mean calculation by comparison

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Thishan Dharshana Karandana Gamalathge 2015-4-25

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评论： Mohammad Abouali 2015-4-25

Hi, In the following vectors, I need to get the "mean" of y values for each set of similar x values. What I mean is I need the avg of Y for each group of x. I also need to save those mean Y values.

x=[3 3 3 4 4 5 5 5 5 3 11]; y=[1 2 4 5 7 1 1 8 10 10 1];

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Answer 1

Star Strider 2015-4-25

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编辑：Star Strider 2015-4-25

在 MATLAB Online 中打开

This looks like an application for accumarray:

x=[3 3 3 4 4 5 5 5 5 3 11]; 
y=[1 2 4 5 7 1 1 8 10 10 1];
y_means = accumarray(x', y, [], @mean);

EDIT:

To get a summary of the x-values (first column) and the corresponding y_means values (second column:

xu = unique(x);
ys_means = [xu' y_means(xu)]

produces:

ys_means =
              3         4.25
              4            6
              5            5
             11            1

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Star Strider 2015-4-25

在 MATLAB Online 中打开

My pleasure.

I must have misunderstood your intent with ‘group’. To me, that meant grouping them even if they were not contiguous. The alternative to accumarray would likely be a series of loops. If x-values are negative, it would not be possible to use accumarray, without offsetting them temporarily to make them all positive integers.

If single isolated values are to be counted separately, they would have to be identified as such and then given separate designations in x, such as their corresponding index values. I did that here.

I don’t know how robust this would be (it works here), but it solves the isolated non-grouped index problem by assigning that value by its index (here, 10):

x=[3 3 3 4 4 5 5 5 5 3 11]; 
y=[1 2 4 5 7 1 1 8 10 10 1];
[xu,ia,ic] = unique(x);
xm = arrayfun(@isequal, repmat(unique(ic),1,length(ic))', repmat(ic,1,length(xu)));
xd = diff([zeros(1,size(xm,2)); xm]);           % Find Transitions
iv = 1:size(y,2);                               % Index Vector
tpos = xd>0;                                    % Positive Transitions
tneg = xd<0;                                    % Negative Transitions
cntg = sum(tpos + circshift(tneg, [-1 0]),2);   % Isolated Values = 2
x(cntg == 2) = iv(cntg == 2);                   % Set Isolated Values = Index
y_means = accumarray(x', y, [], @mean);         % Accumulate ‘y’ By ‘x’
ys_means = [xu' y_means(xu)]                    % Isolate Non-Zeros

producing:

ys_means =
              3       2.3333
              4            6
              5            5
             10           10
             11            1

The code is not as efficient as I would like it to be, but then this problem does not lend itself to concise solutions.

Star Strider 2015-4-25

I may not fully understand exactly what you want.

There is an Answer to your other, somewhat revised and updated version of this Question, that appears to be more efficient than my revised code, but it is not obvious to me how he would also supply the corresponding indices. Before delving into this myself, I’ll see what he comes up with.

Meanwhile, I will think about this and consider an entirely new effort and a new approach in light of your new requirements.

I do appreciate your already having Accepted my Answer.

Mohammad Abouali 2015-4-25

Well, updated the answer, to include both start and end index of the block, as well as x-new.

As star strider mentioned, my answer on the other post is solving the same problem but using a different approach.

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