Removing repeated in order elements from array?

I have an array like [ 1 1 1 2 2 3 3 3 1 1 1 2 2 4 4 4 5 5 6]

Is there a method which will produce array [ 1 2 3 1 2 4 5 6] and remove all repeating orderly values but not remove all duplicates?

have tried code below with no luck

for k=length(array):-1:1
    if array(k)==array(k+1)
        array(k)=[]
    end
end

Answer 1

[Rik]

No need for a loop (even if you should have replaced length by numel):

data=[ 1 1 1 2 2 3 3 3 1 1 1 2 2 4 4 4 5 5 6];
L=diff(data)==0; % note: this is 1 element shorter that data itself
data(L)=[]

data = 1×8

1 2 3 1 2 4 5 6

Answer 2

[Voss]

Your approach will work; you just have to start at the second-to-last element instead of the last:

array = [ 1 1 1 2 2 3 3 3 1 1 1 2 2 4 4 4 5 5 6];
for k=length(array)-1:-1:1
   if array(k)==array(k+1)
       array(k)=[];
   end
end
array

array = 1×8

1 2 3 1 2 4 5 6

Here's another approach:

array = [ 1 1 1 2 2 3 3 3 1 1 1 2 2 4 4 4 5 5 6];
array(diff(array) == 0) = [];
array

array = 1×8

1 2 3 1 2 4 5 6

[Nicholas Kavouris]

i take it this approach will not work with only one value in the array ex: [2 2 2 2 2 2 2]

[Torsten]

Why not ? I get

array = [2]

It's because diff(array) has one element less than array.

[Voss]

As @Torsten says, it should still work (both ways work):

array = [2 2 2 2 2 2 2];
for k=length(array)-1:-1:1
   if array(k)==array(k+1)
       array(k)=[];
   end
end
array

array = 2

array = [2 2 2 2 2 2 2];
array(diff(array) == 0) = [];
array

array = 2