Hey, You can always do `itertools.chain.from_iterable(zip(iterable, itertools.repeat(sep)))` but I agree that it is verbose.
Cheers, E On Wed, 9 Dec 2020 at 04:16, <aurelien.lambert...@gmail.com> wrote: > Hi > > I like using itertools for creating long strings while not paying the cost > of intermediate strings (by eventually calling str.join on the whole > iterator). > However, one missing feature is to mimic the behavior of str.join as an > iterator: an iterator that returns the items of an iterable, separated by > the separator. > I suggest name "interleave" or "join" (whichever is the most clear / least > ambigous). > > def interleave(sep, iterable): > """ > Makes an iterator that returns elements from an iterable, > separated by the separator. > """ > notfirst = False > for i in iterable: > if notfirst: > yield sep > else: > notfirst = True > yield i > > Could imagine a more elaborate implementation that can take several > iterators, and would be equivalent to > lambda chain_zip_interleave sep, *iterables: > itertools.chain.from_iterable(interleave((sep,), zip(*iterables))) > But that may be seriously overkill, and I have hard time describing it. > _______________________________________________ > Python-ideas mailing list -- python-ideas@python.org > To unsubscribe send an email to python-ideas-le...@python.org > https://mail.python.org/mailman3/lists/python-ideas.python.org/ > Message archived at > https://mail.python.org/archives/list/python-ideas@python.org/message/YWT5BVGPNO3UBW4DZYYPXVCJY2JH7B4H/ > Code of Conduct: http://python.org/psf/codeofconduct/ >
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