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Python: How to add path separator to beginning of path

I'm looking for the cleanest way to add a path separator to the beginning of a relative path if it's not already there.
So for example my/path should result in /my/path/.
The way I do it now is the following:
import os
os.sep+'my/path'
This approach works but when a non relative path is passed it will also add the separator which is something I want to avoid.
Suggestions?

Try os.path.join with the root directory as its first argument.
>>> import os
>>> os.path.join('/', '/tmp')
/tmp
>>> os.path.join('/', 'tmp')
/tmp

Excel file name with wildcard [duplicate]

I want get a list of filenames with a search pattern with a wildcard. Like:
getFilenames.py c:\PathToFolder\*
getFilenames.py c:\PathToFolder\FileType*.txt
getFilenames.py c:\PathToFolder\FileTypeA.txt
How can I do this?

You can do it like this:
>>> import glob
>>> glob.glob('./[0-9].*')
['./1.gif', './2.txt']
>>> glob.glob('*.gif')
['1.gif', 'card.gif']
>>> glob.glob('?.gif')
['1.gif']
Note:
If the directory contains files starting with . they won’t be matched by default. For example, consider a directory containing card.gif and .card.gif:
>>> import glob
>>> glob.glob('*.gif')
['card.gif']
>>> glob.glob('.c*')
['.card.gif']
This comes straight from here: http://docs.python.org/library/glob.html

glob is useful if you are doing this in within python, however, your shell may not be passing in the * (I'm not familiar with the windows shell).
For example, when I do the following:
import sys
print sys.argv
On my shell, I type:
$ python test.py *.jpg
I get this:
['test.py', 'test.jpg', 'wasp.jpg']
Notice that argv does not contain "*.jpg"
The important lesson here is that most shells will expand the asterisk at the shell, before it is passed to your application.
In this case, to get the list of files, I would just do sys.argv[1:]. Alternatively, you could escape the *, so that python sees the literal *. Then, you can use the glob module.
$ getFileNames.py "*.jpg"
or
$ getFileNames.py \*.jpg

from glob import glob
import sys
files = glob(sys.argv[1])

If you're on Python 3.5+, you can use pathlib's glob() instead of the glob module alone.
Getting all files in a directory looks like this:
from pathlib import Path
for path in Path("/path/to/directory").glob("*"):
print(path)
Or, to just get a list of all .txt files in a directory, you could do this:
from pathlib import Path
for path in Path("/path/to/directory").glob("*.txt"):
print(path)
Finally, you can search recursively (i.e., to find all .txt files in your target directory and all subdirectories) using a wildcard directory:
from pathlib import Path
for path in Path("/path/to/directory").glob("**/*.txt"):
print(path)

I am adding this to the previous because I found this very useful when you want your scripts to work on multiple shell and with multiple parameters using *.
If you want something that works on every shells, you can do the following (still using glob):
>>> import glob
>>> from functools import reduce # if using python 3+
>>> reduce(lambda r, x: r + glob.glob(x), sys.argv[1:], [])
Note that it can produce duplicate (if you have a test file and you give t* and te*), but you can simply remove them using a set:
>>> set(reduce(lambda r, x: r + glob.glob(x), sys.argv[1:], []))

How to convert path formats to use them whatever the platform or the module they came from?

I'm using
abspath = os.path.dirname(os.path.abspath(__file__))
to get the folder of the current scipt that is excecuted which give me :
'C:\\Users\\Me\\PycharmProjects\\Model_HIPP_ATN_Reu'
Then I'm using a QFileDialogto get a save file path which give me :
savefile : 'C:/Users/Me/PycharmProjects/Model_HIPP_ATN_Reu/PopsManager_auto/Model.py'
What I would like to do is find the relative path between those two paths with path.relpath So something like:
savefile=savefile.replace('/','\\')
os.path.relpath(savefile,abspath)
which give me the correct folder path:
'PopsManager_auto/Model.py'
But my problem with that is if I get the savefile or the abspath on different system this code won't work because path formats are not the same. How can I have path in the same format whatever the platform and whatever the way I get paths (Qt or Os modules for instance)? I don't want use .replace('/','\\') because this will only work on Window, right?

First thing to know: os.path is a shortcut to your system's specific path module, but you can still access other systems path modules by their names, ie import ntpath, posixpath, macpath (but I doubt you'll have a need for macpath - it's the pre OSX path system). You can use this to test what happens when switching from one system to another.
Also, "/" is a valid path sep for Windows, and lives as ntpath.altsep ( while posixpath.altsep and macpath.altsep are None), and ntpath knows how to deal with both separators so you actually don't have to do any replacement on abspath before using os.path.relpath():
>>> import ntpath
>>> abspath = "C:\\Users\\Me\\PycharmProjects\\Model_HIPP_ATN_Reu"
>>> savefile= 'C:/Users/Me/PycharmProjects/Model_HIPP_ATN_Reu/PopsManager_auto/Model.py'
>>> relative = ntpath.relpath(savefile, abspath)
>>> relative
'PopsManager_auto\\Model.py'
At this point I notice that I don't get the same result as you for the relative path (I get the above result with either py2.7.x and py3.4.3 - I don't have py3.6.x installed here). What exact python version are you actually using ?
Now if you want to "normalize" your relative path, you can just test what the current os.path.altsep value is and if it's not None, use it for a replace:
>>> # Q&D hack to use `ntpath` as `os.path` on a posix system
>>> import os, ntpath; os.path = ntpath
>>> # now pretend we're on a nt system
>>> if os.path.altsep:
... relative = relative.replace(os.path.sep, os.path.altsep)
...
>>> relative
'PopsManager_auto/Model.py'
so now you have a relative path that is valid for both posixpath and ntpath.
When it comes to rebuild an absolute path from this relative path and a root directory path, all you have to do is to apply os.path.normpath() to both the root path and the relative path and then os.path.join() them:
>>> relative
'PopsManager_auto/Model.py'
>>> os.path.normpath(relative)
'PopsManager_auto\\Model.py'
>>> root = "C:\\Users\\Me\\PycharmProjects\\Model_HIPP_ATN_Reu"
>>> relative = os.path.normpath(relative)
>>> relative
'PopsManager_auto\\Model.py'
>>> os.path.join(root, relative)
'C:\\Users\\Me\\PycharmProjects\\Model_HIPP_ATN_Reu\\PopsManager_auto\\Model.py'

