我正在嘗試?yán)L制這樣的圖：我不知道如何在 for 循環(huán)中找到較小圓圈的中心。首先，我嘗試用較少數(shù)量的圓圈（例如2個）來繪制它，但我不知道為什么較小的圓圈是半圓？我的嘗試：import numpy as npimport matplotlib.pyplot as pltr = 2, h = 1, k = 1axlim = r + np.max((abs(h),np.max(abs(k))))x = np.linspace(-axlim, axlim, 100)X,Y = np.meshgrid(x,x)F = (X-h)**2 + (Y-k)**2 - r**2plt.contour(X,Y,F,0)F1 = (X-(h+r))**2 + (Y-k)**2 - (r/3)**2plt.contour(X,Y,F1,0)F2 = (X-h)**2 + (Y-(k+r))**2 - (r/3)**2plt.contour(X,Y,F2,0)plt.gca().set_aspect('equal')plt.axis([-4*r, 4*r, -4*r,4*r])# plt.axis('off')plt.show()輸出：
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皈依舞
TA貢獻(xiàn)1851條經(jīng)驗 獲得超3個贊
0, 2pi可以使用正弦、余弦和在范圍內(nèi)均勻劃分的角度：
import numpy as np
import matplotlib.pyplot as plt
num_circ = 7
rad_large = 7
rad_small = 6
thetas = np.linspace(0, 2 * np.pi, num_circ, endpoint=False)
fig, ax = plt.subplots()
ax.add_patch(plt.Circle((0, 0), rad_large, fc='none', ec='navy'))
for theta in thetas:
ax.add_patch(plt.Circle((rad_large * np.cos(theta), rad_large * np.sin(theta),), rad_small, fc='none', ec='crimson'))
ax.autoscale_view() # calculate the limits for the x and y axis
ax.set_aspect('equal') # show circles as circles
plt.show()
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