A missing iterator on itertools module?
ast
none at none.fr
Thu Mar 28 13:12:54 EDT 2024
Le 28/03/2024 à 18:07, Stefan Ram a écrit :
> ast <none at none.fr> wrote or quoted:
>> 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.
>> So I came up with this solution:
>> list(chain.from_iterable(zip("AZERTY", "QSDFGH", "WXCVBN")))
>
> Maybe you meant "zip(s1,s2,s3)" as the definition of s1, s2,
> and s3 otherwise would not be required. Also the "list" is not
> necessary because "chain.from_iterable" already is an iterable.
> You could also use "*" instead of "list" to print it. So,
>
> import itertools as _itertools
> s =[ "AZERTY", "QSDFGH", "WXCVBN" ]
> print( *_itertools.chain.from_iterable( zip( *s )))
>
> . But these are only minor nitpicks; you have found a nice solution!
Why did you renamed itertools as _itertools ?
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