Python 实用宝典

Question 1

I’m using Matplotlib to plot a histogram. Using tips from my previous question: Matplotlib – label each bin, I’ve more or less go the kinks worked out.

There’s one final issue – previously – the x-axis label (“Time (in milliseconds)”) was being rendered underneath the x-axis tickmarks (0.00, 0.04, 0.08, 0.12 etc.)

Using the advice from Joe Kingston (see question above), I tried using:

ax.tick_params(axis='x', pad=30)

However, this moves both the x-axis tickmarks (0.00, 0.04, 0.08, 0.12 etc.), as well as the x-axis label (“Time (in milliseconds)”):

Is there any way to move only the x-axis label to underneath the three rows of figures?

Nb: You may need to open the PNGs below directly – Right Click on the image, then View Image (in FF), or Open image in new tab (Chrome). The image resize done by SO has rendered them nigh unreadable

Question 2

use labelpad parameter:

pl.xlabel("...", labelpad=20)

or set it after:

ax.xaxis.labelpad = 20

Question 3

If the variable ax.xaxis._autolabelpos = True, matplotlib sets the label position in function _update_label_position in axis.py according to (some excerpts):

    bboxes, bboxes2 = self._get_tick_bboxes(ticks_to_draw, renderer)
    bbox = mtransforms.Bbox.union(bboxes)
    bottom = bbox.y0
    x, y = self.label.get_position()
    self.label.set_position((x, bottom - self.labelpad * self.figure.dpi / 72.0))

You can set the label position independently of the ticks by using:

    ax.xaxis.set_label_coords(x0, y0)

that sets _autolabelpos to False or as mentioned above by changing the labelpad parameter.

Question 4

How can I save Python plots at very high quality?

That is, when I keep zooming in on the object saved in a PDF file, why isn’t there any blurring?

Also, what would be the best mode to save it in?

png, eps? Or some other? I can’t do pdf, because there is a hidden number that happens that mess with Latexmk compilation.

Question 5

If you are using Matplotlib and are trying to get good figures in a LaTeX document, save as an EPS. Specifically, try something like this after running the commands to plot the image:

plt.savefig('destination_path.eps', format='eps')

I have found that EPS files work best and the dpi parameter is what really makes them look good in a document.

To specify the orientation of the figure before saving, simply call the following before the plt.savefig call, but after creating the plot (assuming you have plotted using an axes with the name ax):

ax.view_init(elev=elevation_angle, azim=azimuthal_angle)

Where elevation_angle is a number (in degrees) specifying the polar angle (down from vertical z axis) and the azimuthal_angle specifies the azimuthal angle (around the z axis).

I find that it is easiest to determine these values by first plotting the image and then rotating it and watching the current values of the angles appear towards the bottom of the window just below the actual plot. Keep in mind that the x, y, z, positions appear by default, but they are replaced with the two angles when you start to click+drag+rotate the image.

Question 6

Just to add my results, also using Matplotlib.

.eps made all my text bold and removed transparency. .svg gave me high-resolution pictures that actually looked like my graph.

import matplotlib.pyplot as plt
fig, ax = plt.subplots()
# Do the plot code
fig.savefig('myimage.svg', format='svg', dpi=1200)

I used 1200 dpi because a lot of scientific journals require images in 1200 / 600 / 300 dpi, depending on what the image is of. Convert to desired dpi and format in GIMP or Inkscape.

Obviously the dpi doesn’t matter since .svg are vector graphics and have “infinite resolution”.

Question 7

Okay, I found spencerlyon2’s answer working. However, in case anybody would find himself/herself not knowing what to do with that one line, I had to do it this way:

beingsaved = plt.figure()

# Some scatter plots
plt.scatter(X_1_x, X_1_y)
plt.scatter(X_2_x, X_2_y)

beingsaved.savefig('destination_path.eps', format='eps', dpi=1000)

Question 8

In case you are working with seaborn plots, instead of Matplotlib, you can save a .png image like this:

Let’s suppose you have a matrix object (either Pandas or NumPy), and you want to take a heatmap:

import seaborn as sb

image = sb.heatmap(matrix)   # This gets you the heatmap
image.figure.savefig("C:/Your/Path/ ... /your_image.png")   # This saves it

This code is compatible with the latest version of Seaborn. Other code around Stack Overflow worked only for previous versions.

Another way I like is this. I set the size of the next image as follows:

plt.subplots(figsize=(15,15))

And then later I plot the output in the console, from which I can copy-paste it where I want. (Since Seaborn is built on top of Matplotlib, there will not be any problem.)

Question 9

You can save to a figure that is 1920×1080 (or 1080p) using:

fig = plt.figure(figsize=(19.20,10.80))

You can also go much higher or lower. The above solutions work well for printing, but these days you want the created image to go into a PNG/JPG or appear in a wide screen format.

Question 10

I am using matplotlib to make some graphs and unfortunately I cannot export them without the white background.

In other words, when I export a plot like this and position it on top of another image, the white background hides what is behind it rather than allowing it to show through. How can I export plots with a transparent background instead?

Question 11

Use the matplotlib savefig function with the keyword argument transparent=True to save the image as a png file.

In [30]: x = np.linspace(0,6,31)

In [31]: y = np.exp(-0.5*x) * np.sin(x)

In [32]: plot(x, y, 'bo-')
Out[32]: [<matplotlib.lines.Line2D at 0x3f29750>]            

In [33]: savefig('demo.png', transparent=True)

Result:

Of course, that plot doesn’t demonstrate the transparency. Here’s a screenshot of the PNG file displayed using the ImageMagick display command. The checkerboard pattern is the background that is visible through the transparent parts of the PNG file.

Question 12

Png files can handle transparency. So you could use this question Save plot to image file instead of displaying it using Matplotlib so as to save you graph as a png file.

And if you want to turn all white pixel transparent, there’s this other question : Using PIL to make all white pixels transparent?

If you want to turn an entire area to transparent, then there’s this question: And then use the PIL library like in this question Python PIL: how to make area transparent in PNG? so as to make your graph transparent.

