Find max values of n previous values in a matrix without a loop.

Hello,

I have a single column matrix and need to add a second column containing on each row the highest value of n previous row.

For example, my matrix would have this column:

Column1

1

4

3

6

3

5

4

7

2

5

6

1

I am looking to add a column containing the high value of 3 previous rows (including the current row) in the first column, so that I would obtain:

Column1 Column2

1 NaN

4 NaN

3 4

6 6

3 6

5 6

4 5

7 7

2 7

5 7

6 6

1 6

Is it possible in Matlab without a for to loop?

Many thanks,

Libor

Answer 1

[José-Luis]

a = randi(10,10,1);
your_result = max([a, circshift(a,1), circshift(a,2)],[],2);
your_result(1:2) = NaN;

[Libor Cerny]

Thank you so much, Jose-Luis!

Am I correct to think that this would be faster than using a for to loop?

[José-Luis]

My pleasure.

I don't know if this would be faster than a loop. Best way to find out is to test it. There are probably more efficient ways, in any case. arrayfun() is probably not one of them.

[Libor Cerny]

OK, thanks.

Also, how would this scale to an arbitrary number of values to compare? I gave the value of 3 as an example, but realistically, I would need this code to work with a lookback of up to 50 or so...

[Torsten]

https://de.mathworks.com/help/matlab/ref/movmax.html

Best wishes

Torsten.

[Libor Cerny]

Thank you Thorsten, this is actually the best answer, I believe. I don't think I can accept it as such, though, as you entered it in a comment and not as an answer...

[José-Luis]

If you want to set the number of shifts programatically;

numRep = 50;
a = randi(10,10,1);
xx = size(a,1);
idx = bsxfun(@minus,[2:xx+1]',1:numRep);
idx(idx<1) = xx - abs(mod(idx(idx<1), -xx))
your_array = max(a(idx),2);

[Libor Cerny]

Thank you again, Jose-Luis! I really appreciate your help!

[José-Luis]

My pleasure. I had no idea something like movmax() existed, and it is indeed a better answer.

It was fun coming up with this anyway.

Answer 2

[Stephen23]

The simplest solution is to download Jos's excellent FEX submission slidefun, which works for all MATLAB versions (movmax was introduced in R2016a):

https://www.mathworks.com/matlabcentral/fileexchange/12550-slidefun

V = [1;4;3;6;3;5;4;7;2;5;6;1];
N = 3; % window size
W = slidefun(@max,N,V,'backward');
W(1:N-1) = NaN;

And checking the answer:

 >> [V,W]
 ans =
     1   NaN
     4   NaN
     3     4
     6     6
     3     6
     5     6
     4     5
     7     7
     2     7
     5     7
     6     6
     1     6

[Libor Cerny]

Thank you very much Stephen, this also does exactly what I need, although as per Torsten's comment above, there are apparently built-in and optimized functions in Matlab now. Thanks again!

Answer 3

[Guillaume]

Actually, the simplest solution is to use movmax (since R2016a):

v = [1 4 3 6 3 5 4 7 2 5 6 1].';
n = 3;
[v, [nan(n-1, 1); movmax(v, n, 'EndPoints', 'discard')]]

edit: which I see was already suggested by Torsten in the hidden comments

[Libor Cerny]

Yes. :)

But thank you anyhow, Guillaume! I really appreciate all help. Et... Bonne annee! ;)