如何在Matplotlib中在绘图框外部绘制矩形 [英] How to draw rectangle outside of the plot frame in Matplotlib

问题描述

我想按照下图的样式生成子图的标题:

I want to generate the subfigure's title in the style of followed figure:

标题下方应有一个灰色框，该框位于散点的顶部.

A gray box should be beneath the title which are at the top of the scatter point.

这是我尝试过的代码:

x = random.sample(range(50), 50)
y= random.sample(range(50), 50)

fig = pyplot.figure()
ax = pyplot.subplot(111)
ax.scatter(x,y,label='a')
ax.set_aspect('equal')
ax.set_xlim(0,60)
ax.set_ylim(0,60)
ax.plot([0,60], [0, 60], color='k', linestyle='-', linewidth=1.25)

ax.add_patch(patches.Rectangle((0,60),60, 10,facecolor='silver',linewidth = 0))
TITLE = ax.text(26,61, r'$\mathregular{Title}$',fontsize = 14,zorder = 5,color = 'k')

结果显示如下:

作为标题背景框的矩形无法显示在我的结果中

The rectangle as the background box of title can't be shown in my result

任何建议或更好的解决方案都将不胜感激！

Any advice or better solution are appreciate!

推荐答案

我认为，更好的方法是将 clip_on = False 选项用于

I think a better way is to use the clip_on=False option for Rectangle:

import random
import matplotlib.pyplot as pyplot

x = random.sample(range(50), 50)
y= random.sample(range(50), 50)

fig = pyplot.figure()
ax = pyplot.subplot(111)
ax.scatter(x,y,label='a')
ax.set_aspect('equal')
ax.set_xlim(0,60)
ax.set_ylim(0,60)
ax.plot([0,60], [0, 60], color='k', linestyle='-', linewidth=1.25)

ax.add_patch(pyplot.Rectangle((0,60),60, 10,facecolor='silver',
                              clip_on=False,linewidth = 0))
TITLE = ax.text(26,61, r'$\mathregular{Title}$',fontsize = 14,zorder = 5,
                color = 'k')
pyplot.show()

这将产生一个在轴外绘制的矩形，而不必求助于多余的空间:

This yields a rectangle drawn outside of the axes, without having to resort to extra spaces:

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