Python | os.makedirs () method

makedirs | Python Methods and Functions

All functions in the os module raise OSError for invalid or unreachable file names and paths or other arguments that are of the correct type but are not accepted by the operating system.

os.makedirs () in Python is used to create a directory recursively. This means that when creating the target directory, if any intermediate tier directory is missing, os.makedirs() will create them all. 
For example, consider the following path:

 / home / User / Documents / GeeksForGeeks / Authors / ihritik 

Suppose we want to create the "ihritik" directory, but the " GeeksForGeeks "and" Credits "are not available while in transit. Then os.makedirs() will create the entire inaccessible / missing directory at the specified path. First GeeksForGeeks and Authors will be created, then the ihritik directory will be created.

Syntax: os.makedirs (path , mode = 0o777, exist_ok = False)

Parameter:
path : A path-like object representing a file system path. A path-like object is either a string or bytes object representing a path.
mode (optional): A Integer value representing mode of the newly created directory..If this parameter is omitted then the default value Oo777 is used.
exist_ok (optional): A default value False is used for this parameter. If the target directory already exists an OSError is raised if its value is False otherwise not.

Return Type: This method does not return any value.

Code # 1: Using the os.makedirs () method to create a directory

# Python program to explain the os.makedirs () method

 
# module import os

import os

  
# Leaf directory

directory = "ihritik"

 
# Parent directories

parent_di r = "/ home / User / Documents / GeeksForGeeks / Authors"

 
# Path

path = os.path.join ( (parent_dir, directory)

  
# Create directory
# & # 39; ihritik & # 39;
os.makedirs (path)

print ( "Directory'% s' created " % directory)

 
# Directories & # 39; GeeksForGeeks & # 39; and & # 39; Authors & # 39; will
# be created too
# if it doesn't exist

 

 

 
# Leaf directory

directory = " c "

  
# Parent directories

parent_dir = " / home / User / Documents / Python.Engineering / a / b "

  
# Mode

mode = 0o666

 

path = os.path.join ( (parent_dir, directory)

 
# Create directory
# & # 39; c & # 39;

  
os.makedirs (path, mode)

print ( "Directory'% s' created " % directory)

  

  
# & # 39; GeeksForGeeks & # 39 ;, & # 39; a & # 39; and & # 39; b & # 39;
# will also be generated if
# does not exist

 
# If any of the intermediate level
# no catalog
# os.makedirs () method
# create them

  
The # os.makedirs () method can be
# used to create a directory tree

Exit :

 Directory 'ihritik' created Directory' c' created 

Code # 2: Errors when using os.makedirs () method

# Python program to explain met os.makedirs () code

 
# import of the os module

import os

 
The # os.makedirs () method will call
# OSE error if directory
# already exists to create

 

 
# Directory

directory = "ihritik"

 
# Path to parent directory

parent_dir = "/ home / User / Documents / GeeksForGeeks"

  
# Path

path = os.path.join ( (parent_dir, directory)

 
# Create directory
# & # 39; ihritik & # 39;
os.makedirs (path)

print ( " Directory '% s' created" % directory)

Exit :

 Traceback (most recent call last): File "makedirs.py", line 21, in os.makedirs (path) File "/usr/lib/python3.6/os.py", line 220, in makedirs mkdir (name, mode) FileExistsError: [Errno 17] File exists: '/ home / User / Documents / GeeksForGeeks / ihritik' 

Code # 3: Error handling when using the os.makedirs () method

 

# Python program to explain the os.makedirs () method

 
# import of the os module

import os

 
The # os.makedirs () method will call
# OSE error if directory
# already exists to create
# But it may be suppressed
# setting parameter value
# exist_ok as True

 
# Directory

directory = "ihritik"

 
# Path to parent directory

parent_dir = "/ home / ihritik / Desktop / GeeksForGeeks"

 
# Path

path = os.path.join ( (parent_dir, directory)

 
# Create directory
# & # 39; ihritik & # 39 ;

try :

  os.makedirs (path, exist_ok = True )

print (< / code> "Directory'% s' created successfully " % directory)

except OSError as error:

print ( " Directory '% s' can not be created " )

  

 
# Setting Existence_of_s as True
# an error already thrown
# existing directory may be suppressed
# but another OSError may be raised
# because of another error like
# invalid path

Exit:

 Directory 'ihritik' created successfully 

Link: https:// docs. python.org/3/library/os.html





Python | os.makedirs () method: StackOverflow Questions

Python"s os.makedirs doesn"t understand "~" in my path

I have a little problem with ~ in my paths.

