I have a m x n tensor (Tensor 1) and another k x 2 tensor (Tensor 2) and I wish to extract all the values of Tensor 1 using indices based on Tensor 2. For example;
Tensor1
1 2 3 4 5
6 7 8 9 10
11 12 13 14 15
16 17 18 19 20
[torch.DoubleTensor of size 4x5]
Tensor2
2 1
3 5
1 1
4 3
[torch.DoubleTensor of size 4x2]
And the function would yield;
6
15
1
18
The first solution that comes into mind is to simply loop through indexes and pick the correspoding values:
function get_elems_simple(tensor, indices)
local res = torch.Tensor(indices:size(1)):typeAs(tensor)
local i = 0
res:apply(
function ()
i = i + 1
return tensor[indices[i]:clone():storage()]
end)
return res
end
Here tensor[indices[i]:clone():storage()]
is just a generic way to pick an element from a multi-dimensional tensor. In k-dimensional case this is exactly analogous to tensor[{indices[i][1], ... , indices[i][k]}]
.
This method works fine if you don't have to extract lots of values (the bottleneck is :apply
method which is not able to use many optimization techniques and SIMD instructions because the function it executes is a black box). The job can be done way more efficiently: the method :index
does exactly what you need... with a one-dimensional tensor. Multi-dimensional target/index tensors need to be flattened:
function flatten_indices(sp_indices, shape)
sp_indices = sp_indices - 1
local n_elem, n_dim = sp_indices:size(1), sp_indices:size(2)
local flat_ind = torch.LongTensor(n_elem):fill(1)
local mult = 1
for d = n_dim, 1, -1 do
flat_ind:add(sp_indices[{{}, d}] * mult)
mult = mult * shape[d]
end
return flat_ind
end
function get_elems_efficient(tensor, sp_indices)
local flat_indices = flatten_indices(sp_indices, tensor:size())
local flat_tensor = tensor:view(-1)
return flat_tensor:index(1, flat_indices)
end
The difference is drastic:
n = 500000
k = 100
a = torch.rand(n, k)
ind = torch.LongTensor(n, 2)
ind[{{}, 1}]:random(1, n)
ind[{{}, 2}]:random(1, k)
elems1 = get_elems_simple(a, ind) # 4.53 sec
elems2 = get_elems_efficient(a, ind) # 0.05 sec
print(torch.all(elems1:eq(elems2))) # true