Question

I have read the docs but it doesn’t tell how we enter the file path in open() function.

If file path is:

/opt/myapp/report/sample.txt 

or

$MYPATH/report/sample.txt (where $MYPATH=/opt/myapp)

Is it ok to write the statement this way:

f = open('/opt/myapp/report/sample.txt', "r")

or

f = open('$MYPATH/report/sample.txt', "r")

Answer 1

What you want to do here is expand the environment variables in the path, which can be done with os.path.expandvars():

Return the argument with environment variables expanded. Substrings of the form $name or ${name} are replaced by the value of
environment variable name. Malformed variable names and references to
non-existing variables are left unchanged.

On Windows, %name% expansions are supported in addition to $name and ${name}.

E.g:

with open(os.path.expandvars(path), "r") as f:
    ...

Note my use of the with statement here, which is the best way to open files, as it ensures they are closed correctly, even when there is an exception, and reads nicely.