方法一;
numregex20 =re.compile(r'^d{1,3}(,d{3})*$') #^和$来锚定开头和结尾
input_values=['42', '1,234', '6,368,745','12,34,567', '1234']
for value in input_values:
if numregex20.search(value) is not None:
print_value=numregex20.search(value).group()
print(print_value)
方法二;
numregex20_2 = re.compile(r'(?方法三; 使用空白边界来匹配正确格式化的千位逗号
numregex20_2 = re.compile(r"(?方法四; 试图匹配不需要的模式
s = '42 1,234 6,368,745 12,34,567 1234' rgx = re.compile(',[0-9]{1,2},|[0-9]{4,}') nums = [x for x in s.split() if not rgx.search(x)] print(nums)



