Matlab过滤两个字符串以使它们相等

Question

我需要比较两个字符数组并使它们完全相同。同时，我需要我保存的角色的坐标。

比如我有两个字符数组

A=['a';'b';'b';'b';'c';'d'];
B=['a';'b';'b';'d'];

我希望结果是

Anew=['a';'b';'b';'d'];
Bnew=['a';'b';'b';'d'];

我想知道职位

APos=[1;2;3;6];
BPos=[1;2;3;4];

这样

Anew=A(APos);
Bnew=B(BPos);

我尝试了以下代码

[lmod1,lmod2]=size(A);
[lobs1,lobs2]=size(B);

PosB=zeros(lobs1,1);

for i=1:lobs1
    for j=1:lmod1
        if B(i,:)==A(j,:)
            PosB(i,:)=i; %collect the ones that are in there, the zeros should then be the ones that are not 
        end
    end
end

PosB2=PosB(PosB~=0);

PosA=zeros(lmod1,1);
for i=1:lmod1
    for j=1:lobs1
        if A(i,:)==B(j,:)% if this is true at the end of the loop, it means that the value of namesobs is in namesw
            PosA(i,:)=i; %collect the ones that are in there, the zeros should then be the ones that are not 
        end
    end
end

PosA2=PosA(PosA~=0);

但这只会检查是否存在，而不是如果这些值重复了不同的次数。如何添加一个额外的过滤器，以便字符串在两个字符串数组中重复相同的次数？

仅供参考，我正在使用的实际字符串如下所示：

>> ODates(1:10,:)

ans =

  10×12 char array

    '20100202_186'
    '20100202_186'
    '20100202_190'
    '20100202_190'
    '20100202_190'
    '20100202_190'
    '20100202_190'
    '20100202_191'
    '20100202_191'
    '20100202_191'

EDates(1:10,:)

ans =

  10×12 char array

    '20100202_186'
    '20100202_190'
    '20100202_190'
    '20100202_190'
    '20100202_190'

然后答案是：

NewDates(1:10,:)

ans =

  10×12 char array

    '20100202_186'
    '20100202_190'
    '20100202_190'
    '20100202_190'
    '20100202_190'

Answer 1

通过循环解决这个问题，你走在了正确的轨道上，但我觉得你的想法有点太复杂了。看看这个解决方案

A=['a';'b';'b';'b';'c';'d'];
B=['a';'b';'b';'d'];

% allocate logical vectors
lgA = false(size(A));
lgB = false(size(B));

% initialize loop control counter
ia0 = 1;
% looping
for ib = 1:length(B)
    for ia = ia0:length(A)
        % check if A == B and stop the looping over A if it is true
        if A(ia) == B(ib)
            % store 
            lgA(ia) = true;
            lgB(ib) = true;
            % update loop control counter to start from the current
            % index the next time
            ia0 = ia+1;
            break
        end
    end
end
% validity check
assert( all(A(lgA) == B(lgB)) )
% output
APos = find(lgA);
BPos = find(lgB);