how to find average value of floating numbers ??

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Lee
Lee 2014-3-11
评论： Lee 2014-3-11
Hello all,
I have an n-by-2 array comprised of floating numbers.
In the second column of this array, there are some arrays having same floating numbers.
I want to sort this array with respect to the second column and find the average value of the arrays having same floating numbers.
For example,
A = [0.33,0;0.53,0.2;0.433,0.2;0.11,0.5;0.95,0.4;0.99,0.5;0.32,0;0.44,0.3;0.38,0.4;0.62,0.1;0.23,0.4;0.44,0.5;0.76,0.5]
[B, I] = sort(A(:,2));
C = A(I,:)
Sorting was done successfully, and I tried to find the average of the arrays, using accumarray(subs,val). But, because ‘subs’ always requires positive integer, I can’t use this syntax in my case of accumulation of floating numbers.
Is there any solution for this? You may as well to use other syntax…
Thanks in advance

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采纳的回答

Sean de Wolski
Sean de Wolski 2014-3-11
编辑：Sean de Wolski 2014-3-11
Use unique to generate indices of unique floating point values
A = [0.33,0;0.53,0.2;0.433,0.2;0.11,0.5;0.95,0.4;0.99,0.5;0.32,0;0.44,0.3;0.38,0.4;0.62,0.1;0.23,0.4;0.44,0.5;0.76,0.5];
[uA,~,idxA] = unique(A(:,2)); % unique values and index
muA = accumarray(idxA,A(:,1),[],@mean); % mean of values at each index
[uA muA] % display
Also, you can use sortrows() to save yourself a step up above:
C = sortrows(A,2); %sort along second column

更多回答（1 个）

Andrei Bobrov
Andrei Bobrov 2014-3-11
编辑：Andrei Bobrov 2014-3-11
A = [0.33,0;0.53,0.2;0.433,0.2;0.11,0.5;0.95,0.4;0.99,0.5;0.32,0;0.44,0.3;0.38,0.4;0.62,0.1;0.23,0.4;0.44,0.5;0.76,0.5]
[a,b,c] = unique(A(:,2),'first');
[~,i0] = sort(b);
[~,i1] = sort(i0);
anew = a(i0);
c1 = i1(c);
out = [accumarray(c1,A(:,1),[],@mean),anew]

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