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Python os.getcwd() is not working on subfolders in VSCODE

I have a python file, converted from a Jupiter Notebook, and there is a subfolder called 'datasets' inside this file folder. When I'm trying to open a file that is inside that 'datasets' folder, with this code:
import pandas as pd
# Load the CSV data into DataFrames
super_bowls = pd.read_csv('/datasets/super_bowls.csv')
It says that there is no such file or folder. Then I add this line
os.getcwd()
And the output is the top-level folder of the project, and not the subfolder when is this python file. And I think maybe that's the reason why it's not working.
So, how can I open that csv file with relative paths? I don't want to use absolute path because this code is going to be used in another computers.
Why os.getcwd() is not getting the actual folder path?

My observation, the dot (.) notation to move to the parent directory sometimes does not work depending on the operating system. What I generally do to make it os agnostic is this:
import pandas as pd
import os
__location__ = os.path.realpath(os.path.join(os.getcwd(), os.path.dirname(__file__)))
super_bowls = pd.read_csv(__location__ + '/datasets/super_bowls.csv')
This works on my windows and ubantu machine equally well.
I am not sure if there are other and better ways to achieve this. Would like to hear back if there are.

(edited)
Per your comment below, the current working directory is
/Users/ivanparra/AprendizajePython/
while the file is in
/Users/ivanparra/AprendizajePython/Jupyter/datasets/super_bowls.csv
For that reason, going to the datasets subfolder of the current working directory (CWD) takes you to /Users/ivanparra/AprendizajePython/datasets which either doesn't exist or doesn't contain the file you're looking for.
You can do one of two things:
(1) Use an absolute path to the file, as in
super_bowls = pd.read_csv("/Users/ivanparra/AprendizajePython/Jupyter/datasets/super_bowls.csv")
(2) use the right relative path, as in
super_bowls = pd.read_csv("./Jupyter/datasets/super_bowls.csv")
There's also (3) - use os.path.join to contact the CWD to the relative path - it's basically the same as (2).
(you can also use

The answer really lies in the response by user2357112:
os.getcwd() is working fine. The problem is in your expectations. The current working directory is the directory where Python is running, not the directory of any particular source file. – user2357112 supports Monica May 22 at 6:03
The solution is:
data_dir = os.path.dirname(__file__)

Try this code
super_bowls = pd.read_csv( os.getcwd() + '/datasets/super_bowls.csv')

I noticed this problem a few years ago. I think it's a matter of design style. The problem is that: your workspace folder is just a folder, not a project folder. Most of the time, your relative reference is based on the current file.
VSCode actually supports the dynamic setting of cwd, but that's not the default. If your work folder is not a rigorous and professional project, I recommend you adding the following settings to launch.json. This is the simplest answer you need.
"cwd": "${fileDirname}"

Thanks to everyone that tried to help me. Thanks to the Roy2012 response, I got a code that works for me.
import pandas as pd
import os
currentPath = os.path.dirname(__file__)
# Load the CSV data into DataFrames
super_bowls = pd.read_csv(currentPath + '/datasets/super_bowls.csv')
The os.path.dirname gives me the path of the current file, and let me work with relative paths.
'/Users/ivanparra/AprendizajePython/Jupyter'
and with that it works like a charm!!
P.S.: As a side note, the behavior of os.getcwd() is quite different in a Jupyter Notebook than a python file. Inside the notebook, that function gives the current file path, but in a python file, gives the top folder path.

Accessing networked file locations with pathlib

I'm trying to test a program using Python's pathlib module. With the os module, you used to be able to access networked drives by just following the same url-like form. But for some reason, you can't do this with pathlib. Or at least I can't figure out how to do it.
With the os module, all one would have to do is:
path = os.path.join(r'//server-01', 'directory', 'filename.txt')
But if you try to do this with the pathlib module, one could try something like:
path = Path('//server-01', 'directory', 'filename.txt')
If I'm on a Windows machine, path will resolve to:
>> WindowsPath('/server-01/directory/filename.txt)
And if I were to say path.exists() I will of course get False. Because yes, /server-01 does NOT exist, however //server-01 does exist.
Ideally of course, the result I expect to get when I run path.exists() is True and if I were to display path it would look something like:
>> WindowsPath('//server-01/directory/filename.txt')
Update
It's kind of hacky, but it works I guess, regardless I'd like to know the right way to do it.
In order to get to the network location you can:
os.chdir(join(r'//server-01', 'directory', 'filename.txt'))
path = Path()
path = path.resolve()
The result is something like:
>> WindowsPath('//server-01/directory/filename.txt')
path.exists()
>> True
If anyone knows the better way to do it, let me know.

If you create your path as:
path = Path('//server-01/directory/filename.txt')
instead of comma separating each directory it will work.

The server name by itself is not a valid component of a UNC path. You must also include a share. So path = Path('//server-01/directory', 'file') will work. It should resolve and return True when you run path.exists().
Microsoft docs here: https://learn.microsoft.com/en-us/openspecs/windows_protocols/ms-dfsc/149a3039-98ce-491a-9268-2f5ddef08192

after several attempts, I think you can visit the smb folder/file with pathlib by:
folder = pathlib.Path('//server/')
file = pathlib.Path('//server/') / 'relative/path/to/file'
# or
file = pathlib.Path('//server/relative/path/to/file')
the key is that if you want to visit a smb folder, the arg should end with '/'.

