How to select complementary elements from a vector?
显示 更早的评论
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];
2 个评论
John D'Errico
2021-4-7
Your example is flawed, since 7 is in I.
Shadman Shahid
2021-4-20
编辑：Shadman Shahid
2021-4-20
采纳的回答
d = [33 20 4 5 6 75 8 9 0];
I = [1 3 7];
d(setdiff(1:end,I))
更多回答（4 个）
Bruno Luong
2021-4-7
>> d(~ismember(d,I))
ans =
2 5 6 8 9 0
Khalid Mahmood
2021-4-7
编辑：Khalid Mahmood
2021-4-19
%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))
d = [3 2 1 5 6 7 8 9 0];
I = [1 3 7];
setdiff(d,I)
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))
d = [33 20 4 5 6 75 8 9 0];
I = [1 3 7];
d(I) = []
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