Does something similar to 'intersect' command exists for more than 2 vectors?

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Peeyush
Peeyush 2011 年 6 月 5 日
Hi,
There are 5 row vectors with different(or same) number of elements. The problem is to pick out the 'intersecting' elements from these 5 vectors.
'intersect(A,B)' works with only 2 vectors. Is there a command/procedure for more than 2 vectors?
Thanks.

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Matt Fig
Matt Fig 2011 年 6 月 5 日
You could call INTERSECT multiple times:
A = [1 2 3 4 5];B = [3 7 8 9 5];C = [5 0 22 3 77];
F = intersect(intersect(A,B),C)
Or download this file, or this collection.
  3 件のコメント
Matt Fig
Matt Fig 2011 年 6 月 6 日
But you don't have to write a program! The two I pointed you to on the FEX are already written...
Matt Fig
Matt Fig 2011 年 6 月 6 日
Here is an example of another way to do it for the many variables case. Note that if you have variables in the workspace which you do not want to be a part of the intersection, just call SAVE with the 3+ argument format, passing only the variables you wish compared:
clear all % Start with a clean slate.
A = round(rand(1,50)*20);
B = round(rand(1,50)*20);
C = round(rand(1,50)*20);
D = round(rand(1,50)*20);
E = round(rand(1,50)*20);
F = round(rand(1,50)*20);
save 'myvars'
X = load('myvars');
F = fieldnames(X);
H = X.(F{1});
for ii = 2:length(F)
H = intersect(H,X.(F{ii}));
end
H % Show the intersection.

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その他の回答 (1 件)

Morteza Darvish Morshedi
Morteza Darvish Morshedi 2019 年 4 月 2 日
編集済み: Morteza Darvish Morshedi 2019 年 4 月 2 日
Hi,
For three input, you can simply do this:
function [Com,ia,ib,ic] = intersect3(A,B,C)
[C1,ia,ib] = intersect(A,B);
[Com,ic1,ic] = intersect(C1,C);
%~ ic is okay
ia = ia(ic1);
ib = ib(ic1);
end
Going from 3 input sets to 5 input sets or more, you would need to follow same procedure, each time on the outputs from the last step. Like (for 4 inputs):
function [Com,ia,ib,ic,id] = intersect4(A,B,C,D)
[C1,ia,ib] = intersect(A,B);
[C2,ic1,ic] = intersect(C1,C);
ia = ia(ic1);
ib = ib(ic1);
% or [C2,ia,ib,ic] = intersect3(A,B,C);
% Now D
[Com,id1,id] = intersect(C2,D);
%~ id is okay
ia = ia(id1);
ib = ib(id1);
ic = ic(id1);
end
  4 件のコメント
Cesar Daniel Castro
Cesar Daniel Castro 2020 年 3 月 4 日
Could you please give more details about your code?
Morteza Darvish Morshedi
Morteza Darvish Morshedi 2020 年 3 月 7 日
Cesar Daniel Castro Having three sets of A,B and C, you always start with finding intersection of two of them, e.g. intersection of A and B as C1. Next, you take intersection between C and C1, as C2. When it comes to indeces of elemtns of C2 in A, B and C, you can directly have those of C from built-in function 'intersect' as ic. To find 'ia' and 'ib', you take a subset of the first intersection C1 that exist in the second intersection C2 as ia=ia(ic1) and ib = ib(ic1). Generalization of this procedure as one function is what Stephen Cobeldick mentioned.

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