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Checking if a folder is in the current directory

I want to check if there exist a folder in the current folder my code is in, how can this be done?
What I mean is, we have folder called code_folder, which has the python code, i want to check in this python code if in this code_folder we have another folder called folder_1, if is it exists we can store something in it, if not we creat the folder and store something in it.

try this code
import os
path = "code_folder"
# Check whether the defined path exists or not
if not os.path.exists(path):
# Create a new directory
os.makedirs(path)
print("The new directory is created!")
else:
pass

You can use pathlib for this.
if pathlib.Path("path/to/code_folder/folder_1").isdir():
print("the directory exists")
But really, you don't need to check anything, you can just use the exists_ok argument of Path.mkdir, so just:
pathlib.Path("path/to/code_folder/foler_1").mkdir(exists_ok=True)

Hi. I'd like to ask a question about directory in python [duplicate]

This question already has answers here:
How do you properly determine the current script directory?
(16 answers)
How to know/change current directory in Python shell?
(7 answers)
Closed 5 years ago.
How do I determine:
the current directory (where I was in the shell when I ran the Python script), and
where the Python file I am executing is?

To get the full path to the directory a Python file is contained in, write this in that file:
import os
dir_path = os.path.dirname(os.path.realpath(__file__))
(Note that the incantation above won't work if you've already used os.chdir() to change your current working directory, since the value of the __file__ constant is relative to the current working directory and is not changed by an os.chdir() call.)
To get the current working directory use
import os
cwd = os.getcwd()
Documentation references for the modules, constants and functions used above:
The os and os.path modules.
The __file__ constant
os.path.realpath(path) (returns "the canonical path of the specified filename, eliminating any symbolic links encountered in the path")
os.path.dirname(path) (returns "the directory name of pathname path")
os.getcwd() (returns "a string representing the current working directory")
os.chdir(path) ("change the current working directory to path")

Current working directory: os.getcwd()
And the __file__ attribute can help you find out where the file you are executing is located. This Stack Overflow post explains everything: How do I get the path of the current executed file in Python?

You may find this useful as a reference:
import os
print("Path at terminal when executing this file")
print(os.getcwd() + "\n")
print("This file path, relative to os.getcwd()")
print(__file__ + "\n")
print("This file full path (following symlinks)")
full_path = os.path.realpath(__file__)
print(full_path + "\n")
print("This file directory and name")
path, filename = os.path.split(full_path)
print(path + ' --> ' + filename + "\n")
print("This file directory only")
print(os.path.dirname(full_path))

The pathlib module, introduced in Python 3.4 (PEP 428 — The pathlib module — object-oriented filesystem paths), makes the path-related experience much much better.
pwd
/home/skovorodkin/stack
tree
.
└── scripts
├── 1.py
└── 2.py
In order to get the current working directory, use Path.cwd():
from pathlib import Path
print(Path.cwd()) # /home/skovorodkin/stack
To get an absolute path to your script file, use the Path.resolve() method:
print(Path(__file__).resolve()) # /home/skovorodkin/stack/scripts/1.py
And to get the path of a directory where your script is located, access .parent (it is recommended to call .resolve() before .parent):
print(Path(__file__).resolve().parent) # /home/skovorodkin/stack/scripts
Remember that __file__ is not reliable in some situations: How do I get the path of the current executed file in Python?.
Please note, that Path.cwd(), Path.resolve() and other Path methods return path objects (PosixPath in my case), not strings. In Python 3.4 and 3.5 that caused some pain, because open built-in function could only work with string or bytes objects, and did not support Path objects, so you had to convert Path objects to strings or use the Path.open() method, but the latter option required you to change old code:
File scripts/2.py
from pathlib import Path
p = Path(__file__).resolve()
with p.open() as f: pass
with open(str(p)) as f: pass
with open(p) as f: pass
print('OK')
Output
python3.5 scripts/2.py
Traceback (most recent call last):
File "scripts/2.py", line 11, in <module>
with open(p) as f:
TypeError: invalid file: PosixPath('/home/skovorodkin/stack/scripts/2.py')
As you can see, open(p) does not work with Python 3.5.
PEP 519 — Adding a file system path protocol, implemented in Python 3.6, adds support of PathLike objects to the open function, so now you can pass Path objects to the open function directly:
python3.6 scripts/2.py
OK

