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:\python\t\temp.txt"
file = open(fPath, "r")
for line in file:
print (line)
How can I turn fPath via some function or method from:
"P:\python\t\temp.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:\python\t\temp.txt"))
>>> P:/python/t/temp.txt
See the documentation of os.path here.
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.
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)
if you would like to do replace then do
replace("\\","/")
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!
Related
I have three files
My python file running in an unimportant different folder: C:\DD\CC\BB\AA\code.py
A playlist file "C:\ZZ\XX\Playlist.pls" which points to ....\mp3\song.mp3
The C:\mp3\song.mp3 file.
What I want is to get the location of the mp3 as an absolute path. But every attemp I try I get everything related to whenever the code.py file is.
import pathlib
plMaster = pathlib.Path(r"C:\ZZ\XX\Playlist.pls")
plSlave = pathlib.Path(r"..\..\mp3\song.mp3")
I have tried plSlave.absolute() and gives me "C:\DD\CC\BB\AA....\mp3\song.mp3"
Using relative_to doesn't work. I feel like I am doing such an easy task but I must be missing something because I can't find any function that lets me set the reference to compute the relative path.
Note: I already have parsed the pls file, and have the string r"....\mp3\song.mp3" extracted. I just need to get the path "C:\mp3\song.mp3" knowing that they are relative to the pls. (Not relative to the code.py)
If you're using a Windows version of Python, this is fairly easy. You can join the directory of plMaster (plMaster.parent) with the relative path of plSlave, then resolve the path using resolve(). You can use strict=False to force the resolve even if the path components aren't found.
This worked for me:
>>> plMaster = pathlib.Path(r"C:\ZZ\XX\Playlist.pls")
>>> plSlave = pathlib.Path(r"..\..\mp3\song.mp3")
>>> plMaster.parent.joinpath(plSlave).resolve(strict=False)
WindowsPath('C:/mp3/song.mp3')
If you're on a Unix version of Python, using Windows paths, I couldn't get this to work no matter what I tried, even using pathlib.PureWindowsPath().
Might well be a better method here, but you can use pathlib.Path.parents and pathlib.Path.parts to extract some useful info here and get where you are going
new_relative_path = r"..\..\mp3\song.mp3" #however you got this from reading your .pls file or whatever
pls_path = pathlib.Path(r'C:\ZZ\XX\Playlist.pls')
relative_save = pathlib.Path(new_relativePath)
n = relative_save.parts.count('..')
new_path = pls_path.parents[n-1].joinpath(*relative_save.parts[n:])
The key thing here is that you are going to navigate up the original path (the pls_path) n times (so n-1 since we start at 0), and then you are going to append to that whatever your new relative path is, stripping the '..' segments from the beginning of it.
Whilst I was waiting for other answers I manage to figure it out ditching pathlib and using os instead.
import os
plMaster = r"C:\ZZ\XX\Playlist.pls"
plSlave = r"..\..\mp3\song.mp3"
os.chdir(os.path.dirname(plMaster))
os.path.abspath(plSlave)
I have a given file path. For example, "C:\Users\cobyk\Downloads\GrassyPath.jpg". I would like to pull in a separate string, the image file name.
I'm assuming the best way to do that is to start from the back end of the string, find the final slash and then take the characters following that slash. Is there a method to do this already or will I have search through the string via a for loop, find the last slash myself, and then do the transferring manually?
The pathlib module makes it very easy to access individual parts of a file path like the final path component:
from pathlib import Path
image_path = Path(r"C:\Users\cobyk\Downloads\GrassyPath.jpg")
print(image_path.name) # -> GrassyPath.jpg
You can certainly search manually as you've suggested. However, the Python standard library already has, as you suspected, a function which does this for you.
import os
file_name = os.path.basename(r'C:\Users\cobyk\Downloads\GrassyPath.jpg')
I'm trying to load a simple text file with an array of numbers into Python. A MWE is
import numpy as np
BASE_FOLDER = 'C:\\path\\'
BASE_NAME = 'DATA.txt'
fname = BASE_FOLDER + BASE_NAME
data = np.loadtxt(fname)
However, this gives an error while running:
OSError: C:\path\DATA.txt not found.
I'm using VSCode, so in the debug window the link to the path is clickable. And, of course, if I click it the file opens normally, so this tells me that the path is correct.
Also, if I do print(fname), VSCode also gives me a valid path.
Is there anything I'm missing?
EDIT
As per your (very helpful for future reference) comments, I've changed my code using the os module and raw strings:
BASE_FOLDER = r'C:\path_to_folder'
BASE_NAME = r'filename_DATA.txt'
fname = os.path.join(BASE_FOLDER, BASE_NAME)
Still results in error.
