Finding closest points to a given range in matrix

Hi,

I have a 3 column matrix, x,y,and z respectively. I want to obtain the value of z corresponds to (closest point to 0.00 in x), and closest point to 0.29 in y) (both conditions must be satisfied in the same row).

To sum , for example, x range is between 0.00 and 0.02, and that of y is between 0.28 and 0.29). At the end, I may obtain the values as x=0.0005, its corresponding y=0.28 and (z=-0.3355)

I investigated in the forum but could not find a solution. Could you kindly give me some advice for this?

Thanks in advance.

Answer 1

[Sean de Wolski]

A =[ 0.375 0.279 0.366 
0.004 0.256 0.321 
0.004 0.266 0.322 
0.004 0.276 0.333 
0.004 0.286 0.338 
0.004 -0.687 0.211 
0.004 -0.677 0.216 
0.486 -0.687 0.201 
0.787 -0.697 0.146 
1.168 -1.229 0.050 
-0.588 -0.587 0.080 
-0.678 -0.988 0.036 
-0.839 0.065 0.062];

To find the minimum sum of absolute differences.

goal = [0 0.28];
[~, idx] = min(sum(abs(bsxfun(@(minus,A(:,1:2),goal)),2));
A(idx,:)

To find the minimum vector magnitude in 2d space:

goal = [0 0.28];
r = bsxfun(@minus,A(:,1:2),goal);
[~, idx] = min(hypot(r(:,1),r(:,2)));
A(idx,:)

[tethys]

it is working well. Thank you for your suggestions and time!

Answer 2

[Fangjun Jiang]

The closet point to 0 in x might have one point. The closet point to 0.29 might also have one point. What if they are not in the same row? Does't it mean you don't have a solution.

xyz=rand(20,3);
[MinX,Indx]=min(abs(xyz(:,1)))
[MinY,IndY]=min(abs(xyz(:,2)-0.29));
if MinX~=MinY
    disp('no solutin');
else
    FoundZ=xyz(IndX,3);
end

Answer 3

[Mads]

If A is your matrix you could use:

[~,b] = min(abs(A(:,1))+abs(A(:,2)-0.29))

where b is the rows position which satisfies being the minimum absolute distance to the point.

Answer 4

[tethys]

First of all, thank you for your time.

I tried both of them , but "there is no solution". Yet, for instance when I evaluate my data I have [0,0040 0,2759 -0,3329], I want to obtain data like this x is closest to zero, y is closest to 0.29 (intersection)which give me z. Regarding values they do not need to be same.

I am still trying to develop your codes.

[Fangjun Jiang]

Do you understand my point in my answer? Or maybe you didn't describe your question correctly. Could you provide a set of example data (the nx3 matrix) and then explain the expected output?

[tethys]

Maybe I think I could not describe my question properly,Here is the set of data:

0,375 0,279 0,366

0,004 0,256 0,321

0,004 0,266 0,322

0,004 0,276 0,333

0,004 0,286 0,338

0,004 -0,687 0,211

0,004 -0,677 0,216

0,486 -0,687 0,201

0,787 -0,697 0,146

1,168 -1,229 0,050

-0,588 -0,587 0,080

-0,678 -0,988 0,036

-0,839 0,065 0,062

For me, I want to find "0,004 0,276 0,333", because x is closest to zero, and y is closest to 0.28.

For "0,375 0,279 0,366", y is closest to 0.28 but x is not closest to zero. It does not make sense to me.

I hope I can explain better now. Thank you again for your valuable time.

[tethys]

The code you suggested me can find the closest for x and y for the range I aspire; x has to be closest one to zero, than y has to be closest to 0.28 but within minimum x vector.

[Fangjun Jiang]

So are you looking for a point of (x,y) which is the closet to (0,0.28), which means sqrt(x^2+(y-0.28)^2)? That is different than "x close to 0" and "y close to 0.28".

[Sean de Wolski]

BSXFUN is in kaan's future!

[tethys]

Yes, definetely I am looking for closest to (0, 0.28) and its corresponding z. Sorry for the confusing