python numpy: insert array on a specifc columns at a specific index nr

Question

Lets say that we have the following numpy array:

[[ 0  0  0]
 [ 1  0  0]
 [ 2  0  0]
 [ 3  0  0]
 [ 4  0  0]
 [ 5  0  0]
 [ 6  0  0]
 [ 7  0  0]
 [ 8  0  0]
 [ 9  0  0]
 [10  0  0]
 [11  0  0]
 [12  0  0]
 [13  0  0]]

How can i insert this np.array

[[0 45]
 [1 34]
 [2 23]
 [3 56]
 [4 45]
 [5 34]]

staring from index nr 3 on column 1 and column 2 so that in the end it will look like this:

[[ 0  0  0]
 [ 1  0  0]
 [ 2  0  45]
 [ 3  1  34]
 [ 4  2  23]
 [ 5  3  56]
 [ 6  4  45]
 [ 7  5  34]
 [ 8  0  0]
 [ 9  0  0]
 [10  0  0]
 [11  0  0]
 [12  0  0]]

The idea is that i would like to specify the index nr where the second array should be placed in the first one. I would a appreciate a solution which takes into account the speed of execution. Both arrays are have a few million rows, the first array is always bigger than the second one.

Hi @PaulPanzer :),yes they are coming from the GPU solution. With the CPU solution i didn't have to split data1. but with the GPU, i have to send data1 in blocks of around 3mil rows and so after that i must put the result back in the original array to get the original index nr — RaduS
– RaduS, Commented Feb 9, 2017 at 7:47
What? Did you account for these overheads in your benchmarks? I demand a recount! :-P — Paul Panzer
– Paul Panzer, Commented Feb 9, 2017 at 7:53
The overheads are insignificant because data1 is splited in max 3 or 4 blocks, so the only overhead i have is when i put the result back in the original index array which is again only 3-4 blocks per data1 and the speed of doing that is close to 0 — RaduS
– RaduS, Commented Feb 9, 2017 at 7:56

hpaulj · Accepted Answer · 2017-02-09 00:26:59Z

3

Just define the recipient space with sliced indexing:

In [79]: arr = np.zeros((10,3),int)    
In [80]: b = np.ones((4,2), int)
In [81]: nr = 3
In [82]: arr[nr:nr+b.shape[0], 1:] = b

In [83]: arr
Out[83]: 
array([[0, 0, 0],
       [0, 0, 0],
       [0, 0, 0],
       [0, 1, 1],
       [0, 1, 1],
       [0, 1, 1],
       [0, 1, 1],
       [0, 0, 0],
       [0, 0, 0],
       [0, 0, 0]])

Just make sure the shapes match:

In [84]: arr[nr:nr+b.shape[0], 1:].shape
Out[84]: (4, 2)

In [85]: b.shape
Out[85]: (4, 2)

You could refine the indexing to handle the case where nr is too large to fit all of b in the arr.

answered Feb 9, 2017 at 0:26

hpaulj

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