Count number of bouts separated by zeros

问题

I have a vector like this:

A = [1 2 1 1 1 4 5 0 0 1 2 0 2 3 2 2 2 0 0 0 0 33]

I would like to count how many GROUPS of non zero elements it contains and save them.

so I want to isolate:

[1 2 1 1 1 4 5]

[1 2]

[2 3 2 2 2]

[33]

and then count the groups (they should be 4) :)

Can you help me please?

Thanks

回答1:

To count your groups, a fast vectorized method using logical indexing is:

count = sum(diff([A 0]==0)==1)

This assumes that A is a row vector as in your example. This works with no zeros, all zeros, the empty vector, and several other test cases I tried.

To obtain your groups of values themselves, you can use a variation to my answer to a similar question:

a0 = (A~=0);
d = diff(a0);
start = find([a0(1) d]==1)           % Start index of each group
len = find([d -a0(end)]==-1)-start+1 % Length, number of indexes in each group

In your case it might make sense to replace len with

finish = find([d -a0(end)]==-1) % Last index of each group

The length of start, len, and finish should be the same as the value of count so you could just use this if you need to do the breaking up. You can then use start and len (or finish) to store your groups in a cell array or struct or some other ragged array. For example:

count = length(start);
B = cell(count,1);
for i = 1:count
    B{i} = A(start(i):finish(i));
end

来源：https://stackoverflow.com/questions/17870481/count-number-of-bouts-separated-by-zeros