Need to remove repeated adjacent elements in an array

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Matthew Rademacher
Matthew Rademacher 2015-5-15
编辑： Bruno Luong 2020-11-21
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?

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采纳的回答

Star Strider
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.
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显示 3更早的评论
Juan Sierra
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
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.
The answer by posted by Michael Cappello in the comment is still better.

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更多回答（2 个）

Joseph Cheng
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
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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