This code example creates some directories called "~/some_dir" and do not understand that I wanted to create some_dir in my home directory.

my_dir = "~/some_dir"
if not os.path.exists(my_dir):
    os.makedirs(my_dir)

Note this is on a Linux-based system.

Answer #1

os.listdir() - list in the current directory

With listdir in os module you get the files and the folders in the current dir

 import os
 arr = os.listdir()
 print(arr)
 
 >>> ["$RECYCLE.BIN", "work.txt", "3ebooks.txt", "documents"]

Looking in a directory

arr = os.listdir("c:\files")

glob from glob

with glob you can specify a type of file to list like this

import glob

txtfiles = []
for file in glob.glob("*.txt"):
    txtfiles.append(file)

glob in a list comprehension

mylist = [f for f in glob.glob("*.txt")]

get the full path of only files in the current directory

import os
from os import listdir
from os.path import isfile, join

cwd = os.getcwd()
onlyfiles = [os.path.join(cwd, f) for f in os.listdir(cwd) if 
os.path.isfile(os.path.join(cwd, f))]
print(onlyfiles) 

["G:\getfilesname\getfilesname.py", "G:\getfilesname\example.txt"]

Getting the full path name with os.path.abspath

You get the full path in return

 import os
 files_path = [os.path.abspath(x) for x in os.listdir()]
 print(files_path)
 
 ["F:\documentiapplications.txt", "F:\documenticollections.txt"]

Walk: going through sub directories

os.walk returns the root, the directories list and the files list, that is why I unpacked them in r, d, f in the for loop; it, then, looks for other files and directories in the subfolders of the root and so on until there are no subfolders.

import os

# Getting the current work directory (cwd)
thisdir = os.getcwd()

# r=root, d=directories, f = files
for r, d, f in os.walk(thisdir):
    for file in f:
        if file.endswith(".docx"):
            print(os.path.join(r, file))

os.listdir(): get files in the current directory (Python 2)

In Python 2, if you want the list of the files in the current directory, you have to give the argument as "." or os.getcwd() in the os.listdir method.

 import os
 arr = os.listdir(".")
 print(arr)
 
 >>> ["$RECYCLE.BIN", "work.txt", "3ebooks.txt", "documents"]

To go up in the directory tree

# Method 1
x = os.listdir("..")

# Method 2
x= os.listdir("/")

Get files: os.listdir() in a particular directory (Python 2 and 3)

 import os
 arr = os.listdir("F:\python")
 print(arr)
 
 >>> ["$RECYCLE.BIN", "work.txt", "3ebooks.txt", "documents"]

Get files of a particular subdirectory with os.listdir()

import os

x = os.listdir("./content")

os.walk(".") - current directory

 import os
 arr = next(os.walk("."))[2]
 print(arr)
 
 >>> ["5bs_Turismo1.pdf", "5bs_Turismo1.pptx", "esperienza.txt"]

next(os.walk(".")) and os.path.join("dir", "file")

 import os
 arr = []
 for d,r,f in next(os.walk("F:\_python")):
     for file in f:
         arr.append(os.path.join(r,file))

 for f in arr:
     print(files)

>>> F:\_python\dict_class.py
>>> F:\_python\programmi.txt

next(os.walk("F:\") - get the full path - list comprehension

 [os.path.join(r,file) for r,d,f in next(os.walk("F:\_python")) for file in f]
 
 >>> ["F:\_python\dict_class.py", "F:\_python\programmi.txt"]

os.walk - get full path - all files in sub dirs**

x = [os.path.join(r,file) for r,d,f in os.walk("F:\_python") for file in f]
print(x)

>>> ["F:\_python\dict.py", "F:\_python\progr.txt", "F:\_python\readl.py"]

os.listdir() - get only txt files

 arr_txt = [x for x in os.listdir() if x.endswith(".txt")]
 print(arr_txt)
 
 >>> ["work.txt", "3ebooks.txt"]

Using glob to get the full path of the files

If I should need the absolute path of the files:

from path import path
from glob import glob
x = [path(f).abspath() for f in glob("F:\*.txt")]
for f in x:
    print(f)

>>> F:acquistionline.txt
>>> F:acquisti_2018.txt
>>> F:ootstrap_jquery_ecc.txt

Using os.path.isfile to avoid directories in the list

import os.path
listOfFiles = [f for f in os.listdir() if os.path.isfile(f)]
print(listOfFiles)

>>> ["a simple game.py", "data.txt", "decorator.py"]

Using pathlib from Python 3.4

import pathlib

flist = []
for p in pathlib.Path(".").iterdir():
    if p.is_file():
        print(p)
        flist.append(p)

 >>> error.PNG
 >>> exemaker.bat
 >>> guiprova.mp3
 >>> setup.py
 >>> speak_gui2.py
 >>> thumb.PNG

With list comprehension:

flist = [p for p in pathlib.Path(".").iterdir() if p.is_file()]

Alternatively, use pathlib.Path() instead of pathlib.Path(".")

