One way: dot-productwith a range array 2-powered-
b.dot(2**np.arange(b.size)[::-1])
Run Example -
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 the left-shift bitwise operator to create an array of ranges and thus get the desired result, for example:
b.dot(1 << np.arange(b.size)[::-1])
-
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
, np.binary_repr 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])
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])