Getting a List of Files in a Directory With a Glob

Find all files in a directory with extension .txt in Python

You can use glob:

import glob, os
os.chdir("/mydir")
for file in glob.glob("*.txt"):
    print(file)

or simply os.listdir:

import os
for file in os.listdir("/mydir"):
    if file.endswith(".txt"):
        print(os.path.join("/mydir", file))

or if you want to traverse directory, use os.walk:

import os
for root, dirs, files in os.walk("/mydir"):
    for file in files:
        if file.endswith(".txt"):
             print(os.path.join(root, file))

getting file list using glob in python

Looks like you just need to include your slash to search in the correct directory.

import glob

mydir = "C:\Data"

file_list = glob.glob(mydir + "/*.csv") # Include slash or it will search in the wrong directory!!
print('file_list {}'.format(file_list))

How can I search sub-folders using glob.glob module?

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')]

How can i get a list of filename with different patterns using glob npm module

This should match your required solution

var glob = require("glob")
glob(process.cwd() + "/directory/**/?(*|*-)file-name?(-*).*", {}, function (er, 
 files) {
  console.log(files)
})

Resources:

• Glob Read Me

• Glob Tutorial

Python glob, os, relative path, making filenames into a list

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

Glob doesn't return list of files from specified directory

Glob returns a list of pathnames relative to the root directory. That root directory is assumed to be your current working directory unless the glob pattern specified is an absolute path. In short, because your pattern is an absolute path pattern, the returned files will not be relative, but absolute, including the entire path.

When not using an absolute path pattern, in some cases, you could get just a file name if a file name matches in the current working directory. That file name would of course be relative to the current working directory.

In Python 3.10, you should be able to change the assumed root directory without using an absolute pattern via a new root_dir parameter, but this is not currently available in 3.9 and below: https://docs.python.org/3.10/library/glob.html.

In your case, as mentioned in the comments by othes, os.path.basename should be able to get just the file name if that is what you are after. Alternatively, you could change the current working directory via os.chdir and provide a glob pattern of simply *.jpg and get just the file names relative to the that current working directory, both are reasonable solutions.

Extracting the base name:

mydir = r"C:\Users\viodo\PycharmProjects\pythonProject"
file_list = [os.path.basename(f) for f in glob.glob(mydir + "/*.jpg")]

or returning the files relative to an arbitrary "current working directory":

os.chdir(r"C:\Users\viodo\PycharmProjects\pythonProject")
file_list = glob.glob("*.jpg")

Depending on your requirements, one solution may be better than the other.

Get a filtered list of files in a directory

import glob

jpgFilenamesList = glob.glob('145592*.jpg')

See glob in python documenttion

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