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:

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)

print(sem) print(out)

Results to:

[[ 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 <matteo.niccoli@gmail.com>:

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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