是的，可以将这段代码的一部分转换以提高速度吗？

huangapple
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英文:

It is possible convert part of this code to improve speed?

问题

I am new here,

Please, I need help to convert part of this code to anything to improve speed, and still compatibility with python programs to use this function. It is possible to improve speed with this?

On my system, an image colored 224x224 consumes 360 milliseconds to finish.

  1. #!/usr/bin/env python3
  2. # coding: utf-8
  3. # original code https://github.com/verhovsky/squircle/blob/master/squircle.py
  5. import cv2
  6. import math
  7. import time
  8. import numpy
  10. _epsilon = 0.0000000001
  12. def _sgn(x):
  13.     if x == 0.0:
  14.         return 0.0
  15.     if x < 0:
  16.         return -1.0
  17.     return 1.0
  19. def _pixel_coordinates_to_unit(coordinate, max_value):
  20.     return coordinate / max_value * 2 - 1
  22. def _one_coordinates_to_pixels(coordinate, max_value):
  23.     return (coordinate + 1) / 2 * max_value
  25. def _stretch_square_to_disc(x, y):
  26.     if (abs(x) < _epsilon) or (abs(y) < _epsilon):
  27.         return x, y
  29.     x2 = x * x
  30.     y2 = y * y
  31.     hypotenuse_squared = x * x + y * y
  33.     reciprocal_hypotenuse = 1.0 / math.sqrt(hypotenuse_squared)
  35.     multiplier = 1.0
  37.     if x2 > y2:
  38.         multiplier = _sgn(x) * x * reciprocal_hypotenuse
  39.     else:
  40.         multiplier = _sgn(y) * y * reciprocal_hypotenuse
  42.     return x * multiplier, y * multiplier
  44. def _transform(inp):
  45.     result = numpy.zeros_like(inp)
  46.     for x, row in enumerate(inp):
  48.         unit_x = _pixel_coordinates_to_unit(x, len(inp))
  50.         for y, _ in enumerate(row):
  51.             unit_y = _pixel_coordinates_to_unit(y, len(row))
  53.             try:
  54.                 uv = _stretch_square_to_disc(unit_x, unit_y)
  55.                 if uv is None:
  56.                     continue
  57.                 u, v = uv
  59.                 u = _one_coordinates_to_pixels(u, len(inp))
  60.                 v = _one_coordinates_to_pixels(v, len(row))
  62.                 result[x][y] = inp[math.floor(u)][math.floor(v)]
  63.             except IndexError:
  64.                 pass
  66.     return result
  68. # -- load and test
  70. img = cv2.imread('circle.png')
  72. elapsed = round(time.time() * 1000)
  73. squareImage = _transform(img[0:224, 0:224])
  74. print(str(round(time.time() * 1000) - elapsed)+' ms to squareImage')
  76. cv2.imshow('square', squareImage)
  77. key = cv2.waitKey(0)
  79. cv2.destroyAllWindows()

I expected to try to convert this code or part of this to anything faster, maybe CUDA to run directly on the GPU, or NUMBA, Cython, a library, etc...

英文:

I am new here,

Please, I need help to convert part of this code to anything to improve speed, and still compatibility with python programs to use this function. It is possible to improve speed with this?

On my system, a image colored 224x224 consumes 360 milliseconds to finish.

