Develop Reference

why does python say it cant find the path specified when it made the path?

I made this program yesterday because I am using py2exe, so what this program does is it zips up the folder created by py2exe and names it to app4export so I can send it to my friends. I also added in where if i already have a zip file called app4export then it deletes it before hand, it worked yesterday but now today I get the error
FileNotFoundError: [WinError 3] The system cannot find the path specified: 'C:\\Users\\severna\\Desktop\\Non_Test_Python_Files\\app4export'
but python made this location so I dont get why it cant find it later?
import os
import zipfile
import shutil
def zip(src, dst):
zf = zipfile.ZipFile("%s.zip" % (dst), "w", zipfile.ZIP_DEFLATED)
abs_src = os.path.abspath(src)
for dirname, subdirs, files in os.walk(src):
for filename in files:
absname = os.path.abspath(os.path.join(dirname, filename))
arcname = absname[len(abs_src) + 1:]
print('zipping %s as %s' % (os.path.join(dirname, filename),
arcname))
zf.write(absname, arcname)
zf.close()
source=r"C:\Users\severna\Desktop\Non_Test_Python_Files\dist"
destination=r"C:\Users\severna\Desktop\Non_Test_Python_Files\app4export"
shutil.rmtree(str(destination))
try:
zip(str(source), str(destination))
shutil.rmtree(str(source))
except FileNotFoundError:
print("Source cannot be zipped as it does not exist!")

Your code creates the file C:\Users\severna\Desktop\Non_Test_Python_Files\app4export.zip, but you try to remove the directory C:\Users\severna\Desktop\Non_Test_Python_Files\app4export
So just before the try-block you have
shutil.rmtree(str(destination))
which will throw an FileNotFoundError if the path do not exist. And when you hit that line of code, you still havent created the path. The reason it might have worked yesterday was that you mayby had that path.

after discussion with Cleared I found out that I needed i file extension because it was a file and shutil.rmtree doesnt remove files it removes directories so I need to use this code instead
os.remove(str(destination)+".zip")

Renaming multiple files in a directory using Python

I'm trying to rename multiple files in a directory using this Python script:
import os
path = '/Users/myName/Desktop/directory'
files = os.listdir(path)
i = 1
for file in files:
os.rename(file, str(i)+'.jpg')
i = i+1
When I run this script, I get the following error:
Traceback (most recent call last):
File "rename.py", line 7, in <module>
os.rename(file, str(i)+'.jpg')
OSError: [Errno 2] No such file or directory
Why is that? How can I solve this issue?
Thanks.

You are not giving the whole path while renaming, do it like this:
import os
path = '/Users/myName/Desktop/directory'
files = os.listdir(path)
for index, file in enumerate(files):
os.rename(os.path.join(path, file), os.path.join(path, ''.join([str(index), '.jpg'])))
Edit: Thanks to tavo, The first solution would move the file to the current directory, fixed that.

You have to make this path as a current working directory first.
simple enough.
rest of the code has no errors.
to make it current working directory:
os.chdir(path)

import os
from os import path
import shutil
Source_Path = 'E:\Binayak\deep_learning\Datasets\Class_2'
Destination = 'E:\Binayak\deep_learning\Datasets\Class_2_Dest'
#dst_folder = os.mkdir(Destination)
def main():
for count, filename in enumerate(os.listdir(Source_Path)):
dst = "Class_2_" + str(count) + ".jpg"
# rename all the files
os.rename(os.path.join(Source_Path, filename), os.path.join(Destination, dst))
# Driver Code
if __name__ == '__main__':
main()

As per #daniel's comment, os.listdir() returns just the filenames and not the full path of the file. Use os.path.join(path, file) to get the full path and rename that.
import os
path = 'C:\\Users\\Admin\\Desktop\\Jayesh'
files = os.listdir(path)
for file in files:
os.rename(os.path.join(path, file), os.path.join(path, 'xyz_' + file + '.csv'))

Just playing with the accepted answer define the path variable and list:
path = "/Your/path/to/folder/"
files = os.listdir(path)
and then loop over that list:
for index, file in enumerate(files):
#print (file)
os.rename(path+file, path +'file_' + str(index)+ '.jpg')
or loop over same way with one line as python list comprehension :
[os.rename(path+file, path +'jog_' + str(index)+ '.jpg') for index, file in enumerate(files)]
I think the first is more readable, in the second the first part of the loop is just the second part of the list comprehension

If your files are renaming in random manner then you have to sort the files in the directory first. The given code first sort then rename the files.
import os
import re
path = 'target_folder_directory'
files = os.listdir(path)
files.sort(key=lambda var:[int(x) if x.isdigit() else x for x in re.findall(r'[^0-9]|[0-9]+', var)])
for i, file in enumerate(files):
os.rename(path + file, path + "{}".format(i)+".jpg")

