Question

我有一个n-by-3矩阵。像这样：

  mtx =  [3 1 3;
          2 2 3;
          2 3 2;
          5 4 1]

我希望最终结果是：

mtx2 = [0 0 0;
        2 2 3;
        2 3 2;
        0 0 0]

所以我想在没有第一个数字的行上加零，并且他们的第二个数字不等于另一个数字的第三个数字。

想象一下，任何一行的第一个数字是＆＃39; a＆＃39;，第二个＆＃39;＆＃39;和第三个＆＃39; c＆＃39;。对于第1行，我会对它进行比较＆＃39; a＆＃39;和其他所有人一样。如果没有另一个＆＃39; a＆＃39;等于那，然后第1行变为[0 0 0]。但如果还有另一个＆＃39; a＆＃39;等于那个，对于第2行的瞬间，我将比较＆＃39; b＆＃39; ＃1和＆＃39; c＆＃39; ＃2。如果它们相等则这些行保持不变。如果不是第1行变为[0 0 0]。等等。

Answer 1

看看这是否适合你 -

%// Logical masks of matches satisfying condition - 1 and 2
cond1_matches = bsxfun(@eq,mtx(:,1),mtx(:,1).')  %//'
cond2_matches = bsxfun(@eq,mtx(:,3),mtx(:,2).')  %//'

%// Create output variable as copy of input and use the conditions to set
%// rows that satisfy both conditions as all zeros
mtx2 = mtx;
mtx2(~any( cond1_matches & cond2_matches ,1),:)=0

Answer 2

我通过这种方式搜索，针对每一行测试你的两个标准。

Izeros=[];  % here is where I'll store the rows that I'll zero later
for Irow = 1:size(mtx,1) %step through each row

    %find rows where the first element matches
    I=find(mtx(:,1)==mtx(Irow,1));

    %keep only those matches that are NOT this row
    J=find(I ~= Irow);

    % are there any that meet this requirement
    if ~isempty(J)
        %there are some that meet this requirement.  Keep testing.

        %now apply your second check, does the 2nd element
        %match any other row's third element?
        I=find(mtx(:,3)==mtx(Irow,2));

        %keep only those matches that are NOT this row
        J=find(I ~= Irow);

        % are there any that meet this 2nd requirement
        if ~isempty(J)
            %there are some that meet this 2nd requirement.
            % no action.
        else
            %there are none that meet this 2nd requirement.
            % zero the row.
            Izeros(end+1) = Irow;
        end
    else
        %there are none that meet the first requirement.
        % zero the row.
        Izeros(end+1) = Irow;
    end
end

替换矩阵中的线（Matlab）

2 个答案: