So splitting a list using itertools.groupby() is a fairly easy.
>>> import itertools as it
>>> iterable = it.groupby([1, 2, 3, 4, 5, 2, 3, 4, 2], lambda p: p==2)
>>> for x, y in iterable:
... print(x, list(y))
... next(iterable)
False [1]
False [3, 4, 5]
False [3, 4]
Works as expected. But using a common python idiom of ziping up the iterator multiple times to step through 2 at a time seems to break things.
>>> iterable = it.groupby([1, 2, 3, 4, 5, 2, 3, 4, 2], lambda p: p==2)
>>> for (x, y), _ in zip(iterable, iterable):
... print(x, list(y))
False []
False []
False []
Adding a print(y) shows the expected nested iterable <itertools._grouper object at 0xXXXXXXXX>, but I'm obviously missing something as to why the grouper object is empty. Can anyone shed some light?
I get an even weirder result if I have an uneven list and use itertools.zip_longest:
>>> iterable = it.groupby([1, 2, 3, 4, 5, 2, 3, 4], lambda p: p==2)
>>> for (x, y), _ in it.zip_longest(iterable, iterable, fillvalue=None):
... print(x, list(y))
False []
False []
False [4]
Update: Simple fix is to use itertools.islice():
>>> iterable = it.groupby([1, 2, 3, 4, 5, 2, 3, 4, 2], lambda p: p==2)
>>> for x, y in it.islice(iterable, None, None, 2):
... print(x, list(y))
False [1]
False [3, 4, 5]
False [3, 4]
The groupby documentation warns you that
The returned group is itself an iterator that shares the underlying iterable with groupby(). Because the source is shared, when the groupby() object is advanced, the previous group is no longer visible.
When your zip produces a ((key, group), (key, group)) pair, it advances the groupby iterator past the first group, rendering the first group unusable. You need to materialize the group before advancing:
iterable = ((key, list(group)) for (key, group) in it.groupby([1, 2, 3, 4, 5, 2, 3, 4, 2], lambda p: p==2))
for (x, y), _ in zip(iterable, iterable):
print(x, y)