WebDec 4, 2024 · In Python, you can get the filename (basename), directory (folder) name, and extension from a path string or join the strings to generate the path string with the … WebDec 6, 2024 · The first and the easiest way to extract part of the file path in Python is to use the os.path.basename () function. This function returns the filename from the file path along with its extension. Plus, it works for all the Python versions. import os fpath='c:\Project\input.txt' os.path.basename(fpath) Output:
Get the filename, directory, extension from a path string …
WebSep 23, 2024 · The best and most reliable way to get a filename from a path in Python is to use the os.path.basename () function. The os.path.basename () is a built-in function to get the base name in the specified path. That is it. Further reading Python os.path.getsize () Python os.path.isfile () Python os.path.dimname () WebDec 4, 2011 · 6. If you have a number of files in a directory and want to store those file names into a list. Use the below code. import os as os import glob as glob path = 'mypath' file_list= [] for file in glob.glob (path): data_file_list = os.path.basename (file) … jefferson\\u0027s 15 year old
Get filename when loading whole folder #203 - Github
WebApr 7, 2024 · The seventh statement adds the file name and size to the list of files in a dictionary format with keys 'name' and 'size'. files.append({'name': filename, 'size': size}) WebSep 12, 2024 · Python3 import os path = "D:\ABC" fun = lambda x : os.path.isfile (os.path.join (path,x)) files_list = filter(fun, os.listdir (path)) size_of_file = [ (f,os.stat (os.path.join (path, f)).st_size) for f in files_list ] fun = lambda x : x [1] for f,s in sorted(size_of_file,key = fun): print(" {} : {}MB".format(f,round(s/(1024*1024),3))) Output: WebApr 11, 2024 · 1 There is probably more efficient method using slicing (assuming the filename have a fixed properties). But you can use os.path.basename. It will automatically retrieve the valid filename from the path. data ['filename_clean'] = data ['filename'].apply (os.path.basename) Share Improve this answer Follow answered 3 hours ago sgd 136 3 … oxychill pills