Open file in a relative location in Python

Question

Suppose my python code is executed a directory called main and the application needs to access main/2091/data.txt.

how should I use open(location)? what should the parameter location be?

I found that below simple code will work.. does it have any disadvantages?

file = "\2091\sample.txt"
path = os.getcwd()+file
fp = open(path, 'r+');

Solution 1

With this type of thing you need to be careful what your actual working directory is. For example, you may not run the script from the directory the file is in. In this case, you can't just use a relative path by itself.

If you are sure the file you want is in a subdirectory beneath where the script is actually located, you can use __file__ to help you out here. __file__ is the full path to where the script you are running is located.

So you can fiddle with something like this:

import os
script_dir = os.path.dirname(__file__) #<-- absolute dir the script is in
rel_path = "2091/data.txt"
abs_file_path = os.path.join(script_dir, rel_path)

Solution 2

This code works fine:

import os


def readFile(filename):
    filehandle = open(filename)
    print filehandle.read()
    filehandle.close()



fileDir = os.path.dirname(os.path.realpath('__file__'))
print fileDir

#For accessing the file in the same folder
filename = "same.txt"
readFile(filename)

#For accessing the file in a folder contained in the current folder
filename = os.path.join(fileDir, 'Folder1.1/same.txt')
readFile(filename)

#For accessing the file in the parent folder of the current folder
filename = os.path.join(fileDir, '../same.txt')
readFile(filename)

#For accessing the file inside a sibling folder.
filename = os.path.join(fileDir, '../Folder2/same.txt')
filename = os.path.abspath(os.path.realpath(filename))
print filename
readFile(filename)

Solution 3

I created an account just so I could clarify a discrepancy I think I found in Russ's original response.

For reference, his original answer was:

import os
script_dir = os.path.dirname(__file__)
rel_path = "2091/data.txt"
abs_file_path = os.path.join(script_dir, rel_path)

This is a great answer because it is trying to dynamically creates an absolute system path to the desired file.

Cory Mawhorter noticed that __file__ is a relative path (it is as well on my system) and suggested using os.path.abspath(__file__). os.path.abspath, however, returns the absolute path of your current script (i.e. /path/to/dir/foobar.py)

To use this method (and how I eventually got it working) you have to remove the script name from the end of the path:

import os
script_path = os.path.abspath(__file__) # i.e. /path/to/dir/foobar.py
script_dir = os.path.split(script_path)[0] #i.e. /path/to/dir/
rel_path = "2091/data.txt"
abs_file_path = os.path.join(script_dir, rel_path)

The resulting abs_file_path (in this example) becomes: /path/to/dir/2091/data.txt

Solution 4

It depends on what operating system you're using. If you want a solution that is compatible with both Windows and *nix something like:

from os import path

file_path = path.relpath("2091/data.txt")
with open(file_path) as f:
    <do stuff>

should work fine.

The path module is able to format a path for whatever operating system it's running on. Also, python handles relative paths just fine, so long as you have correct permissions.

Edit:

As mentioned by kindall in the comments, python can convert between unix-style and windows-style paths anyway, so even simpler code will work:

with open("2091/data/txt") as f:
    <do stuff>

That being said, the path module still has some useful functions.

Solution 5

I spend a lot time to discover why my code could not find my file running Python 3 on the Windows system. So I added . before / and everything worked fine:

import os

script_dir = os.path.dirname(__file__)
file_path = os.path.join(script_dir, './output03.txt')
print(file_path)
fptr = open(file_path, 'w')

View more solutions

Comments

[orip]

You're using unescaped backslashes. That's one disadvantage.

[Russ]

Several disadvantages. 1) As per @orip, use forward slashes for paths, even on windows. Your string won't work. Or use raw strings like r"\2091\sample.txt". Or escape them like "\\2091\\sample.txt" (but that is annoying). Also, 2) you are using getcwd() which is the path you were in when you execute the script. I thought you wanted relative to the script location (but now am wondering). And 3), always use os.path functions for manipulating paths. Your path joining line should be os.path.join(os.getcwd(), file) 4) the ; is pointless

[Russ]

And for good measure... 5) use context guards to keep it clean and avoid forgetting to close your file: with open(path, 'r+') as fp:. See here for the best explanation of with statements I've seen.

