more-pythonic-code

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字数统计: 180 | 阅读时长 ≈ 1 
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input = ['Duration', 'F0', 'F1', 'F2', 'F3']
output = {'Duration': 0, 'F0': 1, 'F1': 2, 'F2': 3, 'F3': 4}

如何写出来？

字典推导式

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{f:i for f, i in zip(input, range(len(input)))}

其他

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dict(zip(input, range(len(input))))

list(enumerate(input))

dict(enumerate(input))

{f: i for i, f in enumerate(input)}

字符串翻转的办法

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def reverse_string3(s):
    """Return a reversed copy of `s`"""
    chars = list(s)
    for i in range(len(s) // 2):
        tmp = chars[i]
        chars[i] = chars[len(s) - i - 1]
        chars[len(s) - i - 1] = tmp
    return ''.join(chars)
    
def reverse_string2(s):
    """Return a reversed copy of `s`"""
    return "".join(reversed(s))
    
    
def reverse_string1(s):
    """Return a reversed copy of `s`"""
    return s[::-1]