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Streamlip app not searching files in the good directory

I am trying to run a Streamlit app importing pickle files and a DataFrame. The pathfile for my script is :
/Users/myname/Documents/Master2/Python/Final_Project/streamlit_app.py
And the one for my DataFrame is:
/Users/myname/Documents/Master2/Python/Final_Project/data/metabolic_syndrome.csv
One could reasonably argue that I only need to specify df = pd.read_csv('data/df.csv') yet it does not work as the Streamlit app is unexpectedly not searching in its directory:
FileNotFoundError: [Errno 2] No such file or directory: '/Users/myname/data/metabolic_syndrome.csv'
How can I manage to make the app look for the files in the good directory (the one where it is saved) without having to use absolute pathfiles ?

you can use os.getcwd() to get the Current Working Directory. Read up on what this means exactly, fe here
See the sample script below on how to use it. I'm using os.sep for OS agnostic filepath separators.
import os
print(os.getcwd())
relative_path = "data"
full_path = f"{os.getcwd()}{os.sep}{relative_path}"
print(full_path)
filename = "somefile.csv"
full_file_path = f"{full_path}{os.sep}{filename}"
print(full_file_path)
with open(full_file_path) as infile:
for line in infile.read().splitlines():
print(line)

In which directory are you standing when you are running your code?
From your error message I would assume that you are standing in /Users/myname/ which makes python look for data as a subdirectory of /Users/myname/.
But if you first change directory to /Users/myname/Documents/Master2/Python/Final_Project and then run your code from there I think it would work.

Regarding file io in Python

I mistakenly, typed the following code:
f = open('\TestFiles\'sample.txt', 'w')
f.write('I just wrote this line')
f.close()
I ran this code and even though I have mistakenly typed the above code, it is however a valid code because the second backslash ignores the single-quote and what I should get, according to my knowledge is a .txt file named "\TestFiles'sample" in my project folder. However when I navigated to the project folder, I could not find a file there.
However, if I do the same thing with a different filename for example. Like,
f = open('sample1.txt', 'w')
f.write('test')
f.close()
I find the 'sample.txt' file created in my folder. Is there a reason for the file to not being created even though the first code was valid according to my knowledge?
Also is there a way to mention a file relative to my project folder rather than mentioning the absolute path to a file? (For example I want to create a file called 'sample.txt' in a folder called 'TestFiles' inside my project folder. So without mentioning the absolute path to TestFiles folder, is there a way to mention the path to TestFiles folder relative to the project folder in Python when opening files?)
I am a beginner in Python and I hope someone could help me.
Thank you.

What you're looking for are relative paths, long story short, if you want to create a file called 'sample.txt' in a folder 'TestFiles' inside your project folder, you can do:
import os
f = open(os.path.join('TestFiles', 'sample1.txt'), 'w')
f.write('test')
f.close()
Or using the more recent pathlib module:
from pathlib import Path
f = open(Path('TestFiles', 'sample1.txt'), 'w')
f.write('test')
f.close()
But you need to keep in mind that it depends on where you started your Python interpreter (which is probably why you're not able to find "\TestFiles'sample" in your project folder, it's created elsewhere), to make sure everything works fine, you can do something like this instead:
from pathlib import Path
sample_path = Path(Path(__file__).parent, 'TestFiles', 'sample1.txt')
with open(sample_path, "w") as f:
f.write('test')
By using a [context manager]{https://book.pythontips.com/en/latest/context_managers.html} you can avoid using f.close()

When you create a file you can specify either an absolute filename or a relative filename.
If you start the file path with '\' (on Win) or '/' it will be an absolute path. So in your first case you specified an absolute path, which is in fact:
from pathlib import Path
Path('\Testfile\'sample.txt').absolute()
WindowsPath("C:/Testfile'sample.txt")
Whenever you run some code in python, the relative paths that will be generate will be composed by your current folder, which is the folder from which you started the python interpreter, which you can check with:
import os
os.getcwd()
and the relative path that you added afterwards, so if you specify:
Path('Testfiles\sample.txt').absolute()
WindowsPath('C:/Users/user/Testfiles/sample.txt')
In general I suggest you use pathlib to handle paths. That makes it safer and cross platform. For example let's say that your scrip is under:
project
src
script.py
testfiles
and you want to store/read a file in project/testfiles. What you can do is get the path for script.py with __file__ and build the path to project/testfiles
from pathlib import Path
src_path = Path(__file__)
testfiles_path = src_path.parent / 'testfiles'
sample_fname = testfiles_path / 'sample.txt'
with sample_fname.open('w') as f:
f.write('yo')

As I am running the first code example in vscode, I'm getting a warning
Anomalous backslash in string: '\T'. String constant might be missing an r prefix.
And when I am running the file, it is also creating a file with the name \TestFiles'sample.txt. And it is being created in the same directory where the .py file is.
now, if your working tree is like this:
project_folder
-testfiles
-sample.txt
-something.py
then you can just say: open("testfiles//hello.txt")
I hope you find it helpful.

Making a new file in subdirectories using python

I am trying to create a file in a subdirectory, both of which will not exist when the program is first run. When I do this:
newfile = open('abc.txt','w')
It will create abc.txt just fine but the following will cause an error, saying the file or directory does not exist
newfile = open('folder/abc.txt','w')
I tried using os.makedirs to create the directory first but that failed as well raising the same error. What is the best way to create both the folder and file?
Thanks

>>> import os
>>> os.makedirs('folder')
>>> newfile = open('folder' + os.sep + 'abc.txt', 'w')
>>> newfile.close()
>>> os.listdir('folder')
['abc.txt']
This works for me

A couple of things to check:
os.makedirs takes just the path, not the file. I assume you know this already, but you haven't shown your call to makedirs so I thought I'd mention it.
Consider passing an absolute, not a relative, path to makedirs. Alternatively, use os.chdir first, to change to a directory in which you know you have write permission.
Hope those help!

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.

move a file to the current directory using shutil module

I know this may sound really stupid, but how can I move a file in a directory a user browsed to ( I named mine filedir) to the current directory I am in?
for example: I have a file called "pages.html" in "C:\webs". How can I move that file to the current working directory "."?
This is my code:
shutil.move(filedir, "*.*")
#I got errors using this code..
Is there another way to say current directory, other than "." ?

The second argument of shutil.move specifies a directory, not a glob mask:
import os.path
shutil.move(os.path.join(filedir, "pages.html"), os.getcwd())
should work.

It would be very helpful if you posted the error message and the complete traceback. However, to get the current working directory you can use os.getcwd(). As long as filedir points to a file and not the directory this should work.
filedir = r"C:\webs\pages.html"
shutil.move(filedir, os.getcwd())