numpy.bitwise_and(x1, x2[, out])

Compute bit-wise AND of two arrays, element-wise.

When calculating the bit-wise AND between two elements, x and y, each element is first converted to its binary representation (which works just like the decimal system, only now we’re using 2 instead of 10):

x = \sum_{i=0}^{W-1} a_i \cdot 2^i\\
y = \sum_{i=0}^{W-1} b_i \cdot 2^i,

where W is the bit-width of the type (i.e., 8 for a byte or uint8), and each a_i and b_j is either 0 or 1. For example, 13 is represented as 00001101, which translates to 2^4 + 2^3 + 2.

The bit-wise operator is the result of

z = \sum_{i=0}^{i=W-1} (a_i \wedge b_i) \cdot 2^i,

where \wedge is the AND operator, which yields one whenever both a_i and b_i are 1.


x1, x2 : array_like

Only integer types are handled (including booleans).


out : array_like


See also

bitwise_or, bitwise_xor, logical_and

Return the binary representation of the input number as a string.


We’ve seen that 13 is represented by 00001101. Similary, 17 is represented by 00010001. The bit-wise AND of 13 and 17 is therefore 000000001, or 1:

>>> np.bitwise_and(13, 17)
>>> np.bitwise_and(14, 13)
>>> np.binary_repr(12)
>>> np.bitwise_and([14,3], 13)
array([12,  1])
>>> np.bitwise_and([11,7], [4,25])
array([0, 1])
>>> np.bitwise_and(np.array([2,5,255]), np.array([3,14,16]))
array([ 2,  4, 16])
>>> np.bitwise_and([True, True], [False, True])
array([False,  True], dtype=bool)

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