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Finding highest position of values in two arrays?

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Michael Simonovski
Michael Simonovski 2018-7-21
评论： Rik 2018-7-27
Hello,
i have two matrices with two columns. I want to sort them after one column. After it this matrices should compare one column. For example after the sort of the first row the first row of A should be compared with the first row of B. In this both rows are the same numbers, but in different sequence. The highest number in both of this rows should be copied to a matrix C(1,1). The date from the second row of A to C(2,1) and from B to C(3,1). This should be done for all members of the A matrix.
How could the code look like?

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回答（3 个）

Rik
Rik 2018-7-21
编辑：Rik 2018-7-23
Update: the code below incorporates what you describe and results in the required C.
A=[0.1 3;0.3 1;0.6 2];B=[0.7 2; 0.8 3; 0.9 1];
%first, sort by ID, which makes everything easier
A = sortrows(A, 2);
B = sortrows(B, 2);
%next, find the rank for each ID
[~,rankA]=sort(A(:,1),'descend');
[~,rankB]=sort(B(:,1),'ascend');
%calculate the cost and sort it to get the overall ranking
costFcn=rankA/(numel(rankA)-0.1)+rankB/numel(rankB);
[~,rankC]=sort(costFcn);
%create C
C=[rankC A(rankC,1) B(rankC,1)];
original post:
Something like this?
dummy_A=rand(10,2);
dummy_B=rand(10,2);
[~,order]=sort(dummy_A(:,1));
A=dummy_A(order,:);
[~,order]=sort(dummy_B(:,1));
B=dummy_B(order,:);
C=[max([A B],[],2) A(:,2) B(:,2)];
If this is not what you mean, please provide some way to generate plausible data and a sample output.
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Rik
Rik 2018-7-23
Doesn't my updated answer do exactly what you describe?

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Michael Simonovski
Michael Simonovski 2018-7-23
I have tried it, but it doesnt work, why?
A=[0.1,2;0.6,9;0.7,10;0.4,6;0.8,3;0.9,1];
B=[0.3,1;0.8,10;0.6,3;0.9,6;0.22,9;0.82,2];
a=A(:,2);
b=B(:,2);
number = unique([a; b]);
n = 1;
while isempty(number) ~= 1
for k = 1 : length(number)
summe(k) = find(a == number(k)) + find(b == number(k));
end
C(n,1) = number(find(summe == min(summe),1));
avalue=find(a==C(n,1));
bvalue=find(b==C(n,1));
C(n,2)=A(avalue,1);
C(n,3)=A(bvalue,1);
A(avalue,:)=[];
B(bvalue,:)=[];
n=n+1;
clear summe;
clear number;
a=A(:,2);
b=B(:,2);
number = unique([a; b]);
end
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Michael Simonovski
Michael Simonovski 2018-7-23
a) I didnt know it!
b) My result is:
C =
10.0000 0.7000 0.6000
1.0000 0.9000 0.1000
2.0000 0.1000 0.8000
3.0000 0.8000 0.6000
6.0000 0.4000 0.6000
9.0000 0.6000 0.6000
First problem is, that the value 9 in the first row should be at the position 2 - 4, because it has the same sum value. The second problem is, that the values in the another values do not fit (See A and B).
3) I would like to implement a step, which should make following:
If the sum of more than one value is the same, compare the values of the first row, which have the smallest. If more than one value still have the same values, compare the values of the first coloumn of matrix B?
Rik
Rik 2018-7-27
I'm on mobile, so I can't test my code with this input, but it should yield the correct result. Does it? Otherwise I would suggest using Guillaume's answer.

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Guillaume
Guillaume 2018-7-24
I don't really understand your description of the rule to figure out the ordering. Whatever it is however, it should be trivial to implement.
I would recommend that you switch to using tables, it would allow you to trivially merge your two arrays and sort the result according to whichever rule you want. In addition, it makes clear what each column represent and also makes whatever calculation you want to perform easier to understand
A=[0.1,2;0.6,9;0.7,10;0.4,6;0.8,3;0.9,1];
B=[0.3,1;0.8,10;0.6,3;0.9,6;0.22,9;0.82,2];
%convert arrays to tables
tA = array2table(A, 'VariableNames', {'Friction', 'ID'});
tB = array2table(B, 'VariableNames', {'Force', 'ID'});
%merge tables
tmerged = join(tA, tB)
%create a cost function. Use whatever formula you want
tmerged.Cost = tmerged.Force + tMerged.Friction %e.g sum of force and friction
%sort first according to cost, then if equal according to force, then friction
tsorted = sortrows(tmerged, {'Cost', 'Force', 'Friction'})

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