Dear All,
I would appreciate your help.
I have a vector (e.g., x=[4 6 2 3 3 1 5 3 4] )
I need to calculate the state-transition matrix, that is, how many times does each number follow each number. In other words, what are the chances of 3 to come after 1, what are the chances of 2 to come after 5, etc.
This may seem easy in a small vector, but my vectors each have about 20,000 numbers, so an efficient system will help a lot.
Thanks!

Aren't you basically just calculating a histogram?

TTFN

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A brute force program in Octave for 20000 numbers 1-6 took 2.2 seconds to run and 2 minutes to write.

The only line of interest in the entire exercise is

++x(z(i),z(i+1));

and that is about as interesting as lettuce.

Cheers

Greg Locock

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The same prog in matlab runs in 0.016 s, although I had to rewrite the interesting line because matlab doesn't like ++.

So, how fast, exactly, do you need this algorithm to be?

 

Cheers

Greg Locock

SIG:Please see FAQ731-376: Eng-Tips.com Forum Policies for tips on how to make the best use of Eng-Tips.

Histogram it is. My first thought was this crude code:

clear
clc
x=1:6;
for i=1:2000
    y(i)=randsample(x,1);
end
su=2000/6;
z1=zeros(1,6);
z2=z1;z3=z2;z4=z3;z5=z4;z6=z5;

for i=1:1999
    if y(i)==1
        for k=1:6
            if y(i+1)==k
             z1(k)=z1(k)+1;
            end
        end
    end
    
        if y(i)==2
        for k=1:6
            if y(i+1)==k
             z2(k)=z2(k)+1;
            end
        end
        end
        
        if y(i)==3
        for k=1:6
            if y(i+1)==k
             z3(k)=z3(k)+1;
            end
        end
        end
        
        if y(i)==4
        for k=1:6
            if y(i+1)==k
             z4(k)=z4(k)+1;
            end
        end
        end
        
        if y(i)==5
        for k=1:6
            if y(i+1)==k
             z5(k)=z5(k)+1;
            end
        end
        end
        
        if y(i)==6
        for k=1:6
            if y(i+1)==k
             z6(k)=z6(k)+1;
            end
        end
        end      
end

figure(1)
bar(x,z1), title('followers of 1')
figure(2)
bar(x,z2), title('followers of 2')
figure(3)
bar(x,z3), title('followers of 3')
figure(4)
bar(x,z4), title('followers of 4')
figure(5)
bar(x,z5), title('followers of 5')
figure(6)
bar(x,z6), title('followers of 6')


It's quick and was simple to write. From each z (ie. z1 to z6) you have the number of times each number follows it. Knowing the total population size and the variation in numbers you can easily get a percentage of what number follows what.
Greg- I'm sure your code is better. But I'm not sure how you utilized that interesting line.




 

Fe

Also should be able to put those 6 'if' loops in a single 'for' loop to compress the size.
just a thought,
 

Fe

num_spots=input('How many spots   ');
num_trials=input('How many trials  ');

for i=1:num_spots
    for j=1:num_spots
        x(i,j)=0;
    end
end

for i=1:num_trials
    z(i)=1+floor(num_spots*rand(1));
end

tic
for i=1:(num_trials-1)
    x(z(i),z(i+1))=x(z(i),z(i+1))+1;%was the interesting line
end
toc
x

Cheers

Greg Locock

SIG:Please see FAQ731-376: Eng-Tips.com Forum Policies for tips on how to make the best use of Eng-Tips.

Thanks a lot guys!

Glad to help. By the way, I always thought the term 'state-transition matrix' was used in control engineering.

Fe

That's why I didn't reply.  Had I known it was a regular ML golf problem, I may have played.

- Steve

There's a state; it transitions to another state...

TTFN

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BrCo:

This exact question came up on CSSM last year.  There are many ways to solve these sorts of problems.  My favourite is to use the sparse() function:

% Some data with repeated sequences

x=[1 6 1 6 4 4 4 3 1 2 2 3 4 5 4 5 2 6 2 6 2 6];

sparse(x(1:end-1),x(2:end),1)

ans =

   (3,1)        1
   (6,1)        1
   (1,2)        1
   (2,2)        1
   (5,2)        1
   (6,2)        2
   (2,3)        1
   (4,3)        1
   (3,4)        1
   (4,4)        2
   (5,4)        1
   (6,4)        1
   (4,5)        2
   (1,6)        2
   (2,6)        3

Or if you want it in array form, where A(i,j) holds the number of changes from i to j:

A=full(sparse(x(1:end-1),x(2:end),1))

A =

     0     1     0     0     0     2
     0     1     1     0     0     3
     1     0     0     1     0     0
     0     0     1     2     2     0
     0     1     0     1     0     0
     1     2     0     1     0     0


 

- Steve

Nice and simple. I never used sparse before.
What is CSSM?

Fe

comp.soft-sys.matlab is a USENET newsgroup for matlab discussions.  It can be accessed via any regular newsreader.  You can also access it through a web browser pointed to sites like

http://www.mathworks.com/matlabcentral/newsreader/
http://groups.google.com/group/comp.soft-sys.matlab/topics

It is very active!

- Steve

I was waiting for the one line answer!

Cheers

Greg Locock

SIG:Please see FAQ731-376: Eng-Tips.com Forum Policies for tips on how to make the best use of Eng-Tips.

It was my answer in CSSM, so I'm not plagiarising, just copying.

 

- Steve

Sorry, forgot a star for you. Slow isn't it?

where is that construction of x(1:end-1),x(2:end)

documented? What I mean is how does it 'know' that it can and should increment the pointer in each of the two x() statements together?


 

Cheers

Greg Locock

SIG:Please see FAQ731-376: Eng-Tips.com Forum Policies for tips on how to make the best use of Eng-Tips.

Greg,

Just look through the documentation for sparse().  In the form I've used, x(1:end-1) and x(2:end) are just two vector inputs representing the row and column for each of the elements in the third vector.  For example, this produces a sparse version of eye(3):

sparse([1 2 3],[1 2 3],[1 1 1])

If the values of the elements are all the same, a single value can be (re)used:

sparse([1 2 3],[1 2 3],1)

This sort of assignment can't be done with normal arrays, there is no equivalent syntax (you run into all that sub2ind, ind2sub confusion).


Now the trick:

The neat (documented, but not obvious) thing about sparse()is that if a coordinate pair is repeated, the values assigned to it are summed, so this:

sparse([3 1 2 3 3],[3 1 2 3 3],1) gives

(1,1) = 1
(2,2) = 1
(3,3) = 3

So x(1:end-1) are the "from" coordinates and x(2:end) are the "to" coordinates.  Each (from,to) pair is assigned a value of 1.  Each repeat of a (from,to) pair increments the value at that location.

This is what I love about Matlab, sometimes it all just falls together.

- Steve

Thanks for the info on CSSM.
I wish I knew sparse existed before

TTFN

Fe

Ah, ok it is a 'sparse' thing, not an indexing mode that I have yet to get to grips with (ie almost all of them).

Cheers

Greg Locock

SIG:Please see FAQ731-376: Eng-Tips.com Forum Policies for tips on how to make the best use of Eng-Tips.