fmax

paddle. fmax ( x, y, name=None ) [source]

Compares the elements at the corresponding positions of the two tensors and returns a new tensor containing the maximum value of the element. If one of them is a nan value, the other value is directly returned, if both are nan values, then the first nan value is returned. The equation is:

\[out = fmax(x, y)\]

Note

paddle.fmax supports broadcasting. If you want know more about broadcasting, please refer to Introduction to Tensor .

Parameters
  • x (Tensor) – the input tensor, it’s data type should be float16, float32, float64, int32, int64.

  • y (Tensor) – the input tensor, it’s data type should be float16, float32, float64, int32, int64.

  • name (str, optional) – Name for the operation (optional, default is None). For more information, please refer to Name.

Returns

N-D Tensor. A location into which the result is stored. If x, y have different shapes and are “broadcastable”, the resulting tensor shape is the shape of x and y after broadcasting. If x, y have the same shape, its shape is the same as x and y.

Examples

>>> import paddle

>>> x = paddle.to_tensor([[1, 2], [7, 8]])
>>> y = paddle.to_tensor([[3, 4], [5, 6]])
>>> res = paddle.fmax(x, y)
>>> print(res)
Tensor(shape=[2, 2], dtype=int64, place=Place(cpu), stop_gradient=True,
[[3, 4],
 [7, 8]])

>>> x = paddle.to_tensor([[1, 2, 3], [1, 2, 3]])
>>> y = paddle.to_tensor([3, 0, 4])
>>> res = paddle.fmax(x, y)
>>> print(res)
Tensor(shape=[2, 3], dtype=int64, place=Place(cpu), stop_gradient=True,
[[3, 2, 4],
 [3, 2, 4]])

>>> x = paddle.to_tensor([2, 3, 5], dtype='float32')
>>> y = paddle.to_tensor([1, float("nan"), float("nan")], dtype='float32')
>>> res = paddle.fmax(x, y)
>>> print(res)
Tensor(shape=[3], dtype=float32, place=Place(cpu), stop_gradient=True,
[2., 3., 5.])

>>> x = paddle.to_tensor([5, 3, float("inf")], dtype='float32')
>>> y = paddle.to_tensor([1, -float("inf"), 5], dtype='float32')
>>> res = paddle.fmax(x, y)
>>> print(res)
Tensor(shape=[3], dtype=float32, place=Place(cpu), stop_gradient=True,
[5.  , 3.  , inf.])