Matrix match in python










3















How can I find the best "match" for small matrix in big matrix?
For example:



 small=[[1,2,3],
[4,5,6],
[7,8,9]]



big=[[2,4,2,3,5],
[6,0,1,9,0],
[2,8,2,1,0],
[7,7,4,2,1]]


The match is defined as difference of numbers in matrix, so match in position (1,1) is as if number 5 from small would be on number 0 from big matrix (so the central number from small matrix in coordinates (1,1) of big matrix.



The match value in position (1,1) is:
m(1,1)=|2−1|+|4−2|+|2−3|+|6−4|+|0−5|+|1−6|+|2−7|+|8−8|+|2−9|=28



The goal is to find the lowest difference posible in those matrixes.



The small matrix always has odd number of lines and columns, so it's easy to find it's centre.










share|improve this question



















  • 1





    Such template matching almost certainly exists in opencv already.

    – wim
    Nov 13 '18 at 18:23











  • nicely described - where is what you coded to try to solve that?

    – Patrick Artner
    Nov 13 '18 at 18:28







  • 4





    Please do not vandalize your posts. By posting on the Stack Exchange network, you've granted a non-revocable right for SE to distribute that content (under the CC BY-SA 3.0 license). By SE policy, any vandalism will be reverted.

    – FrankerZ
    Nov 19 '18 at 18:12















3















How can I find the best "match" for small matrix in big matrix?
For example:



 small=[[1,2,3],
[4,5,6],
[7,8,9]]



big=[[2,4,2,3,5],
[6,0,1,9,0],
[2,8,2,1,0],
[7,7,4,2,1]]


The match is defined as difference of numbers in matrix, so match in position (1,1) is as if number 5 from small would be on number 0 from big matrix (so the central number from small matrix in coordinates (1,1) of big matrix.



The match value in position (1,1) is:
m(1,1)=|2−1|+|4−2|+|2−3|+|6−4|+|0−5|+|1−6|+|2−7|+|8−8|+|2−9|=28



The goal is to find the lowest difference posible in those matrixes.



The small matrix always has odd number of lines and columns, so it's easy to find it's centre.










share|improve this question



















  • 1





    Such template matching almost certainly exists in opencv already.

    – wim
    Nov 13 '18 at 18:23











  • nicely described - where is what you coded to try to solve that?

    – Patrick Artner
    Nov 13 '18 at 18:28







  • 4





    Please do not vandalize your posts. By posting on the Stack Exchange network, you've granted a non-revocable right for SE to distribute that content (under the CC BY-SA 3.0 license). By SE policy, any vandalism will be reverted.

    – FrankerZ
    Nov 19 '18 at 18:12













3












3








3








How can I find the best "match" for small matrix in big matrix?
For example:



 small=[[1,2,3],
[4,5,6],
[7,8,9]]



big=[[2,4,2,3,5],
[6,0,1,9,0],
[2,8,2,1,0],
[7,7,4,2,1]]


The match is defined as difference of numbers in matrix, so match in position (1,1) is as if number 5 from small would be on number 0 from big matrix (so the central number from small matrix in coordinates (1,1) of big matrix.



The match value in position (1,1) is:
m(1,1)=|2−1|+|4−2|+|2−3|+|6−4|+|0−5|+|1−6|+|2−7|+|8−8|+|2−9|=28



The goal is to find the lowest difference posible in those matrixes.



The small matrix always has odd number of lines and columns, so it's easy to find it's centre.










share|improve this question
















How can I find the best "match" for small matrix in big matrix?
For example:



 small=[[1,2,3],
[4,5,6],
[7,8,9]]



big=[[2,4,2,3,5],
[6,0,1,9,0],
[2,8,2,1,0],
[7,7,4,2,1]]


The match is defined as difference of numbers in matrix, so match in position (1,1) is as if number 5 from small would be on number 0 from big matrix (so the central number from small matrix in coordinates (1,1) of big matrix.



The match value in position (1,1) is:
m(1,1)=|2−1|+|4−2|+|2−3|+|6−4|+|0−5|+|1−6|+|2−7|+|8−8|+|2−9|=28



The goal is to find the lowest difference posible in those matrixes.



The small matrix always has odd number of lines and columns, so it's easy to find it's centre.







python python-3.x matrix similarity






share|improve this question















share|improve this question













share|improve this question




share|improve this question








edited Nov 26 '18 at 15:31







IHav

















asked Nov 13 '18 at 18:20









IHavIHav

575




575







  • 1





    Such template matching almost certainly exists in opencv already.

    – wim
    Nov 13 '18 at 18:23











  • nicely described - where is what you coded to try to solve that?

    – Patrick Artner
    Nov 13 '18 at 18:28







  • 4





    Please do not vandalize your posts. By posting on the Stack Exchange network, you've granted a non-revocable right for SE to distribute that content (under the CC BY-SA 3.0 license). By SE policy, any vandalism will be reverted.