I wrote a quick and flexible script for renaming files, if you want a working solution without reinventing the wheel.
It renames files in the current directory by passing replacement functions.
Each function specifies a change you want done to all the matching file names. The code will determine the changes that will be done, and displays the differences it would generate using colors, and asks for confirmation to perform the changes.
You can find the source code here, and place it in the folder of which you want to rename files https://gist.github.com/aljgom/81e8e4ca9584b481523271b8725448b8
It works in pycharm, I haven't tested it in other consoles
The interaction will look something like this, after defining a few replacement functions
when it's running the first one, it would show all the differences from the files matching in the directory, and you can confirm to make the replacements or no, like this

This works for me and by increasing the index by 1 we can number the dataset.
import os
path = '/Users/myName/Desktop/directory'
files = os.listdir(path)
index=1
for index, file in enumerate(files):
os.rename(os.path.join(path, file),os.path.join(path,''.join([str(index),'.jpg'])))
index = index+1
But if your current image name start with a number this will not work.

Recovering filenames from a folder in linux using python

I am trying to use the listdir function from the os module in python to recover a list of filenames from a particular folder.
here's the code:
import os
def rename_file():
# extract filenames from a folder
#for each filename, rename filename
list_of_files = os.listdir("/home/admin-pc/Downloads/prank/prank")
print (list_of_files)
I am getting the following error:
OSError: [Errno 2] No such file or directory:
it seems to give no trouble in windows, where you start your directory structure from the c drive.
how do i modify the code to work in linux?

The code is correct. There should be some error with the path you provided.
You could open a terminal and enter into the folder first. In the terminal, just key in pwd, then you could get the correct path.
Hope that works.

You could modify your function to exclude that error with check of existence of file/directory:
import os
def rename_file():
# extract filenames from a folder
#for each filename, rename filename
path_to_file = "/home/admin-pc/Downloads/prank/prank"
if os.exists(path_to_file):
list_of_files = os.listdir(path_to_file)
print (list_of_files)

Open a file without specifying the subdirectory python

Lets say my python script is in a folder "/main". I have a bunch of text files inside subfolders in main. I want to be able to open a file just by specifying its name, not the subdirectory its in.
So open_file('test1.csv') should open test1.csv even if its full path is /main/test/test1.csv.
I don't have duplicated file names so it should no be a problem.
I using windows.

you could use os.walk to find your filename in a subfolder structure
import os
def find_and_open(filename):
for root_f, folders, files in os.walk('.'):
if filename in files:
# here you can either open the file
# or just return the full path and process file
# somewhere else
with open(root_f + '/' + filename) as f:
f.read()
# do something
if you have a very deep folder structure you might want to limit the depth of the search

import os
def get_file_path(file):
for (root, dirs, files) in os.walk('.'):
if file in files:
return os.path.join(root, file)
This should work. It'll return the path, so you should handle opening the file, in your code.

import os
def open_file(filename):
f = open(os.path.join('/path/to/main/', filename))
return f

Python - Need to loop through directories looking for TXT files

I am a total Python Newb
I need to loop through a directory looking for .txt files, and then read and process them individually. I would like to set this up so that whatever directory the script is in is treated as the root of this action. For example if the script is in /bsepath/workDir, then it would loop over all of the files in workDir and its children.
What I have so far is:
#!/usr/bin/env python
import os
scrptPth = os.path.realpath(__file__)
for file in os.listdir(scrptPth)
with open(file) as f:
head,sub,auth = [f.readline().strip() for i in range(3)]
data=f.read()
#data.encode('utf-8')
pth = os.getcwd()
print head,sub,auth,data,pth
This code is giving me an invalid syntax error and I suspect that is because os.listdir does not like file paths in standard string format. Also I dont think that I am doing the looped action right. How do I reference a specific file in the looped action? Is it packaged as a variable?
Any help is appriciated

import os, fnmatch
def findFiles (path, filter):
for root, dirs, files in os.walk(path):
for file in fnmatch.filter(files, filter):
yield os.path.join(root, file)
Use it like this, and it will find all text files somewhere within the given path (recursively):
for textFile in findFiles(r'C:\Users\poke\Documents', '*.txt'):
print(textFile)

os.listdir expects a directory as input. So, to get the directory in which the script resides use:
scrptPth = os.path.dirname(os.path.realpath(__file__))
Also, os.listdir returns just the filenames, not the full path.
So open(file) will not work unless the current working directory happens to be the directory where the script resides. To fix this, use os.path.join:
import os
scrptPth = os.path.dirname(os.path.realpath(__file__))
for file in os.listdir(scrptPth):
with open(os.path.join(scrptPth, file)) as f:
Finally, if you want to recurse through subdirectories, use os.walk:
import os
scrptPth = os.path.dirname(os.path.realpath(__file__))
for root, dirs, files in os.walk(scrptPth):
for filename in files:
filename = os.path.join(root, filename)
with open(filename, 'r') as f:
head,sub,auth = [f.readline().strip() for i in range(3)]
data=f.read()
#data.encode('utf-8')