Use glob method in pathlib.Path()

import pathlib

py = pathlib.Path().glob("*.py")
for file in py:
    print(file)

>>> stack_overflow_list.py
>>> stack_overflow_list_tkinter.py

Get all and only files with os.walk

import os
x = [i[2] for i in os.walk(".")]
y=[]
for t in x:
    for f in t:
        y.append(f)
print(y)

>>> ["append_to_list.py", "data.txt", "data1.txt", "data2.txt", "data_180617", "os_walk.py", "READ2.py", "read_data.py", "somma_defaltdic.py", "substitute_words.py", "sum_data.py", "data.txt", "data1.txt", "data_180617"]

Get only files with next and walk in a directory

 import os
 x = next(os.walk("F://python"))[2]
 print(x)
 
 >>> ["calculator.bat","calculator.py"]

Get only directories with next and walk in a directory

 import os
 next(os.walk("F://python"))[1] # for the current dir use (".")
 
 >>> ["python3","others"]

Get all the subdir names with walk

for r,d,f in os.walk("F:\_python"):
    for dirs in d:
        print(dirs)

>>> .vscode
>>> pyexcel
>>> pyschool.py
>>> subtitles
>>> _metaprogramming
>>> .ipynb_checkpoints

os.scandir() from Python 3.5 and greater

import os
x = [f.name for f in os.scandir() if f.is_file()]
print(x)

>>> ["calculator.bat","calculator.py"]

# Another example with scandir (a little variation from docs.python.org)
# This one is more efficient than os.listdir.
# In this case, it shows the files only in the current directory
# where the script is executed.

import os
with os.scandir() as i:
    for entry in i:
        if entry.is_file():
            print(entry.name)

>>> ebookmaker.py
>>> error.PNG
>>> exemaker.bat
>>> guiprova.mp3
>>> setup.py
>>> speakgui4.py
>>> speak_gui2.py
>>> speak_gui3.py
>>> thumb.PNG

Examples:

Ex. 1: How many files are there in the subdirectories?

In this example, we look for the number of files that are included in all the directory and its subdirectories.

import os

def count(dir, counter=0):
    "returns number of files in dir and subdirs"
    for pack in os.walk(dir):
        for f in pack[2]:
            counter += 1
    return dir + " : " + str(counter) + "files"

print(count("F:\python"))

>>> "F:\python" : 12057 files"

Ex.2: How to copy all files from a directory to another?

A script to make order in your computer finding all files of a type (default: pptx) and copying them in a new folder.

import os
import shutil
from path import path

destination = "F:\file_copied"
# os.makedirs(destination)

def copyfile(dir, filetype="pptx", counter=0):
    "Searches for pptx (or other - pptx is the default) files and copies them"
    for pack in os.walk(dir):
        for f in pack[2]:
            if f.endswith(filetype):
                fullpath = pack[0] + "\" + f
                print(fullpath)
                shutil.copy(fullpath, destination)
                counter += 1
    if counter > 0:
        print("-" * 30)
        print("	==> Found in: `" + dir + "` : " + str(counter) + " files
")

for dir in os.listdir():
    "searches for folders that starts with `_`"
    if dir[0] == "_":
        # copyfile(dir, filetype="pdf")
        copyfile(dir, filetype="txt")


>>> _compiti18Compito Contabilità 1conti.txt
>>> _compiti18Compito Contabilità 1modula4.txt
>>> _compiti18Compito Contabilità 1moduloa4.txt
>>> ------------------------
>>> ==> Found in: `_compiti18` : 3 files

Ex. 3: How to get all the files in a txt file

In case you want to create a txt file with all the file names:

import os
mylist = ""
with open("filelist.txt", "w", encoding="utf-8") as file:
    for eachfile in os.listdir():
        mylist += eachfile + "
"
    file.write(mylist)

