Need to remove repeated adjacent elements in an array

I need to turn

[1 1 1 1 2 2 2 6 6 6 6 2 2 2 2] into [1 2 6 2]

unique() gives [1 2 6], but I want to preserve the second value

any advice?

[Michael Cappello]

x(diff(x)==0) = []

[Matthew Rademacher]

Thanks!

[Ravi Mravi]

Excellent solution

Answer 1

[Star Strider]

Taking advantage of ‘logical indexing’, it is relatively straightforward:

A = [1 1 1 1 2 2 2 6 6 6 6 2 2 2 2];
B = A(diff([0 A])~=0);

The code looks for changes in the differences (from the diff function) in ‘A’, then finds the elements in ‘A’ that correspond to those changes.

[Geetha K]

What if the i/p is like this 1111112233333334555555 And o/p should be 135

[Image Analyst]

How are you getting 135 from that?

[Star Strider]

Geetha K —

Your Comment does not relate to this thread. Please post it as a new Question. If you want to cite to this thread, copy the URL and include it in your Question.

[Juan Sierra]

I'd refine this as

A = [1 1 1 1 2 2 2 6 6 6 6 2 2 2 2];
B = A(diff([A(1)-1, A]) ~= 0)

just in case the first value is 0. Just a small suggestion ;)

[Bruno Luong]

No it's still flawed

>> A=1e20

A =

   1.0000e+20

>> B = A(diff([A(1)-1, A]) ~= 0)

B =

     []
     
>> A=uint8(0)

A =

  uint8

   0

>> B = A(diff([A(1)-1, A]) ~= 0)

B =

  0×0 empty uint8 matrix

Better

B = A([true diff(A)~=0])

Still it does work if A is empty.

The answer by posted by Michael Cappello in the comment is still better.

Answer 2

[Joseph Cheng]

you can use diff to determine the consecutive same value numbers

test = [1 1 1 1 2 2 2 6 6 6 6 2 2 2 2]
mtest = [test test(end)-1];
difftest = diff(mtest)
output = test(difftest~=0)

the mtest is the modified test number to get the last value not the same. if you look at the output of difftest you see that we get the positions of the transitions from one number to another.

Answer 3

[Image Analyst]

Here's one way:

m = [1 1 1 1 2 2 2 6 6 6 6 2 2 2 2]
logicalIndexes = [0, diff(m)] ~= 0
output = [m(1), m(logicalIndexes)]