一行多个正则表达式

Question

list_num=['AYSGS11458630001111252','1234567','LUPUP003164311E0111644','ABCGFD','AFC123A']

对于上面的 list_num 我想写 RE，它给出如下。

预期输出

AYSGS11458630001111252
none
LUPUP003164311E0111644
none
none

输出应该是列表中的两个项目

我已经在网上写了请纠正并建议我。

re.match((r"^(([A-Z]{5}[0-9]{17})|([A-Z]{5}[0-9]{9}[A-Z]{1}[0-9]{7}))$", list_num)

Answer 1

re.match(pattern, string, flags=0)定义为return re.compile(pattern, flags).match(string)它不能处理字符串列表而不是字符串，但这很容易用循环修复. 但是由于我们要多次运行同一个表达式，最好编译一次，而不是在循环内重复编译。

这就是我会做的：

import re
list_num=['AYSGS11458630001111252','1234567','LUPUP003164311E0111644','ABCGFD','AFC123A']

pattern = r"^(([A-Z]{5}[0-9]{17})|([A-Z]{5}[0-9]{9}[A-Z]{1}[0-9]{7}))$"
compiled_pattern = re.compile(pattern)

for m in map(compiled_pattern.match, list_num):
    #you could also use pattern.fullmatch, and not enclose your pattern with ^ and $
    if m is not None:
        print(m.string)
    else:
        print(m)
## shows:
##AYSGS11458630001111252
##None
##LUPUP003164311E0111644
##None
##None

或者这个：

x = [m.string if isinstance(m, re.Match) else m
     for m in map(compiled_pattern.match, list_num)]
print (x)
#['AYSGS11458630001111252', None, 'LUPUP003164311E0111644', None, None]

'or even this:'
def multi_match(pattern, seq):
    compiled_pattern=re.compile(pattern)
    return [m.string if isinstance(m, re.Match) else m
            for m in map(compiled_pattern.match, list_num)]

print(multi_match(pattern, list_num))
#['AYSGS11458630001111252', None, 'LUPUP003164311E0111644', None, None]

或者，如果不需要保留索引位置，例如：

def filter_match(pattern, seq):
    compiled_pattern=re.compile(pattern)
    return list(filter(
        lambda line: compiled_pattern.match(line) is not None,
        list_num))

print(filter_match(pattern, list_num))
#['AYSGS11458630001111252', 'LUPUP003164311E0111644']