Python list directory, subdirectory, and files

Question:

I’m trying to make a script to list all directory, subdirectory, and files in a given directory.
I tried this:

import sys, os

root = "/home/patate/directory/"
path = os.path.join(root, "targetdirectory")

for r, d, f in os.walk(path):
    for file in f:
        print(os.path.join(root, file))

Unfortunatly it doesn’t work properly.
I get all the files, but not their complete paths.

For example if the dir struct would be:

/home/patate/directory/targetdirectory/123/456/789/file.txt

It would print:

/home/patate/directory/targetdirectory/file.txt

What I need is the first result. Any help would be greatly appreciated! Thanks.

Asked By: thomytheyon

||

Answers:

Use os.path.join to concatenate the directory and file name:

for path, subdirs, files in os.walk(root):
    for name in files:
        print(os.path.join(path, name))

Note the usage of path and not root in the concatenation, since using root would be incorrect.


In Python 3.4, the pathlib module was added for easier path manipulations. So the equivalent to os.path.join would be:

pathlib.PurePath(path, name)

The advantage of pathlib is that you can use a variety of useful methods on paths. If you use the concrete Path variant you can also do actual OS calls through them, like changing into a directory, deleting the path, opening the file it points to and much more.

Answered By: Eli Bendersky

Just in case… Getting all files in the directory and subdirectories matching some pattern (*.py for example):

import os
from fnmatch import fnmatch

root = '/some/directory'
pattern = "*.py"

for path, subdirs, files in os.walk(root):
    for name in files:
        if fnmatch(name, pattern):
            print(os.path.join(path, name))
Answered By: Ivan Pirog

You can take a look at this sample I made. It uses the os.path.walk function which is deprecated beware.Uses a list to store all the filepaths

root = "Your root directory"
ex = ".txt"
where_to = "Wherever you wanna write your file to"
def fileWalker(ext,dirname,names):
    '''
    checks files in names'''
    pat = "*" + ext[0]
    for f in names:
        if fnmatch.fnmatch(f,pat):
            ext[1].append(os.path.join(dirname,f))


def writeTo(fList):

    with open(where_to,"w") as f:
        for di_r in fList:
            f.write(di_r + "n")






if __name__ == '__main__':
    li = []
    os.path.walk(root,fileWalker,[ex,li])

    writeTo(li)
Answered By: devsaw

Here is a one-liner:

import os

[val for sublist in [[os.path.join(i[0], j) for j in i[2]] for i in os.walk('./')] for val in sublist]
# Meta comment to ease selecting text

The outer most val for sublist in ... loop flattens the list to be one dimensional. The j loop collects a list of every file basename and joins it to the current path. Finally, the i loop iterates over all directories and sub directories.

This example uses the hard-coded path ./ in the os.walk(...) call, you can supplement any path string you like.

Note: os.path.expanduser and/or os.path.expandvars can be used for paths strings like ~/

Extending this example:

Its easy to add in file basename tests and directoryname tests.

For Example, testing for *.jpg files:

... for j in i[2] if j.endswith('.jpg')] ...

Additionally, excluding the .git directory:

... for i in os.walk('./') if '.git' not in i[0].split('/')]
Answered By: ThorSummoner

A bit simpler one-liner:

import os
from itertools import product, chain

chain.from_iterable([[os.sep.join(w) for w in product([i[0]], i[2])] for i in os.walk(dir)])
Answered By: Daniel

Couldn’t comment so writing answer here. This is the clearest one-line I have seen:

import os
[os.path.join(path, name) for path, subdirs, files in os.walk(root) for name in files]
Answered By: Mong H. Ng

Since every example here is just using walk (with join), i’d like to show a nice example and comparison with listdir:

import os, time

def listFiles1(root): # listdir
    allFiles = []; walk = [root]
    while walk:
        folder = walk.pop(0)+"/"; items = os.listdir(folder) # items = folders + files
        for i in items: i=folder+i; (walk if os.path.isdir(i) else allFiles).append(i)
    return allFiles

def listFiles2(root): # listdir/join (takes ~1.4x as long) (and uses '\' instead)
    allFiles = []; walk = [root]
    while walk:
        folder = walk.pop(0); items = os.listdir(folder) # items = folders + files
        for i in items: i=os.path.join(folder,i); (walk if os.path.isdir(i) else allFiles).append(i)
    return allFiles

def listFiles3(root): # walk (takes ~1.5x as long)
    allFiles = []
    for folder, folders, files in os.walk(root):
        for file in files: allFiles+=[folder.replace("\","/")+"/"+file] # folder+"\"+file still ~1.5x
    return allFiles

def listFiles4(root): # walk/join (takes ~1.6x as long) (and uses '\' instead)
    allFiles = []
    for folder, folders, files in os.walk(root):
        for file in files: allFiles+=[os.path.join(folder,file)]
    return allFiles


for i in range(100): files = listFiles1("src") # warm up

start = time.time()
for i in range(100): files = listFiles1("src") # listdir
print("Time taken: %.2fs"%(time.time()-start)) # 0.28s

start = time.time()
for i in range(100): files = listFiles2("src") # listdir and join
print("Time taken: %.2fs"%(time.time()-start)) # 0.38s

start = time.time()
for i in range(100): files = listFiles3("src") # walk
print("Time taken: %.2fs"%(time.time()-start)) # 0.42s

start = time.time()
for i in range(100): files = listFiles4("src") # walk and join
print("Time taken: %.2fs"%(time.time()-start)) # 0.47s

So as you can see for yourself, the listdir version is much more efficient. (and that join is slow)

