Develop Reference

Python cannot find exiting path

While preparing my dataset for a machine learning model I stumbled upon a bug (at least from my perspective). I was trying to move files around in Windows 10 but my system always told me that my file does not exist (FileNotFoundError: [Errno 2] No such file or directory: '../path/to/file'), but I specifically probed for the file before which returned True.
What I was trying in my IPython console was the following (which can reproduce the bug):
import os
import shutil
path = '../path/to/file' # You can see I use relative paths
out_path = '../path/to/out/'
os.makedirs(out_path) # Make sure paths do exist
_, name = os.path.split(path)
out_path += name
# Produces almost the same error FileNotFoundError: [WinError 3] The system cannot find the given directory: path
# os.rename(path, out_path)
shutil.move(path, out_path)
You obviously need a file to move at '../path/to/file'. I tested whether I use the correct path with os.path.exist(path) which returned True.
Do I simply misread the documentation and the usage is different to what I am thinking? I understood it like the first argument is the path to the file to move and the second one is the path for the file after moving it.
Or is there really a bug in my Python environment?
I also tried moving the file with absolute paths returned by os.path.abspath(), but this did not work either.
FYI: I am using Windows 10 and Python 3.8.12.
Edit: I found the mistake I did... After 2.5 days of labeling image data it appears like I had an underscore mistaken for a hyphen which is why my output directory seems to be malformed... :/

You are not doing a relative path to the current directory, but to the parent directory of the current directory with a double dot '../path' and not single dot './path'. So if your files aren't in the parent directory of the current directory, they won't be found.
When you try to append the file name to the directory path "out_path" you do a string append. which just makes a new file with the combined names rather than move the file into the directory. Use the os.path.join() function for this.
Here is code I made that corrects these two points and works on my end.
import os
import shutil
path = './somefile.txt' # CORRECTED: One dot for current directory
out_path = './somedir'
try:
os.makedirs(out_path) # Make sure paths do exist
except Exception as e:
print(e)
_, name = os.path.split(path)
out_path = os.path.join(out_path, name) # CORRECTED: Use appropriate function
# Produces almost the same error FileNotFoundError: [WinError 3] The system cannot find the given directory: path
# os.rename(path, out_path)
shutil.move(path, out_path)

Os.path gives unexpected output

lately I started working with the Os module in python . And I finally arrived to this Os.path method . So here is my question . I ran this method in one of my kivy project just for testing and it actually didn't returned the correct output.The method consisted of finding if any directory exist and return a list of folders in the directory . otherwise print Invalid Path and return -1 . I passed in an existing directory and it returned -1 but the weird path is that when I run similar program out of my kivy project using the same path present in thesame folder as my python file it return the desired output .here is the image with the python file and the directory name image I have tested which returns invalid path.
and here is my code snippet
def get_imgs(self, img_path):
if not os.path.exists(img_path):
print("Invalid Path...")
return -1
else:
all_files = os.listdir(img_path)
imgs = []
for f in all_files:
if (
f.endswith(".png")
or f.endswith(".PNG")
or f.endswith(".jpg")
or f.endswith(".JPG")
or f.endswith(".jpeg")
or f.endswith(".JPEG")
):
imgs.append("/".join([img_path, f]))
return imgs

It's tough to tell without seeing the code with your function call. Whatever argument you're passing must not be a valid path. I use the os module regularly and have slowly learned a lot of useful methods. I always print out paths that I'm reading or where I'm writing before doing it in case anything unexpected happens, I can see that img_path variable, for example. Copy and paste the path in file explorer up to the directory and make sure that's all good.
Some other useful os.path methods you will find useful, based on your code:
os.join(<directory>, <file_name.ext>) is much more intuitive than imgs.append("/".join([img_path, f]))
os.getcwd() gets your working directory (which I print at the start of scripts in dev to quickly address issues before debugging). I typically use full paths to play it safe because Python pathing can cause differences/issues when running from cmd vs. PyCharm
os.path.basename(f) gives you the file, while os.path.dirname(f) gives you the directory.
It seems like a better approach to this is to use pathlib and glob. You can iterate over directories and use wild cards.
Look at these:
iterating over directories: How can I iterate over files in a given directory?
different file types: Python glob multiple filetypes
Then you don't even need to check whether os.path.exists(img_path) because this will read the files directly from your file system. There's also more wild cards in the glob library such as * for anything/any length, ? for any character, [0-9] for any number, found here: https://docs.python.org/3/library/glob.html

