如何在matplotlib中移动刻度标签 [英] How to move a tick label in matplotlib

问题描述

我想沿 x 轴水平移动一些刻度的标签，而不移动相应的刻度.

更具体地说，当使用 plt.setp 旋转标签时，标签文本的中心与刻度保持对齐.我想将这些标签向右移动，以便标签的近端按照下图的建议对齐.

我知道

现在正如其他人已经指出的那样，您可以使用文本的水平对齐方式.

# 旋转标签并将它们水平向左对齐plt.setp( ax.xaxis.get_majorticklabels(), rotation=-45, ha="left" )

您可以使用 rotation_mode 参数让旋转发生在文本的左上角，在这种情况下会得到更好的结果.

# 旋转标签并将它们水平向左对齐plt.setp( ax.xaxis.get_majorticklabels(), rotation=-45, ha="left", rotation_mode="anchor")

如果这些选项不够细粒度，即您想更准确地定位标签，例如将它移到一边一些点，您可以使用变换.以下将使用 matplotlib.transforms.ScaledTranslation.

在水平方向上将标签偏移 5 个点.

import matplotlib.transformsplt.setp( ax.xaxis.get_majorticklabels(), 旋转=-45)# 在 x 方向创建 5 个点的偏移变换dx = 5/72.；dy = 0/72.偏移量 = matplotlib.transforms.ScaledTranslation(dx, dy, fig.dpi_scale_trans)# 对所有 x 刻度标签应用偏移变换.对于 ax.xaxis.get_majorticklabels() 中的标签:label.set_transform(label.get_transform() + 偏移量)

与例如相比，这样做的优势@explorerDude 提供的解决方案是偏移量独立于图中的数据，因此它通常适用于任何绘图并且对于给定的字体大小看起来相同.

I would like to move some ticks' labels horizontally along the x-axis, without moving the corresponding ticks.

More specifically, when rotating labels with plt.setp, the centers of the labels' text stay aligned with the ticks. I would like to shift those labels to the right, so that the near ends of the labels get aligned instead as suggested on the image below.

I am aware of this post and this one, however the answers are interesting kludges rather than strict answers to the question.

my code:

import matplotlib.pyplot as plt
import numpy as np
import datetime

# my fake data
dates = np.array([datetime.datetime(2000,1,1) + datetime.timedelta(days=i) for i in range(365*5)])
data = np.sin(np.arange(365*5)/365.0*2*np.pi - 0.25*np.pi) + np.random.rand(365*5) /3

# creates fig with 2 subplots
fig = plt.figure(figsize=(10.0, 6.0))
ax = plt.subplot2grid((2,1), (0, 0))
ax2 = plt.subplot2grid((2,1), (1, 0))
## plot dates
ax2.plot_date( dates, data )

# rotates labels 
plt.setp( ax2.xaxis.get_majorticklabels(), rotation=-45 ) 

# try to shift labels to the right
ax2.xaxis.get_majorticklabels()[2].set_y(-.1)
ax2.xaxis.get_majorticklabels()[2].set_x(10**99)

plt.show()

Strangely enough, set_y behaves as expected, but even if I set x to a fantasillion, the labels would not move by one iota. (The use of plot_date may introduce additional confusion, but the same actually happens with plot.)

解决方案

First of all, let's use a mcve to show the problem.

import numpy as np
import datetime
import matplotlib.pyplot as plt
plt.rcParams["date.autoformatter.month"] = "%b %Y"

# my fake data
dates = np.array([datetime.datetime(2000,1,1) + datetime.timedelta(days=i) for i in range(365)])
data = np.sin(np.arange(365)/365.0*2*np.pi - 0.25*np.pi) + np.random.rand(365) /3

# creates fig with 2 subplots
fig, ax = plt.subplots(figsize=(6,2))
## plot dates
ax.plot_date( dates, data )

# rotates labels 
plt.setp( ax.xaxis.get_majorticklabels(), rotation=-45 ) 

plt.tight_layout()
plt.show()

Now as other anwers pointed out already, you may use horizontal alignment of the text.

# rotates labels and aligns them horizontally to left 
plt.setp( ax.xaxis.get_majorticklabels(), rotation=-45, ha="left" )

You may use the rotation_mode argument to let the rotation happen about the top left point of the text, giving a slightly nicer result in this case.

# rotates labels and aligns them horizontally to left 
plt.setp( ax.xaxis.get_majorticklabels(), rotation=-45, ha="left", rotation_mode="anchor") 

In case those options are not fine grained enough, i.e. you want to position the labels more accurately, e.g. shifting it to the side by some points, you may use a transform. The following would offset the label by 5 points in horizontal direction, using a matplotlib.transforms.ScaledTranslation.

import matplotlib.transforms

plt.setp( ax.xaxis.get_majorticklabels(), rotation=-45) 

# Create offset transform by 5 points in x direction
dx = 5/72.; dy = 0/72. 
offset = matplotlib.transforms.ScaledTranslation(dx, dy, fig.dpi_scale_trans)

# apply offset transform to all x ticklabels.
for label in ax.xaxis.get_majorticklabels():
    label.set_transform(label.get_transform() + offset)

The advantage of this, compared to e.g. the solution provided by @explorerDude is that the offset is independent on the data in the graph, such that it is generally applicable to any plot and would look the same for a given fontsize.

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