如何在Python中绘制第一个楔形在顶部的饼图？ [matplotlib]

Question

How can a pie chart be drawn with Matplotlib with a first wedge that starts at noon (ie on the top of the pie)? The default is for pyplot.pie() to place the first edge at three o'clock, and it would be great to be able to customize this.

Answer 1

It's a bit of a hack, but you can do something like this...

import matplotlib.pyplot as plt
from matplotlib.transforms import Affine2D
import numpy as np

x = [5, 20, 10, 10]
labels=['cliffs', 'frogs', 'stumps', 'old men on tractors']

plt.figure()
plt.suptitle("Things I narrowly missed while learning to drive")
wedges, labels = plt.pie(x, labels=labels)
plt.axis('equal')

starting_angle = 90
rotation = Affine2D().rotate(np.radians(starting_angle))

for wedge, label in zip(wedges, labels):
    label.set_position(rotation.transform(label.get_position()))
    if label._x > 0:
        label.set_horizontalalignment('left')
    else:
        label.set_horizontalalignment('right')

    wedge._path = wedge._path.transformed(rotation)

plt.show()

Answer 2

Just because this came up in a Google search for me, I'll add that in the meantime, matplotlib has included just this as an additional argument to the pie function.

Now, one can call plt.pie(data, start_angle=90) to have the first wedge start at noon.

PyPlot documentation on this