Python os.makedirs and shutil.copyfile - Error 13 - Permission denied

In my code, I create a directory like so:
try:
os.makedirs(playlist_name)
except OSError as e:
if e.errno != errno.EEXIST:
raise
Which creates a directory in the place where I run my python script.
Then I wish to copy three files from the original directory where the folder is located into the newly created directory, like so
# Copy FFMPEG files into that folder so youtube dl can download the videos as audio tracks
# Tried using os.getcwd() to get full path, same error
shutil.copyfile(os.getcwd() + '\\ffmpeg.exe', os.getcwd() + "\\" + playlist_name)
shutil.copyfile('ffplay.exe', "/" + playlist_name + "/")
shutil.copyfile('ffprobe.exe', "/" + playlist_name + "/")
However, trying to copy those files throws this error:
PermissionError: [Errno 13] Permission denied: 'C:\\Users\\ME\\Documents\\python\\DIRECTORY\\PLAYLIST_NAME_HERE'
I have tried various shutil copy methods with the same error.
EDIT: This is running on windows

Per the copyfile docs:
dst must be the complete target file name; look at shutil.copy() for a copy that accepts a target directory path.
You can't use it to do what you do in the shell, naming a source file and a target directory and having it deduce the file should be put in the directory with the file's original name. You have to explicitly name the target file, or it thinks you're trying to copy to the same name as the directory, and unlike replacing a file, you can't replace a directory with a file without explicitly renaming the directory or deleting the whole directory tree first. To fix, just make sure to repeat the file name in both source and destination:
for filename in ('ffmpeg.exe', 'ffplay.exe', 'ffprobe.exe'):
shutil.copyfile(filename, os.path.join(playlist_name, filename))
The problem would be more obvious on a UNIX-like system, because those systems would reject the action with EISDIR, causing a Python IsADirectoryError to be raised, but Windows for some reason chose to use more general error codes associated with permission/access problems (EACCES and related Windows specific error codes) that Python translates to PermissionError (because Windows just isn't telling it the real problem, and it would introduce all sorts of race conditions if Python tried to check if the real problem was trying to use a directory as a file to fix the exception type).

Renaming files in Python

I'm doing a Python course on Udacity. And there is a class called Rename Troubles, which presents the following incomplete code:
import os
def rename_files():
file_list = os.listdir("C:\Users\Nick\Desktop\temp")
print (file_list)
for file_name in file_list:
os.rename(file_name, file_name.translate(None, "0123456789"))
rename_files()
As the instructor explains it, this will return an error because Python is not attempting to rename files in the right folder. He then proceeds to check the "current working directory", and then goes on to specify to Python which directory to rename files in.
This makes no sense to me. We are using the for loop to specifically tell Python that we want to rename the contents of file_list, which we have just pointed to the directory we need, in rename_files(). So why does it not attempt to rename in that folder? Why do we still need to figure out cwd and then change it?? The code looks entirely logical without any of that.

Look closely at what os.listdir() gives you. It returns only a list of names, not full paths.
You'll then proceed to os.rename one of those names, which will be interpreted as a relative path, relative to whatever your current working directory is.
Instead of messing with the current working directory, you can os.path.join() the path that you're searching to the front of both arguments to os.rename().

I think your code needs some formatting help.
The basic issue is that os.listdir() returns names relative to the directory specified (in this case an absolute path). But your script can be running from any directory. Manipulate the file names passed to os.rename() to account for this.

Look into relative and absolute paths, listdir returns names relative to the path (in this case absolute path) provided to listdir. os.rename is then given this relative name and unless the app's current working directory (usually the directory you launched the app from) is the same as provided to listdir this will fail.
There are a couple of alternative ways of handling this, changing the current working directory:
os.chdir("C:\Users\Nick\Desktop\temp")
for file_name in os.listdir(os.getcwd()):
os.rename(file_name, file_name.translate(None, "0123456789"))
Or use absolute paths:
directory = "C:\Users\Nick\Desktop\temp"
for file_name in os.listdir(directory):
old_file_path = os.path.join(directory, file_name)
new_file_path = os.path.join(directory, file_name.translate(None, "0123456789"))
os.rename(old_file_path, new_file_path)

You can get a file list from ANY existing directory - i.e.
os.listdir("C:\Users\Nick\Desktop\temp")
or
os.listdir("C:\Users\Nick\Desktop")
or
os.listdir("C:\Users\Nick")
etc.
The instance of the Python interpreter that you're using to run your code is being executed in a directory that is independent of any directory for which you're trying to get information. So, in order to rename the correct file, you need to specify the full path to that file (or the relative path from wherever you're running your Python interpreter).

