python: can I extend the upper bound of the range() method?

Question

What is the upper bound of the range() function and how can I extend it, or alternately what's the best way to do this:

for i in range(1,600851475143):

Answer 1

range(1, 600851475143) wants to generate a very large list in memory, and you'll get an out of memory error. To save memory, use xrange instead of range. Unfortunately, xrange doesn't work with large numbers (it's an implementation restriction) Example (raises OverflowError):

for i in xrange(1, 600851475143):
  print i

You can have large minimum or maximum values in your interval with range, if their difference is small. Example:

x = 1 << 200
print list(xrange(x, x + 3))

Output:

[1606938044258990275541962092341162602522202993782792835301376L, 1606938044258990275541962092341162602522202993782792835301377L, 1606938044258990275541962092341162602522202993782792835301378L]

A fancy solution to your original for loop problem:

def bigrange(a, b = None):
  if b is None:
    b = a
    a = 0
  while a < b:
    yield a
    a += 1

for i in bigrange(1, 600851475143):
  print i

A less fancy solution, which works even if you have continue in the loop body:

i = 1 - 1
while i < 600851475143 - 1:
  i += 1
  print i

Answer 2

pts' answer led me to this in the xrange python docs:

Note

xrange() is intended to be simple and fast. Implementations may impose restrictions to achieve this. The C implementation of Python restricts all arguments to native C longs (“short” Python integers), and also requires that the number of elements fit in a native C long. If a larger range is needed, an alternate version can be crafted using the itertools module: islice(count(start, step), (stop-start+step-1)//step)

looks like it's a limitation of c python in particular.

Answer 3

It depends on which version of Python you're using. I'm using 2.5.2 and xrange raises an OverflowError exception for large numbers. One solution is to write your own generator.

def g(start, stop):
    i = start
    while i < stop:
        yield i
        i += 1

x = 1<<200
for i in g(x, x+3):
    print i

Answer 4

Have you considered just doing this? Or is there some reason you specifically need range()?

x = 1
while x < 600851475143:
    // some code
    x += 1

Answer 5

Here's an answer using itertools. It's a little contrived, but it works:

from itertools import repeat, count, izip
for i,_ in izip(count(1), repeat(1, 600851475143)):
    ...

Another answer would be to write your own generator:

def my_xrange(a, b):
    while a < b:
        yield a
        a += 1

for i in my_xrange(1, 600851475143):
    ...