    – FrankerZ
    Nov 19 '18 at 18:12












  • 1





    Such template matching almost certainly exists in opencv already.

    – wim
    Nov 13 '18 at 18:23











  • nicely described - where is what you coded to try to solve that?

    – Patrick Artner
    Nov 13 '18 at 18:28







  • 4





    Please do not vandalize your posts. By posting on the Stack Exchange network, you've granted a non-revocable right for SE to distribute that content (under the CC BY-SA 3.0 license). By SE policy, any vandalism will be reverted.

    – FrankerZ
    Nov 19 '18 at 18:12







1




1





Such template matching almost certainly exists in opencv already.

– wim
Nov 13 '18 at 18:23





Such template matching almost certainly exists in opencv already.

– wim
Nov 13 '18 at 18:23













nicely described - where is what you coded to try to solve that?

– Patrick Artner
Nov 13 '18 at 18:28






nicely described - where is what you coded to try to solve that?

– Patrick Artner
Nov 13 '18 at 18:28





4




4





Please do not vandalize your posts. By posting on the Stack Exchange network, you've granted a non-revocable right for SE to distribute that content (under the CC BY-SA 3.0 license). By SE policy, any vandalism will be reverted.

– FrankerZ
Nov 19 '18 at 18:12





Please do not vandalize your posts. By posting on the Stack Exchange network, you've granted a non-revocable right for SE to distribute that content (under the CC BY-SA 3.0 license). By SE policy, any vandalism will be reverted.

– FrankerZ
Nov 19 '18 at 18:12












4 Answers
4






active

oldest

votes


















1














You can iterate through the viable rows and columns and zip the slices of big with small to calculate the sum of differences, and use min to find the minimum among the differences:



from itertools import islice
min(
(
sum(
sum(abs(x - y) for x, y in zip(a, b))
for a, b in zip(
(
islice(r, col, col + len(small[0]))
for r in islice(big, row, row + len(small))
),
small
)
),
(row, col)
)
for row in range(len(big) - len(small) + 1)
for col in range(len(big[0]) - len(small[0]) + 1)
)


or in one line:



min((sum(sum(abs(x - y) for x, y in zip(a, b)) for a, b in zip((islice(r, col, col + len(small[0])) for r in islice(big, row, row + len(small))), small)), (row, col)) for row in range(len(big) - len(small) + 1) for col in range(len(big[0]) - len(small[0]) + 1))


This returns: (24, (1, 0))






share|improve this answer

























  • pylint: line too long.

    – Muhammad Ahmad
    Nov 13 '18 at 19:16











  • I've updated the answer with a broken down version then.

    – blhsing
    Nov 13 '18 at 19:21











  • Like your solution! Although having some issues: for matrix

    – IHav
    Nov 13 '18 at 21:59






  • 1





    @IvanHlivan I see. I've put your sample data here: repl.it/repls/FrontFoolhardyExtraction and as you can see (by clicking on the Run button), the smallest difference does occur at (0, 16). How did you come to the conclusion that it should be (5, x)? And what is this "x" in (5, x) anyway?

    – blhsing
    Nov 13 '18 at 22:22






  • 1





    Sorry, my fault. Compared other results.

    – IHav
    Nov 13 '18 at 23:18


















0














Done by hand:



small=[[1,2,3],
[4,5,6],
[7,8,9]]


big=[[2,4,2,3,5],
[6,0,1,9,0],
[2,8,2,1,0],
[7,7,4,2,1]]

# collect all the sums
summs=

# k and j are the offset into big

for k in range(len(big)-len(small)+1):
# add inner list for one row
summs.append()
for j in range(len(big[0])-len(small[0])+1):
s = 0
for row in range(len(small)):
for col in range(len(small[0])):
s += abs(big[k+row][j+col]-small[row][col])
# add to the inner list
summs[-1].append(s)

print(summs)


Output:



[[28, 29, 38], [24, 31, 39]]


If you are just interested in the coords in the bigger one, store tuples of (rowoffset,coloffset,sum) and dont box lists into lists. You can use min() with a key that way:



summs = 
for k in range(len(big)-len(small)+1):
for j in range(len(big[0])-len(small[0])+1):
s = 0
for row in range(len(small)):
for col in range(len(small[0])):
s += abs(big[k+row][j+col]-small[row][col])
summs .append( (k,j,s) ) # row,col, sum

print ("Min value for bigger matrix at ", min(summs , key=lambda x:x[2]) )


Output:



Min value for bigger matrix at (1, 0, 24)


If you had "draws" this would only return the one with minimal row, col offset.






