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Creating new CSV files from inputs in different directory

I'm simply trying to alter a CSV file with Python.
When all the files were in the same dir, everything was fine.
Now that the input files are in a different dir than where the output files will be, everything blows up, apparently b/c the files do not exist?
I first found this:
open() in Python does not create a file if it doesn't exist
Then I learned to change to the directory, which helped me loop over the CSVs in the target dir:
Moving up one directory in Python
When I run the command:
python KWRG.py ../Weekly\ Reports\ -\ Inbound/Search\ Activity/ 8/9/2021
I will get:
Traceback (most recent call last): File "KWRG.py", line 15, in <module> with open(args.input, 'r') as in_file, open(args.output, 'w') as out_file: IsADirectoryError: [Errno 21] Is a directory: '../Weekly Reports - Inbound/Search Activity/'
Sorry If I'm missing the obvious here, but why is the file not being created in the directory that I'm pointing the script to (or at all for that matter)?
The code:
import csv
import argparse
import os
# Create a parser to take arguments
#...snip...
cur_dir = os.getcwd()
reports_dir = os.chdir(os.path.join(cur_dir, args.dir))
for csv_file in os.listdir(reports_dir):
# Shorthand the name of the file
#...snip...
# Open the in and out files
with open(csv_file, 'r') as in_file, open(f'{out_name}-Search-Activity-{args.date}.csv', 'w+') as out_file:
# Re-arrange CSV
# EOF

Your problem is with this line:
reports_dir = os.chdir(os.path.join(cur_dir, args.dir))
os.chdir() doesn't return anything, it just performs the operation requested - changing the current working directory. From an interactive session with the REPL:
>>> import os
>>> result = os.chdir("/Users/mattdmo/")
>>> result
>>>
For your purposes, all you need is
reports_dir = os.path.join(cur_dir, args.dir)
and you'll be all set.

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')

File does not exist, but python is printing filename?

I am using python to merge a number of CSV files.
I use this code to find the CSV files:
with open('C:\TODAY.csv', 'w') as f_obj:
rows = []
files = os.listdir('C:\RAW\\')
I then iterate through files and make a list of the rows using:
for f in files:
if fnmatch.fnmatch(f, '*.csv') and not fnmatch.fnmatch(f, 'TODAY.CSV'):
print f
rows.append(open(f).readlines())
I added the print f for the sake of debugging.
What happens is this:
I get an error saying results1.csv does not exist when trying to do the rows.append function however the print f does print results1.csv and therefore it must exist, and python knows it exists because it can print the filename.
So why, if python can print the filename, does python say on the next line that it doesn't exist?
The traceback is:
Traceback (most recent call last):
File "Exc.py", line 22, in <module>
rows.append(open(f).readlines())
IOError: [Errno 2] No such file or directory: 'result1.csv

os.listdir gives a list of file names, not of file paths. The file names can be used to open a file only if your program's current directory is already the directory you're listing.
Use os.path.join to join the directory path with a file name to give a file path:
rows.append(open(os.path.join('C:\RAW\\', f)).readlines())

Ref https://docs.python.org/2/library/fnmatch.html
fnmatch only tells you if the given filename matches the given pattern, NOT whether a file with that name actually exists.
Try this:
for f in files:
if fnmatch.fnmatch(f, '*.csv') and not fnmatch.fnmatch(f, 'TODAY.CSV'):
print f
try:
rows.append(open(f).readlines())
except IOError:
print "File didn't exist, skipping"

Python, trying to write a CSV file to a specific location.

I am trying to write a file to a specific location, to do this I wrote the following code.
The program is in a folder on an external HDD. I have used the os.path to get the current path (I think..)
the "fileName" var is = hello
the "savePath" var is = data
When I run the code I get the following error...
IOError: [Errno 13] Permission denied: 'data\hello_23-04-2014_13-37-55.csv'
Do I need to set permissions for the file before I try to write to it? If so.. how do you do this>?
def writeData(fileName, savePath, data):
# Create a filename
thisdate = time.strftime("%d-%m-%Y")
thistime = time.strftime("%H-%M-%S")
name = fileName + "_" + thisdate + "_" + thistime + ".csv"
# Create the complete filename including the absolute path
completeName = os.path.join(savePath, name)
# Check if directory exists
if not os.path.exists(completeName):
os.makedirs(completeName)
# Write the data to a file
theFile = open(completeName, 'wb')
writer = csv.writer(theFile, quoting=csv.QUOTE_ALL)
writer.writerows(data)

I get a different error (putting permission issues aside) when I try a stripped down version of what you're doing here:
>>> import os
>>> path = "foo/bar/file.txt"
>>> os.makedirs(path)
>>> with open(path, "w") as f:
... f.write("HOWDY!")
...
Traceback (most recent call last):
File "<stdin>", line 1, in <module>
IOError: [Errno 21] Is a directory: 'foo/bar/file.txt'
Note that when you do this:
# Check if directory exists
if not os.path.exists(completeName):
os.makedirs(completeName)
...you're creating a directory that has a name that's both the path you want (good) and the name of the file you're trying to create. Pass the pathname only to makedirs() and then create the file inside that directory when you've done that.

python - open all plain text files in a directory with ".dta" extension and write lines to csv

I have a number of plain-text config files (.dta) that are spread through 27 sub-directories. I am trying to parse some information from all of them into a common document that is easier to work with.
Thus far I have:
import linecache
import csv
import os
csvout = csv.writer(open("dtaCompile.csv","wb"))
directory = os.path.join("c:\\","DirectKey")
for root,dirs,files in os.walk(directory):
for file in files:
if file.endswith(".DTA"):
f=open(file,'r')
lines = f.readlines()
description = lines[1]
articleCode = lines[2]
OS = lines[25]
SMBIOS = lines[32]
pnpID = lines[34]
cmdLine = lines[28]
csvout.writerow([SMBIOS, description, articleCode, pnpID, OS, cmdLine])
f.close()
I'm getting the following error:
Traceback (most recent call last):
File "test.py", line 11, in <module>
f=open(file,'r')
IOError: [Errno 2] No such file or directory: '000003APP.DTA'

Instead of
f=open(file,'r')
Your probaby need
f=open(os.path.join(directory, root, file),'r')
file is just the name of the file, and doesn't say anything about the path to it. you have to use os.path.join with the various components to create the full path

if file.endswith(".DTA"):
file = os.path.join(directory, root, file)

Instead of:
f=open(file,'r')
Try:
f = open(os.path.join(directory, file), "r")

My guess is that the directory you're program is executing in is not the same as the directory you're walking.
Try printing:
os.getcwd()
to see.