I have 2 numpy arrays of shape (5,1) say:
a=[1,2,3,4,5]
b=[2,4,2,3,6]
How can I make a matrix multiplying each i-th element with each j-th? Like:
..a = [1,2,3,4,5]
b
2 2, 4, 6, 8,10
4 4, 8,12,16,20
2 2, 4, 6, 8,10
3 3, 6, 9,12,15
6 6,12,18,24,30
Without using forloops? is there any combination of reshape, reductions or multiplications that I can use?
Right now I create a a*b tiling of each array along rows and along colums and then multiply element wise, but it seems to me there must be an easier way.
With numpy.outer() and numpy.transpose() routines:
import numpy as np
a = [1,2,3,4,5]
b = [2,4,2,3,6]
c = np.outer(a,b).transpose()
print(c)
Or just with swapped array order:
c = np.outer(b, a)
The output;
[[ 2 4 6 8 10]
[ 4 8 12 16 20]
[ 2 4 6 8 10]
[ 3 6 9 12 15]
[ 6 12 18 24 30]]
For some reason np.multiply.outer seems to be faster than np.outer for small inputs. And broadcasting is faster still - but for bigger arrays they are all pretty much equal.
%timeit np.outer(a,b)
%timeit np.multiply.outer(a,b)
%timeit a[:, None]*b
100000 loops, best of 3: 5.97 µs per loop
100000 loops, best of 3: 3.27 µs per loop
1000000 loops, best of 3: 1.38 µs per loop
a = np.random.randint(0,10,100)
b = np.random.randint(0,10,100)
%timeit np.outer(a,b)
%timeit np.multiply.outer(a,b)
%timeit a[:, None]*b
100000 loops, best of 3: 15.5 µs per loop
100000 loops, best of 3: 14 µs per loop
100000 loops, best of 3: 13.5 µs per loop
a = np.random.randint(0,10,10000)
b = np.random.randint(0,10,10000)
%timeit np.outer(a,b)
%timeit np.multiply.outer(a,b)
%timeit a[:, None]*b
10 loops, best of 3: 154 ms per loop
10 loops, best of 3: 154 ms per loop
10 loops, best of 3: 152 ms per loop
