I want to count the number of 0s in a binary sequence which occurs for n number of times, where n=1:25.

Suppose I have a binary sequence A=[0 0 0 1 0 0 1 0 1 0 0 0 0 0 1 0 0 0 1]

Then I want to count in how many cases 0 occurs consecutively for n number of times preceded and succeeded by 1.

Here the desired output would be-

for n=1- noofzeros=1

n=2 - noofzeros=1

n=3- noof zeros=2

n=4- no of zeros=0

n=5- noofzeros=1

and so on....

[sai charan sampara]

Hello Dwijraj, For n = 3 is the count 1 or 2 ? There is only one "10001" in the array A. Do you want to also count "0001" in the starting as a case of n=3?

[Naren]

I guess so, you should consider it as a circular array.

Answer 1

[Image Analyst]

If you have the Image Processing Toolbox you can use a few functions in there:

% Define sample data.
A = [0 0 0 1 0 0 1 0 1 0 0 0 0 0 1 0 0 0 1]

A = 1x19

0 0 0 1 0 0 1 0 1 0 0 0 0 0 1 0 0 0 1

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% Put a row of zeros on top and bottom so we can use imclearborder
% to remove runs that don't have a 1 on both ends of the run.
mask = logical(padarray(A == 0, 1, 0))

mask = 3x19 logical array
   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0
   1   1   1   0   1   1   0   1   0   1   1   1   1   1   0   1   1   1   0
   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0

% Get rid of runs that touch the border.
mask = imclearborder(mask)

mask = 3x19 logical array
   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0
   0   0   0   0   1   1   0   1   0   1   1   1   1   1   0   1   1   1   0
   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0

props = regionprops(mask(2,:), 'Area', 'PixelIdxList');
% Get the run lengths of each stretch of zeros into one vector.
allRunLengths = [props.Area]

allRunLengths = 1x4

2 1 5 3

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% PixelIdxList is the indexes that are in each run.
% If you want the starting index of each run of zeros, 
% use the PixelIdxList and take the first index.
startingIndexes = zeros(1, numel(props)); % Preallocate
for k = 1 : numel(props)
    startingIndexes(k) = props(k).PixelIdxList(1);
end
% Display it in the command window
startingIndexes

startingIndexes = 1x4

5 8 10 16

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Answer 2

[sai charan sampara]

Hello Dwijraj,

You can use "find" to locate the positions of "1" in the array and then the difference in position of 2 consecutive "1"s is the number of continuous zeros. Then number of cases for each value of "n" can be counted. If we consider the array to be circular one more case is to be counted . It can be done similar to the code shown below:

A=[0 0 0 1 0 0 1 0 1 0 0 0 0 0 1 0 0 0 1];
idx=find(A,length(A))

idx = 1x5

4 7 9 15 19

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arr=diff(idx)-1

arr = 1x4

2 1 5 3

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arr=[arr,length(A)-idx(end)+idx(1)-1] % Extra case when we consider the array to be circular

arr = 1x5

2 1 5 3 3

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no_of_zeros=zeros(25,1);
for i=1:length(arr)
    no_of_zeros(arr(i))=no_of_zeros(arr(i))+1;
end
no_of_zeros

no_of_zeros = 25x1

1 1 2 0 1 0 0 0 0 0

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[Dwijaraj]

My actual binary sequence is of the size 7671x1. This is not working if the array size is greater than 25. What if the value of n extends to 50 or 100 even.

[sai charan sampara]

You can add a variable "n" that can store the value of required array size and then define the size of the array "no_of_zeros" in terms of this variable "n". It can be done similar to the code below:

n=100;
A=randi([0,1],[1,50]);
idx=find(A,length(A))

idx = 1x23

1 3 6 7 8 14 22 25 27 28 29 30 31 34 35 36 37 38 40 41 42 46 49

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arr=diff(idx)-1

arr = 1x22

1 2 0 0 5 7 2 1 0 0 0 0 2 0 0 0 0 1 0 0 3 2

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arr=[arr,length(A)-idx(end)+idx(1)-1] % Extra case when we consider the array to be circular

arr = 1x23

1 2 0 0 5 7 2 1 0 0 0 0 2 0 0 0 0 1 0 0 3 2 1

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no_of_zeros=zeros(n,1);
for i=1:length(arr)
    if(arr(i)>0)
    no_of_zeros(arr(i))=no_of_zeros(arr(i))+1;
    end
end
no_of_zeros

no_of_zeros = 100x1

4 4 1 0 1 0 1 0 0 0

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[Matlab Pro]

Hi @Dwijaraj

Here is a solution without Image Processing TBX

The result is a histogram with bins 1:25 with occurancies

x=[0 0 0 1 0 0 1 0 1 0 0 0 0 0 1 0 0 0 1 1 0 0 1 0 1 0 1 0 1 0 0 0 ];
x = randi([0, 1],[1,300]);
idx1 = x;
idx2 = [idx1(2:end), true];

f1 = find(idx1);
f2 = find(idx2);
f2 = f2(2:end);

if length(f1) > length(f2) % x starts with '1'
    f2 = [f1(2)-1, f2]; % Missing 1s section's end
end
d = diff([f1;f2]);
[s,n] = hist(d,1:25)

s = 1x25

113 19 7 2 4 3 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0

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n = 1x25

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25

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