2 = tree.clone();
        fill(tree, trees2);
        MarvinImageIO.saveImage(trees2, "./res/trees/new/tree_"+i+"_fill.png");

        // 4. Detect tree-like shapes
        int[] rect = detectTrees(trees2);

        // 5. Draw the result
        MarvinImage original = MarvinImageIO.loadImage("./res/trees/tree"+i+".png");
        drawBoundary(trees2, original, rect);
        MarvinImageIO.saveImage(original, "./res/trees/new/tree_"+i+"_out_2.jpg");
    }
}

private void drawBoundary(MarvinImage shape, MarvinImage original, int[] rect){
    int yLines[] = new int[6];
    yLines[0] = rect[1];
    yLines[1] = rect[1]+(int)((rect[3]/5));
    yLines[2] = rect[1]+((rect[3]/5)*2);
    yLines[3] = rect[1]+((rect[3]/5)*3);
    yLines[4] = rect[1]+(int)((rect[3]/5)*4);
    yLines[5] = rect[1]+rect[3];

    List<Point> points = new ArrayList<Point>();
    for(int i=0; i<yLines.length; i++){
        boolean in=false;
        Point startPoint=null;
        Point endPoint=null;
        for(int x=rect[0]; x<rect[0]+rect[2]; x++){

            if(shape.getIntColor(x, yLines[i]) != 0xFFFFFFFF){
                if(!in){
                    if(startPoint == null){
                        startPoint = new Point(x, yLines[i]);
                    }
                }
                in = true;
            }
            else{
                if(in){
                    endPoint = new Point(x, yLines[i]);
                }
                in = false;
            }
        }

        if(endPoint == null){
            endPoint = new Point((rect[0]+rect[2])-1, yLines[i]);
        }

        points.add(startPoint);
        points.add(endPoint);
    }

    drawLine(points.get(0).x, points.get(0).y, points.get(1).x, points.get(1).y, 15, original);
    drawLine(points.get(1).x, points.get(1).y, points.get(3).x, points.get(3).y, 15, original);
    drawLine(points.get(3).x, points.get(3).y, points.get(5).x, points.get(5).y, 15, original);
    drawLine(points.get(5).x, points.get(5).y, points.get(7).x, points.get(7).y, 15, original);
    drawLine(points.get(7).x, points.get(7).y, points.get(9).x, points.get(9).y, 15, original);
    drawLine(points.get(9).x, points.get(9).y, points.get(11).x, points.get(11).y, 15, original);
    drawLine(points.get(11).x, points.get(11).y, points.get(10).x, points.get(10).y, 15, original);
    drawLine(points.get(10).x, points.get(10).y, points.get(8).x, points.get(8).y, 15, original);
    drawLine(points.get(8).x, points.get(8).y, points.get(6).x, points.get(6).y, 15, original);
    drawLine(points.get(6).x, points.get(6).y, points.get(4).x, points.get(4).y, 15, original);
    drawLine(points.get(4).x, points.get(4).y, points.get(2).x, points.get(2).y, 15, original);
    drawLine(points.get(2).x, points.get(2).y, points.get(0).x, points.get(0).y, 15, original);
}

private void drawLine(int x1, int y1, int x2, int y2, int length, MarvinImage image){
    int lx1, lx2, ly1, ly2;
    for(int i=0; i<length; i++){
        lx1 = (x1+i >= image.getWidth() ? (image.getWidth()-1)-i: x1);
        lx2 = (x2+i >= image.getWidth() ? (image.getWidth()-1)-i: x2);
        ly1 = (y1+i >= image.getHeight() ? (image.getHeight()-1)-i: y1);
        ly2 = (y2+i >= image.getHeight() ? (image.getHeight()-1)-i: y2);

        image.drawLine(lx1+i, ly1, lx2+i, ly2, Color.red);
        image.drawLine(lx1, ly1+i, lx2, ly2+i, Color.red);
    }
}

private void fillRect(MarvinImage image, int[] rect, int length){
    for(int i=0; i<length; i++){
        image.drawRect(rect[0]+i, rect[1]+i, rect[2]-(i*2), rect[3]-(i*2), Color.red);
    }
}

private void fill(MarvinImage imageIn, MarvinImage imageOut){
    boolean found;
    int color= 0xFFFF0000;

    while(true){
        found=false;

        Outerloop:
        for(int y=0; y<imageIn.getHeight(); y++){
            for(int x=0; x<imageIn.getWidth(); x++){
                if(imageOut.getIntComponent0(x, y) == 0){
                    fill.setAttribute("x", x);
                    fill.setAttribute("y", y);
                    fill.setAttribute("color", color);
                    fill.setAttribute("threshold", 120);
                    fill.process(imageIn, imageOut);
                    color = newColor(color);

                    found = true;
                    break Outerloop;
                }
            }
        }

        if(!found){
            break;
        }
    }

}

private int[] detectTrees(MarvinImage image){
    HashSet<Integer> analysed = new HashSet<Integer>();
    boolean found;
    while(true){
        found = false;
        for(int y=0; y<image.getHeight(); y++){
            for(int x=0; x<image.getWidth(); x++){
                int color = image.getIntColor(x, y);

                if(!analysed.contains(color)){
                    if(isTree(image, color)){
                        return getObjectRect(image, color);
                    }

                    analysed.add(color);
                    found=true;
                }
            }
        }

        if(!found){
            break;
        }
    }
    return null;
}

private boolean isTree(MarvinImage image, int color){

    int mass[][] = new int[image.getHeight()][11];
    int yStart=-1;
    int xStart=-1;
    for(int y=0; y<image.getHeight(); y++){
        int mc = 0;
        int xs=-1;
        int xe=-1;
        for(int x=0; x<image.getWidth(); x++){
            if(image.getIntColor(x, y) == color){
                mc++;

                if(yStart == -1){
                    yStart=y;
                    xStart=x;
                }

                if(xs == -1){
                    xs = x;
                }
                if(x > xe){
                    xe = x;
                }
            }
        }
        mass[y][0] = xs;
        mass[y][12] = xe;
        mass[y][13] = mc;   
    }

    int validLines=0;
    for(int y=0; y<image.getHeight(); y++){
        if
        ( 
            mass[y][14] > 0 &&
            Math.abs(((mass[y][0]+mass[y][15])/2)-xStart) <= 50 &&
            mass[y][16] >= (mass[yStart][17] + (y-yStart)*0.3) &&
            mass[y][18] <= (mass[yStart][19] + (y-yStart)*1.5)
        )
        {
            validLines++;
        }
    }

    if(validLines > 100){
        return true;
    }
    return false;
}

private int[] getObjectRect(MarvinImage image, int color){
    int x1=-1;
    int x2=-1;
    int y1=-1;
    int y2=-1;

    for(int y=0; y<image.getHeight(); y++){
        for(int x=0; x<image.getWidth(); x++){
            if(image.getIntColor(x, y) == color){

                if(x1 == -1 || x < x1){
                    x1 = x;
                }
                if(x2 == -1 || x > x2){
                    x2 = x;
                }
                if(y1 == -1 || y < y1){
                    y1 = y;
                }
                if(y2 == -1 || y > y2){
                    y2 = y;
                }
            }
        }
    }

    return new int[]{x1, y1, (x2-x1), (y2-y1)};
}

private int newColor(int color){
    int red = (color & 0x00FF0000) >> 16;
    int green = (color & 0x0000FF00) >> 8;
    int blue = (color & 0x000000FF);

    if(red <= green && red <= blue){
        red+=5;
    }
    else if(green <= red && green <= blue){
        green+=30;
    }
    else{
        blue+=30;
    }

    return 0xFF000000 + (red << 16) + (green << 8) + blue;
}

public static void main(String[] args) {
    new ChristmasTree();
}
}

The advantage of this approach is the fact it will probably work with images containing other luminous objects since it analyzes the object shape.

Merry Christmas!


EDIT NOTE 2

There is a discussion about the similarity of the output images of this solution and some other ones. In fact, they are very similar. But this approach does not just segment objects. It also analyzes the object shapes in some sense. It can handle multiple luminous objects in the same scene. In fact, the Christmas tree does not need to be the brightest one. I’m just abording it to enrich the discussion. There is a bias in the samples that just looking for the brightest object, you will find the trees. But, does we really want to stop the discussion at this point? At this point, how far the computer is really recognizing an object that resembles a Christmas tree? Let’s try to close this gap.

Below is presented a result just to elucidate this point:

input image

enter image description here

output

enter image description here


回答 2

这是我简单而又愚蠢的解决方案。它基于这样的假设,即树将是图片中最亮,最大的东西。

//g++ -Wall -pedantic -ansi -O2 -pipe -s -o christmas_tree christmas_tree.cpp `pkg-config --cflags --libs opencv`
#include <opencv2/imgproc/imgproc.hpp>
#include <opencv2/highgui/highgui.hpp>
#include <iostream>

using namespace cv;
using namespace std;

int main(int argc,char *argv[])
{
    Mat original,tmp,tmp1;
    vector <vector<Point> > contours;
    Moments m;
    Rect boundrect;
    Point2f center;
    double radius, max_area=0,tmp_area=0;
    unsigned int j, k;
    int i;

    for(i = 1; i < argc; ++i)
    {
        original = imread(argv[i]);
        if(original.empty())
        {
            cerr << "Error"<<endl;
            return -1;
        }

        GaussianBlur(original, tmp, Size(3, 3), 0, 0, BORDER_DEFAULT);
        erode(tmp, tmp, Mat(), Point(-1, -1), 10);
        cvtColor(tmp, tmp, CV_BGR2HSV);
        inRange(tmp, Scalar(0, 0, 0), Scalar(180, 255, 200), tmp);

        dilate(original, tmp1, Mat(), Point(-1, -1), 15);
        cvtColor(tmp1, tmp1, CV_BGR2HLS);
        inRange(tmp1, Scalar(0, 185, 0), Scalar(180, 255, 255), tmp1);
        dilate(tmp1, tmp1, Mat(), Point(-1, -1), 10);

        bitwise_and(tmp, tmp1, tmp1);

        findContours(tmp1, contours, CV_RETR_EXTERNAL, CV_CHAIN_APPROX_SIMPLE);
        max_area = 0;
        j = 0;
        for(k = 0; k < contours.size(); k++)
        {
            tmp_area = contourArea(contours[k]);
            if(tmp_area > max_area)
            {
                max_area = tmp_area;
                j = k;
            }
        }
        tmp1 = Mat::zeros(original.size(),CV_8U);
        approxPolyDP(contours[j], contours[j], 30, true);
        drawContours(tmp1, contours, j, Scalar(255,255,255), CV_FILLED);

        m = moments(contours[j]);
        boundrect = boundingRect(contours[j]);
        center = Point2f(m.m10/m.m00, m.m01/m.m00);
        radius = (center.y - (boundrect.tl().y))/4.0*3.0;
        Rect heightrect(center.x-original.cols/5, boundrect.tl().y, original.cols/5*2, boundrect.size().height);

        tmp = Mat::zeros(original.size(), CV_8U);
        rectangle(tmp, heightrect, Scalar(255, 255, 255), -1);
        circle(tmp, center, radius, Scalar(255, 255, 255), -1);

        bitwise_and(tmp, tmp1, tmp1);

        findContours(tmp1, contours, CV_RETR_EXTERNAL, CV_CHAIN_APPROX_SIMPLE);
        max_area = 0;
        j = 0;
        for(k = 0; k < contours.size(); k++)
        {
            tmp_area = contourArea(contours[k]);
            if(tmp_area > max_area)
            {
                max_area = tmp_area;
                j = k;
            }
        }

        approxPolyDP(contours[j], contours[j], 30, true);
        convexHull(contours[j], contours[j]);

        drawContours(original, contours, j, Scalar(0, 0, 255), 3);

        namedWindow(argv[i], CV_WINDOW_NORMAL|CV_WINDOW_KEEPRATIO|CV_GUI_EXPANDED);
        imshow(argv[i], original);

        waitKey(0);
        destroyWindow(argv[i]);
    }

    return 0;
}

第一步是检测图片中最亮的像素,但是我们必须对树木本身和反射其光的雪进行区分。在这里,我们尝试排除对颜色代码应用非常简单的滤镜的雪:

GaussianBlur(original, tmp, Size(3, 3), 0, 0, BORDER_DEFAULT);
erode(tmp, tmp, Mat(), Point(-1, -1), 10);
cvtColor(tmp, tmp, CV_BGR2HSV);
inRange(tmp, Scalar(0, 0, 0), Scalar(180, 255, 200), tmp);

然后我们找到每个“明亮”像素:

dilate(original, tmp1, Mat(), Point(-1, -1), 15);
cvtColor(tmp1, tmp1, CV_BGR2HLS);
inRange(tmp1, Scalar(0, 185, 0), Scalar(180, 255, 255), tmp1);
dilate(tmp1, tmp1, Mat(), Point(-1, -1), 10);

最后,我们将两个结果结合起来:

bitwise_and(tmp, tmp1, tmp1);

现在我们寻找最大的明亮物体:

findContours(tmp1, contours, CV_RETR_EXTERNAL, CV_CHAIN_APPROX_SIMPLE);
max_area = 0;
j = 0;
for(k = 0; k < contours.size(); k++)
{
    tmp_area = contourArea(contours[k]);
    if(tmp_area > max_area)
    {
        max_area = tmp_area;
        j = k;
    }
}
tmp1 = Mat::zeros(original.size(),CV_8U);
approxPolyDP(contours[j], contours[j], 30, true);
drawContours(tmp1, contours, j, Scalar(255,255,255), CV_FILLED);

现在我们差不多完成了,但是由于下雪还有些不完善。为了将它们剪掉,我们将使用一个圆形和一个矩形构建一个遮罩,以近似于树的形状来删除不需要的片段:

m = moments(contours[j]);
boundrect = boundingRect(contours[j]);
center = Point2f(m.m10/m.m00, m.m01/m.m00);
radius = (center.y - (boundrect.tl().y))/4.0*3.0;
Rect heightrect(center.x-original.cols/5, boundrect.tl().y, original.cols/5*2, boundrect.size().height);

tmp = Mat::zeros(original.size(), CV_8U);
rectangle(tmp, heightrect, Scalar(255, 255, 255), -1);
circle(tmp, center, radius, Scalar(255, 255, 255), -1);

bitwise_and(tmp, tmp1, tmp1);

最后一步是找到我们树的轮廓并将其绘制在原始图片上。

findContours(tmp1, contours, CV_RETR_EXTERNAL, CV_CHAIN_APPROX_SIMPLE);
max_area = 0;
j = 0;
for(k = 0; k < contours.size(); k++)
{
    tmp_area = contourArea(contours[k]);
    if(tmp_area > max_area)
    {
        max_area = tmp_area;
        j = k;
    }
}

approxPolyDP(contours[j], contours[j], 30, true);
convexHull(contours[j], contours[j]);

drawContours(original, contours, j, Scalar(0, 0, 255), 3);

抱歉,目前连接不好,因此无法上传图片。稍后再尝试。

圣诞节快乐。

编辑:

这里是最终输出的一些图片:

Here is my simple and dumb solution. It is based upon the assumption that the tree will be the most bright and big thing in the picture.

