假设有一个火炬张量，例如以下形状：
x = torch.rand(20, 1, 120, 120)
x.squeeze()将其形状变为[20, 120, 120]。然后我想获得torch张量，它是一个形状为[20, 2]的索引列表。如何快速实现？
6个回答
19
torch.topk() 是你正在寻找的函数。根据文档，
torch.topk(input, k, dim=None, largest=True, sorted=True, out=None) -> (Tensor, LongTensor)
返回给定维度上前 k 大的元素及其对应的索引。
如果未指定参数
dim，则默认选择最后一维。如果参数
largest为False，则返回前k小的元素。该函数返回一个由两个张量元素组成的命名元组 (values, indices)，其中
indices表示原始张量中元素的位置索引。当布尔型参数
sorted等于True，返回的前k个元素是已排序状态。
- tejasvi88
2
7这是一个有用的函数，但它并没有回答原始问题。OP想要获得每个20个120x120矩阵中最大元素的索引。也就是说，她想要20个二维坐标，一个对应于一个矩阵。topk仅返回在被最大化的维度中最大元素的索引。 - user118967
请注意，topk函数的文档在返回索引的含义方面存在混淆。它给人的印象是该函数提供了原始张量的索引，而实际上它仅返回最大化维度的索引。请参阅pytorch问题https://github.com/pytorch/pytorch/issues/50331#issue-782748956以澄清此问题。 - user118967
10
如果我理解正确，您不需要值，而是需要索引。不幸的是，没有现成的解决方案。存在一个argmax()函数，但我无法看到如何让它完全符合您的要求。
因此，这里有一个小的解决方法，效率也应该可以，因为我们只是在分割张量：
我希望这是你想要的内容！:)
编辑:
这是稍微修改过的代码，可能会更快一些（我猜不会太多），但它更简单、更美观:
与之前的代码不同，使用以下方式:
而不是：
因此，这里有一个小的解决方法，效率也应该可以，因为我们只是在分割张量：
n = torch.tensor(4)
d = torch.tensor(4)
x = torch.rand(n, 1, d, d)
m = x.view(n, -1).argmax(1)
# since argmax() does only return the index of the flattened
# matrix block we have to calculate the indices by ourself
# by using / and % (// would also work, but as we are dealing with
# type torch.long / works as well
indices = torch.cat(((m / d).view(-1, 1), (m % d).view(-1, 1)), dim=1)
print(x)
print(indices)
n 代表您的第一维，d 代表最后两个维度。我这里使用较小的数字来展示结果。但当然也适用于 n=20 和 d=120：
n = torch.tensor(20)
d = torch.tensor(120)
x = torch.rand(n, 1, d, d)
m = x.view(n, -1).argmax(1)
indices = torch.cat(((m / d).view(-1, 1), (m % d).view(-1, 1)), dim=1)
#print(x)
print(indices)
这里是 n=4 和 d=4 的输出：
tensor([[[[0.3699, 0.3584, 0.4940, 0.8618],
[0.6767, 0.7439, 0.5984, 0.5499],
[0.8465, 0.7276, 0.3078, 0.3882],
[0.1001, 0.0705, 0.2007, 0.4051]]],
[[[0.7520, 0.4528, 0.0525, 0.9253],
[0.6946, 0.0318, 0.5650, 0.7385],
[0.0671, 0.6493, 0.3243, 0.2383],
[0.6119, 0.7762, 0.9687, 0.0896]]],
[[[0.3504, 0.7431, 0.8336, 0.0336],
[0.8208, 0.9051, 0.1681, 0.8722],
[0.5751, 0.7903, 0.0046, 0.1471],
[0.4875, 0.1592, 0.2783, 0.6338]]],
[[[0.9398, 0.7589, 0.6645, 0.8017],
[0.9469, 0.2822, 0.9042, 0.2516],
[0.2576, 0.3852, 0.7349, 0.2806],
[0.7062, 0.1214, 0.0922, 0.1385]]]])
tensor([[0, 3],
[3, 2],
[1, 1],
[1, 0]])
我希望这是你想要的内容！:)
编辑:
这是稍微修改过的代码，可能会更快一些（我猜不会太多），但它更简单、更美观:
与之前的代码不同，使用以下方式:
而不是：
m = x.view(n, -1).argmax(1)
indices = torch.cat(((m // d).view(-1, 1), (m % d).view(-1, 1)), dim=1)
已经对argmax值进行了必要的重塑：
m = x.view(n, -1).argmax(1).view(-1, 1)
indices = torch.cat((m // d, m % d), dim=1)
但是正如评论中提到的那样，我认为很难从中获得更多的信息。
如果你真的非常重视性能提升的最后一点，你可以将上面的函数实现为pytorch的低级扩展（例如C++）。
这将给你一个可以调用的函数，并避免慢速的Python代码。
- MBT
5
是的，那就是我想要的输出。我修改了它，用
.float()将m转换，然后在除以d时使用//。你提出的是一种展开方式，类似于numpy.unravel_indices()。如果你能想到更快的方法，当然会更好。 - Chris@Chris 我刚刚进行了一个短时间测试。实际上，我认为它非常高效，我想目前没有更快的方法：调用
argmax()本身需要大约10倍的时间来计算下一行中的索引 - 在CPU上，我稍后也可以检查GPU。但是这些操作非常简单和直接，因此即使这是一个解决方法，从理论上讲它应该也相当高效。 - MBT1不，它绝不会慢。 在Telsa Volta上我只需要大约5.5毫秒。 我只是需要将其最大化，但是我同意，argmax是一个线性操作，因为张量是无序的。 可能这就是最慢的组件，而且不能加速。 - Chris
@Chris 我在结尾处进行了小的编辑，得到了稍微更好的版本。但是我不会期望在性能方面有什么显著的改进，可能只会比原来快半纳秒左右。如果你真的很重视性能，可能需要使用C++编写自定义扩展。但考虑到代码片段很小，所获得的收益也可能不会太大。 - MBT
谢谢，运行良好。我在评估中也犯了一个错误，看起来只有0.5毫秒而不是5毫秒。 - Chris
4
这是torch中的
unravel_index实现：def unravel_index(
indices: torch.LongTensor,
shape: Tuple[int, ...],
) -> torch.LongTensor:
r"""Converts flat indices into unraveled coordinates in a target shape.
