numpy.maximum¶

numpy.
maximum
(x1, x2, /, out=None, *, where=True, casting='same_kind', order='K', dtype=None, subok=True[, signature, extobj]) = <ufunc 'maximum'>¶ Elementwise maximum of array elements.
Compare two arrays and returns a new array containing the elementwise maxima. If one of the elements being compared is a NaN, then that element is returned. If both elements are NaNs then the first is returned. The latter distinction is important for complex NaNs, which are defined as at least one of the real or imaginary parts being a NaN. The net effect is that NaNs are propagated.
Parameters:  x1, x2 : array_like
The arrays holding the elements to be compared. If
x1.shape != x2.shape
, they must be broadcastable to a common shape (which becomes the shape of the output). out : ndarray, 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 freshlyallocated array is returned. A tuple (possible only as a keyword argument) must have length equal to the number of outputs.
 where : array_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. **kwargs
For other keywordonly arguments, see the ufunc docs.
Returns:  y : ndarray or scalar
The maximum of x1 and x2, elementwise. This is a scalar if both x1 and x2 are scalars.
See also
Notes
The maximum is equivalent to
np.where(x1 >= x2, x1, x2)
when neither x1 nor x2 are nans, but it is faster and does proper broadcasting.Examples
>>> np.maximum([2, 3, 4], [1, 5, 2]) array([2, 5, 4])
>>> np.maximum(np.eye(2), [0.5, 2]) # broadcasting array([[ 1. , 2. ], [ 0.5, 2. ]])
>>> np.maximum([np.nan, 0, np.nan], [0, np.nan, np.nan]) array([nan, nan, nan]) >>> np.maximum(np.Inf, 1) inf