Question 13

I’m trying to make a square plot (using imshow), i.e. aspect ratio of 1:1, but I can’t. None of these work:

import matplotlib.pyplot as plt

ax = fig.add_subplot(111,aspect='equal')
ax = fig.add_subplot(111,aspect=1.0)
ax.set_aspect('equal')
plt.axes().set_aspect('equal')

It seems like the calls are just being ignored (a problem I often seem to have with matplotlib).

Question 14

Third times the charm. My guess is that this is a bug and Zhenya’s answer suggests it’s fixed in the latest version. I have version 0.99.1.1 and I’ve created the following solution:

import matplotlib.pyplot as plt
import numpy as np

def forceAspect(ax,aspect=1):
    im = ax.get_images()
    extent =  im[0].get_extent()
    ax.set_aspect(abs((extent[1]-extent[0])/(extent[3]-extent[2]))/aspect)

data = np.random.rand(10,20)

fig = plt.figure()
ax = fig.add_subplot(111)
ax.imshow(data)
ax.set_xlabel('xlabel')
ax.set_aspect(2)
fig.savefig('equal.png')
ax.set_aspect('auto')
fig.savefig('auto.png')
forceAspect(ax,aspect=1)
fig.savefig('force.png')

This is ‘force.png’:

Below are my unsuccessful, yet hopefully informative attempts.

Second Answer:

My ‘original answer’ below is overkill, as it does something similar to axes.set_aspect(). I think you want to use axes.set_aspect('auto'). I don’t understand why this is the case, but it produces a square image plot for me, for example this script:

import matplotlib.pyplot as plt
import numpy as np

data = np.random.rand(10,20)

fig = plt.figure()
ax = fig.add_subplot(111)
ax.imshow(data)
ax.set_aspect('equal')
fig.savefig('equal.png')
ax.set_aspect('auto')
fig.savefig('auto.png')

Produces an image plot with ‘equal’ aspect ratio: and one with ‘auto’ aspect ratio:

The code provided below in the ‘original answer’ provides a starting off point for an explicitly controlled aspect ratio, but it seems to be ignored once an imshow is called.

Original Answer:

Here’s an example of a routine that will adjust the subplot parameters so that you get the desired aspect ratio:

import matplotlib.pyplot as plt

def adjustFigAspect(fig,aspect=1):
    '''
    Adjust the subplot parameters so that the figure has the correct
    aspect ratio.
    '''
    xsize,ysize = fig.get_size_inches()
    minsize = min(xsize,ysize)
    xlim = .4*minsize/xsize
    ylim = .4*minsize/ysize
    if aspect < 1:
        xlim *= aspect
    else:
        ylim /= aspect
    fig.subplots_adjust(left=.5-xlim,
                        right=.5+xlim,
                        bottom=.5-ylim,
                        top=.5+ylim)

fig = plt.figure()
adjustFigAspect(fig,aspect=.5)
ax = fig.add_subplot(111)
ax.plot(range(10),range(10))

fig.savefig('axAspect.png')

This produces a figure like so:

I can imagine if your having multiple subplots within the figure, you would want to include the number of y and x subplots as keyword parameters (defaulting to 1 each) to the routine provided. Then using those numbers and the hspace and wspace keywords, you can make all the subplots have the correct aspect ratio.

Question 15

What is the matplotlib version you are running? I have recently had to upgrade to 1.1.0, and with it, add_subplot(111,aspect='equal') works for me.

Question 16

After many years of success with the answers above, I have found this not to work again – but I did find a working solution for subplots at

https://jdhao.github.io/2017/06/03/change-aspect-ratio-in-mpl

With full credit of course to the author above (who can perhaps rather post here), the relevant lines are:

ratio = 1.0
xleft, xright = ax.get_xlim()
ybottom, ytop = ax.get_ylim()
ax.set_aspect(abs((xright-xleft)/(ybottom-ytop))*ratio)

The link also has a crystal clear explanation of the different coordinate systems used by matplotlib.

Thanks for all great answers received – especially @Yann’s which will remain the winner.

Question 17

you should try with figaspect. It works for me. From the docs:

Create a figure with specified aspect ratio. If arg is a number, use that aspect ratio. > If arg is an array, figaspect will determine the width and height for a figure that would fit array preserving aspect ratio. The figure width, height in inches are returned. Be sure to create an axes with equal with and height, eg

Example usage:

  # make a figure twice as tall as it is wide
  w, h = figaspect(2.)
  fig = Figure(figsize=(w,h))
  ax = fig.add_axes([0.1, 0.1, 0.8, 0.8])
  ax.imshow(A, **kwargs)

  # make a figure with the proper aspect for an array
  A = rand(5,3)
  w, h = figaspect(A)
  fig = Figure(figsize=(w,h))
  ax = fig.add_axes([0.1, 0.1, 0.8, 0.8])
  ax.imshow(A, **kwargs)

Edit: I am not sure of what you are looking for. The above code changes the canvas (the plot size). If you want to change the size of the matplotlib window, of the figure, then use:

In [68]: f = figure(figsize=(5,1))

this does produce a window of 5×1 (wxh).

Question 18

This answer is based on Yann’s answer. It will set the aspect ratio for linear or log-log plots. I’ve used additional information from https://stackoverflow.com/a/16290035/2966723 to test if the axes are log-scale.

def forceAspect(ax,aspect=1):
    #aspect is width/height
    scale_str = ax.get_yaxis().get_scale()
    xmin,xmax = ax.get_xlim()
    ymin,ymax = ax.get_ylim()
    if scale_str=='linear':
        asp = abs((xmax-xmin)/(ymax-ymin))/aspect
    elif scale_str=='log':
        asp = abs((scipy.log(xmax)-scipy.log(xmin))/(scipy.log(ymax)-scipy.log(ymin)))/aspect
    ax.set_aspect(asp)

Obviously you can use any version of log you want, I’ve used scipy, but numpy or math should be fine.

