python 任何（）を使った少しトリッキーな判定方法※见やすいが，短络评価（短路评价）しないので遅い

Posted 2021-05-09

# all()関数のパターン
MATCH_WORDS = ['fuga', 'foobar']

def is_match(word):
    judge = (
        word in MATCH_WORDS,
        word is 'hoge',
    )
    return any(judge)

#  if文のパターン

MATCH_WORDS = ['fuga', 'foobar']

def is_match(word):
    if word in MATCH_WORDS:
        return True
    elif word is 'hoge':
        return True
    else:
        return False