numpy vectorize / more efficient for loop












1















I am performing erosion on an image. The image has been padded accordingly.
In a nutshell, I have a cross element(+) that I put on every pixel of the image and pick the lowest value for that pixel from pixel above, below, right, left and itself.



It is inefficient and I can't figure out a vectorized version. It must be possible since all calculations are done independently of each other.



for y in range(t,image.shape[0]-b):
for x in range(l,image.shape[1]-r):
a1 = numpy.copy(str_ele)
for filter_y in range(a1.shape[0]):
for filter_x in range(a1.shape[1]):
if (not (numpy.isnan(a1[filter_y][filter_x]))):
a1[filter_y][filter_x] = str_ele[filter_y][filter_x]*image[y+(filter_y-str_ele_center_y)][x+(filter_x-str_ele_center_x)]
eroded_image[y][x] = numpy.nanmin(a1)


basically:



Every pixel in final image = min(pixel, above, below, left, right) from the original image



 for y in range(len(eroded_image)):
for x in range(len(eroded_image[1])):
eroded_image2[y][x] = numpy.nanmin(str_ele*image2[y:y+len(str_ele),x:x+(len(str_ele[1]))])


This is what I now have. Still 2 loops.










share|improve this question





























    1















    I am performing erosion on an image. The image has been padded accordingly.
    In a nutshell, I have a cross element(+) that I put on every pixel of the image and pick the lowest value for that pixel from pixel above, below, right, left and itself.



    It is inefficient and I can't figure out a vectorized version. It must be possible since all calculations are done independently of each other.



    for y in range(t,image.shape[0]-b):
    for x in range(l,image.shape[1]-r):
    a1 = numpy.copy(str_ele)
    for filter_y in range(a1.shape[0]):
    for filter_x in range(a1.shape[1]):
    if (not (numpy.isnan(a1[filter_y][filter_x]))):
    a1[filter_y][filter_x] = str_ele[filter_y][filter_x]*image[y+(filter_y-str_ele_center_y)][x+(filter_x-str_ele_center_x)]
    eroded_image[y][x] = numpy.nanmin(a1)


    basically:



    Every pixel in final image = min(pixel, above, below, left, right) from the original image



     for y in range(len(eroded_image)):
    for x in range(len(eroded_image[1])):
    eroded_image2[y][x] = numpy.nanmin(str_ele*image2[y:y+len(str_ele),x:x+(len(str_ele[1]))])


    This is what I now have. Still 2 loops.










    share|improve this question



























      1












      1








      1


      1






      I am performing erosion on an image. The image has been padded accordingly.
      In a nutshell, I have a cross element(+) that I put on every pixel of the image and pick the lowest value for that pixel from pixel above, below, right, left and itself.



      It is inefficient and I can't figure out a vectorized version. It must be possible since all calculations are done independently of each other.



      for y in range(t,image.shape[0]-b):
      for x in range(l,image.shape[1]-r):
      a1 = numpy.copy(str_ele)
      for filter_y in range(a1.shape[0]):
      for filter_x in range(a1.shape[1]):
      if (not (numpy.isnan(a1[filter_y][filter_x]))):
      a1[filter_y][filter_x] = str_ele[filter_y][filter_x]*image[y+(filter_y-str_ele_center_y)][x+(filter_x-str_ele_center_x)]
      eroded_image[y][x] = numpy.nanmin(a1)


      basically:



      Every pixel in final image = min(pixel, above, below, left, right) from the original image



       for y in range(len(eroded_image)):
      for x in range(len(eroded_image[1])):
      eroded_image2[y][x] = numpy.nanmin(str_ele*image2[y:y+len(str_ele),x:x+(len(str_ele[1]))])


      This is what I now have. Still 2 loops.










      share|improve this question
















      I am performing erosion on an image. The image has been padded accordingly.
      In a nutshell, I have a cross element(+) that I put on every pixel of the image and pick the lowest value for that pixel from pixel above, below, right, left and itself.