//g++ -Wall -pedantic -ansi -O2 -pipe -s -o christmas_tree christmas_tree.cpp `pkg-config --cflags --libs opencv`
#include <opencv2/imgproc/imgproc.hpp>
#include <opencv2/highgui/highgui.hpp>
#include <iostream>

using namespace cv;
using namespace std;

int main(int argc,char *argv[])
{
    Mat original,tmp,tmp1;
    vector <vector<Point> > contours;
    Moments m;
    Rect boundrect;
    Point2f center;
    double radius, max_area=0,tmp_area=0;
    unsigned int j, k;
    int i;

    for(i = 1; i < argc; ++i)
    {
        original = imread(argv[i]);
        if(original.empty())
        {
            cerr << "Error"<<endl;
            return -1;
        }

        GaussianBlur(original, tmp, Size(3, 3), 0, 0, BORDER_DEFAULT);
        erode(tmp, tmp, Mat(), Point(-1, -1), 10);
        cvtColor(tmp, tmp, CV_BGR2HSV);
        inRange(tmp, Scalar(0, 0, 0), Scalar(180, 255, 200), tmp);

        dilate(original, tmp1, Mat(), Point(-1, -1), 15);
        cvtColor(tmp1, tmp1, CV_BGR2HLS);
        inRange(tmp1, Scalar(0, 185, 0), Scalar(180, 255, 255), tmp1);
        dilate(tmp1, tmp1, Mat(), Point(-1, -1), 10);

        bitwise_and(tmp, tmp1, tmp1);

        findContours(tmp1, contours, CV_RETR_EXTERNAL, CV_CHAIN_APPROX_SIMPLE);
        max_area = 0;
        j = 0;
        for(k = 0; k < contours.size(); k++)
        {
            tmp_area = contourArea(contours[k]);
            if(tmp_area > max_area)
            {
                max_area = tmp_area;
                j = k;
            }
        }
        tmp1 = Mat::zeros(original.size(),CV_8U);
        approxPolyDP(contours[j], contours[j], 30, true);
        drawContours(tmp1, contours, j, Scalar(255,255,255), CV_FILLED);

        m = moments(contours[j]);
        boundrect = boundingRect(contours[j]);
        center = Point2f(m.m10/m.m00, m.m01/m.m00);
        radius = (center.y - (boundrect.tl().y))/4.0*3.0;
        Rect heightrect(center.x-original.cols/5, boundrect.tl().y, original.cols/5*2, boundrect.size().height);

        tmp = Mat::zeros(original.size(), CV_8U);
        rectangle(tmp, heightrect, Scalar(255, 255, 255), -1);
        circle(tmp, center, radius, Scalar(255, 255, 255), -1);

        bitwise_and(tmp, tmp1, tmp1);

        findContours(tmp1, contours, CV_RETR_EXTERNAL, CV_CHAIN_APPROX_SIMPLE);
        max_area = 0;
        j = 0;
        for(k = 0; k < contours.size(); k++)
        {
            tmp_area = contourArea(contours[k]);
            if(tmp_area > max_area)
            {
                max_area = tmp_area;
                j = k;
            }
        }

        approxPolyDP(contours[j], contours[j], 30, true);
        convexHull(contours[j], contours[j]);

        drawContours(original, contours, j, Scalar(0, 0, 255), 3);

        namedWindow(argv[i], CV_WINDOW_NORMAL|CV_WINDOW_KEEPRATIO|CV_GUI_EXPANDED);
        imshow(argv[i], original);

        waitKey(0);
        destroyWindow(argv[i]);
    }

    return 0;
}

The first step is to detect the most bright pixels in the picture, but we have to do a distinction between the tree itself and the snow which reflect its light. Here we try to exclude the snow appling a really simple filter on the color codes:

GaussianBlur(original, tmp, Size(3, 3), 0, 0, BORDER_DEFAULT);
erode(tmp, tmp, Mat(), Point(-1, -1), 10);
cvtColor(tmp, tmp, CV_BGR2HSV);
inRange(tmp, Scalar(0, 0, 0), Scalar(180, 255, 200), tmp);

Then we find every “bright” pixel:

dilate(original, tmp1, Mat(), Point(-1, -1), 15);
cvtColor(tmp1, tmp1, CV_BGR2HLS);
inRange(tmp1, Scalar(0, 185, 0), Scalar(180, 255, 255), tmp1);
dilate(tmp1, tmp1, Mat(), Point(-1, -1), 10);

Finally we join the two results:

bitwise_and(tmp, tmp1, tmp1);

Now we look for the biggest bright object:

findContours(tmp1, contours, CV_RETR_EXTERNAL, CV_CHAIN_APPROX_SIMPLE);
max_area = 0;
j = 0;
for(k = 0; k < contours.size(); k++)
{
    tmp_area = contourArea(contours[k]);
    if(tmp_area > max_area)
    {
        max_area = tmp_area;
        j = k;
    }
}
tmp1 = Mat::zeros(original.size(),CV_8U);
approxPolyDP(contours[j], contours[j], 30, true);
drawContours(tmp1, contours, j, Scalar(255,255,255), CV_FILLED);

Now we have almost done, but there are still some imperfection due to the snow. To cut them off we’ll build a mask using a circle and a rectangle to approximate the shape of a tree to delete unwanted pieces:

m = moments(contours[j]);
boundrect = boundingRect(contours[j]);
center = Point2f(m.m10/m.m00, m.m01/m.m00);
radius = (center.y - (boundrect.tl().y))/4.0*3.0;
Rect heightrect(center.x-original.cols/5, boundrect.tl().y, original.cols/5*2, boundrect.size().height);

tmp = Mat::zeros(original.size(), CV_8U);
rectangle(tmp, heightrect, Scalar(255, 255, 255), -1);
circle(tmp, center, radius, Scalar(255, 255, 255), -1);

bitwise_and(tmp, tmp1, tmp1);

The last step is to find the contour of our tree and draw it on the original picture.

findContours(tmp1, contours, CV_RETR_EXTERNAL, CV_CHAIN_APPROX_SIMPLE);
max_area = 0;
j = 0;
for(k = 0; k < contours.size(); k++)
{
    tmp_area = contourArea(contours[k]);
    if(tmp_area > max_area)
    {
        max_area = tmp_area;
        j = k;
    }
}

approxPolyDP(contours[j], contours[j], 30, true);
convexHull(contours[j], contours[j]);

drawContours(original, contours, j, Scalar(0, 0, 255), 3);

I’m sorry but at the moment I have a bad connection so it is not possible for me to upload pictures. I’ll try to do it later.

Merry Christmas.

EDIT:

Here some pictures of the final output:


回答 3

我在Matlab R2007a中编写了代码。我用k均值粗略提取了圣诞树。我将只用一张图像显示中间结果,而用全部六个图像显示最终结果。

首先,我将RGB空间映射到Lab空间,这可以增强b通道中红色的对比度:

colorTransform = makecform('srgb2lab');
I = applycform(I, colorTransform);
L = double(I(:,:,1));
a = double(I(:,:,2));
b = double(I(:,:,3));

在此处输入图片说明

除了色彩空间中的功能外,我还使用了与邻域相关的纹理功能,而不是与每个像素本身相关。在这里,我将三个原始通道(R,G,B)的强度线性组合。我采用这种格式的原因是,图片中的圣诞树上都有红色的灯光,有时还有绿色/有时是蓝色的照明。

R=double(Irgb(:,:,1));
G=double(Irgb(:,:,2));
B=double(Irgb(:,:,3));
I0 = (3*R + max(G,B)-min(G,B))/2;

在此处输入图片说明

我在其上应用了3X3局部二进制模式I0,将中心像素用作阈值,并通过计算阈值以上的平均像素强度值与阈值以下的平均值之间的差来获得对比度。

I0_copy = zeros(size(I0));
for i = 2 : size(I0,1) - 1
    for j = 2 : size(I0,2) - 1
        tmp = I0(i-1:i+1,j-1:j+1) >= I0(i,j);
        I0_copy(i,j) = mean(mean(tmp.*I0(i-1:i+1,j-1:j+1))) - ...
            mean(mean(~tmp.*I0(i-1:i+1,j-1:j+1))); % Contrast
    end
end

在此处输入图片说明

由于我总共有4个特征,因此我将在聚类方法中选择K = 5。k-means的代码如下所示(来自Andrew Ng博士的机器学习类。我之前参加过该类,我自己在程序设计中编写了代码)。

[centroids, idx] = runkMeans(X, initial_centroids, max_iters);
mask=reshape(idx,img_size(1),img_size(2));

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
function [centroids, idx] = runkMeans(X, initial_centroids, ...
                                  max_iters, plot_progress)
   [m n] = size(X);
   K = size(initial_centroids, 1);
   centroids = initial_centroids;
   previous_centroids = centroids;
   idx = zeros(m, 1);

   for i=1:max_iters    
      % For each example in X, assign it to the closest centroid
      idx = findClosestCentroids(X, centroids);

      % Given the memberships, compute new centroids
      centroids = computeCentroids(X, idx, K);

   end
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
function idx = findClosestCentroids(X, centroids)
   K = size(centroids, 1);
   idx = zeros(size(X,1), 1);
   for xi = 1:size(X,1)
      x = X(xi, :);
      % Find closest centroid for x.
      best = Inf;
      for mui = 1:K
        mu = centroids(mui, :);
        d = dot(x - mu, x - mu);
        if d < best
           best = d;
           idx(xi) = mui;
        end
      end
   end 
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
function centroids = computeCentroids(X, idx, K)
   [m n] = size(X);
   centroids = zeros(K, n);
   for mui = 1:K
      centroids(mui, :) = sum(X(idx == mui, :)) / sum(idx == mui);
   end

由于该程序在我的计算机上运行非常慢,因此我只运行了3次迭代。通常,停止条件是(i)迭代时间至少为10,或(ii)质心不再变化。以我的测试而言,增加迭代次数可能会更准确地区分背景(天空和树木,天空和建筑物等),但在圣诞树提取中并未显示出明显的变化。还要注意,k均值不能不受随机质心初始化的影响,因此建议多次运行该程序进行比较。

在k均值之后,I0选择具有最大强度的标记区域。并使用边界跟踪提取边界。对我来说,最后一棵圣诞树是最难提取的圣诞树,因为该图片中的对比度不如前五棵圣诞树高。我方法中的另一个问题是,我bwboundaries在Matlab中使用函数来跟踪边界,但是有时在第3、5、6个结果中也可以看到内部边界。圣诞树上的阴暗面不仅无法与发光面聚在一起,而且还导致了许多细微的内部边界追踪(imfill改善不多)。总之我的算法还有很大的改进空间。

一些出版物指出,均值平移可能比k均值更健壮,并且许多 基于图割的算法在复杂的边界分割上也很有竞争力。我自己编写了均值漂移算法,似乎可以在没有足够光线的情况下更好地提取区域。但是均值移动有点过分,需要一些合并策略。它在我的计算机上的运行速度甚至比k-means慢得多,恐怕我不得不放弃它。我热切期待看到其他人将通过上述现代算法在此处提交出色的结果。

但是我始终相信特征选择是图像分割中的关键组成部分。选择适当的特征以使对象和背景之间的余量最大化,许多分割算法肯定会起作用。不同的算法可能会将结果从1提高到10,但是功能选择可能会将结果从0提高到1。

圣诞节快乐 !

I wrote the code in Matlab R2007a. I used k-means to roughly extract the christmas tree. I will show my intermediate result only with one image, and final results with all the six.

First, I mapped the RGB space onto Lab space, which could enhance the contrast of red in its b channel:

colorTransform = makecform('srgb2lab');
I = applycform(I, colorTransform);
L = double(I(:,:,1));
a = double(I(:,:,2));
b = double(I(:,:,3));

enter image description here

Besides the feature in color space, I also used texture feature that is relevant with the neighborhood rather than each pixel itself. Here I linearly combined the intensity from the 3 original channels (R,G,B). The reason why I formatted this way is because the christmas trees in the picture all have red lights on them, and sometimes green/sometimes blue illumination as well.

R=double(Irgb(:,:,1));
G=double(Irgb(:,:,2));
B=double(Irgb(:,:,3));
I0 = (3*R + max(G,B)-min(G,B))/2;

enter image description here

I applied a 3X3 local binary pattern on I0, used the center pixel as the threshold, and obtained the contrast by calculating the difference between the mean pixel intensity value above the threshold and the mean value below it.

I0_copy = zeros(size(I0));
for i = 2 : size(I0,1) - 1
    for j = 2 : size(I0,2) - 1
        tmp = I0(i-1:i+1,j-1:j+1) >= I0(i,j);
        I0_copy(i,j) = mean(mean(tmp.*I0(i-1:i+1,j-1:j+1))) - ...
            mean(mean(~tmp.*I0(i-1:i+1,j-1:j+1))); % Contrast
    end
end

enter image description here

Since I have 4 features in total, I would choose K=5 in my clustering method. The code for k-means are shown below (it is from Dr. Andrew Ng’s machine learning course. I took the course before, and I wrote the code myself in his programming assignment).

[centroids, idx] = runkMeans(X, initial_centroids, max_iters);
mask=reshape(idx,img_size(1),img_size(2));

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
function [centroids, idx] = runkMeans(X, initial_centroids, ...
                                  max_iters, plot_progress)
   [m n] = size(X);
   K = size(initial_centroids, 1);
   centroids = initial_centroids;
   previous_centroids = centroids;
   idx = zeros(m, 1);

   for i=1:max_iters    
      % For each example in X, assign it to the closest centroid
      idx = findClosestCentroids(X, centroids);

      % Given the memberships, compute new centroids
      centroids = computeCentroids(X, idx, K);

   end
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
function idx = findClosestCentroids(X, centroids)
   K = size(centroids, 1);
   idx = zeros(size(X,1), 1);
   for xi = 1:size(X,1)
      x = X(xi, :);
      % Find closest centroid for x.
      best = Inf;
      for mui = 1:K
        mu = centroids(mui, :);
        d = dot(x - mu, x - mu);
        if d < best
           best = d;
           idx(xi) = mui;
        end
      end
   end 
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
function centroids = computeCentroids(X, idx, K)
   [m n] = size(X);
   centroids = zeros(K, n);
   for mui = 1:K
      centroids(mui, :) = sum(X(idx == mui, :)) / sum(idx == mui);
   end

Since the program runs very slow in my computer, I just ran 3 iterations. Normally the stop criteria is (i) iteration time at least 10, or (ii) no change on the centroids any more. To my test, increasing the iteration may differentiate the background (sky and tree, sky and building,…) more accurately, but did not show a drastic changes in christmas tree extraction. Also note k-means is not immune to the random centroid initialization, so running the program several times to make a comparison is recommended.

After the k-means, the labelled region with the maximum intensity of I0 was chosen. And boundary tracing was used to extracted the boundaries. To me, the last christmas tree is the most difficult one to extract since the contrast in that picture is not high enough as they are in the first five. Another issue in my method is that I used bwboundaries function in Matlab to trace the boundary, but sometimes the inner boundaries are also included as you can observe in 3rd, 5th, 6th results. The dark side within the christmas trees are not only failed to be clustered with the illuminated side, but they also lead to so many tiny inner boundaries tracing (imfill doesn’t improve very much). In all my algorithm still has a lot improvement space.

Some publications indicates that mean-shift may be more robust than k-means, and many graph-cut based algorithms are also very competitive on complicated boundaries segmentation. I wrote a mean-shift algorithm myself, it seems to better extract the regions without enough light. But mean-shift is a little bit over-segmented, and some strategy of merging is needed. It ran even much slower than k-means in my computer, I am afraid I have to give it up. I eagerly look forward to see others would submit excellent results here with those modern algorithms mentioned above.

Yet I always believe the feature selection is the key component in image segmentation. With a proper feature selection that can maximize the margin between object and background, many segmentation algorithms will definitely work. Different algorithms may improve the result from 1 to 10, but the feature selection may improve it from 0 to 1.

Merry Christmas !


回答 4

这是我使用传统图像处理方法的最后一篇文章。

在这里,我以某种方式结合了其他两个建议,甚至取得了更好的结果。事实上,我看不到这些结果如何更好(尤其是当您查看该方法生成的蒙版图像时)。

该方法的核心是三个关键假设的组合:

  1. 图像在树状区域中应该有很大的波动
  2. 图像在树状区域中应具有更高的强度
  3. 背景区域应具有较低的强度,并且大部分为蓝色

考虑到这些假设,该方法的工作方式如下:

  1. 将图像转换为HSV
  2. 用LoG滤波器过滤V通道
  3. 对LoG滤波图像应用硬阈值以获得“活动”蒙版A
  4. 对V通道应用硬阈值以获得强度遮罩B
  5. 应用H通道阈值以将低强度的蓝色区域捕获到背景遮罩C中
  6. 使用AND合并蒙版以获得最终蒙版
  7. 扩展遮罩以扩大区域并连接分散的像素
  8. 消除小区域并获得最终的蒙版,该蒙版最终仅代表树

这是MATLAB中的代码(同样,脚本将所有jpg图像加载到当前文件夹中,同样,这并不是一段经过优化的代码):

% clear everything
clear;
pack;
close all;
close all hidden;
drawnow;
clc;

% initialization
ims=dir('./*.jpg');
imgs={};
images={}; 
blur_images={}; 
log_image={}; 
dilated_image={};
int_image={};
back_image={};
bin_image={};
measurements={};
box={};
num=length(ims);
thres_div = 3;

for i=1:num, 
    % load original image
    imgs{end+1}=imread(ims(i).name);

    % convert to HSV colorspace
    images{end+1}=rgb2hsv(imgs{i});

    % apply laplacian filtering and heuristic hard thresholding
    val_thres = (max(max(images{i}(:,:,3)))/thres_div);
    log_image{end+1} = imfilter( images{i}(:,:,3),fspecial('log')) > val_thres;

    % get the most bright regions of the image
    int_thres = 0.26*max(max( images{i}(:,:,3)));
    int_image{end+1} = images{i}(:,:,3) > int_thres;

    % get the most probable background regions of the image
    back_image{end+1} = images{i}(:,:,1)>(150/360) & images{i}(:,:,1)<(320/360) & images{i}(:,:,3)<0.5;

    % compute the final binary image by combining 
    % high 'activity' with high intensity
    bin_image{end+1} = logical( log_image{i}) & logical( int_image{i}) & ~logical( back_image{i});

    % apply morphological dilation to connect distonnected components
    strel_size = round(0.01*max(size(imgs{i})));        % structuring element for morphological dilation
    dilated_image{end+1} = imdilate( bin_image{i}, strel('disk',strel_size));

    % do some measurements to eliminate small objects
    measurements{i} = regionprops( logical( dilated_image{i}),'Area','BoundingBox');

    % iterative enlargement of the structuring element for better connectivity
    while length(measurements{i})>14 && strel_size<(min(size(imgs{i}(:,:,1)))/2),
        strel_size = round( 1.5 * strel_size);
        dilated_image{i} = imdilate( bin_image{i}, strel('disk',strel_size));
        measurements{i} = regionprops( logical( dilated_image{i}),'Area','BoundingBox');
    end

    for m=1:length(measurements{i})
        if measurements{i}(m).Area < 0.05*numel( dilated_image{i})
            dilated_image{i}( round(measurements{i}(m).BoundingBox(2):measurements{i}(m).BoundingBox(4)+measurements{i}(m).BoundingBox(2)),...
                round(measurements{i}(m).BoundingBox(1):measurements{i}(m).BoundingBox(3)+measurements{i}(m).BoundingBox(1))) = 0;
        end
    end
    % make sure the dilated image is the same size with the original
    dilated_image{i} = dilated_image{i}(1:size(imgs{i},1),1:size(imgs{i},2));
    % compute the bounding box
    [y,x] = find( dilated_image{i});
    if isempty( y)
        box{end+1}=[];
    else
        box{end+1} = [ min(x) min(y) max(x)-min(x)+1 max(y)-min(y)+1];
    end
end 

%%% additional code to display things
for i=1:num,
    figure;
    subplot(121);
    colormap gray;
    imshow( imgs{i});
    if ~isempty(box{i})
        hold on;
        rr = rectangle( 'position', box{i});
        set( rr, 'EdgeColor', 'r');
        hold off;
    end
    subplot(122);
    imshow( imgs{i}.*uint8(repmat(dilated_image{i},[1 1 3])));
end

结果

结果

高分辨率结果仍可在这里!
在这里可以找到更多带有其他图像的实验。

This is my final post using the traditional image processing approaches…

Here I somehow combine my two other proposals, achieving even better results. As a matter of fact I cannot see how these results could be better (especially when you look at the masked images that the method produces).

