How can I grab the value of i for which out(i) is equal to s(2)?

Hi everyone,

I am trying to grab the value of i for which out(i) is equal to s(2). The segment is marked below by '% facing problem here'. Correct value of d is the answer. Can anyone please help me to figure that out? thanks a lot.

function main

n = 6; 
long_min = 1.;
lat_min  = 1.;
w = 2.;
h = 1.;
bound.xmin = long_min;
bound.xmax = long_min + w;

bound.ymin = lat_min;
bound.ymax = lat_min + h;


% generating the sample points 
long  = long_min + w * rand(1,n);
lat   = lat_min  + h * rand(1,n); 

%structure arrays

pts = struct('num',{},'x',{},'y',{});
for i=1:n
    pts(i).num=i;
    pts(i).x=long(i);
    pts(i).y=lat(i);
end

a = 5;
    
for i = 1:n
    out(i) = near_pt(pts(i).x, pts(i).y, pts(a).x, pts(a).y)  
end

% facing problem here
s = sort(out(:));
if (out(i)== s(2))
    d = [i]; % return the value of i for which out(i)== s(2)
end;
 disp(d);

end

function out = near_pt(p, q, r, s)
    out = sqrt((r - p)^2+(s - q)^2);    
end

Answer 1

[Star Strider]

What you want to do is not obvious.

If you want to know the index of the second value of ‘s’ (the second lowest value of ‘out’ with ‘out’ sorted ascending), that is striaghtforward:

[s,idx] = sort(out(:))

since ‘s’ will be the sorted values of ’out’ and ‘idx’ will be their original locations in the ‘out’ vector.

In one run of your code:

s =
     0.0000e+000
     1.2888e+000
     1.5399e+000
     1.6480e+000
     1.8415e+000
     1.8834e+000


idx =
     5
     4
     2
     6
     1
     3

so the second value of ‘s’ was originally ‘out(4)’.

EDIT —

d = idx(2)

Is that the result you want?

[Preyanka Dey]

@Star Strider Thanks a lot for the solution. Yes I want idx(2).

[Star Strider]

As always, my pleasure!

Answer 2

[David Hill]

function main
n = 6; 
long_min = 1.;
lat_min  = 1.;
w = 2.;
h = 1.;
bound.xmin = long_min;
bound.xmax = long_min + w;
bound.ymin = lat_min;
bound.ymax = lat_min + h;
% generating the sample points 
long  = long_min + w * rand(1,n);
lat   = lat_min  + h * rand(1,n); 
%structure arrays
pts = struct('num',{},'x',{},'y',{});
for i=1:n
    pts(i).num=i;
    pts(i).x=long(i);
    pts(i).y=lat(i);
end
a = 5;
    
for i = 1:n
    out(i) = near_pt(pts(i).x, pts(i).y, pts(a).x, pts(a).y); 
end
% facing problem here
s = sort(out(:));
d=find(out==s(2));
disp(d);
end
function out = near_pt(p, q, r, s)
    out = sqrt((r - p)^2+(s - q)^2);    
end

[Preyanka Dey]

@David Hill ..yes ....That works. Thank you a lot.