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Python script that creates new file and returns list of files

I'm trying to create a python script called script.py with new_directory function that creates a new directory inside the current working directory, then creates a new empty file inside the new directory, and returns the list of files in that directory.
The output I get is ["script.py"] which looks correct but gives me this error:
RuntimeErrorElement(RuntimeError,Error on line 5:
directory = os.mkdir("/home/PythonPrograms")
FileExistsError: [Errno 17] File exists: '/home/PythonPrograms'
)
import os
def new_directory(directory, filename):
if os.path.isdir(directory) == False:
directory = os.mkdir("/home/PythonPrograms")
os.chdir("PythonPrograms")
with open("script.py", "w") as file:
pass
# Return the list of files in the new directory
return os.listdir("/home/PythonPrograms")
print(new_directory("PythonPrograms", "script.py"))
How do I correct and why is this wrong?

As others have said, it is hard to debug without the error. In the right cercumstances, your code will work without errors. As #Jack suggested, I suspect you're current directory is not /home. This means you've created a directory called PythonPrograms in /home directory. os.chdir("PythonPrograms") is trying to change the directory to <currentDirectory>/PythonPrograms, which doesn't exist.
I have tried to rework your code (without completely changing it), into something that should work in all cases. I think the lesson here is, work with the variables you already have (i.e. directory), rather than hardcoding it into the function.
import os
def new_directory(directory, filename):
if not os.path.isdir(directory):
# Create directory within current directory
# This is working off the relative path (from your current directory)
directory = os.mkdir(directory)
# Create file if does not exist
# this is a one-liner version of you with...pass statement
open(os.path.join(directory, filename), 'a').close()
# Return the list of files in the new directory
return os.listdir(directory)
print(new_directory("PythonPrograms", "script.py"))
I hope that helps.

I'm guessing the error you're getting is because you're not able to switch directories to PythonPrograms? This would be because your python current working directory does not contain it. If you more explicitly write out the directory you want to switch to, for example putting os.chdir("/home/PythonPrograms"), then it may work for you.
Ideally you should give us any stack traces or more info on the errors, though

I'm not sure why in your code you have with open("script.py", "w") as file: pass,
but here is mt way:
import os
os.mkdir('.\\Newfolder') # Create a new folder called Newfolder in the current directory
open('.\\Newfolder\\file.txt','w').close() # Create a new file called file.txt into Newfolder
print(os.listdir('.')) # Print out all the files in the current directory

How can you harvest the current user and script's working directory in python?

Currently, I am statically specifying a filepath for the source data for a script using the snippet below:
def get_files():
global thedir
thedir = 'C:\\Users\\username\\Documents'
list = os.listdir(thedir)
for i in list:
if i.endswith('.txt'):
print("\n\n"+i)
eat_file(thedir+'\\'+i)
The reason that I am statically assigning the location is that the script is failing to execute correctly when it is executed in a debugging environment such as Eclipse & Visual Studio Code. These debuggers assume that the script is being ran from their working directories.
Since I'm unable to modify the local settings of every system that might be running this script, is there a recommended module to force the script to harvest either the active user (linux and windows) and/or the script's local directory pragmatically?

The new-ish pathlib module (available in Python >= 3.4) is great for working with path-like objects (both Windows and for other OSes). Use it. Don't bother with the outdated os module. And don't bother trying to use naked strings for representing path-like objects.
It's Paths - Paths all the way down
To simplify: you can build up any path (directory and file path objects are treated exactly the same) as an object, which can be an absolute path object or a relative path object.
Simple displaying of some useful paths- such as the current working directory and the user home- works like this:
from pathlib import Path
# Current directory (relative):
cwd = Path() # or Path('.')
print(cwd)
# Current directory (absolute):
cwd = Path.cwd()
print(cwd)
# User home directory:
home = Path.home()
print(home)
# Something inside the current directory
file_path = Path('some_file.txt') # relative path; or
file_path = Path()/'some_file.txt' # also relative path
file_path = Path().resolve()/Path('some_file.txt') # absolute path
print(file_path)
To navigate down the file tree, you can do things like this. Note that the first object, home, is a Path and the rest are just strings:
file_path = home/'Documents'/'project'/'data.txt' # or
file_path = home.join('Documents', 'project', 'data.txt')
To read a file located at a path, you can use its open method rather than the open function:
with file_path.open() as f:
dostuff(f)
But you can also just grab the text directly!
contents = file_path.read_text()
content_lines = contents.split('\n')
...and WRITE text directly!
data = '\n'.join(content_lines)
file_path.write_text(data) # overwrites existing file
Check to see if it is a file or a directory (and exists) this way:
file_path.is_dir() # False
file_path.is_file() # True
Make a new, empty file without opening it like this (silently replaces any existing file):
file_path.touch()
To make the file only if it doesn't exist, use exist_ok=False:
try:
file_path.touch(exist_ok=False)
except FileExistsError:
# file exists
Make a new directory (under the current directory, Path()) like this:
Path().mkdir('new/dir') # get errors if Path()/`new` doesn't exist
Path().mkdir('new/dir', parents=True) # will make Path()/`new` if it doesn't exist
Path().mkdir('new/dir', exist_ok=True) # errors ignored if `dir` already exists
Get the file extension or filename of a path this way:
file_path.suffix # empty string if no extension
file_path.stem # note: works on directories too
Use name for the entire last part of the path (stem and extension if they are there):
file_path.name # note: works on directories too
Rename a file using the with_name method (which returns the same path object but with a new filename):
new_path = file_path.with_name('data.txt')
You can iterate through all the "stuff' in a directory like so using iterdir:
all_the_things = list(Path().iterdir()) # returns a list of Path objects

