numpy.fmax

numpy.fmax(x1, x2, /, out=None, *, where=True, casting='same_kind', order='K', dtype=None, subok=True[, signature, extobj]) = <ufunc 'fmax'>

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 the non-nan 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 ignored when possible.

Parameters:
x1, x2 : array_like

The arrays holding the elements to be compared. They must have the same shape.

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 freshly-allocated array is returned. A tuple (possible only as a keyword argument) must have length equal to the number of outputs.

where : array_like, optional

Values of True indicate to calculate the ufunc at that position, values of False indicate to leave the value in the output alone.

**kwargs

For other keyword-only arguments, see the ufunc docs.

Returns:
y : ndarray or scalar

The maximum of x1 and x2, element-wise. This is a scalar if both x1 and x2 are scalars.

参见

fmin
Element-wise minimum of two arrays, ignores NaNs.
maximum
Element-wise maximum of two arrays, propagates NaNs.
amax
The maximum value of an array along a given axis, propagates NaNs.
nanmax
The maximum value of an array along a given axis, ignores NaNs.

minimum, amin, nanmin

Notes

1.3.0 新版功能.

The fmax 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.fmax([2, 3, 4], [1, 5, 2])
array([ 2.,  5.,  4.])
>>> np.fmax(np.eye(2), [0.5, 2])
array([[ 1. ,  2. ],
       [ 0.5,  2. ]])
>>> np.fmax([np.nan, 0, np.nan],[0, np.nan, np.nan])
array([  0.,   0.,  NaN])