numpy vectorize / more efficient for loop
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
add a comment |
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
add a comment |
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
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
python numpy image-processing vectorization image-morphology
edited Dec 10 '18 at 17:20
Cris Luengo
22.1k52253
22.1k52253
asked Nov 25 '18 at 21:34
GaryGary
83
83
add a comment |
add a comment |
1 Answer
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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
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
add a comment |
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1 Answer
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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
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
add a comment |
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
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
add a comment |
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
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
edited Nov 26 '18 at 6:33
answered Nov 25 '18 at 23:00
unutbuunutbu
559k10512031257
559k10512031257
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
add a comment |
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
add a comment |
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