find stretches of Trues in numpy array

Question

Is there a good way to find stretches of Trues in a numpy boolean array? If I have an array like:

x = numpy.array([True,True,False,True,True,False,False])

Can I get an array of indices like:

starts = [0,3]
ends = [1,4]

or any other appropriate way to store this information. I know this can be done with some complicated while loops, but I'm looking for a better way.

This may be of interest to you https://stackoverflow.com/questions/6352425/whats-the-most-pythonic-way-to-identify-consecutive-duplicates-in-a-list — Cory Kramer, Apr 24 '15 at 17:01
the starts and ends arrays record the starts and ends of the stretches of Trues. Could also be recorded in like : `stretches = [(0,1),(3,4)]` — user3266890, Apr 24 '15 at 17:08
for the record, I'm surprised there are 2 "too broad" close votes. The question seems to be specific and succinct. — shx2, Apr 25 '15 at 05:33

shx2 · Accepted Answer · 2015-04-24T17:29:00.887

You can pad x with Falses (one at the beginning and one at the end), and use np.diff. A "diff" of 1 means transition from False to True, and of -1 means transition from True to False.

The convention is to represent range's end as the index one after the last. This example complies with the convention (you can easily use ends-1 instead of ends to get the array in your question):

x1 = np.hstack([ [False], x, [False] ])  # padding
d = np.diff(x1.astype(int))
starts = np.where(d == 1)[0]
ends = np.where(d == -1)[0]
starts, ends
=> (array([0, 3]), array([2, 5]))

find stretches of Trues in numpy array

1 Answers1