To get the current directory full path
>>import os
>>print os.getcwd()
Output: "C :\Users\admin\myfolder"
To get the current directory folder name alone
>>import os
>>str1=os.getcwd()
>>str2=str1.split('\\')
>>n=len(str2)
>>print str2[n-1]
Output: "myfolder"

Pathlib can be used this way to get the directory containing the current script:
import pathlib
filepath = pathlib.Path(__file__).resolve().parent

If you are trying to find the current directory of the file you are currently in:
OS agnostic way:
dirname, filename = os.path.split(os.path.abspath(__file__))

If you're using Python 3.4, there is the brand new higher-level pathlib module which allows you to conveniently call pathlib.Path.cwd() to get a Path object representing your current working directory, along with many other new features.
More info on this new API can be found here.

To get the current directory full path:
os.path.realpath('.')

Answer to #1:
If you want the current directory, do this:
import os
os.getcwd()
If you want just any folder name and you have the path to that folder, do this:
def get_folder_name(folder):
'''
Returns the folder name, given a full folder path
'''
return folder.split(os.sep)[-1]
Answer to #2:
import os
print os.path.abspath(__file__)

I think the most succinct way to find just the name of your current execution context would be:
current_folder_path, current_folder_name = os.path.split(os.getcwd())

If you're searching for the location of the currently executed script, you can use sys.argv[0] to get the full path.

For question 1, use os.getcwd() # Get working directory and os.chdir(r'D:\Steam\steamapps\common') # Set working directory
I recommend using sys.argv[0] for question 2 because sys.argv is immutable and therefore always returns the current file (module object path) and not affected by os.chdir(). Also you can do like this:
import os
this_py_file = os.path.realpath(__file__)
# vvv Below comes your code vvv #
But that snippet and sys.argv[0] will not work or will work weird when compiled by PyInstaller, because magic properties are not set in __main__ level and sys.argv[0] is the way your executable was called (it means that it becomes affected by the working directory).

Python, Creating a new file/folder in the same directory as the script

I've written a script that basically creates two new folders when it's run the first time, if the folders don't already exist.
import os
try:
os.makedirs('results/graphs')
except OSError:
pass
And everytime the script is run, the graphs are produced in the results/graphs folder.
But I noticed recently that if the script is run from another directory, (eg. script is in home/user/script/ but I run it from: home/user/programs/), the new folders are created in home/user/programs/.
My goal is ultimately that the folders are created only in the script folder and all eventual graphs that are produced will thus be destined to home/user/script/results/graphs.
Is there someway to achieve this using python?
I'm using Debian 8 and python 2.7.13. The graphs are produced using matplotlib.

It's a solution for me, check it (I tried it on windows):
import os
d = os.path.dirname(__file__) # directory of script
p = r'{}/results/graphs'.format(d) # path to be created
try:
os.makedirs(p)
except OSError:
pass

In your script results/graphs refers to the results folder within whatever folder you ran the script from. If you always want the folders to be created in /home/user/script/results/graphs use ~/script/results/graphs in your script instead. Alternatively you could hard code the whole path, /home/user/script/results/graphs.

There is another way, apart from hard coding which is mentioned earlier. You can use __file__ to extract the path of the file where the script is present. And then you can use it to append as parent path to results/graphs

All you have to do is to define the path where your script is. This can be done within the code as:
path = 'home/user/script/'
or by getting the input from the user using a filedialog box.
Then you can use the path variable to make the directories.
os.makedirs(path + 'results/graphs')

