The glob.glob gained the root_dir and dir_fd parameters in Python 3.10.

See Package description

You could do something like this :

def search_xlsx(rootdir):
    file_list = []
    for root, directories, file in os.walk(rootdir):
        for file in file:
            if(file.endswith(".xlsx")):
                file_list.append(file)
    return file_list

There are a couple of ways:

pathlib.Path().rglob()

Use pathlib.Path().rglob() from the pathlib module, which was introduced in Python 3.5.

from pathlib import Path

for path in Path('src').rglob('*.c'):
    print(path.name)

glob.glob()

If you don't want to use pathlib, use glob.glob():

from glob import glob

for filename in glob('src/**/*.c', recursive=True):
    print(filename)   

For cases where matching files beginning with a dot (.); like files in the current directory or hidden files on Unix based system, use the os.walk() solution below.

os.walk()

For older Python versions, use os.walk() to recursively walk a directory and fnmatch.filter() to match against a simple expression:

import fnmatch
import os

matches = []
for root, dirnames, filenames in os.walk('src'):
    for filename in fnmatch.filter(filenames, '*.c'):
        matches.append(os.path.join(root, filename))

This version should also be faster depending on how many files you have, as the pathlib module has a bit of overhead over os.walk().

glob will return paths in a format matching your query, so that

glob.glob("/home/usr/dir/*.root")
# ['home/usr/dir/foo.root', 'home/usr/dir/bar.root', ...]

glob.glob("*.root")
# ['foo.root', 'bar.root', ...]

glob.glob("./*.root")
# ['./foo.root', './bar.root', ...]

...and so forth.

To get only the filename, you can use path.basename of the os module, something like this:

from glob import glob
from os import path

pattern = "/home/usr/dir/*.root"
files = [path.basename(x) for x in glob(pattern)]
# ['foo.root', 'bar.root', ...]

...or, if you want to prepend the dir part:

pattern = "/home/usr/dir/*.root"
files = [path.join('dir', path.basename(x)) for x in glob(pattern)]
# ['dir/foo.root', 'dir/bar.root', ...]

...or, if you really want the path separator at the start:

from glob import glob
import os

pattern = "/home/usr/dir/*.root"
files = [os.sep + os.path.join('dir', os.path.basename(x)) for x in glob(pattern)]
# ['/dir/foo.root', '/dir/bar.root', ...]

Using path.join and path.sep will make sure that the correct path syntax is used, depending on your OS (i.e. / or \ as a separator).

Depending on what you are really trying to do here, you might want to look at os.path.relpath, for the relative path. The title of your question indicates that relative paths might be what you are actually after:

pattern = "/home/usr/dir/*.root"
files = [os.path.relpath(x) for x in glob(pattern)]
# files will now contain the relative path to each file, from the current working directory

In Python 3.5 and newer use the new recursive **/ functionality:

configfiles = glob.glob('C:/Users/sam/Desktop/file1/**/*.txt', recursive=True)

When recursive is set, ** followed by a path separator matches 0 or more subdirectories.

In earlier Python versions, glob.glob() cannot list files in subdirectories recursively.

In that case I'd use os.walk() combined with fnmatch.filter() instead:

import os
import fnmatch

path = 'C:/Users/sam/Desktop/file1'

configfiles = [os.path.join(dirpath, f)
    for dirpath, dirnames, files in os.walk(path)
    for f in fnmatch.filter(files, '*.txt')]

This'll walk your directories recursively and return all absolute pathnames to matching .txt files. In this specific case the fnmatch.filter() may be overkill, you could also use a .endswith() test:

import os

path = 'C:/Users/sam/Desktop/file1'

configfiles = [os.path.join(dirpath, f)
    for dirpath, dirnames, files in os.walk(path)
    for f in files if f.endswith('.txt')]

glob also works with relative paths so you can simply use pdb/*.pdb and it will work -- glob will return results relative to your current directory:

>>> glob.glob('test/*.txt')
['test/c.txt', 'test/b.txt', 'test/a.txt']

And if you need the result without the directory you can always use os.path.basename to get just the filename:

>>> [os.path.basename(p) for p in glob.glob('test/*.txt')]
['c.txt', 'b.txt', 'a.txt']