how to calculate similatity

i have 2 matrix i want to calculate the similarty

i tried this code but they give me false values(negative values)

x=[2 33;2 50;2 88;3 89;4 3;4 15;6 10;6 133;8 134;8 247;8 249;8 289;9 103;10 34;10 189;11 49];
X=[10 278;11 77;11 97;11 165;11 226;11 279;12 179;14 97;14 143;16 181;16 211;17 52;17 212;19 145];
m=length(x)
for i=1:m
s1=sum(x(i)-X(i));
s2=sqrt(sum((x(i)-X(i)).^2));
S(i)=s1/s2;
end

[Jos (10584)]

How do your variables m1 and m2 relate to x and X?

Also note that m1(i) is only one value, equal to sum(m1(i))! You want to calculate m1(1)+m1(2)+m1(3)+..., so sum(m1(1:i)) ...

[kmla]

i modified it

Answer 1

[Walter Roberson]

You have arrays with 2 columns. Why are you indexing them with only a single index?

The formula you show appears to be valid only for two vectors that are the same size. You appear to have two 2D arrays that are of different size.

It is difficult to interpret that formula because it uses i as the variable of summation in both places, making it unclear which i in the second summation is the i from the first summation and which is from the second summation. This is important because the convention is that a summation continues over all terms in the same linear subexpression unless there are brackets that limit the summation the way that [] limit the second summation so that it is clear that the ^0.5 applies to the result of the second summation.

My guess is that the formula has not been written properly and that there should be [] around the first summation ending before the / . If I am correct then one way of writing the expression for the first column would be

m = min(size(m1,1), size(m2,1));
dot(m1(1:n,1),m2(1:n,1)) ./ sqrt(dot(m1(1:n,1).^2,m2(1:n,1).^2))

[kmla]

what is n

[Walter Roberson]

m = min(size(m1,1), size(m2,1));
dot(m1(1:m,1),m2(1:m,1)) ./ sqrt(dot(m1(1:m,1).^2,m2(1:m,1).^2))

Answer 2

[elham kreem]

first : if you change name of variables as x ,y then two variables must the same linghth ,they are not the same try this code :

x=[2 33;2 50;2 88;3 89;4 3;4 15;6 10;6 133;8 134;8 247;8 249;8 289;9 103;10 34;10 189;11 49];

y=[10 278;11 77;11 97;11 165;11 226;11 279;12 179;14 97;14 143;16 181;16 211;17 52;17 212;19 145];

for i = 1 : 14

s = (sum(x(i)*y(i)) ) /  (sum((x(i)^2)*(y(i)^2)))^(0.5)

end

with best

[Walter Roberson]

This code overwrites all of s each iteration of the for i loop.

For any given i, x(i)*y(i) is a scalar so there is no point in using sum(x(i)*y(i))

[elham kreem]

what about :

s=sum(sum(x'*y))/ sum ((sum(x.^2'*y.^2)).^(0.5))

[Walter Roberson]

You have to modify that to account for the columns.

I compare some versions of the code:

m1=[2 33;2 50;2 88;3 89;4 3;4 15;6 10;6 133;8 134;8 247;8 249;8 289;9 103;10 34;10 189;11 49];
m2=[10 278;11 77;11 97;11 165;11 226;11 279;12 179;14 97;14 143;16 181;16 211;17 52;17 212;19 145];

m = min(size(m1,1), size(m2,1));
cols = size(m1,2);

Sdot = zeros(1,cols);
for C = 1 : cols
    Sdot(C) = dot(m1(1:m,C),m2(1:m,C)) ./ sqrt(dot(m1(1:m,C).^2,m2(1:m,C).^2));
end
disp('dot (WDR)')
Sdot %#ok<NOPTS>

Slong = zeros(1,cols);
for C = 1 : cols
    tnum = 0;
    tden = 0;
    for i = 1 : m
        tnum = tnum + m1(i, C) .* m2(i, C);
        tden = tden + m1(i, C).^2 .* m2(i, C).^2;
    end
    Slong(C) = tnum ./ sqrt(tden);
end
disp('long form')
Slong %#ok<NOPTS>

Sshort = zeros(1,cols);
for C = 1 : cols
    Sshort(C) = sum(sum(m1(1:m,C)'*m2(1:m,C)))/ sum ((sum(m1(1:m,C).^2'*m2(1:m,C).^2)).^(0.5));
end
disp('Short form (Elham Kreem)')
Sshort %#ok<NOPTS>

Sshorter = zeros(1,cols);
for C = 1 : cols
    Sshorter(C) = m1(1:m,C).'*m2(1:m,C) ./ sqrt(m1(1:m,C).^2.'*m2(1:m,C).^2);
end
disp('Shorter form (WDR)')
Sshorter %#ok<NOPTS>

Sloopless = diag( m1(1:m,:).'*m2(1:m,:) ./ sqrt(m1(1:m,:).^2.'*m2(1:m,:).^2)).';
disp('loopless (WDR)');
Sloopless  %#ok<NOPTS>

[elham kreem]

you are a gentleman and intelligent