我有一个 pytorch 稀疏张量，我需要使用此切片 [idx][:,idx] 对行/列进行切片，其中 idx 是索引列表，使用提到的切片在普通浮点张量上产生我想要的结果。是否可以在稀疏张量上应用相同的切片？这里的例子:

#constructing sparse matrix
i = np.array([[0,1,2,2],[0,1,2,1]])
v = np.ones(4)
i = torch.from_numpy(i.astype("int64"))
v = torch.from_numpy(v.astype("float32"))
test1 = torch.sparse.FloatTensor(i, v)

#constructing float tensor
test2 = np.array([[1,0,0],[0,1,0],[0,1,1]])
test2 = autograd.Variable(torch.cuda.FloatTensor(test2), requires_grad=False)

#slicing
idx = [1,2]
print(test2[idx][:,idx])

输出:

Variable containing:
 1  0
 1  1
[torch.cuda.FloatTensor of size 2x2 (GPU 0)]

我持有一个 250.000 x 250.000 的邻接矩阵，我需要在其中切片 n 行和 n 列，使用随机 idx，通过简单地采样 n 随机 idx。由于数据集太大，转换为更方便的数据类型是不现实的。

我能在test1上得到同样的切片结果吗？有可能吗？如果没有，是否有任何解决方法？

现在我正在使用以下解决方案的“hack”运行我的模型:

idx = sorted(random.sample(range(0, np.shape(test1)[0]), 9000))
test1 = test1AsCsr[idx][:,idx].todense().astype("int32")
test1 = autograd.Variable(torch.cuda.FloatTensor(test1), requires_grad=False)

其中 test1AsCsr 是我的 test1 转换为 numpy CSR 矩阵。该解决方案有效，但速度非常慢，并且使我的 GPU 利用率非常低，因为它需要不断地从 CPU 内存读取/写入。

编辑:结果是非稀疏张量没问题

最佳答案

二维稀疏索引的可能答案

在下面找到答案，玩几个pytorch方法(torch.eq(), torch.unique(), torch.sort() 等)以输出形状为 (len(idx), len(idx)) 的紧凑切片张量。

我测试了几个边缘情况(无序 idx、v 和 0、i 和多个相同的索引对等)，虽然我可能忘记了一些。还应检查性能。

import torch
import numpy as np

def in1D(x, labels):
    """
    Sub-optimal equivalent to numpy.in1D().
    Hopefully this feature will be properly covered soon
    c.f. https://github.com/pytorch/pytorch/issues/3025
    Snippet by Aron Barreira Bordin
    Args:
        x (Tensor):             Tensor to search values in
        labels (Tensor/list):   1D array of values to search for

    Returns:
        Tensor: Boolean tensor y of same shape as x, with y[ind] = True if x[ind] in labels

    Example:
        >>> in1D(torch.FloatTensor([1, 2, 0, 3]), [2, 3])
        FloatTensor([False, True, False, True])
    """
    mapping = torch.zeros(x.size()).byte()
    for label in labels:
        mapping = mapping | x.eq(label)
    return mapping


def compact1D(x):
    """
    "Compact" values 1D uint tensor, so that all values are in [0, max(unique(x))].
    Args:
        x (Tensor): uint Tensor

    Returns:
        Tensor: uint Tensor of same shape as x

    Example:
        >>> densify1D(torch.ByteTensor([5, 8, 7, 3, 8, 42]))
        ByteTensor([1, 3, 2, 0, 3, 4])
    """
    x_sorted, x_sorted_ind = torch.sort(x, descending=True)
    x_sorted_unique, x_sorted_unique_ind = torch.unique(x_sorted, return_inverse=True)
    x[x_sorted_ind] = x_sorted_unique_ind
    return x

# Input sparse tensor:
i = torch.from_numpy(np.array([[0,1,4,3,2,1],[0,1,3,1,4,1]]).astype("int64"))
v = torch.from_numpy(np.arange(1, 7).astype("float32"))
test1 = torch.sparse.FloatTensor(i, v)
print(test1.to_dense())
# tensor([[ 1.,  0.,  0.,  0.,  0.],
#         [ 0.,  8.,  0.,  0.,  0.],
#         [ 0.,  0.,  0.,  0.,  5.],
#         [ 0.,  4.,  0.,  0.,  0.],
#         [ 0.,  0.,  0.,  3.,  0.]])