File path in python

I'm trying to load the json file but it gives me an error saying No such file or directory:
with open ('folder1/sub1/sub2/sub2/sub3/file.json') as f:
data = json.load(f)
print data
The above file main.py is kept outside the folder1. All of this is kept under project folder.
So, the directory structure is Project/folder1/sub1/sub2/sub2/sub3/file.json
Where am I going wrong?

I prefer to point pathes starting from file directory
import os
script_dir = os.path.dirname(__file__)
file_path = os.path.join(script_dir, 'relative/path/to/file.json')
with open(file_path, 'r') as fi:
pass
this allows not to care about working directory changes. And also this allows to run script from any directory using it's full path.
python script/inner/script.py
or
python script.py

I would use os.path.join method to form the complete path starting from the current directory.
Something like:
json_filepath = os.path.join('.', 'folder1', 'sub1', 'sub2', 'sub3', 'file.json')

As always, an initial slash indicates that the path starts from the root. Omit the initial slash to indicate that it is a relative path.

Directory is not being recognized in Python

I'm uploading a zipped folder that contains a folder of text files, but it's not detecting that the folder that is zipped up is a directory. I think it might have something to do with requiring an absolute path in the os.path.isdir call, but can't seem to figure out how to implement that.
zipped = zipfile.ZipFile(request.FILES['content'])
for libitem in zipped.namelist():
if libitem.startswith('__MACOSX/'):
continue
# If it's a directory, open it
if os.path.isdir(libitem):
print "You have hit a directory in the zip folder -- we must open it before continuing"
for item in os.listdir(libitem):

The file you've uploaded is a single zip file which is simply a container for other files and directories. All of the Python os.path functions operate on files on your local file system which means you must first extract the contents of your zip before you can use os.path or os.listdir.
Unfortunately it's not possible to determine from the ZipFile object whether an entry is for a file or directory.
A rewrite or your code which does an extract first may look something like this:
import tempfile
# Create a temporary directory into which we can extract zip contents.
tmpdir = tempfile.mkdtemp()
try:
zipped = zipfile.ZipFile(request.FILES['content'])
zipped.extractall(tmpdir)
# Walk through the extracted directory structure doing what you
# want with each file.
for (dirpath, dirnames, filenames) in os.walk(tmpdir):
# Look into subdirectories?
for dirname in dirnames:
full_dir_path = os.path.join(dirpath, dirname)
# Do stuff in this directory
for filename in filenames:
full_file_path = os.path.join(dirpath, filename)
# Do stuff with this file.
finally:
# ... Clean up temporary diretory recursively here.

Usually to make things handle relative paths etc when running scripts you'd want to use os.path.
It seems to me that you're reading from a Zipfile the items you've not actually unzipped it so why would you expect the file/dirs to exist?
Usually I'd print os.getcwd() to find out where I am and also use os.path.join to join with the root of the data directory, whether that is the same as the directory containing the script I can't tell. Using something like scriptdir = os.path.dirname(os.path.abspath(__file__)).
I'd expect you would have to do something like
libitempath = os.path.join(scriptdir, libitem)
if os.path.isdir(libitempath):
....
But I'm guessing at what you're doing as it's a little unclear for me.

Create path and filename from string in Python

Given the following strings:
dir/dir2/dir3/dir3/file.txt
dir/dir2/dir3/file.txt
example/directory/path/file.txt
I am looking to create the correct directories and blank files within those directories.
I imported the os module and I saw that there is a mkdir function, but I am not sure what to do to create the whole path and blank files. Any help would be appreciated. Thank you.

Here is the answer on all your questions (directory creation and blank file creation)
import os
fileList = ["dir/dir2/dir3/dir3/file.txt",
"dir/dir2/dir3/file.txt",
"example/directory/path/file.txt"]
for file in fileList:
dir = os.path.dirname(file)
# create directory if it does not exist
if not os.path.exists(dir):
os.makedirs(dir)
# Create blank file if it does not exist
with open(file, "w"):
pass

First of all, given that try to create directory under a directory that doesn't exist, os.mkdir will raise an error. As such, you need to walk through the paths and check whether each of the subdirectories has or has not been created (and use mkdir as required). Alternative, you can use os.makedirs to handle this iteration for you.
A full path can be split into directory name and filename with os.path.split.
Example:
import os
(dirname, filename) = os.path.split('dir/dir2/dir3/dir3/file.txt')
os.makedirs(dirname)
Given we have a set of dirs we want to create, simply use a for loop to iterate through them. Schematically:
dirlist = ['dir1...', 'dir2...', 'dir3...']
for dir in dirlist:
os.makedirs( ... )