CODEWITHSUNDEEP
pythonarraysnumpy
Akavall
Akavall

Reputation: 86306

sum parts of numpy.array

Let's say I have the following array:

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

I want to sum the first two values of the first row: 1+2 = 3, then next two values: 3+4 = 7, and then 5+6 = 11, and so on for every row. My desired output is this:

array([[ 3,  7, 11],
       [15, 19, 23],
       [ 8, 13, 17]])

I have the following solution:

def sum_chunks(x, chunk_size):
    rows, cols = x.shape
    x = x.reshape(x.size / chunk_size, chunk_size)
    return x.sum(axis=1).reshape(rows, cols/chunk_size)

But it feels unnecessarily complicated, is there a better way? Perhaps a built-in?

Upvotes: 8

Views: 9185

Answers (3)

Jaime
Jaime

Reputation: 67497

When I have to do this kind of stuff, I prefer reshaping the 2D array into a 3D array, then collapse the extra dimension with np.sum. Generalizing it to n-dimensional arrays, you could do something like this:

def sum_chunk(x, chunk_size, axis=-1):
    shape = x.shape
    if axis < 0:
        axis += x.ndim
    shape = shape[:axis] + (-1, chunk_size) + shape[axis+1:]
    x = x.reshape(shape)
    return x.sum(axis=axis+1)

>>> a = np.arange(24).reshape(4, 6)
>>> a
array([[ 0,  1,  2,  3,  4,  5],
       [ 6,  7,  8,  9, 10, 11],
       [12, 13, 14, 15, 16, 17],
       [18, 19, 20, 21, 22, 23]])
>>> sum_chunk(a, 2)
array([[ 1,  5,  9],
       [13, 17, 21],
       [25, 29, 33],
       [37, 41, 45]])
>>> sum_chunk(a, 2, axis=0)
array([[ 6,  8, 10, 12, 14, 16],
       [30, 32, 34, 36, 38, 40]])

Upvotes: 7

nneonneo
nneonneo

Reputation: 179697

Just use slicing:

a[:,::2] + a[:,1::2]

This takes the array formed by every even-indexed column (::2), and adds it to the array formed by every odd-indexed column (1::2).

Upvotes: 8

senderle
senderle

Reputation: 151147

Here's one way:

>>> a[:,::2] + a[:,1::2]
array([[ 3,  7, 11],
       [15, 19, 23],
       [ 8, 13, 17]])

Upvotes: 1

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