Getting a new matrix after comparing two matrices in some conditions

Hello,

I have an issue which has bothered me for a while. Please give me some suggestions. Thanks in advance. I have two matrices:

 K1 =

     1     4     2     0     1
     4     3     0     0     1
     1     4     0     0     1
     3     4     0     0     2
     1     4     3     0     2
     4     1     0     0     2
     2     4     0     0     1
     3     4     0     0     1
     4     1     0     0     1
     4     2     0     0     1
     1     4     3     0     1
     2     4     3     0     1
     2     4     1     0     1
     3     4     2     0     1
     3     4     1     0     1

 L1 =

     1     4     0     0     1
     2     4     0     0     1
     3     4     0     0     1
     4     1     0     0     1
     4     2     0     0     1
     4     3     0     0     1
     1     4     0     0     2
     3     4     0     0     2
     4     1     0     0     2
     4     3     0     0     2

What I want to do is that how many times of every row in L1 are included in matrix K1 and the corresponding indices of K1. Then if the times of every row in L1 included in K1 are over 2 times, I will keep only 2 rows of K1 and formulate a new matrix R. I just want to get this matrix R.

I will explain it in details. For example, the first line of L1 is [1 4 0 0 1] which is included in rows of K1,[1 4 2 0 1;1 4 0 0 1;1 4 3 0 1]. Then I want to get the indices:[1;3;11]. Obviously, the times of 1st row in L1 is over 2, So I will just keep the first two indices(1 and 3) and delete the 3rd(11). Then put them in a new matrix R. Now, R=[1 4 2 0 1;1 4 0 0 1]. Then the 2nd line of L1 follows this process. I will take the 5th line of L1 as another example, which is [4 2 0 0 1]. It is included in rows of K1,[1 4 2 0 1; 4 2 0 0 1;3 4 2 0 1]. The number of indices is also over three. Then delete the last one. Then R=[1 4 2 0 1;1 4 0 0 1;...;1 4 2 0 1; 3 4 2 0 1].

What I've done now is as follows

 R=[];

 for i=1:nbL1

l=L1(i,[1,2,5])

 for j=1:nbK1
   k1=K1(j,[1,2,5])  
   k2=K1(j,[2,3,5])
  if K1(j,3)==0 & K1(j,4)==0 %(4 3 0 0 1)or(1 4 0 0 1)
      disp('0-0')
    [match,index]=ismember(l,k1,'rows')
    if index~=0
       R=[R;K1(index,:)]
    end
  elseif K1(j,3)~=0 & K1(j,4)==0 %(1 4 2 0 1) 
      if L1(i,1)==4 %(4 3 0 0 1)
       disp('0-1')
       [match,index]=ismember(l,k2,'rows')
       if index~=0
       R=[R;K1(index,:)]
       end
      else
       disp('0-2')
       [match,index]=ismember(l,k1,'rows')
       if index~=0
       R=[R;K1(index,:)]
       end
      end
  end

end
end

I only can check whether the rows of L1 are included in K1. I can't count the times and formulate new matrix R that I want. Could anyone help me, please?

[Guillaume]

First thing: you need to explain what your definition of include is. You're certainly not using the standard mathematical definition of set membership.

I fail to see how [1 4 2 0 1] includes [1 4 0 0 1] (unless you ignore the 0 maybe?)

I also fail to see how [1 4 2 0 1] includes [4 2 0 0 1] (unless you ignore the 0 and the order maybe?)

[Clémentine OU]

@Guillaume: Thanks for your replying. Because every row in P1 is made by rows of L1. For example, the row in P1([1 4 2 0 1] ) is made by [1 4 0 0 1] and [4 2 0 0 1]. [1 4 0 0 1] formulates the first element of [1 4 2 0 1] and [4 2 0 0 1] formulates the third element of [1 4 2 0 1]. Because they both have '4' and '1'. So put '4' as the second element and '1' as the fifth element.

[Guillaume]

I'm afraid that I don't really understand how you define your relationship between your sets of numbers.

Do you mean that a set of numbers A "includes" a set of numbers B if there exists another set C such that (C ∪ B) ∋ A? (i.e. the union of B and C contains A)

If so, that is a strange definition. If not, please explain using exact mathematical terms to express that relationship?

[dpb]

As Guillaume says, this is perfectly unintelligible as to how you managed to get from the L1 array entry of [1 4 0 0 1] to the others that you say are to be included.

Your followup response introduces yet another array P1 out of the blue that doesn't appear anywhere?

[Clémentine OU]

@Guillaume: Sorry I type the wrong things, 'P1' should be modified as 'K1'. Every row of K1([1 4 2 0 1] ) is made by [1 4 0 0 1] and [4 2 0 0 1]. The logic is that: If the 5th element of every row in L1 is the same, like it's '1' and put it as the 5th element. Then we check the first two elements of these rows. If the 1st element is not 4, put it as the 1st element. If the 1st element is 4, put it as the 2rd element. In this case, put the second element as the 3rd new element. The rest location is 0. That's why K1 formulates. I have no idea expressed in mathematics.

[Clémentine OU]

@dpb: Sorry, I want to say 'K1'. I type the wrong things.

[dpb]

Well, I'll leave the manipulations to build the 'K1' array to you; I still don't follow the explanation (what position is "2rd"? is only one confusing issue from not proofing your postings well enough in the last).

If your code does what you want as you imply, then the Answer I provided below will return the maximum of 2 rows of the matching value for each.

Answer 1

[dpb]

"... can check whether the rows of L1 are included in K1. I can't count the times and formulate new matrix R"

Well, if you understand the process of building the row to check for in the K1 array that we're unable to discern the logic behind, then

ia=ismember(L1,K1_fixedup(i));

is the logical array of which rows (if any) are in L1 of the i th fixed-up K1 row. Since logical true is 1 and false 0, the number is then

nK1_i = sum(ia);

To select those simply check for

L1_Want=L1(find(ia==1,2),:);

to return two (or fewer, find will return the number found up to the maximum requested. Consequently, you don't really even need to compute the number found explicitly.

I'd guess if the logic for building the expanded rows from the initial array were made more clear the whole thing could be reduced to an arrayfun or accumarray expression but the above should let you implement the accumulation of the desired output array from your existing code if it does what you expect except for the selection.

[Clémentine OU]

@dpb:Thanks for your inspiration. I made the code finally for what I need.

R=[];
match=0;
for i=1:nbL1
  l=L1(i,[1,2,5]);
  S=[];
  aa=0;
  for j=1:nbK1
    k=K1(j,[1,2,5]);
    if K1(j,3)==0 & K1(j,4)==0 %(4 3 0 0 1)or(1 4 0 0 1)
      disp('*-----*');
      [match,loc]=ismember(l,k,'rows');
      if match==1
        R=[R;K1(j,:)];
      end
    elseif  K1(j,3)~=0  %(1 4 2 0 1)
      disp('#-----#');
      [match,loc]=ismember(l,k,'rows');
      if match==1
        S=[S;K1(j,:)];
        aa=aa+match 
      end
    end
  end
  if aa>2       
    R=[R;S([1:2],:)];  
  elseif aa<=2 
    R=[R;S]; 
  end
end