Python Convert Windows File path in a variable
Question:
Given is a variable that contains a windows file path. I have to then go and read this file. The problem here is that the path contains escape characters, and I can’t seem to get rid of it. I checked os.path and pathlib, but all expect the correct text formatting already, which I can’t seem to construct.
For example this. Please note that fPath is given, so I cant prefix it with r for a rawpath.
#this is given, I cant rawpath it with r
fPath = "P:pythonttemp.txt"
file = open(fPath, "r")
for line in file:
print (line)
How can I turn fPath via some function or method from:
"P:pythonttemp.txt"
to
"P:/python/t/temp.txt"
I’ve tried also tried .replace(“”,”/”), which doesnt work.
I’m using Python 3.7 for this.
Answers:
You can use os.path.abspath() to convert it:
print(os.path.abspath("P:pythonttemp.txt"))
>>> P:/python/t/temp.txt
See the documentation of os.path here.
if you would like to do replace then do
replace("\","/")
You can use Path function from pathlib library.
from pathlib import Path
docs_folder = Path("some_folder/some_folder/")
text_file = docs_folder / "some_file.txt"
f = open(text_file)
I’ve solved it.
The issues lies with the python interpreter. t and all the others don’t exist as such data, but are interpretations of nonprint characters.
So I got a bit lucky and someone else already faced the same problem and solved it with a hard brute-force method:
http://code.activestate.com/recipes/65211/
I just had to find it.
After that I have a raw string without escaped characters, and just need to run the simple replace() on it to get a workable path.
When using python version >= 3.4, the class Path from module pathlib offers a function called as_posix, which will sort of convert a path to *nix style path. For example, if you were to build Path object via p = pathlib.Path('C:\Windows\SysWOW64\regedit.exe'), asking it for p.as_posix() it would yield C:/Windows/SysWOW64/regedit.exe. So to obtain a complete *nix style path, you’d need to convert the drive letter manually.
I came across similar problem with Windows file paths. This is what is working for me:
import os
file = input(str().split('\')
file = '/'.join(file)
This gave me the input from this:
"D:test.txt"
to this:
"D:/test.txt"
Basically when trying to work with the Windows path, python tends to replace ” to ”. It goes for every backslash. When working with filepaths, you won’t have double slashes since those are splitting folder names.
This way you can list all folders by order by splitting ” and then rejoining them by .join function with frontslash.
Hopefully this helps!
Use below function, this will pass most of the cases
def resolve_path(path):
parent_replace=['t','n','r','f','v','a','b',' 00','\']
child_replace=['/t','/n','/r','/f','/v','/a','/b','/000','/']
for i in range(len(parent_replace)):
path=path.replace(parent_replace[i],child_replace[i])
return path
Given is a variable that contains a windows file path. I have to then go and read this file. The problem here is that the path contains escape characters, and I can’t seem to get rid of it. I checked os.path and pathlib, but all expect the correct text formatting already, which I can’t seem to construct.
For example this. Please note that fPath is given, so I cant prefix it with r for a rawpath.
#this is given, I cant rawpath it with r
fPath = "P:pythonttemp.txt"
file = open(fPath, "r")
for line in file:
print (line)
How can I turn fPath via some function or method from:
"P:pythonttemp.txt"
to
"P:/python/t/temp.txt"
I’ve tried also tried .replace(“”,”/”), which doesnt work.
I’m using Python 3.7 for this.
You can use os.path.abspath() to convert it:
print(os.path.abspath("P:pythonttemp.txt"))
>>> P:/python/t/temp.txt
See the documentation of os.path here.
if you would like to do replace then do
replace("\","/")
You can use Path function from pathlib library.
from pathlib import Path
docs_folder = Path("some_folder/some_folder/")
text_file = docs_folder / "some_file.txt"
f = open(text_file)
I’ve solved it.
The issues lies with the python interpreter. t and all the others don’t exist as such data, but are interpretations of nonprint characters.
So I got a bit lucky and someone else already faced the same problem and solved it with a hard brute-force method:
http://code.activestate.com/recipes/65211/
I just had to find it.
After that I have a raw string without escaped characters, and just need to run the simple replace() on it to get a workable path.
When using python version >= 3.4, the class Path from module pathlib offers a function called as_posix, which will sort of convert a path to *nix style path. For example, if you were to build Path object via p = pathlib.Path('C:\Windows\SysWOW64\regedit.exe'), asking it for p.as_posix() it would yield C:/Windows/SysWOW64/regedit.exe. So to obtain a complete *nix style path, you’d need to convert the drive letter manually.
I came across similar problem with Windows file paths. This is what is working for me:
import os
file = input(str().split('\')
file = '/'.join(file)
This gave me the input from this:
"D:test.txt"
to this:
"D:/test.txt"
Basically when trying to work with the Windows path, python tends to replace ” to ”. It goes for every backslash. When working with filepaths, you won’t have double slashes since those are splitting folder names.
This way you can list all folders by order by splitting ” and then rejoining them by .join function with frontslash.
Hopefully this helps!
Use below function, this will pass most of the cases
def resolve_path(path):
parent_replace=['t','n','r','f','v','a','b',' 00','\']
child_replace=['/t','/n','/r','/f','/v','/a','/b','/000','/']
for i in range(len(parent_replace)):
path=path.replace(parent_replace[i],child_replace[i])
return path