Error while using listdir in Python

I'm trying to get the list of files in a particular directory and count the number of files in the directory. I always get the following error:
WindowsError: [Error 3] The system cannot find the path specified: '/client_side/*.*'
My code is:
print len([name for name in os.listdir('/client_side/') if os.path.isfile(name)])
I followed the code example given here.
I am running the Python script on Pyscripter and the directory /client_side/ do exists. My python code is in the root folder and has a sub-folder called "client_side". Can someone help me out on this?

This error occurs when you use os.listdir on a path which does not refer to an existing path.
For example:
>>> os.listdir('Some directory does not exist')
Traceback (most recent call last):
File "<interactive input>", line 1, in <module>
WindowsError: [Error 3] : 'Some directory does not exist/*.*'
If you want to use os.listdir, you need to either guarantee the existence of the path that you would use, or use os.path.exists to check the existence first.
if os.path.exists('/client_side/'):
do something
else:
do something
Suppose your current working directory is c:\foobar, os.listdir('/client_side/') is equivalent to os.listdir('c:/client_side'), while os.listdir('client_side/') is equivalent to os.listdir('c:/foobar/client_side'). If your client_side directory is not in the root, such error will occur when using os.listdir.
For your 0 ouput problem, let us recall os.listdir(path)
Return a list containing the names of the entries in the directory given by path. The list is in arbitrary order. It does not include the special entries '.' and '..' even if they are present in the directory.
and os.path.isfile(path).
Return True if path is an existing regular file. This follows symbolic links, so both islink() and isfile() can be true for the same path.
listdir returns neither the absolute paths nor relative paths, but a list of the name of your files, while isfile requires path. Therefore, all of those names would yield False.
To obtain the path, we can either use os.path.join , concat two strings directly.
print ([name for name in os.listdir(path)
if os.path.isfile(os.path.join(path, name))])
Or
print ([name for name in os.listdir('client_side/')
if os.path.isfile('client_side/' + name)])

I decided to change the code into:
def numOfFiles(path):
return len(next(os.walk(path))[2])
and use the following the call the code:
print numOfFiles("client_side")
Many thanks to everyone who told me how to pass the windows directory correctly in Python and to nrao91 in here for providing the function code.
EDIT: Thank you eryksun for correcting my code!

Two things:
os.listdir() does not do a glob pattern matching, use the glob module for that
probably you do not have a directory called '/client_side/*.*', but maybe one
without the . in the name
The syntax you used works fine, if the directory you look for exists, but there is no directory called '/client_side/.'.
In addition, be careful if using Python 2.x and os.listdir, as the results on windows are different when you use u'/client_side/' and just '/client_side'.

You can do just
os.listdir('client_side')
without slashes.

As I can see a WindowsError, Just wondering if this has something to do with the '/' in windows ! Ideally, on windows, you should have something like os.path.join('C:','client_side')

You want:
print len([name for name in os.listdir('./client_side/') if os.path.isfile(name)])
with a "." before "/client_side/".
The dot means the current path where you are working (i.e. from where you are calling your code), so "./client_side/" represents the path you want, which is specified relatively to your current directory.
If you write only "/client_side/", in unix, the program would look for a folder in the root of the system, instead of the folder that you want.

If you just want to see all the files in the directory where your script is located, you can use os.path.dirname(sys.argv[0]). This will give the path of the directory where your script is.
Then, with fnmatch function you can obtain the list of files in that directory with a name and/or extension specified in the filenamevariable.
import os,sys
from fnmatch import fnmatch
directory = os.path.dirname(sys.argv[0]) #this determines the directory
file_name= "*" #if you want the python files only, change "*" to "*.py"
for path, subdirs, files in os.walk(directory):
for name in files:
if fnmatch(name, file_name):
print (os.path.join(path, name))
I hope this helps.

Checking for existence is subject to a race. Better to handle the error (beg forgiveness instead of ask permission). Plus, in Python 3 you can suppress errors. Use suppress from contextlib:
with suppress(FileNotFoundError):
for name in os.listdir('foo'):
print(name)