How to undo or reverse np.meshgrid?











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In an image resizing interpolation problem, one could use np.meshgrid on row and col indices before operating on the meshed indices:



nrows = 600
ncols = 800
image_in = np.random.randint(0, 256, size=(nrows, ncols, 3))
scale_factor = 1.5

r = np.arange(nrows, dtype=float) * scale_factor
c = np.arange(ncols, dtype=float) * scale_factor

rr, cc = np.meshgrid(r, c, indexing='ij')

# Nearest Neighbor Interpolation
# np.floor if scale_factor >= 1. np.ceil otherwise
rr = np.floor(rr).astype(int).clip(0, nrows-1)
cc = np.floor(cc).astype(int).clip(0, ncols-1)

image_out = image_in[rr, cc, :]


Now, how would I reverse this process? Say given rr_1, cc_1 (product of np.meshgrid) that's processed in an unknown manner (here illustrated by np.random.randint), how do I get the r_1 and c_1, that is, the inputs to np.meshgrid (preferably with ij indexing)?



# Suppose rr_1, cc_1 = np.meshgrid(r_1, c_1, indexing='ij')
rr_1 = np.random.randint(0, nrows, size=(nrows, ncols, 3))
cc_1 = np.random.randint(0, ncols, size=(nrows, ncols, 3))

r_1 = ?
c_1 = ?


UPDATE:



I figured it out immediately after posting. The answer is:



# Suppose rr_1, cc_1 = np.meshgrid(r_1, c_1, indexing='ij')
rr_1 = np.random.randint(0, nrows, size=(nrows, ncols, 3))
cc_1 = np.random.randint(0, ncols, size=(nrows, ncols, 3))

r_1 = rr_1[:, 0]
c_1 = cc_1[0]









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    up vote
    1
    down vote

    favorite












    In an image resizing interpolation problem, one could use np.meshgrid on row and col indices before operating on the meshed indices:



    nrows = 600
    ncols = 800
    image_in = np.random.randint(0, 256, size=(nrows, ncols, 3))
    scale_factor = 1.5

    r = np.arange(nrows, dtype=float) * scale_factor
    c = np.arange(ncols, dtype=float) * scale_factor

    rr, cc = np.meshgrid(r, c, indexing='ij')

    # Nearest Neighbor Interpolation
    # np.floor if scale_factor >= 1. np.ceil otherwise
    rr = np.floor(rr).astype(int).clip(0, nrows-1)
    cc = np.floor(cc).astype(int).clip(0, ncols-1)

    image_out = image_in[rr, cc, :]


    Now, how would I reverse this process? Say given rr_1, cc_1 (product of np.meshgrid) that's processed in an unknown manner (here illustrated by np.random.randint), how do I get the r_1 and c_1, that is, the inputs to np.meshgrid (preferably with ij indexing)?



    # Suppose rr_1, cc_1 = np.meshgrid(r_1, c_1, indexing='ij')
    rr_1 = np.random.randint(0, nrows, size=(nrows, ncols, 3))
    cc_1 = np.random.randint(0, ncols, size=(nrows, ncols, 3))

    r_1 = ?
    c_1 = ?


    UPDATE:



    I figured it out immediately after posting. The answer is:



    # Suppose rr_1, cc_1 = np.meshgrid(r_1, c_1, indexing='ij')
    rr_1 = np.random.randint(0, nrows, size=(nrows, ncols, 3))
    cc_1 = np.random.randint(0, ncols, size=(nrows, ncols, 3))

    r_1 = rr_1[:, 0]
    c_1 = cc_1[0]









    share|improve this question


























      up vote
      1
      down vote

      favorite









      up vote
      1
      down vote

      favorite











      In an image resizing interpolation problem, one could use np.meshgrid on row and col indices before operating on the meshed indices:



      nrows = 600
      ncols = 800
      image_in = np.random.randint(0, 256, size=(nrows, ncols, 3))
      scale_factor = 1.5

      r = np.arange(nrows, dtype=float) * scale_factor
      c = np.arange(ncols, dtype=float) * scale_factor

      rr, cc = np.meshgrid(r, c, indexing='ij')

      # Nearest Neighbor Interpolation
      # np.floor if scale_factor >= 1. np.ceil otherwise
      rr = np.floor(rr).astype(int).clip(0, nrows-1)
      cc = np.floor(cc).astype(int).clip(0, ncols-1)

      image_out = image_in[rr, cc, :]


      Now, how would I reverse this process? Say given rr_1, cc_1 (product of np.meshgrid) that's processed in an unknown manner (here illustrated by np.random.randint), how do I get the r_1 and c_1, that is, the inputs to np.meshgrid (preferably with ij indexing)?



      # Suppose rr_1, cc_1 = np.meshgrid(r_1, c_1, indexing='ij')
      rr_1 = np.random.randint(0, nrows, size=(nrows, ncols, 3))
      cc_1 = np.random.randint(0, ncols, size=(nrows, ncols, 3))

      r_1 = ?
      c_1 = ?


