通过经验测试和观察，以及https://en.wikipedia.org/wiki/Postcodes_in_the_United_Kingdom#Validation的确认，以下是我的Python正则表达式版本，可以正确地解析和验证英国邮政编码:

UK_POSTCODE_REGEX = r ' (? P < postcode_area > [a - z] {1,2}) (? P <区> (?:[0 - 9]{1,2})| (?:[0 - 9][a - z])) (? P <部门> [0 - 9])(? P <邮编> [a - z]{2})”

这个正则表达式很简单，并且有捕获组。它不包括所有合法的英国邮政编码的验证，而只考虑字母与数字的位置。

下面是我在代码中如何使用它:

@dataclass
class UKPostcode:
    postcode_area: str
    district: str
    sector: int
    postcode: str

    # https://en.wikipedia.org/wiki/Postcodes_in_the_United_Kingdom#Validation
    # Original author of this regex: @jontsai
    # NOTE TO FUTURE DEVELOPER:
    # Verified through empirical testing and observation, as well as confirming with the Wiki article
    # If this regex fails to capture all valid UK postcodes, then I apologize, for I am only human.
    UK_POSTCODE_REGEX = r'(?P<postcode_area>[A-Z]{1,2})(?P<district>(?:[0-9]{1,2})|(?:[0-9][A-Z]))(?P<sector>[0-9])(?P<postcode>[A-Z]{2})'

    @classmethod
    def from_postcode(cls, postcode):
        """Parses a string into a UKPostcode

        Returns a UKPostcode or None
        """
        m = re.match(cls.UK_POSTCODE_REGEX, postcode.replace(' ', ''))

        if m:
            uk_postcode = UKPostcode(
                postcode_area=m.group('postcode_area'),
                district=m.group('district'),
                sector=m.group('sector'),
                postcode=m.group('postcode')
            )
        else:
            uk_postcode = None

        return uk_postcode


def parse_uk_postcode(postcode):
    """Wrapper for UKPostcode.from_postcode
    """
    uk_postcode = UKPostcode.from_postcode(postcode)
    return uk_postcode

下面是单元测试:

@pytest.mark.parametrize(
    'postcode, expected', [
        # https://en.wikipedia.org/wiki/Postcodes_in_the_United_Kingdom#Validation
        (
            'EC1A1BB',
            UKPostcode(
                postcode_area='EC',
                district='1A',
                sector='1',
                postcode='BB'
            ),
        ),
        (
            'W1A0AX',
            UKPostcode(
                postcode_area='W',
                district='1A',
                sector='0',
                postcode='AX'
            ),
        ),
        (
            'M11AE',
            UKPostcode(
                postcode_area='M',
                district='1',
                sector='1',
                postcode='AE'
            ),
        ),
        (
            'B338TH',
            UKPostcode(
                postcode_area='B',
                district='33',
                sector='8',
                postcode='TH'
            )
        ),
        (
            'CR26XH',
            UKPostcode(
                postcode_area='CR',
                district='2',
                sector='6',
                postcode='XH'
            )
        ),
        (
            'DN551PT',
            UKPostcode(
                postcode_area='DN',
                district='55',
                sector='1',
                postcode='PT'
            )
        )
    ]
)
def test_parse_uk_postcode(postcode, expected):
    uk_postcode = parse_uk_postcode(postcode)
    assert(uk_postcode == expected)

接受的答案反映了皇家邮政给出的规则，尽管正则表达式中有一个拼写错误。这个错字似乎在gov.uk网站上也有(就像在XML存档页面中一样)。

在格式A9A 9AA中，规则允许在第三个位置出现P字符，而正则表达式不允许这样。正确的正则表达式应该是:

(GIR 0AA)|((([A-Z-[QVX]][0-9][0-9]?)|(([A-Z-[QVX]][A-Z-[IJZ]][0-9][0-9]?)|(([A-Z-[QVX]][0-9][A-HJKPSTUW])|([A-Z-[QVX]][A-Z-[IJZ]][0-9][ABEHMNPRVWXY])))) [0-9][A-Z-[CIKMOV]]{2}) 

将其缩短为以下正则表达式(使用Perl/Ruby语法):

(GIR 0AA)|([A-PR-UWYZ](([0-9]([0-9A-HJKPSTUW])?)|([A-HK-Y][0-9]([0-9ABEHMNPRVWXY])?))\s?[0-9][ABD-HJLNP-UW-Z]{2})

它还在第一个和第二个块之间包含一个可选的空格。

我需要一个可以在SAS中使用PRXMATCH和相关函数的版本，所以我想到了这个:

