How to group blocks of identical Booleans?

Question

Suppose I have the following list:

a = [True, True, True, False, False, False, False, True, True]

How could get them best grouped by either returning only 0,3, 7 or a grouping like the following?

[True, True, True]
[False, False, False, False]
[True, True]

Background: I am trying to find plateaus in my NumPy arrays, and while setting the derivative to zero is a good start, I still need to sort the array into chunks. I think this basically boils down to the problem above.

I looked up NumPy and itertools (trying to get a solution from the question NumPy grouping using itertools.groupby performance) but I did not succeed. I guess one might use a combination of itertools.takewhile and filtfalse (see the documentation here), but I am out of my depth there. Or maybe I am just thinking way too complicated.

Answer 1

We could get the indices with a sliced array comparison, and it should be good with performance with large size lists/arrays -

a_ext = np.r_[~a[0],a]
out = np.flatnonzero(a_ext[:-1]!=a_ext[1:])

As a one-liner, we could use np.diff + np.flatnonzero -

np.flatnonzero(np.diff(np.r_[~a[0],a]))
# compact alternative : np.where(np.diff(np.r_[~a[0],a]))[0]

Answer 2

The simplest way maybe would be this:

a = [True, True, True, False, False, False, False, True, True]

res = [0] + [i+1 for i, (x, y) in enumerate(zip(a, a[1:])) if x!=y]
print(res)  # -> [0, 3, 7]

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As far as the groupby solution goes, you could do:

from itertools import groupby

groups = [list(g) for _, g in groupby(a)]
print(groups)  # -> [[True, True, True], [False, False, False, False], [True, True]]