Question

我有[句子*单词]矩阵，如下所示

out = 0 1 1 0 1
      1 1 0 0 1
      1 0 1 1 0
      0 0 0 1 0

我想以一种应该告诉W1＆amp;的方式来处理这个矩阵。 W2在＆＃34;句号2＆＃34;和＃34;句号4＆＃34;以相同的值出现，即1 1和0 0。输出应如下：

output{1,2}= 2 4

输出{1,2}告诉单词编号1和2出现在句子编号2和4中，具有相同的值。

比较W1＆amp; W2下一位候选人应为W1＆amp; W3在sentence 3＆amp; sentence 4

output{1,3}= 3 4

依此类推，直到每个nth字与其他所有单词进行比较并保存。

Answer 1

这将是一个vectorized方法 -

%// Get number of columns in input array for later usage 
N = size(out,2);

%//  Get indices for pairwise combinations between columns of input array
[idx2,idx1] = find(bsxfun(@gt,[1:N]',[1:N])); %//'

%// Get indices for matches between out1 and out2. The row indices would
%// represent the occurance values for the final output and columns for the 
%// indices of the final output.
[R,C] = find(out(:,idx1) == out(:,idx2))

%// Form cells off each unique C (these will be final output values)
output_vals = accumarray(C(:),R(:),[],@(x) {x})

%// Setup output cell array
output = cell(N,N)

%// Indices for places in output cell array where occurance values are to be put
all_idx = sub2ind(size(output),idx1,idx2)

%// Finally store the output values at appropriate indices
output(all_idx(1:max(C))) = output_vals

Answer 2

您可以使用bsxfun轻松获得大小＃words-by-＃words-by-＃句子的逻辑矩阵：

coc = bsxfun( @eq, permute( out, [3 2 1]), permute( out, [2 3 1] ) );

如果单词occ( wi, wj, si )和单词wi出现在具有相同值的句子wj中，则

此逻辑数组为si。

从output获取coc单元格数组

nw = size( out, 2 ); %// number of words
output = cell(nw,nw);
for wi = 1:(nw-1)
    for wj = (wi+1):nw
        output{wi,wj} = find( coc(wi,wj,:) );
        output{wj,wi} = output{wi,wj}; %// you can force it to be symmetric if you want
    end
end

如何比较二进制矩阵的列并比较matlab中的元素？

2 个答案: