给3分和一个情节圈 [英] Give 3 points and a plot circle

问题描述

我想将[0,1]，[1,0]和[0，-1]点赋予python并绘制通过它们的圆.是否存在制作此功能的python模块?我试过使用 matplotlib:

I want to give the points [0,1],[1,0] and [0,-1] to python and plot the circle that passes over them. Does exists a python module that make this? I have tried using matplotlib:

import matplotlib.pyplot as plt
plt.plot([0,1,0],[1,0,-1])
plt.show()

但只给了我两行.

推荐答案

有一个与此完全匹配的代码高尔夫"问题(除了要求圆的方程，而不是绘制它)--请参阅 https://codegolf.stackexchange.com/questions/2289/circle-through-three-points.将第一个也是最短的(Python)解决方案分解为更具可读性，更不容易破解的形式，以符合您的确切规格-但保留使用复数进行更简单计算的核心思想:

There was a "code golf" question exactly matching this (except that the circle's equation was requested, rather than plotting it) -- see https://codegolf.stackexchange.com/questions/2289/circle-through-three-points . Unraveling the first and shortest (Python) solution into more readable, less-hacky form to match your exact specs - but keeping the core idea of using complex numbers for simpler calculations:

x, y, z = 0+1j, 1+0j, 0-1j
w = z-x
w /= y-x
c = (x-y)*(w-abs(w)**2)/2j/w.imag-x
print '(x%+.3f)^2+(y%+.3f)^2 = %.3f^2' % (c.real, c.imag, abs(c+x))

好的，这仍然是打印方程式"，而不是绘制圆"，但是，我们越来越接近:-).要在 matplotlib 中实际绘制圆，请参见例如用 pyplot 绘制一个圆——在上面的解决方案中，c 是圆的(否定的)中心(作为一个复数，因此使用 .real 和 .imag 作为x/y坐标)，以及 abs(c + x)半径(实数， abs 如此).

OK, this still "prints the equation" rather than "plotting the circle", but, we're getting close:-). To actually plot the circle in matplotlib, see e.g plot a circle with pyplot -- in the solution above, c is the (negated) center of the circle (as a complex number, so use .real and .imag for the x/y coordinates), and abs(c+x) the radius (a real number, abs makes it so).

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