How to do indexing without moving to a new variable ?

Hi,

I have an variable x=[1 2 3 4 5]

I need to find the mean of the worst half ie, (3+4+5)/2. For that I am sorting x

sort(x,'descend') ----> [5 4 3 2 1]

I need to find the mean of the first 3 numbers with only single statement

mean(sort(x,'descend')(1:3,1)) - This is not working, getting the following error

??? Error: ()-indexing must appear last in an index expression.

Appreciate the help

Karthik

Answer 1

[Matt Fig]

As far as memory goes, if you are o.k. in making the temporary variable through the other methods suggested here then you are o.k. making the temporary variable and overwriting it immediately. This is how I approach the problem when I don't want memory filled with a bunch of non-useful variables. There is no real advantage memory-wise in doing it in one line, and there may even be a penalty if one is forced to create several intermediate structures like padded arrays or cell arrays.

clear 
A = randi(bitmax,1,1e6);
M = sort(A,'descend');  % This will die immediately below.
M = mean(M(1:3))        % your workspace is 'clean' again!

Answer 2

[Azzi Abdelmalek]

sum(sort(x,'descend').*[ones(1,3) zeros(1,2)])/3

[Azzi Abdelmalek]

or using cellfun

cellfun(@(x) mean(x(1:3)),{sort(x,'descend')},'un',0)

Answer 3

[Star Strider]

My contribution:

mx = max(mean(reshape([sort(x,'descend') 0],3,2)).*[1 0]);

Answer 4

[per isakson]

AFAIK: It is not possible. Why do you need a single statement?

[Walter Roberson]

subsref() can allow you to do it, but it isn't clean.

[Karthik KJ]

Ok. The matrix is too large that I thought not to make a temp variable in between.

[Walter Roberson]

The array that is produced from the sort() occupies the same amount of memory whether it is passed on or given a name, except for a small number of bytes in the name table. The data block produced in the expression is internally pointed to both ways; when you assign an expression, a copy of the data is not made if instead the data block can be internally pointed to.

Answer 5

[Matt J]

A slightly more exotic solution using this class

http://www.mathworks.com/matlabcentral/fileexchange/26570-direct-indexing-of-function-calls-oop-exercise

result = mean(flib.S.sort{x,'descend'}(1:3));

Although, I guess if you were allowed to use additional mfiles, there are much more trivial ways...

Answer 6

[Chris A]

Does this work?

m = [5 4 3 2 7 1 6 8 9 10];
sum(sort(m)'.*vertcat(ones(numel(m)-numel(m)/2,1),zeros(numel(m)/2,1))),

It does create another matrix,but it is a single line of code.

Chris

[Matt J]

It works, but it is the same as Azzi's solution, given earlier.