Remove values if they repeat in array [duplicate]

Question

I have an array like

[0,0,0,0,1,1,1,1,0,0,0,0,1,1,0,0]

and I want to determine the number of non-zeros intervals. I know how I can do that of course in a for loop but I wonder if there is a nice numpy solution to it.

The method I am looking for is suppose to "collapse" the array whenever a value repeats itself. So the above array would become for example

[0,1,0,1,0]

for the sake of counting it would of course be sufficient to return just

[1,1]

but I'd like to know a general approach that might also be able to handle more than two different elements such as

[1,1,1,2,2,2,3,3,0,0,1,1,2,2]

or so.

akuiper · Accepted Answer · 2017-04-13 15:59:21Z

1

One option is to pick up the values when there is a change with boolean indexing:

import numpy as np
a = np.array([1,1,1,2,2,2,3,3,0,0,1,1,2,2])

a[np.concatenate(([True], np.diff(a) != 0))]
# array([1, 2, 3, 0, 1, 2])

np.count_nonzero(a[np.concatenate(([True], np.diff(a) != 0))])
# 5

First case:

b = np.array([0,0,0,0,1,1,1,1,0,0,0,0,1,1,0,0])

b[np.concatenate(([True], np.diff(b) != 0))]
# array([0, 1, 0, 1, 0])

np.count_nonzero(b[np.concatenate(([True], np.diff(b) != 0))])
# 2

edited Apr 13, 2017 at 15:59

answered Apr 13, 2017 at 15:53

akuiper

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