Answered By: Puddle

It’s just an addition, with this you can get the data into CSV format

import sys,os
try:
    import pandas as pd
except:
    os.system("pip3 install pandas")
    
root = "/home/kiran/Downloads/MainFolder" # it may have many subfolders and files inside
lst = []
from fnmatch import fnmatch
pattern = "*.csv"      #I want to get only csv files 
pattern = "*.*"        # Note: Use this pattern to get all types of files and folders 
for path, subdirs, files in os.walk(root):
    for name in files:
        if fnmatch(name, pattern):
            lst.append((os.path.join(path, name)))
df = pd.DataFrame({"filePaths":lst})
df.to_csv("filepaths.csv")
Answered By: kiran beethoju

Pretty simple solution would be to run a couple of sub process calls to export the files into CSV format:

import subprocess

# Global variables for directory being mapped

location = '.' # Enter the path here.
pattern = '*.py' # Use this if you want to only return certain filetypes
rootDir = location.rpartition('/')[-1]
outputFile = rootDir + '_directory_contents.csv'

# Find the requested data and export to CSV, specifying a pattern if needed.
find_cmd = 'find ' + location + ' -name ' + pattern +  ' -fprintf ' + outputFile + '  "%Y%M,%n,%u,%g,%s,%A+,%Pn"'
subprocess.call(find_cmd, shell=True)

That command produces comma separated values that can be easily analyzed in Excel.

f-rwxrwxrwx,1,cathy,cathy,2642,2021-06-01+00:22:00.2970880000,content-audit.py

The resulting CSV file doesn’t have a header row, but you can use a second command to add them.

# Add headers to the CSV
headers_cmd = 'sed -i.bak 1i"Permissions,Links,Owner,Group,Size,ModifiedTime,FilePath" ' + outputFile
subprocess.call(headers_cmd, shell=True)

Depending on how much data you get back, you can massage it further using Pandas. Here are some things I found useful, especially if you’re dealing with many levels of directories to look through.

Add these to your imports:

import numpy as np
import pandas as pd

Then add this to your code:

# Create DataFrame from the csv file created above.
df = pd.read_csv(outputFile)
    
# Format columns
# Get the filename and file extension from the filepath 
df['FileName'] = df['FilePath'].str.rsplit("/",1).str[-1]
df['FileExt'] = df['FileName'].str.rsplit('.',1).str[1]

# Get the full path to the files. If the path doesn't include a "/" it's the root directory
df['FullPath'] = df["FilePath"].str.rsplit("/",1).str[0]
df['FullPath'] = np.where(df['FullPath'].str.contains("/"), df['FullPath'], rootDir)

# Split the path into columns for the parent directory and its children
df['ParentDir'] = df['FullPath'].str.split("/",1).str[0]
df['SubDirs'] = df['FullPath'].str.split("/",1).str[1]
# Account for NaN returns, indicates the path is the root directory
df['SubDirs'] = np.where(df.SubDirs.str.contains('NaN'), '', df.SubDirs)

# Determine if the item is a directory or file.
df['Type'] = np.where(df['Permissions'].str.startswith('d'), 'Dir', 'File')

# Split the time stamp into date and time columns
df[['ModifiedDate', 'Time']] = df.ModifiedTime.str.rsplit('+', 1, expand=True)
df['Time'] = df['Time'].str.split('.').str[0]

# Show only files, output includes paths so you don't necessarily need to display the individual directories.
df = df[df['Type'].str.contains('File')]

# Set columns to show and their order.
df=df[['FileName','ParentDir','SubDirs','FullPath','DocType','ModifiedDate','Time', 'Size']]

filesize=[] # Create an empty list to store file sizes to convert them to something more readable.

# Go through the items and convert the filesize from bytes to something more readable.
for items in df['Size'].items():
    filesize.append(convert_bytes(items[1]))
    df['Size'] = filesize 

# Send the data to an Excel workbook with sheets by parent directory
with pd.ExcelWriter("scripts_directory_contents.xlsx") as writer:
    for directory, data in df.groupby('ParentDir'):
    data.to_excel(writer, sheet_name = directory, index=False) 
        

# To convert sizes to be more human readable
def convert_bytes(size):
    for x in ['b', 'K', 'M', 'G', 'T']:
        if size < 1024:
            return "%3.1f %s" % (size, x)
        size /= 1024

    return size
Answered By: CMSG

And this is how you list it in case you want to list the files on SharePoint. Your path will probably start after the "teams" part

    import os
    root = r"\[email protected] and ProceduresDeal DocsMy Deals"
    list = [os.path.join(path, name) for path, subdirs, files in os.walk(root) for name in files]
    print(list)
Answered By: Chadee Fouad

Another option would be using the glob module from the standard lib:

import glob

path = "/home/patate/directory/targetdirectory/**"

for path in glob.glob(path, recursive=True):
    print(path)

If you need an iterator you can use iglob as an alternative:

for file in glob.iglob(my_path, recursive=True):
    # ...
Answered By: Rotareti

Using any supported Python version (3.4+), you should use pathlib.rglob to recusrively list the contents of the current directory and all subdirectories:

from pathlib import Path


def generate_all_files(root: Path, only_files: bool = True):
    for p in root.rglob("*"):
        if only_files and not p.is_file():
            continue
        yield p


for p in generate_all_files(Path("."), only_files=False):
    print(p)

If you want something copy-pasteable:

Example

Folder structure:

$ tree . -a
.
├── a.txt
├── bar
├── b.py
├── collect.py
├── empty
├── foo
│   └── bar.bz.gz2
├── .hidden
│   └── secrect-file
└── martin
    └── thoma
        └── cv.pdf

gives:

$ python collect.py 
bar
empty
.hidden
collect.py
a.txt
b.py
martin
foo
.hidden/secrect-file
martin/thoma
martin/thoma/cv.pdf
foo/bar.bz.gz2
Answered By: Martin Thoma
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