An elegant way to unpool matrix with numpy [duplicate]












2
















This question already has an answer here:




  • Quick way to upsample numpy array by nearest neighbor tiling [duplicate]

    3 answers



  • How to repeat elements of an array along two axes?

    4 answers




I will describe the problem with an example. There is a matrix



[[1 2]

 [3 4]]


I need to resize/unpool it to the next one



[[1 1 2 2]

 [1 1 2 2]

 [3 3 4 4]

 [3 3 4 4]]


The operation is kind of opposite to max pooling in convnets. I have a brute solution with cycles. I would be very happy to see if there is a sophisticated solution with numpy.
My brute force solution is below.



matrix = np.arange(1, 5).reshape(2, 2)

matrix_m, matrix_n = matrix.shape

kernel = np.ones((2, 2), dtype=np.int)

kernel_m, kernel_n = kernel.shape

unpooled_matrix = np.zeros((matrix_m * kernel_m, matrix_n * kernel_n), dtype=np.int)

for i in range(matrix_m):

    pos_i = i * kernel_m

    for j in range(matrix_n):

        pos_j = j * kernel_n

        unpooled_matrix[pos_i: pos_i + kernel_m, pos_j: pos_j + kernel_n] = kernel * matrix[i][j]



print(matrix)

print(kernel)

print(unpooled_matrix)





python numpy







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 Nov 26 '18 at 18:50


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.























    2
















    This question already has an answer here:




    • Quick way to upsample numpy array by nearest neighbor tiling [duplicate]

      3 answers



    • How to repeat elements of an array along two axes?

      4 answers




    I will describe the problem with an example. There is a matrix



    [[1 2]
    
 [3 4]]


    I need to resize/unpool it to the next one



    [[1 1 2 2]
    
 [1 1 2 2]
    
 [3 3 4 4]
    
 [3 3 4 4]]


    The operation is kind of opposite to max pooling in convnets. I have a brute solution with cycles. I would be very happy to see if there is a sophisticated solution with numpy.
    My brute force solution is below.



    matrix = np.arange(1, 5).reshape(2, 2)
    
matrix_m, matrix_n = matrix.shape
    
kernel = np.ones((2, 2), dtype=np.int)
    
kernel_m, kernel_n = kernel.shape
    
unpooled_matrix = np.zeros((matrix_m * kernel_m, matrix_n * kernel_n), dtype=np.int)
    
for i in range(matrix_m):
    
    pos_i = i * kernel_m
    
    for j in range(matrix_n):
    
        pos_j = j * kernel_n
    
        unpooled_matrix[pos_i: pos_i + kernel_m, pos_j: pos_j + kernel_n] = kernel * matrix[i][j]
    

    
print(matrix)
    
print(kernel)
    
print(unpooled_matrix)





    python numpy







    share|improve this question













    marked as duplicate by Divakar numpy
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     Nov 26 '18 at 18:50


    This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.





















      2












      2








      2









      This question already has an answer here:




      • Quick way to upsample numpy array by nearest neighbor tiling [duplicate]

        3 answers



      • How to repeat elements of an array along two axes?

        4 answers




      I will describe the problem with an example. There is a matrix



      [[1 2]
      
 [3 4]]


      I need to resize/unpool it to the next one



      [[1 1 2 2]
      
 [1 1 2 2]
      
 [3 3 4 4]
      
 [3 3 4 4]]


      The operation is kind of opposite to max pooling in convnets. I have a brute solution with cycles. I would be very happy to see if there is a sophisticated solution with numpy.
      My brute force solution is below.



      matrix = np.arange(1, 5).reshape(2, 2)
      
matrix_m, matrix_n = matrix.shape
      
kernel = np.ones((2, 2), dtype=np.int)
      
kernel_m, kernel_n = kernel.shape
      
unpooled_matrix = np.zeros((matrix_m * kernel_m, matrix_n * kernel_n), dtype=np.int)
      
for i in range(matrix_m):
      
    pos_i = i * kernel_m
      
    for j in range(matrix_n):
      
        pos_j = j * kernel_n
      
        unpooled_matrix[pos_i: pos_i + kernel_m, pos_j: pos_j + kernel_n] = kernel * matrix[i][j]
      

      
print(matrix)
      
print(kernel)
      
print(unpooled_matrix)





      python numpy







      share|improve this question















      This question already has an answer here:




      • Quick way to upsample numpy array by nearest neighbor tiling [duplicate]

        3 answers



      • How to repeat elements of an array along two axes?

        4 answers




      I will describe the problem with an example. There is a matrix



      [[1 2]
      
 [3 4]]


      I need to resize/unpool it to the next one



      [[1 1 2 2]
      
 [1 1 2 2]
      
 [3 3 4 4]
      
 [3 3 4 4]]


      The operation is kind of opposite to max pooling in convnets. I have a brute solution with cycles. I would be very happy to see if there is a sophisticated solution with numpy.
      My brute force solution is below.



      matrix = np.arange(1, 5).reshape(2, 2)
      
matrix_m, matrix_n = matrix.shape
      
kernel = np.ones((2, 2), dtype=np.int)
      
kernel_m, kernel_n = kernel.shape
      
unpooled_matrix = np.zeros((matrix_m * kernel_m, matrix_n * kernel_n), dtype=np.int)
      
for i in range(matrix_m):
      
    pos_i = i * kernel_m
      
    for j in range(matrix_n):
      
        pos_j = j * kernel_n
      
        unpooled_matrix[pos_i: pos_i + kernel_m, pos_j: pos_j + kernel_n] = kernel * matrix[i][j]
      

      
print(matrix)
      
print(kernel)
      
print(unpooled_matrix)




      This question already has an answer here:




      • Quick way to upsample numpy array by nearest neighbor tiling [duplicate]

        3 answers



      • How to repeat elements of an array along two axes?

        4 answers







      python numpy




      python numpy






      share|improve this question













      share|improve this question











      share|improve this question




      share|improve this question










      asked Nov 26 '18 at 18:47









      vogdbvogdb

      2,84621821




      2,84621821




      marked as duplicate by Divakar numpy
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       Nov 26 '18 at 18:50


      This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.









      marked as duplicate by Divakar numpy
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       Nov 26 '18 at 18:50


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