A fast way to multiply a NumPy array of scalars by an array of arrays

Question

What is the most concise way to carry out multiplication like this?

# c's are scalars (or arrays like A's in general)
x = np.array([c1, c2, c3])
# A's are NumPy arrays
M = np.array([A1, A2, A3])

to get

x*M = [c1*A1, c2*A2, c3*A3]

c's are scalars, A's are NumPy numerical multidim arrays (let's say, matrices).

Example code:

x = np.array([1,2,3])
A = np.random.rand(2,2)
M = np.array([A,A,A])

What's the shape and datatype of M? Do all sub-arrays in M have the same shapes? — Divakar
– Divakar, Commented Sep 21, 2017 at 19:16
And dtype of M? Is it a 1d array of objects, or a multidimensional array of numbers? — hpaulj
– hpaulj, Commented Sep 21, 2017 at 19:22
@hpaulj Let's suppose they are multidimensional arrays of numbers - for simplicity - 2D matrices stored as np.array. Why does it matter? Scalar can be multiplied by almost anything, can't it? — homocomputeris
– homocomputeris, Commented Sep 22, 2017 at 8:07

akuiper · Accepted Answer · 2017-09-21 19:34:57Z

2

If M is a numpy array of primitive numeric types, (i.e. not objects), to take advantage of the numpy broadcasting, you can add dimensions to x so it has the same number of dimensions as M, and then the element-wise multiplication should work:

x.reshape((-1,) + (1,)*(M.ndim - 1)) * M

x = np.array([1,2,3])

2D case:

M = np.arange(12).reshape(3,4)    
x.reshape((-1,) + (1,)*(M.ndim - 1)) * M
#array([[ 0,  1,  2,  3],
#       [ 8, 10, 12, 14],
#       [24, 27, 30, 33]])

3D case:

M = np.arange(12).reshape(3,2,2)
x.reshape((-1,) + (1,)*(M.ndim - 1)) * M
#array([[[ 0,  1],
#        [ 2,  3]],

#       [[ 8, 10],
#        [12, 14]],

#       [[24, 27],
#        [30, 33]]])

answered Sep 21, 2017 at 19:34

akuiper

216k33 gold badges362 silver badges379 bronze badges

Sign up to request clarification or add additional context in comments.

1 Comment

homocomputeris Over a year ago

Thanks, this one works. But the readability... It took me a while to understand what's happening :)

Acccumulation · Accepted Answer · 2017-09-21 19:30:33Z

-1

product = [x[i]*M[i] for i in range(len(x))]

answered Sep 21, 2017 at 19:30

Acccumulation

3,6311 gold badge11 silver badges13 bronze badges

1 Comment

homocomputeris Over a year ago

Or like this: [A*c for (A,c) in zip(M,x)]. Anyway, I believe that comprehensions are slower.

Collectives™ on Stack Overflow

A fast way to multiply a NumPy array of scalars by an array of arrays

2 Answers 2

1 Comment

1 Comment

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

2 Answers 2

1 Comment

1 Comment

Your Answer

Sign up or log in

Post as a guest

Related