# note: test1[1, 1] = v[i[1,:]] + v[i[6,:]] = 2 + 6 = 8
#       since both i[1,:] and i[6,:] are [1,1]

# Input slicing indices:
idx = [4,1,3]

# Getting the elements in `i` which correspond to `idx`:
v_idx = in1D(i, idx).byte()
v_idx = v_idx.sum(dim=0).squeeze() == i.size(0) # or `v_idx.all(dim=1)` for pytorch 0.5+
v_idx = v_idx.nonzero().squeeze()

# Slicing `v` and `i` accordingly:
v_sliced = v[v_idx]
i_sliced = i.index_select(dim=1, index=v_idx)

# Building sparse result tensor:
i_sliced[0] = compact1D(i_sliced[0])
i_sliced[1] = compact1D(i_sliced[1])

# To make sure to have a square dense representation:
size_sliced = torch.Size([len(idx), len(idx)])
res = torch.sparse.FloatTensor(i_sliced, v_sliced, size_sliced)

print(res)
# torch.sparse.FloatTensor of size (3,3) with indices:
# tensor([[ 0,  2,  1,  0],
#         [ 0,  1,  0,  0]])
# and values:
# tensor([ 2.,  3.,  4.,  6.])

print(res.to_dense())
# tensor([[ 8.,  0.,  0.],
#         [ 4.,  0.,  0.],
#         [ 0.,  3.,  0.]])

一维稀疏索引的上一个答案

根据相关 open issue 中分享的直觉，这是一个(可能不是最优的并且没有涵盖所有边缘情况)解决方案(希望这个功能很快就会被正确覆盖):

# Constructing a sparse tensor a bit more complicated for the sake of demo:
i = torch.LongTensor([[0, 1, 5, 2]])
v = torch.FloatTensor([[1, 3, 0], [5, 7, 0], [9, 9, 9], [1,2,3]])
test1 = torch.sparse.FloatTensor(i, v)

# note: if you directly have sparse `test1`, you can get `i` and `v`:
# i, v = test1._indices(), test1._values()

# Getting the slicing indices:
idx = [1,2]

# Preparing to slice `v` according to `idx`.
# For that, we gather the list of indices `v_idx` such that i[v_idx[k]] == idx[k]:
i_squeeze = i.squeeze()
v_idx = [(i_squeeze == j).nonzero() for j in idx] # <- doesn't seem optimal...
v_idx = torch.cat(v_idx, dim=1)

# Slicing `v` accordingly:
v_sliced = v[v_idx.squeeze()][:,idx]

# Now defining your resulting sparse tensor.
# I'm not sure what kind of indexing you want, so here are 2 possibilities:
# 1) "Dense" indixing:
test1x = torch.sparse.FloatTensor(torch.arange(v_idx.size(1)).long().unsqueeze(0), v_sliced)
print(test1x)
# torch.sparse.FloatTensor of size (3,2) with indices:
#
#  0  1
# [torch.LongTensor of size (1,2)]
# and values:
#
#  7  0
#  2  3
# [torch.FloatTensor of size (2,2)]

# 2) "Sparse" indixing using the original `idx`:
test1x = torch.sparse.FloatTensor(autograd.Variable(torch.LongTensor(idx)).unsqueeze(0), v_sliced)
# note: this indexing would fail if elements of `idx` were not in `i`.
print(test1x)
# torch.sparse.FloatTensor of size (3,2) with indices:
#
#  1  2
# [torch.LongTensor of size (1,2)]
# and values:
#
#  7  0
#  2  3
# [torch.FloatTensor of size (2,2)]

关于python - 列/行切片 torch 稀疏张量，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/50666440/