multiply numpy ndarray with 1d array along a given axis

Question

It seems I am getting lost in something potentially silly. I have an n-dimensional numpy array, and I want to multiply it with a vector (1d array) along some dimension (whi

Answer 1

You could build a slice object, and select the desired dimension in that:

import numpy as np

a = np.arange(18).reshape((3,2,3))
b = np.array([1,3])

ss = [None for i in range(a.ndim)]
ss[1] = slice(None)    # set the dimension along which to broadcast

print ss  #  [None, slice(None, None, None), None]

c = a*b[ss]

Answer 2

Utilizing casting and views, instead of actually copying data N times into a new array with appropiate shape (as existing answers do) is way more memory efficient. Here is such a method (based on @ShuxuanXU's code):

def mult_along_axis(A, B, axis):

    # ensure we're working with Numpy arrays
    A = np.array(A)
    B = np.array(B)

    # shape check
    if axis >= A.ndim:
        raise AxisError(axis, A.ndim)
    if A.shape[axis] != B.size:
        raise ValueError(
            "Length of 'A' along the given axis must be the same as B.size"
            )

    # np.broadcast_to puts the new axis as the last axis, so 
    # we swap the given axis with the last one, to determine the
    # corresponding array shape. np.swapaxes only returns a view
    # of the supplied array, so no data is copied unneccessarily.
    shape = np.swapaxes(A, A.ndim-1, axis).shape

    # Broadcast to an array with the shape as above. Again, 
    # no data is copied, we only get a new look at the existing data.
    B_brc = np.broadcast_to(B, shape)

    # Swap back the axes. As before, this only changes our "point of view".
    B_brc = np.swapaxes(B_brc, A.ndim-1, axis)

    return A * B_brc

Answer 3

Solution Code -

import numpy as np

# Given axis along which elementwise multiplication with broadcasting 
# is to be performed
given_axis = 1

# Create an array which would be used to reshape 1D array, b to have 
# singleton dimensions except for the given axis where we would put -1 
# signifying to use the entire length of elements along that axis  
dim_array = np.ones((1,a.ndim),int).ravel()
dim_array[given_axis] = -1

# Reshape b with dim_array and perform elementwise multiplication with 
# broadcasting along the singleton dimensions for the final output
b_reshaped = b.reshape(dim_array)
mult_out = a*b_reshaped

Sample run for a demo of the steps -

In [149]: import numpy as np

In [150]: a = np.random.randint(0,9,(4,2,3))

In [151]: b = np.random.randint(0,9,(2,1)).ravel()

In [152]: whos
Variable   Type       Data/Info
-------------------------------
a          ndarray    4x2x3: 24 elems, type `int32`, 96 bytes
b          ndarray    2: 2 elems, type `int32`, 8 bytes

In [153]: given_axis = 1

Now, we would like to perform elementwise multiplications along given axis = 1. Let's create dim_array:

In [154]: dim_array = np.ones((1,a.ndim),int).ravel()
     ...: dim_array[given_axis] = -1
     ...: 

In [155]: dim_array
Out[155]: array([ 1, -1,  1])

Finally, reshape b & perform the elementwise multiplication:

In [156]: b_reshaped = b.reshape(dim_array)
     ...: mult_out = a*b_reshaped
     ...: 

Check out the whos info again and pay special attention to b_reshaped & mult_out:

In [157]: whos
Variable     Type       Data/Info
---------------------------------
a            ndarray    4x2x3: 24 elems, type `int32`, 96 bytes
b            ndarray    2: 2 elems, type `int32`, 8 bytes
b_reshaped   ndarray    1x2x1: 2 elems, type `int32`, 8 bytes
dim_array    ndarray    3: 3 elems, type `int32`, 12 bytes
given_axis   int        1
mult_out     ndarray    4x2x3: 24 elems, type `int32`, 96 bytes

Answer 4

You could also use a simple matrices trick

c = np.matmul(a,diag(b))

basically just doing matrix multiplication between a and a matrix whose diagonals are the elements of b. Maybe not as efficient but it's a nice single line solution

Answer 5

I got a similar demand when I was working on some numerical calculation.

Let's assume we have two arrays (A and B) and a user-specified 'axis'. A is a multi-dimensional array. B is a 1-d array.

The basic idea is to expand B so that A and B have the same shape. Here is the solution code

import numpy as np
from numpy.core._internal import AxisError

def multiply_along_axis(A, B, axis):
    A = np.array(A)
    B = np.array(B)
    # shape check
    if axis >= A.ndim:
        raise AxisError(axis, A.ndim)
    if A.shape[axis] != B.size:
        raise ValueError("'A' and 'B' must have the same length along the given axis")
    # Expand the 'B' according to 'axis':
    # 1. Swap the given axis with axis=0 (just need the swapped 'shape' tuple here)
    swapped_shape = A.swapaxes(0, axis).shape
    # 2. Repeat:
    # loop through the number of A's dimensions, at each step:
    # a) repeat 'B':
    #    The number of repetition = the length of 'A' along the 
    #    current looping step; 
    #    The axis along which the values are repeated. This is always axis=0,
    #    because 'B' initially has just 1 dimension
    # b) reshape 'B':
    #    'B' is then reshaped as the shape of 'A'. But this 'shape' only 
    #     contains the dimensions that have been counted by the loop
    for dim_step in range(A.ndim-1):
        B = B.repeat(swapped_shape[dim_step+1], axis=0)\
             .reshape(swapped_shape[:dim_step+2])
    # 3. Swap the axis back to ensure the returned 'B' has exactly the 
    # same shape of 'A'
    B = B.swapaxes(0, axis)
    return A * B

And here is an example

In [33]: A = np.random.rand(3,5)*10; A = A.astype(int); A
Out[33]: 
array([[7, 1, 4, 3, 1],
       [1, 8, 8, 2, 4],
       [7, 4, 8, 0, 2]])

In [34]: B = np.linspace(3,7,5); B
Out[34]: array([3., 4., 5., 6., 7.])

In [35]: multiply_along_axis(A, B, axis=1)
Out[34]: 
array([[21.,  4., 20., 18.,  7.],
       [ 3., 32., 40., 12., 28.],
       [21., 16., 40.,  0., 14.]])