python check if bit in sequence is true or false

Without the bit shifting:

if bits & 0b1000:
    ...

EDIT: Actually, (1 << 3) is optimized out by the compiler.

>>> dis.dis(lambda x: x & (1 << 3))
  1           0 LOAD_FAST                0 (x)
              3 LOAD_CONST               3 (8)
              6 BINARY_AND          
              7 RETURN_VALUE        
>>> dis.dis(lambda x: x & 0b1000)
  1           0 LOAD_FAST                0 (x)
              3 LOAD_CONST               1 (8)
              6 BINARY_AND          
              7 RETURN_VALUE    

The two solutions are equivalent, choose the one that looks more readable in your context.

You can use bit shifting

>>> 0b10010101 >> 4 & 1
1
>>> 0b10010101 >> 3 & 1
0

bits = 0b11010011

if bits & (1 << 3):
    ...

bits = '1110111'

if bits[-4] == '0':
    print "......false"
else:
    print  "....true"