With numpy's where() function I'm getting the indices of an array by applying a mask. For example, an index list, where (1, 2, 3, 10, 11, 12, 13) where True:
[1, 2, 3, 10, 11, 12, 13]
Is there a way to transform this (or directly get it from a similar function) in separate ranges, for my example:
[ range(1, 4), range(10, 14) ]
Assuming all the ranges include unit differences only, it's worth to start with a search of indexes where to split:
split_idx = np.flatnonzero(np.diff(ls, prepend = ls[0], append=ls[-1])!=1)
Note that it is improved slightly in order to insert starting and ending indexes as well. A further work could be done like so:
bounding_points = np.transpose([split_idx[:-1], split_idx[1:]])
out = [range(ls[n], ls[m-1]+1) for n, m in bounding_points.tolist()]
Sample run:
>>> ls = [1, 2, 3, 10, 11, 12, 13, 16, 18, 19]
>>> split_idx
array([ 0, 3, 7, 8, 10], dtype=int32)
>>> bounding_points
array([[ 0, 3],
[ 3, 7],
[ 7, 8],
[ 8, 10]], dtype=int32)
>>> out
[range(1, 4), range(10, 14), range(16, 17), range(18, 20)]
n=2 to np.diff(), and comparing result of np.diff() to 0 rather than 1. That would probably make the solution work even if the range steps are greater than 1. I would try it out myself, but my numpy is old, and doesn't even accept the prepend arg !!n=2 would produce second order differences (differences of differences). (Hence that suggestion)
step > 1 is adding a bit too much complexity based on the questionreading the link I sent you I got this answer:
from operator import itemgetter
from itertools import groupby
data = [1, 2, 3, 10, 11, 12, 13]
ranges = []
for key, group in groupby(enumerate(data), lambda x: x[0] - x[1]):
group = list(map(itemgetter(1), group))
if len(group) > 1:
ranges.append(range(group[0], group[-1]+1))
else:
ranges.append(group[0])
print(ranges)
the output was
[range(1, 4), range(10, 14)]
numpy.[1, 3, 15, 5, 6], one possible answer could be[range(1,4,2), range(15,16,1), range(5,7,1)], and another possible answer could be[range(1,4,2), range(15,4,-10), range(6,7,1)]. Are you ok with either of these answers?