      It is inefficient and I can't figure out a vectorized version. It must be possible since all calculations are done independently of each other.



      for y in range(t,image.shape[0]-b):
      for x in range(l,image.shape[1]-r):
      a1 = numpy.copy(str_ele)
      for filter_y in range(a1.shape[0]):
      for filter_x in range(a1.shape[1]):
      if (not (numpy.isnan(a1[filter_y][filter_x]))):
      a1[filter_y][filter_x] = str_ele[filter_y][filter_x]*image[y+(filter_y-str_ele_center_y)][x+(filter_x-str_ele_center_x)]
      eroded_image[y][x] = numpy.nanmin(a1)


      basically:



      Every pixel in final image = min(pixel, above, below, left, right) from the original image



       for y in range(len(eroded_image)):
      for x in range(len(eroded_image[1])):
      eroded_image2[y][x] = numpy.nanmin(str_ele*image2[y:y+len(str_ele),x:x+(len(str_ele[1]))])


      This is what I now have. Still 2 loops.







      python numpy image-processing vectorization image-morphology






      share|improve this question















      share|improve this question













      share|improve this question




      share|improve this question








      edited Dec 10 '18 at 17:20









      Cris Luengo

      20.2k52249




      20.2k52249










      asked Nov 25 '18 at 21:34









      GaryGary

      83




      83
























          1 Answer
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          oldest

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          2














          If image is a NaN-padded array, and you are eroding with a cross-shaped footprint,
          you could remove the for-loops by stacking slices of image (to effectively shift the image up, left, right, and down)
          and then apply np.nanmin to the stack of slices.



          import numpy as np

          def orig(image):
          t, l, b, r = 1, 1, 1, 1
          str_ele = np.array([[np.nan, 1, np.nan], [1, 1, 1], [np.nan, 1, np.nan]], dtype='float')
          str_ele_center_x, str_ele_center_y = 1, 1
          eroded_image = np.full_like(image, dtype='float', fill_value=np.nan)
          for y in range(t,image.shape[0]-b):
          for x in range(l,image.shape[1]-r):
          a1 = np.copy(str_ele)
          for filter_y in range(a1.shape[0]):
          for filter_x in range(a1.shape[1]):
          if (not (np.isnan(a1[filter_y][filter_x]))):
          a1[filter_y][filter_x] = str_ele[filter_y][filter_x]*image[y+(filter_y-str_ele_center_y)][x+(filter_x-str_ele_center_x)]
          eroded_image[y][x] = np.nanmin(a1)
          return eroded_image

          def erode(image):
          result = np.stack([image[1:-1, 1:-1], image[2:, 1:-1], image[:-2, 1:-1], image[1:-1, 2:], image[1:-1, :-2]])
          eroded_image = np.full_like(image, dtype='float', fill_value=np.nan)
          eroded_image[1:-1, 1:-1] = np.nanmin(result, axis=0)
          return eroded_image

          image = np.arange(24).reshape(4,6)
          image = np.pad(image.astype(float), 1, mode='constant', constant_values=np.nan)


          yields



          In [228]: erode(image)
          Out[228]:
          array([[nan, nan, nan, nan, nan, nan, nan, nan],
          [nan, 0., 0., 1., 2., 3., 4., nan],
          [nan, 0., 1., 2., 3., 4., 5., nan],
          [nan, 6., 7., 8., 9., 10., 11., nan],
          [nan, 12., 13., 14., 15., 16., 17., nan],
          [nan, nan, nan, nan, nan, nan, nan, nan]])




          For the small example image above, erode appears to be around 33x faster than orig:



          In [23]: %timeit erode(image)
          10000 loops, best of 3: 35.6 µs per loop

          In [24]: %timeit orig(image)
          1000 loops, best of 3: 1.19 ms per loop

          In [25]: 1190/35.6
          Out[25]: 33.42696629213483





          share|improve this answer





















          • 1





            For a proper image (scipy.misc.face().mean(2) I get a more than twofold speedup by stacking (and taking the nanmin) along axis 0 instead of -1.

            – Paul Panzer
            Nov 26 '18 at 0:13











          • @PaulPanzer: Thanks very much for the improvement.

