I already read some useful infos here about moving files with Python. But those are not working on my machine. I use eclipse to run python and the program should move files within windows.
I used os.rename, shutil.move, shutil.copy and so on....
Here is my simple code.
import os
import shutil
source = os.listdir("C:/test/")
dest_dkfrontend = "C:/test1/"
for files in source:
if files.startswith("Detail"):
print('Files found ' + files)
shutil.copy(files, dest_dkfrontend)
else:
print('File name not matching')
I receive an error like:
with open(src, 'rb') as fsrc:
IOError: [Errno 2] No such file or directory:
Could you please help to address this?
first you have to check if your destination directory exists or not.
and shutil.copy requires 1st parameter as source of file with file name and 2nd parameter as destination of file where to copy with new file name.
try this.
import os
import shutil
source = os.listdir("C:/test/")
dest_dkfrontend = "C:/test1/"
if not os.path.exists(dest_dkfrontend):
os.makedirs(dest_dkfrontend)
for files in source:
if files.startswith("Detail"):
print('Files found ' + files)
shutil.copy(source+files, dest_dkfrontend+files)
else:
print('File name not matching')
Related
This is a school program to learn how to use file and directory in Python. So to do my best I create a function to open, set it as a variable and close properly my file.
But I got the error of the title:
FileNotFoundError: [Errno 2] No such file or directory: 'codedata.pkl'
def load_db():
""" load data base properly
And get ready for later use
Return:
-------
cd : (list) list of tuples
"""
file = open('codedata.pkl', 'rb')
codedata = pickle.loads(file)
file.close()
return codedata
From the interpreter, this is the line
file = open('codedata.pkl', 'rb')
Which is the problem, but I don't see where is the source of the problem.
Can anyone help me?
Can you check what is the location of the file?
If your file is located at /Users/abc/Desktop/, then the code to open the file on python would be as shown below
file = open('/Users/abc/Desktop/codedata.pkl', 'rb')
codedata = pickle.load(file)
file.close()
You can also check if the file exists at the desired path by doing something like this
import os
filepath = '/Users/abc/Desktop/codedata.pkl'
if os.path.exists(filepath):
file = open('/Users/abc/Desktop/codedata.pkl', 'rb')
codedata = pickle.load(file)
file.close()
else:
print("File not present at desired location")
it happens when you run the script without determining it's the current working directory (example in vs code if you go to Explorer Tape )
You do not work from the same directory that your data.pkl in that's why No file exists
You can know the current directory from getcwd() usually it will be the C/User/.
print(os.getcwd())
filepath=""
if os.path.exists(r"D:\research\StleGAN\karras2019stylegan-ffhq-1024x1024.pkl"):
print("yes")
else:
print("no")
The solution is to open a directory that contains the script or to add the full path.
I'm having an issue trying to open a file that is definitely saved to my computer ('NYT-bestsellers.txt'), but whenever I try opening it with my code I get the error
FileNotFoundError: [Errno 2] No such file or directory: 'NYT-bestsellers.txt'
I thought about using the method where you use the full path to open the file… but this is part of an assignment that I'll be submitting later this week. If I open the file using a specific path from my laptop, I'm worried that it won't open for the marker. Please advise!
with open('NYT-bestsellers.txt', 'r') as file:
file = file.splitlines()
As Ryan said, every time you open a file by a relative name, you need to make clear for the current work path.
import sys
import os
current_work_directory = os.getcwd() # Return a string representing the current working directory.
print('Current work directory: {}'.format(current_work_directory))
# Make sure it's an absolute path.
abs_work_directory = os.path.abspath(current_work_directory)
print('Current work directory (full path): {}'.format(abs_work_directory))
print()
filename = 'NYT-bestsellers.txt'
# Check whether file exists.
if not os.path.isfile(filename):
# Stop with leaving a note to the user.
print('It seems file "{}" not exists in directory: "{}"'.format(filename, current_work_directory))
sys.exit(1)
# File exists, go on!
with open(filename, 'r') as file:
file = file.splitlines()
If you confirm that the file will be along with your python script file, you can do some preparatory work before opening the file:
script_directory = os.path.split(os.path.abspath(__file__))[0]
print(script_directory)
abs_filename = os.path.join(script_directory, filename)
print(abs_filename)
with open(abs_filename, 'r') as file:
file = file.splitlines()
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")
I'm trying to rename an audio file but I keep getting OSError: [Errno 2] No such file or directory.
In my program, each user has a directory that holds the users files. I obtain the path for each user by doing the following:
current_user_path = os.path.join(current_app.config['UPLOAD_FOLDER'], user.username)
/Users/johnsmith/Documents/pythonprojects/project/files/john
Now I want to obtain the path for the existing file I want to rename:
current_file_path = os.path.join(current_user_path,form.currentTitle.data)
/Users/johnsmith/Documents/pythonprojects/project/files/john/File1
The path for the rename file will be:
new_file_path = os.path.join(current_user_path, form.newTitle.data)
/Users/johnsmith/Documents/pythonprojects/project/files/john/One
Now I'm just running the rename command:
os.rename(current_file_path, new_file_path)
you can use os.rename for rename single file.
to avoid
OSError: [Errno 2] No such file or directory.
check if file exist or not.
here is the working example:
import os
src = "D:/test/Untitled003.wav"
dst = "D:/test/Audio001.wav"
if os.path.isfile(src):
os.rename(src, dst)
If the OS says there's no such file or directory, that's the gospel truth. You're making a lot of assumptions about where the file is, constructing a path to it, and renaming it. It's a safe bet there's no such file as the one named by current_file_path, or no directory to new_file_path.
