numpy subtract/add 1d array from 2d array

Question

I have the following 2D-array:

a = array([[ 1,  2,  3],
           [ 4,  5,  6],
           [ 7,  8,  9],
           [10, 11, 12],
           [13, 14, 15]])

and another 1D-array:

b = array([ 1,  2,  3,  4,  5])

then I want to calculate something like

c = a - b

with the intent of getting:

c = array([[0, 1,  2],
           [2, 3,  4],
           [4, 5,  6],
           [6, 7,  8],
           [8, 9, 10]])

but instead I get the error message:

Traceback (most recent call last):
  Python Shell, prompt 79, line 1
ValueError: operands could not be broadcast together with shapes (5,3) (5,)

I read the broadcasting rules but didn´t get any wiser. I could do a workaround with for-loops or similar but there should be a direct way. Thanks

Do a - b[:,None] or a - b[:,np.newaxis], that is convert b to a 2D array by adding a dimension and then do subtraction. — Divakar
– Divakar, Commented Oct 23, 2015 at 13:21
Isn't a basic question like this covered before on SO? Dup-hammer? :) — Divakar
– Divakar, Commented Oct 23, 2015 at 13:35

Sakib Ahammed · Accepted Answer · 2015-10-25 10:53:52Z

34

You need to convert array b to a (2, 1) shape array, use None or numpy.newaxis in the index tuple. Here is the Indexing of Numpy array.

You can do it Like:

import numpy

a = numpy.array([[ 1,  2,  3],
           [ 4,  5,  6],
           [ 7,  8,  9],
           [10, 11, 12],
           [13, 14, 15]])

b = numpy.array([ 1,  2,  3,  4,  5])
c=a - b[:,None]
print c

Output:

Out[2]: 
array([[ 0,  1,  2],
       [ 2,  3,  4],
       [ 4,  5,  6],
       [ 6,  7,  8],
       [ 8,  9, 10]])

edited Oct 25, 2015 at 10:53

answered Oct 23, 2015 at 13:30

Sakib Ahammed

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1 Comment

mLstudent33 Over a year ago

@Sakib, I am able to do this without any conversion and it's confusing me. I have A.shape = (100,1) and B.shape = (100) and when I do A-B I get C of shape (100,100. What is going on?

P. Camilleri · Accepted Answer · 2015-10-23 14:18:17Z

4

As Divakar specified in the comments, just add a new axis to b.

I suggest you read more about broadcasting which is very often useful to vectorize computations in numpy: interestingly enough, a.transpose() - b wouldn't have raised an error (you'd need to transpose the result again to obtain your desired output).

In this computaion, the first array's shape is (3, 5), and b.shape is (5,). So the shape of b corresponds to the tail of the shape of a, and broadcasting can happen. This is not the case when the shape of the first array is (5, 3), hence the error you obtained.

Here are some runtime tests to compare the speeds of the suggested answers, with your values for a and b : you can see that the differences are not really significant

In [9]: %timeit (a.T - b).T
Out[9]: 1000000 loops, best of 3: 1.32 µs per loop

In [10]: %timeit a - b[:,None]
Out[10]: 1000000 loops, best of 3: 1.25 µs per loop

In [11]: %timeit a - b[None].T
Out[11]: 1000000 loops, best of 3: 1.3 µs per loop

edited Oct 23, 2015 at 14:18

answered Oct 23, 2015 at 13:33

P. Camilleri

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6 Comments

Divakar Over a year ago

a.transpose() - b would require another transpose to get the desired output, right?

P. Camilleri Over a year ago

@Divakar Yes, yes, I'm not proposing a solution, I'm explicitly telling to use yours. Since your solution was a comment, I added the reason why broadcasting didn't work. If you'd posted an answer, I'd have commented it.

Divakar Over a year ago

I was stating for the purpose of completeness of this post on how to get to the desired output and shape :)

Divakar Over a year ago

Or another way would be a - b[None].T? Quick runtime tests show this to be marginal better than [:,None] and I wasn't expecting this. But make sure to test it out at your end if you are incorporating runtime tests, but definitely worth checking out!

Divakar Over a year ago

Actually that "marginal" thing is really marginal and goes either way across many runs, but nevertheless those runtime tests might be of some interest somewhere.

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