Question 19

I’d like to create a colorbar legend for a heatmap, such that the labels are in the center of each discrete color. Example borrowed from here:

import matplotlib.pyplot as plt
import numpy as np
from matplotlib.colors import ListedColormap

#discrete color scheme
cMap = ListedColormap(['white', 'green', 'blue','red'])

#data
np.random.seed(42)
data = np.random.rand(4, 4)
fig, ax = plt.subplots()
heatmap = ax.pcolor(data, cmap=cMap)

#legend
cbar = plt.colorbar(heatmap)
cbar.ax.set_yticklabels(['0','1','2','>3'])
cbar.set_label('# of contacts', rotation=270)

# put the major ticks at the middle of each cell
ax.set_xticks(np.arange(data.shape[1]) + 0.5, minor=False)
ax.set_yticks(np.arange(data.shape[0]) + 0.5, minor=False)
ax.invert_yaxis()

#labels
column_labels = list('ABCD')
row_labels = list('WXYZ')
ax.set_xticklabels(column_labels, minor=False)
ax.set_yticklabels(row_labels, minor=False)

plt.show()

This generates the following plot:

Ideally I’d like to generate a legend bar which has the four colors and for each color, a label in its center: 0,1,2,>3. How can this be achieved?

Question 20

import matplotlib.pyplot as plt
import numpy as np
from matplotlib.colors import ListedColormap

#discrete color scheme
cMap = ListedColormap(['white', 'green', 'blue','red'])

#data
np.random.seed(42)
data = np.random.rand(4, 4)
fig, ax = plt.subplots()
heatmap = ax.pcolor(data, cmap=cMap)

#legend
cbar = plt.colorbar(heatmap)

cbar.ax.get_yaxis().set_ticks([])
for j, lab in enumerate(['$0$','$1$','$2$','$>3$']):
    cbar.ax.text(.5, (2 * j + 1) / 8.0, lab, ha='center', va='center')
cbar.ax.get_yaxis().labelpad = 15
cbar.ax.set_ylabel('# of contacts', rotation=270)


# put the major ticks at the middle of each cell
ax.set_xticks(np.arange(data.shape[1]) + 0.5, minor=False)
ax.set_yticks(np.arange(data.shape[0]) + 0.5, minor=False)
ax.invert_yaxis()

#labels
column_labels = list('ABCD')
row_labels = list('WXYZ')
ax.set_xticklabels(column_labels, minor=False)
ax.set_yticklabels(row_labels, minor=False)

plt.show()

You were very close. Once you have a reference to the color bar axis, you can do what ever you want to it, including putting text labels in the middle. You might want to play with the formatting to make it more visible.

Question 21

To add to tacaswell’s answer, the colorbar() function has an optional cax input you can use to pass an axis on which the colorbar should be drawn. If you are using that input, you can directly set a label using that axis.

import matplotlib.pyplot as plt
from mpl_toolkits.axes_grid1 import make_axes_locatable

fig, ax = plt.subplots()
heatmap = ax.imshow(data)
divider = make_axes_locatable(ax)
cax = divider.append_axes('bottom', size='10%', pad=0.6)
cb = fig.colorbar(heatmap, cax=cax, orientation='horizontal')

cax.set_xlabel('data label')  # cax == cb.ax

Question 22

我0.19.1在Python 3上使用Pandas 。我在这些代码行上收到警告。我正在尝试获取一个包含所有Peter在column处存在string的行号的列表Unnamed: 5。

df = pd.read_excel(xls_path)
myRows = df[df['Unnamed: 5'] == 'Peter'].index.tolist()

它产生一个警告：

"\Python36\lib\site-packages\pandas\core\ops.py:792: FutureWarning: elementwise 
comparison failed; returning scalar, but in the future will perform 
elementwise comparison 
result = getattr(x, name)(y)"

这是什么FutureFarning，由于它似乎起作用，因此我应该忽略它。

Question 23

I am using Pandas 0.19.1 on Python 3. I am getting a warning on these lines of code. I’m trying to get a list that contains all the row numbers where string Peter is present at column Unnamed: 5.

df = pd.read_excel(xls_path)
myRows = df[df['Unnamed: 5'] == 'Peter'].index.tolist()

It produces a Warning:

"\Python36\lib\site-packages\pandas\core\ops.py:792: FutureWarning: elementwise 
comparison failed; returning scalar, but in the future will perform 
elementwise comparison 
result = getattr(x, name)(y)"

What is this FutureWarning and should I ignore it since it seems to work.

Question 24

此FutureWarning并非来自Pandas，而是来自numpy，并且该错误也影响了matplotlib和其他漏洞，以下是在更接近问题根源的位置重现警告的方法：

import numpy as np
print(np.__version__)   # Numpy version '1.12.0'
'x' in np.arange(5)       #Future warning thrown here

FutureWarning: elementwise comparison failed; returning scalar instead, but in the 
future will perform elementwise comparison
False

使用double equals运算符重现此错误的另一种方法：

import numpy as np
np.arange(5) == np.arange(5).astype(str)    #FutureWarning thrown here

受此FutureWarning影响的Matplotlib示例在其颤动图实施下：https ://matplotlib.org/examples/pylab_examples/quiver_demo.html

这里发生了什么？

当您将字符串与numpy的数字类型进行比较时，Numpy和本机python之间会发生什么分歧。请注意，左操作数是python的草皮，是原始字符串，中间操作是python的草皮，而右操作数是numpy的草皮。您应该返回Python样式的Scalar还是Numpy样式的ndarray布尔值？Numpy说布尔的ndarray，Pythonic开发人员不同意。经典对峙。

如果数组中存在item，应该是元素比较还是标量？

如果您的代码或库使用in或==运算符将python字符串与numpy ndarrays比较，则它们不兼容，因此，当您尝试使用它时，它将返回标量，但仅在现在。警告表示将来这种行为可能会改变，因此，如果python / numpy决定采用Numpy样式，则您的代码会全程吐槽。

提交的错误报告：

Numpy和Python处于僵持状态，目前操作返回标量，但将来可能会改变。

https://github.com/numpy/numpy/issues/6784

https://github.com/pandas-dev/pandas/issues/7830

两种解决方法：

无论您锁定Python和numpy的版本，忽略这些警告并期望行为不改变，或转换的左侧和右侧的操作数==，并in从一个numpy的类型或原始数值Python类型。

全局禁止警告：

import warnings
import numpy as np
warnings.simplefilter(action='ignore', category=FutureWarning)
print('x' in np.arange(5))   #returns False, without Warning

