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C = intersect(A,B) for vectors A and B, returns the values common to
the two vectors with no repetitions. C will be sorted.
C = intersect(A,B,'rows') for matrices A and B with the same
number of columns, returns the rows common to the two matrices. The
rows of the matrix C will be in sorted order.
[C,IA,IB] = intersect(A,B) also returns index vectors IA and IB such
that C = A(IA) and C = B(IB). If there are repeated common values in
A or B then the index of the last occurrence of each repeated value is
returned.
[C,IA,IB] = intersect(A,B,'rows') also returns index vectors IA and IB
such that C = A(IA,:) and C = B(IB,:).
[C,IA,IB(A,B,'stable') for arrays A and B, returns the
values of C in the same order that they appear in A.
[C,IA,IB(A,B,'sorted') returns the values od
order.
If A and B are row vectors, then C will be a row vector as well,
otherwise C will be a column vector. IA and IB are column vector
If there are repeated common values in A or B then the index of
first occurrence of each repeated valued.
[C,IA,IB] = intersect(A,B,'rows','stable') returns the rows of C in the
same order than A.
[C,IA,IB] = intersec','sorted') returns the rows
sorted order
In a future release, the bllowing syntaxes will change
inc
- occurrence of indices in IA and IB will switch from last to first
- orientation of vector C
- IA and IB will always be column index vectors
- tighter restrictions on combinations of classes
In order to see what impact those changes will have on your code, use:
[C,IA,IB] = intersect(A,B,'R2012a')
[C,IA,IB] = intersect(A,B,'rows','R2012a')
If the changes in behavior adversely affect your code, you may preserve
the current behavior with:
[C,IA,IB] = intersect(A,B,'legacy')
[C,IA,IB] = intersect(A,B,'rows','legacy')
Examples:
a = [9 9 9 9 9 9 8 8 8 8 7 7 7 6 6 6 5 5 4 2 1]
b = [1 1 1 3 3 3 3 3 4 4 4 4 4 10 10 10]
[c1,ia1,ib1] = intersect(a,b)
% returns
c1 = [1 4], ia1 = [21 19], ib1 = [3 13]
[c2,ia2,ib2] = intersect(a,b,'stable')
% returns
c2 = [4 1], ia2 = [19 21]', ib2 = [9 1]'
c = intersect([1 NaN 2 3],[3 4 NaN 1])
% NaNs compare as not equal, so this returns
c = [1 3]
Class support for inputs A and B, where A and B must be of the same
class unless stated otherwise:
- logical, char, all numeric classes (may combine with double arrays)
- cell arrays of strings (may combine with char arrays)
-- 'rows' option is not supported for cell arrays
- objects with methods SORT (SORTROWS for the 'rows' option), EQ and NE
-- including heterogeneous arrays derived from the same root class 2 简单示例：集合A： A =
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15 3 求A∩B为：intersect（A，B） END setdiff补集运算 1 setdiff Set difference.
C = setdiff(A,B) for vectors A and B, returns the values in A that
are not in B with no repetitions. C will be sorted.
C = setdiff(A,B,'rows') for matrices A and B with the same number of
columns, returns the rows from A that are not in B. The rows of the
matrix C will be in sorted order.
[C,IA] = setdiff(A,B) also returns an index vector IA such that
C = A(IA). If there are repeated values in A that are not in B, then
the index of the last occurrence of each repeated value is returned.
[C,IA] = setdiff(A,B,'rows') also returns an index vector IA such that
C = A(IA,:).
[C,IA] = setdiff(A,B,'stable') for arrays A and B, returns the values
of C in the order that they appear in A.
[C,IA] = setdiff(A,B,'sorted') returns the values of C in sorted order.
If A is a row vector, then C will be a row vector as well, otherwise C
will be a column vector. IA is a column vector. If there are repeated
values in A that are not in B, then the index of the first occurrence of
each repeated value is returned.
[C,IA] = setdiff(A,B,'rows','stable') returns the rows of C in the
same order that they appear in A.
[C,IA] = setdiff(A,B,'rows','sorted') returns the rows of C in sorted
order.
In a future release, the behavior of the following syntaxes will change
including:
- occurrence of indices in IA will switch from last to first
- orientation of vector C
- IA will always be a column index vector
- tighter restrictions on combinations of classes
In order to see what impact those changes will have on your code, use:
[C,IA] = setdiff(A,B,'R2012a')
[C,IA] = setdiff(A,B,'rows','R2012a') 2 例如，求上例中在集合A中，不在集合B中的元素。
命令为setdiff(A,B) END setxor去两个集合交集的补集 1 例如setxor(A,B)= setxor(A,B) ans =
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15 END union取两者的并集 1 union（A,B）= > union(A,B) ans =
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15 内容来自网友回答
补集及其运算
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