calculating Kernel density for each column
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Hi
Please how do I need a short code that will calculate the KDE of each column in the R.length data below
the KDE is given as = I/n*h sum ( K * (( v - i )/h) which is computed for each column
where h = 1.06 * variance * (n^(-0.2)) for each colum
n is the number of each column
i = first, second, third, fourth, fifth, sixth number of each column
v =pv is given as 3, 4, 5, 6 for each column
Thanks in advance
jonathan
R = [ 0.6164 3.4161 0.9950 3.4117;
3.1654 0.4123 4.2391 1.0198;
0.5745 3.0364 1.3191 3.1129;
2.9883 0.7348 3.8730 0.4123;
0.9381 3.3749 2.0421 3.5014;
2.1817 1.0630 3.0643 0.9487];
5 个评论
David Wilson
2019-4-24
Would ksdensity work? (From stats toolbox)
Rik
2019-4-24
This is what I have done so far and the answers are below
thanks
n = 6;
K = 3;
h1 = 1.06 * var(Z(:,1)) * (n ^ 0.2);
h2 = 1.06 * var(Z(:,2)) * (n ^ 0.2);
h3 = 1.06 * var(Z(:,3)) * (n ^ 0.2);
h4 = 1.06 * var(Z(:,4)) * (n ^ 0.2);
z1 = 1/ (n * h1);
z2 = 1/ (n * h2);
z3 = 1/ (n * h3);
z4 = 1/ (n * h4);
Ec11 = (K * (v1 - Z(1,1))/h1) ;
Ec12 = (K * (v1 - Z(2,1))/h1) ;
Ec13 = (K * (v1 - Z(3,1))/h1) ;
Ec14 = (K * (v1 - Z(4,1))/h1) ;
Ec15 = (K * (v1 - Z(5,1))/h1) ;
Ec16 = (K * (v1 - Z(6,1))/h1) ;
e1 = [Ec11; Ec12; Ec13; Ec14; Ec15; Ec16];
e1 = sum (e1);
Ec21 = K * (v2 - Z(1,2)/h2);
Ec22 = (K * (v2 - Z(2,2))/h2) ;
Ec23 = (K * (v2 - Z(3,2))/h2) ;
Ec24 = (K * (v2 - Z(4,2))/h2) ;
Ec25 = (K * (v2 - Z(4,2))/h2) ;
Ec26 = (K * (v2 - Z(4,2))/h2) ;
e2 = [ Ec21; Ec22; Ec23; Ec24; Ec25; Ec26];
e2 = sum (e2);
Ec31 = K * (v3 - Z(1,3)/h3);
Ec32 = (K * (v3 - Z(2,3))/h3) ;
Ec33 = (K * (v3 - Z(3,3))/h3) ;
Ec34 = (K * (v3 - Z(4,3))/h3) ;
Ec35 = (K * (v3 - Z(5,3))/h3) ;
Ec36 = (K * (v3 - Z(6,3))/h3) ;
e3 = [Ec31; Ec32; Ec33; Ec34; Ec35; Ec36];
e3 = sum (e3);
Ec41 = K * (v4 - Z(1,4)/h4);
Ec42 = (K * (v4 - Z(2,4))/h4) ;
Ec43 = (K * (v4 - Z(3,4))/h4) ;
Ec44 = (K * (v4 - Z(4,4))/h4) ;
Ec45 = (K * (v4 - Z(5,4))/h4) ;
Ec46 = (K * (v4 - Z(6,4))/h4) ;
e4 = [Ec41; Ec42; Ec43; Ec44; Ec45; Ec46];
e4 = sum(e4);
k = e1;
l = e2;
m = e3;
b = e4;
% the kernal density estimation
KDE1 = k * z1;
KDE2 = l * z2;
KDE3 = m * z3;
KDE4 = b * z4;
answer
-0.4881
-0.1668
-0.7734
-0.3972
Rik
2019-4-24
Instead of using numbered variables, why don't you process the columns in a loop?
Tino
2019-4-24
采纳的回答
Rik
2019-4-24
I am assuming the v values are the same as the column index, and that you made a mistake with the code for the second column.
Z = [ 0.6164 3.4161 0.9950 3.4117;
3.1654 0.4123 4.2391 1.0198;
0.5745 3.0364 1.3191 3.1129;
2.9883 0.7348 3.8730 0.4123;
0.9381 3.3749 2.0421 3.5014;
2.1817 1.0630 3.0643 0.9487];
n = 6;
K = 3;
z=zeros(1,size(Z,2));e=zeros(size(z));
for col=1:size(Z,2)
v=col;%is this what you mean?
h = 1.06 * var(Z(:,col)) * (n ^ 0.2);
z(col) = 1/ (n * h);
E = K * (v - Z(:,col))/h;
if col~=1
%did you mean for this to be different?
E(1)= K * (v - Z(1,col)/h);
end
e(col) = sum(E) ;
end
% the kernal density estimation
KDE = e .* z;
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