CODEWITHSUNDEEP
python
moorecats
moorecats

Reputation: 3492

Using integer argument in python for a for loop

Newbie to python here, I am trying to use an integer as an argument for a for loop. The code I use 

from sys import argv

script, numberofposts = argv
postsint = int(numberofposts)

for x in range(0,"%d" % postsint):
    print 'on time'

print "number %d" % postsint

Gives me this error - 

Traceback (most recent call last):
  File "forarg.py", line 6, in <module>
    for x in range(0,"%d" % postsint):
TypeError: range() integer end argument expected, got str.

What am I doing wrong here? I assumed that it was a syntax issue but the error seems to indicate that the for loop is expecting an integer which I attempted to force as you can see.

Upvotes: 2

Views: 8591

Answers (3)

Ashwini Chaudhary
Ashwini Chaudhary

Reputation: 251001

it's because "%d" is used for string formatting and it returns string only:

In [18]: type( "%d" % 3)
Out[18]: <type 'str'>

use just ,range(0,4) or xrange(0,4) if you're on python 2.x:

In [19]: range(0,4)  #0 is the default value of the first argument, so range(0,4)==range(4)
Out[19]: [0, 1, 2, 3]

Upvotes: 3

sean
sean

Reputation: 3985

You are doing a formatted string as the second param to range.

In the case that you are using you are saying start at 1 and go up to a string with the number postint in it.

This is wrong. You want to do range( 1, postint )

Upvotes: 1

dm03514
dm03514

Reputation: 55962

for x in range(postsint): # start at beginning as mgilson suggested

should work fine. You already casted it an integer, You shouldn't be creating a string from it

Upvotes: 3

Related Questions