逐行抑制警告。

import warnings
import numpy as np

with warnings.catch_warnings():
    warnings.simplefilter(action='ignore', category=FutureWarning)
    print('x' in np.arange(2))   #returns False, warning is suppressed

print('x' in np.arange(10))   #returns False, Throws FutureWarning

只需按名称隐藏警告，然后在其旁边添加一个大声注释，提及python和numpy的当前版本，并说此代码很脆弱，并且需要这些版本，并在此处添加了链接。踢罐子的路。

TLDR： pandas是绝地武士；numpy是小屋并且python是银河帝国。 https://youtu.be/OZczsiCfQQk?t=3

Question 25

This FutureWarning isn’t from Pandas, it is from numpy and the bug also affects matplotlib and others, here’s how to reproduce the warning nearer to the source of the trouble:

import numpy as np
print(np.__version__)   # Numpy version '1.12.0'
'x' in np.arange(5)       #Future warning thrown here

FutureWarning: elementwise comparison failed; returning scalar instead, but in the 
future will perform elementwise comparison
False

Another way to reproduce this bug using the double equals operator:

import numpy as np
np.arange(5) == np.arange(5).astype(str)    #FutureWarning thrown here

An example of Matplotlib affected by this FutureWarning under their quiver plot implementation: https://matplotlib.org/examples/pylab_examples/quiver_demo.html

What’s going on here?

There is a disagreement between Numpy and native python on what should happen when you compare a strings to numpy’s numeric types. Notice the left operand is python’s turf, a primitive string, and the middle operation is python’s turf, but the right operand is numpy’s turf. Should you return a Python style Scalar or a Numpy style ndarray of Boolean? Numpy says ndarray of bool, Pythonic developers disagree. Classic standoff.

Should it be elementwise comparison or Scalar if item exists in the array?

If your code or library is using the in or == operators to compare python string to numpy ndarrays, they aren’t compatible, so when if you try it, it returns a scalar, but only for now. The Warning indicates that in the future this behavior might change so your code pukes all over the carpet if python/numpy decide to do adopt Numpy style.

Submitted Bug reports:

Numpy and Python are in a standoff, for now the operation returns a scalar, but in the future it may change.

https://github.com/numpy/numpy/issues/6784

https://github.com/pandas-dev/pandas/issues/7830

Two workaround solutions:

Either lockdown your version of python and numpy, ignore the warnings and expect the behavior to not change, or convert both left and right operands of == and in to be from a numpy type or primitive python numeric type.

Suppress the warning globally:

import warnings
import numpy as np
warnings.simplefilter(action='ignore', category=FutureWarning)
print('x' in np.arange(5))   #returns False, without Warning

Suppress the warning on a line by line basis.

import warnings
import numpy as np

with warnings.catch_warnings():
    warnings.simplefilter(action='ignore', category=FutureWarning)
    print('x' in np.arange(2))   #returns False, warning is suppressed

print('x' in np.arange(10))   #returns False, Throws FutureWarning

Just suppress the warning by name, then put a loud comment next to it mentioning the current version of python and numpy, saying this code is brittle and requires these versions and put a link to here. Kick the can down the road.

TLDR: pandas are Jedi; numpy are the hutts; and python is the galactic empire. https://youtu.be/OZczsiCfQQk?t=3

Question 26

当我尝试将index_col读取文件设置为Panda的数据帧时，出现相同的错误：

df = pd.read_csv('my_file.tsv', sep='\t', header=0, index_col=['0'])  ## or same with the following
df = pd.read_csv('my_file.tsv', sep='\t', header=0, index_col=[0])

我以前从未遇到过这样的错误。我仍然试图找出背后的原因（使用@Eric Leschinski的解释和其他解释）。

无论如何，在我找出原因之前，以下方法可以立即解决该问题：

df = pd.read_csv('my_file.tsv', sep='\t', header=0)  ## not setting the index_col
df.set_index(['0'], inplace=True)

一旦弄清这种行为的原因，我将立即更新。

Question 27

I get the same error when I try to set the index_col reading a file into a Panda‘s data-frame:

df = pd.read_csv('my_file.tsv', sep='\t', header=0, index_col=['0'])  ## or same with the following
df = pd.read_csv('my_file.tsv', sep='\t', header=0, index_col=[0])

I have never encountered such an error previously. I still am trying to figure out the reason behind this (using @Eric Leschinski explanation and others).

Anyhow, the following approach solves the problem for now until I figure the reason out:

df = pd.read_csv('my_file.tsv', sep='\t', header=0)  ## not setting the index_col
df.set_index(['0'], inplace=True)

I will update this as soon as I figure out the reason for such behavior.

Question 28

我对同一条警告消息的体验是由TypeError引起的。

TypeError：类型比较无效

因此，您可能要检查 Unnamed: 5

for x in df['Unnamed: 5']:
  print(type(x))  # are they 'str' ?

这是我可以复制警告消息的方法：

import pandas as pd
import numpy as np
df = pd.DataFrame(np.random.randn(3, 2), columns=['num1', 'num2'])
df['num3'] = 3
df.loc[df['num3'] == '3', 'num3'] = 4  # TypeError and the Warning
df.loc[df['num3'] == 3, 'num3'] = 4  # No Error

希望能帮助到你。

Question 29

My experience to the same warning message was caused by TypeError.

TypeError: invalid type comparison

So, you may want to check the data type of the Unnamed: 5

for x in df['Unnamed: 5']:
  print(type(x))  # are they 'str' ?

Here is how I can replicate the warning message:

import pandas as pd
import numpy as np
df = pd.DataFrame(np.random.randn(3, 2), columns=['num1', 'num2'])
df['num3'] = 3
df.loc[df['num3'] == '3', 'num3'] = 4  # TypeError and the Warning
df.loc[df['num3'] == 3, 'num3'] = 4  # No Error

Hope it helps.

