counting & indexing sequences of consecutive integers

Given a vector of ordered (increasing) integers, I want to identify each sequence of consecutive integers & count them. Thus, I want to extract three pieces of information from any such vector:

1) the number of sequences of consecutive integers; 2) the starting indices [in the original vector] of the sequences; 3) the length of each sequence.

The below code works but I had to include the step "Ldx(end) = Ldx(end) + 1" to make it work because the difference function I created (Idiff) gives a duplicate value at the end instead of the 'next' [non-existent] index, which would be ideal. (I included the step "Idx = Idx(1:end-1)" here to rid myself of the extra difference value but it's unnecessary; I get the number of sequences (ndx) from the Ldx vector and I can just ignore that last value in Idx.)

I've included two sample vectors for anyone who wants to try this out. Have fun.

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% First: define a necessary "differencing" function

x = []'; Idiff = @(x) [1; find(diff(x)-1)+1; length(x)];

samp = [11 13 14 15 16]'; smp2 = [11 13 14 15 16 20]';

Idx = Idiff(samp); Ldx = diff(Idx); Ldx(end) = Ldx(end) + 1; Idx = Idx(1:end-1); ndx = length(Ldx);

Id2 = Idiff(smp2); Ld2 = diff(Id2); Ld2(end) = Ld2(end) + 1; Id2 = Id2(1:end-1); nd2 = length(Ld2);

[Adam]

What is the question you are asking exactly? You seem to have just posted code that you say works even if it isn't 100% ideal.

[Dean Ranmar]

Sorry! My question is: is there a way to re-write the difference function (Idiff) OR a way to use it to get my desired results (#1-3 as shown) without having to resort to adding 1 to the last Ldx value? (You're right that I'm bothered by it not being 'ideal.') I thought maybe someone could improve on what I have. Thanks.

Answer 1

[Jan]

With FEX: RunLength (if you have problems compiling the C-Mex file, use RunLength_M instead):

[B, N, Pos] = RunLength(diff(samp));
conseq = (B == 1);
Number = sum(conseq)
Index  = Pos(conseq)
Length = N(conseq) + 1