在一个numpy数组中相乘。

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:

我要把二维数组中的每一项乘以一维数组中的相应项。如果我想把每一列都乘以一维数组，这很简单，如numpy所示。用函数。但是我想做相反的事情，把这一行中的每一项相乘。换句话说，我想乘以

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

and get

并获得

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

but instead I get

而是我

[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

有谁知道是否有一种优雅的方式来处理numpy?非常感谢,亚历克斯

5 个解决方案

#1

Normal multiplication like you showed:

就像你展示的一般乘法:

>>> 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:

如果你加上一个轴，它会以你想要的方式相乘:

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

You could also transpose twice:

你也可以转置两次

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

#2

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

你也可以使用矩阵乘法(即点积):

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.

哪个更优雅可能是品味的问题。

#3

Yet another trick (as of v1.6)

还有一个技巧(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.

我精通numpy广播(newaxis)，但我仍在使用这个新的einsum工具。为了找到这个解，我做了一些尝试。

Timings (using Ipython timeit):

计时时间(使用Ipython):

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.

顺便说一句，把a i换成j, np。einsum('ij,j->ij' a,b)会生成Alex不想要的矩阵。而np。einsum('ji,j->ji,A,b)实际上是，双转置。

#4

I've compared the different options for speed and found that – much to my surprise – all options (except einsum for small n and diag for large n) are equally fast:

我已经比较了不同的速度选项，并且发现——让我惊讶的是——所有的选项(除了小n和diag对于大n)都是一样快的:

Code to reproduce the plot:

复制情节代码:

import numpy
import perfplot


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


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.show(
    setup=lambda n: (numpy.random.rand(n, n), numpy.random.rand(n)),
    kernels=[
        newaxis, double_transpose, double_transpose_contiguous, diag_dot,
        einsum
        ],
    n_range=[2**k for k in range(10)],
    logx=True,
    logy=True,
    xlabel='len(A), len(b)'
    )

#5

-3

Why don't you just do

你为什么不这么做呢

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

? ?

#1