            – unutbu
            Nov 26 '18 at 6:33











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          2














          If image is a NaN-padded array, and you are eroding with a cross-shaped footprint,
          you could remove the for-loops by stacking slices of image (to effectively shift the image up, left, right, and down)
          and then apply np.nanmin to the stack of slices.



          import numpy as np

          def orig(image):
          t, l, b, r = 1, 1, 1, 1
          str_ele = np.array([[np.nan, 1, np.nan], [1, 1, 1], [np.nan, 1, np.nan]], dtype='float')
          str_ele_center_x, str_ele_center_y = 1, 1
          eroded_image = np.full_like(image, dtype='float', fill_value=np.nan)
          for y in range(t,image.shape[0]-b):
          for x in range(l,image.shape[1]-r):
          a1 = np.copy(str_ele)
          for filter_y in range(a1.shape[0]):
          for filter_x in range(a1.shape[1]):
          if (not (np.isnan(a1[filter_y][filter_x]))):
          a1[filter_y][filter_x] = str_ele[filter_y][filter_x]*image[y+(filter_y-str_ele_center_y)][x+(filter_x-str_ele_center_x)]
          eroded_image[y][x] = np.nanmin(a1)
          return eroded_image

          def erode(image):
          result = np.stack([image[1:-1, 1:-1], image[2:, 1:-1], image[:-2, 1:-1], image[1:-1, 2:], image[1:-1, :-2]])
          eroded_image = np.full_like(image, dtype='float', fill_value=np.nan)
          eroded_image[1:-1, 1:-1] = np.nanmin(result, axis=0)
          return eroded_image

          image = np.arange(24).reshape(4,6)
          image = np.pad(image.astype(float), 1, mode='constant', constant_values=np.nan)


          yields



          In [228]: erode(image)
          Out[228]:
          array([[nan, nan, nan, nan, nan, nan, nan, nan],
          [nan, 0., 0., 1., 2., 3., 4., nan],
          [nan, 0., 1., 2., 3., 4., 5., nan],
          [nan, 6., 7., 8., 9., 10., 11., nan],
          [nan, 12., 13., 14., 15., 16., 17., nan],
          [nan, nan, nan, nan, nan, nan, nan, nan]])




          For the small example image above, erode appears to be around 33x faster than orig:



          In [23]: %timeit erode(image)
          10000 loops, best of 3: 35.6 µs per loop

          In [24]: %timeit orig(image)
          1000 loops, best of 3: 1.19 ms per loop

          In [25]: 1190/35.6
          Out[25]: 33.42696629213483





          share|improve this answer





















          • 1





            For a proper image (scipy.misc.face().mean(2) I get a more than twofold speedup by stacking (and taking the nanmin) along axis 0 instead of -1.

            – Paul Panzer
            Nov 26 '18 at 0:13











          • @PaulPanzer: Thanks very much for the improvement.

            – unutbu
            Nov 26 '18 at 6:33
















          2














          If image is a NaN-padded array, and you are eroding with a cross-shaped footprint,
          you could remove the for-loops by stacking slices of image (to effectively shift the image up, left, right, and down)
          and then apply np.nanmin to the stack of slices.



          import numpy as np

          def orig(image):
          t, l, b, r = 1, 1, 1, 1
          str_ele = np.array([[np.nan, 1, np.nan], [1, 1, 1], [np.nan, 1, np.nan]], dtype='float')
          str_ele_center_x, str_ele_center_y = 1, 1
          eroded_image = np.full_like(image, dtype='float', fill_value=np.nan)
          for y in range(t,image.shape[0]-b):
          for x in range(l,image.shape[1]-r):
          a1 = np.copy(str_ele)
          for filter_y in range(a1.shape[0]):
          for filter_x in range(a1.shape[1]):
          if (not (np.isnan(a1[filter_y][filter_x]))):
          a1[filter_y][filter_x] = str_ele[filter_y][filter_x]*image[y+(filter_y-str_ele_center_y)][x+(filter_x-str_ele_center_x)]
          eroded_image[y][x] = np.nanmin(a1)
          return eroded_image

