We Keep Coding

can't find .nii files

I'm trying to load .nii-files with the nibabel library and am running into the following error with the .nii-files:
>>> path = '..\\..\\data\\raw\\data\\images\\AD\\female\\002_S_0938\\MPR__GradWarp__B1_Correction__N3__Scaled\\2006-10-05_15_54_26.0\\S19852\\ADNI_002_S_0938_MR_MPR__GradWarp__B1_Correction__N3__Scaled_Br_20070219175406282_S19852_I40980.nii'
>>> with open(Path(path), 'rb') as f:
>>> print('great success!')
FileNotFoundError: [Errno 2] No such file or directory: '..\\..\\data\\raw\\data\\images\\AD\\female\\005_S_1341\\MPR__GradWarp__B1_Correction__N3__Scaled\\2007-03-07_12_44_49.0\\S27673\\ADNI_005_S_1341_MR_MPR__GradWarp__B1_Correction__N3__Scaled_Br_20070717180348670_S27673_I60417.nii'
And with the test-file:
>>> path = '..\\..\\data\\raw\\data\\images\\AD\\female\\002_S_0938\\MPR__GradWarp__B1_Correction__N3__Scaled\\2006-10-05_15_54_26.0\\S19852\\test.txt'
>>> with open(Path(path), 'rb') as f:
>>> print('great success!')
great success!
Alternatively, the problem still exists with the library designed to read .nii-files:
>>> import nibabel as nib
>>> path = '..\\..\\data\\raw\\data\\images\\AD\\female\\002_S_0938\\MPR__GradWarp__B1_Correction__N3__Scaled\\2006-10-05_15_54_26.0\\S19852\\ADNI_002_S_0938_MR_MPR__GradWarp__B1_Correction__N3__Scaled_Br_20070219175406282_S19852_I40980.nii'
>>> nib.load(path)
FileNotFoundError: No such file or no access: '..\..\data\raw\data\images\AD\female\005_S_1341\MPR__GradWarp__B1_Correction__N3__Scaled\2007-03-07_12_44_49.0\S27673\ADNI_005_S_1341_MR_MPR__GradWarp__B1_Correction__N3__Scaled_Br_20070717180348670_S27673_I60417.nii'
And with the test file:
>>> path = '..\\..\\data\\raw\\data\\images\\AD\\female\\002_S_0938\\MPR__GradWarp__B1_Correction__N3__Scaled\\2006-10-05_15_54_26.0\\S19852\\test.txt'
>>> nib.load(path)
ImageFileError: Empty file: '..\..\data\raw\data\images\AD\female\002_S_0938\MPR__GradWarp__B1_Correction__N3__Scaled\2006-10-05_15_54_26.0\S19852\test.txt'
which in my opinion is strange behavior as the library finds the text file but not the .nii file, which exists in the same directory.
Note:
the .nii files are not corruputed as I can open them in a separate program with no problem.
I operate a windows OS
Update:
Current file structure:
Full traceback:

Since os.listdir('..\\..\\data\\raw\\data\\images\\AD\\female\\002_S_0938\\MPR__GradWarp__B1_Correction__N3__Scaled\\2006-10-05_15_54_26.0\\S19852\\') returns both files, and you can open the file test.txt but not 'ADNI_002_S_0938_MR_MPR__GradWarp__B1_Correction__N3__Scaled_Br_20070219175406282_S19852_I40980.nii', it suggests the issue might be related to the long filepath.
The 260 character limit can be disabled by setting this registry key to 1: Computer\HKEY_LOCAL_MACHINE\SYSTEM\CurrentControlSet\Control\FileSystem\LongPathsEnabled
See Enable Long Paths in Windows 10, Version 1607, and Later

How to Fix "Python permission denied error" [duplicate]

