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Relative path in Python

I'm writing some python code to generate the relative path. Situation need to be considered:
Under the same folder. I want "." or ".\", both of tham are ok for me.
Other folder. I want like ".\xxx\" and "..\xxx\xxx\"
os.path.relpath() will generate the relative path, but without .\ at the beginning and \ in the end. We can add \ in the end by using os.path.join(dirname, ""). But i can't figure out how to add ".\" at the beginning without impacting the first case when they are under the same folder and "..\xxx\xxx\".

It will give you relative path
import os
dir = os.path.dirname(__file__)
filename = os.path.join(dir,'Path')

The relpath() function will produce the ".." syntax given the appropriate base to start from (second parameter). For instance, supposing you were writing something like a script generator that produces code using relative paths, if the working directory is as the second parameter to relpath() as below indicates, and you want to reference in your code another file in your project under a directory one level up and two deep, you'll get "../blah/blah".. In the case where you want to prefix paths in the same folder, you can simply do a join with ".". That will produce a path with the correct OS specific separator.
print(os.path.relpath("/foo/bar/blah/blah", "/foo/bar/baz"))
>>> ../blah/blah
print(os.path.join('.', 'blah'))
>>> ./blah

python abspath returning path twice

I'm trying to get an absolute path of a relative path string with python but it keeps printing the path twice. For example:
self.path = 'Users/abdulahmad/Desktop'
self.actual_path = os.path.abspath(self.path)
print self.actual_path
my console prints
/Users/abdulahmad/Desktop/Users/abdulahmad/Desktop
and if I change the path to:
self.path = 'Desktop'
my console prints:
/Users/abdulahmad/Desktop/Desktop
shouldn't it just print /Users/abdulahmad/Desktop in both cases?

Probably because the current working directory is /Users/abdulahmad/Desktop.
If you write for example path/to/file it means relative to current working directory and relative to /Users/abdulahmad/Desktop it would mean /Users/abdulahmad/Desktop/path/to/file.
If you read the python3 manual it actually shows an implementation of os.abspath(path) as being the same as os.path.normpath(os.path.join(os.getcwd(), path)). This can be used to get a path relative to arbitrarily provided path. (It also shows that you actually basically joins the current working directory and the supplied (relative) path)

How to resolve relative paths in python?

I have Directory structure like this
projectfolder/fold1/fold2/fold3/script.py
now I'm giving script.py a path as commandline argument of a file which is there in
fold1/fold_temp/myfile.txt
So basically I want to be able to give path in this way
../../fold_temp/myfile.txt
>>python somepath/pythonfile.py -input ../../fold_temp/myfile.txt
Here problem is that I might be given full path or relative path so I should be able to decide and based on that I should be able to create absolute path.
I already have knowledge of functions related to path.
Question 1
Question 2
Reference questions are giving partial answer but I don't know how to build full path using the functions provided in them.

try os.path.abspath, it should do what you want ;)
Basically it converts any given path to an absolute path you can work with, so you do not need to distinguish between relative and absolute paths, just normalize any of them with this function.
Example:
from os.path import abspath
filename = abspath('../../fold_temp/myfile.txt')
print(filename)
It will output the absolute path to your file.
EDIT:
If you are using Python 3.4 or newer you may also use the resolve() method of pathlib.Path. Be aware that this will return a Path object and not a string. If you need a string you can still use str() to convert it to a string.
Example:
from pathlib import Path
filename = Path('../../fold_temp/myfile.txt').resolve()
print(filename)

A practical example:
sys.argv[0] gives you the name of the current script
os.path.dirname() gives you the relative directory name
thus, the next line, gives you the absolute working directory of the current executing file.
cwd = os.path.abspath(os.path.dirname(sys.argv[0]))
Personally, I always use this instead of os.getcwd() since it gives me the script absolute path, independently of the directory from where the script was called.

For Python3, you can use pathlib's resolve functionality to resolve symlinks and .. components.
You need to have a Path object however it is very simple to do convert between str and Path.
I recommend for anyone using Python3 to drop os.path and its messy long function names and stick to pathlib Path objects.

import os
dir = os.path.dirname(__file__)
path = raw_input()
if os.path.isabs(path):
print "input path is absolute"
else:
path = os.path.join(dir, path)
print "absolute path is %s" % path
Use os.path.isabs to judge if input path is absolute or relative, if it is relative, then use os.path.join to convert it to absolute

How to generate path of a directory in python

I have a file abc.py under the workspace dir.
I am using os.listdir('/home/workspace/tests') in abc.py to list all the files (test1.py, test2.py...)
I want to generate the path '/home/workspace/tests' or even '/home/workspace' instead of hardcoding it.
I tried os.getcwd() and os.path.dirname(os.path.abspath(____file____)) but this instead generates the path where the test script is being run.
How to go about it?

The only way you can refer to a specific folder from which you don't relate in any way and you don't want to hardcode it, is to pass it as a parameter to the script (search for: command line argument)

I think you are asking about how to get the relative path instead of absolute one.
Absolute path is the one like: "/home/workspace"
Relative looks like the following "./../workspace"
You should construct the relative path from the dir where your script is (/home/workspace/tests) to the dir that you want to acces (/home/workspace) that means, in this case, to go one step up in the directory tree.
You can get this by executing:
os.path.dirname(os.path.join("..", os.path.abspath(__file__)))
The same result may be achieved if you go two steps up and one step down to workspace dir:
os.path.dirname(os.path.join("..", "..", "workspace", os.path.abspath(__file__)))
In this manner you actually can access any directory without knowing it's absolute path, but only knowing where it resides relatively to your executed file.

How to set current working directory in python in a automatic way

How can I set the current path of my python file "myproject.py" to the file itself?
I do not want something like this:
path = "the path of myproject.py"
In mathematica I can set:
SetDirectory[NotebookDirectory[]]
The advantage with the code in Mathematica is that if I change the path of my Mathematica file, for example if I give it to someone else or I put it in another folder, I do not need to do anything extra. Each time Mathematica automatically set the directory to the current folder.
I want something similar to this in Python.

The right solution is not to change the current working directory, but to get the full path to the directory containing your script or module then use os.path.join to build your files path:
import os
ROOT_PATH = os.path.dirname(os.path.abspath(__file__))
# then:
myfile_path = os.path.join(ROOT_PATH, "myfile.txt")
This is safer than messing with current working directory (hint : what would happen if another module changes the current working directory after you did but before you access your files ?)

I want to set the directory in which the python file is, as working directory
There are two step:
Find out path to the python file
Set its parent directory as the working directory
The 2nd is simple:
import os
os.chdir(module_dir) # set working directory
The 1st might be complex if you want to support a general case (python file that is run as a script directly, python file that is imported in another module, python file that is symlinked, etc). Here's one possible solution:
import inspect
import os
module_path = inspect.getfile(inspect.currentframe())
module_dir = os.path.realpath(os.path.dirname(module_path))

Use the os.getcwd() function from the built in os module also there's os.getcwdu() which returns a unicode object of the current working directory
Example usage:
import os
path = os.getcwd()
print path
#C:\Users\KDawG\Desktop\Python