Python "List" object is not callable

Question

I'm writing a program that looks through CSVs in a directory and appends the contents of each CSV to a list. Here's a snippet of the offending code:

import glob
import re 

c = glob.glob("*.csv")
print c
archive = []

for element in c:
    look = open(element, "r").read()
    open = re.split("\n+", look)

    for n in open:
        n = re.split(",", n)[0]
        archive.append(n)

However, when I run this script, I get a TypeError: 'list' object is not callable. Can somebody please explain what's going on?

Bastien Léonard · Accepted Answer · 2010-07-02 01:47:02Z

10

I think it's because you redefine open as a list and call it in the next loop iteration. Just give the list another name.

Note that strings have a split() method for when you don't need a regex.

edited Jul 2, 2010 at 1:47

answered Jul 2, 2010 at 0:45

Bastien Léonard

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John Machin · Accepted Answer · 2010-07-02 01:12:10Z

5

The fact that open is a builtin function is irrelevant. It could have been a function defined in the same module.

The basic problem is that the same name was used to refer to two different objects (a function and a list), both of which were needed again. When the first object was needed again, the name referred to the second object. Result: splat. The golden rule is: don't re-use names unthinkingly.

answered Jul 2, 2010 at 1:12

John Machin

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Jack · Accepted Answer · 2010-07-02 00:47:06Z

2

The gold rule is: never use the name of a builtin thing for your variables!

It's a matter of aliasing, just don't call the list open..

answered Jul 2, 2010 at 0:47

Jack

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tcarobruce · Accepted Answer · 2010-07-02 01:03:53Z

2

Agree with previous answers: never call a variable open or any other builtin.

You may be interested int the Python csv module, which will correctly parse csv files that re.split(',', line) won't.

Also, you can use the file object as a line-by-line iterator like so:

for line in open('data.csv'):
    do_something(line)

answered Jul 2, 2010 at 1:03

tcarobruce

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