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

Reading a file using a relative path in a Python project

Say I have a Python project that is structured as follows:
project
/data
test.csv
/package
__init__.py
module.py
main.py
__init__.py:
from .module import test
module.py:
import csv
with open("..data/test.csv") as f:
test = [line for line in csv.reader(f)]
main.py:
import package
print(package.test)
When I run main.py I get the following error:
C:\Users\Patrick\Desktop\project>python main.py
Traceback (most recent call last):
File "main.py", line 1, in <module>
import package
File "C:\Users\Patrick\Desktop\project\package\__init__.py", line 1, in <module>
from .module import test
File "C:\Users\Patrick\Desktop\project\package\module.py", line 3, in <module>
with open("../data/test.csv") as f:
FileNotFoundError: [Errno 2] No such file or directory: '../data/test.csv'
However, if I run module.py from the package directory, I don’t get any errors. So it seems that the relative path used in open(...) is only relative to where the originating file is being run from (i.e __name__ == "__main__")? I don't want to use absolute paths. What are some ways to deal with this?

Relative paths are relative to current working directory.
If you do not want your path to be relative, it must be absolute.
But there is an often used trick to build an absolute path from current script: use its __file__ special attribute:
from pathlib import Path
path = Path(__file__).parent / "../data/test.csv"
with path.open() as f:
test = list(csv.reader(f))
This requires python 3.4+ (for the pathlib module).
If you still need to support older versions, you can get the same result with:
import csv
import os.path
my_path = os.path.abspath(os.path.dirname(__file__))
path = os.path.join(my_path, "../data/test.csv")
with open(path) as f:
test = list(csv.reader(f))
[2020 edit: python3.4+ should now be the norm, so I moved the pathlib version inspired by jpyams' comment first]

For Python 3.4+:
import csv
from pathlib import Path
base_path = Path(__file__).parent
file_path = (base_path / "../data/test.csv").resolve()
with open(file_path) as f:
test = [line for line in csv.reader(f)]

This worked for me.
with open('data/test.csv') as f:

My Python version is Python 3.5.2 and the solution proposed in the accepted answer didn't work for me. I've still were given an error
FileNotFoundError: [Errno 2] No such file or directory
when I was running my_script.py from the terminal. Although it worked fine when I run it through Run/Debug Configurations from the PyCharm IDE (PyCharm 2018.3.2 (Community Edition)).
Solution:
instead of using:
my_path = os.path.abspath(os.path.dirname(__file__)) + some_rel_dir_path
as suggested in the accepted answer, I used:
my_path = os.path.abspath(os.path.dirname(os.path.abspath(__file__))) + some_rel_dir_path
Explanation:
Changing os.path.dirname(__file__) to os.path.dirname(os.path.abspath(__file__))
solves the following problem:
When we run our script like that: python3 my_script.py
the __file__ variable has a just a string value of "my_script.py" without path leading to that particular script. That is why method dirname(__file__) returns an empty string "". That is also the reason why my_path = os.path.abspath(os.path.dirname(__file__)) + some_rel_dir_path is actually the same thing as my_path = some_rel_dir_path. Consequently FileNotFoundError: [Errno 2] No such file or directory is given when trying to use open method because there is no directory like "some_rel_dir_path".
Running script from PyCharm IDE Running/Debug Configurations worked because it runs a command python3 /full/path/to/my_script.py (where "/full/path/to" is specified by us in "Working directory" variable in Run/Debug Configurations) instead of justpython3 my_script.py like it is done when we run it from the terminal.

Try
with open(f"{os.path.dirname(sys.argv[0])}/data/test.csv", newline='') as f:

I was surprised when the following code worked.
import os
for file in os.listdir("../FutureBookList"):
if file.endswith(".adoc"):
filename, file_extension = os.path.splitext(file)
print(filename)
print(file_extension)
continue
else:
continue
So, I checked the documentation and it says:
Changed in version 3.6: Accepts a path-like object.
path-like object:
An object representing a file system path. A path-like object is
either a str or...
I did a little more digging and the following also works:
with open("../FutureBookList/file.txt") as file:
data = file.read()

Python Permission Error when reading

import os
import rarfile
file = input("Password List Directory: ")
rarFile = input("Rar File: ")
passwordList = open(os.path.dirname(file+'.txt'),"r")
with this code I am getting the error:
Traceback (most recent call last):
File "C:\Users\Nick L\Desktop\Programming\PythonProgramming\RarCracker.py", line 7, in <module>
passwordList = open(os.path.dirname(file+'.txt'),"r")
PermissionError: [Errno 13] Permission denied: 'C:\\Users\\Nick L\\Desktop'
This is weird because I have full permission to this file as I can edit it and do whatever I want, and I am only trying to read it. Every other question I read on stackoverflow was regarding writing to a file and getting a permissions error.