This is a `torch` implementation of `numpy.unravel_index`.
Args:
indices: A tensor of (flat) indices, (*, N).
shape: The targeted shape, (D,).
Returns:
The unraveled coordinates, (*, N, D).
"""
coord = []
for dim in reversed(shape):
coord.append(indices % dim)
indices = indices // dim
coord = torch.stack(coord[::-1], dim=-1)
return coord
接下来，您可以使用torch.argmax函数来获取“扁平化”张量的索引。
y = x.view(20, -1)
indices = torch.argmax(y)
indices.shape # (20,)
使用unravel_index函数来展开索引。
indices = unravel_index(indices, x.shape[-2:])
indices.shape # (20, 2)
- Donshel
2
这是最接近真实、通用答案的回答！更直接地回答原始问题，即如何获取最大值的索引，您可能希望编辑以显示如何首先使用argmax获取索引，然后展开它们。 - user118967
最终我不得不编写与argmax的连接代码，请检查我的答案。请随意在您的答案中加入我的代码。 - user118967
0
被接受的答案仅适用于给定的示例。
tejasvi88的答案很有趣，但无法帮助回答原始问题（如我在那里的评论中所解释的）。
我认为Francois的答案最接近，因为它处理更通用的情况（任意数量的维度）。但是，它没有连接到argmax，并且所示示例未说明该函数处理批处理的能力。
因此，我将在Francois的答案基础上构建代码以连接到argmax。我编写了一个新函数batch_argmax，它返回批次中最大值的索引。批次可以组织在多个维度中。我还包括一些测试用例以供说明：
def batch_argmax(tensor, batch_dim=1):
"""
Assumes that dimensions of tensor up to batch_dim are "batch dimensions"
and returns the indices of the max element of each "batch row".
More precisely, returns tensor `a` such that, for each index v of tensor.shape[:batch_dim], a[v] is
the indices of the max element of tensor[v].
"""
if batch_dim >= len(tensor.shape):
raise NoArgMaxIndices()
batch_shape = tensor.shape[:batch_dim]
non_batch_shape = tensor.shape[batch_dim:]
flat_non_batch_size = prod(non_batch_shape)
tensor_with_flat_non_batch_portion = tensor.reshape(*batch_shape, flat_non_batch_size)
dimension_of_indices = len(non_batch_shape)
# We now have each batch row flattened in the last dimension of tensor_with_flat_non_batch_portion,
# so we can invoke its argmax(dim=-1) method. However, that method throws an exception if the tensor
# is empty. We cover that case first.
if tensor_with_flat_non_batch_portion.numel() == 0:
# If empty, either the batch dimensions or the non-batch dimensions are empty
batch_size = prod(batch_shape)
if batch_size == 0: # if batch dimensions are empty
# return empty tensor of appropriate shape
batch_of_unraveled_indices = torch.ones(*batch_shape, dimension_of_indices).long() # 'ones' is irrelevant as it will be empty
else: # non-batch dimensions are empty, so argmax indices are undefined
raise NoArgMaxIndices()
else: # We actually have elements to maximize, so we search for them
indices_of_non_batch_portion = tensor_with_flat_non_batch_portion.argmax(dim=-1)
batch_of_unraveled_indices = unravel_indices(indices_of_non_batch_portion, non_batch_shape)
if dimension_of_indices == 1:
