How do I open a file, passed to open as a variable?

Question

def readfile(file): edges = [] # to contain tuples of all edges with open(file) as f: for line in f:

I'm trying to pass in a text file called file, then read it, but it doesn't work. Even if I cast file to a string it doesn't work. Python says

with open(file) as f: IOError: [Errno 22] invalid mode ('r') or filename: "<type 'file'>"

How your are calling function readfile? seems you are not passing value for file argument. — NPatel
– NPatel, Commented Feb 4, 2015 at 10:00
first of all, rename variable file. Then, it seems that you are passing a file object to your function, and open expects a string name. So there is an inconsistency here. — MariusSiuram
– MariusSiuram, Commented Feb 4, 2015 at 10:04

Jérôme Radix · Accepted Answer · 2015-02-04 10:03:24Z

1

Rename file by filename as file is a built-in name for python :

def readfile(filename): edges = [] # to contain tuples of all edges with open(filename) as f: for line in f:

open() use 'r' as default mode.

answered Feb 4, 2015 at 10:03

Jérôme Radix

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2 Comments

georg Over a year ago

No, file is not a reserved word.

Sam Over a year ago

Thanks Jerome, Lafada, Kasra, georg, Marius (done), stupid mistake, not passing anything in. Doh!

Kasravnd · Accepted Answer · 2015-02-04 10:07:39Z

First of all , as file is a type in python you shouldn't use it as a variable name or file name , second you need to put the file name in quote inside the open function and note that open function use read mod as default ! :

with open('file_name.txt','r') as f: for line in f:

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