share|improve this answer
































    0














    Another possible solution would be this, returning the minimum difference and the coordinates in the big matrix:



    small=[[1,2,3],
    [4,5,6],
    [7,8,9]]

    big=[[2,4,2,3,5],
    [6,0,1,9,0],
    [2,8,2,1,0],
    [7,7,4,2,1]]

    def difference(small, matrix):
    l = len(small)
    return sum([abs(small[i][j] - matrix[i][j]) for i in range(l) for j in range(l)])

    def getSubmatrices(big, smallLength):
    submatrices =
    bigLength = len(big)
    step = (bigLength // smallLength) + 1
    for i in range(smallLength):
    for j in range(step):
    tempMatrix = [big[j+k][i:i+smallLength] for k in range(smallLength)]
    submatrices.append([i+1,j+1,tempMatrix])
    return submatrices

    def minDiff(small, big):
    submatrices = getSubmatrices(big, len(small))
    diffs = [(x,y, difference(small, submatrix)) for x, y, submatrix in submatrices]
    minDiff = min(diffs, key=lambda elem: elem[2])
    return minDiff

    y, x, diff = minDiff(small, big)

    print("Minimum difference: ", diff)
    print("X = ", x)
    print("Y = ", y)


    Output:



    Minimum difference: 24
    X = 1
    Y = 2





    share|improve this answer
































      0














      I would use numpy to help with this.



      To start I would convert the arrays to numpy arrays



      import numpy as np

      small = np.array([[1,2,3], [4,5,6], [7,8,9]])
      big = np.array([[2,4,2,3,5], [6,0,1,9,0], [2,8,2,1,0], [7,7,4,2,1]])


      then I would initialize an array to store the results of the test (optional: a dictionary as well)



      result_shape = np.array(big.shape) - np.array(small.shape) + 1
      results = np.zeros((result_shape[0], result_shape[1]))
      result_dict =


      Then iterate over the positions in which the small matrix can be positioned over the large matrix and calculate the difference:



      insert = np.zeros(big.shape)
      for i in range(results.shape[0]):
      for j in range(results.shape):
      insert[i:small.shape[0] + i, j:small.shape[1] + j] = small
      results[i, j] = np.sum(np.abs(big - insert)[i:3+i, j:3+j])
      # Optional dictionary
      result_dict[''.format(i, j)] = np.sum(np.abs(big - insert)[i:3+i, j:3+j])


      Then you can print(results) and obtain:



      [[ 28. 29. 38.]
      [ 24. 31. 39.]]


      and/or because the position of the small matrix over the big matrix is stored in the keys of the dictionary, you can get the position of the small matrix over the large matrix where the difference is smallest by key manipulation:



      pos_min = [int(i) for i in list(min(result_dict, key=result_dict.get))]


      and if you print(pos_min), you obtain:



      [1, 0]


      then if you need the index for anything you can iterate over it if required. Hope this helps!






      share|improve this answer
























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        4 Answers
        4






        active

        oldest

        votes








        4 Answers
        4






        active

        oldest

        votes









        active

        oldest

        votes






        active

        oldest

        votes









        1














        You can iterate through the viable rows and columns and zip the slices of big with small to calculate the sum of differences, and use min to find the minimum among the differences:



        from itertools import islice
        min(
        (
        sum(
        sum(abs(x - y) for x, y in zip(a, b))
        for a, b in zip(
        (
        islice(r, col, col + len(small[0]))
        for r in islice(big, row, row + len(small))
        ),
        small
        )
        ),
        (row, col)
        )
        for row in range(len(big) - len(small) + 1)
        for col in range(len(big[0]) - len(small[0]) + 1)
        )


        or in one line:



        min((sum(sum(abs(x - y) for x, y in zip(a, b)) for a, b in zip((islice(r, col, col + len(small[0])) for r in islice(big, row, row + len(small))), small)), (row, col)) for row in range(len(big) - len(small) + 1) for col in range(len(big[0]) - len(small[0]) + 1))


        This returns: (24, (1, 0))






        share|improve this answer

























        • pylint: line too long.

          – Muhammad Ahmad
          Nov 13 '18 at 19:16











        • I've updated the answer with a broken down version then.

          – blhsing
          Nov 13 '18 at 19:21











        • Like your solution! Although having some issues: for matrix

          – IHav
          Nov 13 '18 at 21:59






        • 1





          @IvanHlivan I see. I've put your sample data here: repl.it/repls/FrontFoolhardyExtraction and as you can see (by clicking on the Run button), the smallest difference does occur at (0, 16). How did you come to the conclusion that it should be (5, x)? And what is this "x" in (5, x) anyway?

          – blhsing
          Nov 13 '18 at 22:22






        • 1





          Sorry, my fault. Compared other results.

          – IHav
          Nov 13 '18 at 23:18















        1














        You can iterate through the viable rows and columns and zip the slices of big with small to calculate the sum of differences, and use min to find the minimum among the differences:



        from itertools import islice
        min(
        (
        sum(
        sum(abs(x - y) for x, y in zip(a, b))
        for a, b in zip(
        (
        islice(r, col, col + len(small[0]))
        for r in islice(big, row, row + len(small))
        ),
        small
        )
        ),
        (row, col)
        )
        for row in range(len(big) - len(small) + 1)
        for col in range(len(big[0]) - len(small[0]) + 1)
        )


        or in one line:



        min((sum(sum(abs(x - y) for x, y in zip(a, b)) for a, b in zip((islice(r, col, col + len(small[0])) for r in islice(big, row, row + len(small))), small)), (row, col)) for row in range(len(big) - len(small) + 1) for col in range(len(big[0]) - len(small[0]) + 1))


        This returns: (24, (1, 0))






        share|improve this answer

























        • pylint: line too long.

