Numpy 运算符用于具有矩阵单独行乘法的每个向量元素

Question

Is there a numpy operator that will result in the individual vector element multiplying with the corresponding matrix row?

For eg,

    import numpy
    a,b=numpy.array([1,2]), numpy.array([[1,2,3,4],[5,6,7,8]])

When I multiply a and b, I want the result to be

[[1,2,3,4],[10,12,14,16]]

where each vector element is multiplied with the corresponding matrix row elements.

I know how to implement this using loops, but I just wanted to know whether an in-built function exists in numpy for this, especially when b is an extremely large, but sparse matrix ?

Thank you.

Answer 1

You could use multiply like the following:

import numpy
a,b=numpy.array([1,2]), numpy.array([[1,2,3,4],[5,6,7,8]])
print(numpy.multiply(a,b.T).T)
# [[ 1  2  3  4]
# [10 12 14 16]]

Other option is to use * and transpose like the following:

import numpy
a,b=numpy.array([1,2]), numpy.array([[1,2,3,4],[5,6,7,8]])
print(a*b.T)
# [[ 1 10]
# [ 2 12]
# [ 3 14]
# [ 4 16]]

Answer 2

You can use:

a[:,None]*b

This should be fairly fast with no extra calculation cost.

output:

[[ 1  2  3  4]
 [10 12 14 16]]