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comp / comp.lang.python / RE: A missing iterator on itertools module?

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o RE: A missing iterator on itertools module?<avi.e.gross

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Subject: RE: A missing iterator on itertools module?
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Subject: RE: A missing iterator on itertools module?
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Antoon,

Even if the suggested solution offers a partial result, you would need
specific requirements to determine what should be done if one or more of the
parts being cycled is shorter than the others. Stopping at that point is one
option. Another is to continue but only interleave ones still producing and
in the same order.

There is a function in itertools called zip_longest() that might be
considered as it keeps going but substitutes a customizable value for
"missing" parts. You could then, perhaps, make a change so that sentinel is
not passed along.

-----Original Message-----
From: Python-list <python-list-bounces+avi.e.gross=gmail.com@python.org> On
Behalf Of Antoon Pardon via Python-list
Sent: Wednesday, April 3, 2024 5:11 AM
To: python-list@python.org
Subject: Re: A missing iterator on itertools module?

Op 28/03/2024 om 17:45 schreef ast via Python-list:
> Hello
>
> Suppose I have these 3 strings:
>
> s1 = "AZERTY"
> s2 = "QSDFGH"
> s3 = "WXCVBN"
>
> and I need an itertor who delivers
>
> A Q W Z S C E D C ...
>
> I didn't found anything in itertools to do the job.

The documentation mentions a roundrobin recipe.
>
> So I came up with this solution:
>
>
> list(chain.from_iterable(zip("AZERTY", "QSDFGH", "WXCVBN")))
>
> ['A', 'Q', 'W', 'Z', 'S', 'X', 'E', 'D', 'C', 'R', 'F', 'V', 'T', 'G',
> 'B', 'Y', 'H', 'N']

But if your strings are not equal, this will only produce a partial result.
--
https://mail.python.org/mailman/listinfo/python-list

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