我需要提取 + 符号或字符串开头的所有字母，如下所示:
formula = "X+BC+DAF"
我试过了，但我不想在结果中看到 + 符号。我希望只看到 ['X', 'B', 'D']。
>>> re.findall("^[A-Z]|[+][A-Z]", formula)
['X', '+B', '+D']
当我用括号分组时，我得到了这个奇怪的结果:
re.findall("^([A-Z])|[+]([A-Z])", formula)
[('X', ''), ('', 'B'), ('', 'D')]
为什么在我尝试分组时它创建了元组？如何直接编写正则表达式以使其返回 ['X', 'B', 'D'] ？
最佳答案
如果正则表达式中有任何捕获组，则 re.findall仅返回组捕获的值。如果没有组，则返回整个匹配的字符串。
re.findall(pattern, string, flags=0)Return all non-overlapping matches of pattern in string, as a list of strings. The string is scanned left-to-right, and matches are returned in the order found. If one or more groups are present in the pattern, return a list of groups; this will be a list of tuples if the pattern has more than one group. Empty matches are included in the result unless they touch the beginning of another match.
How to write the regexp directly such that it returns ['X', 'B', 'D'] ?
您可以使用非捕获组代替捕获组:
>>> re.findall(r"(?:^|\+)([A-Z])", formula)
['X', 'B', 'D']
或者对于这种特定情况，您可以尝试使用单词边界的更简单的解决方案:
>>> re.findall(r"\b[A-Z]", formula)
['X', 'B', 'D']
或者使用不使用正则表达式的 str.split 的解决方案:
>>> [s[0] for s in formula.split('+')]
['X', 'B', 'D']
关于python - Python中findall和括号的使用，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/13840883/