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

Asked by Yoanna Ivanova on 22 Jun 2019 at 9:40
Latest activity Commented on by Bruno Luong
on 22 Jun 2019 at 12:02
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

  3 Comments

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

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4 Answers

Answer by madhan ravi
on 22 Jun 2019 at 11:22

Simpler:
nnz(~diff(vector))
Note: Taking into account that we only deal with integers.

  5 Comments

"Note: Taking into account that we only deal with integers."
Why ?
Huh? What kind of question is that?
Why you are saying "Taking into account that we only deal with integers."

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Answer by KALYAN ACHARJYA on 22 Jun 2019 at 10:01
Edited by KALYAN ACHARJYA on 22 Jun 2019 at 10:15

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?

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Answer by Bruno Luong
on 22 Jun 2019 at 10:27
Edited by Bruno Luong
on 22 Jun 2019 at 10:29

>> 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
>>

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Answer by Kilian Liss on 22 Jun 2019 at 10:32

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

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