^[A-PR-UWYZ](([A-HK-Y]?\d\d?)|(\d[A-HJKPSTUW])|([A-HK-Y]\d[ABEHMNPRV-Y]))\s?\d[ABD-HJLNP-UW-Z]{2}$

测试用例和注意事项:

/* 
Notes
The letters QVX are not used in the 1st position.
The letters IJZ are not used in the second position.
The only letters to appear in the third position are ABCDEFGHJKPSTUW when the structure starts with A9A.
The only letters to appear in the fourth position are ABEHMNPRVWXY when the structure starts with AA9A.
The final two letters do not use the letters CIKMOV, so as not to resemble digits or each other when hand-written.
*/

/*
    Bits and pieces
    1st position (any):         [A-PR-UWYZ]         
    2nd position (if letter):   [A-HK-Y]
    3rd position (A1A format):  [A-HJKPSTUW]
    4th position (AA1A format): [ABEHMNPRV-Y]
    Last 2 positions:           [ABD-HJLNP-UW-Z]    
*/


data example;
infile cards truncover;
input valid 1. postcode &$10. Notes &$100.;
flag = prxmatch('/^[A-PR-UWYZ](([A-HK-Y]?\d\d?)|(\d[A-HJKPSTUW])|([A-HK-Y]\d[ABEHMNPRV-Y]))\s?\d[ABD-HJLNP-UW-Z]{2}$/',strip(postcode));
cards;
1  EC1A 1BB  Special case 1
1  W1A 0AX   Special case 2
1  M1 1AE    Standard format
1  B33 8TH   Standard format
1  CR2 6XH   Standard format
1  DN55 1PT  Standard format
0  QN55 1PT  Bad letter in 1st position
0  DI55 1PT  Bad letter in 2nd position
0  W1Z 0AX   Bad letter in 3rd position
0  EC1Z 1BB  Bad letter in 4th position
0  DN55 1CT  Bad letter in 2nd group
0  A11A 1AA  Invalid digits in 1st group
0  AA11A 1AA  1st group too long
0  AA11 1AAA  2nd group too long
0  AA11 1AAA  2nd group too long
0  AAA 1AA   No digit in 1st group
0  AA 1AA    No digit in 1st group
0  A 1AA     No digit in 1st group
0  1A 1AA    Missing letter in 1st group
0  1 1AA     Missing letter in 1st group
0  11 1AA    Missing letter in 1st group
0  AA1 1A    Missing letter in 2nd group
0  AA1 1     Missing letter in 2nd group
;
run;

看看本页的python代码:

http://www.brunningonline.net/simon/blog/archives/001292.html

I've got some postcode parsing to do. The requirement is pretty simple; I have to parse a postcode into an outcode and (optional) incode. The good new is that I don't have to perform any validation - I just have to chop up what I've been provided with in a vaguely intelligent manner. I can't assume much about my import in terms of formatting, i.e. case and embedded spaces. But this isn't the bad news; the bad news is that I have to do it all in RPG. :-( Nevertheless, I threw a little Python function together to clarify my thinking.

我用它来处理邮政编码。

这个允许两边有空格和制表符，以防你不想验证失败，然后在另一边修剪它。

^\s*(([Gg][Ii][Rr] 0[Aa]{2})|((([A-Za-z][0-9]{1,2})|(([A-Za-z][A-Ha-hJ-Yj-y][0-9]{1,2})|(([A-Za-z][0-9][A-Za-z])|([A-Za-z][A-Ha-hJ-Yj-y][0-9]?[A-Za-z])))) {0,1}[0-9][A-Za-z]{2})\s*$)

我们得到了一个说明:

UK postcodes must be in one of the following forms (with one exception, see below): 
    § A9 9AA 
    § A99 9AA
    § AA9 9AA
    § AA99 9AA
    § A9A 9AA
    § AA9A 9AA
where A represents an alphabetic character and 9 represents a numeric character.
Additional rules apply to alphabetic characters, as follows:
    § The character in position 1 may not be Q, V or X
    § The character in position 2 may not be I, J or Z
    § The character in position 3 may not be I, L, M, N, O, P, Q, R, V, X, Y or Z
    § The character in position 4 may not be C, D, F, G, I, J, K, L, O, Q, S, T, U or Z
    § The characters in the rightmost two positions may not be C, I, K, M, O or V
The one exception that does not follow these general rules is the postcode "GIR 0AA", which is a special valid postcode.

我们想出了这个:

/^([A-PR-UWYZ][A-HK-Y0-9](?:[A-HJKS-UW0-9][ABEHMNPRV-Y0-9]?)?\s*[0-9][ABD-HJLNP-UW-Z]{2}|GIR\s*0AA)$/i

但是注意-这允许组之间有任意数量的空格。