Second EDIT
I've tried again with another file. Very basic path and filename
BASE_FOLDER = r'Z:\Data\Enzo\Waste_Code'
BASE_NAME = r'run3b.txt'
And again, I get the same error.
If I try an alternative approach,
os.chdir(BASE_FOLDER)
a = os.listdir()
then select the right file,
fname = a[1]
I still get the error when trying to import it. Even though I'm retrieving it directly from listdir.
>> os.path.isfile(a[1])
False
Using the module os you can check the existence of the file within python by running
import os
os.path.isfile(fname)
If it returns False, that means that your file doesn't exist in the specified fname. If it returns True, it should be read by np.loadtxt().
Extra: good practice working with files and paths
When working with files it is advisable to use the amazing functionality built in the Base Library, specifically the module os. Where os.path.join() will take care of the joins no matter the operating system you are using.
fname = os.path.join(BASE_FOLDER, BASE_NAME)
In addition it is advisable to use raw strings by adding an r to the beginning of the string. This will be less tedious when writing paths, as it allows you to copy-paste from the navigation bar. It will be something like BASE_FOLDER = r'C:\path'. Note that you don't need to add the latest '\' as os.path.join takes care of it.
You may not have the full permission to read the downloaded file. Use
sudo chmod -R a+rwx file_name.txt
in the command prompt to give yourself permission to read if you are using Ubuntu.
For me the problem was that I was using the Linux home symbol in the link (~/path/file). Replacing it with the absolute path /home/user/etc_path/file worked like charm.
Say that I have this string "D:\Users\Zache\Downloads\example.obj" and I want to copy another file to the same directory as example.obj. How do I do this in a way that´s not hardcoded?
"example" can also be something else (user input). I'm using filedialog2 to get the big string.
This is for an exporter with a basic GUI.
os.path.dirname() gives you the directory portion of a given filename:
>>> import os.path
>>> os.path.dirname(r"D:\Users\Zache\Downloads\example.obj")
'D:\\Users\\Zache\\Downloads'
You can solve it with str.split but this should be solved with os.path.split
I have the python code which will download the .war file and put it in a path which is specified by the variable path.
Now I wish to extract a specific file from that war to a specific folder.
But I got struck up here :
os.system(jar -xvf /*how to give the path varible here*/ js/pay.js)
I'm not sure how to pass on the variable path to os.system command.
I'm very new to python, kindly help me out.
If you really want to use os.system, the shell command line is passed as a string, and you can pass any string you want. So:
os.system('jar -xvf "' + pathvariable + '" js/pay.js)
Or you can use {} or %s formatting, etc.
However, you probably do not want to use os.system.
First, if you want to run other programs, it's almost always better to use the subprocess module. For example:
subprocess.check_call(['jar', '-xvf', pathvariable, 'js/pay.js'])
As you can see, you can pass a list of arguments instead of trying to work out how to put a string together (and deal with escaping and quoting and all that mess). And there are lots of other advantages, mostly described in the documentation itself.
However, you probably don't want to run the war tool at all. As jimhark says, a WAR file is just a special kind of JAR file, which is just a special kind of ZIP file. For creating them, you generally want to use JAR/WAR-specific tools (you need to verify the layout, make sure the manifest is the first entry in the ZIP directory, take care of the package signature, etc.), but for expanding them, any ZIP tool will work. And Python has ZIP support built in. What you want to do is probably as simple as this:
import zipfile
with zipfile.ZipFile(pathvariable, 'r') as zf:
zf.extract('js/pay.js', destinationpathvariable)
IIRC, you can only directly use ZipFile in a with statement in 2.7 and 3.2+, so if you're on, say, 2.6 or 3.1, you have to do it indirectly:
from contextlib import closing
import zipfile
with closing(zipfile.ZipFile(pathvariable, 'r')) as zf:
zf.extract('js/pay.js', destinationpathvariable)
Or, if this is just a quick&dirty script that quits as soon as it's done, you can get away with:
import zipfile
zf = zipfile.ZipFile(pathvariable, 'r')
zf.extract('js/pay.js', destinationpathvariable)
But I try to always use with statements whenever possible, because it's a good habit to have.
Isn't a war file a type of zip file? Python has zipfile support (click link for docs page).
You can use os.environ, it holds all environment variables on it.
It is a dict, so you can just use it like:
pypath = os.environ['PYTHONPATH']
now if you mean it's a common python variable, just use it like:
var1 = 'pause'
os.system('#echo & %s' % var1)