Finding a row in a matrix that is closest to two separate conditions

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Paul Vedier 2020-9-27

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评论： Paul Vedier 2020-9-27

I'm familiar with using:

min(abs(1-x))

to find the value closest to whatever "x" is when specifying a target or condition. However, I'm trying to evaluate a matrix where a real part and imaginary part (each in an individual column) are the closest to a target value for each. An example snippet of my matrix would look like

    %Delay    Voltage   Length   Frequency   Real    Imaginary 
3500    1.1500    0.1429   20.0000    0.9678   -2.2574
4000    1.1000    0.1429   20.0000    0.9181   -2.5046
4500    1.0000    0.1429   20.0000    0.9296   -3.0746
4500    1.0500    0.1429   20.0000    0.9342   -2.7859
4500    1.1000    0.1429   20.0000    1.0023   -2.5453
3500    1.1500    0.1429   21.2000    0.9369   -2.2832
4000    1.1000    0.1429   21.2000    0.9022   -2.5270
4000    1.1500    0.1429   21.2000    1.0067   -2.3389

In reality, I have around 800 rows that I have sorted by the Frequency column. I've already written:

freqVector = COMSOL_raw(1:50,freqColumn);   % The number of frequencies found
for eachf = 1:length(freqVector)
    eachChunk = COMSOL_sorted_RealImag((COMSOL_sorted_RealImag(:,freqColumn(1))==freqVector(1)),:)
end

eachChunk finds all the rows that match each Frequency and separates them into its own "chunk," if you will. From this, I want to add a line that will find which row has the closest "Real" value to 1.00 and also the closest "Imaginary" value to -3.14, and spit out the singular row that comes closest to matching those two simultaneous conditions. Thus far, my best efforts have been the following lines:

% ***Within The For Loop***
best(eachf,:) = eachChunk(:,min(abs(1 - eachChunk(:,realColumn)))) %Returns an position 2 index error
%OR
best = min(abs(1 - eachChunk(:,realColumn))) & min(abs(-3.14 - eachChunk(:,imagColumn)))% Returns Boolean 1/True

For the first line, last night I could get MATLAB to output the whole line that came closest for one condition (but i deleted it and can't figure out how to do it again). For the second line, I know it finds something to satisfy both conditions, but I can't tell if it's doing it for the array or if it is actually finding one line that satisfies both. And if it's the latter, I very much need help writing a line that will output the whole line from the array eachChunk. Thank you!

EDIT: I also attempted to create a complex number from the last two columns and evaluate the closest complex number to a single complex condition (i.e. 1 - 3.14i) but I couldn't get that to work either.

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Image Analyst 2020-9-27

What if one criteria points to one line, and the other criteria points to a different line? Which do you pick? You have to define a "cost function" that somehow uses the two distances to come up with one number that you can use to determine "closeness".

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Answer 1

Bruno Luong 2020-9-27

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https://ww2.mathworks.cn/matlabcentral/answers/600763-finding-a-row-in-a-matrix-that-is-closest-to-two-separate-conditions#answer_501265

编辑：Bruno Luong 2020-9-27

在 MATLAB Online 中打开

Assuming the data is in yourarr

rx = yourarr(:,5);
iy = yourarr(:,6);
z = rx + 1i*iy;
ztarget = 1;
[absdz, locmin] = min(abs(ztarget-z))

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Paul Vedier 2020-9-27

Hi, thanks!

Your exact code doesn't quite work because if i leave ztarget=1, I only get something that corresponds to the real part of z. However, if I modify it to be ztarget=1-3.14i, the locmin variable equals the row number corresponding to the closest match.

Is this what locmin should be doing? Spot checking, it seems to work perfectly for all but 1 Frequency array (yourarr).

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