We Keep Coding

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.

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.

Create file in python dynamically

I know it's a noob question but I have some difficulties to make it works
def create(file):
f = open(file,'w')
it returns "IOError: [Errno 2] No such file or directory: "
If I do that it works of course:
file ="myfile"
f = open(file,'w')
But I can't figure out how to create my file from the function parameter
Sorry for the noob question, thanks in advance for your help.

when you pass the "http://somesite.com/" as file to your function python treats it as a directory structure.
As soon as python gets to "http:/" it presumes we have a directory. Using forward slashes in unix is not allowed and I imagine it is the same for windows.
To turn the name into something useable you can use some variation of urlparse.urlsplit:
import urlparse
import urlparse
def parse(f):
prse = urlparse.urlsplit(f)
return prse.netloc if f.startswith("http") else prse.path.split("/",1)[0]

Sites can look like paths to directories to the operating system. for instance: stackoverflow.com/something will be interpreted as a directory stackoverflow.com in which there is a file something.
You can see this when you use os.path.dirname:
>>> os.path.dirname('stackoverflow.com/something')
'stackoverflow.com'
If this is indeed the case, and you still want to proceed, you're passing a path to a location in a directory and not just a file name.
You have to make sure the directory stackoverflow.com exists first:
file_path = 'stackoverflow.com/something'
dirname = os.path.dirname(file_path)
if not os.path.exists(dirname):
# if stackoverflow.com directory does not exist it will be created
os.makedirs(dirname)
# .. carry on to open file_path and use it.
Watch out from http:// and the likes and consider using a real url parser.
tip: file is already defined in python, you shouldn't override it by using it to name a variable.

Editing:
def create(file):
f = open(file,'w')
f.close()
If you call this function using:
create('myfile.txt')
It will create a file named myfile.txt in whatever directory the code is being run from. Note that you are passing in a string not an object.
Since I now see you are passing in a string similar to http://www.google.com, you are trying to create a file named www.google.com in the http: folder. You are going to have to truncate or change the / since Windows files cannot contain that character in their names.
We'll use everything after the last / in this example:
def create(filename):
filename = re.sub(r'.*//*', '', filename)
f = open(filename, 'w')
f.close()
So calling: create('www.google.com/morestuff/things') will create a file called things

Absolute Path and Relative Path issue in python

How to remove path related problems in python?
For e.g. I have a module test.py inside a directory TEST
**test.py**
import os
file_path = os.getcwd() + '/../abc.txt'
f = open(file_path)
lines = f.readlines()
f.close
print lines
Now, when I execute the above program outside TEST directory, it gives me error:-
Traceback (most recent call last):
File "TEST/test.py", line 4, in ?
f = open(file_path)
IOError: [Errno 2] No such file or directory: 'abc.txt'
how to resolve this kind of problem. Basically this is just a small example that I have given up.
I am dealing with a huge problem of this kind.
I am using existing packages, which needs to be run only from that directory where it exists, how to resolve such kind of problems, so that I can run the program from anywhere I want.
Or able to deal with the above example either running inside TEST directory or outside TEST directory.
Any help.?

I think the easiest thing is to change the current working directory to the one of the script file:
import os
os.chdir(os.path.dirname(__file__))
This may cause problems, however, if the script is also working with files in the original working directory.

Your code is looking at the current working directory, and uses that as a basis for finding the files it needs. This is almost never a good idea, as you are now finding out.
The solution mentioned in the answer by Emil Vikström is a quickfix solution, but a more correct solution would be to not use current working directory as a startingpoint.
As mentioned in the other answer, __file__ isn't available in the interpreter, but it's an excellent solution for your code.
Rewrite your second line to something like this:
file_path = os.path.join(os.path.dirname(__file__), "..", "abc.txt")
This will take the directory the current file is in, join it first with .. and then with abc.txt, to create the path you want.
You should fix similar usage of os.getcwd() elsewhere in your code in the same way.

How do I create a file at a specific path?

In python I´m creating a file doing:
f = open("test.py", "a")
where is the file created? How can I create a file on a specific path?
f = open("C:\Test.py", "a")
returns error.

The file path "c:\Test\blah" will have a tab character for the `\T'. You need to use either:
"C:\\Test"
or
r"C:\Test"

I recommend using the os module to avoid trouble in cross-platform. (windows,linux,mac)
Cause if the directory doesn't exists, it will return an exception.
import os
filepath = os.path.join('c:/your/full/path', 'filename')
if not os.path.exists('c:/your/full/path'):
os.makedirs('c:/your/full/path')
f = open(filepath, "a")
If this will be a function for a system or something, you can improve it by adding try/except for error control.

where is the file created?
In the application's current working directory. You can use os.getcwd to check it, and os.chdir to change it.
Opening file in the root directory probably fails due to lack of privileges.

It will be created once you close the file (with or without writing). Use os.path.join() to create your path eg
filepath = os.path.join("c:\\","test.py")

The file is created wherever the root of the python interpreter was started.
Eg, you start python in /home/user/program, then the file "test.py" would be located at /home/user/program/test.py

f = open("test.py", "a") Will be created in whatever directory the python file is run from.
I'm not sure about the other error...I don't work in windows.

The besty practice is to use '/' and a so called 'raw string' to define file path in Python.
path = r"C:/Test.py"
However, a normal program may not have the permission to write in the C: drive root directory. You may need to allow your program to do so, or choose something more reasonable since you probably not need to do so.

Use os module
filename = "random.txt"
x = os.path.join("path", "filename")
with open(x, "w") as file:
file.write("Your Text")
file.close