Finding a maximum/minimum

Dear all,

I got a matrix (31225x1) which consists of closing price values (each row is one working day, so one closing price value in each row).

Now, at time "t", I want to find the maximum (minimum) which is determined as the most recent closing price (so nearest to time "t") that is greater (less) than the "c" previous closing prices (so before time "t"), where "c"=[5 10 25 50];.

I'm struggling with getting it working (efficiently) with the "max" and "min" functions. Does somebody has a suggestion how to do this?

Kind regards, Wildman

Answer 1

[Matt J]

Something like this perhaps

        %%fake data
        cvals=[5, 10,25,50];
        Prices=[0;sort(rand(31225,1));1]*100; %fake prices
          N=length(Prices);
        days=1:N;
        t=days(1:end-1)+rand(1,N-1);

   tic;

      %%algorithm
      T=toeplitz(Prices,[Prices(1),nan(1,49)]);
      Tmin=cummin(T,2); Tmin=Tmin(:,cvals);
      Tmax=cummax(T,2); Tmax=Tmax(:,cvals);

      [~, ~, bin]=histcounts(t,days);
           %
           %This might be a better way to compute 'bin':
           %     bin=interp1(days,t,'previous');

      minvals=Tmin(bin,:);  %the result
      maxvals=Tmax(bin,:); 

   toc;

[Wildman]

Thanks for your time and reply!

If I run the code I get the following error:

>>[~, ~, bin]=histcounts(t,A);
Undefined function 'histcounts' for input arguments of type 'double'.

The same goes for the "cummin" and "cummax" functions. I'm not sure what the code exactly does?

To clarify more, I made a graphical example of what values I need to find (see attachment).

- On time "t1" the code has to produce the (maximum) value of "max1", because it's the most recent closing price that is greater than the previous "c" closing prices. And as well the (minimum) value of "min1", because it is the most recent closing price that is less than the previous "c" closing prices.

"min1" will be in this case just the row before time "t1". "max1" will be the row where the first derivative of the price graph is zero, so were the graph is flat (the "real" maximum in this case).

Same procedure goes for t2.

I need to find those values because I'm coding a Support and Resistance trading rule: Buy when the closing price exceeds the maximum price and sell when the closing price is less than the minimum price. ("t1" will be a sell and "t2" will be a buy). But it's tricky to find the values of the maximum/minimums.

[Matt J]

Yes, I understood what you are trying to do. Can you upgrade to a more recent version of MATLAB (R2015)? If efficiency and speed is what you're looking for, that's probably best, and it will make histcounts, cummin, cummax available to you.

[Matt J]

If you can't upgrade to R2015, here are provisional definitions of histcounts, cummin, cummax that you can supply.

    function Amin = cummin(A,~)

        Amin=A;
        for i=2:size(A,2)

            Amin(:,i)=min(A(:,1:i),[],2);

        end

     function Amax = cummax(A,~)

        Amax=A;
        for i=2:size(A,2)

            Amax(:,i)=max(A(:,1:i),[],2);

        end   

     function [p,q,bin] = histcounts(x,e)
        q=[];
        [p,bin]=histc(x,e);

EDIT:

If your days(i) are consecutive integers, then instead of histcounts, you could also just do

   bin=floor(t);

This is compatible with even very old MATLAB versions.

[Wildman]

At the moment I only have R2013 to my disposal. I made use of the provisional definitions of the functions, thanks!

I'm not sure how to interpret the code, for example the "minvals" result output shows a matrix which is the original "price" matrix but shifted "c" rows down. I will look into this deeper and how to combine this to the rest of my code.

[Matt J]

It may be more convincing if you re-run the code, but with just one t value, randomly chosen from somewhere in the middle of the list of days(i). For example,

   t=rand*(N-100)+100;

Once you do this, you can check that days(bin-49:bin) and Prices(bin-49:bin) are the 50 days and prices immediately preceeding t. Then you can verify that minvals(:,1) gives the minimum Price over the first 5 preceding days, minvals(:,2) over the preceding 10 days etc...

[Wildman]

Thanks for your time again!

bin=floor(t-1);

If I change the part "t" in "t-1" (otherwise it also takes the current closing price in consideration) the code seems to find the correct maximum value where the value is greater than the "c" previous values.

However I'm not sure what the code does when taking a minimum value. In some cases it seems to take the "minimum" value of the "c" next values after that "minimum" (it needs to take the most recent closing price that is less than the "c" previous closing prices) and in other cases it seems a bit random which value it takes?

[Matt J]

You'll need to attach a .mat file with an illustration of what's going wrong on actual data. As you can see from the code, the min. operation operates on the very same data, T, as the max. operation. No reason you shouldn't be getting analogous behavior. If you look at the rows of T, you will see that they represent the stream of previous Prices starting from a given day T and extending back a further c=1,2,...49 days.

If you think the max. operation is working to your needs, however, you could just apply the same maximization steps to -Prices. In other words, flip the sign of the Prices and then flip the sign of maxvals correspondingly, minvals=-maxvals.