Question

我有这段代码：

string = """a = 10 + 15
b = 50 + b
c = a + b
d = c + 50"""


letter = "([a-z])"
signs = "(\+|\-|\*|\/)"
regex = re.compile(r"{0} = (\d+) {1} (\d+)|"
                   r"{0} = (\d+) {1} {0}|"
                   r"{0} = {0} {1} {0}|"
                   r"{0} = {0} {1} (\d+)".format(letter, signs))signs))(\d+)".format(letter,signs))

如果我做re.search（正则表达式，字符串）.groups（）我最终得到

('a', '10', '+', '15', None, None, None, None, None, None, None, None, None, None, None, None)
(None, None, None, None, 'b', '50', '+', 'b', None, None, None, None, None, None, None, None)
(None, None, None, None, None, None, None, None, 'c', 'a', '+', 'b', None, None, None, None)
(None, None, None, None, None, None, None, None, None, None, None, None, 'd', 'c', '+', '50')

但我只想要4组。 [VAR，VAL1，操作者，VAL2]

我正在使用列表理解

[r for r in re.search(regex,string).groups() if r != None]

但我想知道在正则表达式本身是否有办法做到这一点。

Answer 1

在这种情况下，最好将正则表达式从四个单独的语句简化为一个稍微重载的语句，这需要修改letter：

letter = "[a-z]"
signs = "(\+|\-|\*|\/)"
regex = re.compile(r"({0}) = (\d+|{0}) {1} (\d+|{0})".format(letter, signs))signs))(\d+)".format(letter,signs))

Answer 2

您可以将(?:...)用作非捕获组，以用于需要分组但不捕获的内容。 E.g：

signs = "(?:\+|\-|\*|\/)"

但是，你可以通过而不是首先制作signs和letter群来摆脱大量的群体：

letter = "[a-z]"
signs = "[+*/-]"

正则表达式避免不必要的组

2 个答案: