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Save a CSV in same directory as python file, using 'to_csv' and 'os.path'?

I want this line to save the csv in my current directory alongside my python file:
df.to_csv(./"test.csv")
My python file is in "C:\Users\Micheal\Desktop\VisualStudioCodes\Q1"
Unfortunately it saves it in "C:\Users\Micheal" instead.
I have tried import os path to use os.curdir but i get nothing but errors with that.
Is there even a way to save the csv alongside the python file using os.curdir?
Or is there a simpler way to just do this in python without importing anything?

import os
directory_of_python_script = os.path.dirname(os.path.abspath(__file__))
df.to_csv(os.path.join(directory_of_python_script, "test.csv"))
And if you want to read same .csv file later,
pandas.read_csv(os.path.join(directory_of_python_script, "test.csv"))
Here, __file__ gives the relative location(path) of the python script being runned. We get the absolute path by os.path.abspath() and then convert it to the name of the parent directory.
os.path.join() joins two paths together considering the operating system defaults for path seperators, '\' for Windows and '/' for Linux, for example.
This kind of an approach should work, I haven't tried, if does not work, let me know.

Rename and move file with Python

I have a Python script that compares existing file names in a folder to a reference table and then determines if it needs to be renamed or not.
As it loops through each filename:
'oldname' = the current file name
'newname' = what it needs to be renamed to
I want rename the file and move it to a new folder "..\renamedfiles"
Can I do the rename and the move at the same time as it iterates through the loop?

Yes you can do this. In Python you can use the move function in shutil library to achieve this.
Let's say on Linux, you have a file in /home/user/Downloads folder named "test.txt" and you want to move it to /home/user/Documents and also change the name to "useful_name.txt". You can do both things in the same line of code:
import shutil
shutil.move('/home/user/Downloads/test.txt', '/home/user/Documents/useful_name.txt')
In your case you can do this:
import shutil
shutil.move('oldname', 'renamedfiles/newname')

os.rename (and os.replace) won't work if the source and target locations are on different partitions/drives/devices. If that's the case, you need to use shutil.move, which will use atomic renaming if possible, and fallback to copy-then-delete if the destination is not on the same file system. It's perfectly happy to both move and rename in the same operation; the operation is the same regardless.

To do both of the operations, you can use the os.rename(src, dest) function.
You should have the wanted directory to save the file in, and the new file name. You can do this for every file you run across in your loop.
For example:
# In Windows
dest_dir = "tmp\\2"
new_name = "bar.txt"
current_file_name = "tmp\\1\\foo.txt"
os.rename(current_file_name, os.path.join(dest_dir, new_name))
The rename function allows you to change the name of the file and it's folder at the same time.
To prevent any errors in renaming and moving of the file, use shutil.move.

Since Python 3.4, working with paths is done easily with pathlib. Moving/renaming a file is done with rename or replace (will unconditionally do the replace). So combining with the parent attribute and the concat operator, you can do:
from pathlib import Path
source = Path("path/to/file/oldname")
target = source.replace(source.parent / "renames" / "newname")

Create a Python file in your desired directory and write something like that :
import os
for filename in os.listdir("."):
if(filename ...):
newFilename = ...
os.rename(filename, newFilename)

Errno13, Permission denied when trying to read file

I have created a small python script. With that I am trying to read a txt file but my access is denied resolving to an no.13 error, here is my code:
import time
import os
destPath = 'C:\Users\PC\Desktop\New folder(13)'
for root, dirs, files in os.walk(destPath):
f=open(destPath, 'r')
.....

Based on the name, I'm guessing that destPath is a directory, not a file. You can do a os.walk or a os.listdir on the directory, but you can't open it for reading. You can only call open on a file.
Maybe you meant to call open on one or more of the items from files

1:
I take it you are trying to access a file to get what's inside but don't want to use a direct path and instead want a variable to denote the path. This is why you did the destPath I'm assuming.
From what I've experienced the issue is that you are skipping a simple step. What you have to do is INPUT the location then use os.CHDIR to go to that location. and finally you can use your 'open()'.
From there you can either use open('[direct path]','r') or destPath2 = 'something' then open(destPath2, 'r').
To summarize: You want to get the path then NAVIGATE to the path, then get the 'filename' (can be done sooner or not at all if using a direct path for this), then open the file.
2: You can also try adding an "r" in front of your path. r'[path]' for the raw line in case python is using the "\" for something else.
3: Try deleting the "c:/" and switching the / to \ or vice versa.
That's all I got, hope one of them helps! :-)

I got this issue when trying to create a file in the path -C:/Users/anshu/Documents/Python_files/Test_files . I discovered python couldn't really access the directory that was under the user's name.
So, I tried creating the file under the directory - C:/Users/anshu/Desktop .
I was able to create files in this directory through python without any issue.

How to loop through the list of .tar.gz files using linux command in python

Using python 2.7
I have a list of *.tat.gz files on a linux box. Using python, I want to loop through the files and extract those files in a different location, under their respective folders.
For example: if my file name is ~/TargetData/zip/1440198002317590001.tar.gz
then I want to untar and ungzip this file in a different location under its
respective folder name i.e. ~/TargetData/unzip/1440198002317590001.
I have written some code but I am not able to loop through the files. In a command line I am able to untar using $ tar -czf 1440198002317590001.tar.gz 1440198002317590001 command. But I want to be able to loop through the .tar.gz files. The code is mentioned below. Here, I’m not able to loop just the files Or print only the files. Can you please help?
import os
inF = []
inF = str(os.system('ls ~/TargetData/zip/*.tar.gz'))
#print(inF)
if inF is not None:
for files in inF[:-1]:
print files
"""
os.system('tar -czf files /unzip/files[:-7]')
# This is what i am expecting here files = "1440198002317590001.tar.gz" and files[:-7]= "1440198002317590001"
"""
Have you ever worked on this type of use case? Your help is greatly appreciated!! Thank you!

I think you misunderstood the meaning of os.system(), that will do the job, but its return value was not expected by you, it returns 0 for successful done, you can not directly assign its output to a variable. You may consider the module [subprocess], see doc here. However, I DO NOT recommend that way to list files (actually, it returns string instead of list, see doc find the detail by yourself).
The best way I think would be glob module, see doc here. Use glob.glob(pattern), you can put all files match the pattern in a list, then you can loop it easily.
Of course, if you are familiar with os module, you also can use os.listdir(), os.path.join(), or even os.paht.expanduser() to do this. (Unlike glob, it only put filenames without fully path into a list, you need to reconstruct file path).
By the way, for you purpose here, there is no need to declare an empty list first (i.e. inF = [])
For unzip file part, you can do it by os.system, but I also recommend to use subprocess module instead of os.system, you will find the reason in the doc of subprocess.
DO NOT see the following code, ONLY see them after you really can not solve this by yourself.
import os
import glob
inF = glob.glob('~/TargetData/zip/*.tar.gz')
if inF:
for files in inF:
# consider subprocess.call() instead of os.system
unzip_name = files.replace('zip', 'unzip')[:-7]
# get directory name and make sure it exists, otherwise create it
unzip_dir = os.path.dirname(unzip_name)
if not os.path.exists(unzip_dir):
os.mkdir(unzip_dir)
subprocess.call(['tar -xzf', files, '-C', unzip_name])
# os.system('tar -czf files /unzip/files[:-7]')

How do I decide where to put the file while creating it in python?

With the code open, I can create a new file in the computer. How do i decide which folder it goes? I need to put them in a certain folder when i am creating them. Should I put sth in the brackets?
e.g. open("apple juice. txt", "a")

If you don't specify a path, then the file will be created in the current directory. Where exactly that is depends on how you started the interpreter. For example, when you start Python 3.4 from the Windows Start Menu, then the file will be saved in C:\Python34\.
If you want to specify a certain path, then do so:
f = open(r"C:\Users\David\Python Files\apple juice.txt", "a")

Give the full path:
with open("path_where/to_save/apple_juice.txt", "a") as f:
# do work
with will automatically close your file.

If you are looking for a place for temp files, use the module tempfile.
You can use the function tempfile.gettempdir() to get a path to a folder directory.
You can use tempfile.TemporaryFile() to generate a full path to the place where the temp files are usually stored on your OS.
The method used in either case to generate the temp path is explained here.