如何在 Python 中绘制从域区域到辅域区域的函数？

Question

我要计算绘制的图就是上图

from matplotlib import pyplot as plt
from numpy import *

def f(z):
    return (1/z)
    
#r is radius, p is radian
#below is strange part
for r in linspace(0,1,50):
    for p in linspace(0,2*pi,50):
        Z1=array([complex(r*cos(p),r*sin(p))])
        W1 = f(Z1)
    
plt.figure(figsize=(8,8))
plt.xlim(-pi/2,pi,2);plt.ylim(-pi/2,pi/2)
for i in range(0,2499):
     plt.plot(real(W1[i]), imag(W1[i]), c='b', lw=1.5, ls='-')
    
plt.axvline(x=0,color='k', lw=1)
plt.axhline(y=0,color='k', lw=1)
plt.show()

code是圆心为(0,0)，半径为1的圆

我是新手Python，所以代码还不完整，

但是有些地方很奇怪，所以请修复它。

或者如果有更好的方法，请告诉我。

Answer 1

可以从这里对您的代码进行一些编辑。

我使用 numpy 向量化操作重写了 for 循环。另外，我不确定将 rs 设置为 linspace 是否正确。由于您正在尝试绘制一个圆，因此半径应该相同，因此您可以改用 np.full。

from matplotlib import pyplot as plt
from numpy import *

def f(z):
    return (1/z)
    
#r is radius, p is radian
# for r in linspace(0,1,50):
#     for p in linspace(0,2*pi,50):
#         Z1=array([complex(r*cos(p),r*sin(p))])
#         W1 = f(Z1)

# rewriting the above code in a vectorized style
rs = linspace(0, 1, 50)
ps = linspace(0, 2*pi, 50)
Z1 = rs * cos(ps) + 1j * rs * sin(ps)
W1 = f(Z1) 

# gussing that you want to plot a circle of radius 1 centered at (1, 0)
# in this case, rs should be constant
rs = full(50, 1.0)
Z1 = rs * cos(ps) + 1j * rs * sin(ps)
# shift the real component
Z1.real += 1
W1 = f(Z1)
    
plt.figure(figsize=(8,8))
plt.xlim(-pi/2,pi,2);plt.ylim(-pi/2,pi/2)
plt.plot(Z1.real, Z1.imag, c='b', lw=1.5, ls="-")
plt.plot(W1.real, W1.imag, c='r', lw=1.5, ls="-")
# for i in range(0,2499):
#      plt.plot(real(W1[i]), imag(W1[i]), c='b', lw=1.5, ls='-')
    
plt.axvline(x=0,color='k', lw=1)
plt.axhline(y=0,color='k', lw=1)
plt.show()

如何在 Python 中绘制从域区域到辅域区域的函数？

How can I draw a function that from domain region to codomain region in Python?

python

numpy

matplotlib