How do you Efficiently Multiply a Numpy Vector by an Array of Numbers?

Question

If I have an array of values:

numbers = np.array([1, 2, 4, 5])

and a vector:

vector = np.array([1, 0, 1])

How do I multiply the vector by the value array to get the following:

vector_array = np.array([[1, 0, 1], [2, 0, 2], [4, 0, 4], [5, 0, 5]])

I have tried to do this using matmul by doing the following:

vector_array = vector[..., None]@numbers

and:

vector_array = vector.T@numbers

I expect to get column vectors which I can then transpose, however instead I get this output:

Option 1:

vector_array = vector[..., None]@numbers
ValueError: matmul: Input operand 1 has a mismatch in its core dimension 0, with gufunc signature (n?,k),(k,m?)->(n?,m?) (size 2 is different from 1)

Option 2:

vector_array = vector.T@numbers
ValueError: matmul: Input operand 1 has a mismatch in its core dimension 0, with gufunc signature (n?,k),(k,m?)->(n?,m?) (size 2 is different from 3)

How can I force matmul to behave in the expected way and multiply the column vector by the row vector to give me a matrix? Is there another function I should be using?

In your mind, what's the difference between an array of numbers and a vector. Both are 1d numpy arrays. — hpaulj
– hpaulj, Commented Feb 21, 2022 at 17:39

user17242583 · Accepted Answer · 2022-02-21 17:13:11Z

2

Use numpy broadcasting:

vector_array = vector * numbers[:, None]

Output:

>>> vector_array
array([[1, 0, 1],
       [2, 0, 2],
       [4, 0, 4],
       [5, 0, 5]])

To understand it, look at numbers[:, None]:

>>> numbers
array([1, 2, 4, 5])

>>> numbers[:, None]
array([[1],
       [2],
       [4],
       [5]])

So basically vector * numbers[:, None] multiplies vector by each element of numbers.

answered Feb 21, 2022 at 17:13

user17242583

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5 Comments

Connor Over a year ago

How does this compare, speed wise, to a built in function like numpy.outer?

user17242583 Over a year ago

Interesting question. I'm guessing this is faster but I'll run some benchmarks.

user17242583 Over a year ago

(by the way, numpy broadcasting is builtin, because it's not a Python feature, it's a numpy feature.)

user17242583 Over a year ago

@Connor My benchmarks indicate that for small amounts of data, mine is a bit faster, and for larger data, ours are about the same.

Connor Over a year ago

Thanks so much! I would give you a second upvote if I could!

Connor · Accepted Answer · 2022-02-23 16:58:25Z

0

One possibility different from the one given by @richardec is to use numpy.outer:

numpy.outer(numbers, normal) = np.array([[1, 0, 1], [2, 0, 2], [4, 0, 4], [5, 0, 5]])

as required. However, I am not sure which method is faster.

edited Feb 23, 2022 at 16:58

answered Feb 21, 2022 at 17:36

Connor

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1 Comment

hpaulj Over a year ago

outer is more focused; while the broadcasted approach can be used with other operators. Many of the ufunc, (e.g. np.add) have an outer method that does the same thing. Overall I don't think speed differences are significant.

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