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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.

How to read a file whose name includes '/' in python?

Now I have a file named Land/SeaMask and I want to open it, but it cannot be recognized as a filename by programme, but as a directory, how to do it?

First of all I recommend you to find out how Python interpreter displays yours file name. You can do this simply using os built-in module:
import os
os.listdir('path/to/directory')
You'll get a list of directories and files in directory you passed as argument in listdir method. In this list you can find something like Land:SeaMask. After recognizing this, open('path/to/Land:SeaMask') will work for you.

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 to extract a specific war file to a specific folder

I have the python code which will download the .war file and put it in a path which is specified by the variable path.
Now I wish to extract a specific file from that war to a specific folder.
But I got struck up here :
os.system(jar -xvf /*how to give the path varible here*/ js/pay.js)
I'm not sure how to pass on the variable path to os.system command.
I'm very new to python, kindly help me out.

If you really want to use os.system, the shell command line is passed as a string, and you can pass any string you want. So:
os.system('jar -xvf "' + pathvariable + '" js/pay.js)
Or you can use {} or %s formatting, etc.
However, you probably do not want to use os.system.
First, if you want to run other programs, it's almost always better to use the subprocess module. For example:
subprocess.check_call(['jar', '-xvf', pathvariable, 'js/pay.js'])
As you can see, you can pass a list of arguments instead of trying to work out how to put a string together (and deal with escaping and quoting and all that mess). And there are lots of other advantages, mostly described in the documentation itself.
However, you probably don't want to run the war tool at all. As jimhark says, a WAR file is just a special kind of JAR file, which is just a special kind of ZIP file. For creating them, you generally want to use JAR/WAR-specific tools (you need to verify the layout, make sure the manifest is the first entry in the ZIP directory, take care of the package signature, etc.), but for expanding them, any ZIP tool will work. And Python has ZIP support built in. What you want to do is probably as simple as this:
import zipfile
with zipfile.ZipFile(pathvariable, 'r') as zf:
zf.extract('js/pay.js', destinationpathvariable)
IIRC, you can only directly use ZipFile in a with statement in 2.7 and 3.2+, so if you're on, say, 2.6 or 3.1, you have to do it indirectly:
from contextlib import closing
import zipfile
with closing(zipfile.ZipFile(pathvariable, 'r')) as zf:
zf.extract('js/pay.js', destinationpathvariable)
Or, if this is just a quick&dirty script that quits as soon as it's done, you can get away with:
import zipfile
zf = zipfile.ZipFile(pathvariable, 'r')
zf.extract('js/pay.js', destinationpathvariable)
But I try to always use with statements whenever possible, because it's a good habit to have.

Isn't a war file a type of zip file? Python has zipfile support (click link for docs page).

You can use os.environ, it holds all environment variables on it.
It is a dict, so you can just use it like:
pypath = os.environ['PYTHONPATH']
now if you mean it's a common python variable, just use it like:
var1 = 'pause'
os.system('#echo & %s' % var1)

Use python tarfile add without normpath being applied to arcname

I'm using python's tarfile library to create a gzipped tar file.
The tar file needs to have absolute pathnames for reasons that I've got no control over. (I'm aware that this isn't normal practice.)
When I call
tarobject.add("/foo/xx1", "/bar/xx1")
the arcname argument "/bar/xx1" is run through os.path.normpath() and converted to "bar/xx1"
How do I avoid this and end up with "/bar/xx1" as I require?
I've read that I can replace normpath somewhere, but I'm fairly new to Python and I'm not sure how to do this or what the wider implications would be.
edit
After looking at this question I had a closer look at the tarinfo object, and this seems to work:
my_tarinfo = tarobject.gettarinfo("/foo/xx1")
my_tarinfo.name = "/bar/xx1"
tarobject.addfile(my_tarinfo, file("/foo/xx1"))

You really do strange things. You need to copy /usr/lib64/python/tarfile.py to a directory in your PYTHONPATH. Then you can modify this file. But you need to bundle you own tarfile module with your code.