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python Path from pathlib module doesn't add path separator

I have the following issue using Path from the pathlib library:
from pathlib import Path
import os
path_list = ["path","to","directory1","directory2"]
join_str = os.path.sep
my_path = join_str.join(path_list)
my_path = str(Path(my_path))
dir_exists = os.path.isdir(my_path)
Output:
my_path = "path/to/directory1/directory2"
dir_exists = False
I have tried with Path itself and manually giving it a path that I know exists:
(Pdb) test2 = str(Path("Users","userK","my_directory"))
(Pdb) os.path.isdir(test2)
False
(Pdb) test2
Users/userK/my_directory
I was wondering why Path doesn't simply add the backslash at the beginning? Is there a better way to build my path with the needed path separator from a list?
I didn't notice it before, because I use os.getcwd() and add to it to build my paths:
(Pdb) my_cwd = os.getcwd()
(Pdb) my_path = str(Path(my_cwd,"directory1"))
(Pdb) os.path.isdir(my_path)
True
However, I am not in the position code-wise to explicitly build the path I need, hence the issue.
Thanks!

Path cannot assume how the result will be used further, so it constructs the path object from the input data as is.
I will demonstrate with examples.
To begin with, prepare the directories in /tmp:
zyzop#localhost:~$ cd /tmp/
zyzop#localhost:/tmp$ pwd
/tmp
zyzop#localhost:/tmp$ mkdir -p path/to/directory1/directory2
zyzop#localhost:/tmp$ ls -ld path/to/directory1/directory2
drwxr-xr-x 2 zyzop zyzop 4096 Jul 8 03:28 path/to/directory1/directory2
zyzop#localhost:/tmp$ python3
Let's create a relative path from the list of its parts
>>> from pathlib import Path
>>> path_list = ["path", "to", "directory1", "directory2"]
>>> my_path = Path(*path_list)
>>> my_path
PosixPath('path/to/directory1/directory2')
Since the relative path exists in the current directory, we can check its existence
>>> my_path.exists()
True
>>> my_path.is_absolute()
False
>>> my_path.resolve()
PosixPath('/tmp/path/to/directory1/directory2')
There are two ways to create an absolute path. The first is to attach the path to the current directory to the beginning of the path
>>> my_path1 = Path(Path().cwd(), *path_list)
>>> my_path1
PosixPath('/tmp/path/to/directory1/directory2')
>>> my_path1.is_absolute()
True
>>> my_path1.exists()
True
The second way is to explicitly specify the path from the root
>>> path_list2 = ["/", "tmp", "path", "to", "directory1", "directory2"]
>>> my_path2 = Path(*path_list2)
>>> my_path2
PosixPath('/tmp/path/to/directory1/directory2')
>>> my_path2.exists()
True
>>> my_path2.is_absolute()
True
For example, let's see what happens to relative paths in another directory. Let's move to the home directory
>>> import os
>>> os.chdir(Path.home())
>>> Path().cwd()
PosixPath('/home/zyzop')
Let's check the existence of paths
>>> my_path
PosixPath('path/to/directory1/directory2')
>>> my_path.exists()
False
>>> my_path2
PosixPath('/tmp/path/to/directory1/directory2')
>>> my_path2.exists()
True
Now let's see how the absolute paths have changed. Since my_path is a relative path, it will always resolve relative to the current directory
>>> my_path.resolve()
PosixPath('/home/zyzop/path/to/directory1/directory2')
my_path 2 is an absolute path, so it will always be unchanged
>>> my_path2.resolve()
PosixPath('/tmp/path/to/directory1/directory2')
More details can be found here: https://docs.python.org/3/library/pathlib.html

I do not know why pathlib does not add a slash at the beginning. If I had to guess, maybe sometimes it is not obvious what is the "root" of the file system (think windows with multiple disks) and also it is probably better this way.
Anyway, easy to add it yourself.
Path("/path/to/directory1/directory2")
Path("/Users/userK/my_directory")
Path("./directory1") # or Path.cwd() / "directory1"

To get the root, you can use this syntax :
from pathlib import Path
Path.cwd().root
Another approach could be :
from pathlib import Path
Path("/") # Or Path("/").root
It works on UNIX, to test on Windows

Proper way to remove last element of delimeter while it's in a string? [duplicate]

I need to extract the name of the parent directory of a certain path. This is what it looks like:
C:\stuff\directory_i_need\subdir\file.jpg
I would like to extract directory_i_need.

import os
## first file in current dir (with full path)
file = os.path.join(os.getcwd(), os.listdir(os.getcwd())[0])
file
os.path.dirname(file) ## directory of file
os.path.dirname(os.path.dirname(file)) ## directory of directory of file
...
And you can continue doing this as many times as necessary...
Edit: from os.path, you can use either os.path.split or os.path.basename:
dir = os.path.dirname(os.path.dirname(file)) ## dir of dir of file
## once you're at the directory level you want, with the desired directory as the final path node:
dirname1 = os.path.basename(dir)
dirname2 = os.path.split(dir)[1] ## if you look at the documentation, this is exactly what os.path.basename does.

