1-liner: set(string.ascii_lowercase).isdisjoint clear solution for All Upper I by Phil15 - python coding challenges

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1-liner: set(string.ascii_lowercase).isdisjoint solution in Clear category for All Upper I by Phil15

is_all_upper = set(__import__('string').ascii_lowercase).isdisjoint

Dec. 7, 2020

Comments:

Stensen on Dec. 8, 2020, 5:33 p.m.
Genuis, I liked it!

Phil15 on Dec. 8, 2020, 6:17 p.m.
Thanks to <a href="https://py.checkio.org/user/Phil15/practice/">practice page</a>, I did go into set methods and realized that the argument we give to the method does not always need to be a set too. Like here where it receives a text. I usually prefer operators, but using an operator require the argument to be set. Example: <pre class='brush: python'>>>> not set('123') & '53' # "isdisjoint" with & TypeError: unsupported operand type(s) for &: 'set' and 'str' >>> set('123').isdisjoint('53') False >>> set('123').difference('53') {'1', '2'} </pre> It's a bit tricky but nice to know.

Stensen on Dec. 8, 2020, 11:34 p.m.
That's truly amazing to know. Thanks for explaining it deeply for me. You're the best!

Strings and Integers

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