python for loop range(bigint)

Question

In Python, is there some short way to do something like

"for i in range(n)"

when n is too big for Python to actually create the array range(n)?

(short because otherwise I'd just use a while loop)

Answer 1

You could use xrange()... although that is restricted to "short" integers in CPython:

CPython implementation detail: 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: takewhile(lambda x: x<stop, (start+i*step for i in count())).

I don't know whether that restriction also applies to other implementations (or which ones) - but there's a workaround listed...

I know you mention bigint in your question title, but the question body talks about the number being too big to create the array - I suspect there are plenty of numbers which are small enough for xrange to work, but big enough to cause you memory headaches with range.