          – Muhammad Ahmad
          Nov 13 '18 at 19:16











        • I've updated the answer with a broken down version then.

          – blhsing
          Nov 13 '18 at 19:21











        • Like your solution! Although having some issues: for matrix

          – IHav
          Nov 13 '18 at 21:59






        • 1





          @IvanHlivan I see. I've put your sample data here: repl.it/repls/FrontFoolhardyExtraction and as you can see (by clicking on the Run button), the smallest difference does occur at (0, 16). How did you come to the conclusion that it should be (5, x)? And what is this "x" in (5, x) anyway?

          – blhsing
          Nov 13 '18 at 22:22






        • 1





          Sorry, my fault. Compared other results.

          – IHav
          Nov 13 '18 at 23:18













        1












        1








        1







        You can iterate through the viable rows and columns and zip the slices of big with small to calculate the sum of differences, and use min to find the minimum among the differences:



        from itertools import islice
        min(
        (
        sum(
        sum(abs(x - y) for x, y in zip(a, b))
        for a, b in zip(
        (
        islice(r, col, col + len(small[0]))
        for r in islice(big, row, row + len(small))
        ),
        small
        )
        ),
        (row, col)
        )
        for row in range(len(big) - len(small) + 1)
        for col in range(len(big[0]) - len(small[0]) + 1)
        )


        or in one line:



        min((sum(sum(abs(x - y) for x, y in zip(a, b)) for a, b in zip((islice(r, col, col + len(small[0])) for r in islice(big, row, row + len(small))), small)), (row, col)) for row in range(len(big) - len(small) + 1) for col in range(len(big[0]) - len(small[0]) + 1))


        This returns: (24, (1, 0))






        share|improve this answer















        You can iterate through the viable rows and columns and zip the slices of big with small to calculate the sum of differences, and use min to find the minimum among the differences:



        from itertools import islice
        min(
        (
        sum(
        sum(abs(x - y) for x, y in zip(a, b))
        for a, b in zip(
        (
        islice(r, col, col + len(small[0]))
        for r in islice(big, row, row + len(small))
        ),
        small
        )
        ),
        (row, col)
        )
        for row in range(len(big) - len(small) + 1)
        for col in range(len(big[0]) - len(small[0]) + 1)
        )


        or in one line:



        min((sum(sum(abs(x - y) for x, y in zip(a, b)) for a, b in zip((islice(r, col, col + len(small[0])) for r in islice(big, row, row + len(small))), small)), (row, col)) for row in range(len(big) - len(small) + 1) for col in range(len(big[0]) - len(small[0]) + 1))


        This returns: (24, (1, 0))







        share|improve this answer














        share|improve this answer



        share|improve this answer








        edited Nov 13 '18 at 22:03

























        answered Nov 13 '18 at 19:06









        blhsingblhsing

        30k41336




        30k41336












        • pylint: line too long.

          – Muhammad Ahmad
          Nov 13 '18 at 19:16











        • I've updated the answer with a broken down version then.

          – blhsing
          Nov 13 '18 at 19:21











        • Like your solution! Although having some issues: for matrix

          – IHav
          Nov 13 '18 at 21:59






        • 1





          @IvanHlivan I see. I've put your sample data here: repl.it/repls/FrontFoolhardyExtraction and as you can see (by clicking on the Run button), the smallest difference does occur at (0, 16). How did you come to the conclusion that it should be (5, x)? And what is this "x" in (5, x) anyway?

          – blhsing
          Nov 13 '18 at 22:22






        • 1





          Sorry, my fault. Compared other results.

          – IHav
          Nov 13 '18 at 23:18

















        • pylint: line too long.

          – Muhammad Ahmad
          Nov 13 '18 at 19:16











        • I've updated the answer with a broken down version then.

          – blhsing
          Nov 13 '18 at 19:21











        • Like your solution! Although having some issues: for matrix

          – IHav
          Nov 13 '18 at 21:59






        • 1





          @IvanHlivan I see. I've put your sample data here: repl.it/repls/FrontFoolhardyExtraction and as you can see (by clicking on the Run button), the smallest difference does occur at (0, 16). How did you come to the conclusion that it should be (5, x)? And what is this "x" in (5, x) anyway?

          – blhsing
          Nov 13 '18 at 22:22






        • 1





          Sorry, my fault. Compared other results.

          – IHav
          Nov 13 '18 at 23:18
















        pylint: line too long.

        – Muhammad Ahmad
        Nov 13 '18 at 19:16





        pylint: line too long.

        – Muhammad Ahmad
        Nov 13 '18 at 19:16













        I've updated the answer with a broken down version then.