Question 30

无法击败Eric Leschinski的详细答案，但这是针对我认为尚未提及的原始问题的快速解决方法-将字符串放在列表中并使用.isin而不是==

例如：

import pandas as pd
import numpy as np

df = pd.DataFrame({"Name": ["Peter", "Joe"], "Number": [1, 2]})

# Raises warning using == to compare different types:
df.loc[df["Number"] == "2", "Number"]

# No warning using .isin:
df.loc[df["Number"].isin(["2"]), "Number"]

Question 31

Can’t beat Eric Leschinski’s awesomely detailed answer, but here’s a quick workaround to the original question that I don’t think has been mentioned yet – put the string in a list and use .isin instead of ==

For example:

import pandas as pd
import numpy as np

df = pd.DataFrame({"Name": ["Peter", "Joe"], "Number": [1, 2]})

# Raises warning using == to compare different types:
df.loc[df["Number"] == "2", "Number"]

# No warning using .isin:
df.loc[df["Number"].isin(["2"]), "Number"]

Question 32

一个快速的解决方法是使用numpy.core.defchararray。我也遇到了同样的警告消息，并且能够使用上述模块来解决它。

import numpy.core.defchararray as npd
resultdataset = npd.equal(dataset1, dataset2)

Question 33

A quick workaround for this is to use numpy.core.defchararray. I also faced the same warning message and was able to resolve it using above module.

import numpy.core.defchararray as npd
resultdataset = npd.equal(dataset1, dataset2)

Question 34

埃里克（Eric）的回答很有帮助，说明了麻烦来自将Pandas系列（包含NumPy数组）与Python字符串进行比较。不幸的是，他的两个解决方法都只是抑制了警告。

要首先编写不会引起警告的代码，请显式地将字符串与Series的每个元素进行比较，并为每个元素获取单独的布尔值。例如，您可以使用map和匿名函数。

myRows = df[df['Unnamed: 5'].map( lambda x: x == 'Peter' )].index.tolist()

Question 35

Eric’s answer helpfully explains that the trouble comes from comparing a Pandas Series (containing a NumPy array) to a Python string. Unfortunately, his two workarounds both just suppress the warning.

To write code that doesn’t cause the warning in the first place, explicitly compare your string to each element of the Series and get a separate bool for each. For example, you could use map and an anonymous function.

myRows = df[df['Unnamed: 5'].map( lambda x: x == 'Peter' )].index.tolist()

Question 36

如果数组不太大或数组不太多，则可以通过将其左侧强制==为字符串来摆脱困境：

myRows = df[str(df['Unnamed: 5']) == 'Peter'].index.tolist()

但这如果df['Unnamed: 5']是字符串则要慢约1.5倍，如果df['Unnamed: 5']是小的numpy数组（长度= 10）则要慢25-30倍，如果是长度为100的numpy数组则要慢150-160倍（时间超过500次试验）。

a = linspace(0, 5, 10)
b = linspace(0, 50, 100)
n = 500
string1 = 'Peter'
string2 = 'blargh'
times_a = zeros(n)
times_str_a = zeros(n)
times_s = zeros(n)
times_str_s = zeros(n)
times_b = zeros(n)
times_str_b = zeros(n)
for i in range(n):
    t0 = time.time()
    tmp1 = a == string1
    t1 = time.time()
    tmp2 = str(a) == string1
    t2 = time.time()
    tmp3 = string2 == string1
    t3 = time.time()
    tmp4 = str(string2) == string1
    t4 = time.time()
    tmp5 = b == string1
    t5 = time.time()
    tmp6 = str(b) == string1
    t6 = time.time()
    times_a[i] = t1 - t0
    times_str_a[i] = t2 - t1
    times_s[i] = t3 - t2
    times_str_s[i] = t4 - t3
    times_b[i] = t5 - t4
    times_str_b[i] = t6 - t5
print('Small array:')
print('Time to compare without str conversion: {} s. With str conversion: {} s'.format(mean(times_a), mean(times_str_a)))
print('Ratio of time with/without string conversion: {}'.format(mean(times_str_a)/mean(times_a)))

print('\nBig array')
print('Time to compare without str conversion: {} s. With str conversion: {} s'.format(mean(times_b), mean(times_str_b)))
print(mean(times_str_b)/mean(times_b))

print('\nString')
print('Time to compare without str conversion: {} s. With str conversion: {} s'.format(mean(times_s), mean(times_str_s)))
print('Ratio of time with/without string conversion: {}'.format(mean(times_str_s)/mean(times_s)))

结果：

Small array:
Time to compare without str conversion: 6.58464431763e-06 s. With str conversion: 0.000173756599426 s
Ratio of time with/without string conversion: 26.3881526541

Big array
Time to compare without str conversion: 5.44309616089e-06 s. With str conversion: 0.000870866775513 s
159.99474375821288

String
Time to compare without str conversion: 5.89370727539e-07 s. With str conversion: 8.30173492432e-07 s
Ratio of time with/without string conversion: 1.40857605178

Question 37

If your arrays aren’t too big or you don’t have too many of them, you might be able to get away with forcing the left hand side of == to be a string:

myRows = df[str(df['Unnamed: 5']) == 'Peter'].index.tolist()

But this is ~1.5 times slower if df['Unnamed: 5'] is a string, 25-30 times slower if df['Unnamed: 5'] is a small numpy array (length = 10), and 150-160 times slower if it’s a numpy array with length 100 (times averaged over 500 trials).

a = linspace(0, 5, 10)
b = linspace(0, 50, 100)
n = 500
string1 = 'Peter'
string2 = 'blargh'
times_a = zeros(n)
times_str_a = zeros(n)
times_s = zeros(n)
times_str_s = zeros(n)
times_b = zeros(n)
times_str_b = zeros(n)
for i in range(n):
    t0 = time.time()
    tmp1 = a == string1
    t1 = time.time()
    tmp2 = str(a) == string1
    t2 = time.time()
    tmp3 = string2 == string1
    t3 = time.time()
    tmp4 = str(string2) == string1
    t4 = time.time()
    tmp5 = b == string1
    t5 = time.time()
    tmp6 = str(b) == string1
    t6 = time.time()
    times_a[i] = t1 - t0
    times_str_a[i] = t2 - t1
    times_s[i] = t3 - t2
    times_str_s[i] = t4 - t3
    times_b[i] = t5 - t4
    times_str_b[i] = t6 - t5
print('Small array:')
print('Time to compare without str conversion: {} s. With str conversion: {} s'.format(mean(times_a), mean(times_str_a)))
print('Ratio of time with/without string conversion: {}'.format(mean(times_str_a)/mean(times_a)))

print('\nBig array')
print('Time to compare without str conversion: {} s. With str conversion: {} s'.format(mean(times_b), mean(times_str_b)))
print(mean(times_str_b)/mean(times_b))

print('\nString')
print('Time to compare without str conversion: {} s. With str conversion: {} s'.format(mean(times_s), mean(times_str_s)))
print('Ratio of time with/without string conversion: {}'.format(mean(times_str_s)/mean(times_s)))

