Splitting a vector into 'on' periods

Hi,

I'm investigating data from a force plate (study of balance in people with movement disorders). I have a raw signal such as:

vector = 1 2 3 4 0 0 0 0 5 6 7 4 7 8 0 0 0 1 2 5 4 0 0 0 0 0 5 6

I need to split this signal into separate vectors that only contain sequences of numbers greater than 0. So the ideal result for the above would be:

vector 1 = 1 2 3 4
vector 2 = 5 6 7 4 7 8
vector 3 = 1 2 5 4
vector 4 = 5 6

I hope the question is clear enough. Any help would be greatly appreciated!

Answer 1

[Star Strider]

I did my best to make this as robust as possible:

vector = [1 2 3 4 0 0 0 0 5 6 7 4 7 8 0 0 0 1 2 5 4 0 0 0 0 0 5 6];
vec_nz = vector > 0;                                        % Non-Zero Elements
dv = diff([0 vec_nz 0]);                                    % Index Vector
on  = find(dv > 0);
off = find(dv < 0);
sections = abs(on-off);                                     % Divide ‘vector’ Here
VectorCell = mat2cell(vector(vec_nz), 1, sections);         % Create Cell
VectorCell{:}                                               % Display Results (Can Be Deleted)

ans =
     1     2     3     4
ans =
     5     6     7     4     7     8
ans =
     1     2     5     4
ans =
     5     6

[CarrotCakeIsYum]

Thank you! This was a great help.

[Star Strider]

My pleasure!

Answer 2

[Thorsten]

I found a solution here: http://stackoverflow.com/questions/27076618/how-to-split-vector-by-zeros-in-matlab

w = [false v~=0 false]; % "close" v with zeros, and transform to logical
starts = find(w(2:end) & ~w(1:end-1)); % find starts of runs of non-zeros
ends = find(~w(2:end) & w(1:end-1))-1; % find ends of runs of non-zeros
result = arrayfun(@(s,e) v(s:e), starts, ends, 'uniformout', false); % build result

You can also use diff to compute starts and ends indices

starts = find(diff(w) == 1);
ends = find((diff(w) == -1)) - 1;

Answer 3

[Stephen23]

This is fast and simple using accumarray:

V = [1,2,3,4,0,0,0,0,5,6,7,4,7,8,0,0,0,1,2,5,4,0,0,0,0,0,5,6];
X = V>0;
Y = diff([false,X]);
Z = cumsum(Y(X));
out = accumarray(Z(:),V(X),[],@(n){n})

Answer 4

[Jos (10584)]

% A one-liner, works when vector only contains values as in the example:

vector = [1 2 3 4 0 0 0 0 5 6 7 4 7 8 0 0 0 1 2 5 4 0 0 0 0 0 5 6] ;
OUT = cellfun(@(x) double(x),strsplit(char(vector),char(0)),'un',0) % cellarray

Answer 5

[CarrotCakeIsYum]

Thanks everyone - the advice was brilliant and all is working perfectly.