Hi Egor For soem reason had not occurred to me to think of it as a mean filter. I like your ndimage + skimage solution, which will allow using circle, maybe I'll even try a diamond. Thanks a lot, Matteo On Thursday, May 26, 2016 at 9:40:40 AM UTC-6, Egor Panfilov wrote:

By the way :), we do have such filter in `skimage`: http://scikit-image.org/docs/dev/api/skimage.filters.rank.html#mean , but it is selective to input dtype and doesn't support many boundary conditions' alternatives.

Cheers, Egor

четверг, 26 мая 2016 г., 18:29:59 UTC+3 пользователь Egor Panfilov написал:

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Hi Matteo!

The function you're looking for is as simple as convolution [1] with a typical structuring element [2] (we have square, circle and many more). I suppose, this can also be called as 'mean filter':

Here is an example:

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import numpy as np import scipy.ndimage as ndi from skimage.morphology import square

img = np.random.uniform(size=(5, 5)) img = (img > 0.5).astype(np.float) sem = square(3)/3**2 # Don't forget to normalize out = ndi.convolve(img, sem, mode='wrap')

print(img)

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print(sem) print(out)

Results to:

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[[ 1. 0. 0. 1. 0.] [ 1. 0. 0. 0. 0.] [ 0. 0. 1. 1. 0.] [ 0. 0. 0. 0. 1.] [ 1. 0. 1. 0. 0.]] [[ 0.11111111 0.11111111 0.11111111] [ 0.11111111 0.11111111 0.11111111] [ 0.11111111 0.11111111 0.11111111]] [[ 0.33333333 0.44444444 0.22222222 0.22222222 0.44444444] [ 0.22222222 0.33333333 0.33333333 0.33333333 0.44444444] [ 0.22222222 0.22222222 0.22222222 0.33333333 0.33333333] [ 0.22222222 0.33333333 0.33333333 0.44444444 0.33333333] [ 0.33333333 0.33333333 0.22222222 0.33333333 0.44444444]]

Cheers, Egor

[1] http://docs.scipy.org/doc/scipy/reference/generated/scipy.ndimage.convolve.h...

[2] http://scikit-image.org/docs/dev/api/skimage.morphology.html#module-skimage....

2016-05-26 18:11 GMT+03:00 Matteo javascript:>:

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Is there a function in scikit-image to calculate the the local density of 1s in a binary image?

By local density of 1s I mean: 1) count the number of pixels that have value of 1 in a running/sliding window (preferrably circular but square would work too) 2) divide by the total number of pixels in the running window

Thanks Matteo

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