用于检查字符串中单词的开头和结尾的python正则表达式

Question

I am working on a script to rename files. In this scenario there are three possibilities.

1.file does not exist: Create new file

2.File exists: create new file with filename '(number of occurence of file)'.eg filename(1)

3.Duplicate of file already exists: create new file with filename '(number of occurence of file)'.eg filename(2)

I have the filename in a string. I can check the last character of filename using regex but how to check the last characters from '(' to ')' and get the number inside it?

Answer 1

You just need something like this:

(?<=\()(\d+)(?=\)[^()]*$)

Demo

Explanation:

• (?<=\\() must be preceded by a literal (
• (\\d+) match and capture the digits
• (?=\\)[^()]+$) must be followed by ) and then no more ( or ) until the end of the string.

Example: if the file name is Foo (Bar) Baz (23).jpg , the regex above matches 23

Answer 2

Here is the code and tests to get a filename based on existing filenames:

import re

def get_name(filename, existing_names):
    exist = False
    index = 0

    p = re.compile("^%s(\((?P<idx>\d+)\))?$" % filename)

    for name in existing_names:
        m = p.match(name)
        if m:
            exist = True
            idx = m.group('idx')
            if idx and int(idx) > index:
                index = int(idx)
    if exist:
        return "%s(%d)" % (filename, index + 1)
    else:
        return filename

# test data
exists = ["abc(1)", "ab", "abc", "abc(2)", "ab(1)", "de", "ab(5)"]
tests = ["abc", "ab", "de", "xyz"]
expects = ["abc(3)", "ab(6)", "de(1)", "xyz"]

print exists
for name, exp in zip(tests, expects):
    new_name = get_name(name, exists)
    print "%s -> %s" % (name, new_name)
    assert new_name == exp

Look at this line for the regex to get the number in (*) :

p = re.compile("^%s(\\((?P<idx>\\d+)\\))?$" % filename)

Here it uses a named capture ?P<idx>\\d+ for the number \\d+ , and access the capture later with m.group('idx') .