The Python Oracle

Finding False-True transitions in a numpy array

--------------------------------------------------
Rise to the top 3% as a developer or hire one of them at Toptal: https://topt.al/25cXVn
--------------------------------------------------

Music by Eric Matyas
https://www.soundimage.org
Track title: Light Drops

--

Chapters
00:00 Finding False-True Transitions In A Numpy Array
00:42 Accepted Answer Score 9
01:21 Answer 2 Score 2
01:52 Thank you

--

Full question
https://stackoverflow.com/questions/4775...

--

Content licensed under CC BY-SA
https://meta.stackexchange.com/help/lice...

--

Tags
#python #numpy

#avk47



ACCEPTED ANSWER

Score 9


Get one-off slices - x[:-1] (starting from the first elem and ending in second last elem) and x[1:] (starting from the second elem and going on until the end), then look for the first slice being lesser than the second one, i.e. catch the pattern of [False, True] and finally get the count with ndarray.sum() or np.count_nonzero() -

(x[:-1] < x[1:]).sum()
np.count_nonzero(x[:-1] < x[1:])

Another way would be to look for the first slice being False and the second one as True, the idea again being to catch that pattern of [False, True] -

(~x[:-1] & x[1:]).sum()
np.count_nonzero(~x[:-1] & x[1:])



ANSWER 2

Score 2


I kind of like to use numpy method "roll" for this kind of problems... "roll" rotates the array to left some step length : (-1,-2,...) or to right (1,2,...)

import numpy as np
np.roll(x,-1)

...this will give x but shifted one step to the left:

array([ True,  True, False, False, False, False, False,  True, False, False], 
dtype=bool)

A False followed by a True can then be expressed as:

~x & np.roll(x,-1)

array([ True, False, False, False, False, False, False,  True, False, False], 
dtype=bool)