How to match points in two unequal list?

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Rishi Kiran Shankar 2019-8-15

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https://ww2.mathworks.cn/matlabcentral/answers/476293-how-to-match-points-in-two-unequal-list

编辑： Adam Danz 2019-8-19

Hi I have two lists. For example, A(1.1 2.2 3.1515 4.92121 1.1) and B = (1.25 6.754 2.332 3.15454 5.01214 1.25).

In above you can see that A has 5 elements and B has 6 elements and I have to match an element from both the list which are of closest value.

I have to match each elements from both the list from starting but B(2) is a new unrelated value and must be omitted and the matching should continue with next elements.

A(1) = B(1), A(2) = B(3), A(3) = B(4), A(4) = B(5), A(5) = B(6)

Each elements should be matched with only one elemnents of other list.

Finally I need the elements which didn't match with any other element.

Can anyone help me please?

I am a beginner and finding it difficult to find a logic.

Thanks in advance

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Adam Danz 2019-8-15

I didn't understand this part "Finally I need the omitted elements in the list. " but my answer addresses the rest.

Rishi Kiran Shankar 2019-8-15

Hi,

I want the element which is not matched with any other element.

For above example the answer should be B(2) = 6.754.

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采纳的回答

Adam Danz 2019-8-15

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https://ww2.mathworks.cn/matlabcentral/answers/476293-how-to-match-points-in-two-unequal-list#answer_387728

编辑：Adam Danz 2019-8-15

在 MATLAB Online 中打开

Here's how to find the closest value in B for each value in A.

A = [1.1  2.2  3.1515  4.92121 1.1];  % Row vector!
B = [1.25  6.754  2.332  3.15454  5.01214 1.25]; % Row vector!
[~, minIdx] = min(abs(A - B'),[],1);
% Summary Table
T = table(A', B(minIdx)', (1:numel(A))', minIdx', 'VariableNames', {'A', 'B_pair', 'A_index','B_index'});

Result

T =
  5×4 table
      A       B_pair    A_index    B_index
    ______    ______    _______    _______
       1.1      1.25       1          1   
       2.2     2.332       2          3   
    3.1515    3.1545       3          4   
    4.9212    5.0121       4          5   
       1.1      1.25       5          1   

[addendum]

For each value of A, here's how to find the closest value in B that hasn't already been chosen.

A = [1.1  2.2  3.1515  4.92121 1.1];  % Row vector!
B = [1.25  6.754  2.332  3.15454  5.01214 1.25]; % Row vector!
B2 = B; 
minIdx = zeros(size(A)); 
for i = 1:numel(A)
    [~, minIdx(i)] = min(abs(A(i)-B2)); 
    B2(minIdx(i)) = NaN; 
end
% Summary Table
T = table(A', B(minIdx)', (1:numel(A))', minIdx', 'VariableNames', {'A', 'B_pair', 'A_index','B_index'});

Result

T =
  5×4 table
      A       B_pair    A_index    B_index
    ______    ______    _______    _______
       1.1      1.25       1          1   
       2.2     2.332       2          3   
    3.1515    3.1545       3          4   
    4.9212    5.0121       4          5   
       1.1      1.25       5          6   

And here's how to find which indices of B have not been selected.

Bidx_notChosen = find(~isnan(B2));  %omit find() for logical index (more optimal)

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Bruno Luong 2019-8-19

编辑：Bruno Luong 2019-8-19

Sorry, but you stuck with a flawed algorithm.

I show one more time that my code below actually works.

Adam Danz 2019-8-19

编辑：Adam Danz 2019-8-19

在 MATLAB Online 中打开

@Rishi ,

Please let me know if this goal description is corrected:

For each 'A(i)', match the nearest value in 'B' that is at least within +/-5% of A(i).Each B can only be matched once. Then return any B that is unmatched.

If that's the case, here's an updated version

tol = 0.05; %tolerance (+/- 5% of A)
B2 = B;
minIdx = zeros(size(A));
for i = 1:numel(A)
    ABdiff = abs(A(i)-B2);
    [~, minIdx(i)] = min(ABdiff);
    % If the closest match is within tolerance, eliminate the B value 
    % from being selected again. 
    if ABdiff(minIdx(i)) < A(i)*tol*2
        B2(minIdx(i)) = NaN;
    end
end
% Summary Table
T = table(A', B(minIdx)', (1:numel(A))', minIdx', 'VariableNames', {'A', 'B_pair', 'A_index','B_index'});

To find the indices of B that have not been matched,

Bidx_notMatched = ~ismember(1:numel(B),T.B_index);
B(Bidx_notMatched)  %B values that are not matched
sum(Bidx_notMatched)  % total amount of unmatched in B (=11)

BTW, one reason there is so much back-and-forth on this is because the goal is not clearly defined. I know it's hard sometimes to communicate the goal but that's the bottleneck here. The good news is that I think we're close and the solution should just involve a few lines of code.

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