Result:

Small array:
Time to compare without str conversion: 6.58464431763e-06 s. With str conversion: 0.000173756599426 s
Ratio of time with/without string conversion: 26.3881526541

Big array
Time to compare without str conversion: 5.44309616089e-06 s. With str conversion: 0.000870866775513 s
159.99474375821288

String
Time to compare without str conversion: 5.89370727539e-07 s. With str conversion: 8.30173492432e-07 s
Ratio of time with/without string conversion: 1.40857605178

Question 38

就我而言，发出警告的原因仅仅是布尔索引的常规类型-因为该系列只有np.nan。示范（熊猫1.0.3）：

>>> import pandas as pd
>>> import numpy as np
>>> pd.Series([np.nan, 'Hi']) == 'Hi'
0    False
1     True
>>> pd.Series([np.nan, np.nan]) == 'Hi'
~/anaconda3/envs/ms3/lib/python3.7/site-packages/pandas/core/ops/array_ops.py:255: FutureWarning: elementwise comparison failed; returning scalar instead, but in the future will perform elementwise comparison
  res_values = method(rvalues)
0    False
1    False

我认为对于pandas 1.0，他们确实希望您使用'string'允许pd.NA值的新数据类型：

>>> pd.Series([pd.NA, pd.NA]) == 'Hi'
0    False
1    False
>>> pd.Series([np.nan, np.nan], dtype='string') == 'Hi'
0    <NA>
1    <NA>
>>> (pd.Series([np.nan, np.nan], dtype='string') == 'Hi').fillna(False)
0    False
1    False

不喜欢他们在何时开始使用布尔索引等日常功能。

Question 39

In my case, the warning occurred because of just the regular type of boolean indexing — because the series had only np.nan. Demonstration (pandas 1.0.3):

>>> import pandas as pd
>>> import numpy as np
>>> pd.Series([np.nan, 'Hi']) == 'Hi'
0    False
1     True
>>> pd.Series([np.nan, np.nan]) == 'Hi'
~/anaconda3/envs/ms3/lib/python3.7/site-packages/pandas/core/ops/array_ops.py:255: FutureWarning: elementwise comparison failed; returning scalar instead, but in the future will perform elementwise comparison
  res_values = method(rvalues)
0    False
1    False

I think with pandas 1.0 they really want you to use the new 'string' datatype which allows for pd.NA values:

>>> pd.Series([pd.NA, pd.NA]) == 'Hi'
0    False
1    False
>>> pd.Series([np.nan, np.nan], dtype='string') == 'Hi'
0    <NA>
1    <NA>
>>> (pd.Series([np.nan, np.nan], dtype='string') == 'Hi').fillna(False)
0    False
1    False

Don’t love at which point they tinkered with every-day functionality such as boolean indexing.

Question 40

我收到此警告是因为我认为我的列包含空字符串，但是在检查时，它包含了np.nan！

if df['column'] == '':

将我的列更改为空字符串有帮助:)

Question 41

I got this warning because I thought my column contained null strings, but on checking, it contained np.nan!

if df['column'] == '':

Changing my column to empty strings helped :)

Question 42

我已经比较了几种可能的方法，包括熊猫，几种numpy方法和列表理解方法。

首先，让我们从基线开始：

>>> import numpy as np
>>> import operator
>>> import pandas as pd

>>> x = [1, 2, 1, 2]
>>> %time count = np.sum(np.equal(1, x))
>>> print("Count {} using numpy equal with ints".format(count))
CPU times: user 52 µs, sys: 0 ns, total: 52 µs
Wall time: 56 µs
Count 2 using numpy equal with ints

因此，我们的基准是该计数应该正确2，并且我们应该大约50 us。

现在，我们尝试使用朴素的方法：

>>> x = ['s', 'b', 's', 'b']
>>> %time count = np.sum(np.equal('s', x))
>>> print("Count {} using numpy equal".format(count))
CPU times: user 145 µs, sys: 24 µs, total: 169 µs
Wall time: 158 µs
Count NotImplemented using numpy equal
/Library/Frameworks/Python.framework/Versions/3.6/lib/python3.6/site-packages/ipykernel_launcher.py:1: FutureWarning: elementwise comparison failed; returning scalar instead, but in the future will perform elementwise comparison
  """Entry point for launching an IPython kernel.

在这里，我们得到了错误的答案（NotImplemented != 2），这花了我们很长时间，并且引发了警告。

因此，我们将尝试另一种幼稚的方法：

>>> %time count = np.sum(x == 's')
>>> print("Count {} using ==".format(count))
CPU times: user 46 µs, sys: 1 µs, total: 47 µs
Wall time: 50.1 µs
Count 0 using ==

同样，错误答案（0 != 2）。这更加隐蔽，因为没有后续警告（0可以像一样传递2）。

现在，让我们尝试一个列表理解：

>>> %time count = np.sum([operator.eq(_x, 's') for _x in x])
>>> print("Count {} using list comprehension".format(count))
CPU times: user 55 µs, sys: 1 µs, total: 56 µs
Wall time: 60.3 µs
Count 2 using list comprehension

我们在这里得到正确的答案，而且速度很快！

另一种可能性pandas：

>>> y = pd.Series(x)
>>> %time count = np.sum(y == 's')
>>> print("Count {} using pandas ==".format(count))
CPU times: user 453 µs, sys: 31 µs, total: 484 µs
Wall time: 463 µs
Count 2 using pandas ==

慢，但是正确！

最后，我将使用的选项是：将numpy数组转换为object类型：

>>> x = np.array(['s', 'b', 's', 'b']).astype(object)
>>> %time count = np.sum(np.equal('s', x))
>>> print("Count {} using numpy equal".format(count))
CPU times: user 50 µs, sys: 1 µs, total: 51 µs
Wall time: 55.1 µs
Count 2 using numpy equal

快速正确！

Question 43

I’ve compared a few of the methods possible for doing this, including pandas, several numpy methods, and a list comprehension method.

