火炬/lua:从Tensor中检索n个最佳子集 [英] torch / lua: retrieving n-best subset from Tensor
问题描述
我现在有以下代码，该代码将每个问题的最高分的索引存储在pred中，并将其转换为字符串.
我想对每个问题的n个最佳索引执行相同的操作，而不仅仅是具有最高分数的单个索引，然后将它们转换为字符串.我还想显示每个索引(或每个转换后的字符串)的分数.
因此，必须对scores进行排序，并且pred必须具有多个行/列，而不是1 x nqs.并且pred中每个条目的相应score值必须可检索.
我对lua/torch语法一无所知，任何帮助将不胜感激.
nqs=dataset['question']:size(1);
scores=torch.Tensor(nqs,noutput);
qids=torch.LongTensor(nqs);
for i=1,nqs,batch_size do
xlua.progress(i, nqs)
r=math.min(i+batch_size-1,nqs);
scores[{{i,r},{}}],qids[{{i,r}}]=forward(i,r);
end
tmp,pred=torch.max(scores,2);
answer=json_file['ix_to_ans'][tostring(pred[{i,1}])]
print(answer)
这是我的尝试，我使用简单的随机scores张量来演示其行为:
> scores=torch.floor(torch.rand(4,10)*100)
> =scores
9 1 90 12 62 1 62 86 46 27
7 4 7 4 71 99 33 48 98 63
82 5 73 84 61 92 81 99 65 9
33 93 64 77 36 68 89 44 19 25
[torch.DoubleTensor of size 4x10]
现在，由于您想要每个问题(行)的最佳索引，因此让我们对张量的每一行进行排序:
> values,indexes=scores:sort(2)
现在，让我们看一下返回张量包含什么:
> =values
1 1 9 12 27 46 62 62 86 90
4 4 7 7 33 48 63 71 98 99
5 9 61 65 73 81 82 84 92 99
19 25 33 36 44 64 68 77 89 93
[torch.DoubleTensor of size 4x10]
> =indexes
2 6 1 4 10 9 5 7 8 3
2 4 1 3 7 8 10 5 9 6
2 10 5 9 3 7 1 4 6 8
9 10 1 5 8 3 6 4 7 2
[torch.LongTensor of size 4x10]
如您所见，values的i-th行是scores的i-th行的排序版本(升序)，并且indexes中的每一行都为您提供相应的索引. /p>
您可以使用
获得每个问题(即行)的N最佳值/索引.
> N_best_indexes=indexes[{{},{indexes:size(2)-N+1,indexes:size(2)}}]
> N_best_values=values[{{},{values:size(2)-N+1,values:size(2)}}]
让我们用N=3看到给定示例的值:
> return N_best_indexes
7 8 3
5 9 6
4 6 8
4 7 2
[torch.LongTensor of size 4x3]
> return N_best_values
62 86 90
71 98 99
84 92 99
77 89 93
[torch.DoubleTensor of size 4x3]
因此，问题j的k-th最佳值是N_best_values[{{j},{values:size(2)-k+1}]]，并且在scores矩阵中的相应索引由以下row, column值给出:
row=j
column=N_best_indexes[{{j},indexes:size(2)-k+1}}].
例如，第二个问题的第一个最佳值(k=1)是99，它位于scores中的2nd行和6th列.您会看到values[{{2},values:size(2)}}]是99，并且indexes[{{2},{indexes:size(2)}}]给您6，这是scores矩阵中的列索引.
希望我能很好地解释我的解决方案.
I have following code now, which stores the indices with the maximum score for each question in pred, and convert it to string.
I want to do the same for n-best indices for each question, not just single index with the maximum score, and convert them to string. I also want to display the score for each index (or each converted string).
So scores will have to be sorted, and pred will have to be multiple rows/columns instead of 1 x nqs. And corresponding score value for each entry in pred must be retrievable.
I am clueless as to lua/torch syntax, and any help would be greatly appreciated.
nqs=dataset['question']:size(1);
scores=torch.Tensor(nqs,noutput);
qids=torch.LongTensor(nqs);
for i=1,nqs,batch_size do
xlua.progress(i, nqs)
r=math.min(i+batch_size-1,nqs);
scores[{{i,r},{}}],qids[{{i,r}}]=forward(i,r);
end
tmp,pred=torch.max(scores,2);
answer=json_file['ix_to_ans'][tostring(pred[{i,1}])]
print(answer)
Here is my attempt, I demonstrate its behavior using a simple random scores tensor:
> scores=torch.floor(torch.rand(4,10)*100)
> =scores
9 1 90 12 62 1 62 86 46 27
7 4 7 4 71 99 33 48 98 63
82 5 73 84 61 92 81 99 65 9
33 93 64 77 36 68 89 44 19 25
[torch.DoubleTensor of size 4x10]
Now, since you want the N best indexes for each question (row), let's sort each row of the tensor:
> values,indexes=scores:sort(2)
Now, let's look at what the return tensors contain:
> =values
1 1 9 12 27 46 62 62 86 90
4 4 7 7 33 48 63 71 98 99
5 9 61 65 73 81 82 84 92 99
19 25 33 36 44 64 68 77 89 93
[torch.DoubleTensor of size 4x10]
> =indexes
2 6 1 4 10 9 5 7 8 3
2 4 1 3 7 8 10 5 9 6
2 10 5 9 3 7 1 4 6 8
9 10 1 5 8 3 6 4 7 2
[torch.LongTensor of size 4x10]
As you see, the i-th row of values is the sorted version (in increasing order) of the i-th row of scores, and each row in indexes gives you the corresponding indexes.
You can get the N best values/indexes for each question (i.e. row) with
> N_best_indexes=indexes[{{},{indexes:size(2)-N+1,indexes:size(2)}}]
> N_best_values=values[{{},{values:size(2)-N+1,values:size(2)}}]
Let's see their values for the given example, with N=3:
> return N_best_indexes
7 8 3
5 9 6
4 6 8
4 7 2
[torch.LongTensor of size 4x3]
> return N_best_values
62 86 90
71 98 99
84 92 99
77 89 93
[torch.DoubleTensor of size 4x3]
So, the k-th best value for question j is N_best_values[{{j},{values:size(2)-k+1}]], and its corresponding index in the scores matrix is given by this row, column values:
row=j
column=N_best_indexes[{{j},indexes:size(2)-k+1}}].
For example, the first best value (k=1) for the second question is 99, which lies at the 2nd row and 6th column in scores. And you can see that values[{{2},values:size(2)}}] is 99, and that indexes[{{2},{indexes:size(2)}}] gives you 6, which is the column index in the scores matrix.
Hope that I explained my solution well.
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