Возможный ответ для 2-мерных разреженных индексов
Найдите ответ ниже, играя с несколькими методами pytorch (torch.eq(), torch.unique(), torch.sort() и т. Д.), Чтобы вывести компакт,нарезанный тензор формы (len(idx), len(idx)).
Я протестировал несколько краевых случаев (неупорядоченные idx, v с 0 s, i с несколькими одинаковыми индексными парами и т. д.), хотя я могузабыл некоторые.Производительность также должна быть проверена.
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.]])
Предыдущий ответ для одномерных разреженных индексов
Вот решение (вероятно, неоптимальное и не охватывающее все граничные случаи), следующееИнтуиции, которыми поделились в связанном открытом выпуске (надеюсь, эта функция скоро будет раскрыта):
# 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)]