绘制带填充的甜甜圈图

Question

我想画一个甜甜圈图，我的代码是

import numpy as np
import matplotlib.pyplot as plt
pi,sin,cos = np.pi,np.sin,np.cos

r1 = 1
r2 = 2

theta = np.linspace(0,2*pi,36)

x1 = r1*cos(theta)
y1 = r1*sin(theta)

x2 = r2*cos(theta)
y2 = r2*sin(theta)

怎么才能让这个甜甜圈的填充区域变成红色呢？

Answer 1

tom10给出的答案是十 ;) 但如果你想定义圆形（甜甜圈）的原点，其实很简单，只需要在x和y的位置上加上xI、yI、yO、-yO和-yI就可以了：

    ax.fill_between(x+pos[0], yI+pos[1], yO+pos[1], color=color)
    ax.fill_between(x+pos[0], -yO+pos[1], -yI+pos[1], color=color)

如下所示：

import numpy as np
import matplotlib.pyplot as plt
import math
 
 
def plot_circle_donut(pos, inner, outer, color):
    """
     REF: https://stackoverflow.com/questions/22789356/plot-a-donut-with-fill-or-fill-between-use-pyplot-in-matplotlib
     ton10's answer
    """
    x = np.linspace(-outer, outer, 300, endpoint=True)
 
    yO = outer * np.sin(np.arccos(x/ outer )) # x-axis values -> outer circle
    yI = inner * np.sin(np.arccos(x/ inner )) # x-axis values -> inner circle (with nan's beyond circle)
    yI[np.isnan(yI)] = 0.                 # yI now looks like a boulder hat, meeting yO at the outer points
 
    ax = plt.subplot(111)
    ax.fill_between(x+pos[0], yI+pos[1], yO+pos[1], color=color)
    ax.fill_between(x+pos[0], -yO+pos[1], -yI+pos[1], color=color)
 
    plt.show()
#
 
def plot_circle(r, pos):
    """ REF: https://math.stackexchange.com/questions/260096/find-the-coordinates-of-a-point-on-a-circle """
    arrx = []
    arry = []
    for theta in xrange(1000):
        x,y = r * math.sin(theta), r * math.cos(theta)
        arrx.append(x)
        arry.append(y)
    #
    plt.plot(arrx, arry, color='red')
 
    plt.show()
#
 
 
#r = 3
#pos = 2,2
#plot_circle(r, pos)
 
r1, r2 = 2, 2.1
position = [4,2]
color = 'b'
plot_circle_donut(position, r1, r2, color)

参考示例： https://pastebin.com/8Ew4Vthb

Answer 2

这有点像小技巧，但下面这个方法可以实现：

import numpy as np
import matplotlib.pyplot as plt
pi,sin,cos = np.pi,np.sin,np.cos

r1 = 1
r2 = 2

theta = np.linspace(0,2*pi,36)

x1 = r1*cos(theta)
y1 = r1*sin(theta)

x2 = r2*cos(theta)
y2 = r2*sin(theta)

fig, ax = plt.subplots()

ax.fill_between(x2, -y2, y2, color='red')
ax.fill_between(x1, y1, -y1, color='white')

plt.show()

它会把你甜甜圈的整个区域画成红色，然后再把中间的“洞”画成白色。

Answer 3

你可以通过分别绘制上半部分和下半部分来实现这个效果：

import numpy as np
import matplotlib.pyplot as plt

inner = 5.
outer = 10.

x = np.linspace(-outer, outer, 1000, endpoint=True)

yO = outer*np.sin(np.arccos(x/outer)) # x-axis values -> outer circle
yI = inner*np.sin(np.arccos(x/inner)) # x-axis values -> inner circle (with nan's beyond circle)
yI[np.isnan(yI)] = 0.                 # yI now looks like a boulder hat, meeting yO at the outer points

ax = plt.subplot(111)
ax.fill_between(x, yI, yO, color="red")
ax.fill_between(x, -yO, -yI, color="red")

plt.show()

或者你也可以使用极坐标来做，不过这是否有好处要看具体情况：

import numpy as np
import matplotlib.pyplot as plt

theta = np.linspace(0., 2.*np.pi, 80, endpoint=True)
ax = plt.subplot(111, polar=True)
ax.fill_between(theta, 5., 10., color="red")

plt.show()

Answer 4

你可以沿着一个封闭曲线的边界进行遍历，然后使用 fill 方法来填充这个封闭区域内部的空间：

import numpy as np
import matplotlib.pyplot as plt

n, radii = 50, [.7, .95]
theta = np.linspace(0, 2*np.pi, n, endpoint=True)
xs = np.outer(radii, np.cos(theta))
ys = np.outer(radii, np.sin(theta))

# in order to have a closed area, the circles
# should be traversed in opposite directions
xs[1,:] = xs[1,::-1]
ys[1,:] = ys[1,::-1]

ax = plt.subplot(111, aspect='equal')
ax.fill(np.ravel(xs), np.ravel(ys), edgecolor='#348ABD')

plt.show()

这个方法可以很容易地应用到任何形状上，比如说一个五边形可以在圆的内部或外部：