Approximate match between two matrices

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Mohammed Rasul Iqbal
评论： Voss 2022-2-5
i have two vectors C1 is calculated using code and A1 is input vector. C1(1x4) = [0.4311 0.123 0.011 0.441] , A1(5x4)=
X A B C
0.11 0.13 13.46 0.87
0.01 0.10 10.47 0.90
0.44 0.12 12.19 0.88
0.43 0.09 9.24 0.91
0.06 0.07 7.15 0.93
Need to find approximate match between individual C1 value with A1(:,1) and should give the coresponding output of A(:,2),A(:,3) & A(:,4).

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Arif Hoq
Arif Hoq 2022-2-5
Ran in:
if there is no approximate value then it will return the minimum value of that column.
C1= [0.4311 0.123 0.011 0.441];
A1=[0.11,0.13,13.46,0.87;0.01,0.10,10.47,0.90;0.44,0.12,12.19,0.88;...
0.43,0.09,9.24,0.91;0.06,0.07,7.15,0.93 ];
A2=A1(:,1);
A3=A1(:,2);
A4=A1(:,3);
A5=A1(:,4);
[minValue1,nearestIndex] = min(abs(A2-C1(1,1)));
C1_first = A2(nearestIndex) % for the value C1=0.4311
C1_first = 0.4300
[minValue2,nearestIndex] = min(abs(A3-C1(1,2)));
C1_second = A3(nearestIndex) % for the value C1=0.123
C1_second = 0.1200
[minValue3,nearestIndex] = min(abs(A4-C1(1,3)));
C1_third = A4(nearestIndex) % for the value C1=0.011
C1_third = 7.1500
[minValue4,nearestIndex] = min(abs(A5-C1(1,4)));
C1_fourth = A5(nearestIndex) % for the value C1=0.441
C1_fourth = 0.8700
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Voss
Voss 2022-2-5
Ran in:
As an alternative to the first method:
C1 = [0.4311 0.123 0.011 0.441];
A1 = [0.11,0.13,13.46,0.87; ...
0.01,0.10,10.47,0.90; ...
0.44,0.12,12.19,0.88;...
0.43,0.09,9.24,0.91; ...
0.06,0.07,7.15,0.93];
[min_diff,idx] = min(abs(A1-C1),[],1)
min_diff = 1×4
0.0011 0.0030 7.1390 0.4290
idx = 1×4
4 3 5 1
nearest_A1 = A1(sub2ind(size(A1),idx,1:size(A1,2)))
nearest_A1 = 1×4
0.4300 0.1200 7.1500 0.8700

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