How to use numpy instead of for loop with different vectors

Question

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))

Asked By: nico.wagner

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Source

Answer 1

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.

Answered By: Vishnu Balaji

Answer 2

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

Answered By: Pranav Hosangadi

How to use numpy instead of for loop with different vectors

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