You're trying to open a directory, not a file, because of the call to dirname on this line:
passwordList = open(os.path.dirname(file+'.txt'),"r")
To open the file instead of the directory containing it, you want something like:
passwordList = open(file + '.txt', 'r')
Or better yet, use the with construct to guarantee that the file is closed after you're done with it.
with open(file + '.txt', 'r') as passwordList:
# Use passwordList here.
...
# passwordList has now been closed for you.
On Linux, trying to open a directory raises an IsADirectoryError in Python 3.5, and an IOError in Python 3.1:
IsADirectoryError: [Errno 21] Is a directory: '/home/kjc/'
I don't have a Windows box to test this on, but according to Daoctor's comment, at least one version of Windows raises a PermissionError when you try to open a directory.
PS: I think you should either trust the user to enter the whole directory-and-file name him- or herself --- without you appending the '.txt' to it --- or you should ask for just the directory, and then append a default filename to it (like os.path.join(directory, 'passwords.txt')).
Either way, asking for a "directory" and then storing it in a variable named file is guaranteed to be confusing, so pick one or the other.

os.path.dirname() will return the Directory in which the file is present not the file path. For example if file.txt is in path= 'C:/Users/Desktop/file.txt' then os.path.dirname(path)wil return 'C:/Users/Desktop' as output, while the open() function expects a file path.
You can change the current working directory to file location and open the file directly.
os.chdir(<File Directory>)
open(<filename>,'r')
or
open(os.path.join(<fileDirectory>,<fileName>),'r')

Relative paths break when executing Python script from Windows batch?

My Python script works perfectly if I execute it directly from the directory it's located in. However if I back out of that directory and try to execute it from somewhere else (without changing any code or file locations), all the relative paths break and I get a FileNotFoundError.
The script is located at ./scripts/bin/my_script.py. There is a directory called ./scripts/bin/data/. Like I said, it works absolutely perfectly as long as I execute it from the same directory... so I'm very confused.
Successful Execution (in ./scripts/bin/): python my_script.py
Failed Execution (in ./scripts/): Both python bin/my_script.py and python ./bin/my_script.py
Failure Message:
Traceback (most recent call last):
File "./bin/my_script.py", line 87, in <module>
run()
File "./bin/my_script.py", line 61, in run
load_data()
File "C:\Users\XXXX\Desktop\scripts\bin\tables.py", line 12, in load_data
DATA = read_file("data/my_data.txt")
File "C:\Users\XXXX\Desktop\scripts\bin\fileutil.py", line 5, in read_file
with open(filename, "r") as file:
FileNotFoundError: [Errno 2] No such file or directory: 'data/my_data.txt'
Relevant Python Code:
def read_file(filename):
with open(filename, "r") as file:
lines = [line.strip() for line in file]
return [line for line in lines if len(line) == 0 or line[0] != "#"]
def load_data():
global DATA
DATA = read_file("data/my_data.txt")

Yes, that is logical. The files are relative to your working directory. You change that by running the script from a different directory.
What you could do is take the directory of the script you are running at run time and build from that.
import os
def read_file(filename):
#get the directory of the current running script. "__file__" is its full path
path, fl = os.path.split(os.path.realpath(__file__))
#use path to create the fully classified path to your data
full_path = os.path.join(path, filename)
with open(full_path, "r") as file:
#etc

Your resource files are relative to your script. This is OK, but you need to use
os.path.realpath(__file__)
or
os.path.dirname(sys.argv[0])
to obtain the directory where the script is located. Then use os.path.join() or other function to generate the paths to the resource files.