查找以相同的大写字母开头和结尾的子字符串

Question

我有一个家庭作业问题，我需要使用正则表达式从大字符串中解析出子字符串。

目标是选择与以下参数匹配的子字符串：

子字符串以相同的大写字符开头和结尾，我需要忽略前面带有数字0的任何大写字符实例。

例如，ZAp0ZuZAuX0AZA将包含匹配项ZAp0ZuZ和AuX0AZA

我已经弄了几个小时，说实话还没来得及...

我已经尝试了一些类似下面的代码的方法，但是它将选择从第一个大写字母到最后一个大写字母的所有内容。我也

[A-Z]{1}[[:alnum:]]*[A-Z]{1} <--- this selects the whole string
[A-Z]{1}[[:alnum:]][A-Z]{1} <--- this gives me strings like ZuZ, AuX

真的很感谢您的帮助，对此我深感困惑。

Answer 1

这可能有效

(?<!0)([A-Z]).*?(?<!0)\1

https://regex101.com/r/nES9FP/1

解释

 (?<! 0 )      # Ignore Upper case with zero in front of it
 ( [A-Z] )     # (1), This Upper case is to be found down stream
 .*?           # Lazy, any character
 (?<! 0 )      # Ignore Upper case with zero in front of it
 \1            # Backref to what is in group (1)

Answer 2

您可以使用

(?<!0)([A-Z]).*?(?<!0)\1

请参见regex demo。

详细信息

• (?<!0)([A-Z])-第1组：一个不以零开头的ASCII大写字母
• .*?-除换行符外，任何字符都应尽可能少
• (?<!0)\1-与第1组相同的字母，紧随0之后。

请参见Python demo：

import re
s="ZAp0ZuZAuX0AZA"
for m in re.finditer(r'(?<!0)([A-Z]).*?(?<!0)\1', s):
    print(m.group()) # => ['ZAp0ZuZ', 'AuX0AZA']

Answer 3

使用正则表达式执行此操作可能不是最好的主意，因为您可以拆分它们。但是，如果您希望这样做，this expression可能会告诉您，当字符列表扩展时，您可能会遇到什么问题：

(?=.[A-Z])([A-Z])(.*?)\1

我添加了必须包含一个大写字母的(?=.[A-Z])。您可以删除它，它将起作用。但是，为了安全起见，可以在表达式中添加此类边界。

JavaScript测试

const regex = /([A-Z])(.*?)\1/gm;
const str = `ZAp0ZuZAuX0AZA
ZApxxZuZAuXxafaAZA
ZApxaf09xZuZAuX090xafaAZA
abcZApxaf09xZuZAuX090xafaAZA`;
let m;

while ((m = regex.exec(str)) !== null) {
    // This is necessary to avoid infinite loops with zero-width matches
    if (m.index === regex.lastIndex) {
        regex.lastIndex++;
    }
    
    // The result can be accessed through the `m`-variable.
    m.forEach((match, groupIndex) => {
        console.log(`Found match, group ${groupIndex}: ${match}`);
    });
}

Python测试

# coding=utf8
# the above tag defines encoding for this document and is for Python 2.x compatibility

import re

regex = r"([A-Z])(.*?)\1"

test_str = ("ZAp0ZuZAuX0AZA\n"
    "ZApxxZuZAuXxafaAZA\n"
    "ZApxaf09xZuZAuX090xafaAZA\n"
    "abcZApxaf09xZuZAuX090xafaAZA")

matches = re.finditer(regex, test_str, re.MULTILINE)

for matchNum, match in enumerate(matches, start=1):

    print ("Match {matchNum} was found at {start}-{end}: {match}".format(matchNum = matchNum, start = match.start(), end = match.end(), match = match.group()))

    for groupNum in range(0, len(match.groups())):
        groupNum = groupNum + 1

        print ("Group {groupNum} found at {start}-{end}: {group}".format(groupNum = groupNum, start = match.start(groupNum), end = match.end(groupNum), group = match.group(groupNum)))

# Note: for Python 2.7 compatibility, use ur"" to prefix the regex and u"" to prefix the test string and substitution.