          def erode(image):
          result = np.stack([image[1:-1, 1:-1], image[2:, 1:-1], image[:-2, 1:-1], image[1:-1, 2:], image[1:-1, :-2]])
          eroded_image = np.full_like(image, dtype='float', fill_value=np.nan)
          eroded_image[1:-1, 1:-1] = np.nanmin(result, axis=0)
          return eroded_image

          image = np.arange(24).reshape(4,6)
          image = np.pad(image.astype(float), 1, mode='constant', constant_values=np.nan)


          yields



          In [228]: erode(image)
          Out[228]:
          array([[nan, nan, nan, nan, nan, nan, nan, nan],
          [nan, 0., 0., 1., 2., 3., 4., nan],
          [nan, 0., 1., 2., 3., 4., 5., nan],
          [nan, 6., 7., 8., 9., 10., 11., nan],
          [nan, 12., 13., 14., 15., 16., 17., nan],
          [nan, nan, nan, nan, nan, nan, nan, nan]])




          For the small example image above, erode appears to be around 33x faster than orig:



          In [23]: %timeit erode(image)
          10000 loops, best of 3: 35.6 µs per loop

          In [24]: %timeit orig(image)
          1000 loops, best of 3: 1.19 ms per loop

          In [25]: 1190/35.6
          Out[25]: 33.42696629213483





          share|improve this answer





















          • 1





            For a proper image (scipy.misc.face().mean(2) I get a more than twofold speedup by stacking (and taking the nanmin) along axis 0 instead of -1.

            – Paul Panzer
            Nov 26 '18 at 0:13











          • @PaulPanzer: Thanks very much for the improvement.

            – unutbu
            Nov 26 '18 at 6:33














          2












          2








          2







          If image is a NaN-padded array, and you are eroding with a cross-shaped footprint,
          you could remove the for-loops by stacking slices of image (to effectively shift the image up, left, right, and down)
          and then apply np.nanmin to the stack of slices.



          import numpy as np

          def orig(image):
          t, l, b, r = 1, 1, 1, 1
          str_ele = np.array([[np.nan, 1, np.nan], [1, 1, 1], [np.nan, 1, np.nan]], dtype='float')
          str_ele_center_x, str_ele_center_y = 1, 1
          eroded_image = np.full_like(image, dtype='float', fill_value=np.nan)
          for y in range(t,image.shape[0]-b):
          for x in range(l,image.shape[1]-r):
          a1 = np.copy(str_ele)
          for filter_y in range(a1.shape[0]):
          for filter_x in range(a1.shape[1]):
          if (not (np.isnan(a1[filter_y][filter_x]))):
          a1[filter_y][filter_x] = str_ele[filter_y][filter_x]*image[y+(filter_y-str_ele_center_y)][x+(filter_x-str_ele_center_x)]
          eroded_image[y][x] = np.nanmin(a1)
          return eroded_image

          def erode(image):
          result = np.stack([image[1:-1, 1:-1], image[2:, 1:-1], image[:-2, 1:-1], image[1:-1, 2:], image[1:-1, :-2]])
          eroded_image = np.full_like(image, dtype='float', fill_value=np.nan)
          eroded_image[1:-1, 1:-1] = np.nanmin(result, axis=0)
          return eroded_image

          image = np.arange(24).reshape(4,6)
          image = np.pad(image.astype(float), 1, mode='constant', constant_values=np.nan)


          yields



          In [228]: erode(image)
          Out[228]:
          array([[nan, nan, nan, nan, nan, nan, nan, nan],
          [nan, 0., 0., 1., 2., 3., 4., nan],
          [nan, 0., 1., 2., 3., 4., 5., nan],
          [nan, 6., 7., 8., 9., 10., 11., nan],
          [nan, 12., 13., 14., 15., 16., 17., nan],
          [nan, nan, nan, nan, nan, nan, nan, nan]])




          For the small example image above, erode appears to be around 33x faster than orig:



          In [23]: %timeit erode(image)
          10000 loops, best of 3: 35.6 µs per loop

          In [24]: %timeit orig(image)
          1000 loops, best of 3: 1.19 ms per loop

          In [25]: 1190/35.6
          Out[25]: 33.42696629213483





          share|improve this answer















          If image is a NaN-padded array, and you are eroding with a cross-shaped footprint,
          you could remove the for-loops by stacking slices of image (to effectively shift the image up, left, right, and down)
          and then apply np.nanmin to the stack of slices.



