CODEWITHSUNDEEP
matlab
Frank Meulenaar
Frank Meulenaar

Reputation: 1219

Get average y value per x position in Matlab

I don't know how to phrase this, but an example:

x = [1 4 4 5 5 5];
y = [5 1 3 3 4 5];

and then I'd like output

xs          = [1 4 5];
ys          = [5 2 4];
frequencies = [1 2 3]

(because the average 'y' at x=1 is 5, and the average 'y' at x=4 is (1+3)/2 = 2, and the average 'y' at x=5 is (3+4+5)/3 = 4).

I can compute this in a clumsy way but maybe there's a nice solution.

Upvotes: 3

Views: 1958

Answers (7)

ioums
ioums

Reputation: 1395

x = [1 4 4 5 5 5];
y = [5 1 3 3 4 5];
xs = unique(x);
[frequencies,bin] = histc(x,xs);
ys = arrayfun(@(i) mean(y(bin==i)), 1:length(xs));

Upvotes: 3

George
George

Reputation: 784

@ioum 's answer worked great for me, there was a small mistake though in the last line, that came up when I gave as an input other vectors than the ones posted here. For example, after deleting the last element of each vector the answer should be: 

ys = [5 2 3.5]

The slightly corrected code is:

x = [1 4 4 5 5 5];
y = [5 1 3 3 4 5];
xs = unique(x);
[frequencies,bin] = histc(x,xs);
ys = arrayfun(@(i) mean(y(bin==i)), 1:length(xs));

I tried to edit @ioum 's post, but the edit did not go through.

Upvotes: 2

Dennis Jaheruddin
Dennis Jaheruddin

Reputation: 21561

Though I would recommend one of the histogram approaches, here is how I would do it in a loop. Not that much different from some other solutions but I believe it is just a little nicer so I will post it anyway.

xs = unique(x)
for t = 1:length(xs)
   idx = x == xs(t);
   ys(t) = mean(y(idx));
   frequencies(t) = sum(idx);
end

Upvotes: 0

Muhammad
Muhammad

Reputation: 89

here is my code, hope it helps... 

   x=sort(x);
   ind=1;
    for i=1:length(x)
        if (i>1 && x(i)==x(i-1))
           continue;
        end
        xs(ind)=x(i);
        freq(ind)=sum((x==x(i)));
        ys(ind)=sum((x==x(i)).*y)/freq(ind);
        ind=ind+1;
    end

Upvotes: 0

Floris
Floris

Reputation: 46445

You can use the histogramming function histc to get each of the categories:

x = [ 1 4 4 5 5 5];
y = [ 5 1 3 3 4 5];
xs = unique(x);
[frequencies xb] = histc(x, xs); % counts the number of each unique occurrence
ysp = sparse(1:numel(x), xb, y); % a sparse matrix is a good way to organize the numbers
ys = full(sum(ysp)./sum(ysp>0)); % each column in the matrix corresponds to a "index"

This gives you the three arrays you wanted. I think this is quite clean and efficient - no looping, only four lines of code.

Upvotes: 4

user2182839
user2182839

Reputation: 25

x = [1 4 4 5 5 5]';
y = [5 1 3 3 4 5]';

%this can probably be done smarter...

indexlong=accumarray(x,x,[],@mean)'
meanlong=accumarray(x,y,[],@mean)'
frequencieslong=accumarray(x,1)'

%leave out zeros

takethese=(indexlong>0);
xs=indexlong(takethese)
ys=meanlong(takethese)
frequencies=frequencieslong(takethese)

Upvotes: 0

Roney Michael
Roney Michael

Reputation: 3994

I'm not sure if this solution would be considered elegant enough, but this should work:

x = [1 4 4 5 5 5];
y = [5 1 3 3 4 5];
[xs,I,J] = unique(x);    %The value of the index vector I is not required here.
ys = zeros(size(xs));
frequencies = zeros(size(xs));
for i = 1:max(J)
    I = find(J==i);
    ys(i) = mean(y(I));
    frequencies(i) = length(I);
end
xs,ys,frequencies

The output would be:

xs =

     1     4     5


ys =

     5     2     4


frequencies =

     1     2     3

I hope this helps.

Upvotes: 0

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