        – blhsing
        Nov 13 '18 at 19:21





        I've updated the answer with a broken down version then.

        – blhsing
        Nov 13 '18 at 19:21













        Like your solution! Although having some issues: for matrix

        – IHav
        Nov 13 '18 at 21:59





        Like your solution! Although having some issues: for matrix

        – IHav
        Nov 13 '18 at 21:59




        1




        1





        @IvanHlivan I see. I've put your sample data here: repl.it/repls/FrontFoolhardyExtraction and as you can see (by clicking on the Run button), the smallest difference does occur at (0, 16). How did you come to the conclusion that it should be (5, x)? And what is this "x" in (5, x) anyway?

        – blhsing
        Nov 13 '18 at 22:22





        @IvanHlivan I see. I've put your sample data here: repl.it/repls/FrontFoolhardyExtraction and as you can see (by clicking on the Run button), the smallest difference does occur at (0, 16). How did you come to the conclusion that it should be (5, x)? And what is this "x" in (5, x) anyway?

        – blhsing
        Nov 13 '18 at 22:22




        1




        1





        Sorry, my fault. Compared other results.

        – IHav
        Nov 13 '18 at 23:18





        Sorry, my fault. Compared other results.

        – IHav
        Nov 13 '18 at 23:18













        0














        Done by hand:



        small=[[1,2,3],
        [4,5,6],
        [7,8,9]]


        big=[[2,4,2,3,5],
        [6,0,1,9,0],
        [2,8,2,1,0],
        [7,7,4,2,1]]

        # collect all the sums
        summs=

        # k and j are the offset into big

        for k in range(len(big)-len(small)+1):
        # add inner list for one row
        summs.append()
        for j in range(len(big[0])-len(small[0])+1):
        s = 0
        for row in range(len(small)):
        for col in range(len(small[0])):
        s += abs(big[k+row][j+col]-small[row][col])
        # add to the inner list
        summs[-1].append(s)

        print(summs)


        Output:



        [[28, 29, 38], [24, 31, 39]]


        If you are just interested in the coords in the bigger one, store tuples of (rowoffset,coloffset,sum) and dont box lists into lists. You can use min() with a key that way:



        summs = 
        for k in range(len(big)-len(small)+1):
        for j in range(len(big[0])-len(small[0])+1):
        s = 0
        for row in range(len(small)):
        for col in range(len(small[0])):
        s += abs(big[k+row][j+col]-small[row][col])
        summs .append( (k,j,s) ) # row,col, sum

        print ("Min value for bigger matrix at ", min(summs , key=lambda x:x[2]) )


        Output:



        Min value for bigger matrix at (1, 0, 24)


        If you had "draws" this would only return the one with minimal row, col offset.






        share|improve this answer





























          0














          Done by hand:



          small=[[1,2,3],
          [4,5,6],
          [7,8,9]]


          big=[[2,4,2,3,5],
          [6,0,1,9,0],
          [2,8,2,1,0],
          [7,7,4,2,1]]

          # collect all the sums
          summs=

          # k and j are the offset into big

          for k in range(len(big)-len(small)+1):
          # add inner list for one row
          summs.append()
          for j in range(len(big[0])-len(small[0])+1):
          s = 0
          for row in range(len(small)):
          for col in range(len(small[0])):
          s += abs(big[k+row][j+col]-small[row][col])
          # add to the inner list
          summs[-1].append(s)

          print(summs)


          Output:



          [[28, 29, 38], [24, 31, 39]]


          If you are just interested in the coords in the bigger one, store tuples of (rowoffset,coloffset,sum) and dont box lists into lists. You can use min() with a key that way:



          summs = 
          for k in range(len(big)-len(small)+1):
          for j in range(len(big[0])-len(small[0])+1):
          s = 0
          for row in range(len(small)):
          for col in range(len(small[0])):
          s += abs(big[k+row][j+col]-small[row][col])
          summs .append( (k,j,s) ) # row,col, sum

          print ("Min value for bigger matrix at ", min(summs , key=lambda x:x[2]) )


          Output:



          Min value for bigger matrix at (1, 0, 24)


          If you had "draws" this would only return the one with minimal row, col offset.






          share|improve this answer



























            0












            0








            0







            Done by hand:



            small=[[1,2,3],
            [4,5,6],
            [7,8,9]]


            big=[[2,4,2,3,5],
            [6,0,1,9,0],
            [2,8,2,1,0],
            [7,7,4,2,1]]

            # collect all the sums
            summs=

            # k and j are the offset into big

            for k in range(len(big)-len(small)+1):
            # add inner list for one row
            summs.append()
            for j in range(len(big[0])-len(small[0])+1):
            s = 0
            for row in range(len(small)):
            for col in range(len(small[0])):
            s += abs(big[k+row][j+col]-small[row][col])
            # add to the inner list
            summs[-1].append(s)

            print(summs)


            Output:



            [[28, 29, 38], [24, 31, 39]]