Example: txt with all the files of an hard drive

"""
We are going to save a txt file with all the files in your directory.
We will use the function walk()
"""

import os

# see all the methods of os
# print(*dir(os), sep=", ")
listafile = []
percorso = []
with open("lista_file.txt", "w", encoding="utf-8") as testo:
    for root, dirs, files in os.walk("D:\"):
        for file in files:
            listafile.append(file)
            percorso.append(root + "\" + file)
            testo.write(file + "
")
listafile.sort()
print("N. of files", len(listafile))
with open("lista_file_ordinata.txt", "w", encoding="utf-8") as testo_ordinato:
    for file in listafile:
        testo_ordinato.write(file + "
")

with open("percorso.txt", "w", encoding="utf-8") as file_percorso:
    for file in percorso:
        file_percorso.write(file + "
")

os.system("lista_file.txt")
os.system("lista_file_ordinata.txt")
os.system("percorso.txt")

All the file of C: in one text file

This is a shorter version of the previous code. Change the folder where to start finding the files if you need to start from another position. This code generate a 50 mb on text file on my computer with something less then 500.000 lines with files with the complete path.

import os

with open("file.txt", "w", encoding="utf-8") as filewrite:
    for r, d, f in os.walk("C:\"):
        for file in f:
            filewrite.write(f"{r + file}
")

How to write a file with all paths in a folder of a type

With this function you can create a txt file that will have the name of a type of file that you look for (ex. pngfile.txt) with all the full path of all the files of that type. It can be useful sometimes, I think.

import os

def searchfiles(extension=".ttf", folder="H:\"):
    "Create a txt file with all the file of a type"
    with open(extension[1:] + "file.txt", "w", encoding="utf-8") as filewrite:
        for r, d, f in os.walk(folder):
            for file in f:
                if file.endswith(extension):
                    filewrite.write(f"{r + file}
")

# looking for png file (fonts) in the hard disk H:
searchfiles(".png", "H:\")

>>> H:4bs_18Dolphins5.png
>>> H:4bs_18Dolphins6.png
>>> H:4bs_18Dolphins7.png
>>> H:5_18marketing htmlassetsimageslogo2.png
>>> H:7z001.png
>>> H:7z002.png

(New) Find all files and open them with tkinter GUI

I just wanted to add in this 2019 a little app to search for all files in a dir and be able to open them by doubleclicking on the name of the file in the list. enter image description here

import tkinter as tk
import os

def searchfiles(extension=".txt", folder="H:\"):
    "insert all files in the listbox"
    for r, d, f in os.walk(folder):
        for file in f:
            if file.endswith(extension):
                lb.insert(0, r + "\" + file)

def open_file():
    os.startfile(lb.get(lb.curselection()[0]))

root = tk.Tk()
root.geometry("400x400")
bt = tk.Button(root, text="Search", command=lambda:searchfiles(".png", "H:\"))
bt.pack()
lb = tk.Listbox(root)
lb.pack(fill="both", expand=1)
lb.bind("<Double-Button>", lambda x: open_file())
root.mainloop()

Answer #2

a fresh answer to a very old question:

starting from python 3.2 you can do this:

import os
path = "/home/dail/first/second/third"
os.makedirs(path, exist_ok=True)

thanks to the exist_ok flag this will not even complain if the directory exists (depending on your needs....).


starting from python 3.4 (which includes the pathlib module) you can do this:

from pathlib import Path
path = Path("/home/dail/first/second/third")
path.mkdir(parents=True)

starting from python 3.5 mkdir also has an exist_ok flag - setting it to True will raise no exception if the directory exists:

path.mkdir(parents=True, exist_ok=True)

Answer #3

Starting from Python 3.5, pathlib.Path.mkdir has an exist_ok flag:

from pathlib import Path
path = Path("/my/directory/filename.txt")
path.parent.mkdir(parents=True, exist_ok=True) 
# path.parent ~ os.path.dirname(path)

This recursively creates the directory and does not raise an exception if the directory already exists.

(just as os.makedirs got an exist_ok flag starting from python 3.2 e.g os.makedirs(path, exist_ok=True))


Note: when i posted this answer none of the other answers mentioned exist_ok...