          import numpy as np

          def orig(image):
          t, l, b, r = 1, 1, 1, 1
          str_ele = np.array([[np.nan, 1, np.nan], [1, 1, 1], [np.nan, 1, np.nan]], dtype='float')
          str_ele_center_x, str_ele_center_y = 1, 1
          eroded_image = np.full_like(image, dtype='float', fill_value=np.nan)
          for y in range(t,image.shape[0]-b):
          for x in range(l,image.shape[1]-r):
          a1 = np.copy(str_ele)
          for filter_y in range(a1.shape[0]):
          for filter_x in range(a1.shape[1]):
          if (not (np.isnan(a1[filter_y][filter_x]))):
          a1[filter_y][filter_x] = str_ele[filter_y][filter_x]*image[y+(filter_y-str_ele_center_y)][x+(filter_x-str_ele_center_x)]
          eroded_image[y][x] = np.nanmin(a1)
          return eroded_image

          def erode(image):
          result = np.stack([image[1:-1, 1:-1], image[2:, 1:-1], image[:-2, 1:-1], image[1:-1, 2:], image[1:-1, :-2]])
          eroded_image = np.full_like(image, dtype='float', fill_value=np.nan)
          eroded_image[1:-1, 1:-1] = np.nanmin(result, axis=0)
          return eroded_image

          image = np.arange(24).reshape(4,6)
          image = np.pad(image.astype(float), 1, mode='constant', constant_values=np.nan)


          yields



          In [228]: erode(image)
          Out[228]:
          array([[nan, nan, nan, nan, nan, nan, nan, nan],
          [nan, 0., 0., 1., 2., 3., 4., nan],
          [nan, 0., 1., 2., 3., 4., 5., nan],
          [nan, 6., 7., 8., 9., 10., 11., nan],
          [nan, 12., 13., 14., 15., 16., 17., nan],
          [nan, nan, nan, nan, nan, nan, nan, nan]])




          For the small example image above, erode appears to be around 33x faster than orig:



          In [23]: %timeit erode(image)
          10000 loops, best of 3: 35.6 µs per loop

          In [24]: %timeit orig(image)
          1000 loops, best of 3: 1.19 ms per loop

          In [25]: 1190/35.6
          Out[25]: 33.42696629213483






          share|improve this answer














          share|improve this answer



          share|improve this answer








          edited Nov 26 '18 at 6:33

























          answered Nov 25 '18 at 23:00









          unutbuunutbu

          549k10111781237




          549k10111781237








          • 1





            For a proper image (scipy.misc.face().mean(2) I get a more than twofold speedup by stacking (and taking the nanmin) along axis 0 instead of -1.

            – Paul Panzer
            Nov 26 '18 at 0:13











          • @PaulPanzer: Thanks very much for the improvement.

            – unutbu
            Nov 26 '18 at 6:33














          • 1





            For a proper image (scipy.misc.face().mean(2) I get a more than twofold speedup by stacking (and taking the nanmin) along axis 0 instead of -1.

            – Paul Panzer
            Nov 26 '18 at 0:13











          • @PaulPanzer: Thanks very much for the improvement.

            – unutbu
            Nov 26 '18 at 6:33








          1




          1





          For a proper image (scipy.misc.face().mean(2) I get a more than twofold speedup by stacking (and taking the nanmin) along axis 0 instead of -1.

          – Paul Panzer
          Nov 26 '18 at 0:13





          For a proper image (scipy.misc.face().mean(2) I get a more than twofold speedup by stacking (and taking the nanmin) along axis 0 instead of -1.

          – Paul Panzer
          Nov 26 '18 at 0:13













          @PaulPanzer: Thanks very much for the improvement.

          – unutbu
          Nov 26 '18 at 6:33





          @PaulPanzer: Thanks very much for the improvement.

          – unutbu
          Nov 26 '18 at 6:33


















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