one line clear solution for Surjection Strings by kazuki.h - python coding challenges

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one line solution in Clear category for Surjection Strings by kazuki.h

def isometric_strings(str1: str, str2: str) -> bool:
    return str2 == "".join([{str1[i]: str2[i] for i in range(len(str1))}[str1_c] for str1_c in str1])

April 3, 2021

Comments:

jakubbortlik on Dec. 6, 2022, 6:53 a.m.
Oh, a dictionary comprehension in a list comprehension, nice. Maybe this would be a bit clearer: <pre class='brush: python'> return str2 == "".join({s1: s2 for s1, s2 in zip(str1, str2)}[char] for char in str1) </pre> The square brackets of the list comprehension are not needed, actually this is now a generator expression which is the only expression in the <pre class='brush: python'>''.join()</pre> call and does not need an extra pair of parentheses. Also, <pre class='brush: python'>zip()</pre> is more elegant than indexing when looping over two iterables at the same time.

kazuki.h on Dec. 6, 2022, 10:59 a.m.
Thanks for your idea. <a href="https://py.checkio.org/mission/isometric-strings/publications/kazuki.h/python-3/second/">https://py.checkio.org/mission/isometric-strings/publications/kazuki.h/python-3/second/</a>

kazuki.h on Dec. 6, 2022, 10:55 a.m.
my new solution(12/06/2022) <a href="https://py.checkio.org/mission/isometric-strings/publications/kazuki.h/python-3/second/">https://py.checkio.org/mission/isometric-strings/publications/kazuki.h/python-3/second/</a>

Strings Theory

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