            If you are just interested in the coords in the bigger one, store tuples of (rowoffset,coloffset,sum) and dont box lists into lists. You can use min() with a key that way:



            summs = 
            for k in range(len(big)-len(small)+1):
            for j in range(len(big[0])-len(small[0])+1):
            s = 0
            for row in range(len(small)):
            for col in range(len(small[0])):
            s += abs(big[k+row][j+col]-small[row][col])
            summs .append( (k,j,s) ) # row,col, sum

            print ("Min value for bigger matrix at ", min(summs , key=lambda x:x[2]) )


            Output:



            Min value for bigger matrix at (1, 0, 24)


            If you had "draws" this would only return the one with minimal row, col offset.






            share|improve this answer















            Done by hand:



            small=[[1,2,3],
            [4,5,6],
            [7,8,9]]


            big=[[2,4,2,3,5],
            [6,0,1,9,0],
            [2,8,2,1,0],
            [7,7,4,2,1]]

            # collect all the sums
            summs=

            # k and j are the offset into big

            for k in range(len(big)-len(small)+1):
            # add inner list for one row
            summs.append()
            for j in range(len(big[0])-len(small[0])+1):
            s = 0
            for row in range(len(small)):
            for col in range(len(small[0])):
            s += abs(big[k+row][j+col]-small[row][col])
            # add to the inner list
            summs[-1].append(s)

            print(summs)


            Output:



            [[28, 29, 38], [24, 31, 39]]


            If you are just interested in the coords in the bigger one, store tuples of (rowoffset,coloffset,sum) and dont box lists into lists. You can use min() with a key that way:



            summs = 
            for k in range(len(big)-len(small)+1):
            for j in range(len(big[0])-len(small[0])+1):
            s = 0
            for row in range(len(small)):
            for col in range(len(small[0])):
            s += abs(big[k+row][j+col]-small[row][col])
            summs .append( (k,j,s) ) # row,col, sum

            print ("Min value for bigger matrix at ", min(summs , key=lambda x:x[2]) )


            Output:



            Min value for bigger matrix at (1, 0, 24)


            If you had "draws" this would only return the one with minimal row, col offset.







            share|improve this answer














            share|improve this answer



            share|improve this answer








            edited Nov 13 '18 at 18:48

























            answered Nov 13 '18 at 18:40









            Patrick ArtnerPatrick Artner

            23.2k62343




            23.2k62343





















                0














                Another possible solution would be this, returning the minimum difference and the coordinates in the big matrix:



                small=[[1,2,3],
                [4,5,6],
                [7,8,9]]

                big=[[2,4,2,3,5],
                [6,0,1,9,0],
                [2,8,2,1,0],
                [7,7,4,2,1]]

                def difference(small, matrix):
                l = len(small)
                return sum([abs(small[i][j] - matrix[i][j]) for i in range(l) for j in range(l)])

                def getSubmatrices(big, smallLength):
                submatrices =
                bigLength = len(big)
                step = (bigLength // smallLength) + 1
                for i in range(smallLength):
                for j in range(step):
                tempMatrix = [big[j+k][i:i+smallLength] for k in range(smallLength)]
                submatrices.append([i+1,j+1,tempMatrix])
                return submatrices

                def minDiff(small, big):
                submatrices = getSubmatrices(big, len(small))
                diffs = [(x,y, difference(small, submatrix)) for x, y, submatrix in submatrices]
                minDiff = min(diffs, key=lambda elem: elem[2])
                return minDiff

                y, x, diff = minDiff(small, big)

                print("Minimum difference: ", diff)
                print("X = ", x)
                print("Y = ", y)


                Output:



                Minimum difference: 24
                X = 1
                Y = 2





                share|improve this answer





























                  0














                  Another possible solution would be this, returning the minimum difference and the coordinates in the big matrix:



                  small=[[1,2,3],
                  [4,5,6],
                  [7,8,9]]

                  big=[[2,4,2,3,5],
                  [6,0,1,9,0],
                  [2,8,2,1,0],
                  [7,7,4,2,1]]

                  def difference(small, matrix):
                  l = len(small)
                  return sum([abs(small[i][j] - matrix[i][j]) for i in range(l) for j in range(l)])

                  def getSubmatrices(big, smallLength):
                  submatrices =
                  bigLength = len(big)
                  step = (bigLength // smallLength) + 1
                  for i in range(smallLength):
                  for j in range(step):
                  tempMatrix = [big[j+k][i:i+smallLength] for k in range(smallLength)]
                  submatrices.append([i+1,j+1,tempMatrix])
                  return submatrices

                  def minDiff(small, big):
                  submatrices = getSubmatrices(big, len(small))
                  diffs = [(x,y, difference(small, submatrix)) for x, y, submatrix in submatrices]
                  minDiff = min(diffs, key=lambda elem: elem[2])
                  return minDiff

                  y, x, diff = minDiff(small, big)

                  print("Minimum difference: ", diff)
                  print("X = ", x)
                  print("Y = ", y)


                  Output:



