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

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Preyanka Dey
Preyanka Dey 2020-9-4
评论： Star Strider 2020-9-4
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

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采纳的回答

Star Strider
Star Strider 2020-9-4
编辑：Star Strider 2020-9-4
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?

更多回答（1 个）

David Hill
David Hill 2020-9-4
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

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