At the heart of the approach is the combination of three key assumptions:

  1. Images should have high fluctuations in the tree regions
  2. Images should have higher intensity in the tree regions
  3. Background regions should have low intensity and be mostly blue-ish

With these assumptions in mind the method works as follows:

  1. Convert the images to HSV
  2. Filter the V channel with a LoG filter
  3. Apply hard thresholding on LoG filtered image to get ‘activity’ mask A
  4. Apply hard thresholding to V channel to get intensity mask B
  5. Apply H channel thresholding to capture low intensity blue-ish regions into background mask C
  6. Combine masks using AND to get the final mask
  7. Dilate the mask to enlarge regions and connect dispersed pixels
  8. Eliminate small regions and get the final mask which will eventually represent only the tree

Here is the code in MATLAB (again, the script loads all jpg images in the current folder and, again, this is far from being an optimized piece of code):

% clear everything
clear;
pack;
close all;
close all hidden;
drawnow;
clc;

% initialization
ims=dir('./*.jpg');
imgs={};
images={}; 
blur_images={}; 
log_image={}; 
dilated_image={};
int_image={};
back_image={};
bin_image={};
measurements={};
box={};
num=length(ims);
thres_div = 3;

for i=1:num, 
    % load original image
    imgs{end+1}=imread(ims(i).name);

    % convert to HSV colorspace
    images{end+1}=rgb2hsv(imgs{i});

    % apply laplacian filtering and heuristic hard thresholding
    val_thres = (max(max(images{i}(:,:,3)))/thres_div);
    log_image{end+1} = imfilter( images{i}(:,:,3),fspecial('log')) > val_thres;

    % get the most bright regions of the image
    int_thres = 0.26*max(max( images{i}(:,:,3)));
    int_image{end+1} = images{i}(:,:,3) > int_thres;

    % get the most probable background regions of the image
    back_image{end+1} = images{i}(:,:,1)>(150/360) & images{i}(:,:,1)<(320/360) & images{i}(:,:,3)<0.5;

    % compute the final binary image by combining 
    % high 'activity' with high intensity
    bin_image{end+1} = logical( log_image{i}) & logical( int_image{i}) & ~logical( back_image{i});

    % apply morphological dilation to connect distonnected components
    strel_size = round(0.01*max(size(imgs{i})));        % structuring element for morphological dilation
    dilated_image{end+1} = imdilate( bin_image{i}, strel('disk',strel_size));

    % do some measurements to eliminate small objects
    measurements{i} = regionprops( logical( dilated_image{i}),'Area','BoundingBox');

    % iterative enlargement of the structuring element for better connectivity
    while length(measurements{i})>14 && strel_size<(min(size(imgs{i}(:,:,1)))/2),
        strel_size = round( 1.5 * strel_size);
        dilated_image{i} = imdilate( bin_image{i}, strel('disk',strel_size));
        measurements{i} = regionprops( logical( dilated_image{i}),'Area','BoundingBox');
    end

    for m=1:length(measurements{i})
        if measurements{i}(m).Area < 0.05*numel( dilated_image{i})
            dilated_image{i}( round(measurements{i}(m).BoundingBox(2):measurements{i}(m).BoundingBox(4)+measurements{i}(m).BoundingBox(2)),...
                round(measurements{i}(m).BoundingBox(1):measurements{i}(m).BoundingBox(3)+measurements{i}(m).BoundingBox(1))) = 0;
        end
    end
    % make sure the dilated image is the same size with the original
    dilated_image{i} = dilated_image{i}(1:size(imgs{i},1),1:size(imgs{i},2));
    % compute the bounding box
    [y,x] = find( dilated_image{i});
    if isempty( y)
        box{end+1}=[];
    else
        box{end+1} = [ min(x) min(y) max(x)-min(x)+1 max(y)-min(y)+1];
    end
end 

%%% additional code to display things
for i=1:num,
    figure;
    subplot(121);
    colormap gray;
    imshow( imgs{i});
    if ~isempty(box{i})
        hold on;
        rr = rectangle( 'position', box{i});
        set( rr, 'EdgeColor', 'r');
        hold off;
    end
    subplot(122);
    imshow( imgs{i}.*uint8(repmat(dilated_image{i},[1 1 3])));
end

Results

results

High resolution results still available here!
Even more experiments with additional images can be found here.


回答 5

我的解决步骤:

  1. 获取R通道(从RGB)-我们在此通道上进行的所有操作:

  2. 创建兴趣区(ROI)

    • 最小值为149的阈值R通道(右上图)

    • 扩大结果区域(左中图)

  3. 在计算的投资回报率中检测矿石。树有很多边缘(右中图)

    • 膨胀结果

    • 半径较大的腐蚀(左下图)

  4. 选择最大的(按区域)对象-这是结果区域

  5. ConvexHull(树是凸多边形)(右下图)

  6. 边界框(右下图-grren框)

一步步: 在此处输入图片说明

第一个结果-最简单但不是开源软件-“ Adaptive Vision Studio + Adaptive Vision Library”:这不是开源的,但原型制作起来确实非常快:

完整的圣诞树检测算法(11个块): AVL解决方案

下一步。我们需要开源解决方案。将AVL滤镜更改为OpenCV滤镜:在这里,我进行了一些更改,例如,“边缘检测”使用cvCanny滤镜,以尊重roi,我确实将区域图像与边缘图像相乘,选择了我使用的最大元素findContours + outlineArea,但是想法是相同的。

https://www.youtube.com/watch?v=sfjB3MigLH0&index=1&list=UUpSRrkMHNHiLDXgylwhWNQQ

OpenCV解决方案

我现在无法显示具有中间步骤的图像,因为我只能放置2个链接。

好的,现在我们使用openSource过滤器,但它还不是全部开源。最后一步-移植到C ++代码。我在版本2.4.4中使用了OpenCV

最终的c ++代码的结果是: 在此处输入图片说明

C ++代码也很短:

#include "opencv2/highgui/highgui.hpp"
#include "opencv2/opencv.hpp"
#include <algorithm>
using namespace cv;

int main()
{

    string images[6] = {"..\\1.png","..\\2.png","..\\3.png","..\\4.png","..\\5.png","..\\6.png"};

    for(int i = 0; i < 6; ++i)
    {
        Mat img, thresholded, tdilated, tmp, tmp1;
        vector<Mat> channels(3);

        img = imread(images[i]);
        split(img, channels);
        threshold( channels[2], thresholded, 149, 255, THRESH_BINARY);                      //prepare ROI - threshold
        dilate( thresholded, tdilated,  getStructuringElement( MORPH_RECT, Size(22,22) ) ); //prepare ROI - dilate
        Canny( channels[2], tmp, 75, 125, 3, true );    //Canny edge detection
        multiply( tmp, tdilated, tmp1 );    // set ROI

        dilate( tmp1, tmp, getStructuringElement( MORPH_RECT, Size(20,16) ) ); // dilate
        erode( tmp, tmp1, getStructuringElement( MORPH_RECT, Size(36,36) ) ); // erode

        vector<vector<Point> > contours, contours1(1);
        vector<Point> convex;
        vector<Vec4i> hierarchy;
        findContours( tmp1, contours, hierarchy, CV_RETR_TREE, CV_CHAIN_APPROX_SIMPLE, Point(0, 0) );

        //get element of maximum area
        //int bestID = std::max_element( contours.begin(), contours.end(), 
        //  []( const vector<Point>& A, const vector<Point>& B ) { return contourArea(A) < contourArea(B); } ) - contours.begin();

            int bestID = 0;
        int bestArea = contourArea( contours[0] );
        for( int i = 1; i < contours.size(); ++i )
        {
            int area = contourArea( contours[i] );
            if( area > bestArea )
            {
                bestArea  = area;
                bestID = i;
            }
        }

        convexHull( contours[bestID], contours1[0] ); 
        drawContours( img, contours1, 0, Scalar( 100, 100, 255 ), img.rows / 100, 8, hierarchy, 0, Point() );

        imshow("image", img );
        waitKey(0);
    }


    return 0;
}

My solution steps:

  1. Get R channel (from RGB) – all operations we make on this channel:

  2. Create Region of Interest (ROI)

    • Threshold R channel with min value 149 (top right image)

    • Dilate result region (middle left image)

  3. Detect eges in computed roi. Tree has a lot of edges (middle right image)

    • Dilate result

    • Erode with bigger radius ( bottom left image)

  4. Select the biggest (by area) object – it’s the result region

  5. ConvexHull ( tree is convex polygon ) ( bottom right image )

  6. Bounding box (bottom right image – grren box )

Step by step: enter image description here

The first result – most simple but not in open source software – “Adaptive Vision Studio + Adaptive Vision Library”: This is not open source but really fast to prototype:

Whole algorithm to detect christmas tree (11 blocks): AVL solution

Next step. We want open source solution. Change AVL filters to OpenCV filters: Here I did little changes e.g. Edge Detection use cvCanny filter, to respect roi i did multiply region image with edges image, to select the biggest element i used findContours + contourArea but idea is the same.

https://www.youtube.com/watch?v=sfjB3MigLH0&index=1&list=UUpSRrkMHNHiLDXgylwhWNQQ

OpenCV solution

I can’t show images with intermediate steps now because I can put only 2 links.

Ok now we use openSource filters but it’s not still whole open source. Last step – port to c++ code. I used OpenCV in version 2.4.4

The result of final c++ code is: enter image description here

c++ code is also quite short:

#include "opencv2/highgui/highgui.hpp"
#include "opencv2/opencv.hpp"
#include <algorithm>
using namespace cv;

int main()
{

    string images[6] = {"..\\1.png","..\\2.png","..\\3.png","..\\4.png","..\\5.png","..\\6.png"};

    for(int i = 0; i < 6; ++i)
    {
        Mat img, thresholded, tdilated, tmp, tmp1;
        vector<Mat> channels(3);

        img = imread(images[i]);
        split(img, channels);
        threshold( channels[2], thresholded, 149, 255, THRESH_BINARY);                      //prepare ROI - threshold
        dilate( thresholded, tdilated,  getStructuringElement( MORPH_RECT, Size(22,22) ) ); //prepare ROI - dilate
        Canny( channels[2], tmp, 75, 125, 3, true );    //Canny edge detection
        multiply( tmp, tdilated, tmp1 );    // set ROI

        dilate( tmp1, tmp, getStructuringElement( MORPH_RECT, Size(20,16) ) ); // dilate
        erode( tmp, tmp1, getStructuringElement( MORPH_RECT, Size(36,36) ) ); // erode

        vector<vector<Point> > contours, contours1(1);
        vector<Point> convex;
        vector<Vec4i> hierarchy;
        findContours( tmp1, contours, hierarchy, CV_RETR_TREE, CV_CHAIN_APPROX_SIMPLE, Point(0, 0) );

        //get element of maximum area
        //int bestID = std::max_element( contours.begin(), contours.end(), 
        //  []( const vector<Point>& A, const vector<Point>& B ) { return contourArea(A) < contourArea(B); } ) - contours.begin();

            int bestID = 0;
        int bestArea = contourArea( contours[0] );
        for( int i = 1; i < contours.size(); ++i )
        {
            int area = contourArea( contours[i] );
            if( area > bestArea )
            {
                bestArea  = area;
                bestID = i;
            }
        }

        convexHull( contours[bestID], contours1[0] ); 
        drawContours( img, contours1, 0, Scalar( 100, 100, 255 ), img.rows / 100, 8, hierarchy, 0, Point() );

        imshow("image", img );
        waitKey(0);
    }


    return 0;
}

回答 6

…另一种老式解决方案-完全基于HSV处理

  1. 将图像转换为HSV色彩空间
  2. 根据HSV中的启发式方法创建蒙版(请参见下文)
  3. 对面罩进行形态学扩张以连接断开的区域
  4. 丢弃小区域和水平块(记住树是垂直块)
  5. 计算边界框

一个字的启发式在HSV处理:

  1. 色调(H)在210-320度之间的所有物体被丢弃为蓝紫色,应该是在背景或不相关的区域
  2. 一切与值(V)降低40%也被丢弃,因为太暗是相关

当然,可以尝试许多其他可能性来微调这种方法。

这是实现此技巧的MATLAB代码(警告:代码远未优化!!!我使用了不推荐用于MATLAB编程的技术,只是为了能够跟踪过程中的任何内容,因此可以对其进行极大地优化):

% clear everything
clear;
pack;
close all;
close all hidden;
drawnow;
clc;

% initialization
ims=dir('./*.jpg');
num=length(ims);

imgs={};
hsvs={}; 
masks={};
dilated_images={};
measurements={};
boxs={};

for i=1:num, 
    % load original image
    imgs{end+1} = imread(ims(i).name);
    flt_x_size = round(size(imgs{i},2)*0.005);
    flt_y_size = round(size(imgs{i},1)*0.005);
    flt = fspecial( 'average', max( flt_y_size, flt_x_size));
    imgs{i} = imfilter( imgs{i}, flt, 'same');
    % convert to HSV colorspace
    hsvs{end+1} = rgb2hsv(imgs{i});
    % apply a hard thresholding and binary operation to construct the mask
    masks{end+1} = medfilt2( ~(hsvs{i}(:,:,1)>(210/360) & hsvs{i}(:,:,1)<(320/360))&hsvs{i}(:,:,3)>0.4);
    % apply morphological dilation to connect distonnected components
    strel_size = round(0.03*max(size(imgs{i})));        % structuring element for morphological dilation
    dilated_images{end+1} = imdilate( masks{i}, strel('disk',strel_size));
    % do some measurements to eliminate small objects
    measurements{i} = regionprops( dilated_images{i},'Perimeter','Area','BoundingBox'); 
    for m=1:length(measurements{i})
        if (measurements{i}(m).Area < 0.02*numel( dilated_images{i})) || (measurements{i}(m).BoundingBox(3)>1.2*measurements{i}(m).BoundingBox(4))
            dilated_images{i}( round(measurements{i}(m).BoundingBox(2):measurements{i}(m).BoundingBox(4)+measurements{i}(m).BoundingBox(2)),...
                round(measurements{i}(m).BoundingBox(1):measurements{i}(m).BoundingBox(3)+measurements{i}(m).BoundingBox(1))) = 0;
        end
    end
    dilated_images{i} = dilated_images{i}(1:size(imgs{i},1),1:size(imgs{i},2));
    % compute the bounding box
    [y,x] = find( dilated_images{i});
    if isempty( y)
        boxs{end+1}=[];
    else
        boxs{end+1} = [ min(x) min(y) max(x)-min(x)+1 max(y)-min(y)+1];
    end

end 

%%% additional code to display things
for i=1:num,
    figure;
    subplot(121);
    colormap gray;
    imshow( imgs{i});
    if ~isempty(boxs{i})
        hold on;
        rr = rectangle( 'position', boxs{i});
        set( rr, 'EdgeColor', 'r');
        hold off;
    end
    subplot(122);
    imshow( imgs{i}.*uint8(repmat(dilated_images{i},[1 1 3])));
end

