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我有一组数据，其中包含我想删除存在子集的超集的集合，如下所示：

a{1} = [5]
a{2} = [4 11 14]
a{3} = [1]
a{4} = [5 16]
a{5} = [5]
a{6} = [11 16]
a{7} = [11]
a{8} = [16]
a{9} = [9 14 17]
a{10} = [14]

[ii, jj] = ndgrid(1:numel(a));
s = cellfun(@(x,y) all(ismember(x,y)), a(ii), a(jj));
s = triu(s,1); %// count each pair just once, and remove self-pairs
similarity = a(~any(s,1));
celldisp(similarity)

结果如下：

a{1} = [5]
a{2} = [4 11 14]
a{3} = [1]
a{4} = [11 16]
a{5} = [11]
a{6} = [16]
a{7} = [9 14 17]
a{8} = [14]

正如输出所示，仍然有应该删除的超集，即a{2}因为a{5}contains11是它的子集，a{4}应该删除因为a{5}contains11和a{6}contains16以及a{7}应该删除，因为a{8}contains subset 14。

预期输出是

a{1} = [5] 
a{2} = [1]
a{3} = [11]
a{4} = [16]
a{5} = [14]

任何人都可以帮助如何修复此代码，以便我可以获得准确的结果集。谢谢

4

1 回答 1

2

我认为您需要使用下三角部分而不是上三角部分：

s = tril(s,-1); % instead of s = triu(s,1);

编辑

仅当超集始终出现在子集之前时，保留下三角形部分才有效。这是一个应该始终可以正常工作的通用版本。

[ii, jj] = ndgrid(1:numel(a));
s = cellfun(@(x,y) all(ismember(x,y)), a(ii), a(jj));
% Set diagonal to zero.
s = s - diag(diag(s));
% Indicator matrix for sets that are exactly equal.
same = s & s';
% For equal sets keep only the first occurence.
keep = triu(same) | ~same.*s;
% Delete supersets.
similarity = a(~any(keep,1));
celldisp(similarity)

顺便说一句，运行双循环而不是上述矩阵运算可能更容易。

于 2015-04-16T16:06:59.623 回答