Just one "with open file as f", based on a conditional

Question

I am building a function that needs to open two type of files (plain text files and .gz files)

I want to use a conditional so I am able to use only one "with open" statement, instead of having duplicated code.

This is what I do want to achieve (but obviously it does not compile)

if ".gz" in f:  # gzipped version of the file, we need gzip.open
    with gzip.open(f, "rt") as file:
else:
    with open(f,"rt") as file: # normal open

for line in file: # processing of the lines of the file
    ...

I want to be able to just use one with open, instead of having to create two "with open" statements with two "for line in file"

Reason: I want the less amount of code possible.

How can it be done in a Pythonic way?

Answer 1

I would do the opening like this (assign open() or gzip.open() functions to variable, based on your filename):

import gzip

f = 'myfile.gz'
opener = open

if ".gz" in f:  # gzipped version of the file, we need gzip.open
    opener = gzip.open

with opener(f, "rt") as file:
    pass # your code