Python Glob without the whole path - only the filename
PythonGlobPython Problem Overview
Is there a way I can use glob on a directory, to get files with a specific extension, but only the filename itself, not the whole path?
Python Solutions
Solution 1 - Python
Use os.path.basename(path) to get the filename.
Solution 2 - Python
This might help someone:
names = [os.path.basename(x) for x in glob.glob('/your_path')]
Solution 3 - Python
map(os.path.basename, glob.glob("your/path"))
Returns an iterable with all the file names and extensions.
Solution 4 - Python
os.path.basename works for me.
Here is Code example:
import sys,glob
import os
expectedDir = sys.argv[1] ## User input for directory where files to search
for fileName_relative in glob.glob(expectedDir+"**/*.txt",recursive=True): ## first get full file name with directores using for loop
print("Full file name with directories: ", fileName_relative)
fileName_absolute = os.path.basename(fileName_relative) ## Now get the file name with os.path.basename
print("Only file name: ", fileName_absolute)
Output :
Full file name with directories: C:\Users\erinksh\PycharmProjects\EMM_Test2\venv\Lib\site-packages\wheel-0.33.6.dist-info\top_level.txt
Only file name: top_level.txt
Solution 5 - Python
I keep rewriting the solution for relative globbing (esp. when I need to add items to a zipfile) - this is what it usually ends up looking like.
# Function
def rel_glob(pattern, rel):
"""glob.glob but with relative path
"""
for v in glob.glob(os.path.join(rel, pattern)):
yield v[len(rel):].lstrip("/")
# Use
# For example, when you have files like: 'dir1/dir2/*.py'
for p in rel_glob("dir2/*.py", "dir1"):
# do work
pass
Solution 6 - Python
If you are looking for CSV file:
file = [os.path.basename(x) for x in glob.glob(r'C:\Users\rajat.prakash\Downloads//' + '*.csv')]
If you are looking for EXCEL file:
file = [os.path.basename(x) for x in glob.glob(r'C:\Users\rajat.prakash\Downloads//' + '*.xlsx')]
Solution 7 - Python
for f in glob.glob(gt_path + "/*.png"): # find all png files
exc_name = f.split('/')[-1].split(',')[0]
Then the exc_name is like myphoto.png
Solution 8 - Python
None of the existing answers mention using the new pathlib module, which is what I was searching for, so I'll add a new answer here.
Path.glob produces Path objects containing the full path including any directories. If you only need the file names, use the Path.name property.
If you find yourself frequently converting between pathlib and os.path, check out this handy table converting functions between the two libraries.
Solution 9 - Python
Or using pathlib:
from pathlib import Path
dir_URL = Path("your_directory_URL") # e.g. Path("/tmp")
filename_list = [file.name for file in dir_URL.glob("your_pattern")]