How to make a +5 (or any +number) jump in array values inside a for loop?

So when line 0 condition is met, I want the for loop to jump 5 rows from the given data set of a single column. So when count1 ==5 , I want the loop to skip 5 rows from the column and take the 6th element as the new reference. [8,8,8,8,8,9,6,546,6,5,6,4,5,5,5,5,5]. In this dataset, the count value should be 2. Now what the issue is that the code given below does not skip to the 6th reference value to compare the next 5 values. I tried line 1 and 2 but those did not work or have any effect on the code.

count=0;
count1=0;
for i=1:length(data)-4
for j=i+1:i+4
if data(i)~=data(j)
count1=0;
break;
else
count1=count1+1;
end
if count1==5    % line 0
count=count+1;
%data(i,1)=data(i+5,1); //line 1
%data(i)=data(i+5);     //line 2
else
continue;
end
end
end

[KALYAN ACHARJYA]

This seems to be a simple issue, please redefine your problem ??

data=[8,8,8,8,8,9,6,546,6,5,6,4,5,5,5,5,5];
for count1=1:length(data)
if data(count1)==5
count1=count1+5;
end
end 

[Karthik Garimella]

So you replied to one of my other questions which is linked here => https://in.mathworks.com/matlabcentral/answers/748792-looping-query-where-i-need-the-count-of-the-consecutive-same-number-of-occurrences-equal-to-5. Now i had asked another question which i will link where you will get an elbaoration of what i needed and what i got => https://in.mathworks.com/matlabcentral/answers/757439-looping-error-where-i-want-to-get-5-instances-of-any-same-number-in-a-column#comment_1357719 . Now the problem that i'm stating is when i get the 5 consecutive occurrences of a number that is equal to the reference value. So let's say a[1] =8. and a[2] to a[5] =8 as well. My code will take a[1] as the reference value and compare 5 consecutive values which are equal and increment my global count by 1. Now i want to take my a[i] where i==1 to i==6 as i got 5 consecutive equal values when compared to a[1]. So my reference array element should point to a[6] so that i can start my new iteration for starting my comparison again. so if a[7] is not equal to a[6], the code will take a[7] as the reference value to start comparison with a[8],a[9],etc.

[Jan]

Does this mean, that you are looking for the number of blocks with equal value and more than 4 elements? What is the wanted answer for ones(1, 10) ?

[Karthik Garimella]

So the first one only has 5 consecutive 8's in the manner i specified. Hence the first one will have an output count =1 . In the second one, there are 4 of the 5 consecutive numbers of the same starting number. Hence the output would be 4, Another example i can give is 8,8,8,8,8,8,8,8,8,8(ten 8's),9,9,9,9,4,5,6,4,6,6,6,6,6. In this , the count should be 3 as it has 10 8's which increments the count to 2 and another 5 consecutive 6 which increments count one more time. Total count would be 3. My problem is that i'm not able to skip/jump the array index from x to x+5 in the for loop when my condition gets satisfied. for ones(1,10) i'm assuming that the array would be of 10 ones [1,1,1,1,1,1,1,1,1,1]? If this is the case then the count would be 2.

[Jan]

I do not know, what "the first one" and "the second one" are.

What does "there are 4 of the 5 consecutive numbers of the same starting number" mean?

As far as I understand, you want to:

count the number of blocks, which have 5 equal elements.

This is a much shorter description, isn't it?

Answer 1

[Jan]

You can use FEX: RunLength , or this part taken from it:

function n = RunLength_count(x)
d = [true, diff(x(:).') ~= 0, true];  % TRUE if values change
n = diff(find(d));                    % Number of repetitions
end

Then:

x      = [8,8,8,8,8,9,6,546,6,5,6,4,5,5,5,5,5];
n      = RunLength_count(x);
Result = sum(floor(n / 5))

USing floor(n / 5) let all blocks with less then 5 element be counted as zeros. 9 consecutive elements are counted as 1 block and 10 as 2 blocks.

You can do this with a loop also, but this is less elegant and efficient:

count = 0;
block = 0;
q     = nan;  % Initial element to compare with
for k = 1:numel(data)
    if data(k) == q  % Element equals former one
        block = block + 1;
        if block == 5
            count = count + 1;
            block = 0;
        end
    else  % A new value
        q     = data(k);
        block = 1;
    end
end

You do not need a 2nd loop.