First, let’s start with a baseline:

>>> import numpy as np
>>> import operator
>>> import pandas as pd

>>> x = [1, 2, 1, 2]
>>> %time count = np.sum(np.equal(1, x))
>>> print("Count {} using numpy equal with ints".format(count))
CPU times: user 52 µs, sys: 0 ns, total: 52 µs
Wall time: 56 µs
Count 2 using numpy equal with ints

So, our baseline is that the count should be correct 2, and we should take about 50 us.

Now, we try the naive method:

>>> x = ['s', 'b', 's', 'b']
>>> %time count = np.sum(np.equal('s', x))
>>> print("Count {} using numpy equal".format(count))
CPU times: user 145 µs, sys: 24 µs, total: 169 µs
Wall time: 158 µs
Count NotImplemented using numpy equal
/Library/Frameworks/Python.framework/Versions/3.6/lib/python3.6/site-packages/ipykernel_launcher.py:1: FutureWarning: elementwise comparison failed; returning scalar instead, but in the future will perform elementwise comparison
  """Entry point for launching an IPython kernel.

And here, we get the wrong answer (NotImplemented != 2), it takes us a long time, and it throws the warning.

So we’ll try another naive method:

>>> %time count = np.sum(x == 's')
>>> print("Count {} using ==".format(count))
CPU times: user 46 µs, sys: 1 µs, total: 47 µs
Wall time: 50.1 µs
Count 0 using ==

Again, the wrong answer (0 != 2). This is even more insidious because there’s no subsequent warnings (0 can be passed around just like 2).

Now, let’s try a list comprehension:

>>> %time count = np.sum([operator.eq(_x, 's') for _x in x])
>>> print("Count {} using list comprehension".format(count))
CPU times: user 55 µs, sys: 1 µs, total: 56 µs
Wall time: 60.3 µs
Count 2 using list comprehension

We get the right answer here, and it’s pretty fast!

Another possibility, pandas:

>>> y = pd.Series(x)
>>> %time count = np.sum(y == 's')
>>> print("Count {} using pandas ==".format(count))
CPU times: user 453 µs, sys: 31 µs, total: 484 µs
Wall time: 463 µs
Count 2 using pandas ==

Slow, but correct!

And finally, the option I’m going to use: casting the numpy array to the object type:

>>> x = np.array(['s', 'b', 's', 'b']).astype(object)
>>> %time count = np.sum(np.equal('s', x))
>>> print("Count {} using numpy equal".format(count))
CPU times: user 50 µs, sys: 1 µs, total: 51 µs
Wall time: 55.1 µs
Count 2 using numpy equal

Fast and correct!

Question 44

我有导致错误的此代码：

for t in dfObj['time']:
  if type(t) == str:
    the_date = dateutil.parser.parse(t)
    loc_dt_int = int(the_date.timestamp())
    dfObj.loc[t == dfObj.time, 'time'] = loc_dt_int

我将其更改为：

for t in dfObj['time']:
  try:
    the_date = dateutil.parser.parse(t)
    loc_dt_int = int(the_date.timestamp())
    dfObj.loc[t == dfObj.time, 'time'] = loc_dt_int
  except Exception as e:
    print(e)
    continue

为了避免比较，它会发出警告-如上所述。我只需要避免这种异常，因为dfObj.loc在for循环中，也许有一种方法可以告诉它不要检查已更改的行。

Question 45

I had this code which was causing the error:

for t in dfObj['time']:
  if type(t) == str:
    the_date = dateutil.parser.parse(t)
    loc_dt_int = int(the_date.timestamp())
    dfObj.loc[t == dfObj.time, 'time'] = loc_dt_int

I changed it to this:

for t in dfObj['time']:
  try:
    the_date = dateutil.parser.parse(t)
    loc_dt_int = int(the_date.timestamp())
    dfObj.loc[t == dfObj.time, 'time'] = loc_dt_int
  except Exception as e:
    print(e)
    continue

to avoid the comparison, which is throwing the warning – as stated above. I only had to avoid the exception because of dfObj.loc in the for loop, maybe there is a way to tell it not to check the rows it has already changed.

Question 46

So I have a little problem. I have a data set in scipy that is already in the histogram format, so I have the center of the bins and the number of events per bin. How can I now plot is as a histogram. I tried just doing

bins, n=hist()

but it didn’t like that. Any recommendations?

Question 47

import matplotlib.pyplot as plt
import numpy as np

mu, sigma = 100, 15
x = mu + sigma * np.random.randn(10000)
hist, bins = np.histogram(x, bins=50)
width = 0.7 * (bins[1] - bins[0])
center = (bins[:-1] + bins[1:]) / 2
plt.bar(center, hist, align='center', width=width)
plt.show()

The object-oriented interface is also straightforward:

fig, ax = plt.subplots()
ax.bar(center, hist, align='center', width=width)
fig.savefig("1.png")

If you are using custom (non-constant) bins, you can pass compute the widths using np.diff, pass the widths to ax.bar and use ax.set_xticks to label the bin edges:

import matplotlib.pyplot as plt
import numpy as np

mu, sigma = 100, 15
x = mu + sigma * np.random.randn(10000)
bins = [0, 40, 60, 75, 90, 110, 125, 140, 160, 200]
hist, bins = np.histogram(x, bins=bins)
width = np.diff(bins)
center = (bins[:-1] + bins[1:]) / 2

fig, ax = plt.subplots(figsize=(8,3))
ax.bar(center, hist, align='center', width=width)
ax.set_xticks(bins)
fig.savefig("/tmp/out.png")

plt.show()

Question 48

If you don’t want bars you can plot it like this:

import numpy as np
import matplotlib.pyplot as plt

mu, sigma = 100, 15
x = mu + sigma * np.random.randn(10000)

bins, edges = np.histogram(x, 50, normed=1)
left,right = edges[:-1],edges[1:]
X = np.array([left,right]).T.flatten()
Y = np.array([bins,bins]).T.flatten()

plt.plot(X,Y)
plt.show()

Question 49

I know this does not answer your question, but I always end up on this page, when I search for the matplotlib solution to histograms, because the simple histogram_demo was removed from the matplotlib example gallery page.

Here is a solution, which doesn’t require numpy to be imported. I only import numpy to generate the data x to be plotted. It relies on the function hist instead of the function bar as in the answer by @unutbu.

import numpy as np
mu, sigma = 100, 15
x = mu + sigma * np.random.randn(10000)

import matplotlib.pyplot as plt
plt.hist(x, bins=50)
plt.savefig('hist.png')

Also check out the matplotlib gallery and the matplotlib examples.

