counting the number of times a number appears next to the same one in a row?

Hi everyone,

I am trying to generate a random sequence and for this I have a row vector, which contains the values 1 to 6 in a random order 4 times (so my vector has 24 elements). I need a way to find how many times the same number appears next to the same number - I have a hard time explaining what I mean but here is an example:

1 2 3 4 5 6 -- no same number appears next to the same number so answer should be 0

1 1 2 3 4 5 -- here 1 is repeated once, so answer should be 1

1 1 2 3 4 4 - here 1 and 4 are repeated and so the answer should be 2

[Kilian Liss]

Probably not the most ellegant solution, but the following code seems to work:

x1 = [1 1 2 3 4 4];
count = 0;

for i = 1:length(x1) - 1
    if(x1(i) == x1(i + 1))
        count = count + 1;
    end
end

[madhan ravi]

vector = [1,1,1,5,6,2,2,3,1]
result = ???

[Andrei Bobrov]

3

Answer 1

[madhan ravi]

Simpler:

 nnz(~diff(vector))

Note: Taking into account that we only deal with integers.

[Andrei Bobrov]

+1

[madhan ravi]

Thank you Andrei!

[Bruno Luong]

"Note: Taking into account that we only deal with integers."

Why ?

[madhan ravi]

Huh? What kind of question is that?

[Bruno Luong]

Why you are saying "Taking into account that we only deal with integers."

Answer 2

[KALYAN ACHARJYA]

num=[1 2 3 4 5 6]; % Change this one and test
uniq_num=unique(num);
digit_repeat=length(num)-length(uniq_num)

Its works right?

Answer 3

[Bruno Luong]

>> A=[1 1 1 2 3 4 4 2]

A =

     1     1     1     2     3     4     4     2

>> sum(diff(A)==0 & diff([NaN, A(1:end-1)])~=0)

ans =

     2

>> 

Answer 4

[Kilian Liss]

Probably not the most ellegant solution, but the following code seems to work:

x1 = [1 1 2 3 4 4];
count = 0;
for i = 1:length(x1) - 1
    if(x1(i) == x1(i + 1))
        count = count + 1;
    end
end