Le 28/03/2024 à 18:07, Stefan Ram a écrit :
ast <[email protected]> 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 ? -- https://mail.python.org/mailman/listinfo/python-list
