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

1 view (last 30 days)
Preyanka Dey on 4 Sep 2020
Commented: Star Strider on 4 Sep 2020
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

Star Strider on 4 Sep 2020
Edited: Star Strider on 4 Sep 2020
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 on 4 Sep 2020
@Star Strider Thanks a lot for the solution. Yes I want idx(2).
Star Strider on 4 Sep 2020
As always, my pleasure!

David Hill on 4 Sep 2020
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

#### 1 Comment

Preyanka Dey on 4 Sep 2020
@David Hill ..yes ....That works. Thank you a lot.