In Slack, @mastrof wanted the equivalent of the following Python 3 code which returns
n items at a time and pads with
fillvalue if there are less than
n items remaining
def f(iterable, n, fillvalue=None): args = [iter(iterable)] * n return zip_longest(*args, fillvalue=fillvalue)
We currently do not have
zip_longest in IterTools.jl although there is a PR:
Additionally, the iteration protocol is distinct in Julia from Python.