Remove unecssary directories in path name constructed with os.join

In Python when I print directory path constructed with os.join I get something like this :
rep/rep2/../rep1
Is there a way to get only this :
rep/rep1

Yes, os.path.normpath() collapses redundant separators and up-references.
os.path.realpath() converts the path to a canonical path, which includes eliminating '..' components, but it also eliminates symlinks.
See https://docs.python.org/2/library/os.path.html.

Use os.path.relpath:
>>> import os
>>> os.path.relpath("rep/rep2/../rep1", start="")
'rep/rep1'
Or os.path.normpath:
>>> import os
>>> os.path.normpath("rep/rep2/../rep1")
'rep/rep1'

How to get an absolute file path in Python

Given a path such as "mydir/myfile.txt", how do I find the file's absolute path in Python? E.g. on Windows, I might end up with:
"C:/example/cwd/mydir/myfile.txt"

>>> import os
>>> os.path.abspath("mydir/myfile.txt")
'C:/example/cwd/mydir/myfile.txt'
Also works if it is already an absolute path:
>>> import os
>>> os.path.abspath("C:/example/cwd/mydir/myfile.txt")
'C:/example/cwd/mydir/myfile.txt'

You could use the new Python 3.4 library pathlib. (You can also get it for Python 2.6 or 2.7 using pip install pathlib.) The authors wrote: "The aim of this library is to provide a simple hierarchy of classes to handle filesystem paths and the common operations users do over them."
To get an absolute path in Windows:
>>> from pathlib import Path
>>> p = Path("pythonw.exe").resolve()
>>> p
WindowsPath('C:/Python27/pythonw.exe')
>>> str(p)
'C:\\Python27\\pythonw.exe'
Or on UNIX:
>>> from pathlib import Path
>>> p = Path("python3.4").resolve()
>>> p
PosixPath('/opt/python3/bin/python3.4')
>>> str(p)
'/opt/python3/bin/python3.4'
Docs are here: https://docs.python.org/3/library/pathlib.html

import os
os.path.abspath(os.path.expanduser(os.path.expandvars(PathNameString)))
Note that expanduser is necessary (on Unix) in case the given expression for the file (or directory) name and location may contain a leading ~/(the tilde refers to the user's home directory), and expandvars takes care of any other environment variables (like $HOME).

Install a third-party path module (found on PyPI), it wraps all the os.path functions and other related functions into methods on an object that can be used wherever strings are used:
>>> from path import path
>>> path('mydir/myfile.txt').abspath()
'C:\\example\\cwd\\mydir\\myfile.txt'

Update for Python 3.4+ pathlib that actually answers the question:
from pathlib import Path
relative = Path("mydir/myfile.txt")
absolute = relative.absolute() # absolute is a Path object
If you only need a temporary string, keep in mind that you can use Path objects with all the relevant functions in os.path, including of course abspath:
from os.path import abspath
absolute = abspath(relative) # absolute is a str object

This always gets the right filename of the current script, even when it is called from within another script. It is especially useful when using subprocess.
import sys,os
filename = sys.argv[0]
from there, you can get the script's full path with:
>>> os.path.abspath(filename)
'/foo/bar/script.py'
It also makes easier to navigate folders by just appending /.. as many times as you want to go 'up' in the directories' hierarchy.
To get the cwd:
>>> os.path.abspath(filename+"/..")
'/foo/bar'
For the parent path:
>>> os.path.abspath(filename+"/../..")
'/foo'
By combining "/.." with other filenames, you can access any file in the system.

Today you can also use the unipath package which was based on path.py: http://sluggo.scrapping.cc/python/unipath/
>>> from unipath import Path
>>> absolute_path = Path('mydir/myfile.txt').absolute()
Path('C:\\example\\cwd\\mydir\\myfile.txt')
>>> str(absolute_path)
C:\\example\\cwd\\mydir\\myfile.txt
>>>
I would recommend using this package as it offers a clean interface to common os.path utilities.

You can use this to get absolute path of a specific file.
from pathlib import Path
fpath = Path('myfile.txt').absolute()
print(fpath)

Given a path such as mydir/myfile.txt, how do I find the file's absolute path relative to the current working directory in Python?
I would do it like this,
import os.path
os.path.join( os.getcwd(), 'mydir/myfile.txt' )
That returns '/home/ecarroll/mydir/myfile.txt'

if you are on a mac
import os
upload_folder = os.path.abspath("static/img/users")
this will give you a full path:
print(upload_folder)
will show the following path:
>>>/Users/myUsername/PycharmProjects/OBS/static/img/user

In case someone is using python and linux and looking for full path to file:
>>> path=os.popen("readlink -f file").read()
>>> print path
abs/path/to/file