Python for loop offset (Itertools.product)

Question

The following code generates all possible combinations using 0 and 1, where four digits have to be used.

import itertools
for i in itertools.product([0, 1], repeat=4):
    print i

Output:

(0, 0, 0, 0)(0, 0, 0, 1)(0, 0, 1, 0)(0, 0, 1, 1)(0, 1, 0, 0)(0, 1, 0, 1)(0, 1, 1, 0)(0, 1, 1, 1)(1, 0, 0, 0)(1, 0, 0, 1)(1, 0, 1, 0)(1, 0, 1, 1)(1,1, 0, 0)(1, 1, 0, 1)(1, 1, 1, 0)(1, 1, 1, 1)

I would like to be able to set an offset for the for loop. Example:

import itertools
offSet = 10
for i in itertools.product([0, 1], repeat=4):
    # Some code that applies the offset
    print i

Which would then output:

(1, 0, 1, 0)(1, 0, 1, 1)(1,1, 0, 0)(1, 1, 0, 1)(1, 1, 1, 0)(1, 1, 1, 1)

How can I apply a such offset to this for loop?

Note: The code being used is simplified. As I am actually using a very large value for repeat, performance matters. I cannot afford the possible combinations before the offset to be calculated.

Answer 1

How about this:

In [29]: offSet = 10

In [30]: repeat = 4

In [31]: for i in xrange(offSet, 2**repeat):
    print tuple(int(x) for x in bin(i)[2:])
   ....:     
(1, 0, 1, 0)
(1, 0, 1, 1)
(1, 1, 0, 0)
(1, 1, 0, 1)
(1, 1, 1, 0)
(1, 1, 1, 1)