使用正则表达式删除字符串中间的前导零

Question

I have a large number of strings on the format YYYYYYYYXXXXXXXXZZZZZZZZ, where X, Y, and Z are numbers of fix length, eight digits. Now, the problem is that I need to parse out the middle sequence of integers and remove any leading zeroes. Unfortunately is the only way to determine where each of the three sequences begins/ends is to count the number of digits.

I am currently doing it in two steps, ie:

m = re.match(
    r"(?P<first_sequence>\d{8})"
    r"(?P<second_sequence>\d{8})"
    r"(?P<third_sequence>\d{8})",
    string)
second_secquence = m.group(2)
second_secquence.lstrip(0)

Which does work, and gives me the right results, eg:

112233441234567855667788 --> 12345678
112233440012345655667788 --> 123456
112233001234567855667788 --> 12345678
112233000012345655667788 --> 123456

But is there a better method? Is is possible to write a single regex expression which matches against the second sequence, sans the leading zeros?

I guess I am looking for a regex which does the following:

1. Skips over the first eight digits.
2. Skips any leading zeros.
3. Captures anything after that, up to the point where there's sixteen characters behind/eight infront.

The above solution does work, as mentioned, so the purpose of this problem is more to improve my knowledge of regex. I appreciate any pointers.

Answer 1

This is a typical case of "useless use of regular expressions".

Your strings are fixed-length. Just cut them at the appropriate positions.

s = "112233440012345655667788"
int(s[8:16])
# -> 123456

Answer 2

我认为不使用正则表达式更简单。

result = my_str[8:16].lstrip('0')

Answer 3

Agree with the other answers here that regex isn't really required. If you really want to use regex, then \\d{8}0*(\\d*)\\d{8} should do it.

Answer 4

Just to show that it is possible with regex:

https://regex101.com/r/8RSxaH/2

# CODE AUTO GENERATED BY REGEX101.COM (SEE LINK ABOVE)
# coding=utf8
# the above tag defines encoding for this document and is for Python 2.x compatibility

import re

regex = r"(?<=\d{8})((?:0*)(\d{,8}))(?=\d{8})"

test_str = ("112233441234567855667788\n"
    "112233440012345655667788\n"
    "112233001234567855667788\n"
    "112233000012345655667788")

matches = re.finditer(regex, test_str)

for matchNum, match in enumerate(matches):
    matchNum = matchNum + 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.

Although you don't really need it to do what you're asking