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suppose I have an array called view:

array([[[[ 7,  9],
         [10, 11]],

        [[19, 18],
         [20, 16]]],


       [[[24,  5],
         [ 6, 10]],

        [[18, 11],
         [45, 12]]]])

as you may know from maxpooling, this is a view of the original input, and the kernel size is 2x2:

[[ 7,  9],  [[19, 18],
 [10, 11]],  [20, 16]]], ....

The goal is to find both max values and their indices. However, argmax only works on single axis, so I need to flatten view, i.e. using flatten=view.reshape(2,2,4):

array([[[ 7,  9, 10, 11], [19, 18, 20, 16]],

       [[24,  5,  6, 10], [18, 11, 45, 12]]])

Now, with the help I get from my previous question, I can find indices of max using inds = flatten.argmax(-1):

array([[3, 2],
       [0, 2]])

and values of max:

i, j = np.indices(flatten.shape[:-1])
flatten[i, j, inds]

>>> array([[11, 20],
           [24, 45]])

The problem
the problem arise when I flatten the view array. Since view array is a view of the original array i.e. view = as_strided(original, newshape, newstrides), so view and original shares the same data. However, reshape breaks it, so any change on view is not reflected on original. This is problematical during backpropagation.

My question
Given the array view and indices ind, I'd like to change max values in view to 1000, without using reshape, or any operation that breaks the 'bond' between view and original. Thanks for any help!!!

reproducible example

import numpy as np
from numpy.lib.stride_tricks import as_strided

original=np.array([[[7,9,19,18],[10,11,20,16]],[[24,5,18,11],[6,10,45,12]]],dtype=np.float64)
view=as_strided(original, shape=(2,1,2,2,2),strides=(64,32*2,8*2,32,8))

I'd like to change max values of each kernel in view to, say, 1000, that can be reflected on original, i.e. if I run view[0,0,0,0,0]=1000, then the first element of both view and original are 1000.

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4
  • You can aid the people trying to answer your question by preparing a runnable minimal reproducible example: an array, a view, and then what you want to do with it. Commented Oct 4, 2021 at 7:40
  • Hi Andras I've added an example :) Commented Oct 4, 2021 at 8:08
  • And as you can see it worked :P Yet again you don't need (and should not use) as_strided: view = original.reshape(2, 1, 2, 2, 2).transpose(0, 1, 3, 2, 4). Commented Oct 4, 2021 at 10:14
  • yep you are right. although sometimes when stride, dilation and groups are involved, like in pytorch conv2d, I don't know if there are any other options for me. I've just finished writing this maxpooling using numpy, can't believe it's harder than a conv2d, indexing has given me so much headaches smh. Commented Oct 4, 2021 at 12:37

1 Answer 1

Reset to default
1

how about this:

import numpy as np
view = np.array(
    [[[[ 7,  9],
       [10, 11]],
      [[19, 18],
       [20, 16]]],
     [[[24,  5],
       [ 6, 10]],
      [[18, 11],
       [45, 12]]]]
)
# Getting the indices of the max values
max0 = view.max(-2)
idx2 = view.argmax(-2)
idx2 = idx2.reshape(-1, idx2.shape[1])
max1 = max0.max(-1)
idx3 = max0.argmax(-1).flatten()
idx2 = idx2[np.arange(idx3.size), idx3]

idx0 = np.arange(view.shape[0]).repeat(view.shape[1])
idx1 = np.arange(view.shape[1]).reshape(1, -1).repeat(view.shape[0], 0).flatten()

# Replacing the maximal vlues with 1000
view[idx0, idx1, idx2, idx3] = 1000
print(f'view = \n{view}')

output:

view = 
[[[[   7    9]
   [  10 1000]]

  [[  19   18]
   [1000   16]]]


 [[[1000    5]
   [   6   10]]

  [[  18   11]
   [1000   12]]]]

Basically, idx{n} is the index of the maximal value in the last two dimensions for every matrix contained in the first two dimensions.

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3 Comments

Thanks Yann this looks promising! could you please adapt your answer a bit based on my original and view from reproducible example?
@Sam-gege. woudl you kindly adapt your reproducible example to include a single, clear tuple of inputs as well as a clear and full expected output? as it stands, i don't understand what is the input or what is wrong with the output in my answer.
Hi Yann, the input is just my view array from reproducible example section, which is one more dimension than your view. But anyways, I've made the necessary changes. you just need to change this idx2 = idx2.reshape(-1, idx2.shape[ - 1]) i.e. idx2.shape(-1) rather than shape(1). Other than that, there should be 3 axis in view[ i, j, k, idx2, idx3] = 1000, where i,j,k=np.indices(view.shape[:-2]) to save some work. anyways, I've accept your answer, thanks man!

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