Need to remove repeated adjacent elements in an array
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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 个评论
Michael Cappello
2015-5-15
x(diff(x)==0) = []
Matthew Rademacher
2015-5-16
Ravi Mravi
2017-10-30
Excellent solution
采纳的回答
Star Strider
2015-5-15
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 个评论
Geetha K
2020-4-28
What if the i/p is like this 1111112233333334555555 And o/p should be 135
Image Analyst
2020-4-28
How are you getting 135 from that?
Star Strider
2020-4-28
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
2020-11-21
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
2020-11-21
编辑：Bruno Luong
2020-11-21
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.
更多回答（2 个）
Joseph Cheng
2015-5-15
编辑：Joseph Cheng
2015-5-15
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.
Image Analyst
2015-5-15
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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