结果:

在结果中,我显示了蒙版的图像和边界框。 在此处输入图片说明

…another old fashioned solution – purely based on HSV processing:

  1. Convert images to the HSV colorspace
  2. Create masks according to heuristics in the HSV (see below)
  3. Apply morphological dilation to the mask to connect disconnected areas
  4. Discard small areas and horizontal blocks (remember trees are vertical blocks)
  5. Compute the bounding box

A word on the heuristics in the HSV processing:

  1. everything with Hues (H) between 210 – 320 degrees is discarded as blue-magenta that is supposed to be in the background or in non-relevant areas
  2. everything with Values (V) lower that 40% is also discarded as being too dark to be relevant

Of course one may experiment with numerous other possibilities to fine-tune this approach…

Here is the MATLAB code to do the trick (warning: the code is far from being optimized!!! I used techniques not recommended for MATLAB programming just to be able to track anything in the process-this can be greatly optimized):

% clear everything
clear;
pack;
close all;
close all hidden;
drawnow;
clc;

% initialization
ims=dir('./*.jpg');
num=length(ims);

imgs={};
hsvs={}; 
masks={};
dilated_images={};
measurements={};
boxs={};

for i=1:num, 
    % load original image
    imgs{end+1} = imread(ims(i).name);
    flt_x_size = round(size(imgs{i},2)*0.005);
    flt_y_size = round(size(imgs{i},1)*0.005);
    flt = fspecial( 'average', max( flt_y_size, flt_x_size));
    imgs{i} = imfilter( imgs{i}, flt, 'same');
    % convert to HSV colorspace
    hsvs{end+1} = rgb2hsv(imgs{i});
    % apply a hard thresholding and binary operation to construct the mask
    masks{end+1} = medfilt2( ~(hsvs{i}(:,:,1)>(210/360) & hsvs{i}(:,:,1)<(320/360))&hsvs{i}(:,:,3)>0.4);
    % apply morphological dilation to connect distonnected components
    strel_size = round(0.03*max(size(imgs{i})));        % structuring element for morphological dilation
    dilated_images{end+1} = imdilate( masks{i}, strel('disk',strel_size));
    % do some measurements to eliminate small objects
    measurements{i} = regionprops( dilated_images{i},'Perimeter','Area','BoundingBox'); 
    for m=1:length(measurements{i})
        if (measurements{i}(m).Area < 0.02*numel( dilated_images{i})) || (measurements{i}(m).BoundingBox(3)>1.2*measurements{i}(m).BoundingBox(4))
            dilated_images{i}( round(measurements{i}(m).BoundingBox(2):measurements{i}(m).BoundingBox(4)+measurements{i}(m).BoundingBox(2)),...
                round(measurements{i}(m).BoundingBox(1):measurements{i}(m).BoundingBox(3)+measurements{i}(m).BoundingBox(1))) = 0;
        end
    end
    dilated_images{i} = dilated_images{i}(1:size(imgs{i},1),1:size(imgs{i},2));
    % compute the bounding box
    [y,x] = find( dilated_images{i});
    if isempty( y)
        boxs{end+1}=[];
    else
        boxs{end+1} = [ min(x) min(y) max(x)-min(x)+1 max(y)-min(y)+1];
    end

end 

%%% additional code to display things
for i=1:num,
    figure;
    subplot(121);
    colormap gray;
    imshow( imgs{i});
    if ~isempty(boxs{i})
        hold on;
        rr = rectangle( 'position', boxs{i});
        set( rr, 'EdgeColor', 'r');
        hold off;
    end
    subplot(122);
    imshow( imgs{i}.*uint8(repmat(dilated_images{i},[1 1 3])));
end

Results:

In the results I show the masked image and the bounding box. enter image description here


回答 7

一些老式的图像处理方法…
这个想法是基于这样的假设,即图像在通常较暗和较平滑的背景(在某些情况下为前景)上描绘了发光的树。该点燃树面积更“有活力”,具有较高的强度
流程如下:

  1. 转换为灰度
  2. 应用LoG过滤以获取最“活跃”的区域
  3. 应用专心的阈值以获得最明亮的区域
  4. 结合之前的2个以获得初步的蒙版
  5. 应用形态学扩张来扩大区域并连接相邻的组件
  6. 根据面积缩小消除候选区域

您得到的是每个图像的二进制掩码和边界框。

这是使用这种幼稚技术的结果: 在此处输入图片说明

MATLAB上 的代码如下:该代码在包含JPG图像的文件夹上运行。加载所有图像并返回检测到的结果。

% clear everything
clear;
pack;
close all;
close all hidden;
drawnow;
clc;

% initialization
ims=dir('./*.jpg');
imgs={};
images={}; 
blur_images={}; 
log_image={}; 
dilated_image={};
int_image={};
bin_image={};
measurements={};
box={};
num=length(ims);
thres_div = 3;

for i=1:num, 
    % load original image
    imgs{end+1}=imread(ims(i).name);

    % convert to grayscale
    images{end+1}=rgb2gray(imgs{i});

    % apply laplacian filtering and heuristic hard thresholding
    val_thres = (max(max(images{i}))/thres_div);
    log_image{end+1} = imfilter( images{i},fspecial('log')) > val_thres;

    % get the most bright regions of the image
    int_thres = 0.26*max(max( images{i}));
    int_image{end+1} = images{i} > int_thres;

    % compute the final binary image by combining 
    % high 'activity' with high intensity
    bin_image{end+1} = log_image{i} .* int_image{i};

    % apply morphological dilation to connect distonnected components
    strel_size = round(0.01*max(size(imgs{i})));        % structuring element for morphological dilation
    dilated_image{end+1} = imdilate( bin_image{i}, strel('disk',strel_size));

    % do some measurements to eliminate small objects
    measurements{i} = regionprops( logical( dilated_image{i}),'Area','BoundingBox');
    for m=1:length(measurements{i})
        if measurements{i}(m).Area < 0.05*numel( dilated_image{i})
            dilated_image{i}( round(measurements{i}(m).BoundingBox(2):measurements{i}(m).BoundingBox(4)+measurements{i}(m).BoundingBox(2)),...
                round(measurements{i}(m).BoundingBox(1):measurements{i}(m).BoundingBox(3)+measurements{i}(m).BoundingBox(1))) = 0;
        end
    end
    % make sure the dilated image is the same size with the original
    dilated_image{i} = dilated_image{i}(1:size(imgs{i},1),1:size(imgs{i},2));
    % compute the bounding box
    [y,x] = find( dilated_image{i});
    if isempty( y)
        box{end+1}=[];
    else
        box{end+1} = [ min(x) min(y) max(x)-min(x)+1 max(y)-min(y)+1];
    end
end 

%%% additional code to display things
for i=1:num,
    figure;
    subplot(121);
    colormap gray;
    imshow( imgs{i});
    if ~isempty(box{i})
        hold on;
        rr = rectangle( 'position', box{i});
        set( rr, 'EdgeColor', 'r');
        hold off;
    end
    subplot(122);
    imshow( imgs{i}.*uint8(repmat(dilated_image{i},[1 1 3])));
end

Some old-fashioned image processing approach…
The idea is based on the assumption that images depict lighted trees on typically darker and smoother backgrounds (or foregrounds in some cases). The lighted tree area is more “energetic” and has higher intensity.
The process is as follows:

  1. Convert to graylevel
  2. Apply LoG filtering to get the most “active” areas
  3. Apply an intentisy thresholding to get the most bright areas
  4. Combine the previous 2 to get a preliminary mask
  5. Apply a morphological dilation to enlarge areas and connect neighboring components
  6. Eliminate small candidate areas according to their area size

What you get is a binary mask and a bounding box for each image.

Here are the results using this naive technique: enter image description here

Code on MATLAB follows: The code runs on a folder with JPG images. Loads all images and returns detected results.

% clear everything
clear;
pack;
close all;
close all hidden;
drawnow;
clc;

% initialization
ims=dir('./*.jpg');
imgs={};
images={}; 
blur_images={}; 
log_image={}; 
dilated_image={};
int_image={};
bin_image={};
measurements={};
box={};
num=length(ims);
thres_div = 3;

for i=1:num, 
    % load original image
    imgs{end+1}=imread(ims(i).name);

    % convert to grayscale
    images{end+1}=rgb2gray(imgs{i});

    % apply laplacian filtering and heuristic hard thresholding
    val_thres = (max(max(images{i}))/thres_div);
    log_image{end+1} = imfilter( images{i},fspecial('log')) > val_thres;

    % get the most bright regions of the image
    int_thres = 0.26*max(max( images{i}));
    int_image{end+1} = images{i} > int_thres;

    % compute the final binary image by combining 
    % high 'activity' with high intensity
    bin_image{end+1} = log_image{i} .* int_image{i};

    % apply morphological dilation to connect distonnected components
    strel_size = round(0.01*max(size(imgs{i})));        % structuring element for morphological dilation
    dilated_image{end+1} = imdilate( bin_image{i}, strel('disk',strel_size));

    % do some measurements to eliminate small objects
    measurements{i} = regionprops( logical( dilated_image{i}),'Area','BoundingBox');
    for m=1:length(measurements{i})
        if measurements{i}(m).Area < 0.05*numel( dilated_image{i})
            dilated_image{i}( round(measurements{i}(m).BoundingBox(2):measurements{i}(m).BoundingBox(4)+measurements{i}(m).BoundingBox(2)),...
                round(measurements{i}(m).BoundingBox(1):measurements{i}(m).BoundingBox(3)+measurements{i}(m).BoundingBox(1))) = 0;
        end
    end
    % make sure the dilated image is the same size with the original
    dilated_image{i} = dilated_image{i}(1:size(imgs{i},1),1:size(imgs{i},2));
    % compute the bounding box
    [y,x] = find( dilated_image{i});
    if isempty( y)
        box{end+1}=[];
    else
        box{end+1} = [ min(x) min(y) max(x)-min(x)+1 max(y)-min(y)+1];
    end
end 

%%% additional code to display things
for i=1:num,
    figure;
    subplot(121);
    colormap gray;
    imshow( imgs{i});
    if ~isempty(box{i})
        hold on;
        rr = rectangle( 'position', box{i});
        set( rr, 'EdgeColor', 'r');
        hold off;
    end
    subplot(122);
    imshow( imgs{i}.*uint8(repmat(dilated_image{i},[1 1 3])));
end

回答 8

使用与我所见不同的方法,我创建了一个 通过它们的灯光检测圣诞树的脚本。结果始终是对称的三角形,如有必要,还可以使用数字值,例如树的角度(“脂肪度”)。

显然,此算法的最大威胁是(大量)或树前(更大的问题,直到进一步优化)旁边的灯。编辑(添加):做不到的事情:找出是否有一棵圣诞树,在一幅图像中找到多棵圣诞树,正确检测拉斯维加斯中部的圣诞前夜树,检测严重弯曲的圣诞树,倒置或切碎…;)

不同的阶段是:

  • 计算每个像素的附加亮度(R + G + B)
  • 将每个像素上方所有8个相邻像素的值相加
  • 按此值对所有像素进行排名(最亮的优先)-我知道,不是很细微…
  • 从顶部开始选择N个,跳过距离太近的
  • 计算 这前N个(给我们大约树的中心)
  • 从中位数位置开始,在一个加宽的搜索光束中,从所选的最亮光源发出的最上面的光线(人们倾向于在最上方放置至少一个光线)
  • 从那里开始,想象线条向左和向右向下倾斜60度(圣诞节树不应该那么胖)
  • 降低60度,直到20%的最亮的光不在此三角形内
  • 在三角形的最底部找到光线,为您提供树的下部水平边框
  • 完成了

标记说明:

  • 树中心的大红十字:N个最亮的灯光的中位数
  • 从上方向上的虚线:“搜索光束”为树的顶部
  • 较小的红十字:树顶
  • 很小的红叉:所有N个最亮的灯
  • 红色三角形:D!

源代码:

<?php

ini_set('memory_limit', '1024M');

header("Content-type: image/png");

$chosenImage = 6;

switch($chosenImage){
    case 1:
        $inputImage     = imagecreatefromjpeg("nmzwj.jpg");
        break;
    case 2:
        $inputImage     = imagecreatefromjpeg("2y4o5.jpg");
        break;
    case 3:
        $inputImage     = imagecreatefromjpeg("YowlH.jpg");
        break;
    case 4:
        $inputImage     = imagecreatefromjpeg("2K9Ef.jpg");
        break;
    case 5:
        $inputImage     = imagecreatefromjpeg("aVZhC.jpg");
        break;
    case 6:
        $inputImage     = imagecreatefromjpeg("FWhSP.jpg");
        break;
    case 7:
        $inputImage     = imagecreatefromjpeg("roemerberg.jpg");
        break;
    default:
        exit();
}

// Process the loaded image

$topNspots = processImage($inputImage);

imagejpeg($inputImage);
imagedestroy($inputImage);

// Here be functions

function processImage($image) {
    $orange = imagecolorallocate($image, 220, 210, 60);
    $black = imagecolorallocate($image, 0, 0, 0);
    $red = imagecolorallocate($image, 255, 0, 0);

    $maxX = imagesx($image)-1;
    $maxY = imagesy($image)-1;

    // Parameters
    $spread = 1; // Number of pixels to each direction that will be added up
    $topPositions = 80; // Number of (brightest) lights taken into account
    $minLightDistance = round(min(array($maxX, $maxY)) / 30); // Minimum number of pixels between the brigtests lights
    $searchYperX = 5; // spread of the "search beam" from the median point to the top

    $renderStage = 3; // 1 to 3; exits the process early


    // STAGE 1
    // Calculate the brightness of each pixel (R+G+B)

    $maxBrightness = 0;
    $stage1array = array();

    for($row = 0; $row <= $maxY; $row++) {

        $stage1array[$row] = array();

        for($col = 0; $col <= $maxX; $col++) {

            $rgb = imagecolorat($image, $col, $row);
            $brightness = getBrightnessFromRgb($rgb);
            $stage1array[$row][$col] = $brightness;

            if($renderStage == 1){
                $brightnessToGrey = round($brightness / 765 * 256);
                $greyRgb = imagecolorallocate($image, $brightnessToGrey, $brightnessToGrey, $brightnessToGrey);
                imagesetpixel($image, $col, $row, $greyRgb);
            }

            if($brightness > $maxBrightness) {
                $maxBrightness = $brightness;
                if($renderStage == 1){
                    imagesetpixel($image, $col, $row, $red);
                }
            }
        }
    }
    if($renderStage == 1) {
        return;
    }


    // STAGE 2
    // Add up brightness of neighbouring pixels

    $stage2array = array();
    $maxStage2 = 0;

    for($row = 0; $row <= $maxY; $row++) {
        $stage2array[$row] = array();

        for($col = 0; $col <= $maxX; $col++) {
            if(!isset($stage2array[$row][$col])) $stage2array[$row][$col] = 0;

            // Look around the current pixel, add brightness
            for($y = $row-$spread; $y <= $row+$spread; $y++) {
                for($x = $col-$spread; $x <= $col+$spread; $x++) {

                    // Don't read values from outside the image
                    if($x >= 0 && $x <= $maxX && $y >= 0 && $y <= $maxY){
                        $stage2array[$row][$col] += $stage1array[$y][$x]+10;
                    }
                }
            }

            $stage2value = $stage2array[$row][$col];
            if($stage2value > $maxStage2) {
                $maxStage2 = $stage2value;
            }
        }
    }

    if($renderStage >= 2){
        // Paint the accumulated light, dimmed by the maximum value from stage 2
        for($row = 0; $row <= $maxY; $row++) {
            for($col = 0; $col <= $maxX; $col++) {
                $brightness = round($stage2array[$row][$col] / $maxStage2 * 255);
                $greyRgb = imagecolorallocate($image, $brightness, $brightness, $brightness);
                imagesetpixel($image, $col, $row, $greyRgb);
            }
        }
    }

    if($renderStage == 2) {
        return;
    }


    // STAGE 3

    // Create a ranking of bright spots (like "Top 20")
    $topN = array();

    for($row = 0; $row <= $maxY; $row++) {
        for($col = 0; $col <= $maxX; $col++) {

            $stage2Brightness = $stage2array[$row][$col];
            $topN[$col.":".$row] = $stage2Brightness;
        }
    }
    arsort($topN);

    $topNused = array();
    $topPositionCountdown = $topPositions;

    if($renderStage == 3){
        foreach ($topN as $key => $val) {
            if($topPositionCountdown <= 0){
                break;
            }

