Python Numpy - Complex Numbers - Is there a function for Polar to Rectangular conversion?

问题

Is there a built-in Numpy function to convert a complex number in polar form, a magnitude and an angle (degrees) to one in real and imaginary components?

Clearly I could write my own but it seems like the type of thing for which there is an optimised version included in some module?

More specifically, I have an array of magnitudes and an array of angles:

>>> a
array([1, 1, 1, 1, 1])
>>> b
array([120, 121, 120, 120, 121])

And what I would like is:

>>> c
[(-0.5+0.8660254038j),(-0.515038074+0.8571673007j),(-0.5+0.8660254038j),(-0.5+0.8660254038j),(-0.515038074+0.8571673007j)]

回答1:

There isn't a function to do exactly what you want, but there is angle, which does the hardest part. So, for example, one could define two functions:

def P2R(radii, angles):
    return radii * exp(1j*angles)

def R2P(x):
    return abs(x), angle(x)

These functions are using radians for input and output, and for degrees, one would need to do the conversion to radians in both functions.

In the numpy reference there's a section on handling complex numbers, and this is where the function you're looking for would be listed (so since they're not there, I don't think they exist within numpy).

回答2:

There's an error in the previous answer that uses numpy.vectorize - cmath.rect is not a module that can be imported. Numpy also provides the deg2rad function that provides a cleaner piece of code for the angle conversion. Another version of that code could be:

import numpy as np
from cmath import rect

nprect = np.vectorize(rect)

c = nprect(a, np.deg2rad(b))

The code uses numpy's vectorize function to return a numpy style version of the standard library's cmath.rect function that can be applied element wise across numpy arrays.

回答3:

I used cmath with itertools:

from cmath import rect,pi
from itertools import imap
b = b*pi/180                   # convert from deg to rad
c = [x for x in imap(rect,a,b)]

回答4:

import numpy as np
import cmath.rect

nprect = np.vectorize(rect)

c = nprect(a,b*np.pi/180)

回答5:

tom10 answer works fine... you can also expand the Euler's formula to:

def P2R(A, phi):
    return A * ( np.cos(phi) + np.sin(phi)*1j )

来源：https://stackoverflow.com/questions/16444719/python-numpy-complex-numbers-is-there-a-function-for-polar-to-rectangular-co

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