Question 50

If you’re willing to use pandas:

pandas.DataFrame({'x':hist[1][1:],'y':hist[0]}).plot(x='x',kind='bar')

Question 51

I think this might be useful for someone.

Numpy’s histogram function, to my annoyance (although, I appreciate there is a good reason for it), returns back the edges of each bin, rather than the value of the bin. While, this makes sense for floating-point numbers, which can lie within an interval (i.e. the center value is not super meaningful), this is not the desired output when dealing with discrete values or integers (0, 1, 2, etc). In particular, the length of bins returned from np.histogram is not equal to the length of the counts / density.

To get around this, I used np.digitize to quantize the input, and return a discrete number of bins, along with fraction of counts for each bin. You could easily edit to get the integer number of counts.

def compute_PMF(data)
    import numpy as np
    from collections import Counter
    _, bins = np.histogram(data, bins='auto', range=(data.min(), data.max()), density=False)
    h = Counter(np.digitize(data,bins) - 1)
    weights = np.asarray(list(h.values())) 
    weights = weights / weights.sum()
    values = np.asarray(list(h.keys()))
    return weights, values
####

Refs:

[1] https://docs.scipy.org/doc/numpy/reference/generated/numpy.histogram.html

[2] https://docs.scipy.org/doc/numpy/reference/generated/numpy.digitize.html

Question 52

When plotting heatmaps with seaborn (and correlation matrices with matplotlib) the first and the last row is cut in halve. This happens also when I run this minimal code example which I found online.

import pandas as pd
import seaborn as sns
import matplotlib.pyplot as plt

data = pd.read_csv('https://raw.githubusercontent.com/resbaz/r-novice-gapminder-files/master/data/gapminder-FiveYearData.csv')
plt.figure(figsize=(10,5))
sns.heatmap(data.corr())
plt.show()

The labels at the y axis are on the correct spot, but the rows aren’t completely there.

A few days ago, it work as intended. Since then, I installed texlive-xetex so I removed it again but it didn’t solve my problem.

Any ideas what I could be missing?

Question 53

Unfortunately matplotlib 3.1.1 broke seaborn heatmaps; and in general inverted axes with fixed ticks.
This is fixed in the current development version; you may hence

revert to matplotlib 3.1.0
use matplotlib 3.1.2 or higher
set the heatmap limits manually (ax.set_ylim(bottom, top) # set the ylim to bottom, top)

Question 54

Its a bug in the matplotlib regression between 3.1.0 and 3.1.1 You can correct this by:

import seaborn as sns
df_corr = someDataFrame.corr()
ax = sns.heatmap(df_corr, annot=True) #notation: "annot" not "annote"
bottom, top = ax.get_ylim()
ax.set_ylim(bottom + 0.5, top - 0.5)

Question 55

Fixed using the above and setting the heatmap limits manually.

First

ax = sns.heatmap(...

checked the current axes with

ax.get_ylim()
(5.5, 0.5)

Fixed with

ax.set_ylim(6.0, 0)

Question 56

I solved it by adding this line in my code, with matplotlib==3.1.1:

ax.set_ylim(sorted(ax.get_xlim(), reverse=True))

NB. The only reason this works is because the x-axis isn’t changed, so use at your own risk with future mpl versions

Question 57

matplotlib 3.1.2 is out – It is available in the Anaconda cloud via conda-forge but I was not able to install it via conda install. The manual alternative worked: Download matplotlib 3.1.2 from github and install via pip

 % curl https://codeload.github.com/matplotlib/matplotlib/tar.gz/v3.1.2 --output matplotlib-3.1.2.tar.gz
 % pip install matplotlib-3.1.2.tar.gz

Question 58

It happens with matplotlib version 3.1.1 as suggested by importanceofbeingernest

Following solved my problem

pip install matplotlib==3.1.0

Question 59

rustyDev is right about conda-forge, but I did not need to do a manual pip install from a github download. For me, on Windows, it worked directly. And the plots are all nice again.

https://anaconda.org/conda-forge/matplotlib

conda install -c conda-forge matplotlib

optional points, not needed for the answer:

Afterwards, I tried other steps, but they are not needed: In conda prompt: conda search matplotlib –info showed no new version info, the most recent info was for 3.1.1. Thus I tried pip using pip install matplotlib==3.1.2 But pip says “Requirement already satisfied”

Then getting the version according to medium.com/@rakshithvasudev/… python - import matplotlib - matplotlib.__version__ shows that 3.1.2 was successfully installed

Btw, I had this error directly after updating Spyder to v4.0.0. The error was in a plot of a confusion matrix. This was mentioned already some months ago. stackoverflow.com/questions/57225685/… which is already linked to this seaborn question.

Question 60

conda install matplotlib=3.1.0

This worked for me and downgraded matplotlib from 3.1.1 to 3.1.0 and the heatmaps started to behave correctly

Question 61

I solved this problem with the following code:

Question 62

I am drawing a plot using matplotlib and python like the sample code below.

x = array([0,1,2,3])
y = array([20,21,22,23])
plot(x,y)
show()

As it is the code above on the x axis I will see drawn values 0.0, 0.5, 1.0, 1.5 i.e. the same values of my reference x values.

Is there anyway to map each point of x to a different string? So for example I want x axis to show months names( strings Jun, July,...) or other strings like people names ( "John", "Arnold", ... ) or clock time ( "12:20", "12:21", "12:22", .. ).

Do you know what I can do or what function to have a look at?
For my purpose could it be matplotlib.ticker of help?

Question 63

You can manually set xticks (and yticks) using pyplot.xticks:

import matplotlib.pyplot as plt
import numpy as np

x = np.array([0,1,2,3])
y = np.array([20,21,22,23])
my_xticks = ['John','Arnold','Mavis','Matt']
plt.xticks(x, my_xticks)
plt.plot(x, y)
plt.show()

Question 64

This worked for me. Each month on X axis

str_month_list = ['January','February','March','April','May','June','July','August','September','October','November','December']
ax.set_xticks(range(0,12))
ax.set_xticklabels(str_month_list)

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