For Python 3.4+, try the pathlib module:
>>> from pathlib import Path
>>> p = Path('C:\\Program Files\\Internet Explorer\\iexplore.exe')
>>> str(p.parent)
'C:\\Program Files\\Internet Explorer'
>>> p.name
'iexplore.exe'
>>> p.suffix
'.exe'
>>> p.parts
('C:\\', 'Program Files', 'Internet Explorer', 'iexplore.exe')
>>> p.relative_to('C:\\Program Files')
WindowsPath('Internet Explorer/iexplore.exe')
>>> p.exists()
True

All you need is parent part if you use pathlib.
from pathlib import Path
p = Path(r'C:\Program Files\Internet Explorer\iexplore.exe')
print(p.parent)
Will output:
C:\Program Files\Internet Explorer
Case you need all parts (already covered in other answers) use parts:
p = Path(r'C:\Program Files\Internet Explorer\iexplore.exe')
print(p.parts)
Then you will get a list:
('C:\\', 'Program Files', 'Internet Explorer', 'iexplore.exe')
Saves tone of time.

First, see if you have splitunc() as an available function within os.path. The first item returned should be what you want... but I am on Linux and I do not have this function when I import os and try to use it.
Otherwise, one semi-ugly way that gets the job done is to use:
>>> pathname = "\\C:\\mystuff\\project\\file.py"
>>> pathname
'\\C:\\mystuff\\project\\file.py'
>>> print pathname
\C:\mystuff\project\file.py
>>> "\\".join(pathname.split('\\')[:-2])
'\\C:\\mystuff'
>>> "\\".join(pathname.split('\\')[:-1])
'\\C:\\mystuff\\project'
which shows retrieving the directory just above the file, and the directory just above that.

import os
directory = os.path.abspath('\\') # root directory
print(directory) # e.g. 'C:\'
directory = os.path.abspath('.') # current directory
print(directory) # e.g. 'C:\Users\User\Desktop'
parent_directory, directory_name = os.path.split(directory)
print(directory_name) # e.g. 'Desktop'
parent_parent_directory, parent_directory_name = os.path.split(parent_directory)
print(parent_directory_name) # e.g. 'User'
This should also do the trick.

This is what I did to extract the piece of the directory:
for path in file_list:
directories = path.rsplit('\\')
directories.reverse()
line_replace_add_directory = line_replace+directories[2]
Thank you for your help.

You have to put the entire path as a parameter to os.path.split. See The docs. It doesn't work like string split.

How to get parent folder name of current directory?

I know there are functions for finding parent directory or path such as.
os.path.dirname(os.path.realpath(__file__))
'C:\Users\jahon\Desktop\Projects\CAA\Result\caa\project_folder'
Is there a function that just returns the parent folder name? In this case it should be project_folder.

You can achieve this easily with os
import os
os.path.basename(os.getcwd())

You can get the last part of any path using basename (from os.path):
>>> from os.path import basename
>>> basename('/path/to/directory')
'directory'
Just to note, if your path ends with / then the last part of the path is empty:
>>> basename('/path/to/directory/')
''

Yes, you can use PurePath.
PurePath(__file__).parent.name == 'parent_dir'

You can use split and os.path.sep to get the list of path elements and then call the last element of the list:
import os
path = 'C:\\Users\\jahon\\Desktop\\Projects\\CAA\\Result\\caa\\project_folder'
if path.split(os.path.sep)[-1]:
parent_folder = path.split(os.path.sep)[-1] # if no backslashes at the end
else:
parent_folder = path.split(os.path.sep)[-2] # with backslashes at the end

How to move down to a parent directory in Python?

The following code in Python gives me the current path.
import os
DIR = os.path.dirname(os.path.dirname(__file__))
How can I now use the variable DIR to go down one more directory? I don't want to change the value of DIR as it is used elsewhere.
I have tried this:
DIR + "../path/"
But it does not seems to work.

Call one more dirname:
os.path.join(os.path.dirname(DIR), 'path')

Try:
import os.path
print(os.path.abspath(os.path.join(DIR, os.pardir)))

When you join a path via '+' you have to add a 'r':
path = r'C:/home/' + r'user/dekstop'
or write double backslashes:
path = 'C://home//' + 'user//dekstop'
Anyway you should never use that!
That's the best way:
import os
path = os.path.join('C:/home/', 'user/dekstop')

How do I get the full path of the current file's directory?