                  Minimum difference: 24
                  X = 1
                  Y = 2





                  share|improve this answer



























                    0












                    0








                    0







                    Another possible solution would be this, returning the minimum difference and the coordinates in the big matrix:



                    small=[[1,2,3],
                    [4,5,6],
                    [7,8,9]]

                    big=[[2,4,2,3,5],
                    [6,0,1,9,0],
                    [2,8,2,1,0],
                    [7,7,4,2,1]]

                    def difference(small, matrix):
                    l = len(small)
                    return sum([abs(small[i][j] - matrix[i][j]) for i in range(l) for j in range(l)])

                    def getSubmatrices(big, smallLength):
                    submatrices =
                    bigLength = len(big)
                    step = (bigLength // smallLength) + 1
                    for i in range(smallLength):
                    for j in range(step):
                    tempMatrix = [big[j+k][i:i+smallLength] for k in range(smallLength)]
                    submatrices.append([i+1,j+1,tempMatrix])
                    return submatrices

                    def minDiff(small, big):
                    submatrices = getSubmatrices(big, len(small))
                    diffs = [(x,y, difference(small, submatrix)) for x, y, submatrix in submatrices]
                    minDiff = min(diffs, key=lambda elem: elem[2])
                    return minDiff

                    y, x, diff = minDiff(small, big)

                    print("Minimum difference: ", diff)
                    print("X = ", x)
                    print("Y = ", y)


                    Output:



                    Minimum difference: 24
                    X = 1
                    Y = 2





                    share|improve this answer















                    Another possible solution would be this, returning the minimum difference and the coordinates in the big matrix:



                    small=[[1,2,3],
                    [4,5,6],
                    [7,8,9]]

                    big=[[2,4,2,3,5],
                    [6,0,1,9,0],
                    [2,8,2,1,0],
                    [7,7,4,2,1]]

                    def difference(small, matrix):
                    l = len(small)
                    return sum([abs(small[i][j] - matrix[i][j]) for i in range(l) for j in range(l)])

                    def getSubmatrices(big, smallLength):
                    submatrices =
                    bigLength = len(big)
                    step = (bigLength // smallLength) + 1
                    for i in range(smallLength):
                    for j in range(step):
                    tempMatrix = [big[j+k][i:i+smallLength] for k in range(smallLength)]
                    submatrices.append([i+1,j+1,tempMatrix])
                    return submatrices

                    def minDiff(small, big):
                    submatrices = getSubmatrices(big, len(small))
                    diffs = [(x,y, difference(small, submatrix)) for x, y, submatrix in submatrices]
                    minDiff = min(diffs, key=lambda elem: elem[2])
                    return minDiff

                    y, x, diff = minDiff(small, big)

                    print("Minimum difference: ", diff)
                    print("X = ", x)
                    print("Y = ", y)


                    Output:



                    Minimum difference: 24
                    X = 1
                    Y = 2






                    share|improve this answer














                    share|improve this answer



                    share|improve this answer








                    edited Nov 13 '18 at 18:57

























                    answered Nov 13 '18 at 18:52









                    Vasilis G.Vasilis G.

                    3,5022722




                    3,5022722





















                        0














                        I would use numpy to help with this.



                        To start I would convert the arrays to numpy arrays



                        import numpy as np

                        small = np.array([[1,2,3], [4,5,6], [7,8,9]])
                        big = np.array([[2,4,2,3,5], [6,0,1,9,0], [2,8,2,1,0], [7,7,4,2,1]])


                        then I would initialize an array to store the results of the test (optional: a dictionary as well)



                        result_shape = np.array(big.shape) - np.array(small.shape) + 1
                        results = np.zeros((result_shape[0], result_shape[1]))
                        result_dict =


                        Then iterate over the positions in which the small matrix can be positioned over the large matrix and calculate the difference:



                        insert = np.zeros(big.shape)
                        for i in range(results.shape[0]):
                        for j in range(results.shape):
                        insert[i:small.shape[0] + i, j:small.shape[1] + j] = small
                        results[i, j] = np.sum(np.abs(big - insert)[i:3+i, j:3+j])
                        # Optional dictionary
                        result_dict[''.format(i, j)] = np.sum(np.abs(big - insert)[i:3+i, j:3+j])


                        Then you can print(results) and obtain:



                        [[ 28. 29. 38.]
                        [ 24. 31. 39.]]


                        and/or because the position of the small matrix over the big matrix is stored in the keys of the dictionary, you can get the position of the small matrix over the large matrix where the difference is smallest by key manipulation:



                        pos_min = [int(i) for i in list(min(result_dict, key=result_dict.get))]


                        and if you print(pos_min), you obtain:



                        [1, 0]


                        then if you need the index for anything you can iterate over it if required. Hope this helps!






                        share|improve this answer





























                          0














                          I would use numpy to help with this.