            $position = explode(":", $key);

            foreach($topNused as $usedPosition => $usedValue) {
                $usedPosition = explode(":", $usedPosition);
                $distance = abs($usedPosition[0] - $position[0]) + abs($usedPosition[1] - $position[1]);
                if($distance < $minLightDistance) {
                    continue 2;
                }
            }

            $topNused[$key] = $val;

            paintCrosshair($image, $position[0], $position[1], $red, 2);

            $topPositionCountdown--;

        }
    }


    // STAGE 4
    // Median of all Top N lights
    $topNxValues = array();
    $topNyValues = array();

    foreach ($topNused as $key => $val) {
        $position = explode(":", $key);
        array_push($topNxValues, $position[0]);
        array_push($topNyValues, $position[1]);
    }

    $medianXvalue = round(calculate_median($topNxValues));
    $medianYvalue = round(calculate_median($topNyValues));
    paintCrosshair($image, $medianXvalue, $medianYvalue, $red, 15);


    // STAGE 5
    // Find treetop

    $filename = 'debug.log';
    $handle = fopen($filename, "w");
    fwrite($handle, "\n\n STAGE 5");

    $treetopX = $medianXvalue;
    $treetopY = $medianYvalue;

    $searchXmin = $medianXvalue;
    $searchXmax = $medianXvalue;

    $width = 0;
    for($y = $medianYvalue; $y >= 0; $y--) {
        fwrite($handle, "\nAt y = ".$y);

        if(($y % $searchYperX) == 0) { // Modulo
            $width++;
            $searchXmin = $medianXvalue - $width;
            $searchXmax = $medianXvalue + $width;
            imagesetpixel($image, $searchXmin, $y, $red);
            imagesetpixel($image, $searchXmax, $y, $red);
        }

        foreach ($topNused as $key => $val) {
            $position = explode(":", $key); // "x:y"

            if($position[1] != $y){
                continue;
            }

            if($position[0] >= $searchXmin && $position[0] <= $searchXmax){
                $treetopX = $position[0];
                $treetopY = $y;
            }
        }

    }

    paintCrosshair($image, $treetopX, $treetopY, $red, 5);


    // STAGE 6
    // Find tree sides
    fwrite($handle, "\n\n STAGE 6");

    $treesideAngle = 60; // The extremely "fat" end of a christmas tree
    $treeBottomY = $treetopY;

    $topPositionsExcluded = 0;
    $xymultiplier = 0;
    while(($topPositionsExcluded < ($topPositions / 5)) && $treesideAngle >= 1){
        fwrite($handle, "\n\nWe're at angle ".$treesideAngle);
        $xymultiplier = sin(deg2rad($treesideAngle));
        fwrite($handle, "\nMultiplier: ".$xymultiplier);

        $topPositionsExcluded = 0;
        foreach ($topNused as $key => $val) {
            $position = explode(":", $key);
            fwrite($handle, "\nAt position ".$key);

            if($position[1] > $treeBottomY) {
                $treeBottomY = $position[1];
            }

            // Lights above the tree are outside of it, but don't matter
            if($position[1] < $treetopY){
                $topPositionsExcluded++;
                fwrite($handle, "\nTOO HIGH");
                continue;
            }

            // Top light will generate division by zero
            if($treetopY-$position[1] == 0) {
                fwrite($handle, "\nDIVISION BY ZERO");
                continue;
            }

            // Lights left end right of it are also not inside
            fwrite($handle, "\nLight position factor: ".(abs($treetopX-$position[0]) / abs($treetopY-$position[1])));
            if((abs($treetopX-$position[0]) / abs($treetopY-$position[1])) > $xymultiplier){
                $topPositionsExcluded++;
                fwrite($handle, "\n --- Outside tree ---");
            }
        }

        $treesideAngle--;
    }
    fclose($handle);

    // Paint tree's outline
    $treeHeight = abs($treetopY-$treeBottomY);
    $treeBottomLeft = 0;
    $treeBottomRight = 0;
    $previousState = false; // line has not started; assumes the tree does not "leave"^^

    for($x = 0; $x <= $maxX; $x++){
        if(abs($treetopX-$x) != 0 && abs($treetopX-$x) / $treeHeight > $xymultiplier){
            if($previousState == true){
                $treeBottomRight = $x;
                $previousState = false;
            }
            continue;
        }
        imagesetpixel($image, $x, $treeBottomY, $red);
        if($previousState == false){
            $treeBottomLeft = $x;
            $previousState = true;
        }
    }
    imageline($image, $treeBottomLeft, $treeBottomY, $treetopX, $treetopY, $red);
    imageline($image, $treeBottomRight, $treeBottomY, $treetopX, $treetopY, $red);


    // Print out some parameters

    $string = "Min dist: ".$minLightDistance." | Tree angle: ".$treesideAngle." deg | Tree bottom: ".$treeBottomY;

    $px     = (imagesx($image) - 6.5 * strlen($string)) / 2;
    imagestring($image, 2, $px, 5, $string, $orange);

    return $topN;
}

/**
 * Returns values from 0 to 765
 */
function getBrightnessFromRgb($rgb) {
    $r = ($rgb >> 16) & 0xFF;
    $g = ($rgb >> 8) & 0xFF;
    $b = $rgb & 0xFF;

    return $r+$r+$b;
}

function paintCrosshair($image, $posX, $posY, $color, $size=5) {
    for($x = $posX-$size; $x <= $posX+$size; $x++) {
        if($x>=0 && $x < imagesx($image)){
            imagesetpixel($image, $x, $posY, $color);
        }
    }
    for($y = $posY-$size; $y <= $posY+$size; $y++) {
        if($y>=0 && $y < imagesy($image)){
            imagesetpixel($image, $posX, $y, $color);
        }
    }
}

// From http://www.mdj.us/web-development/php-programming/calculating-the-median-average-values-of-an-array-with-php/
function calculate_median($arr) {
    sort($arr);
    $count = count($arr); //total numbers in array
    $middleval = floor(($count-1)/2); // find the middle value, or the lowest middle value
    if($count % 2) { // odd number, middle is the median
        $median = $arr[$middleval];
    } else { // even number, calculate avg of 2 medians
        $low = $arr[$middleval];
        $high = $arr[$middleval+1];
        $median = (($low+$high)/2);
    }
    return $median;
}


?>

图片: 左上 下中心 左下 右上方 上中 右下

奖励:来自维基百科的德国人Weihnachtsbaum,网址:http://commons.wikimedia.org/wiki/File:Weihnachtsbaum_R%C3%B6merberg.jpg 罗默贝格

Using a quite different approach from what I’ve seen, I created a script that detects christmas trees by their lights. The result ist always a symmetrical triangle, and if necessary numeric values like the angle (“fatness”) of the tree.

The biggest threat to this algorithm obviously are lights next to (in great numbers) or in front of the tree (the greater problem until further optimization). Edit (added): What it can’t do: Find out if there’s a christmas tree or not, find multiple christmas trees in one image, correctly detect a cristmas tree in the middle of Las Vegas, detect christmas trees that are heavily bent, upside-down or chopped down… ;)

The different stages are:

  • Calculate the added brightness (R+G+B) for each pixel
  • Add up this value of all 8 neighbouring pixels on top of each pixel
  • Rank all pixels by this value (brightest first) – I know, not really subtle…
  • Choose N of these, starting from the top, skipping ones that are too close
  • Calculate the of these top N (gives us the approximate center of the tree)
  • Start from the median position upwards in a widening search beam for the topmost light from the selected brightest ones (people tend to put at least one light at the very top)
  • From there, imagine lines going 60 degrees left and right downwards (christmas trees shouldn’t be that fat)
  • Decrease those 60 degrees until 20% of the brightest lights are outside this triangle
  • Find the light at the very bottom of the triangle, giving you the lower horizontal border of the tree
  • Done

Explanation of the markings:

  • Big red cross in the center of the tree: Median of the top N brightest lights
  • Dotted line from there upwards: “search beam” for the top of the tree
  • Smaller red cross: top of the tree
  • Really small red crosses: All of the top N brightest lights
  • Red triangle: D’uh!

Source code:

<?php

ini_set('memory_limit', '1024M');

header("Content-type: image/png");

$chosenImage = 6;

switch($chosenImage){
    case 1:
        $inputImage     = imagecreatefromjpeg("nmzwj.jpg");
        break;
    case 2:
        $inputImage     = imagecreatefromjpeg("2y4o5.jpg");
        break;
    case 3:
        $inputImage     = imagecreatefromjpeg("YowlH.jpg");
        break;
    case 4:
        $inputImage     = imagecreatefromjpeg("2K9Ef.jpg");
        break;
    case 5:
        $inputImage     = imagecreatefromjpeg("aVZhC.jpg");
        break;
    case 6:
        $inputImage     = imagecreatefromjpeg("FWhSP.jpg");
        break;
    case 7:
        $inputImage     = imagecreatefromjpeg("roemerberg.jpg");
        break;
    default:
        exit();
}

// Process the loaded image

$topNspots = processImage($inputImage);

imagejpeg($inputImage);
imagedestroy($inputImage);

// Here be functions

function processImage($image) {
    $orange = imagecolorallocate($image, 220, 210, 60);
    $black = imagecolorallocate($image, 0, 0, 0);
    $red = imagecolorallocate($image, 255, 0, 0);

    $maxX = imagesx($image)-1;
    $maxY = imagesy($image)-1;

    // Parameters
    $spread = 1; // Number of pixels to each direction that will be added up
    $topPositions = 80; // Number of (brightest) lights taken into account
    $minLightDistance = round(min(array($maxX, $maxY)) / 30); // Minimum number of pixels between the brigtests lights
    $searchYperX = 5; // spread of the "search beam" from the median point to the top

    $renderStage = 3; // 1 to 3; exits the process early


    // STAGE 1
    // Calculate the brightness of each pixel (R+G+B)

    $maxBrightness = 0;
    $stage1array = array();

    for($row = 0; $row <= $maxY; $row++) {

        $stage1array[$row] = array();

        for($col = 0; $col <= $maxX; $col++) {

            $rgb = imagecolorat($image, $col, $row);
            $brightness = getBrightnessFromRgb($rgb);
            $stage1array[$row][$col] = $brightness;

            if($renderStage == 1){
                $brightnessToGrey = round($brightness / 765 * 256);
                $greyRgb = imagecolorallocate($image, $brightnessToGrey, $brightnessToGrey, $brightnessToGrey);
                imagesetpixel($image, $col, $row, $greyRgb);
            }

            if($brightness > $maxBrightness) {
                $maxBrightness = $brightness;
                if($renderStage == 1){
                    imagesetpixel($image, $col, $row, $red);
                }
            }
        }
    }
    if($renderStage == 1) {
        return;
    }


    // STAGE 2
    // Add up brightness of neighbouring pixels

    $stage2array = array();
    $maxStage2 = 0;

    for($row = 0; $row <= $maxY; $row++) {
        $stage2array[$row] = array();

        for($col = 0; $col <= $maxX; $col++) {
            if(!isset($stage2array[$row][$col])) $stage2array[$row][$col] = 0;

            // Look around the current pixel, add brightness
            for($y = $row-$spread; $y <= $row+$spread; $y++) {
                for($x = $col-$spread; $x <= $col+$spread; $x++) {

                    // Don't read values from outside the image
                    if($x >= 0 && $x <= $maxX && $y >= 0 && $y <= $maxY){
                        $stage2array[$row][$col] += $stage1array[$y][$x]+10;
                    }
                }
            }

            $stage2value = $stage2array[$row][$col];
            if($stage2value > $maxStage2) {
                $maxStage2 = $stage2value;
            }
        }
    }

    if($renderStage >= 2){
        // Paint the accumulated light, dimmed by the maximum value from stage 2
        for($row = 0; $row <= $maxY; $row++) {
            for($col = 0; $col <= $maxX; $col++) {
                $brightness = round($stage2array[$row][$col] / $maxStage2 * 255);
                $greyRgb = imagecolorallocate($image, $brightness, $brightness, $brightness);
                imagesetpixel($image, $col, $row, $greyRgb);
            }
        }
    }

    if($renderStage == 2) {
        return;
    }


    // STAGE 3

    // Create a ranking of bright spots (like "Top 20")
    $topN = array();

    for($row = 0; $row <= $maxY; $row++) {
        for($col = 0; $col <= $maxX; $col++) {

            $stage2Brightness = $stage2array[$row][$col];
            $topN[$col.":".$row] = $stage2Brightness;
        }
    }
    arsort($topN);

    $topNused = array();
    $topPositionCountdown = $topPositions;

    if($renderStage == 3){
        foreach ($topN as $key => $val) {
            if($topPositionCountdown <= 0){
                break;
            }

            $position = explode(":", $key);

            foreach($topNused as $usedPosition => $usedValue) {
                $usedPosition = explode(":", $usedPosition);
                $distance = abs($usedPosition[0] - $position[0]) + abs($usedPosition[1] - $position[1]);
                if($distance < $minLightDistance) {
                    continue 2;
                }
            }

            $topNused[$key] = $val;

            paintCrosshair($image, $position[0], $position[1], $red, 2);

            $topPositionCountdown--;

        }
    }


    // STAGE 4
    // Median of all Top N lights
    $topNxValues = array();
    $topNyValues = array();

    foreach ($topNused as $key => $val) {
        $position = explode(":", $key);
        array_push($topNxValues, $position[0]);
        array_push($topNyValues, $position[1]);
    }

    $medianXvalue = round(calculate_median($topNxValues));
    $medianYvalue = round(calculate_median($topNyValues));
    paintCrosshair($image, $medianXvalue, $medianYvalue, $red, 15);


    // STAGE 5
    // Find treetop

    $filename = 'debug.log';
    $handle = fopen($filename, "w");
    fwrite($handle, "\n\n STAGE 5");

    $treetopX = $medianXvalue;
    $treetopY = $medianYvalue;

    $searchXmin = $medianXvalue;
    $searchXmax = $medianXvalue;

    $width = 0;
    for($y = $medianYvalue; $y >= 0; $y--) {
        fwrite($handle, "\nAt y = ".$y);

        if(($y % $searchYperX) == 0) { // Modulo
            $width++;
            $searchXmin = $medianXvalue - $width;
            $searchXmax = $medianXvalue + $width;
            imagesetpixel($image, $searchXmin, $y, $red);
            imagesetpixel($image, $searchXmax, $y, $red);
        }

        foreach ($topNused as $key => $val) {
            $position = explode(":", $key); // "x:y"

            if($position[1] != $y){
                continue;
            }

            if($position[0] >= $searchXmin && $position[0] <= $searchXmax){
                $treetopX = $position[0];
                $treetopY = $y;
            }
        }

    }

    paintCrosshair($image, $treetopX, $treetopY, $red, 5);


    // STAGE 6
    // Find tree sides
    fwrite($handle, "\n\n STAGE 6");

    $treesideAngle = 60; // The extremely "fat" end of a christmas tree
    $treeBottomY = $treetopY;

    $topPositionsExcluded = 0;
    $xymultiplier = 0;
    while(($topPositionsExcluded < ($topPositions / 5)) && $treesideAngle >= 1){
        fwrite($handle, "\n\nWe're at angle ".$treesideAngle);
        $xymultiplier = sin(deg2rad($treesideAngle));
        fwrite($handle, "\nMultiplier: ".$xymultiplier);

        $topPositionsExcluded = 0;
        foreach ($topNused as $key => $val) {
            $position = explode(":", $key);
            fwrite($handle, "\nAt position ".$key);

            if($position[1] > $treeBottomY) {
                $treeBottomY = $position[1];
            }

            // Lights above the tree are outside of it, but don't matter
            if($position[1] < $treetopY){
                $topPositionsExcluded++;
                fwrite($handle, "\nTOO HIGH");
                continue;
            }

            // Top light will generate division by zero
            if($treetopY-$position[1] == 0) {
                fwrite($handle, "\nDIVISION BY ZERO");
                continue;
            }

            // Lights left end right of it are also not inside
            fwrite($handle, "\nLight position factor: ".(abs($treetopX-$position[0]) / abs($treetopY-$position[1])));
            if((abs($treetopX-$position[0]) / abs($treetopY-$position[1])) > $xymultiplier){
                $topPositionsExcluded++;
                fwrite($handle, "\n --- Outside tree ---");
            }
        }

        $treesideAngle--;
    }
    fclose($handle);

