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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];

Bruno Luong
on 20 Apr 2021 at 18:14

d = [33 20 4 5 6 75 8 9 0];

I = [1 3 7];

d(setdiff(1:end,I))

Khalid Mahmood
on 7 Apr 2021

Edited: Khalid Mahmood
on 19 Apr 2021 at 0:30

%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))

John D'Errico
on 7 Apr 2021

Edited: John D'Errico
on 7 Apr 2021

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))

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