How to open a file (inside function) provided by function parameter

Question

I am learning Python 3 at the moment, and I have a problem with opening a file provided as a function parameter.

This is my code:

def make_list_from_file(file_name):

    with open(file_name,"r") as provided_file:
        temp_list = [line.strip() for line in provided_file]
    detailed_list = [ item.split("\t") for item in temp_list ]
    return detailed_list

make_list_from_file(game_stat.txt)

This gives me:

NameError: name 'Game_Stat' is not defined.

The file is in the same directory. I would be grateful for any help.

pass a string to the function, like this make_list_from_file("game_stat.txt") — Omar Einea
– Omar Einea, Commented Jan 6, 2018 at 18:13

Roel Gerrits · Accepted Answer · 2018-01-06 18:15:13Z

2

make_list_from_file(game_stat.txt)

strings should be enclosed with single (') or double (") quotes.

So:

make_list_from_file("game_stat.txt")

answered Jan 6, 2018 at 18:15

Roel Gerrits

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Comments

user9158931 · Accepted Answer · 2018-01-06 18:52:41Z

0

Change this line:

with open(file_name,"r") as provided_file:

to :

with open(file_name + '.txt',"r") as provided_file:

answered Jan 6, 2018 at 18:52

user9158931

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