Compare all elements of a vector against each other

I have a vector that includes some values

v=[2.4 3.5 7.4 3.6 4.5]

each value is the mean value of a row in a matrix called y

now I want to compare each value with the other values in the vector and if they are similar (with a deviation of 10%) then delete the row in y of the larger value.

I managed to compare just the neighborhood values

for i=1:length(v)-1
    if v(i)==v(i+1)
        y(i,:)=[];
    end
end

I am struggling to find the right way to compare the values against each other and to include this 10%

Any help is much appreciated !

[José-Luis]

A deviation of 10% with respect to what? Would the comparison be recursive? How about if all values in your vector fall within that 10%?

[monmatlab]

in respect to the current value

so basically if we take the first value, which is 2.4 , to delete a row in y, the value that is compared to must be in the range of 2.16 to 2.64 . I hope it is somehow clear

[José-Luis]

Still not clear to me. How do you go over all values?

[monmatlab]

so basically you compare 2.4 to 3.5 2.4 to 7.4 2.4 to 3.6 2.4 to 4.5

then you move on to the next value 3.5 to 7.4 3.5 to 3.6 3.5 to 4.5

and so on

[jahan jahan]

Dear.

I need checking numbers inside the vector if they inside specific limit. If yes the results equal to 1 otherwise 0. The results will be new vector

Thank you

Answer 1

[Adam]

    relativeDiffs = abs( ( v - v' ) ) ./ v;  % R2016b syntax only
    indicatorMatrix = relativeDiffs <= 0.1;

will give you an indicator matrix of values to delete. Obviously you would ignore the leading diagonal as this is each value against itself. The values are normalised against the original v by row so the upper right triangle and lower left triangle will differ by the fact that e.g.

(3.5 - 2.4) / 2.4 ~= (3.5 - 2.4) / 3.5

You will get both of these answers from the relativeDiffs matrix. In some cases only one of these will highlight in the final indicator matrix as being < 10%.

relativeDiffs = abs( bsxfun( @minus, v, v' ) ) ./ v

should work for pre R2016b syntax.

Answer 2

[Guillaume]

If v is the mean of the rows (e.g. obtained with v = mean(y, 2)), shouldn't it be a column vector rather than a row vector as in your example.

In any case, if I understood correctly, this should work:

v = mean(y, 2);
inrange = v >= 0.9*v.' & v <= 1.1*v.';  %In r2016b only
inrange = inrange .* ~eye(numel(v));  %diagonal will always be 1 as a number is in range of itself. set to 0 instead.
y(any(inrange, 2), :) = [];  %delete rows of y whose mean is in range of any other row mean

In versions prior to R2016b, replace the relevant line by:

inrange = bsxfun(@ge, v, 0.9*v.') & bsxfun(@le, v, 1.1*v.');

[Pavan teja]

How to compute Mean along x and y direction