native_acceptable_password_iii clear solution for Acceptable Password III by Jon_Red - python coding challenges

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native_acceptable_password_iii solution in Clear category for Acceptable Password III by Jon_Red

def is_acceptable_password(password:str)->bool:
    if not len(password)>6:return False
    if not any(x.isdigit()for x in password):return False
    if not any(x.isalpha()for x in password):return False
    return True


if __name__=='__main__':
    # self-checks
    assert is_acceptable_password('short')==False
    assert is_acceptable_password('muchlonger')==False
    assert is_acceptable_password('ashort')==False
    assert is_acceptable_password('muchlonger5')==True
    assert is_acceptable_password('sh5')==False
    assert is_acceptable_password('1234567')==False

Aug. 3, 2020

Comments:

Jonethan on Sept. 24, 2020, 12:12 p.m.
You can use str.isdecimal() instead of str.isdigit()

Jon_Red on Sept. 28, 2020, 6:46 p.m.
str.isnumeric() would also work. Since I've never once encountered a real world issue where the distinction between these three string methods mattered, I typically choose to use the one with the shortest function name. :)

jakubbortlik on Nov. 29, 2022, 12:03 p.m.
What I like about this solution is that it's easily extensible.

Strings Theory

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