                          To start I would convert the arrays to numpy arrays



                          import numpy as np

                          small = np.array([[1,2,3], [4,5,6], [7,8,9]])
                          big = np.array([[2,4,2,3,5], [6,0,1,9,0], [2,8,2,1,0], [7,7,4,2,1]])


                          then I would initialize an array to store the results of the test (optional: a dictionary as well)



                          result_shape = np.array(big.shape) - np.array(small.shape) + 1
                          results = np.zeros((result_shape[0], result_shape[1]))
                          result_dict =


                          Then iterate over the positions in which the small matrix can be positioned over the large matrix and calculate the difference:



                          insert = np.zeros(big.shape)
                          for i in range(results.shape[0]):
                          for j in range(results.shape):
                          insert[i:small.shape[0] + i, j:small.shape[1] + j] = small
                          results[i, j] = np.sum(np.abs(big - insert)[i:3+i, j:3+j])
                          # Optional dictionary
                          result_dict[''.format(i, j)] = np.sum(np.abs(big - insert)[i:3+i, j:3+j])


                          Then you can print(results) and obtain:



                          [[ 28. 29. 38.]
                          [ 24. 31. 39.]]


                          and/or because the position of the small matrix over the big matrix is stored in the keys of the dictionary, you can get the position of the small matrix over the large matrix where the difference is smallest by key manipulation:



                          pos_min = [int(i) for i in list(min(result_dict, key=result_dict.get))]


                          and if you print(pos_min), you obtain:



                          [1, 0]


                          then if you need the index for anything you can iterate over it if required. Hope this helps!






                          share|improve this answer



























                            0












                            0








                            0







                            I would use numpy to help with this.



                            To start I would convert the arrays to numpy arrays



                            import numpy as np

                            small = np.array([[1,2,3], [4,5,6], [7,8,9]])
                            big = np.array([[2,4,2,3,5], [6,0,1,9,0], [2,8,2,1,0], [7,7,4,2,1]])


                            then I would initialize an array to store the results of the test (optional: a dictionary as well)



                            result_shape = np.array(big.shape) - np.array(small.shape) + 1
                            results = np.zeros((result_shape[0], result_shape[1]))
                            result_dict =


                            Then iterate over the positions in which the small matrix can be positioned over the large matrix and calculate the difference:



                            insert = np.zeros(big.shape)
                            for i in range(results.shape[0]):
                            for j in range(results.shape):
                            insert[i:small.shape[0] + i, j:small.shape[1] + j] = small
                            results[i, j] = np.sum(np.abs(big - insert)[i:3+i, j:3+j])
                            # Optional dictionary
                            result_dict[''.format(i, j)] = np.sum(np.abs(big - insert)[i:3+i, j:3+j])


                            Then you can print(results) and obtain:



                            [[ 28. 29. 38.]
                            [ 24. 31. 39.]]


                            and/or because the position of the small matrix over the big matrix is stored in the keys of the dictionary, you can get the position of the small matrix over the large matrix where the difference is smallest by key manipulation:



                            pos_min = [int(i) for i in list(min(result_dict, key=result_dict.get))]


                            and if you print(pos_min), you obtain:



                            [1, 0]


                            then if you need the index for anything you can iterate over it if required. Hope this helps!






                            share|improve this answer















                            I would use numpy to help with this.



                            To start I would convert the arrays to numpy arrays



                            import numpy as np

                            small = np.array([[1,2,3], [4,5,6], [7,8,9]])
                            big = np.array([[2,4,2,3,5], [6,0,1,9,0], [2,8,2,1,0], [7,7,4,2,1]])


                            then I would initialize an array to store the results of the test (optional: a dictionary as well)



                            result_shape = np.array(big.shape) - np.array(small.shape) + 1
                            results = np.zeros((result_shape[0], result_shape[1]))
                            result_dict =


                            Then iterate over the positions in which the small matrix can be positioned over the large matrix and calculate the difference:



                            insert = np.zeros(big.shape)
                            for i in range(results.shape[0]):
                            for j in range(results.shape):
                            insert[i:small.shape[0] + i, j:small.shape[1] + j] = small
                            results[i, j] = np.sum(np.abs(big - insert)[i:3+i, j:3+j])
                            # Optional dictionary
                            result_dict[''.format(i, j)] = np.sum(np.abs(big - insert)[i:3+i, j:3+j])


                            Then you can print(results) and obtain:



                            [[ 28. 29. 38.]
                            [ 24. 31. 39.]]


                            and/or because the position of the small matrix over the big matrix is stored in the keys of the dictionary, you can get the position of the small matrix over the large matrix where the difference is smallest by key manipulation:



                            pos_min = [int(i) for i in list(min(result_dict, key=result_dict.get))]


                            and if you print(pos_min), you obtain:



                            [1, 0]


                            then if you need the index for anything you can iterate over it if required. Hope this helps!







                            share|improve this answer














                            share|improve this answer



                            share|improve this answer








                            edited Nov 13 '18 at 20:21

























                            answered Nov 13 '18 at 19:53









                            J.AlukoJ.Aluko

                            163




                            163



























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