# How to select complementary elements from a vector?

조회 수: 39(최근 30일)
답변: Bruno Luong 2021년 4월 20일
I have a vector
d = [33 20 4 5 6 75 8 9 0];
and another vector containing the indices whose complement i have to select from d.
I = [1 3 7];
so, the output I want is a vector which is - d minus the elements contained in d(I)
i.e
ans = [20 5 6 75 9 0];
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No. Its correct, I have changed the values to make the example clearer.
I is the vector of the index.
So, d(I) = [3 1 8];
What I want is to get rid of these elements from d. Not the values [1,3,7]. I is the index of the vector d.

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### 채택된 답변

Bruno Luong 2021년 4월 20일
d = [33 20 4 5 6 75 8 9 0];
I = [1 3 7];
d(setdiff(1:end,I))
ans = 1×6
20 5 6 75 9 0

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### 추가 답변(4개)

Bruno Luong 2021년 4월 7일
>> d(~ismember(d,I))
ans =
2 5 6 8 9 0
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Khalid Mahmood 2021년 4월 7일
편집: Khalid Mahmood 2021년 4월 19일
%Another but lengthy way is as follows:
%Vector 1
d = [3 2 1 5 6 7 8 9 0];
%Another vector containing the indices, which must be removed from d .
I = [1 3 7];
%so, the output is a vector which removes those values, i.e output= [2 5 6 9 0];
n1=size(d,2);
n2=size(I,2);
%A=zeros(1,n1-n2)
k=1;i=1;
for n=1:n1
if n~=I(k)
A(i)=d(n);
i=i+1;
else
if k<n2
k=k+1;
end
end
end
A
%same as A=d(~ismember(d,I))
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John D'Errico 2021년 4월 7일
편집: John D'Errico 2021년 4월 7일
d = [3 2 1 5 6 7 8 9 0];
I = [1 3 7];
setdiff(d,I)
ans = 1×6
0 2 5 6 8 9
Note that your example is actually incorrect, in that you claim 7 should be in the final result. But since 7 is a member of I, that is not the case.
Also, it depends on if you want elements that remain in the original order. setdiff will return a sorted set, and if any elements of d were repeated, then only one copy will remain in the result.
So if setdiff does not do as you wish, in that case, you need to use ismember, deleting the elements found. Thus...
d(~ismember(d,I))
ans = 1×6
2 5 6 8 9 0
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Bruno Luong 2021년 4월 20일
d = [33 20 4 5 6 75 8 9 0];
I = [1 3 7];
d(I) = []
d = 1×6
20 5 6 75 9 0

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