Count sequence of zeros with For loop

Let's say I have the following matrix:

a= 3 0 7 8 0 0 1 5 
   1 0 5 0 0 2 3 0
   0 2 0 0 1 4 7 0 
   2 0 5 9 0 0 0 0 
   3 0 1 0 0 0 1 5

I wish to count the number of continuous 0s every time a 0 exists in each column. So for the above example, I want my output to be

zeros= 1 2 1 2 2 1 1 3 
         2   1 2 2

Is it possible to do this using a For loop? The output for each column can be made into a separate file if necessary.

Thank you!

Answer 1

[Roman Müller-Hainbach]

It is not pretty, but it does what you want:

function zeros = contzeros(A)
  zeros = cell( 1, size(A,2) );
  for colind = 1:length(zeros)
      column = A(:,colind)';
      if column(1) == 0
          zeros{colind} = max([0,find(column~=0,1)])-1;
      end
      [~,remains] = strtok(column,0);
      while ~isempty(remains)
          n = max([0,find(remains~=0,1)])-1;
          if n == -1 && remains(end) == 0
              zeros{colind}(end+1,1) = length(remains);
          elseif n >= 1
              zeros{colind}(end+1,1) = n;
          end
          [~,remains] = strtok(remains,0); %#ok<STTOK>
      end
  end
end

Notice that the output is a cell-array.

Answer 2

[Stephen23]

Simply use diff and then run over the columns:

a = [...
   3 0 7 8 0 0 1 5
   1 0 5 0 0 2 3 0
   0 2 0 0 1 4 7 0
   2 0 5 9 0 0 0 0
   3 0 1 0 0 0 1 5
];
idx = a([1,1:end,end],:)==0;
idx([1,end],:) = false;
tmp = num2cell(diff(idx,1,1),1);
fun = @(v)find(v<0)-find(v>0);
out = cellfun(fun,tmp,'uni',0);

giving the output in a cell array:

>> out{:}
ans =  1
ans =
   2
   2
ans =  1
ans =
   2
   1
ans =
   2
   2
ans =
   1
   2
ans =  1
ans =  3

Answer 3

[Jan]

With FEX: RunLength:

a = [3 0 7 8 0 0 1 5; ...
     1 0 5 0 0 2 3 0; ...
     0 2 0 0 1 4 7 0 ; ...
     2 0 5 9 0 0 0 0 ; ...
     3 0 1 0 0 0 1 5];

nCol = size(a, 2);
List = cell(1, nCol);
for iCol = 1:nCol
  [B, N]     = RunLength(a(:, iCol));
  List{iCol} = N(B == 0);
end

You can create a cheap version of RunLength also:

function [b, n] = cheapRunLength(x)
    d = [true; diff(x) ~= 0];   % TRUE if values change
    b = x(d);                   % Elements without repetitions
    k = find([d', true]);       % Indices of changes
    n = diff(k);                % Number of repetitions
end

With the C-Mex this needs 40-50% of the runtime, but for small data, this might be not important.