如何在Matplotlib中制作自定义图例

Question

我目前以这种方式使用matplotlib生成图例：if t==25:    l1,l2 = ax2.plot(x320,vTemp320,'or',x320,vAnaTemp320,'-r')elif t==50:    l3,l4 = ax2.plot(x320,vTemp320,'ob',x320,vAnaTemp320,'-b')else:    l5,l6 = ax2.plot(x320,vTemp320,'og',x320,vAnaTemp320,'-g')plt.legend((l1,l2,l3,l4,l5,l6), ('t=25 Simulation', 't=25 Analytical','t=50 Simulation', 't=50 Analytical','t=500 Simulation', 't=500 Analytical'),   bbox_to_anchor=(-.25, 1), loc=2, borderaxespad=0.,prop={'size':12})哪个工作原理见1。但是我的传说中有重复的信息。我宁愿把传说分开。这样我就有了与时间t对应的不同色线。一条法线作为我的解析解决方案，它为模拟结果提供了一个点。像这样-（红线）t = 25-（蓝线）t = 50-（绿线）t = 500o Simulaton-分析解决方案现在有人可以使用matplotlib实现此目标吗？

Answer 1

您可以如下选择艺术家和标签以显示在图例中。您需要为图例中未实际绘制的元素创建自定义艺术家。

import matplotlib.pyplot as plt

import numpy as np

x = np.linspace(0,10,31)

fig = plt.figure()

ax = fig.add_subplot(1,1,1)

#Plot analytic solution

ax.plot(x,1*x**2, color='r', label="t = 25")

ax.plot(x,2*x**2, color='b', label="t = 50")

ax.plot(x,3*x**2, color='g', label="t = 500")

#Plot simulation

ax.plot(x,1*x**2, color='r', linestyle='', marker='o')

ax.plot(x,2*x**2, color='b', linestyle='', marker='o')

ax.plot(x,3*x**2, color='g', linestyle='', marker='o')

#Get artists and labels for legend and chose which ones to display

handles, labels = ax.get_legend_handles_labels()

display = (0,1,2)

#Create custom artists

simArtist = plt.Line2D((0,1),(0,0), color='k', marker='o', linestyle='')

anyArtist = plt.Line2D((0,1),(0,0), color='k')

#Create legend from custom artist/label lists

ax.legend([handle for i,handle in enumerate(handles) if i in display]+[simArtist,anyArtist],

          [label for i,label in enumerate(labels) if i in display]+['Simulation', 'Analytic'])

plt.show()