44

I have those arrays:

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

b = np.array([1,2,3,4])

and I want them to multiply like so:

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

... basically out[i] = a[i] * b[i], where a[i].shape is (2,) and b[i] then is a scalar.

What's the trick? np.multiply seems not to work:

>>> np.multiply(a, b)
ValueError: operands could not be broadcast together with shapes (4,2) (4)
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2
  • 3
    The answer below uses a feature called broadcasting. You can read about it here, here and here. It's also more standard to use the operator * rather than multiply Commented Apr 8, 2014 at 11:20
  • 1
    @YXD. You're right, although there are 2 things at play here - first reshape, then broadcast together. Nevertheless, I would rather insert a link to this question in the documentation, than the other way round - the theory behind broadcasting sounds very complicated, and seeing a simple example like this one, or e.g. a multiplication table on 2 aranges (outer product) gives a good concrete example. Commented Nov 6, 2019 at 19:31

6 Answers 6

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54

add an axis to b:

>>> np.multiply(a, b[:, np.newaxis])
array([[ 1,  2],
       [ 6,  8],
       [15, 18],
       [28, 32]])
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2 Comments

Perfect, thanks! I think I should spend some time figuring out what's the think with "axis" ...
a * b[:, np.newaxis] does not work for sparse matrices though. Using multiply works.
21

For those who don't want to use np.newaxis or reshape, this is as simple as:

a * b[:, None]

This is because np.newaxis is actually an alias for None.

Read more here.

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12 
>>> a * b.reshape(-1, 1)
array([[ 1,  2],
       [ 6,  8],
       [15, 18],
       [28, 32]])
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1 Comment

So what? Answering on SX not only helps authors but also the people who will reach this page in the future when they've encountered the same problem.
7

What's missing here is the einsum (doc) variant:

np.einsum("ij,i->ij", a, b)

This gives you full control over the indices and a and b are passed blank.

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1 Comment

As np.einsum was already available when the question was asked, I think this should be the accepted answer. The answer above uses additional memory I guess?
0

This is how a*b works when a is transposed so you can also do

np.transpose(np.transpose(a)*b)

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-2

It looks nice, but quite naive, I think, because if you change the dimensions of a or b, the solution

np.mulitply(a, b[:, None])

doesn't work anymore.

I've always had the same doubt about multiplying arrays of arbitrary size row rise, or even, more generally, n-th dimension wise.

I used to do something like

 z = np.array([np.multiply(a, b) for a, b in zip(x,y)])

and that works for x or y that have dimension 1 or 2.

Does it exist with a method with "axis" argument like in other numpy methods? Such like

 z = np.mulitply(x, y, axis=0)
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