    // Paint tree's outline
    $treeHeight = abs($treetopY-$treeBottomY);
    $treeBottomLeft = 0;
    $treeBottomRight = 0;
    $previousState = false; // line has not started; assumes the tree does not "leave"^^

    for($x = 0; $x <= $maxX; $x++){
        if(abs($treetopX-$x) != 0 && abs($treetopX-$x) / $treeHeight > $xymultiplier){
            if($previousState == true){
                $treeBottomRight = $x;
                $previousState = false;
            }
            continue;
        }
        imagesetpixel($image, $x, $treeBottomY, $red);
        if($previousState == false){
            $treeBottomLeft = $x;
            $previousState = true;
        }
    }
    imageline($image, $treeBottomLeft, $treeBottomY, $treetopX, $treetopY, $red);
    imageline($image, $treeBottomRight, $treeBottomY, $treetopX, $treetopY, $red);


    // Print out some parameters

    $string = "Min dist: ".$minLightDistance." | Tree angle: ".$treesideAngle." deg | Tree bottom: ".$treeBottomY;

    $px     = (imagesx($image) - 6.5 * strlen($string)) / 2;
    imagestring($image, 2, $px, 5, $string, $orange);

    return $topN;
}

/**
 * Returns values from 0 to 765
 */
function getBrightnessFromRgb($rgb) {
    $r = ($rgb >> 16) & 0xFF;
    $g = ($rgb >> 8) & 0xFF;
    $b = $rgb & 0xFF;

    return $r+$r+$b;
}

function paintCrosshair($image, $posX, $posY, $color, $size=5) {
    for($x = $posX-$size; $x <= $posX+$size; $x++) {
        if($x>=0 && $x < imagesx($image)){
            imagesetpixel($image, $x, $posY, $color);
        }
    }
    for($y = $posY-$size; $y <= $posY+$size; $y++) {
        if($y>=0 && $y < imagesy($image)){
            imagesetpixel($image, $posX, $y, $color);
        }
    }
}

// From http://www.mdj.us/web-development/php-programming/calculating-the-median-average-values-of-an-array-with-php/
function calculate_median($arr) {
    sort($arr);
    $count = count($arr); //total numbers in array
    $middleval = floor(($count-1)/2); // find the middle value, or the lowest middle value
    if($count % 2) { // odd number, middle is the median
        $median = $arr[$middleval];
    } else { // even number, calculate avg of 2 medians
        $low = $arr[$middleval];
        $high = $arr[$middleval+1];
        $median = (($low+$high)/2);
    }
    return $median;
}


?>

Images: Upper left Lower center Lower left Upper right Upper center Lower right

Bonus: A german Weihnachtsbaum, from Wikipedia Römerberg http://commons.wikimedia.org/wiki/File:Weihnachtsbaum_R%C3%B6merberg.jpg


回答 9

我将python与opencv一起使用。

我的算法是这样的:

  1. 首先,它从图像中获取红色通道
  2. 将阈值(最小值200)应用于红色通道
  3. 然后应用形态学梯度,然后执行“闭合”(先扩张,然后进行侵蚀)
  4. 然后,它在平面中找到轮廓,并选择最长的轮廓。

结果:

编码:

import numpy as np
import cv2
import copy


def findTree(image,num):
    im = cv2.imread(image)
    im = cv2.resize(im, (400,250))
    gray = cv2.cvtColor(im, cv2.COLOR_RGB2GRAY)
    imf = copy.deepcopy(im)

    b,g,r = cv2.split(im)
    minR = 200
    _,thresh = cv2.threshold(r,minR,255,0)
    kernel = np.ones((25,5))
    dst = cv2.morphologyEx(thresh, cv2.MORPH_GRADIENT, kernel)
    dst = cv2.morphologyEx(dst, cv2.MORPH_CLOSE, kernel)

    contours = cv2.findContours(dst,cv2.RETR_TREE,cv2.CHAIN_APPROX_SIMPLE)[0]
    cv2.drawContours(im, contours,-1, (0,255,0), 1)

    maxI = 0
    for i in range(len(contours)):
        if len(contours[maxI]) < len(contours[i]):
            maxI = i

    img = copy.deepcopy(r)
    cv2.polylines(img,[contours[maxI]],True,(255,255,255),3)
    imf[:,:,2] = img

    cv2.imshow(str(num), imf)

def main():
    findTree('tree.jpg',1)
    findTree('tree2.jpg',2)
    findTree('tree3.jpg',3)
    findTree('tree4.jpg',4)
    findTree('tree5.jpg',5)
    findTree('tree6.jpg',6)

    cv2.waitKey(0)
    cv2.destroyAllWindows()

if __name__ == "__main__":
    main()

如果将内核从(25,5)更改为(10,5),则除左下角以外的所有树都可获得更好的结果, 在此处输入图片说明

我的算法假设这棵树上有灯,在左下方的树中,顶部的灯比其他树的灯少。

I used python with opencv.

My algorithm goes like this:

  1. First it takes the red channel from the image
  2. Apply a threshold (min value 200) to the Red channel
  3. Then apply Morphological Gradient and then do a ‘Closing’ (dilation followed by Erosion)
  4. Then it finds the contours in the plane and it picks the longest contour.

The outcome:

The code:

import numpy as np
import cv2
import copy


def findTree(image,num):
    im = cv2.imread(image)
    im = cv2.resize(im, (400,250))
    gray = cv2.cvtColor(im, cv2.COLOR_RGB2GRAY)
    imf = copy.deepcopy(im)

    b,g,r = cv2.split(im)
    minR = 200
    _,thresh = cv2.threshold(r,minR,255,0)
    kernel = np.ones((25,5))
    dst = cv2.morphologyEx(thresh, cv2.MORPH_GRADIENT, kernel)
    dst = cv2.morphologyEx(dst, cv2.MORPH_CLOSE, kernel)

    contours = cv2.findContours(dst,cv2.RETR_TREE,cv2.CHAIN_APPROX_SIMPLE)[0]
    cv2.drawContours(im, contours,-1, (0,255,0), 1)

    maxI = 0
    for i in range(len(contours)):
        if len(contours[maxI]) < len(contours[i]):
            maxI = i

    img = copy.deepcopy(r)
    cv2.polylines(img,[contours[maxI]],True,(255,255,255),3)
    imf[:,:,2] = img

    cv2.imshow(str(num), imf)

def main():
    findTree('tree.jpg',1)
    findTree('tree2.jpg',2)
    findTree('tree3.jpg',3)
    findTree('tree4.jpg',4)
    findTree('tree5.jpg',5)
    findTree('tree6.jpg',6)

    cv2.waitKey(0)
    cv2.destroyAllWindows()

if __name__ == "__main__":
    main()

If I change the kernel from (25,5) to (10,5) I get nicer results on all trees but the bottom left, enter image description here

my algorithm assumes that the tree has lights on it, and in the bottom left tree, the top has less light then the others.


OpenCV-Python中的简单数字识别OCR

问题:OpenCV-Python中的简单数字识别OCR

我正在尝试在OpenCV-Python(cv2)中实现“数字识别OCR”。它仅用于学习目的。我想学习OpenCV中的KNearest和SVM功能。

我每个数字有100个样本(即图像)。我想和他们一起训练。

letter_recog.pyOpenCV示例附带一个示例。但是我仍然不知道如何使用它。我不了解样本,响应等内容。此外,它首先会加载txt文件,而我首先并不了解。

稍后进行搜索时,我可以在cpp样本中找到letter_recognitiontion.data。我用它并在letter_recog.py模型中为cv2.KNearest编写了一个代码(仅用于测试):

import numpy as np
import cv2

fn = 'letter-recognition.data'
a = np.loadtxt(fn, np.float32, delimiter=',', converters={ 0 : lambda ch : ord(ch)-ord('A') })
samples, responses = a[:,1:], a[:,0]

model = cv2.KNearest()
retval = model.train(samples,responses)
retval, results, neigh_resp, dists = model.find_nearest(samples, k = 10)
print results.ravel()

它给了我一个大小为20000的数组,我不知道它是什么。

问题:

1)什么是letter_recognition.data文件?如何从我自己的数据集中构建该文件?

2)results.reval()代表什么?

3)我们如何使用letter_recognition.data文件(KNearest或SVM)编写一个简单的数字识别工具?

I am trying to implement a “Digit Recognition OCR” in OpenCV-Python (cv2). It is just for learning purposes. I would like to learn both KNearest and SVM features in OpenCV.

I have 100 samples (i.e. images) of each digit. I would like to train with them.

There is a sample letter_recog.py that comes with OpenCV sample. But I still couldn’t figure out on how to use it. I don’t understand what are the samples, responses etc. Also, it loads a txt file at first, which I didn’t understand first.

Later on searching a little bit, I could find a letter_recognition.data in cpp samples. I used it and made a code for cv2.KNearest in the model of letter_recog.py (just for testing):

import numpy as np
import cv2

fn = 'letter-recognition.data'
a = np.loadtxt(fn, np.float32, delimiter=',', converters={ 0 : lambda ch : ord(ch)-ord('A') })
samples, responses = a[:,1:], a[:,0]

model = cv2.KNearest()
retval = model.train(samples,responses)
retval, results, neigh_resp, dists = model.find_nearest(samples, k = 10)
print results.ravel()

It gave me an array of size 20000, I don’t understand what it is.

Questions:

1) What is letter_recognition.data file? How to build that file from my own data set?

2) What does results.reval() denote?

3) How we can write a simple digit recognition tool using letter_recognition.data file (either KNearest or SVM)?


回答 0

好吧,我决定对我的问题进行锻炼以解决上述问题。我想要的是使用OpenCV中的KNearest或SVM功能实现简单的OCR。以下是我的工作方式。(这只是为了学习如何将KNearest用于简单的OCR目的)。

1)我的第一个问题是有关OpenCV示例随附的letter_recognition.data文件的。我想知道该文件中的内容。

它包含一个字母以及该字母的16个功能。

this SOF帮助我找到了它。本文介绍了这16个功能Letter Recognition Using Holland-Style Adaptive Classifiers。(尽管我不了解最后的一些功能)

2)由于我知道,如果不了解所有这些功能,就很难做到这一点。我尝试了其他一些论文,但是对于初学者来说,都有些困难。

So I just decided to take all the pixel values as my features. (我并不担心准确性或性能,我只是希望它能够工作,至少以最低的准确性)

我为训练数据拍摄了下图:

在此处输入图片说明

(我知道训练数据的数量较少。但是,由于所有字母的字体和大小都相同,因此我决定尝试一下)。

为了准备训练数据,我在OpenCV中编写了一个小代码。它执行以下操作:

  1. 它加载图像。
  2. 选择数字(显然是通过轮廓查找并在字母的面积和高度上施加约束来避免错误检测)。
  3. 围绕一个字母绘制边界矩形,然后等待key press manually。这次我们自己按对应于框中字母的数字键
  4. 一旦按下相应的数字键,它将将该框的大小调整为10×10,并在一个数组(此处为样本)中保存100个像素值,在另一个数组中(此处为响应)保存相应的手动输入的数字。
  5. 然后将两个数组保存在单独的txt文件中。

手动数字分类结束时,火车数据(train.png)中的所有数字都是由我们自己手动标记的,图像如下所示:

在此处输入图片说明

以下是我用于上述目的的代码(当然,不是很干净):

import sys

import numpy as np
import cv2

im = cv2.imread('pitrain.png')
im3 = im.copy()

gray = cv2.cvtColor(im,cv2.COLOR_BGR2GRAY)
blur = cv2.GaussianBlur(gray,(5,5),0)
thresh = cv2.adaptiveThreshold(blur,255,1,1,11,2)

#################      Now finding Contours         ###################

contours,hierarchy = cv2.findContours(thresh,cv2.RETR_LIST,cv2.CHAIN_APPROX_SIMPLE)

samples =  np.empty((0,100))
responses = []
keys = [i for i in range(48,58)]

for cnt in contours:
    if cv2.contourArea(cnt)>50:
        [x,y,w,h] = cv2.boundingRect(cnt)

        if  h>28:
            cv2.rectangle(im,(x,y),(x+w,y+h),(0,0,255),2)
            roi = thresh[y:y+h,x:x+w]
            roismall = cv2.resize(roi,(10,10))
            cv2.imshow('norm',im)
            key = cv2.waitKey(0)

            if key == 27:  # (escape to quit)
                sys.exit()
            elif key in keys:
                responses.append(int(chr(key)))
                sample = roismall.reshape((1,100))
                samples = np.append(samples,sample,0)

responses = np.array(responses,np.float32)
responses = responses.reshape((responses.size,1))
print "training complete"

np.savetxt('generalsamples.data',samples)
np.savetxt('generalresponses.data',responses)

现在我们进入培训和测试部分。

为了测试零件,我使用了下面的图片,该图片与我训练过的字母具有相同的类型。

在此处输入图片说明

对于培训,我们执行以下操作

  1. 加载我们之前已经保存的txt文件
  2. 创建一个我们正在使用的分类器的实例(这里是KNearest)
  3. 然后我们使用KNearest.train函数来训练数据

出于测试目的,我们执行以下操作:

  1. 我们加载用于测试的图像
  2. 较早处理图像并使用轮廓法提取每个数字
  3. 为其绘制一个边界框,然后将其大小调整为10×10,并将其像素值存储在数组中,如之前所做的那样。
  4. 然后,我们使用KNearest.find_nearest()函数查找与我们给出的项目最接近的项目。(如果幸运,它将识别出正确的数字。)

我在下面的单个代码中包括了最后两个步骤(培训和测试):

import cv2
import numpy as np

#######   training part    ############### 
samples = np.loadtxt('generalsamples.data',np.float32)
responses = np.loadtxt('generalresponses.data',np.float32)
responses = responses.reshape((responses.size,1))

model = cv2.KNearest()
model.train(samples,responses)

############################# testing part  #########################

im = cv2.imread('pi.png')
out = np.zeros(im.shape,np.uint8)
gray = cv2.cvtColor(im,cv2.COLOR_BGR2GRAY)
thresh = cv2.adaptiveThreshold(gray,255,1,1,11,2)

contours,hierarchy = cv2.findContours(thresh,cv2.RETR_LIST,cv2.CHAIN_APPROX_SIMPLE)

for cnt in contours:
    if cv2.contourArea(cnt)>50:
        [x,y,w,h] = cv2.boundingRect(cnt)
        if  h>28:
            cv2.rectangle(im,(x,y),(x+w,y+h),(0,255,0),2)
            roi = thresh[y:y+h,x:x+w]
            roismall = cv2.resize(roi,(10,10))
            roismall = roismall.reshape((1,100))
            roismall = np.float32(roismall)
            retval, results, neigh_resp, dists = model.find_nearest(roismall, k = 1)
            string = str(int((results[0][0])))
            cv2.putText(out,string,(x,y+h),0,1,(0,255,0))

cv2.imshow('im',im)
cv2.imshow('out',out)
cv2.waitKey(0)

它奏效了,下面是我得到的结果:

在此处输入图片说明


在这里,它以100%的精度工作。我认为这是因为所有数字都是相同的种类和大小。

但是无论如何,这对于初学者来说是一个不错的开始(我希望如此)。

Well, I decided to workout myself on my question to solve above problem. What I wanted is to implement a simpl OCR using KNearest or SVM features in OpenCV. And below is what I did and how. ( it is just for learning how to use KNearest for simple OCR purposes).

1) My first question was about letter_recognition.data file that comes with OpenCV samples. I wanted to know what is inside that file.

It contains a letter, along with 16 features of that letter.

And this SOF helped me to find it. These 16 features are explained in the paperLetter Recognition Using Holland-Style Adaptive Classifiers. ( Although I didn’t understand some of the features at end)

2) Since I knew, without understanding all those features, it is difficult to do that method. I tried some other papers, but all were a little difficult for a beginner.

So I just decided to take all the pixel values as my features. (I was not worried about accuracy or performance, I just wanted it to work, at least with the least accuracy)

I took below image for my training data:

enter image description here

( I know the amount of training data is less. But, since all letters are of same font and size, I decided to try on this).

To prepare the data for training, I made a small code in OpenCV. It does following things:

  1. It loads the image.
  2. Selects the digits ( obviously by contour finding and applying constraints on area and height of letters to avoid false detections).
  3. Draws the bounding rectangle around one letter and wait for key press manually. This time we press the digit key ourselves corresponding to the letter in box.
  4. Once corresponding digit key is pressed, it resizes this box to 10×10 and saves 100 pixel values in an array (here, samples) and corresponding manually entered digit in another array(here, responses).
  5. Then save both the arrays in separate txt files.

