If i have a column vector A where A = [5;5;1;2;3;4;5;5;5;5;6;7;8] How can i find in general the length(=l) of the longest continuous string of the same number being repeated. So here in A it would be l=4...for the number 5.
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Here is one way to do it.
A = [5;5;1;2;3;4;5;5;5;5;6;7;8].'; % Transpose to row vector.
[M,V] = regexp(sprintf('%i',[0 diff(A)==0]),'1+','match'); [M,I] = max(cellfun('length',M)); M = M + 1 % Holds the max length. I = V(I) - 1 % Holds the starting index of first run of length M. V = A(I) % The value of the run.
You can use findseq
A = [5;5;1;2;3;4;5;5;5;5;6;7;8]
out = findseq(A); out = 5 1 2 2 5 7 10 4
It tells you that there are two sequences of repeated 5. The first starts at position 1 and ends in pos 2 and it's length is 2. Same interpretation for the other sequence.
A more explicit way of doing it:
lenmax = 1; len = 1;
for n = 2:numel(A) if A(n) == A(n-1); len = len+1; else if len > lenmax lenmax = len; end len = 1; end end
Well, what if you tried differencing the vector?
What happens to repeats? So now, you need only look for the length of the longest string of zeros. How can you do that?
Simplest seems to be to convert it to a logical vector, zeros and ones. So, something like...
diff(A) ~= 0
Now, how do you find strings of consecutive zeros?
What happens if you look for the pattern [1 0]?
How about the pattern [0 1]?
Must these be interleaved? Note, you only need to worry about the end points. So perhaps see what happens when you do this...
[1,(diff(A) ~= 0),1]
Given this, you should be able to write the code yourself now. TRY IT!!!!!!! The way to learn MATLAB is by doing it.
This can be done in one line:
out = max(diff(find([1,diff(A'),1])));
out = 4
This works by looking at the differences between successive values, which is zero within a run, which means that non-zero values mark the beginning and end of a run (Because you have to account for runs terminating at the beginning and end of the array you are querying, you have to append a 1 to the beginning and end of the difference array, hence the transposition of A in the command line above). By finding the difference between the indices that mark the beginning and end of a run, you get the lengths of the run. Basic idea from:
If you need to know the value of the number in the longest run, you need more code:
runs = diff(find([1,diff(A'),1]));
which gives you the lengths of all runs, and
starts = find([1,diff(A')]);
which gives you the index at which each run starts. With this done, this command gives you the value of the numbers in the longest run
top_value = A(starts(find(runs == max(runs))));
top_value = 5