Matplotlib：从头开始制作彩色标记图例

Question

在 Matplotlib 中，我正在尝试用彩色 "markers" 制作一个像这样的图例：

这个是用scatter函数制作的，但不适合我的情节。我想生成一个没有关联数据的图例 "from scratch"。颜色很重要，因此应该是每个标记的属性。

我试过了

import matplotlib.markers as mmark
list_mak = [mmark.MarkerStyle('.'),mmark.MarkerStyle(','),mmark.MarkerStyle('o')]
list_lab = ['Marker 1','Marker 2','Marker 3']

plt.legend(list_mak,list_lab)

但是:

1) MarkerStyle class 不支持颜色信息

2) 我收到警告：

UserWarning: Legend does not support <matplotlib.markers.MarkerStyle object at 0x7fca640c44d0> instances.
A proxy artist may be used instead.

但是如何根据标记定义代理艺术家？

感谢您的帮助！

Answer 1

遵循 legend guide, you can use a Line2D 对象而不是 marker 对象中给出的示例。

与指南中给出的示例的唯一区别是您要设置 linestyle='None'

import matplotlib.lines as mlines
import matplotlib.pyplot as plt

blue_star = mlines.Line2D([], [], color='blue', marker='*', linestyle='None',
                          markersize=10, label='Blue stars')
red_square = mlines.Line2D([], [], color='red', marker='s', linestyle='None',
                          markersize=10, label='Red squares')
purple_triangle = mlines.Line2D([], [], color='purple', marker='^', linestyle='None',
                          markersize=10, label='Purple triangles')

plt.legend(handles=[blue_star, red_square, purple_triangle])

plt.show()

Answer 2

您可以将 HandlerBase 子类化以从 (color, marker) 的元组创建处理程序。

import matplotlib.pyplot as plt
from matplotlib.legend_handler import HandlerBase

list_color  = ["c", "gold", "crimson"]
list_mak    = ["d","s","o"]
list_lab    = ['Marker 1','Marker 2','Marker 3']

ax = plt.gca()

class MarkerHandler(HandlerBase):
    def create_artists(self, legend, tup,xdescent, ydescent,
                        width, height, fontsize,trans):
        return [plt.Line2D([width/2], [height/2.],ls="",
                       marker=tup[1],color=tup[0], transform=trans)]


ax.legend(list(zip(list_color,list_mak)), list_lab, 
          handler_map={tuple:MarkerHandler()}) 

plt.show()

Matplotlib：从头开始制作彩色标记图例

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