Different groups of 1s

Have been trying to solve this, but couldn't

Any help would be appreciated

I have an array A=[-1 1 -1 -1 -1 -1 1 -1 1 -1 1 -1 -1 -1]

i want to get different groups of 1s like group1 contains ones with length 1 (for consecutive ones), second group contains ones with length 2 ,etc.

same approach for -1s

For 1s output shoud be [ 3 0 0 0 0 0 0 0 0 0 0 0 0 0 ]

For -1s outputs shd be [3 0 1 1 0 0 0 0 0 0 0 0 0 0 ].

[Matt J]

Why is the output [3 0 0 0 0 0 0,...]? I see 4 groups of length 1 in A.

Answer 1

[Jan]

See FEX: RunLength

If speed matters, compile the c file according to the instructions. If the problems are such tiny, it is sufficient to use the M-verion RunLength_M:

A = [-1     1    -1    -1    -1    -1     1    -1     1    -1     1    -1    -1    -1];

value  = 1;
[B, N] = RunLength_M(A == value);
Result = histcounts(N(B), 1:numel(A))
%  [4 0 0 0 0 0 0 0 0 0 0 0 0 0]    not [3 0 0 ...]

value  = -1;
[B, N] = RunLength_M(A == value);
Result = histcounts(N(B), 1:numel(A))
%  [3 0 1 1 0 0 0 0 0 0 0 0 0 0]

If RunLength_M is still an overkill for you, use:

function [b, n] = RunLengthEnc(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

And if histcounts is too powerful also, a simple loop prodiuces the same result:

function H = myHist(N, Len)
H = zeros(1, Len);
for k = 1:numel(N)
  H(N(k)) = H(N(k)) + 1;
end
end

Then the the main code looks like:

A = [-1     1    -1    -1    -1    -1     1    -1     1    -1     1    -1    -1    -1];

value  = 1;
[B, N] = RunLengthEnc(A == value);
Result = myHist(N(B), numel(A))

Or with 1 and -1 in the same run:

[B, N] = RunLengthEnc(A);
Result = myHist(N(B ==  1), numel(A))
Result = myHist(N(B == -1), numel(A))

[AB WAHEED LONE]

Without any fucntion like Runlength.

Tried my way back.. seems it is working

[Jan]

Without subfunctions? Then simply copy the code into the main function:

A = [-1     1    -1    -1    -1    -1     1    -1     1    -1     1    -1    -1    -1];

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

Result = histcounts(n(b ==  1), 1:numel(A))
Result = histcounts(n(b == -1), 1:numel(A))

[AB WAHEED LONE]

This seems shortest one,,Got it ,thank You

Answer 2

[Matt J]

Using the tools in this File Exchange submission,

https://www.mathworks.com/matlabcentral/fileexchange/78008-group1s?s_tid=srchtitle

A=[-1     1    -1    -1    -1    -1     1    -1     1    -1     1    -1    -1    -1];

map=(A==-1);

L=groupFcn(@sum, map, groupTrue(map)); 

output=histcounts(L,1:numel(A)+1)

output = 1×14

3 0 1 1 0 0 0 0 0 0 0 0 0 0