numpy: multiplying a 2D array by a 1D array

Question

Let us say one has an array of 2D vectors:

v = np.array([ [1, 1], [1, 1], [1, 1], [1, 1]])
v.shape = (4, 2)

And an array of scalars:

s = np.array( [2, 2, 2, 2] )
s.shape = (4,)

I would like the result:

f(v, s) = np.array([ [2, 2], [2, 2], [2, 2], [2, 2]])

Now, executing v*s is an error. Then, what is the most efficient way to go about implementing f?

nneonneo · Accepted Answer · 2014-06-16 22:18:11Z

10

Add a new singular dimension to the vector:

v*s[:,None]

This is equivalent to reshaping the vector as (len(s), 1). Then, the shapes of the multiplied objects will be (4,2) and (4,1), which are compatible due to NumPy broadcasting rules (corresponding dimensions are either equal to each other or equal to 1).

Note that when two operands have unequal numbers of dimensions, NumPy will insert extra singular dimensions "in front" of the operand with fewer dimensions. This would make your vector (1,4) which is incompatible with (4,2). Therefore, we explicitly specify where the extra dimensions are added, in order to make the shapes compatible.

edited Jun 16, 2014 at 22:18

answered Jun 16, 2014 at 22:12

nneonneo

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3 Comments

bzm3r Over a year ago

This works! Could you explain the rationale behind why this works?

nneonneo Over a year ago

@user89: added an explanation.

usethedeathstar Over a year ago

this syntax with None is the same as with np.newaxis instead of the None, (though personally i prefer the None over the np.newaxis too).

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