python - matplotlib:饼图，变量 pctdistance

Question

我希望能够将每个百分比值定位在距离中心的不同距离处，但 pctdistance 需要是单个值。

对于我的情况，pctdistance 应该是一个列表，其中包含生成的距离(由范围生成)。

import matplotlib.pyplot as plt

fig =plt.figure(figsize = (10,10))
ax11 = fig.add_subplot(111)
# Data to plot
labels = 'Python', 'C++', 'Ruby', 'Java'
sizes = [215, 130, 245, 2000]
colors = ['gold', 'yellowgreen', 'lightcoral', 'lightskyblue']
explode = (0.1, 0, 0, 0)  # explode 1st slice

# Plot
w,l,p = ax11.pie(sizes,  labels=labels, colors=colors,
        autopct='%1.1f%%', startangle=140, pctdistance=0.8, radius = 0.5)
[t.set_rotation(0) for t in p]
[t.set_fontsize(50) for t in p]
plt.axis('equal')
plt.show()

我有: 我想要的是:

Answer 1

pie 函数不将列表或数组作为 pctdistance 参数的输入。

您可以使用预定义的 pctdistances 列表手动定位文本。

import numpy as np
import matplotlib.pyplot as plt

fig =plt.figure(figsize = (4,4))
ax11 = fig.add_subplot(111)
# Data to plot
labels = 'Python', 'C++', 'Ruby', 'Java'
sizes = [215, 130, 245, 2000]
colors = ['gold', 'yellowgreen', 'lightcoral', 'lightskyblue']

# Plot
w,l,p = ax11.pie(sizes,  labels=labels, colors=colors,
        autopct='%1.1f%%', startangle=140, pctdistance=1, radius = 0.5)

pctdists = [.8, .5, .4, .2]

for t,d in zip(p, pctdists):
    xi,yi = t.get_position()
    ri = np.sqrt(xi**2+yi**2)
    phi = np.arctan2(yi,xi)
    x = d*ri*np.cos(phi)
    y = d*ri*np.sin(phi)
    t.set_position((x,y))

plt.axis('equal')
plt.show()