In [1]: s = 'RegularExpression'
In [2]: answer = []
In [3]: breaks = [i for i,char in enumerate(s) if char.isupper()]
In [4]: breaks = breaks[1:]
In [5]: answer.append(s[:breaks[0]])
In [6]: for start,end in zip(breaks, breaks[1:]):
...: answer.append(s[start:end])
...:
In [7]: answer.append(s[breaks[-1]:])
In [8]: answer
Out[8]: ['Regular', 'Expression']
In [9]: print(' '.join(answer))
Regular Expression
尝试使用Lookbehind
"(?<=[a-z])([A-Z])"
Ex:
输出:
您可以通过以下方式执行此操作:
这意味着“在第一个匹配组和第二个匹配组之间留出一个空格”,其中匹配组是一个大写字母,后跟一个或多个非大写字母
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