for loop: range() result has too many items
steven at REMOVE.THIS.cybersource.com.au
Tue Oct 13 23:39:46 CEST 2009
On Tue, 13 Oct 2009 16:17:58 -0500, Peng Yu wrote:
> The following code does not run because range() does not accept a big
>>> range(sys.maxint+2, sys.maxint+5)
[2147483649L, 2147483650L, 2147483651L]
> Is there a way to make the code work. I'm wondering if there is
> a way to write a for-loop in python similar to that of C style.
> for(int i = 0; i < a_big_number; ++ i)
An exact translation of that would be a while loop:
i = 0
while i < a_big_number:
i += 1
but that's not actually your problem. Your actual problem is that you're
trying to generate a list containing 2147483647 (sys.maxint) items. Each
item requires 8 bytes, so this will need a single contiguous block of at
least 16 gigabytes.
On my system, when I try it, I get MemoryError. You get OverflowError.
Try instead using a lazy list:
> def foo():
> for i in range(sys.maxint):
> if i % 100 == 0:
> print i
Rather than iterating through every single integer between 0 and
2147483647, you can skip the ones you don't care about:
for i in xrange(0, sys.maxint, 100):
This will run approximately 100 times faster than your code.
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