Calculate how often a 1 turns into a 0 or 1, and vice versa

Hi all,

I've got some data that characterize two possible states and stored in an array (1s and 0s), and there are 8 repetitions per recording. The data look something like this:

a = [1 1 1 0 1 0 0 1; 1 1 0 1 1 0 1 0];

Where I perform 8 experimental repetitions (2 shown).

I've got a hunch that a 1 is followed by a 0 more often than by a 1, and want to quantify that.

Let's say a 1 has a 65% chance of being followed by a 1, and a 35% chance of being followed by a 0. Additionally, a 0 might have a 90% chance of being followed by a 1, and a 10% chance of being followed by a 0.

What's an efficient way to ask MATLAB if this is actually the case?

So far, I'm thinking maybe use strfind with 4 repetitions, 1 for each possibility, like:

strfind(a,[1,1]);
strfind(a,[1,0]);
strfind(a,[0,1]);
strfind(a,[0,0]);

Thanks for your thoughts.

Answer 1

[Matt J]

   bayes=@(u,v) 100*sum(u&v,2)/sum(v,2);  %conditional prob as percent

   first1=a(:,1:end-1);
   next1=a(:,2:end);
   first0=~first1;
   next0=~next1;

    out_1_1 = bayes( first1 & next1, first1  );
    out_0_0 = bayes( first0 & next0, first0  );
    out_1_0 = bayes( first1 & next0, first1 );
    out_0_1 = bayes( first0 & next1, first0 );

[Stefan]

Thanks! I'll have to think through this. Much appreciated.

Answer 2

[Ahmet Cecen]

 check1=(a+circshift(a,[-1 0]));
 check2=(a-circshift(a,[-1 0]));

 check1==0 % is 0's followed by 0's
 check1==2 % is 1's followed by 1's
 check2==1 % is 1's followed by 0's
 check2==-1 % is 0's followed by 1's

Very fast, elegant, no explicit loops. You will have to ignore the values at the last column, since that doesn't make sense anyways.

[Matt J]

no explicit loops

You could avoid implicit loops too by doing

 check1=conv2(a,[1,1],'valid');
 check2=-diff(a,1,2);

[Stefan]

Learning lots of new functions, thanks!

Answer 3

[Azzi Abdelmalek]

a = [1 1 1 0 1 0 0 1; 1 1 0 1 1 0 1 0]
for k=1:size(a,1)
  out{1,k}=numel(strfind(a(k,:),[1,1]))
  out{2,k}=numel(strfind(a(k,:),[1,0]))
  out{3,k}=numel(strfind(a(k,:),[0,1]))
  out{4,k}=numel(strfind(a(k,:),[0,0]))
end

Answer 4

[Andrei Bobrov]

a = [1 1 1 0 1 0 0 1; 1 1 0 1 1 0 1 0];

n = size(a)-[0 1];
da = diff(a,1,2);
t = [ones(n(1),1) da] == 0;
[r,~] = find(t);
da2 = a.*t;
da2 = da2(t);

out_1_1 = accumarray(r,da2 == 1)/n(2)*100;
out_0_0 = accumarray(r,da2 == 0)/n(2)*100;
out_1_0 = sum(da == -1,2)/n(2)*100;
out_0_1 = sum(da == 1,2)/n(2)*100;

with strfind

z = cellnum([0 0; 1 1; 0 1; 1 0],2)';
n = [size(a,1),numel(z)];
out = zeros(n);
for ii = 1:n
    out(ii,:) = cellfun(@(x)numel(strfind(a(ii,:),x))/(n(2)-1)*100,z);
end