For loop to calculate distance between a 5d dimensional data point and another matrix

I am trying to create a for loop to calculate the distance between my matrix 'T' and another matrix 'centroids'.

I dont know how to add in another loop to include the matrix 'centroids' Can someone help me. please see the code i have completed so far. If someone could explain where i am going wrong and tell me what to add to fix it, then i would really appreciate it.

T1 = [1 4 3 100 12 1000 25 32 500 700 20 9000 25 650 40 3 67 2 305 80 200 50 2 15 4]
T2 = [1 6 2 120 14 900 25 35 550 650 23 7500 25 700 43 6 72 8 360 70 220 60 1 14 8]
T3 = [1.5 5 7 150 13 550 1 40 520 600 27 5500 27 900 44 12 75 9 360 50 190 50 4 13 6]
T4 = [122 120 120 170 140 700 100.5 134 570 600 125 6100 125 1050 122 130 187 109 460 145 330 160 105 113 104]
T5 = [25 25 30 40 50 600 6 33 500 600 26 5500 26 1150 14 32 126 8 360 28 230 60 2 13 5]
T=[T1;T2;T3;T4;T5]
display (T)    % (notes: T is a 5*25 matrix)
centroids=[T1(1:3);T2(1:3);T3(1:3);T4(1:3);T5(1:3)]  (notes: centroids is the first 3 colums of T)
display (centroids)
T=[T1;T2;T3;T4;T5]
[m n] = size(T);
for i=1:(n-1)
   b = T(:,i);
   c = T(:,i+1);
   d = 0;
for j=1:m
 d = d + (b(j)-c(j))^2;
 end
 e(i)=sqrt(d);
 end
display(e)

[KSSV]

How to point the distance ?

[Deborah O'Connor]

I dont know what you mean

[KSSV]

T=[T1;T2;T3;T4;T5]  % 5*25 size 
C = T(:,1:3) ;  % centroids 5*3 size 
P = T(:,4:end) ;   % points 5*22 

There is missmatch in dimensions......how you would expect to get distance?

[Deborah O'Connor]

thats what i dont know. here is the exact question i am being asked:

'use a for loop to look at each column of T in turn

Each of the 25 columns describes a 5 dimensional data point

complete the code so that:

inside the for-loop you calculate the distance between the 5 dimensional data point and each of the 3 centroids

then assign to the corresponding component of the vector cluster_assigned either the number 1, 2 or 3 depending on which cluster is nearest

[Deborah O'Connor]

Can anyone help me with the above query? thanks

Answer 1

[the cyclist]

The distance from the i-th point of T to the j-th centroid is

sqrt(sum((T(:,i) - centroids(:,j)).^2)); % Check my formula here, but I think that's right.

So, if that's the distance for one T point, from one centroid, you just need to loop over all points:

for i = 1:size(T,2)
    for j = 1:size(centroids,2)
        % ...
    end
end

I don't think you need your other loop down the first dimension of T, because that is covered by the sum command above, which is vectorized.

That's a pretty big hint, so I hope that is enough to get you going.

[Deborah O'Connor]

Thank you very much for your help. Your name 'the cyclist' is this because you like to cycle. Im a keen cyclist myself.

I dont know what you mean when you say I dont need the other loop down the first dimension of T. Is it this section you are referring to:

b = T(:,i);

c = T(:,i+1);

d = 0;

if I take this section out i am left with the code:

for i=1:size(T,2)

for j=1:size(centroids,2)

e=sqrt(sum((T(:,i) - centroids(:,j)).^2));

end

display (e)

but this gives me a message: 'parse error' so i guess i need something else in here

[Deborah O'Connor]

Ok I think we are getting somewhere. It has given me 75 data points.

I guess the next stage is to group them into 1 of the 3 clusters

[the cyclist]

I would say the code has calculated 75 distances, not "data points". That was the purpose.

Yes, now you have the distance of each point to each centroid, so you need to figure out which centroid each poin is closest to. (You could do that within the loop over the points.)

My nickname "the cyclist" is very old -- at least 20 years! It referred to three interests of mine that were cyclical in nature: cycling, juggling, and astronomy. None of those are serious hobbies anymore. :-/

[Deborah O'Connor]

Thanks so much for your help. Though Im still not sure the code is correct. Is the code taking the sum of each column in centroids and comparing it to the sum of each column in T?

[the cyclist]

Take the code apart, from the "inside out":

T(:,i)

is the 5-d coordinate vector of the i'th point. Similarly,

centroids(:,j)

is the coordinate vector of the j'th centroid.

T(:,i) - centroids(:,j)

is the vector of the differences of those coordinates.

(T(:,i) - centroids(:,j)).^2

is the square of each of those differences.

sum((T(:,i) - centroids(:,j)).^2)

is the sum of those squares. Finally,

sqrt(sum((T(:,i) - centroids(:,j)).^2))

is the square root of that sum. That's the distance formula (assuming I did not make some silly mistake). It is the 5-d equivalent of the formula

sqrt( (x1-x2).^2 + (y1-y2).^2 ) % distance formula in two dimensions

[Deborah O'Connor]

I hope I did'nt offend you. It is me who is not sure what i am doing. Big big apologies if i offended you, it was not intentional. I am extremely grateful for your help. Thank you so much for explaining this for me. I cannot thank you enough.

[the cyclist]

I'm not offended at all. I'm sorry if I gave that impression.

Answer 2

[Deborah O'Connor]

I am happy you are not offended. Just making sure. I might need your help if i cant work out how to do the next part (assigning the results to clusters 1 2 and 3). Maths is my weakpoint so it takes me a long time to get my head around things