Answer #4

Good news, PyDrive has first class support on CoLab! PyDrive is a wrapper for the Google Drive python client. Here is an example on how you would download ALL files from a folder, similar to using glob + *:

!pip install -U -q PyDrive
import os
from pydrive.auth import GoogleAuth
from pydrive.drive import GoogleDrive
from google.colab import auth
from oauth2client.client import GoogleCredentials

# 1. Authenticate and create the PyDrive client.
auth.authenticate_user()
gauth = GoogleAuth()
gauth.credentials = GoogleCredentials.get_application_default()
drive = GoogleDrive(gauth)

# choose a local (colab) directory to store the data.
local_download_path = os.path.expanduser("~/data")
try:
  os.makedirs(local_download_path)
except: pass

# 2. Auto-iterate using the query syntax
#    https://developers.google.com/drive/v2/web/search-parameters
file_list = drive.ListFile(
    {"q": ""1SooKSw8M4ACbznKjnNrYvJ5wxuqJ-YCk" in parents"}).GetList()

for f in file_list:
  # 3. Create & download by id.
  print("title: %s, id: %s" % (f["title"], f["id"]))
  fname = os.path.join(local_download_path, f["title"])
  print("downloading to {}".format(fname))
  f_ = drive.CreateFile({"id": f["id"]})
  f_.GetContentFile(fname)


with open(fname, "r") as f:
  print(f.read())

Notice that the arguments to drive.ListFile is a dictionary that coincides with the parameters used by Google Drive HTTP API (you can customize the q parameter to be tuned to your use-case).

Know that in all cases, files/folders are encoded by id"s (peep the 1SooKSw8M4ACbznKjnNrYvJ5wxuqJ-YCk) on Google Drive. This requires that you search Google Drive for the specific id corresponding to the folder you want to root your search in.

For example, navigate to the folder "/projects/my_project/my_data" that is located in your Google Drive.

Google Drive

See that it contains some files, in which we want to download to CoLab. To get the id of the folder in order to use it by PyDrive, look at the url and extract the id parameter. In this case, the url corresponding to the folder was:

https://drive.google.com/drive/folders/1SooKSw8M4ACbznKjnNrYvJ5wxuqJ-YCk

Where the id is the last piece of the url: 1SooKSw8M4ACbznKjnNrYvJ5wxuqJ-YCk.

Answer #5

On Python ‚â• 3.5, use pathlib.Path.mkdir:

from pathlib import Path
Path("/my/directory").mkdir(parents=True, exist_ok=True)

For older versions of Python, I see two answers with good qualities, each with a small flaw, so I will give my take on it:

Try os.path.exists, and consider os.makedirs for the creation.

import os
if not os.path.exists(directory):
    os.makedirs(directory)

As noted in comments and elsewhere, there"s a race condition – if the directory is created between the os.path.exists and the os.makedirs calls, the os.makedirs will fail with an OSError. Unfortunately, blanket-catching OSError and continuing is not foolproof, as it will ignore a failure to create the directory due to other factors, such as insufficient permissions, full disk, etc.

One option would be to trap the OSError and examine the embedded error code (see Is there a cross-platform way of getting information from Python’s OSError):

import os, errno

try:
    os.makedirs(directory)
except OSError as e:
    if e.errno != errno.EEXIST:
        raise

Alternatively, there could be a second os.path.exists, but suppose another created the directory after the first check, then removed it before the second one – we could still be fooled.

Depending on the application, the danger of concurrent operations may be more or less than the danger posed by other factors such as file permissions. The developer would have to know more about the particular application being developed and its expected environment before choosing an implementation.

Modern versions of Python improve this code quite a bit, both by exposing FileExistsError (in 3.3+)...

try:
    os.makedirs("path/to/directory")
except FileExistsError:
    # directory already exists
    pass

...and by allowing a keyword argument to os.makedirs called exist_ok (in 3.2+).

os.makedirs("path/to/directory", exist_ok=True)  # succeeds even if directory exists.

Answer #6

Python 3.5+:

import pathlib
pathlib.Path("/my/directory").mkdir(parents=True, exist_ok=True) 

pathlib.Path.mkdir as used above recursively creates the directory and does not raise an exception if the directory already exists. If you don"t need or want the parents to be created, skip the parents argument.

Python 3.2+:

Using pathlib:

If you can, install the current pathlib backport named pathlib2. Do not install the older unmaintained backport named pathlib. Next, refer to the Python 3.5+ section above and use it the same.

If using Python 3.4, even though it comes with pathlib, it is missing the useful exist_ok option. The backport is intended to offer a newer and superior implementation of mkdir which includes this missing option.

Using os:

import os
os.makedirs(path, exist_ok=True)

os.makedirs as used above recursively creates the directory and does not raise an exception if the directory already exists. It has the optional exist_ok argument only if using Python 3.2+, with a default value of False. This argument does not exist in Python 2.x up to 2.7. As such, there is no need for manual exception handling as with Python 2.7.