##creating folder
import tkinter as tk
from tkinter import filedialog
root = tk.Tk()
root.withdraw()
file_path = filedialog.askopenfilename()
dirctry = os.path.dirname(file_path)
toolSharePath = os.path.abspath(dirctry)
final_directory = os.path.join(toolSharePath, r'new_folder')
if not os.path.exists(final_directory):
os.makedirs(final_directory,exist_ok=True)

check if a directory is already created python

A very basic question, I have a module that creates directories on the fly, however, sometimes I want to put more than one file in a dir. If this happens, python rises an exception and says that the dir is already created, how can I avoid this and check if the dir is already created or not?
The save module looks something like this:
def createdirs(realiz):
# Create all the necessary directories
path = "./doubDifferences_probandamp_realiz%d/"%realiz
os.mkdir(path,0755)
directory = os.path.split(path)[0]
return directory
On the main program, I have this:
for realiz in range(1,1000):
for i in range(dim):
for j in range(i+1,i+4):
...
dirspaths = mod_savefile.createdirs(realiz)

You could go for a try except:
try:
os.mkdir(path,0755)
except OSError:
pass
“Easier to ask forgiveness than permission !”
Also this method is more safe that testing the directory before doing mkdir. Indeed, it is fairly possible that between the two os call implied by ispath and mkdir the directory may have been created or destroyed by another thread.

This should cover you. Just test if it is a directory before you try to create it.
if not os.path.isdir(path)
os.mkdir(path,0755)

You have several ways. Either use os.path.isdir function:
import os.path
def createdirs(realiz):
# Create all the necessary directories
path = "./doubDifferences_probandamp_realiz%d/"%realiz
if not os.path.isdir(path): # if not exists
os.mkdir(path,0755) # create it
directory = os.path.split(path)[0]
return directory
Or handle the exception.

Cross-platform way of getting temp directory in Python

Is there a cross-platform way of getting the path to the temp directory in Python 2.6?
For example, under Linux that would be /tmp, while under XP C:\Documents and settings\[user]\Application settings\Temp.

That would be the tempfile module.
It has functions to get the temporary directory, and also has some shortcuts to create temporary files and directories in it, either named or unnamed.
Example:
import tempfile
print tempfile.gettempdir() # prints the current temporary directory
f = tempfile.TemporaryFile()
f.write('something on temporaryfile')
f.seek(0) # return to beginning of file
print f.read() # reads data back from the file
f.close() # temporary file is automatically deleted here
For completeness, here's how it searches for the temporary directory, according to the documentation:
The directory named by the TMPDIR environment variable.
The directory named by the TEMP environment variable.
The directory named by the TMP environment variable.
A platform-specific location:
On RiscOS, the directory named by the Wimp$ScrapDir environment variable.
On Windows, the directories C:\TEMP, C:\TMP, \TEMP, and \TMP, in that order.
On all other platforms, the directories /tmp, /var/tmp, and /usr/tmp, in that order.
As a last resort, the current working directory.

This should do what you want:
print(tempfile.gettempdir())
For me on my Windows box, I get:
c:\temp
and on my Linux box I get:
/tmp

I use:
from pathlib import Path
import platform
import tempfile
tempdir = Path("/tmp" if platform.system() == "Darwin" else tempfile.gettempdir())
This is because on MacOS, i.e. Darwin, tempfile.gettempdir() and os.getenv('TMPDIR') return a value such as '/var/folders/nj/269977hs0_96bttwj2gs_jhhp48z54/T'; it is one that I do not always want.

The simplest way, based on #nosklo's comment and answer:
import tempfile
tmp = tempfile.mkdtemp()
But if you want to manually control the creation of the directories:
import os
from tempfile import gettempdir
tmp = os.path.join(gettempdir(), '.{}'.format(hash(os.times())))
os.makedirs(tmp)
That way you can easily clean up after yourself when you are done (for privacy, resources, security, whatever) with:
from shutil import rmtree
rmtree(tmp, ignore_errors=True)
This is similar to what applications like Google Chrome and Linux systemd do. They just use a shorter hex hash and an app-specific prefix to "advertise" their presence.

Why so many complex answers?
I just use this
(os.getenv("TEMP") if os.name=="nt" else "/tmp") + os.path.sep + "tempfilename.tmp"