Open File in Another Directory (Python)

问题

I've always been sort of confused on the subject of directory traversal in Python, and have a situation I'm curious about: I have a file that I want to access in a directory essentially parallel to the one I'm currently in. Given this directory structure:

\parentDirectory
    \subfldr1
        -testfile.txt
    \subfldr2
        -fileOpener.py

I'm trying to script in fileOpener.py to get out of subfldr2, get into subfldr1, and then call an open() on testfile.txt.

From browsing stackoverflow, I've seen people use os and os.path to accomplish this, but I've only found examples regarding files in subdirectories beneath the script's origin.

Working on this, I realized I could just relocate the script into subfldr1 and then all would be well, but my curiosity is piqued as to how this would be accomplished.

EDIT: This question pertains particularly to a Windows machine, as I don't know how drive letters and backslashes would factor into this.

回答1:

If you know the full path to the file you can just do something similar to this. However if you question directly relates to relative paths, that I am unfamiliar with and would have to research and test.

path = 'C:\\Users\\Username\\Path\\To\\File'

with open(path, 'w') as f:
    f.write(data)

Edit:

Here is a way to do it relatively instead of absolute. Not sure if this works on windows, you will have to test it.

import os

cur_path = os.path.dirname(__file__)

new_path = os.path.relpath('..\\subfldr1\\testfile.txt', cur_path)
with open(new_path, 'w') as f:
    f.write(data)

Edit 2: One quick note about __file__, this will not work in the interactive interpreter due it being ran interactively and not from an actual file.

回答2:

import os
import os.path
import shutil

You find your current directory:

d = os.getcwd() #Gets the current working directory

Then you change one directory up:

os.chdir("..") #Go up one directory from working directory

Then you can get a tupple/list of all the directories, for one directory up:

o = [os.path.join(d,o) for o in os.listdir(d) if os.path.isdir(os.path.join(d,o))] # Gets all directories in the folder as a tuple

Then you can search the tuple for the directory you want and open the file in that directory:

for item in o:
    if os.path.exists(item + '\\testfile.txt'):
    file = item + '\\testfile.txt'

Then you can do stuf with the full file path 'file'

回答3:

from pathlib import Path

data_folder = Path("source_data/text_files/")
file_to_open = data_folder / "raw_data.txt"

f = open(file_to_open)
print(f.read())

回答4:

Its a very old question but I think it will help newbies line me who are learning python. If you have Python 3.4 or above, the pathlib library comes with the default distribution.

To use it, you just pass a path or filename into a new Path() object using forward slashes and it handles the rest. To indicate that the path is a raw string, put r in front of the string with your actual path.

For example,

from pathlib import Path

dataFolder = Path(r'D:\Desktop dump\example.txt')

Source: The easy way to deal with file paths on Windows, Mac and Linux

(unicode error) 'unicodeescape' codec can't decode bytes in position 2-3: truncated \UXXXXXXXX escape

来源：https://stackoverflow.com/questions/32470543/open-file-in-another-directory-python

标签

python

windows

os.path