overlap-> tensor([[0.0000, 0.0000, 0.0000, ..., 0.6466, 0.7945, 0.5389]],
device='cuda:0')
overlap_for_each_prior, object_for_each_prior = overlap.max(dim=0) # (8732)
这个 .max(dim=0) 返回两个返回值。
Tensorflow 2.0有等价方法吗?
tf.math.argmax
。使用它,您还可以获得最大元素:
A = tf.constant([2, 20, 30, 3, 6])
maximum_index = tf.math.argmax(A)
A[maximum_index], maximum_index
# >>> 30, 2
如果我正确理解你的问题,你可以使用
tf.math.reduce_max
功能:
A = tf.constant([[1, 2, 3, 4, 5], [6, 7, 8, 9, 10]])
print('tf->', tf.math.reduce_max(A, axis=d))
tf-> tf.Tensor([ 5 10], shape=(2,), dtype=int32)
A2 = torch.from_numpy(A.numpy())
print('torch', A2.amax(1))
torch-> 张量([ 5, 10], dtype=torch.int32)