Numpy: outer product of n vectors

Question

I\'m trying to do something simple in numpy, and I\'m sure there should be an easy way of doing it.

Basically, I have a list of n vectors with various l

Answer 1

You use use following one line code:

reduce(np.multiply, np.ix_(*vs))

np.ix_() will do the outer broadcast, you need reduce, but you can pass the ufunc np.multiply without lambda function.

Here is the comparing:

import numpy as np
vs = [np.r_[1,2,3.0],np.r_[4,5.0],np.r_[6,7,8.0]]
shape = map(len, vs)

 # specify the orientation of each vector
newshapes = np.diag(np.array(shape)-1)+1
reshaped = [x.reshape(y) for x,y in zip(vs, newshapes)]

# direct product
A = reduce(lambda a,b: a*b, reshaped, 1)
B = reduce(np.multiply, np.ix_(*vs))

np.all(A==B)

The reuslt:

True

Answer 2

There is an alternative line of code:

reduce(np.multiply.outer, vs)

It is more transparent for me than np.ix_(*vs) construction and support multidimensional arrays like in this question.

Timings are the same within a tolerance:

import numpy as np
from functools import reduce

def outer1(*vs):
    return np.multiply.reduce(np.ix_(*vs))
def outer2(*vs):
    return reduce(np.multiply.outer, vs)

v1 = np.random.randn(100)
v2 = np.random.randn(200)
v3 = np.random.randn(300)
v4 = np.random.randn(50)

%timeit outer1(v1, v2, v3, v4)
# 1 loop, best of 3: 796 ms per loop

%timeit outer2(v1, v2, v3, v4)
# 1 loop, best of 3: 795 ms per loop

np.all(outer1(v1, v2, v3, v4) == outer2(v1, v2, v3, v4))
# True