Need to remove repeated adjacent elements in an array
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Matthew Rademacher
on 15 May 2015
Edited: Bruno Luong
on 21 Nov 2020
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?
3 Comments
Accepted Answer
Star Strider
on 15 May 2015
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.
5 Comments
Juan Sierra
on 21 Nov 2020
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
on 21 Nov 2020
Edited: Bruno Luong
on 21 Nov 2020
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.
More Answers (2)
Joseph Cheng
on 15 May 2015
Edited: Joseph Cheng
on 15 May 2015
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.
0 Comments
Image Analyst
on 15 May 2015
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)]
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