How to crate cells contain subvectors from logical vector in MATLAB

user_17087165 0
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For example I have a logial vector in MATLAB:

idx1 = [0 0 0 1 1 1 0 0 1 1 1 0 0 0 0 1 1 1 1 1 1 0 0 0 0 0 0 0 1 1 1 0 0]

And I want to get numbers (count each block) of such blocks: 1 1 1, i.e. such block contain N elements == 1 (N "1"). idx1 - array, and his dimension can be any, for example 3820000.

How count many blocks (sequences of ones) occur in the entire array idx1?


counts_idx = 0;
init_counts_idx = 0;
arr = 0;

for i = 1:length(idx1) -1
for kk = 1 : length(idx1) - 1
    if idx1(kk + 1) == 1 
       init_counts_idx = init_counts_idx + 1;
       arr = init_counts_idx;
    else
       init_counts_idx =  counts_idx;  
    end
    C = {i,arr};
end
end

I try to using cells...

indexing matlab
2 Answers
user_31811040

You can calculate the start and end indices of each block by diff([0 idx1 0]). Then, use this information to calculate block lengths Ns. Finally express the result as a cell array using the function C = mat2cell(A,rowDist).

idx1 = [0 0 0 1 1 1 0 0 1 1 1 0 0 0 0 1 1 1 1 1 1 0 0 0 0 0 0 0 1 1 1 0 0];
diffs = diff([0 idx1 0]);

% find start index of the blocks
loc = find(diffs == 1);
% calc block lengths by subtracting end - start indices
Ns = find(diffs == -1) - loc;

C = mat2cell([loc' Ns'],ones(size(loc)))

  4×1 cell array

    {[ 4 3]}
    {[ 9 3]}
    {[16 6]}
    {[29 3]}

If you are interested only in the number of such blocks, length(loc) will give you the answer, it is similar to bwconncomp(idx1).NumObjects.

user_39431700
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