Try os.stat(current_file_path), and similarly double-check the new file path with os.stat(os.posixpath.dirname(new_file_path)). Once you've got them right, os.rename will work if you've got permissions.
Try changing the current working directory to the one you want to work with. This code below should give you a simple walk through of how you should go about it:
import os
print (os.getcwd())
os.chdir(r"D:\Python\Example")
print (os.getcwd())
print ("start")
def rename_files():
file_list= os.listdir(r"D:\Python\Example")
print(file_list)
for file_name in file_list:
os.rename(file_name,file_name.translate(None,"0123456789")) rename_files()
print("stop")
print (os.getcwd())
I have two folders: In, Out - it is not system folder on disk D: - Windows 7. Out contain "myfile.txt" I run the following command in python:
>>> shutil.copyfile( r"d:\Out\myfile.txt", r"D:\In" )
Traceback (most recent call last):
File "<pyshell#39>", line 1, in <module>
shutil.copyfile( r"d:\Out\myfile.txt", r"D:\In" )
File "C:\Python27\lib\shutil.py", line 82, in copyfile
with open(dst, 'wb') as fdst:
IOError: [Errno 13] Permission denied: 'D:\\In'
What's the problem?
Read the docs:
shutil.copyfile(src, dst)
Copy the contents (no metadata) of the file named src to a file
named dst. dst must be the complete target file name; look at copy()
for a copy that accepts a target directory path.
use
shutil.copy instead of shutil.copyfile
example:
shutil.copy(PathOf_SourceFileName.extension,TargetFolderPath)
Use shutil.copy2 instead of shutil.copyfile
import shutil
shutil.copy2('/src/dir/file.ext','/dst/dir/newname.ext') # file copy to another file
shutil.copy2('/src/file.ext', '/dst/dir') # file copy to diff directory
I solved this problem, you should be the complete target file name for destination
destination = pathdirectory + filename.*
I use this code fir copy wav file with shutil :
# open file with QFileDialog
browse_file = QFileDialog.getOpenFileName(None, 'Open file', 'c:', "wav files (*.wav)")
# get file name
base = os.path.basename(browse_file[0])
os.path.splitext(base)
print(os.path.splitext(base)[1])
# make destination path with file name
destination= "test/" + os.path.splitext(base)[0] + os.path.splitext(base)[1]
shutil.copyfile(browse_file[0], destination)
First of all, make sure that your files aren't locked by Windows, some applications, like MS Office, locks the oppened files.
I got erro 13 when i was is trying to rename a long file list in a directory, but Python was trying to rename some folders that was at the same path of my files. So, if you are not using shutil library, check if it is a directory or file!
import os
path="abc.txt"
if os.path.isfile(path):
#do yor copy here
print("\nIt is a normal file")
Or
if os.path.isdir(path):
print("It is a directory!")
else:
#do yor copy here
print("It is a file!")
Visual Studio 2019
Solution : Administrator provided full Access to this folder "C:\ProgramData\Docker"
it is working.
ERROR: File IO error seen copying files to volume: edgehubdev. Errno: 13, Error Permission denied : [Errno 13] Permission denied: 'C:\ProgramData\Docker\volumes\edgehubdev\_data\edge-chain-ca.cert.pem'
[ERROR]: Failed to run 'iotedgehubdev start -d "C:\Users\radhe.sah\source\repos\testing\AzureIotEdgeApp1\config\deployment.windows-amd64.json" -v' with error: WARNING! Using --password via the CLI is insecure. Use --password-stdin.
ERROR: File IO error seen copying files to volume: edgehubdev. Errno: 13, Error Permission denied : [Errno 13] Permission denied: 'C:\ProgramData\Docker\volumes\edgehubdev\_data\edge-chain-ca.cert.pem'
use
> from shutil import copyfile
>
> copyfile(src, dst)
for src and dst use:
srcname = os.path.join(src, name)
dstname = os.path.join(dst, name)
This works for me:
import os
import shutil
import random
dir = r'E:/up/2000_img'
output_dir = r'E:/train_test_split/out_dir'
files = [file for file in os.listdir(dir) if os.path.isfile(os.path.join(dir, file))]
if len(files) < 200:
# for file in files:
# shutil.copyfile(os.path.join(dir, file), dst)
pass
else:
# Amount of random files you'd like to select
random_amount = 10
for x in range(random_amount):
if len(files) == 0:
break
else:
file = random.choice(files)
shutil.copyfile(os.path.join(dir, file), os.path.join(output_dir, file))
Make sure you aren't in (locked) any of the the files you're trying to use shutil.copy in.
This should assist in solving your problem
I avoid this error by doing this:
Import lib 'pdb' and insert 'pdb.set_trace()' before 'shutil.copyfile', it would just like this:
import pdb
...
print(dst)
pdb.set_trace()
shutil.copyfile(src,dst)
run the python file in a terminal, it will execute to the line 'pdb.set_trace()', and now the 'dst' file will print out.
copy the 'src' file by myself, and substitute and remove the 'dst' file which has been created by the above code.
Then input 'c' and click the 'Enter' key in the terminal to execute the following code.
well the questionis old, for new viewer of Python 3.6
use
shutil.copyfile( "D:\Out\myfile.txt", "D:\In" )
instead of
shutil.copyfile( r"d:\Out\myfile.txt", r"D:\In" )
r argument is passed for reading file not for copying