如何从Python中的URL读取图像数据?

问题:如何从Python中的URL读取图像数据?

当我们处理本地文件时,我想做的事情很简单,但是当我尝试使用远程URL时,问题就来了。

基本上,我试图从URL提取的文件中创建一个PIL图像对象。当然,我总是可以仅获取URL并将其存储在临时文件中,然后将其打开到图像对象中,但这感觉效率很低。

这是我所拥有的:

Image.open(urlopen(url))

它抱怨seek()说不可用,所以我尝试了这个:

Image.open(urlopen(url).read())

但这也不起作用。有没有更好的方法可以执行此操作,还是可以将这种方式写入临时文件?

What I’m trying to do is fairly simple when we’re dealing with a local file, but the problem comes when I try to do this with a remote URL.

Basically, I’m trying to create a PIL image object from a file pulled from a URL. Sure, I could always just fetch the URL and store it in a temp file, then open it into an image object, but that feels very inefficient.

Here’s what I have:

Image.open(urlopen(url))

It flakes out complaining that seek() isn’t available, so then I tried this:

Image.open(urlopen(url).read())

But that didn’t work either. Is there a Better Way to do this, or is writing to a temporary file the accepted way of doing this sort of thing?


回答 0

在Python3中,StringIO和cStringIO模块不见了。

在Python3中,您应该使用:

from PIL import Image
import requests
from io import BytesIO

response = requests.get(url)
img = Image.open(BytesIO(response.content))

In Python3 the StringIO and cStringIO modules are gone.

In Python3 you should use:

from PIL import Image
import requests
from io import BytesIO

response = requests.get(url)
img = Image.open(BytesIO(response.content))

回答 1

你可以尝试使用StringIO

import urllib, cStringIO

file = cStringIO.StringIO(urllib.urlopen(URL).read())
img = Image.open(file)

you could try using a StringIO

import urllib, cStringIO

file = cStringIO.StringIO(urllib.urlopen(URL).read())
img = Image.open(file)

回答 2

我使用请求库。它似乎更强大。

from PIL import Image
import requests
from StringIO import StringIO

response = requests.get(url)
img = Image.open(StringIO(response.content))

I use the requests library. It seems to be more robust.

from PIL import Image
import requests
from StringIO import StringIO

response = requests.get(url)
img = Image.open(StringIO(response.content))

回答 3

对于使用Pillow的用户,从2.8.0版开始,您可以:

from PIL import Image
import urllib2

im = Image.open(urllib2.urlopen(url))

或者,如果您使用requests

from PIL import Image
import requests

im = Image.open(requests.get(url, stream=True).raw)

参考文献:

For those of you who use Pillow, from version 2.8.0 you can:

from PIL import Image
import urllib2

im = Image.open(urllib2.urlopen(url))

or if you use requests:

from PIL import Image
import requests

im = Image.open(requests.get(url, stream=True).raw)

References:


回答 4

使用StringIO转读字符串转换为一个类文件对象:

from StringIO import StringIO
import urllib

Image.open(StringIO(urllib.requests.urlopen(url).read()))

Use StringIO to turn the read string into a file-like object:

from StringIO import StringIO
import urllib

Image.open(StringIO(urllib.requests.urlopen(url).read()))

回答 5

对于进行某些sklearn / numpy后处理(即深度学习)的用户,可以使用np.array()包装PIL对象。这样可以避免您像我一样去过Google:

from PIL import Image
import requests
import numpy as np
from StringIO import StringIO

response = requests.get(url)
img = np.array(Image.open(StringIO(response.content)))

For those doing some sklearn/numpy post processing (i.e. Deep learning) you can wrap the PIL object with np.array(). This might save you from having to Google it like I did:

from PIL import Image
import requests
import numpy as np
from StringIO import StringIO

response = requests.get(url)
img = np.array(Image.open(StringIO(response.content)))

回答 6

Python 3

from urllib.request import urlopen
from PIL import Image

img = Image.open(urlopen(url))
img

Jupyter Notebook和IPython

import IPython
url = 'https://newevolutiondesigns.com/images/freebies/colorful-background-14.jpg'
IPython.display.Image(url, width = 250)

与其他方法不同,此方法还可以在for循环中使用!

Python 3

from urllib.request import urlopen
from PIL import Image

img = Image.open(urlopen(url))
img

Jupyter Notebook and IPython

import IPython
url = 'https://newevolutiondesigns.com/images/freebies/colorful-background-14.jpg'
IPython.display.Image(url, width = 250)

Unlike other methods, this method also works in a for loop!


回答 7

如今,可以建议使用的图像输入/输出方法是使用专用的软件包ImageIO。可以使用以下简单代码行直接从URL读取图像数据:

from imageio import imread
image = imread('https://cdn.sstatic.net/Sites/stackoverflow/img/logo.png')

此页面上的许多答案早于该软件包的发布,因此没有提及。ImageIO最初是Scikit-Image工具箱的组件。除了流行的图像处理库PILlow提供的格式外,它还支持多种科学格式。它将所有内容包装在仅关注图像输入/输出的干净API中。实际上,SciPy 取消了自己的图像读取器/写入器,转而使用ImageIO

The arguably recommended way to do image input/output these days is to use the dedicated package ImageIO. Image data can be read directly from a URL with one simple line of code:

from imageio import imread
image = imread('https://cdn.sstatic.net/Sites/stackoverflow/img/logo.png')

Many answers on this page predate the release of that package and therefore do not mention it. ImageIO started out as component of the Scikit-Image toolkit. It supports a number of scientific formats on top of the ones provided by the popular image-processing library PILlow. It wraps it all in a clean API solely focused on image input/output. In fact, SciPy removed its own image reader/writer in favor of ImageIO.


回答 8

选择chrome图像,右键单击它,单击Copy image address,将其粘贴到str变量(my_url)中以读取图像:

import shutil
import requests

my_url = 'https://www.washingtonian.com/wp-content/uploads/2017/06/6-30-17-goat-yoga-congressional-cemetery-1-994x559.jpg'
response = requests.get(my_url, stream=True)
with open('my_image.png', 'wb') as file:
    shutil.copyfileobj(response.raw, file)
del response

打开它;

from PIL import Image

img = Image.open('my_image.png')
img.show()

select the image in chrome, right click on it, click on Copy image address, paste it into a str variable (my_url) to read the image:

import shutil
import requests

my_url = 'https://www.washingtonian.com/wp-content/uploads/2017/06/6-30-17-goat-yoga-congressional-cemetery-1-994x559.jpg'
response = requests.get(my_url, stream=True)
with open('my_image.png', 'wb') as file:
    shutil.copyfileobj(response.raw, file)
del response

open it;

from PIL import Image

img = Image.open('my_image.png')
img.show()