How to sum elements of same position in n dimensional array

[Felipe]

Hi, guys,

I have a numpy array of (2,100,2) dimensions, that I want to sum elements of the same position. In order to clarify my doubts, I wrote the example below where I substituted the numbers by letters with dimensions (2,2,2): 

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[[[a b]   [c d]]  [[a b]   [c d]]]

So, I look for a way to sum a+a, b+b, c+c, d+d, and obtain an array with dimensions (2,2,1). Here's the desired output:

1 2	[[a+a  b+b]  [c+c  d+d]]

I'd like something that I can extrapolate to cases with dimensions (2,100,X), where X can be numbers from 2 to 10.
Thanks !!!

[zivoni]

It seems that you can use .sum() method with axis parameter.

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In [1]: import numpy as np   In [2]: a = np.array([[[1,2], [3,4]], [[10, 20], [30, 40]]])   In [3]: a Out[3]: array([[[ 1,  2],         [ 3,  4]],        [[10, 20],         [30, 40]]])   In [4]: a.sum(axis=0) Out[4]: array([[11, 22],        [33, 44]])

As you can see, result has dimensionality reduced to (2, 2), if you want result with shape (2, 2, 1), you need to reshape it or add new dimension:

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In [5]: a.sum(axis=0)[..., None] Out[5]: array([[[11],         [22]],        [[33],         [44]]])

[Felipe]

It seems that you can use .sum() method with axis parameter.

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In [1]: import numpy as np   In [2]: a = np.array([[[1,2], [3,4]], [[10, 20], [30, 40]]])   In [3]: a Out[3]: array([[[ 1,  2],         [ 3,  4]],        [[10, 20],         [30, 40]]])   In [4]: a.sum(axis=0) Out[4]: array([[11, 22],        [33, 44]])

As you can see, result has dimensionality reduced to (2, 2), if you want result with shape (2, 2, 1), you need to reshape it or add new dimension:

1 2 3 4 5 6

In [5]: a.sum(axis=0)[..., None] Out[5]: array([[[11],         [22]],        [[33],         [44]]])

Thank you !!! This works perfectly for me.