Question

I am given a large integer a, and a (relatively small) integer n.

What is the fastest way to get the nth bit (from the right) of the binary representation of a using native Python?

Answer 1

You asked for the fastest way, presumably using a modern version of Python. Modern versions of Python have variable-length ints, and conventional wisdom does’t apply. Shifting a large number is not cheap. Shifting 1 is cheap. Here are some -mtimeit inputs and corresponding outputs. The first is an abbreviation of

windows command prompt>\python27\python -mtimeit -s"a=10**20;n=3" "(a>>n)&1"
1000000 loops, best of 3: 0.238 usec per loop

-s"a=10**20;n=3" "(a>>n)&1"
0.238 usec 

-s"a=10**20;n=3" "not not(a & (1 << n))"
0.154 usec 

-s"a=10**200;n=3" "(a>>n)&1"
0.382 usec 

-s"a=10**200;n=3" "not not(a & (1 << n))"
0.155 usec 

-s"a=10**10;n=3" "(a>>n)&1"
0.231 usec 

-s"a=10**10;n=3" "not not(a & (1 << n))"
0.156 usec 

-s"a=10**9;n=3" "(a>>n)&1"
0.0801 usec

-s"a=10**9;n=3" "not not(a & (1 << n))"
0.0938 usec

-s"a=2**1000;n=64" "(a>>n)&1"
0.446 usec

-s"a=2**1000;n=64" "not not(a & (1 << n))"
0.255 usec

If not not(foo) freaks you out, or you really want an int answer instead of a bool, you can use 1 if foo else 0; it’s only slightly slower.