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Compute the bit-wise AND of two arrays element-wise.
Computes the bit-wise AND of the underlying binary representation of
the integers in the input arrays. This ufunc implements the C/Python
operator &.
Only integer and boolean types are handled.
If x1.shape != x2.shape, they must be broadcastable to a common
shape (which becomes the shape of the output).
A location into which the result is stored. If provided, it must have a shape that the inputs broadcast to. If not provided or None, a freshly-allocated array is returned. A tuple (possible only as a keyword argument) must have length equal to the number of outputs.
This condition is broadcast over the input. At locations where the
condition is True, the out array will be set to the ufunc result.
Elsewhere, the out array will retain its original value.
Note that if an uninitialized out array is created via the default
out=None, locations within it where the condition is False will
remain uninitialized.
For other keyword-only arguments, see the ufunc docs.
Result. This is a scalar if both x1 and x2 are scalars.
See also
logical_andbitwise_orbitwise_xorbinary_reprReturn the binary representation of the input number as a string.
Examples
>>> import numpy as np
The number 13 is represented by 00001101. Likewise, 17 is
represented by 00010001. The bit-wise AND of 13 and 17 is
therefore 000000001, or 1:
>>> np.bitwise_and(13, 17)
1
>>> np.bitwise_and(14, 13)
12
>>> np.binary_repr(12)
'1100'
>>> 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])
The & operator can be used as a shorthand for np.bitwise_and on
ndarrays.
>>> x1 = np.array([2, 5, 255])
>>> x2 = np.array([3, 14, 16])
>>> x1 & x2
array([ 2, 4, 16])
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