      UPDATE:



      I figured it out immediately after posting. The answer is:



      # Suppose rr_1, cc_1 = np.meshgrid(r_1, c_1, indexing='ij')
      rr_1 = np.random.randint(0, nrows, size=(nrows, ncols, 3))
      cc_1 = np.random.randint(0, ncols, size=(nrows, ncols, 3))

      r_1 = rr_1[:, 0]
      c_1 = cc_1[0]









      share|improve this question















      In an image resizing interpolation problem, one could use np.meshgrid on row and col indices before operating on the meshed indices:



      nrows = 600
      ncols = 800
      image_in = np.random.randint(0, 256, size=(nrows, ncols, 3))
      scale_factor = 1.5

      r = np.arange(nrows, dtype=float) * scale_factor
      c = np.arange(ncols, dtype=float) * scale_factor

      rr, cc = np.meshgrid(r, c, indexing='ij')

      # Nearest Neighbor Interpolation
      # np.floor if scale_factor >= 1. np.ceil otherwise
      rr = np.floor(rr).astype(int).clip(0, nrows-1)
      cc = np.floor(cc).astype(int).clip(0, ncols-1)

      image_out = image_in[rr, cc, :]


      Now, how would I reverse this process? Say given rr_1, cc_1 (product of np.meshgrid) that's processed in an unknown manner (here illustrated by np.random.randint), how do I get the r_1 and c_1, that is, the inputs to np.meshgrid (preferably with ij indexing)?



      # Suppose rr_1, cc_1 = np.meshgrid(r_1, c_1, indexing='ij')
      rr_1 = np.random.randint(0, nrows, size=(nrows, ncols, 3))
      cc_1 = np.random.randint(0, ncols, size=(nrows, ncols, 3))

      r_1 = ?
      c_1 = ?


      UPDATE:



      I figured it out immediately after posting. The answer is:



      # Suppose rr_1, cc_1 = np.meshgrid(r_1, c_1, indexing='ij')
      rr_1 = np.random.randint(0, nrows, size=(nrows, ncols, 3))
      cc_1 = np.random.randint(0, ncols, size=(nrows, ncols, 3))

      r_1 = rr_1[:, 0]
      c_1 = cc_1[0]






      python numpy image-processing






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      edited Nov 20 at 4:57









      Aqueous Carlos

      295213




      295213










      asked Nov 20 at 3:03









      Zhanwen Chen

      8019




      8019
























          1 Answer
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          The numpy.meshgrid creates a higher dimensional array from input arrays in order to create grid-like arrays. So imagine you want to get a 2D grid by using some input 1D vectors r and c. numpy.meshgrid returns rr and cc as 2D arrays which respectively hold the y axis or x axis constant everywhere on the 2D array (this is why it is a grid).



          Here is a test case:



          import numpy as np

          r = np.arange(5) # [0 1 2 3 4]
          c = np.arange(5,10,1) # [5 6 7 8 9]

          rr, cc = np.meshgrid(r,c,indexing='ij')

          r_original = rr[:,0]
          c_original = cc[0,:]

          print(r_original) # [0 1 2 3 4]
          print(c_original) # [5 6 7 8 9]


          Note that the grids we have created for rr and cc are



          rr = [[0 0 0 0 0]
          [1 1 1 1 1]
          [2 2 2 2 2]
          [3 3 3 3 3]
          [4 4 4 4 4]]

          cc = [[5 6 7 8 9]
          [5 6 7 8 9]
          [5 6 7 8 9]
          [5 6 7 8 9]
          [5 6 7 8 9]]


          Since you are using indexing='ij' in your case and the 2D arrays are transposed. Hence, the values that hold constant for rr and cc respectively are the x axis and y axis (contrary to the case where you do not use indexing='ij').






          share|improve this answer























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            1 Answer
            1






            active

            oldest

            votes








            1 Answer
            1






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes








            up vote
            1
            down vote



            accepted










            The numpy.meshgrid creates a higher dimensional array from input arrays in order to create grid-like arrays. So imagine you want to get a 2D grid by using some input 1D vectors r and c. numpy.meshgrid returns rr and cc as 2D arrays which respectively hold the y axis or x axis constant everywhere on the 2D array (this is why it is a grid).