[OPMendeavor]

beside the necessary care on slashes, as just indicated, there is the function os.path.abspath to get easly the full path of the relative path to open. final statement looks like this: os.path.abspath('./2091/sample.txt')

[kindall]

relpath() converts a pathname to a relative path. Since it's already a relative path, it will do nothing.

[Wilduck]

It will convert it from a unix-style path to windows-style path if appropriate. Is there another function in the os.path module that would be a better choice?

[kindall]

Windows will already work fine with a UNIX-style path. At least the NT-based series will (2000, XP, Vista, 7). No conversion is necessary.

[Russ]

Not true... the working directory inside a script is the location you ran the script from, not the location of the script. If you run the script from elsewhere (maybe the script is in your system path) the relative path to the subdirectory will not work. Please see my answer on how to get around this.

[Russ]

This answer is not quite correct and will cause problems. Relative paths are by default relative to the current working directory (path the script was executed from), and NOT the actual script location. You need to use __file__. Please see my answer.

[Admin]

I found that below simple code will work..does it have any disadvantages ? <pre> file="\sample.txt" path=os.getcwd()+str(loc)+file fp=open(path,'r+');<code>

[orip]

@Russ - the OP's example uses getcwd(). I read the original description as "relative to where I run the script, regardless of where the code sits".

[Russ]

@orip - the OP added the getcwd() call 3 hrs after the question. No matter... moving on. :)

[Cory Mawhorter]

@Arash The disadvantage to that is the cwd (current working directory) can be anything, and won't necessarily be where your script is located.

[Cory Mawhorter]

__file__ is a relative path (at least on my setup, for some reason), and you need to call os.path.abspath(__file__) first. osx/homebrew 2.7

[Luke Taylor]

You could even combine both approaches for the simpler os.path.dirname(os.path.abspath(__file__))

[Grant Hulegaard]

@LukeTaylor Indeed that would be better than trying to replicate the os.path.dirname functionality yourself as I did in my answer last year.

[foobarbecue]

Did the author of this answer confuse os.path.relpath with os.path.abspath?

[Soumendra]

os.path.dirname(file) is not working for me in Python 2.7. It is showing NameError: name '__file__' is not defined

[Enkum]

@Soumendra I think you are trying it in the console. Try it in a *.py file.

[M. Paul]

For me accessing the file in the parent folder of the current folder did not worked..the .. is added as string..

[Mr_and_Mrs_D]

Instead of path = "/".join(script_directory) + "/" + rel_path you should use the os module as in path = os.path.join(script_directory, rel_path). Instead of manually parsing the path you should be using script_path = os.path.dirname(__file__)

[Mr_and_Mrs_D]

Better: file_path = os.path.join(script_dir, 'output03.txt')

[Ângelo Polotto]

I tried on Windows OS that but I didn't have success.

[Mr_and_Mrs_D]

Interesting - can you print script_dir? Then turn it to absolute path as in script_dir = os.path.abspath(os.path.dirname(__file__))

[Ângelo Polotto]

I will try that, if I succeed, I will change the answer.

[lonstar]

Did not work on Windows. Path to file is correct but Python states "file not found" and shows the path with \\ separators.

[Ronald]

Backslashes are escape characters for several characters. If you happen to encounter \t like for example \top\directory, than '\t' is interpreted as a tab-character and your 'trick' fails. The best option is to use the raw string format r'C:\Users\chidu\Desktop\Skipper New\Special_Note.txt' which does not try to 'escape' characters.

[pchtsp]

in Ubuntu and python3, I had to take the quotes from '__file__' so it became: fileDir = os.path.dirname(os.path.realpath(__file__))

[Neuron]

please use with data_folder.open() as f instead

[Neuron]

all upper case names should be reserved for constants

[James Wong]

In this case, wouldn't BASE_DIR be a constant? The parent folder of the script being run won't change during runtime.

[FLAW]

shouldn't you use os.path.join for 2091/data.txt ? woudn't it cause problems in window? \` & /`

[Mauricio Gracia Gutierrez]

to avoid using / or \ you can os.path.join(path,'folderA','file.txt')