One way would be using dot-product
with 2-powered
range array -
b.dot(2**np.arange(b.size)[::-1])
Sample run -
In [95]: b = np.array([1,0,1,0,0,0,0,0,1,0,1])
In [96]: b.dot(2**np.arange(b.size)[::-1])
Out[96]: 1285
Alternatively, we could use bitwise left-shift operator to create the range array and thus get the desired output, like so -
b.dot(1 << np.arange(b.size)[::-1])
If timings are of interest -
In [148]: b = np.random.randint(0,2,(50))
In [149]: %timeit b.dot(2**np.arange(b.size)[::-1])
100000 loops, best of 3: 13.1 μs per loop
In [150]: %timeit b.dot(1 << np.arange(b.size)[::-1])
100000 loops, best of 3: 7.92 μs per loop
Reverse process
To retrieve back the binary array, use np.binary_repr
alongwith np.fromstring
-
In [96]: b = np.array([1,0,1,0,0,0,0,0,1,0,1])
In [97]: num = b.dot(2**np.arange(b.size)[::-1]) # integer
In [98]: np.fromstring(np.binary_repr(num), dtype='S1').astype(int)
Out[98]: array([1, 0, 1, 0, 0, 0, 0, 0, 1, 0, 1])