【Paddle Tensor 第二期 API鲁棒性增强】解决paddle.max、paddle.min在输入存在nan时传播不正确问题

[aquagull]

PR Category

User Experience

PR Types

Bug fixes

Description

解决了paddle.max、paddle.min在gpu、cpu环境下输入存在nan时传播不正确问题.

测试代码

import paddle
import numpy as np

def test_reduce_with_nan(func, on_gpu):
    shape = (2, 3)  
    x_np = np.arange(np.prod(shape), dtype=np.float32).reshape(shape)
    x_np[0, 0] = np.nan  
    x = paddle.to_tensor(x_np, place=paddle.CPUPlace())
    if on_gpu:
        x = x.cuda()

    out = func(x)
    print(f"{func.__name__}({x.place}): {out.numpy()}")

test_reduce_with_nan(paddle.min, False)
test_reduce_with_nan(paddle.max, False)

if paddle.is_compiled_with_cuda():
    test_reduce_with_nan(paddle.min, True)
    test_reduce_with_nan(paddle.max, True)

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[HydrogenSulfate]

添加单测

[HydrogenSulfate]

这个逻辑不用这么复杂吧

Suggested change

	def _test_with_nan(self, func, shape, on_gpu, dtype=np.float32):
	with dygraph_guard():
	x_np = np.arange(np.prod(shape), dtype=dtype).reshape(shape)
	x_np[0, 0] = np.nan
	x = paddle.to_tensor(x_np)
	if on_gpu:
	if not paddle.is_compiled_with_cuda():
	return
	x = x.cuda()
	out = func(x)
	self.assertTrue(paddle.isnan(out), "Result should be NaN")
	def _test_with_nan(self, func, shape, place, dtype=np.float32):
	with dygraph_guard():
	x_np = np.arange(np.prod(shape), dtype=dtype).reshape(shape)
	x_np[0, 0] = np.nan
	x = paddle.to_tensor(x_np, place=place)
	out = func(x)
	self.assertTrue(paddle.isnan(out), "Result should be NaN")

[aquagull]

已更改

[HydrogenSulfate]

@aquagull 用SFINAE合并下int32/int64两种模板

[aquagull]

用SFINAE合并下int32/int64两种模板

好的