Open file in a relative location in Python

问题

Suppose python code is executed in not known by prior windows directory say \'main\' , and wherever code is installed when it runs it needs to access to directory \'main/2091/data.txt\' .

how should I use open(location) function? what should be location ?

Edit :

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+\');

回答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)

回答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)

回答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

回答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.

回答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')

回答6:

Code:

import os
script_path = os.path.abspath(__file__) 
path_list = script_path.split(os.sep)
script_directory = path_list[0:len(path_list)-1]
rel_path = "main/2091/data.txt"
path = "/".join(script_directory) + "/" + rel_path

Explanation:

Import library:

import os

Use __file__ to attain the current script's path:

script_path = os.path.abspath(__file__)

Separates the script path into multiple items:

path_list = script_path.split(os.sep)

Remove the last item in the list (the actual script file):

script_directory = path_list[0:len(path_list)-1]

Add the relative file's path:

rel_path = "main/2091/data.txt

Join the list items, and addition the relative path's file:

path = "/".join(script_directory) + "/" + rel_path

Now you are set to do whatever you want with the file, such as, for example:

file = open(path)

回答7:

If the file is in your parent folder, eg. follower.txt, you can simply use open('../follower.txt', 'r').read()

回答8:

Try this:

from pathlib import Path

data_folder = Path("/relative/path")
file_to_open = data_folder / "file.pdf"

f = open(file_to_open)

print(f.read())

Python 3.4 introduced a new standard library for dealing with files and paths called pathlib. It works for me!

回答9:

Not sure if this work everywhere.

I'm using ipython in ubuntu.

If you want to read file in current folder's sub-directory:

/current-folder/sub-directory/data.csv

your script is in current-folder simply try this:

import pandas as pd
path = './sub-directory/data.csv'
pd.read_csv(path)

回答10:

Python just passes the filename you give it to the operating system, which opens it. If your operating system supports relative paths like main/2091/data.txt (hint: it does), then that will work fine.

You may find that the easiest way to answer a question like this is to try it and see what happens.

来源：https://stackoverflow.com/questions/7165749/open-file-in-a-relative-location-in-python