# above function makes each unraveled index of a n-D tensor a n-long tensor
# however indices of 1D tensors are typically represented by scalars, so we squeeze them in this case.
batch_of_unraveled_indices = batch_of_unraveled_indices.squeeze(dim=-1)
return batch_of_unraveled_indices
class NoArgMaxIndices(BaseException):
def __init__(self):
super(NoArgMaxIndices, self).__init__(
"no argmax indices: batch_argmax requires non-batch shape to be non-empty")
以下是测试：
def test_basic():
# a simple array
tensor = torch.tensor([0, 1, 2, 3, 4])
batch_dim = 0
expected = torch.tensor(4)
run_test(tensor, batch_dim, expected)
# making batch_dim = 1 renders the non-batch portion empty and argmax indices undefined
tensor = torch.tensor([0, 1, 2, 3, 4])
batch_dim = 1
check_that_exception_is_thrown(lambda: batch_argmax(tensor, batch_dim), NoArgMaxIndices)
# now a batch of arrays
tensor = torch.tensor([[1, 2, 3], [6, 5, 4]])
batch_dim = 1
expected = torch.tensor([2, 0])
run_test(tensor, batch_dim, expected)
# Now we have an empty batch with non-batch 3-dim arrays' shape (the arrays are actually non-existent)
tensor = torch.ones(0, 3) # 'ones' is irrelevant since this is empty
batch_dim = 1
# empty batch of the right shape: just the batch dimension 0,since indices of arrays are scalar (0D)
expected = torch.ones(0)
run_test(tensor, batch_dim, expected)
# Now we have an empty batch with non-batch matrices' shape (the matrices are actually non-existent)
tensor = torch.ones(0, 3, 2) # 'ones' is irrelevant since this is empty
batch_dim = 1
# empty batch of the right shape: the batch and two dimension for the indices since we have 2D matrices
expected = torch.ones(0, 2)
run_test(tensor, batch_dim, expected)
# a batch of 2D matrices:
tensor = torch.tensor([[[1, 2, 3], [6, 5, 4]], [[2, 3, 1], [4, 5, 6]]])
batch_dim = 1
expected = torch.tensor([[1, 0], [1, 2]]) # coordinates of two 6's, one in each 2D matrix
run_test(tensor, batch_dim, expected)
# same as before, but testing that batch_dim supports negative values
tensor = torch.tensor([[[1, 2, 3], [6, 5, 4]], [[2, 3, 1], [4, 5, 6]]])
batch_dim = -2
expected = torch.tensor([[1, 0], [1, 2]])
run_test(tensor, batch_dim, expected)
# Same data, but a 2-dimensional batch of 1D arrays!
tensor = torch.tensor([[[1, 2, 3], [6, 5, 4]], [[2, 3, 1], [4, 5, 6]]])
batch_dim = 2
expected = torch.tensor([[2, 0], [1, 2]]) # coordinates of 3, 6, 3, and 6
run_test(tensor, batch_dim, expected)
# same as before, but testing that batch_dim supports negative values
tensor = torch.tensor([[[1, 2, 3], [6, 5, 4]], [[2, 3, 1], [4, 5, 6]]])
batch_dim = -1
expected = torch.tensor([[2, 0], [1, 2]])
run_test(tensor, batch_dim, expected)
def run_test(tensor, batch_dim, expected):
actual = batch_argmax(tensor, batch_dim)
print(f"batch_argmax of {tensor} with batch_dim {batch_dim} is\n{actual}\nExpected:\n{expected}")
assert actual.shape == expected.shape
assert actual.eq(expected).all()
def check_that_exception_is_thrown(thunk, exception_type):
if isinstance(exception_type, BaseException):
raise Exception(f"check_that_exception_is_thrown received an exception instance rather than an exception type: "
f"{exception_type}")
try:
thunk()
raise AssertionError(f"Should have thrown {exception_type}")
except exception_type:
pass
except Exception as e:
raise AssertionError(f"Should have thrown {exception_type} but instead threw {e}")
- user118967
0
我有一个直接的解决方法，但不是每个项目批量计算最大值的最佳方案的。简单的解决方法可能是：
# suppose the tensor is of shape (3,2,2),
>>> a = torch.randn(3, 2, 2)
>>> a
tensor([[[ 0.1450, -1.3480],
[-0.3339, -0.5133]],
[[ 0.6867, -0.2972],
[ 0.8768, 0.0844]],
[[-2.3115, -0.4549],
[-1.5074, -0.8706]]])
# then perform batch-wise max
>>> torch.stack([(a[i]==torch.max(a[i])).nonzero() for i in range(a.size(0))], dim=0)
tensor([[[0, 0]],
[[1, 0]],
[[0, 1]]])
- KingLiu
-2
ps=ps.numpy()
ps=ps.tolist()
mx=[max(l) for l in ps]
mx=max(mx)
for i in range(len(ps[0])):
if mx==ps[0][i]:
print("The digit is "+str(i))
break
这对我来说非常有效
- Sihat Afnan
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原文链接
[20, 2]矩阵？你想要在每个120 * 120的矩阵中沿着行和列找到最大值吗？ - Kashyap120 * 120矩阵中的每一个，我想要具有最大值的单元格的[x, y]坐标。 - Chris