            Here is a test case:



            import numpy as np

            r = np.arange(5) # [0 1 2 3 4]
            c = np.arange(5,10,1) # [5 6 7 8 9]

            rr, cc = np.meshgrid(r,c,indexing='ij')

            r_original = rr[:,0]
            c_original = cc[0,:]

            print(r_original) # [0 1 2 3 4]
            print(c_original) # [5 6 7 8 9]


            Note that the grids we have created for rr and cc are



            rr = [[0 0 0 0 0]
            [1 1 1 1 1]
            [2 2 2 2 2]
            [3 3 3 3 3]
            [4 4 4 4 4]]

            cc = [[5 6 7 8 9]
            [5 6 7 8 9]
            [5 6 7 8 9]
            [5 6 7 8 9]
            [5 6 7 8 9]]


            Since you are using indexing='ij' in your case and the 2D arrays are transposed. Hence, the values that hold constant for rr and cc respectively are the x axis and y axis (contrary to the case where you do not use indexing='ij').






            share|improve this answer



























              up vote
              1
              down vote



              accepted










              The numpy.meshgrid creates a higher dimensional array from input arrays in order to create grid-like arrays. So imagine you want to get a 2D grid by using some input 1D vectors r and c. numpy.meshgrid returns rr and cc as 2D arrays which respectively hold the y axis or x axis constant everywhere on the 2D array (this is why it is a grid).



              Here is a test case:



              import numpy as np

              r = np.arange(5) # [0 1 2 3 4]
              c = np.arange(5,10,1) # [5 6 7 8 9]

              rr, cc = np.meshgrid(r,c,indexing='ij')

              r_original = rr[:,0]
              c_original = cc[0,:]

              print(r_original) # [0 1 2 3 4]
              print(c_original) # [5 6 7 8 9]


              Note that the grids we have created for rr and cc are



              rr = [[0 0 0 0 0]
              [1 1 1 1 1]
              [2 2 2 2 2]
              [3 3 3 3 3]
              [4 4 4 4 4]]

              cc = [[5 6 7 8 9]
              [5 6 7 8 9]
              [5 6 7 8 9]
              [5 6 7 8 9]
              [5 6 7 8 9]]


              Since you are using indexing='ij' in your case and the 2D arrays are transposed. Hence, the values that hold constant for rr and cc respectively are the x axis and y axis (contrary to the case where you do not use indexing='ij').






              share|improve this answer

























                up vote
                1
                down vote



                accepted







                up vote
                1
                down vote



                accepted






                The numpy.meshgrid creates a higher dimensional array from input arrays in order to create grid-like arrays. So imagine you want to get a 2D grid by using some input 1D vectors r and c. numpy.meshgrid returns rr and cc as 2D arrays which respectively hold the y axis or x axis constant everywhere on the 2D array (this is why it is a grid).



                Here is a test case:



                import numpy as np

                r = np.arange(5) # [0 1 2 3 4]
                c = np.arange(5,10,1) # [5 6 7 8 9]

                rr, cc = np.meshgrid(r,c,indexing='ij')

                r_original = rr[:,0]
                c_original = cc[0,:]

                print(r_original) # [0 1 2 3 4]
                print(c_original) # [5 6 7 8 9]


                Note that the grids we have created for rr and cc are



                rr = [[0 0 0 0 0]
                [1 1 1 1 1]
                [2 2 2 2 2]
                [3 3 3 3 3]
                [4 4 4 4 4]]

                cc = [[5 6 7 8 9]
                [5 6 7 8 9]
                [5 6 7 8 9]
                [5 6 7 8 9]
                [5 6 7 8 9]]


                Since you are using indexing='ij' in your case and the 2D arrays are transposed. Hence, the values that hold constant for rr and cc respectively are the x axis and y axis (contrary to the case where you do not use indexing='ij').






                share|improve this answer














                The numpy.meshgrid creates a higher dimensional array from input arrays in order to create grid-like arrays. So imagine you want to get a 2D grid by using some input 1D vectors r and c. numpy.meshgrid returns rr and cc as 2D arrays which respectively hold the y axis or x axis constant everywhere on the 2D array (this is why it is a grid).



                Here is a test case:



                import numpy as np

                r = np.arange(5) # [0 1 2 3 4]
                c = np.arange(5,10,1) # [5 6 7 8 9]

                rr, cc = np.meshgrid(r,c,indexing='ij')

                r_original = rr[:,0]
                c_original = cc[0,:]

                print(r_original) # [0 1 2 3 4]
                print(c_original) # [5 6 7 8 9]


                Note that the grids we have created for rr and cc are



                rr = [[0 0 0 0 0]
                [1 1 1 1 1]
                [2 2 2 2 2]
                [3 3 3 3 3]
                [4 4 4 4 4]]

                cc = [[5 6 7 8 9]
                [5 6 7 8 9]
                [5 6 7 8 9]
                [5 6 7 8 9]
                [5 6 7 8 9]]


                Since you are using indexing='ij' in your case and the 2D arrays are transposed. Hence, the values that hold constant for rr and cc respectively are the x axis and y axis (contrary to the case where you do not use indexing='ij').







                share|improve this answer














                share|improve this answer



                share|improve this answer








                edited Nov 20 at 3:38

























                answered Nov 20 at 3:19









                b-fg

                1,80111422




                1,80111422






























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