正则表达式匹配目录，然后匹配子目录

Question

For example, if I have a list of paths (ie. dir1/subdirA, dir2/subdirB, dir1/subdirB, etc..). I have a regex to match some directory names and then another regex to match the subdirectories. What's the best way to get the valid paths. Or is there a way to combine the two regex using the 2 existing regex?

DIR_RE = re.compile(r'somedirname', re.I)

SUB_RE = re.compile(r'^/somesubdir$', re.I)

Answer 1

import re

directories = ["dir1/subdirA", "dir2/subdirB", "dir1/subdirB", "subdir9/dirC"]

expression = re.compile('^dir[1-9]\/subdir[A-Z]$', re.I)

for directory in directories:
    if (re.match(expression, directory)):
        print "Yes the directory path :" +directory+ "  is valid"
        #Do something.
        #Passed cases = ["dir1/subdirA", "dir2/subdirB", "dir1/subdirB"]
    else:
        #Failed cases = ["subdir9/dirC"]
        #Do something here.

NOTE : The regex is created keeping in mind the example provided in case your directory structure is different, you have to change it accordingly.

Answer 2

You can combine both regexps this way. This example is an alternative based on solution proposed by @ZdaR.

import re
directories = ["dir1/subdirA", "dir2/subdirB", "dir1/subdirB", "subdir9/dirC"]

regexp = re.compile('^(dir[1-9])\/(subdir[A-Z])$', re.I)

for path in directories:
    frag = regexp.match(path)
    if frag != None:
        dir_str = frag.group(1)
        subdir_str = frag.group(2)
        entire_match = frag.group(0)
        # Do something with them

Notice the parenthesis in the regexp. These parenthesis allow to define several groups inside the regexp, so fragments of every match can be obtained using group(n) method over the Match object.

This regexp assumes that the length of every path will be only 2

(Eg: 'dir/subdir' and not 'dir/subdir/subdir')

I hope this will help.