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?
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채택된 답변
Star Strider
2015년 5월 15일
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.
댓글 수: 5
Juan Sierra
2020년 11월 21일
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
2020년 11월 21일
편집: Bruno Luong
2020년 11월 21일
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.
추가 답변 (2개)
Joseph Cheng
2015년 5월 15일
편집: Joseph Cheng
2015년 5월 15일
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.
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Image Analyst
2015년 5월 15일
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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