Question

Imagine you have a very long sequence. What is the most efficient way of finding the intervals where the sequence is all zeros (or more precisely the sequence drops to near-zero values abs(X)<eps):

For simplicity, lets assume the following sequence:

sig = [1 1 0 0 0 0 1 1 1 1 1 0 1 0 0 0 1 1 1 1 1 1 1 1 0 0 1 1 1 0];

I’m trying to get the following information:

startIndex   EndIndex    Duration
3            6           4
12           12          1
14           16          3
25           26          2
30           30          1

then using this information, we find the intervals with duration >= to some specified value (say 3), and returning the indices of the values in all these intervals combined:

indices = [3 4 5 6 14 15 16];

That last part is related to a previous question:

MATLAB: vectorized array creation
from a list of start/end indices

This is what I have so far:

sig = [1 1 0 0 0 0 1 1 1 1 1 0 1 0 0 0 1 1 1 1 1 1 1 1 0 0 1 1 1 0];
len = length(sig);
thresh = 3;

%# align the signal with itself successively shifted by one
%# v will thus contain 1 in the starting locations of the zero interval
v = true(1,len-thresh+1);
for i=1:thresh
    v = v & ( sig(i:len-thresh+i) == 0 );
end

%# extend the 1's till the end of the intervals
for i=1:thresh-1
    v(find(v)+1) = true;
end

%# get the final indices
v = find(v);

I’m looking to vectorize/optimize the code, but I’m open to other solutions.
I have to stress that space and time efficiencies are very important, since I’m processing a large number of long bio-signals.

Answer 1

These are the steps I would take to solve your problem in a vectorized way, starting with a given vector sig:

• First, threshold the vector to get a vector tsig of zeros and ones (zeroes where the absolute value of the signal drops close enough to zero, ones elsewhere):

  tsig = (abs(sig) >= eps);  %# Using eps as the threshold
  

• Next, find the starting indices, ending indices, and duration of each string of zeroes using the functions DIFF and FIND:

  dsig = diff([1 tsig 1]);
  startIndex = find(dsig < 0);
  endIndex = find(dsig > 0)-1;
  duration = endIndex-startIndex+1;
  

• Then, find the strings of zeroes with a duration greater than or equal to some value (such as 3, from your example):

  stringIndex = (duration >= 3);
  startIndex = startIndex(stringIndex);
  endIndex = endIndex(stringIndex);
  

• Finally, use the method from my answer to the linked question to generate your final set of indices:

  indices = zeros(1,max(endIndex)+1);
  indices(startIndex) = 1;
  indices(endIndex+1) = indices(endIndex+1)-1;
  indices = find(cumsum(indices));