Multiplying matrices that are stored in an array in Python

Question

This may seem like a silly question, but I am a bloody newby in Python (and in programming). I am running a physics simulation that involves many (~10 000) 2x2 matrices that I store in an array. I call these matrices M and the array T in the code below. Then I simply want to compute the product of all of these matrices. This is what I came up with, but it looks ugly and it would be so much work for 10000+ 2x2 matrices. Is there a simpler way or an inbuilt function that I could use?

import numpy as np
#build matrix M (dont read it, just an example, doesnt matter here)    
def M(k1 , k2 , x):
    a = (k1 + k2) * np.exp(1j * (k2-k1) * x)
    b = (k1 - k2) * np.exp(-1j * (k1 + k2) * x)
    c = (k1 - k2) * np.exp(1j * (k2 + k1) * x)
    d = (k1 + k2) * np.exp(-1j * (k2 - k1) * x)
    M = np.array([[a , b] , [c , d]])
    M *= 1. / (2. * k1)
    return M


#array of test matrices T
T = np.array([M(1,2,3), M(3,3,3), M(54,3,9), M(33,11,42) ,M(12,9,5)])
#compute the matrix product of T[0] * T[1] *... * T[4]
#I originally had this line of code, which is wrong, as pointed out in the comments
#np.dot(T[0],np.dot(T[1], np.dot(T[2], np.dot(T[2],np.dot(T[3],T[4])))))
#it should be:
np.dot(T[0], np.dot(T[1], np.dot(T[2],np.dot(T[3],T[4]))))

Are the matrices always i * M for i from 1 upto some number? — Fred Foo
– Fred Foo, Commented Apr 4, 2013 at 15:36
no, they are really complicated. I actually plan to compute them on the fly in a for loop, if I can figure out how to do it. should I change that in my code snippet? maybe it's confusing. — seb
– seb, Commented Apr 4, 2013 at 15:36
Is there a reason for putting the matrices in an array? I think I would probably just put them in a list ... — mgilson
– mgilson, Commented Apr 4, 2013 at 15:39
Indeed, a list would probably even be faster with a for loop. — Fred Foo
– Fred Foo, Commented Apr 4, 2013 at 15:40
ok, now I'm done editing. the reason why I put them into an array is because I do numerical calculations. I don't know how it would affect the precision if I put it into a list. does it matter? — seb
– seb, Commented Apr 4, 2013 at 15:44

YXD · Accepted Answer · 2013-04-04 15:37:44Z

1

Not very NumPythonic, but you could do:

reduce(lambda x,y: np.dot(x,y), T, np.eye(2))

Or more concisely, as suggested

reduce(np.dot, T, np.eye(2))

answered Apr 4, 2013 at 15:37

YXD

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7 Comments

Fred Foo Over a year ago

lambda x, y: np.dot(x, y) is just np.dot.

mgilson Over a year ago

I was thinking that this seemed like a job for reduce

StuGrey Over a year ago

you dont need the identity matrix, reduce(np.dot, T) works fine

seb Over a year ago

are you sure that this is valid? maybe I'm doing it wrong, but I'm getting different results. with my method above and with your reduce() line mine: [[ 0.24112953-0.1020551j -0.22988481+0.11903133j] [-0.22988481-0.11903133j 0.24112953+0.1020551j ]] . yours: [[ 0.29577303-0.1009189j -0.24969598+0.08683577j] [-0.24969598-0.08683577j 0.29577303+0.1009189j ]]

StuGrey Over a year ago

np.dot(T[0],np.dot(T[1], np.dot(T[2], np.dot(T[2],np.dot(T[3],T[4]))))) you have T[2] twice

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