  1. #!/usr/bin/env python3
  2. # coding: utf-8
  3. # original code https://github.com/verhovsky/squircle/blob/master/squircle.py
  4. import cv2
  5. import math
  6. import time
  7. import numpy
  8. _epsilon = 0.0000000001
  9. def _sgn(x):
  10. if x == 0.0:
  11. return 0.0
  12. if x &lt; 0:
  13. return -1.0
  14. return 1.0
  15. def _pixel_coordinates_to_unit(coordinate, max_value):
  16. return coordinate / max_value * 2 - 1
  17. def _one_coordinates_to_pixels(coordinate, max_value):
  18. return (coordinate + 1) / 2 * max_value
  19. def _stretch_square_to_disc(x, y):
  20. if (abs(x) &lt; _epsilon) or (abs(y) &lt; _epsilon):
  21. return x, y
  22. x2 = x * x
  23. y2 = y * y
  24. hypotenuse_squared = x * x + y * y
  25. reciprocal_hypotenuse = 1.0 / math.sqrt(hypotenuse_squared)
  26. multiplier = 1.0
  27. if x2 &gt; y2:
  28. multiplier = _sgn(x) * x * reciprocal_hypotenuse
  29. else:
  30. multiplier = _sgn(y) * y * reciprocal_hypotenuse
  31. return x * multiplier, y * multiplier
  32. def _transform(inp):
  33. result = numpy.zeros_like(inp)
  34. for x, row in enumerate(inp):
  35. unit_x = _pixel_coordinates_to_unit(x, len(inp))
  36. for y, _ in enumerate(row):
  37. unit_y = _pixel_coordinates_to_unit(y, len(row))
  38. try:
  39. uv = _stretch_square_to_disc(unit_x, unit_y)
  40. if uv is None:
  41. continue
  42. u, v = uv
  43. u = _one_coordinates_to_pixels(u, len(inp))
  44. v = _one_coordinates_to_pixels(v, len(row))
  45. result[x][y] = inp[math.floor(u)][math.floor(v)]
  46. except IndexError:
  47. pass
  48. return result
  49. # -- load and test
  50. img = cv2.imread(&#39;circle.png&#39;)
  51. elapsed = round(time.time() * 1000)
  52. squareImage = _transform(img[0:224, 0:224])
  53. print(str(round(time.time() * 1000) - elapsed)+&#39; ms to squareImage&#39;)
  54. cv2.imshow(&#39;square&#39;, squareImage)
  55. key = cv2.waitKey(0)
  56. cv2.destroyAllWindows()

I expected to try convert this code or part of this to anything more fasted, may be CUDA to run direct on GPU, or NUMBA, Cython, lib etc...

答案1

得分: 2

我使用numba装饰了这些函数（删除了_sgn，使用了np.sign，删除了try..except - 是否需要？）：

  1. import math
  2. import time
  4. import cv2
  5. import numpy
  6. from numba import njit
  8. _epsilon = 0.0000000001
  11. @njit
  12. def _pixel_coordinates_to_unit(coordinate, max_value):
  13.     return coordinate / max_value * 2 - 1
  16. @njit
  17. def _one_coordinates_to_pixels(coordinate, max_value):
  18.     return (coordinate + 1) / 2 * max_value
  21. @njit
  22. def _stretch_square_to_disc(x, y):
  23.     if (abs(x) < _epsilon) or (abs(y) < _epsilon):
  24.         return x, y
  26.     x2 = x * x
  27.     y2 = y * y
  28.     hypotenuse_squared = x * x + y * y
  30.     reciprocal_hypotenuse = 1.0 / np.sqrt(hypotenuse_squared)
  32.     multiplier = 1.0
  34.     if x2 > y2:
  35.         multiplier = np.sign(x) * x * reciprocal_hypotenuse
  36.     else:
  37.         multiplier = np.sign(y) * y * reciprocal_hypotenuse
  39.     return x * multiplier, y * multiplier
  42. @njit
  43. def _transform(inp):
  44.     result = numpy.zeros_like(inp)
  45.     for x, row in enumerate(inp):
  46.         unit_x = _pixel_coordinates_to_unit(x, len(inp))
  48.         for y, _ in enumerate(row):
  49.             unit_y = _pixel_coordinates_to_unit(y, len(row))
  51.             uv = _stretch_square_to_disc(unit_x, unit_y)
  52.             if uv is None:
  53.                 continue
  55.             u, v = uv
  57.             u = _one_coordinates_to_pixels(u, len(inp))
  58.             v = _one_coordinates_to_pixels(v, len(row))
  60.             result[x][y] = inp[math.floor(u)][math.floor(v)]
  62.     return result
  65. # -- load and test
  67. img = cv2.imread("circle.png")
  69. # warm jit
  70. # this is needed to let numba do the JIT optimizations
  71. # if you run the the function "cold", the running time will be larger
  73. # you can use compile-ahead-of-time
  74. # https://numba.pydata.org/numba-doc/dev/user/pycc.html
  76. squareImage = _transform(img[0:224, 0:224])
  78. elapsed = time.perf_counter_ns()
  79. squareImage = _transform(img[0:224, 0:224])
  80. print(str((time.perf_counter_ns() - elapsed) / 1000) + " us to squareImage")
  82. cv2.imwrite("shashed.png", squareImage)