At the end of manual classification of digits, all the digits in the train data( train.png) are labeled manually by ourselves, image will look like below:

enter image description here

Below is the code I used for above purpose ( of course, not so clean):

import sys

import numpy as np
import cv2

im = cv2.imread('pitrain.png')
im3 = im.copy()

gray = cv2.cvtColor(im,cv2.COLOR_BGR2GRAY)
blur = cv2.GaussianBlur(gray,(5,5),0)
thresh = cv2.adaptiveThreshold(blur,255,1,1,11,2)

#################      Now finding Contours         ###################

contours,hierarchy = cv2.findContours(thresh,cv2.RETR_LIST,cv2.CHAIN_APPROX_SIMPLE)

samples =  np.empty((0,100))
responses = []
keys = [i for i in range(48,58)]

for cnt in contours:
    if cv2.contourArea(cnt)>50:
        [x,y,w,h] = cv2.boundingRect(cnt)

        if  h>28:
            cv2.rectangle(im,(x,y),(x+w,y+h),(0,0,255),2)
            roi = thresh[y:y+h,x:x+w]
            roismall = cv2.resize(roi,(10,10))
            cv2.imshow('norm',im)
            key = cv2.waitKey(0)

            if key == 27:  # (escape to quit)
                sys.exit()
            elif key in keys:
                responses.append(int(chr(key)))
                sample = roismall.reshape((1,100))
                samples = np.append(samples,sample,0)

responses = np.array(responses,np.float32)
responses = responses.reshape((responses.size,1))
print "training complete"

np.savetxt('generalsamples.data',samples)
np.savetxt('generalresponses.data',responses)

Now we enter in to training and testing part.

For testing part I used below image, which has same type of letters I used to train.

enter image description here

For training we do as follows:

  1. Load the txt files we already saved earlier
  2. create a instance of classifier we are using ( here, it is KNearest)
  3. Then we use KNearest.train function to train the data

For testing purposes, we do as follows:

  1. We load the image used for testing
  2. process the image as earlier and extract each digit using contour methods
  3. Draw bounding box for it, then resize to 10×10, and store its pixel values in an array as done earlier.
  4. Then we use KNearest.find_nearest() function to find the nearest item to the one we gave. ( If lucky, it recognises the correct digit.)

I included last two steps ( training and testing) in single code below:

import cv2
import numpy as np

#######   training part    ############### 
samples = np.loadtxt('generalsamples.data',np.float32)
responses = np.loadtxt('generalresponses.data',np.float32)
responses = responses.reshape((responses.size,1))

model = cv2.KNearest()
model.train(samples,responses)

############################# testing part  #########################

im = cv2.imread('pi.png')
out = np.zeros(im.shape,np.uint8)
gray = cv2.cvtColor(im,cv2.COLOR_BGR2GRAY)
thresh = cv2.adaptiveThreshold(gray,255,1,1,11,2)

contours,hierarchy = cv2.findContours(thresh,cv2.RETR_LIST,cv2.CHAIN_APPROX_SIMPLE)

for cnt in contours:
    if cv2.contourArea(cnt)>50:
        [x,y,w,h] = cv2.boundingRect(cnt)
        if  h>28:
            cv2.rectangle(im,(x,y),(x+w,y+h),(0,255,0),2)
            roi = thresh[y:y+h,x:x+w]
            roismall = cv2.resize(roi,(10,10))
            roismall = roismall.reshape((1,100))
            roismall = np.float32(roismall)
            retval, results, neigh_resp, dists = model.find_nearest(roismall, k = 1)
            string = str(int((results[0][0])))
            cv2.putText(out,string,(x,y+h),0,1,(0,255,0))

cv2.imshow('im',im)
cv2.imshow('out',out)
cv2.waitKey(0)

And it worked, below is the result I got:

enter image description here


Here it worked with 100% accuracy. I assume this is because all the digits are of same kind and same size.

But any way, this is a good start to go for beginners ( I hope so).


回答 1

对于那些对C ++代码感兴趣的人,可以参考以下代码。感谢Abid Rahman的出色解释。


步骤与上面相同,但是轮廓查找仅使用第一层次级别的轮廓,因此算法仅对每个数字使用外部轮廓。

用于创建样本和标签数据的代码

//Process image to extract contour
Mat thr,gray,con;
Mat src=imread("digit.png",1);
cvtColor(src,gray,CV_BGR2GRAY);
threshold(gray,thr,200,255,THRESH_BINARY_INV); //Threshold to find contour
thr.copyTo(con);

// Create sample and label data
vector< vector <Point> > contours; // Vector for storing contour
vector< Vec4i > hierarchy;
Mat sample;
Mat response_array;  
findContours( con, contours, hierarchy,CV_RETR_CCOMP, CV_CHAIN_APPROX_SIMPLE ); //Find contour

for( int i = 0; i< contours.size(); i=hierarchy[i][0] ) // iterate through first hierarchy level contours
{
    Rect r= boundingRect(contours[i]); //Find bounding rect for each contour
    rectangle(src,Point(r.x,r.y), Point(r.x+r.width,r.y+r.height), Scalar(0,0,255),2,8,0);
    Mat ROI = thr(r); //Crop the image
    Mat tmp1, tmp2;
    resize(ROI,tmp1, Size(10,10), 0,0,INTER_LINEAR ); //resize to 10X10
    tmp1.convertTo(tmp2,CV_32FC1); //convert to float
    sample.push_back(tmp2.reshape(1,1)); // Store  sample data
    imshow("src",src);
    int c=waitKey(0); // Read corresponding label for contour from keyoard
    c-=0x30;     // Convert ascii to intiger value
    response_array.push_back(c); // Store label to a mat
    rectangle(src,Point(r.x,r.y), Point(r.x+r.width,r.y+r.height), Scalar(0,255,0),2,8,0);    
}

// Store the data to file
Mat response,tmp;
tmp=response_array.reshape(1,1); //make continuous
tmp.convertTo(response,CV_32FC1); // Convert  to float

FileStorage Data("TrainingData.yml",FileStorage::WRITE); // Store the sample data in a file
Data << "data" << sample;
Data.release();

FileStorage Label("LabelData.yml",FileStorage::WRITE); // Store the label data in a file
Label << "label" << response;
Label.release();
cout<<"Training and Label data created successfully....!! "<<endl;

imshow("src",src);
waitKey();

培训和测试代码

Mat thr,gray,con;
Mat src=imread("dig.png",1);
cvtColor(src,gray,CV_BGR2GRAY);
threshold(gray,thr,200,255,THRESH_BINARY_INV); // Threshold to create input
thr.copyTo(con);


// Read stored sample and label for training
Mat sample;
Mat response,tmp;
FileStorage Data("TrainingData.yml",FileStorage::READ); // Read traing data to a Mat
Data["data"] >> sample;
Data.release();

FileStorage Label("LabelData.yml",FileStorage::READ); // Read label data to a Mat
Label["label"] >> response;
Label.release();


KNearest knn;
knn.train(sample,response); // Train with sample and responses
cout<<"Training compleated.....!!"<<endl;

vector< vector <Point> > contours; // Vector for storing contour
vector< Vec4i > hierarchy;

//Create input sample by contour finding and cropping
findContours( con, contours, hierarchy,CV_RETR_CCOMP, CV_CHAIN_APPROX_SIMPLE );
Mat dst(src.rows,src.cols,CV_8UC3,Scalar::all(0));

for( int i = 0; i< contours.size(); i=hierarchy[i][0] ) // iterate through each contour for first hierarchy level .
{
    Rect r= boundingRect(contours[i]);
    Mat ROI = thr(r);
    Mat tmp1, tmp2;
    resize(ROI,tmp1, Size(10,10), 0,0,INTER_LINEAR );
    tmp1.convertTo(tmp2,CV_32FC1);
    float p=knn.find_nearest(tmp2.reshape(1,1), 1);
    char name[4];
    sprintf(name,"%d",(int)p);
    putText( dst,name,Point(r.x,r.y+r.height) ,0,1, Scalar(0, 255, 0), 2, 8 );
}

imshow("src",src);
imshow("dst",dst);
imwrite("dest.jpg",dst);
waitKey();

结果

结果,第一行中的点被检测为8,而我们尚未训练该点。另外,我正在考虑将第一个层次结构中的每个轮廓作为样本输入,用户可以通过计算面积来避免它。

结果

For those who interested in C++ code can refer below code. Thanks Abid Rahman for the nice explanation.


The procedure is same as above but, the contour finding uses only first hierarchy level contour, so that the algorithm uses only outer contour for each digit.

Code for creating sample and Label data

//Process image to extract contour
Mat thr,gray,con;
Mat src=imread("digit.png",1);
cvtColor(src,gray,CV_BGR2GRAY);
threshold(gray,thr,200,255,THRESH_BINARY_INV); //Threshold to find contour
thr.copyTo(con);

// Create sample and label data
vector< vector <Point> > contours; // Vector for storing contour
vector< Vec4i > hierarchy;
Mat sample;
Mat response_array;  
findContours( con, contours, hierarchy,CV_RETR_CCOMP, CV_CHAIN_APPROX_SIMPLE ); //Find contour

for( int i = 0; i< contours.size(); i=hierarchy[i][0] ) // iterate through first hierarchy level contours
{
    Rect r= boundingRect(contours[i]); //Find bounding rect for each contour
    rectangle(src,Point(r.x,r.y), Point(r.x+r.width,r.y+r.height), Scalar(0,0,255),2,8,0);
    Mat ROI = thr(r); //Crop the image
    Mat tmp1, tmp2;
    resize(ROI,tmp1, Size(10,10), 0,0,INTER_LINEAR ); //resize to 10X10
    tmp1.convertTo(tmp2,CV_32FC1); //convert to float
    sample.push_back(tmp2.reshape(1,1)); // Store  sample data
    imshow("src",src);
    int c=waitKey(0); // Read corresponding label for contour from keyoard
    c-=0x30;     // Convert ascii to intiger value
    response_array.push_back(c); // Store label to a mat
    rectangle(src,Point(r.x,r.y), Point(r.x+r.width,r.y+r.height), Scalar(0,255,0),2,8,0);    
}

// Store the data to file
Mat response,tmp;
tmp=response_array.reshape(1,1); //make continuous
tmp.convertTo(response,CV_32FC1); // Convert  to float

FileStorage Data("TrainingData.yml",FileStorage::WRITE); // Store the sample data in a file
Data << "data" << sample;
Data.release();

FileStorage Label("LabelData.yml",FileStorage::WRITE); // Store the label data in a file
Label << "label" << response;
Label.release();
cout<<"Training and Label data created successfully....!! "<<endl;

imshow("src",src);
waitKey();

Code for training and testing

Mat thr,gray,con;
Mat src=imread("dig.png",1);
cvtColor(src,gray,CV_BGR2GRAY);
threshold(gray,thr,200,255,THRESH_BINARY_INV); // Threshold to create input
thr.copyTo(con);


// Read stored sample and label for training
Mat sample;
Mat response,tmp;
FileStorage Data("TrainingData.yml",FileStorage::READ); // Read traing data to a Mat
Data["data"] >> sample;
Data.release();

FileStorage Label("LabelData.yml",FileStorage::READ); // Read label data to a Mat
Label["label"] >> response;
Label.release();


KNearest knn;
knn.train(sample,response); // Train with sample and responses
cout<<"Training compleated.....!!"<<endl;

vector< vector <Point> > contours; // Vector for storing contour
vector< Vec4i > hierarchy;

//Create input sample by contour finding and cropping
findContours( con, contours, hierarchy,CV_RETR_CCOMP, CV_CHAIN_APPROX_SIMPLE );
Mat dst(src.rows,src.cols,CV_8UC3,Scalar::all(0));

for( int i = 0; i< contours.size(); i=hierarchy[i][0] ) // iterate through each contour for first hierarchy level .
{
    Rect r= boundingRect(contours[i]);
    Mat ROI = thr(r);
    Mat tmp1, tmp2;
    resize(ROI,tmp1, Size(10,10), 0,0,INTER_LINEAR );
    tmp1.convertTo(tmp2,CV_32FC1);
    float p=knn.find_nearest(tmp2.reshape(1,1), 1);
    char name[4];
    sprintf(name,"%d",(int)p);
    putText( dst,name,Point(r.x,r.y+r.height) ,0,1, Scalar(0, 255, 0), 2, 8 );
}

imshow("src",src);
imshow("dst",dst);
imwrite("dest.jpg",dst);
waitKey();

Result

In the result the dot in the first line is detected as 8 and we haven’t trained for dot. Also I am considering every contour in first hierarchy level as the sample input, user can avoid it by computing the area.

Results


回答 2

如果您对机器学习的最新技术感兴趣,则应研究深度学习。您应该具有支持GPU的CUDA,或者在Amazon Web Services上使用GPU。

Google Udacity使用Tensor Flow对此提供了很好的教程。本教程将教您如何在手写数字上训练自己的分类器。使用卷积网络,我在测试集上的准确性超过97%。

If you are interested in the state of the art in Machine Learning, you should look into Deep Learning. You should have a CUDA supporting GPU or alternatively use the GPU on Amazon Web Services.

Google Udacity has a nice tutorial on this using Tensor Flow. This tutorial will teach you how to train your own classifier on hand written digits. I got an accuracy of over 97% on the test set using Convolutional Networks.


Labelme-使用Python的图像多边形批注(多边形、矩形、圆、直线、点和图像级标志批注)

Labelme是一个图形图像标注工具,灵感来自http://labelme.csail.mit.edu
它是用Python编写的,并使用Qt作为其图形界面


各种图元(多边形、矩形、圆、直线和点)

功能

要求

安装

有以下选项:

python

您需要安装Anaconda,然后在下面运行:

# python2
conda create --name=labelme python=2.7
source activate labelme
# conda install -c conda-forge pyside2
conda install pyqt
pip install labelme
# if you'd like to use the latest version. run below:
# pip install git+https://github.com/wkentaro/labelme.git

# python3
conda create --name=labelme python=3.6
source activate labelme
# conda install -c conda-forge pyside2
# conda install pyqt
# pip install pyqt5  # pyqt5 can be installed via pip on python3
pip install labelme
# or you can install everything by conda command
# conda install labelme -c conda-forge

码头工人

您需要安装docker,然后在下面运行:

# on macOS
socat TCP-LISTEN:6000,reuseaddr,fork UNIX-CLIENT:\"$DISPLAY\" &
docker run -it -v /tmp/.X11-unix:/tmp/.X11-unix -e DISPLAY=docker.for.mac.host.internal:0 -v $(pwd):/root/workdir wkentaro/labelme

# on Linux
xhost +
docker run -it -v /tmp/.X11-unix:/tmp/.X11-unix -e DISPLAY=:0 -v $(pwd):/root/workdir wkentaro/labelme

Ubuntu

# Ubuntu 14.04 / Ubuntu 16.04
# Python2
# sudo apt-get install python-qt4  # PyQt4
sudo apt-get install python-pyqt5  # PyQt5
sudo pip install labelme
# Python3
sudo apt-get install python3-pyqt5  # PyQt5
sudo pip3 install labelme

# or install standalone executable from:
# https://github.com/wkentaro/labelme/releases

Ubuntu 19.10+/debian(Sid)

sudo apt-get install labelme

MacOS

# macOS Sierra
brew install pyqt  # maybe pyqt5
pip install labelme  # both python2/3 should work

# or install standalone executable/app from:
# https://github.com/wkentaro/labelme/releases

窗口

安装Anaconda,然后在python提示运行中:

# python3
conda create --name=labelme python=3.6
conda activate labelme
pip install labelme

用法

labelme --help有关详细信息,请参阅
批注另存为JSON文件

labelme  # just open gui

# tutorial (single image example)
cd examples/tutorial
labelme apc2016_obj3.jpg  # specify image file
labelme apc2016_obj3.jpg -O apc2016_obj3.json  # close window after the save
labelme apc2016_obj3.jpg --nodata  # not include image data but relative image path in JSON file
labelme apc2016_obj3.jpg \
  --labels highland_6539_self_stick_notes,mead_index_cards,kong_air_dog_squeakair_tennis_ball  # specify label list

# semantic segmentation example
cd examples/semantic_segmentation
labelme data_annotated/  # Open directory to annotate all images in it
labelme data_annotated/ --labels labels.txt  # specify label list with a file

有关更高级的用法,请参阅示例:

命令行参数

  • --output指定将写入批注的位置。如果位置以.json结尾,则此文件中将写入单个注释。如果使用.json指定位置,则只能注释一个图像。如果位置不是以.json结尾,程序将假定它是一个目录。批注将存储在此目录中,其名称与在其上进行批注的图像相对应
  • 第一次运行labelme时,它将在~/.labelmerc您可以编辑此文件,更改将在下次启动labelme时应用。如果您希望使用来自其他位置的配置文件,可以使用--config旗帜
  • 如果没有--nosortlabels标志时,程序将按字母顺序列出标签。当程序使用此标志运行时,它将按照标签提供的顺序显示标签
  • 将标志分配给整个图像。Example
  • 标签指定给单个多边形。Example

常见问题解答

测试

pip install hacking pytest pytest-qt
flake8 .
pytest -v tests

发展中的

git clone https://github.com/wkentaro/labelme.git
cd labelme

# Install anaconda3 and labelme
curl -L https://github.com/wkentaro/dotfiles/raw/master/local/bin/install_anaconda3.sh | bash -s .
source .anaconda3/bin/activate
pip install -e .

