I’m trying to multiply each of the terms in a 2D array by the corresponding terms in a 1D array. This is very easy if I want to multiply every column by the 1D array, as shown in the numpy.multiply function. But I want to do the opposite, multiply each term in the row.
In other words I want to multiply:

[1,2,3]   [0]
[4,5,6] * [1]
[7,8,9]   [2]

and get

[0,0,0]
[4,5,6]
[14,16,18]

but instead I get

[0,2,6]
[0,5,12]
[0,8,18]

Does anyone know if there’s an elegant way to do that with numpy?
Thanks a lot,
Alex

Normal multiplication like you showed:

>>> import numpy as np
>>> m = np.array([[1,2,3],[4,5,6],[7,8,9]])
>>> c = np.array([0,1,2])
>>> m * c
array([[ 0,  2,  6],
       [ 0,  5, 12],
       [ 0,  8, 18]])

If you add an axis, it will multiply the way you want:

>>> m * c[:, np.newaxis]
array([[ 0,  0,  0],
       [ 4,  5,  6],
       [14, 16, 18]])

You could also transpose twice:

>>> (m.T * c).T
array([[ 0,  0,  0],
       [ 4,  5,  6],
       [14, 16, 18]])

I’ve compared the different options for speed and found that – much to my surprise – all options (except diag) are equally fast. I personally use

A * b[:, None]

(or (A.T * b).T) because it’s short.

enter image description here


Code to reproduce the plot:

import numpy
import perfplot


def newaxis(data):
    A, b = data
    return A * b[:, numpy.newaxis]


def none(data):
    A, b = data
    return A * b[:, None]


def double_transpose(data):
    A, b = data
    return (A.T * b).T


def double_transpose_contiguous(data):
    A, b = data
    return numpy.ascontiguousarray((A.T * b).T)


def diag_dot(data):
    A, b = data
    return numpy.dot(numpy.diag(b), A)


def einsum(data):
    A, b = data
    return numpy.einsum("ij,i->ij", A, b)


perfplot.save(
    "p.png",
    setup=lambda n: (numpy.random.rand(n, n), numpy.random.rand(n)),
    kernels=[
        newaxis,
        none,
        double_transpose,
        double_transpose_contiguous,
        diag_dot,
        einsum,
    ],
    n_range=[2 ** k for k in range(13)],
    xlabel="len(A), len(b)",
)

You could also use matrix multiplication (aka dot product):

a = [[1,2,3],[4,5,6],[7,8,9]]
b = [0,1,2]
c = numpy.diag(b)

numpy.dot(c,a)

Which is more elegant is probably a matter of taste.

Yet another trick (as of v1.6)

A=np.arange(1,10).reshape(3,3)
b=np.arange(3)

np.einsum('ij,i->ij',A,b)

I’m proficient with the numpy broadcasting (newaxis), but I’m still finding my way around this new einsum tool. So I had play around a bit to find this solution.

Timings (using Ipython timeit):

einsum: 4.9 micro
transpose: 8.1 micro
newaxis: 8.35 micro
dot-diag: 10.5 micro

Incidentally, changing a i to j, np.einsum('ij,j->ij',A,b), produces the matrix that Alex does not want. And np.einsum('ji,j->ji',A,b) does, in effect, the double transpose.

For those lost souls on google, using numpy.expand_dims then numpy.repeat will work, and will also work in higher dimensional cases (i.e. multiplying a shape (10, 12, 3) by a (10, 12)).

>>> import numpy
>>> a = numpy.array([[1,2,3],[4,5,6],[7,8,9]])
>>> b = numpy.array([0,1,2])
>>> b0 = numpy.expand_dims(b, axis = 0)
>>> b0 = numpy.repeat(b0, a.shape[0], axis = 0)
>>> b1 = numpy.expand_dims(b, axis = 1)
>>> b1 = numpy.repeat(b1, a.shape[1], axis = 1)
>>> a*b0
array([[ 0,  2,  6],
   [ 0,  5, 12],
   [ 0,  8, 18]])
>>> a*b1
array([[ 0,  0,  0],
   [ 4,  5,  6],
   [14, 16, 18]])

Why don’t you just do

>>> m = np.array([[1,2,3],[4,5,6],[7,8,9]])
>>> c = np.array([0,1,2])
>>> (m.T * c).T

??