How to count numbers considering all consecutives as one??

Hi, I need to count some numbers like: Data=2, 3, 4, 7,8 9, 10, 20,25,27; Now, counting should be: 2,3,4=1; 7,8,9,10=1; 20=1; 25=1; 27=1; Total count=5 (1+1+1+1+1) i.e. consecutive numbers should be count as one together and any individual value should be one. Any help regarding matlab programming would be highly appreciated. Thanks in advance. Badrul

Answer 1

[Shashank Prasanna]

Data=[2, 3, 4, 7,8 9, 10, 20,25,27];
sum = 1; % The first set is counted.
for i = 2:length(Data)

      if Data(i)~=Data(i-1)+1
          %Data(i)
          sum=sum+1;
      end

end

disp(sum)

[Mohammad]

Thanks a lot. It works for me.

[Azzi Abdelmalek]

Don't use sum and i as variables. sum is the function that sums the elements of an array, and i is used to represent complex numbers

[Shashank Prasanna]

Azzi, Its perfectly fine to use them since MATLAB gives the highest precedence to variables before function names. Avoid them if you plan to use these function in your program or just clear the variables right before you want to use the function:

http://www.mathworks.com/help/matlab/matlab_prog/function-precedence-order.html

However as you mentioned it is safer to avoid using them.

Answer 2

[Cedric]

Here is a funny way to achieve this:

 >> sum(diff(Data)>1)+1

EDIT: see my comment after Azzi's remark.

[Mohammad]

Superb...............thanks a lot

[Azzi Abdelmalek]

Cedric, it don't work if Data is not sorted

[Cedric]

I assumed Data(i+1)>=Data(i) from the wording of the statement, but you are right, without this assumption that would be:

 sum(abs(diff(data))>1)+1

I also assumed that two equal numbers do not break a group. If it is not the case the expression is:

 sum(abs(diff(data))~=1)+1

[Mohammad]

Yes, your data have to be sorted...................

[Azzi Abdelmalek]

It still dont work. Look at this example.

Data=[1 2 3 4 7 8 7 6 2 1]

I suggest

sum(diff(Data)~=0 & diff(Data)~=1)+1

[Cedric]

Well, the OP will pick the one that matches his requirements I guess; I don't test Data(i+1)==Data(i)+1 (consec. elements are consec. integers) in my solutions, but Data(i)==Data(j)+1 for |i-j|=1 (ordered neighbors are consec. integers).

Answer 3

[Roger Stafford]

 count = sum([true,diff(data)~=0]);