How to use numpy instead of for loop with different vectors-CodePudding

I want to improve my code to make it faster and for now, I have a for loop that I don't know how to replace it with numpy functions.

import numpy as np

N = 1000000
d = 2000

p = np.linspace(0,210,211)
alpha = np.linspace(0.00000000000001, np.pi/2, N)
d1 = d*np.cos(alpha)

for i in range(len(p)):
    p1 = p[i]*np.cos(alpha)
    k = 1/((p[i] d)*np.tan(alpha))
    z = np.exp(p1 d1)**k

First, I tried to vectorize the p1, d1 and k to a matrix with right sizes, but I don't know how to calculate the z without a loop. Furthermore, I think this is not an effective way.

import numpy as np

N = 1000000
d = 2000

p = np.linspace(0,210,211)
alpha = np.linspace(0.00000000000001, np.pi/2, N)
d1 = d*np.cos(alpha)


p1 = np.outer(np.cos(alpha),p)
d1 = np.matrix(d1).T * np.matrix(np.ones(len(p)))
k = 1/(np.outer(np.tan(alpha),p) np.outer(np.tan(alpha),d))

CodePudding user response：

If you want one row per element in p, and one column per element in alpha, you just need to add an axis to p so it's a column vector. Numpy's broadcasting takes care of the rest:

import numpy as np

N = 100 # modified to run quickly
d = 20

# reshape p to a column vector
p = np.linspace(0,210,211).reshape((-1, 1))

alpha = np.linspace(0.00000000000001, np.pi/2, N)
d1 = d*np.cos(alpha)

p1 = p*np.cos(alpha)        # shape (211, 100)
k = 1/((p d)*np.tan(alpha)) 
z = np.exp(p1 d1)**k

Try it online

Note that the power operation overflows to infinity, but that's not related to numpy

CodePudding user response：

since your math is complicated why not define your For loop requirements in C or C function and use in python via ctypes module. this will improve execution time.