How do I get the current file's directory path?
I tried:
>>> os.path.abspath(__file__)
'C:\\python27\\test.py'
But I want:
'C:\\python27\\'

The special variable __file__ contains the path to the current file. From that we can get the directory using either pathlib or the os.path module.
Python 3
For the directory of the script being run:
import pathlib
pathlib.Path(__file__).parent.resolve()
For the current working directory:
import pathlib
pathlib.Path().resolve()
Python 2 and 3
For the directory of the script being run:
import os
os.path.dirname(os.path.abspath(__file__))
If you mean the current working directory:
import os
os.path.abspath(os.getcwd())
Note that before and after file is two underscores, not just one.
Also note that if you are running interactively or have loaded code from something other than a file (eg: a database or online resource), __file__ may not be set since there is no notion of "current file". The above answer assumes the most common scenario of running a python script that is in a file.
References
pathlib in the python documentation.
os.path - Python 2.7, os.path - Python 3
os.getcwd - Python 2.7, os.getcwd - Python 3
what does the __file__ variable mean/do?

Using Path from pathlib is the recommended way since Python 3:
from pathlib import Path
print("File Path:", Path(__file__).absolute())
print("Directory Path:", Path().absolute()) # Directory of current working directory, not __file__
Note: If using Jupyter Notebook, __file__ doesn't return expected value, so Path().absolute() has to be used.

In Python 3.x I do:
from pathlib import Path
path = Path(__file__).parent.absolute()
Explanation:
Path(__file__) is the path to the current file.
.parent gives you the directory the file is in.
.absolute() gives you the full absolute path to it.
Using pathlib is the modern way to work with paths. If you need it as a string later for some reason, just do str(path).

Try this:
import os
dir_path = os.path.dirname(os.path.realpath(__file__))

import os
print(os.path.dirname(__file__))

I found the following commands return the full path of the parent directory of a Python 3 script.
Python 3 Script:
#!/usr/bin/env python3
# -*- coding: utf-8 -*-
from pathlib import Path
#Get the absolute path of a Python3.6 and above script.
dir1 = Path().resolve() #Make the path absolute, resolving any symlinks.
dir2 = Path().absolute() #See #RonKalian answer
dir3 = Path(__file__).parent.absolute() #See #Arminius answer
dir4 = Path(__file__).parent
print(f'dir1={dir1}\ndir2={dir2}\ndir3={dir3}\ndir4={dir4}')
REMARKS !!!!
dir1 and dir2 works only when running a script located in the current working directory, but will break in any other case.
Given that Path(__file__).is_absolute() is True, the use of the .absolute() method in dir3 appears redundant.
The shortest command that works is dir4.
Explanation links: .resolve(), .absolute(), Path(file).parent().absolute()

USEFUL PATH PROPERTIES IN PYTHON:
from pathlib import Path
#Returns the path of the current directory
mypath = Path().absolute()
print('Absolute path : {}'.format(mypath))
#if you want to go to any other file inside the subdirectories of the directory path got from above method
filePath = mypath/'data'/'fuel_econ.csv'
print('File path : {}'.format(filePath))
#To check if file present in that directory or Not
isfileExist = filePath.exists()
print('isfileExist : {}'.format(isfileExist))
#To check if the path is a directory or a File
isadirectory = filePath.is_dir()
print('isadirectory : {}'.format(isadirectory))
#To get the extension of the file
fileExtension = mypath/'data'/'fuel_econ.csv'
print('File extension : {}'.format(filePath.suffix))
OUTPUT:
ABSOLUTE PATH IS THE PATH WHERE YOUR PYTHON FILE IS PLACED
Absolute path : D:\Study\Machine Learning\Jupitor Notebook\JupytorNotebookTest2\Udacity_Scripts\Matplotlib and seaborn Part2
File path : D:\Study\Machine Learning\Jupitor Notebook\JupytorNotebookTest2\Udacity_Scripts\Matplotlib and seaborn Part2\data\fuel_econ.csv
isfileExist : True
isadirectory : False
File extension : .csv

works also if __file__ is not available (jupyter notebooks)
import sys
from pathlib import Path
path_file = Path(sys.path[0])
print(path_file)
Also uses pathlib, which is the object oriented way of handling paths in python 3.

IPython has a magic command %pwd to get the present working directory. It can be used in following way:
from IPython.terminal.embed import InteractiveShellEmbed
ip_shell = InteractiveShellEmbed()
present_working_directory = ip_shell.magic("%pwd")
On IPython Jupyter Notebook %pwd can be used directly as following:
present_working_directory = %pwd

I have made a function to use when running python under IIS in CGI in order to get the current folder:
import os
def getLocalFolder():
path=str(os.path.dirname(os.path.abspath(__file__))).split(os.sep)
return path[len(path)-1]

Python 2 and 3
You can simply also do:
from os import sep
print(__file__.rsplit(sep, 1)[0] + sep)
Which outputs something like:
C:\my_folder\sub_folder\

This can be done without a module.
def get_path():
return (__file__.replace(f"<your script name>.py", ""))
print(get_path())