Python: Open () using a variable

Question

I am having a problem opening a file with a randomly generated name in Python 2.6.

import random

random = random.randint(1,10)

localfile = file("%s","wb") % random

Then I get an error on the last line:

TypeError: unsupported operand type(s) for %: 'file' and 'int'

I just can't figure it out myself, nor with Google, but in my opinion there must be a cure for this.

+2

user335417 May 07 '10 at 12:09

2 answers

Tamás · Answer 1 · 2010-05-07T12:11:06+0000

This will probably work:

import random

num = random.randint(1, 10)
localfile = open("%d" % num, "wb")

Please note that I changed a couple of things here:

You should not assign the generated random number to a named variable random

as you are overwriting an existing module reference random

. In other words, you will no longer be able to access random.randint

if you overwrite with a random

randomly generated number.
The formatting operator ( %

) must be applied to the string you are formatting, not to a method call file

.
file

Deprecated in Python 3 I think . It's time to use open

instead file

.
Since you are formatting an integer to a string, you should write "%d"

instead "%s"

(although the latter will work too).

An alternative recording method "%d" % num

is str(num)

, which may be slightly more efficient.

guillermooo · Answer 2 · 2010-05-07T12:11:09+0000

Try:

localfile = file("%s" % random,"wb")