how to count number of repeating's in data series.

for example my matrix A=[22 24 24 36 36 36 48 48 48 48 24 33 22 22]; i need to count how many twos, threes, four time repeating's etc

Answer 1

[Massimo Zanetti]

This code

A = [22 24 24 36 36 36 48 48 48 48 24 33 22 22]';
C = accumarray(A,1);
I = unique(A);
counts = [I,C(I)]

returns

out = 22     3
     24     3
     33     1
     36     3
     48     4

[shashika iresh]

no dear, i need to find the repeating(double, triple, four time, etc of same number. for a example above matrix 2 doubles(24 24,22 22),and only 1 triple(36 36 36) ,and only one four times(48 48 48 48). li

Answer 2

[Andrei Bobrov]

A=[22 24 24 36 36 36 48 48 48 48 24 33 22 22];

V = A(:);
[~,~,c] = unique(V);
t = diff([0;c])~= 0;
ix = cumsum(t);
out = [V(t),accumarray(ix(:),1)];

out2 = [num2cell((1:max(out(:,2)))'),...
           accumarray(out(:,2),out(:,1),[],@(x){sort(x(:)')})];

[Adrian Stannard]

I vote for this answer ;)

[Andrei Bobrov]

Thank you Adrian!

[Sylwia Kaduk]

Dear Andrei, 4 years later your code is very helpful in part of my PhD. Thanks a lot.

Answer 3

[Adrian Stannard]

I think I understand what you want - it is the successive re-occurrences of a number.

Array=[22 24 24 36 36 36 48 48 48 48 24 33 22 22]; 
Array(find(diff(Array)==0))

This returns only the repeated elements from the Array after the first occurrence.

You could go further, for example use:

counts=hist(Array(find(diff(Array)==0)))

This gives you the how many times they are repeated. Alternatively:

Array2 = Array(find(diff(Array)==0));
Array3= zeros (size(Array2));
for i = 1:length(Array2)
Array3(i) = sum(Array2==Array2(i));
end

Array3=Array3+1
Array2=Array2([1,diff(Array2)]~=0);
Array3=Array3([1,diff(Array3)]~=0);

will return:

24 36 48 22

2 3 4 2

[shashika iresh]

dear, your code quite ok, but it give the final result like this ans =

    24    36    36    48    48    48    22

Array3 =

     2     3     3     4     4     4     2
not like your return answer

[shashika iresh]

if there is 5 times 6 times repeating's are there what will will be the code for those also

[Adrian Stannard]

The final two lines I gave should strip out the repeated occurrences:

Array2=Array2([1,diff(Array2)]~=0);
Array3=Array3([1,diff(Array3)]~=0);

This code should work for any number of repetitions:

Array=[22 24 24 36 36 36 48 48 48 48 48 48 48 24 33 22 22 22 22 22 22 22 22 22 22];
Array2 = 24    36    48    22
Array3 = 2     3     7    10

[Guillaume]

@Adrian,

Please use the {}Code format button for the code (or just put two spaces before each line) instead of putting blank lines between each line of code.

edit: Thanks!

[shashika iresh]

please check for this it give wrong answer. A =[24 24 24 24 36 36 36 48 48 48 48 48 48 48 28 24 24 22 22 22 22 22 22 22 22 22 22]; Array2 =

    24    36    48    24    22

Array3 =

     5     3     7     5    10

Answer 4

[Thorsten]

N(i) is the the number of occurrences of series of i numbers in A:

N = zeros(numel(A), 1);
i = 1; 
while(i) < numel(A)
  Ni = 1; val = A(i);
  while i < numel(A) && A(i+1) == A(i)
    i = i + 1; Ni = Ni + 1; 
  end,
N(Ni) = N(Ni) + 1;
i = i + 1;