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

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Stefan
Stefan 2014-8-13
评论： Stefan 2014-8-13
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.

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回答（4 个）

Matt J
Matt J 2014-8-13
编辑：Matt J 2014-8-13
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 );
Ahmet Cecen
Ahmet Cecen 2014-8-13
编辑：Ahmet Cecen 2014-8-13
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.
Azzi Abdelmalek
Azzi Abdelmalek 2014-8-13
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
Andrei Bobrov
Andrei Bobrov 2014-8-13
编辑：Andrei Bobrov 2014-8-13
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

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