Instantiating path as a PureWindowsPath should do the trick:
path = PureWindowsPath("//server-01", "directory", "file") # '\\\\server-01\\directory\\file'

How do go into a directory without being given its full path in python 2?

I am currently trying to go into a folder and call a python 2 script, but I cannot get any answer to go into a folder without using its full path. As example in DOS I would normally type this:
C:\unknownpath\> cd otherpath
C:\unknownpath\otherpath\>
Thanks for any help.

Try this:
import os
os.chdir('otherpath')
This at least matches your DOS example, and will change your working directory to otherpath relative to the directory the command is run from. For example if you are in /home/myusername/, then this will take you to /home/myusername/otherpath/. You can also use . for the current directory or .. to move back one directory. So if you are in /home/myusername/Desktop/, os.chdir('..') would change the working directory to /home/myusername/ and os.chdir('../Documents/ would change you to /home/myusername/Documents/, etc.
Forgive my use of Unix-style paths, but you should be able to easily translate these commands to Windows paths if that is the platform you are on. I don't want to attempt to use Windows paths in my examples because I won't be able to test their efficacy.

os.chdir works on relative path.
>>> os.getcwd()
'C:\\Users\\sba001\\PycharmProjects'
>>> os.listdir('.')
['untitled', 'untitled1', 'untitled2', 'untitled3', 'untitled4', 'untitled5']
>>> os.chdir('untitled')
>>> os.getcwd()
'C:\\Users\\sba001\\PycharmProjects\\untitled'

Running python script with new directories

I have recently begun working on a new computer. All my python files and my data are in the dropbox folder, so having access to the data is not a problem. However, the "user" name on the file has changed. Thus, none of my os.chdir() operations work. Obviously, I can modify all of my scripts using a find and replace, but that won't help if I try using my old computer.
Currently, all the directories called look something like this:
"C:\Users\Old_Username\Dropbox\Path"
and the files I want to access on the new computer look like:
"C:\Users\New_Username\Dropbox\Path"
Is there some sort of try/except I can build into my script so it goes through the various path-name options if the first attempt doesn't work?
Thanks!

Any solution will involve editing your code; so if you are going to edit it anyway - its best to make it generic enough so it works on all platforms.
In the answer to How can I get the Dropbox folder location programmatically in Python? there is a code snippet that you can use if this problem is limited to dropbox.
For a more generic solution, you can use environment variables to figure out the home directory of a user.
On Windows the home directory is location is stored in %UserProfile%, on Linux and OSX it is in $HOME. Luckily Python will take care of all this for you with os.path.expanduser:
import os
home_dir = os.path.expanduser('~')
Using home_dir will ensure that the same path is resolved on all systems.

Thought the file sq.py with these codes(your olds):
C:/Users/Old_Username/Dropbox/Path
for x in range:
#something
def Something():
#something...
C:/Users/Old_Username/Dropbox/Path
Then a new .py file run these codes:
with open("sq.py","r") as f:
for x in f.readlines():
y=x
if re.findall("C:/Users/Old_Username/Dropbox/Path",x) == ['C:/Users/Old_Username/Dropbox/Path']:
x="C:/Users/New_Username/Dropbox/Path"
y=y.replace(y,x)
print (y)
Output is:
C:/Users/New_Username/Dropbox/Path
for x in range:
#something
def Something():
#something...
C:/Users/New_Username/Dropbox/Path
Hope its your solution at least can give you some idea dealing with your problem.

Knowing that eventually I will move or rename my projects or scripts, I always use this code right at the beginning:
import os, inspect
this_dir = os.path.dirname(os.path.abspath(inspect.getfile(inspect.currentframe())))
this_script = inspect.stack()[0][1]
this_script_name = this_script.split('/')[-1]
If you call your script not with the full but a relative path, then this_script will also not contain a full path. this_dir however will always be the full path to the directory.

os.path.join not properly formatting path

I'm writing a command-line directory navigator for Windows in Python and struggling a bit with os.path.join. Here's, in essence, what I'm trying to do:
abspath = "C:\Python32\Projects\ls.py"
abspath = abspath.split('\\')
print(abspath) #this prints ['C:', 'Python32', 'Projects', 'ls.py']
if(options.mFlag):
print(os.path.join(*abspath)) #this prints C:Python32\Projects\ls.py
m = time.ctime(os.path.getmtime(os.path.join(*abspath))) #this throws an exception
The problem is that os.path.join is not inserting a '/' after 'C:' and I can't figure out why. Any help?
Edit: In case anyone in the future comes here looking for a solution, I just added os.sep after "C:" instead of hardcoding a backslash and that worked.

From the documentation:
Note that on Windows, since there is a current directory for each drive, os.path.join("c:", "foo") represents a path relative to the current directory on drive C: (c:foo), not c:\foo.
It's a little hard to tell what you're trying to accomplish, since all your code seems to be aiming at is to split the path and then put it back together exactly the way it was, in which case why split it in the first place? But maybe os.path.splitdrive will help you? It splits the drive letter from the path.

The docs ( http://docs.python.org/2/library/os.path.html) specify this behaviour:
Note that on Windows, since there is a current directory for each drive, os.path.join("c:", "foo") represents a path relative to the current directory on drive C: (c:foo), not c:\foo.