Python 2.7+:

Using pathlib:

If you can, install the current pathlib backport named pathlib2. Do not install the older unmaintained backport named pathlib. Next, refer to the Python 3.5+ section above and use it the same.

Using os:

import os
try: 
    os.makedirs(path)
except OSError:
    if not os.path.isdir(path):
        raise

While a naive solution may first use os.path.isdir followed by os.makedirs, the solution above reverses the order of the two operations. In doing so, it prevents a common race condition having to do with a duplicated attempt at creating the directory, and also disambiguates files from directories.

Note that capturing the exception and using errno is of limited usefulness because OSError: [Errno 17] File exists, i.e. errno.EEXIST, is raised for both files and directories. It is more reliable simply to check if the directory exists.

Alternative:

mkpath creates the nested directory, and does nothing if the directory already exists. This works in both Python 2 and 3.

import distutils.dir_util
distutils.dir_util.mkpath(path)

Per Bug 10948, a severe limitation of this alternative is that it works only once per python process for a given path. In other words, if you use it to create a directory, then delete the directory from inside or outside Python, then use mkpath again to recreate the same directory, mkpath will simply silently use its invalid cached info of having previously created the directory, and will not actually make the directory again. In contrast, os.makedirs doesn"t rely on any such cache. This limitation may be okay for some applications.


With regard to the directory"s mode, please refer to the documentation if you care about it.

Answer #7

For Python ‚â• 3.5, use pathlib.Path.mkdir:

import pathlib
pathlib.Path("/tmp/path/to/desired/directory").mkdir(parents=True, exist_ok=True)

The exist_ok parameter was added in Python 3.5.


For Python ≥ 3.2, os.makedirs has an optional third argument exist_ok that, when True, enables the mkdir -p functionality—unless mode is provided and the existing directory has different permissions than the intended ones; in that case, OSError is raised as previously:

import os
os.makedirs("/tmp/path/to/desired/directory", exist_ok=True)

For even older versions of Python, you can use os.makedirs and ignore the error:

import errno    
import os

def mkdir_p(path):
    try:
        os.makedirs(path)
    except OSError as exc:  # Python ‚â• 2.5
        if exc.errno == errno.EEXIST and os.path.isdir(path):
            pass
        # possibly handle other errno cases here, otherwise finally:
        else:
            raise

Answer #8

The os.makedirs function does this. Try the following:

import os
import errno

filename = "/foo/bar/baz.txt"
if not os.path.exists(os.path.dirname(filename)):
    try:
        os.makedirs(os.path.dirname(filename))
    except OSError as exc: # Guard against race condition
        if exc.errno != errno.EEXIST:
            raise

with open(filename, "w") as f:
    f.write("FOOBAR")

The reason to add the try-except block is to handle the case when the directory was created between the os.path.exists and the os.makedirs calls, so that to protect us from race conditions.


In Python 3.2+, there is a more elegant way that avoids the race condition above:

import os

filename = "/foo/bar/baz.txt"
os.makedirs(os.path.dirname(filename), exist_ok=True)
with open(filename, "w") as f:
    f.write("FOOBAR")

Answer #9

Using try except and the right error code from errno module gets rid of the race condition and is cross-platform:

import os
import errno

def make_sure_path_exists(path):
    try:
        os.makedirs(path)
    except OSError as exception:
        if exception.errno != errno.EEXIST:
            raise

In other words, we try to create the directories, but if they already exist we ignore the error. On the other hand, any other error gets reported. For example, if you create dir "a" beforehand and remove all permissions from it, you will get an OSError raised with errno.EACCES (Permission denied, error 13).

Answer #10

There is no way to create a file without opening it There is os.mknod("newfile.txt") (but it requires root privileges on OSX). The system call to create a file is actually open() with the O_CREAT flag. So no matter how, you"ll always open the file.

So the easiest way to simply create a file without truncating it in case it exists is this:

open(x, "a").close()

Actually you could omit the .close() since the refcounting GC of CPython will close it immediately after the open() statement finished - but it"s cleaner to do it explicitely and relying on CPython-specific behaviour is not good either.

In case you want touch"s behaviour (i.e. update the mtime in case the file exists):

import os
def touch(path):
    with open(path, "a"):
        os.utime(path, None)

You could extend this to also create any directories in the path that do not exist:

basedir = os.path.dirname(path)
if not os.path.exists(basedir):
    os.makedirs(basedir)

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