bitwise_or(x1, x2, /, out=None, *, where=True, casting='same_kind', order='K', dtype=None, subok=True[, signature, extobj]) = <ufunc 'bitwise_or'>¶
Compute the bit-wise OR of two arrays element-wise.
Computes the bit-wise OR of the underlying binary representation of the integers in the input arrays. This ufunc implements the C/Python operator
- x1, x2array_like
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).
- outndarray, None, or tuple of ndarray and None, optional
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.
- wherearray_like, optional
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.
- outndarray or scalar
Result. This is a scalar if both x1 and x2 are scalars.
The number 13 has the binaray representation
00001101. Likewise, 16 is represented by
00010000. The bit-wise OR of 13 and 16 is then
000111011, or 29:
>>> np.bitwise_or(13, 16) 29 >>> np.binary_repr(29) '11101'
>>> np.bitwise_or(32, 2) 34 >>> np.bitwise_or([33, 4], 1) array([33, 5]) >>> np.bitwise_or([33, 4], [1, 2]) array([33, 6])
>>> np.bitwise_or(np.array([2, 5, 255]), np.array([4, 4, 4])) array([ 6, 5, 255]) >>> np.array([2, 5, 255]) | np.array([4, 4, 4]) array([ 6, 5, 255]) >>> np.bitwise_or(np.array([2, 5, 255, 2147483647], dtype=np.int32), ... np.array([4, 4, 4, 2147483647], dtype=np.int32)) array([ 6, 5, 255, 2147483647]) >>> np.bitwise_or([True, True], [False, True]) array([ True, True])