在我的电脑上（AMD 5700X），它打印出：

  1. 528.596 us to squareImage
  2. # without using numba:
  3. # 47928.774 us to squareImage

使用的图像：

circle.png

是的，可以将这段代码的一部分转换以提高速度吗？

结果

是的，可以将这段代码的一部分转换以提高速度吗？

英文:

I've used numba to decorate the functions (removed the _sgn and used np.sign, removed the try..except - is it needed?):

  1. import math
  2. import time
  4. import cv2
  5. import numpy
  6. from numba import njit
  8. _epsilon = 0.0000000001
  11. @njit
  12. def _pixel_coordinates_to_unit(coordinate, max_value):
  13.     return coordinate / max_value * 2 - 1
  16. @njit
  17. def _one_coordinates_to_pixels(coordinate, max_value):
  18.     return (coordinate + 1) / 2 * max_value
  21. @njit
  22. def _stretch_square_to_disc(x, y):
  23.     if (abs(x) &lt; _epsilon) or (abs(y) &lt; _epsilon):
  24.         return x, y
  26.     x2 = x * x
  27.     y2 = y * y
  28.     hypotenuse_squared = x * x + y * y
  30.     reciprocal_hypotenuse = 1.0 / np.sqrt(hypotenuse_squared)
  32.     multiplier = 1.0
  34.     if x2 &gt; y2:
  35.         multiplier = np.sign(x) * x * reciprocal_hypotenuse
  36.     else:
  37.         multiplier = np.sign(y) * y * reciprocal_hypotenuse
  39.     return x * multiplier, y * multiplier
  42. @njit
  43. def _transform(inp):
  44.     result = numpy.zeros_like(inp)
  45.     for x, row in enumerate(inp):
  46.         unit_x = _pixel_coordinates_to_unit(x, len(inp))
  48.         for y, _ in enumerate(row):
  49.             unit_y = _pixel_coordinates_to_unit(y, len(row))
  51.             uv = _stretch_square_to_disc(unit_x, unit_y)
  52.             if uv is None:
  53.                 continue
  55.             u, v = uv
  57.             u = _one_coordinates_to_pixels(u, len(inp))
  58.             v = _one_coordinates_to_pixels(v, len(row))
  60.             result[x][y] = inp[math.floor(u)][math.floor(v)]
  62.     return result
  65. # -- load and test
  67. img = cv2.imread(&quot;circle.png&quot;)
  69. # warm jit
  70. # this is needed to let numba do the JIT optimizations
  71. # if you run the the function &quot;cold&quot;, the running time will be larger
  73. # you can use compile-ahead-of-time
  74. # https://numba.pydata.org/numba-doc/dev/user/pycc.html
  76. squareImage = _transform(img[0:224, 0:224])
  78. elapsed = time.perf_counter_ns()
  79. squareImage = _transform(img[0:224, 0:224])
  80. print(str((time.perf_counter_ns() - elapsed) / 1000) + &quot; us to squareImage&quot;)
  82. cv2.imwrite(&quot;shashed.png&quot;, squareImage)

On my computer (AMD 5700X) it prints:

  1. 528.596 us to squareImage
  2. # without using numba:
  3. # 47928.774 us to squareImage

Images used:

circle.png

是的，可以将这段代码的一部分转换以提高速度吗？

the result

是的，可以将这段代码的一部分转换以提高速度吗？

huangapple
  • 本文由 发表于 2023年8月5日 03:17:12
  • 转载请务必保留本文链接：https://go.coder-hub.com/76838644.html
  • cython
  • numba
  • python
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