numpy.maximum
-
numpy.maximum(x1, x2, /, out=None, *, where=True, casting='same_kind', order='K', dtype=None, subok=True[, signature, extobj]) = <ufunc 'maximum'>
-
Element-wise maximum of array elements.
Compare two arrays and returns a new array containing the element-wise 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, x2array_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). -
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, theout
array will retain its original value. Note that if an uninitializedout
array is created via the defaultout=None
, locations within it where the condition is False will remain uninitialized. - **kwargs
-
For other keyword-only arguments, see the ufunc docs.
-
- Returns
-
-
yndarray or scalar
-
The maximum of
x1
andx2
, element-wise. This is a scalar if bothx1
andx2
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
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https://numpy.org/doc/1.19/reference/generated/numpy.maximum.html