在Matlab中,我正在寻找一种以更优雅的方式解决以下问题的方法:
在三维数组中,在一个维度上(即在我的情况下是时间),从所有值的某个索引等于零,例如,以下示例数组a为0表示第二维的索引为3(即a(:,3:end,: == 0)
):
a(:,:,1) =
1 1 0
1 0 0
1 0 0
a(:,:,2) =
1 1 0
1 0 0
1 1 0
a(:,:,3) =
1 1 0
1 1 0
1 1 0
[编辑,被要求预期结果]
预期结果是:
o(:,:,1) =
1 1
1 0
1 0
o(:,:,2) =
1 1
1 0
1 1
o(:,:,3) =
1 1
1 1
1 1
现在我当然可以检查维度2中的每个索引,到处都是零,这就是我现在正在做的事情,但我觉得有一些更好的方法可以在matlab中解决这个问题一些更优雅的方式(甚至可能对任何多维数组)。谢谢你的帮助!
答案 0 :(得分:1)
function req = removeColwithSpecVal(a,spec_val)
req=num2cell(a,[1 2]); %Converting to cell keeping the order of rows & columns same
req=vertcat(req{:}); %Stacking multidimensional slices vertically
ind= ~all(req==spec_val,1);%Finding logical indices of the required columns
req = req(:,ind); %The matrix after removal of the not required columns
%Finding the size of the input matrix 'a'; stacking with 1 so that
%it is applicable on 1-D/2-D matrices as well
sz =[size(a) 1];
%Finding the # of columns that the required multi-dimensional matrix will have
sz(2)= numel(req)/prod([sz(1), sz(3:end)]);
%Reshaping to the desired result
req=reshape(permute(reshape(req.', sz(2),sz(1),[]),[2 1 3]),sz);
end
示例运行
%1-D Example
spec_val=5;
a=[1 4 5 2 5];
req = removeColwithSpecVal(a,spec_val)
req =
1 4 2
%2-D Example
spec_val=0;
a=[1 1 0 ;
1 0 0 ;
1 0 0 ];
req = removeColwithSpecVal(a,spec_val)
req =
1 1
1 0
1 0
%Your example (3-D)
spec_val=0;
a(:,:,1) = [1 1 0;
1 0 0;
1 0 0];
a(:,:,2) = [1 1 0;
1 0 0;
1 1 0];
a(:,:,3) = [1 1 0;
1 1 0;
1 1 0];
req = removeColwithSpecVal(a,spec_val)
req(:,:,1) =
1 1
1 0
1 0
req(:,:,2) =
1 1
1 0
1 1
req(:,:,3) =
1 1
1 1
1 1
也适用于高维矩阵。