Python list directory, subdirectory, and files

Question

I'm trying to make a script to list all directories, subdirectories, 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))

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

For example, if the directory struct would be:

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

It would print:

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

I need the first result.

Answer 1

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:

It’s 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('/')]

Answer 2

Using any supported Python version (3.4+), you should use pathlib.rglob to recursively 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

Answer 3

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

import os

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.

Answer 4

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)

Answer 5

You can take a look at this sample I made. It uses the os.path.walk function which is deprecated beware. It 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)

Answer 6

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

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):
    # ...

Answer 7

A 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+,%P\n"'
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

Answer 8

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"\\mycompany.sharepoint.com@SSL\DavWWWRoot\teams\MyFolder\Policies and Procedures\Deal Docs\My Deals"
list = [os.path.join(path, name) for path, subdirs, files in os.walk(root) for name in files]
print(list)

Answer 9

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")

Answer 10

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))

Answer 11

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)])