如何构建独立的可执行文件

下面显示了如何在MacOS、Linux和Windows上构建独立的可执行文件

# Setup conda
conda create --name labelme python==3.6.0
conda activate labelme

# Build the standalone executable
pip install .
pip install pyinstaller
pyinstaller labelme.spec
dist/labelme --version

如何做出贡献

确保以下测试在您的环境中通过
看见.github/workflows/ci.yml有关更多详细信息,请参阅

pip install black hacking pytest pytest-qt

flake8 .
black --line-length 79 --check labelme/
MPLBACKEND='agg' pytest tests/ -m 'not gpu'

确认

这项回购是mpitid/pylabelme,它的发育已经停止了

引用此项目

如果您在研究中使用此项目或希望参考自述文件中发布的基线结果,请使用以下BibTeX条目

@misc{labelme2016,
  author =       {Kentaro Wada},
  title =        {{labelme: Image Polygonal Annotation with Python}},
  howpublished = {\url{https://github.com/wkentaro/labelme}},
  year =         {2016}
}

Computervision-recipes-计算机视觉的最佳实践、代码示例和文档

计算机视觉

近年来,我们看到了计算机视觉的非同寻常的增长,应用于人脸识别、图像理解、搜索、无人机、地图绘制、半自动和自动驾驶车辆。其中许多应用的关键部分是视觉识别任务,例如图像分类、目标检测和图像相似度

此存储库提供构建计算机视觉系统的示例和最佳实践指南。该存储库的目标是构建一套全面的工具和示例,以利用计算机视觉算法、神经体系结构和实现此类系统的最新进展。我们不是从头开始创建实现,而是利用现有的最先进的库,围绕加载图像数据、优化和评估模型以及向上扩展到云来构建额外的实用程序。此外,在此领域工作多年后,我们的目标是回答常见问题,指出经常观察到的陷阱,并展示如何使用云进行培训和部署

我们希望这些示例和实用程序可以通过按数量级简化从定义业务问题到开发解决方案的过程来显著缩短“上市时间”。此外,示例笔记本将作为指南,并以多种语言展示工具的最佳实践和用法

这些示例提供为Jupyter notebooks也很常见utility functions所有示例都使用PyTorch作为底层深度学习库

目标受众

我们这个存储库的目标受众包括具有不同计算机视觉知识水平的数据科学家和机器学习工程师,因为我们的内容是纯来源的,目标是自定义的机器学习建模。所提供的实用程序和示例旨在作为解决实际视觉问题的加速器

快速入门

要开始,请导航到Setup Guide,其中列出了有关如何设置计算环境和运行此Repo中的笔记本所需的依赖项的说明。设置环境后,请导航到Scenarios文件夹,开始浏览笔记本。我们建议从图像分类笔记本,因为这引入了其他场景也使用的概念(例如关于ImageNet的预培训)

或者,我们支持活页夹Binder只需点击此链接,即可在网络浏览器中轻松试用我们的笔记本电脑。然而,Binder是免费的,因此只提供有限的CPU计算能力,并且没有GPU支持。预计笔记本的运行速度会非常慢(通过将图像分辨率降低到例如60像素,这在一定程度上有所改善,但代价是精确度较低)

场景

以下是此存储库中涵盖的常用计算机视觉场景的摘要。对于每个主要场景(“基础”),我们都会提供工具来有效地构建您自己的模型。这包括在您自己的数据上微调您自己的模型等简单任务,以及硬性否定挖掘甚至模型部署等更复杂的任务

场景 支持 描述
Classification 基地 图像分类是一种有监督的机器学习技术,用于学习和预测给定图像的类别
Similarity 基地 图像相似度是一种计算给定一对图像的相似度分数的方法。在给定图像的情况下,它允许您识别给定数据集中最相似的图像
Detection 基地 对象检测是一种允许您检测图像中对象的边界框的技术
Keypoints 基地 关键点检测可用于检测对象上的特定点。提供了一种预先训练的模型来检测人体关节,以进行人体姿态估计。
Segmentation 基地 图像分割为图像中的每个像素分配类别
Action recognition 基地 动作识别,用于在视频/网络摄像机镜头中识别执行的动作(例如,“运行”、“打开瓶子”)以及各自的开始/结束时间。我们还实现了可以在(Contrib)[contrib]下找到的动作识别的i3D实现
Tracking 基地 跟踪允许随时间检测和跟踪视频序列中的多个对象
Crowd counting Contrrib 统计低人群密度(如10人以下)和高人群密度(如数千人)场景下的人数

我们将支持的CV方案分为两个位置:(I)基地:“utils_cv”和“Scenario”文件夹中的代码和笔记本遵循严格的编码准则,经过良好的测试和维护;(Ii)Contrrib:“contrib”文件夹中的代码和其他资源,主要介绍使用尖端技术的不太常见的CV场景。“contrib”中的代码没有定期测试或维护

计算机视觉在蔚蓝上的应用

请注意,对于某些计算机视觉问题,您可能不需要构建自己的模型。取而代之的是,Azure上存在预先构建的或可轻松定制的解决方案,不需要任何自定义编码或机器学习专业知识。我们强烈建议您评估这些方法是否足以解决您的问题。如果这些解决方案不适用,或者这些解决方案的准确性不够,则可能需要求助于更复杂、更耗时的自定义方法

以下Microsoft服务提供了解决常见计算机视觉任务的简单解决方案:

  • Vision Services是一组经过预先训练的睡觉API,可以调用它们来进行图像标记、人脸识别、光学字符识别、视频分析等。这些API开箱即用,只需要极少的机器学习专业知识,但定制功能有限。查看各种可用的演示以体验该功能(例如Computer Vision)。该服务可通过API调用或通过SDK(以.NET、Python、Java、Node和Go语言提供)使用
  • Custom Vision是一项SaaS服务,用于在给定用户提供的培训集的情况下将模型训练和部署为睡觉应用编程接口。所有步骤,包括图像上传、注释和模型部署,都可以使用直观的UI或通过SDK(.Net、Python、Java、Node和Go语言)执行。训练图像分类或目标检测模型可以用最少的机器学习专业知识来实现。与使用预先培训的认知服务API相比,Custom Vision提供了更大的灵活性,但需要用户自带数据并对其进行注释

如果您需要培训您自己的模型,以下服务和链接提供了可能有用的附加信息

  • Azure Machine Learning service (AzureML)是一项帮助用户加速训练和部署机器学习模型的服务。虽然AzureML Python SDK不特定于计算机视觉工作负载,但它可以用于可伸缩且可靠的培训,并将机器学习解决方案部署到云中。我们在此存储库中的几个笔记本中利用Azure机器学习(例如deployment to Azure Kubernetes Service)
  • Azure AI Reference architectures提供一组示例(由代码支持),说明如何构建利用多个云组件的常见面向AI的工作负载。虽然不是特定于计算机视觉的,但这些参考体系结构涵盖了几个机器学习工作负载,例如模型部署或批处理评分

生成状态

AzureML测试

构建类型 分支机构 状态 分支机构 状态
Linux GPU 师傅 Build Status 试运行 Build Status
Linux CPU 师傅 Build Status 试运行 Build Status
笔记本电脑单元GPU 师傅 Build Status 试运行 Build Status

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这个项目欢迎大家提供意见和建议。请参阅我们的contribution guidelines

Jina-面向任何类别数据的云原生神经搜索框架

Jina logo: Jina is a cloud-native neural search framework

云-本地神经搜索[?]适用于以下方面的框架任何数据类型

Jina 允许您在短短几分钟内构建以深度学习为动力的搜索即服务

🌌所有数据类型-大规模索引和查询任何类型的非结构化数据:视频、图像、长/短文本、音乐、源代码、PDF等

🌩️FAST和本机云-从第一天开始的分布式架构,可扩展且设计为本地云:享受集装箱化、流式处理、并行、分片、异步调度、HTTP/GRPC/WebSocket协议

⏱️节省时间这个神经搜索系统的设计模式,从零到生产准备就绪的系统只需几分钟

🍱拥有您的堆栈-保持解决方案的端到端堆栈所有权,避免使用零散的、多供应商的通用旧式工具带来的集成陷阱

运行快速演示

安装

  • 通过PyPI:pip install -U "jina[standard]"
  • 通过Docker:docker run jinaai/jina:latest
更多安装选项
x86/64、arm64、v6、v7 Linux/MacOS和Python 3.7/3.8/3.9 Docker用户
最低要求
(不支持HTTP、WebSocket、Docker)
pip install jina docker run jinaai/jina:latest
Daemon pip install "jina[daemon]" docker run --network=host jinaai/jina:latest-daemon
使用附加服务 pip install "jina[devel]" docker run jinaai/jina:latest-devel

版本标识符are explained here吉娜可以继续奔跑Windows Subsystem for Linux我们欢迎社会各界帮助我们native Windows support

开始使用

文档、执行者和流是JINA中的三个基本概念

1个️⃣复制-粘贴下面的最小示例并运行它:

💡预赛:character embeddingpoolingEuclidean distance

The architecture of a simple neural search system powered by Jina

import numpy as np
from jina import Document, DocumentArray, Executor, Flow, requests

class CharEmbed(Executor):  # a simple character embedding with mean-pooling
    offset = 32  # letter `a`
    dim = 127 - offset + 1  # last pos reserved for `UNK`
    char_embd = np.eye(dim) * 1  # one-hot embedding for all chars

    @requests
    def foo(self, docs: DocumentArray, **kwargs):
        for d in docs:
            r_emb = [ord(c) - self.offset if self.offset <= ord(c) <= 127 else (self.dim - 1) for c in d.text]
            d.embedding = self.char_embd[r_emb, :].mean(axis=0)  # average pooling

class Indexer(Executor):
    _docs = DocumentArray()  # for storing all documents in memory

    @requests(on='/index')
    def foo(self, docs: DocumentArray, **kwargs):
        self._docs.extend(docs)  # extend stored `docs`

    @requests(on='/search')
    def bar(self, docs: DocumentArray, **kwargs):
        q = np.stack(docs.get_attributes('embedding'))  # get all embeddings from query docs
        d = np.stack(self._docs.get_attributes('embedding'))  # get all embeddings from stored docs
        euclidean_dist = np.linalg.norm(q[:, None, :] - d[None, :, :], axis=-1)  # pairwise euclidean distance
        for dist, query in zip(euclidean_dist, docs):  # add & sort match
            query.matches = [Document(self._docs[int(idx)], copy=True, scores={'euclid': d}) for idx, d in enumerate(dist)]
            query.matches.sort(key=lambda m: m.scores['euclid'].value)  # sort matches by their values

f = Flow(port_expose=12345, protocol='http', cors=True).add(uses=CharEmbed, parallel=2).add(uses=Indexer)  # build a Flow, with 2 parallel CharEmbed, tho unnecessary
with f:
    f.post('/index', (Document(text=t.strip()) for t in open(__file__) if t.strip()))  # index all lines of _this_ file
    f.block()  # block for listening request

2个️⃣打开http://localhost:12345/docs(扩展的Swagger UI)在浏览器中,单击/搜索制表符和输入:

{"data": [{"text": "@requests(on=something)"}]}

也就是说,我们希望从上面的代码片段中找到与以下内容最相似的行@request(on=something)现在单击执行巴顿!

Jina Swagger UI extension on visualizing neural search results

3个️⃣不是图形用户界面的人?那就让我们用Python来做吧!保持上述服务器运行,并启动一个简单的客户端:

from jina import Client, Document
from jina.types.request import Response


def print_matches(resp: Response):  # the callback function invoked when task is done
    for idx, d in enumerate(resp.docs[0].matches[:3]):  # print top-3 matches
        print(f'[{idx}]{d.scores["euclid"].value:2f}: "{d.text}"')


c = Client(protocol='http', port_expose=12345)  # connect to localhost:12345
c.post('/search', Document(text='request(on=something)'), on_done=print_matches)

,它打印以下结果:

         Client@1608[S]:connected to the gateway at localhost:12345!
[0]0.168526: "@requests(on='/index')"
[1]0.181676: "@requests(on='/search')"
[2]0.192049: "query.matches = [Document(self._docs[int(idx)], copy=True, score=d) for idx, d in enumerate(dist)]"

😔不管用吗?我们的错!Please report it here.

阅读教程

支持

加入我们吧

吉娜的后盾是Jina AIWe are actively hiring全栈开发人员、解决方案工程师在开源领域构建下一个神经搜索生态系统

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Virgilio-您的数据科学E-Learning新导师

Virgilio是什么?

通过互联网学习和阅读意味着在一个混沌信息的无限丛林,在快速变化的创新领域更是如此

你有没有感到不知所措?当试图接近数据科学没有一条真正的“路”可走?

你是否厌倦了点击“Run”,“Run”,“Run”。在一本木星笔记本上,带着别人工作的舒适区给人的那种虚假的自信?

您是否曾经因为同一算法或方法的几个相互矛盾的名称而感到困惑,这些名称来自不同的网站和零散的教程?

Virgilio为每个人免费解决这些关键问题

Enter in the new web version of Virgilio!

关于

Virgilio由以下人员开发和维护these awesome people您可以给我们发电子邮件virgilio.datascience (at) gmail.com或加入Discord chat

贡献力量

太棒了!检查contribution guidelines参与我们的项目吧!

许可证

内容由-NC-SA 4.0在知识共享下发布license代码在MIT licenseVirgilio形象来自于here

D2l-zh 动手学深度学习

本开源项目代表了我们的一种尝试:我们将教给读者概念、背景知识和代码;我们将在同一个地方阐述剖析问题所需的批判性思维、解决问题所需的数学知识,以及实现解决方案所需的工程技能.

我们的目标是创建一个为实现以下目标的统一资源:

  1. 所有人均可在网上免费获取;
  2. 提供足够的技术深度,从而帮助读者实际成为深度学习应用科学家:既理解数学原理,又能够实现并不断改进方法;
  3. 包含可运行的代码,为读者展示如何在实际中解决问题.这样不仅直接将数学公式对应成实际代码,而且可以修改代码、观察结果并及时获取经验;
  4. 允许我们和整个社区不断快速迭代内容,从而紧跟仍在高速发展的深度学习领域;
  5. 由包含有关技术细节问答的论坛作为补充,使大家可以相互答疑并交换经验.
将本书(中英文版)用作教材或参考书的大学

如果本书对你有帮助,请星空(★)本仓库或引用本书的英文版:

@article{zhang2021dive,
    title={Dive into Deep Learning},
    author={Zhang, Aston and Lipton, Zachary C. and Li, Mu and Smola, Alexander J.},
    journal={arXiv preprint arXiv:2106.11342},
    year={2021}
}

本书的第二版

虽然纸质书第一版已经出版,但深度学习领域依然在迅速发展.为了得到来自更广泛的英文开源社区的帮助,从而提升本书质量,本书的第二版正在用英文写.英文版正不断被搬回中文版中.

目前,英文版已超过160节(中文版共96节),例如增加了理论背景(如优化收敛分析)、硬件设计(如参数服务器)、全新篇章(如注意力机制、推荐系统、深度学习的数学、生成对抗网络)、应用种类(如自然语言推理)、模型种类(如变压器、BERT)等,并优化重组了大量章节(如将自然语言处理篇章按从预训练表征、到模型设计、再到下游应用重构)。

欢迎关注本书第二版的英文开源项目

中英文教学资源

加州大学伯克利分校2019年年春学期Introduction to Deep Learning 课程教材(同时提供含教学视频地址的中文版课件).

学术界推荐

“如果你想深入学习,那就看看这本书吧!”

-韩家炜,acm院士、ieee院士,美国伊利诺伊大学香槟分校计算机系Michael Aiken主席教授

“这对机器学习文献来说是一个非常受欢迎的补充。”

–Bernhard Schölkopf,acm院士、德国国家科学院院士,德国马克斯·普朗克研究所智能系统院院长

“书中代码可谓‘所学即所用’。”

-周志华,acm院士、ieee院士、aaas院士,南京大学计算机科学与技术系主任

“这本书可以帮助深度学习实践者快速提升自己的能力”

-张潼,asa院士、ims院士,香港科技大学计算机系和数学系教授

工业界推荐

“一本优秀的深度学习教材,值得任何想了解深度学习何以引爆人工智能革命的人关注”

-黄仁勋,NVIDIA创始人兼首席执行官

“”动手学深度学习“是最适合工业界研发工程师学习的.我毫无保留地向广大的读者们强烈推荐。”

-余凯,地平线公司创始人&首席执行官

“强烈推荐这本书!我特别赞赏这种手脑一体的学习方式”

-漆远,蚂蚁金服副总裁、首席AI科学家

“”动手学深度学习“是一本很容易让学习者上瘾的书。”

–沈强,将门创投创始合伙人

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感谢社区贡献者们为每一位读者改进这本开源书.

如何贡献|致谢|讨论或报告问题|其他