any() 相當(dāng)于或操作，只要有1，就返回1

all() 相當(dāng)于與操作，只要有0，就返回0

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C = union(A,B)：

C為A和B的并集。去掉相同元素。

C = intersect(A,B)

C為A和B的交集. The values of C are in sorted order.

[Lia,Locb] = ismember(A,B)：

判斷A中元素是否為B中的子元素，如果不是則，Lia和Locb中元素值都為0（注意，Lia大小和A一樣，Locb大小和B一樣）。如果A中有元素是B中的，在Lia中對(duì)應(yīng)返回值1，Locb, containing the highest index in B for each value in A that is a member of B. The output array, Locb, contains 0 wherever A is not a member of B.

[C,ia] = setdiff(A,B)：

返回的C元素為在A中出現(xiàn)，但是沒(méi)有在B中出現(xiàn)，ia為其索引值，滿(mǎn)足C = A(ia)。

D = D(:)'; % ensure that D is a row vector

D(:)按列將矩陣變?yōu)榱邢蛄?/p>

reshape是根據(jù)原來(lái)的矩陣創(chuàng)造新的矩陣，repmat是將原來(lái)的矩陣復(fù)制多次，構(gòu)成一個(gè)大矩陣。

>> A=magic(4)

A =

16 2 3 13

5 11 10 8

9 7 6 12

4 14 15 1

B = reshape(A,...,[],...) calculates the length of the dimension represented by the placeholder [], such that the product of the dimensions equals prod(size(A)). The value of prod(size(A)) must be evenly divisible by the product of the specified dimensions. You can use only one occurrence of [].

>> reshape(A,[],2)

ans =

16 3

5 10

9 6

4 15

2 13

11 8

7 12

14 1

max(a)，若a為向量，返回大值，若a為矩陣，返回每列大值作為一個(gè)向量。

C = max(A,B) returns an array the same size as A and B with the largest elements taken from A or B. The dimensions of A and B must match, or they may be scalar.

C = max(A,[],dim) returnsthe largest elements along the dimension of A specifiedby scalar dim. For example, max(A,[],1) producesthe maximum values along the first dimension of A.

[C,I] = max(...) finds the indices of the maximum values of A, and returns them in output vector I. If there are several identical maximum values, the index of the first one found is returned.

K>> c=magic(4)

c =

16 2 3 13

5 11 10 8

9 7 6 12

4 14 15 1

K>> max(c) 等同于max(c,[],1)

ans =

16 14 15 13

K>> max(c,[],2)

ans =

16

11

12

15

K>> [m,n]=max(c,[],2)

m =

16

11

12

15

n =

1

2

4

3

find； sparse；full

[i,j,s] = find(S);
[m,n] = size(S);
S = sparse(i,j,s,m,n);

對(duì)于稀疏矩陣，i,j分別是不為零項(xiàng)的index，s是其值

So does this, if the last row and column have nonzero entries:

[i,j,s] = find(S);
S = sparse(i,j,s);

標(biāo)題名稱(chēng)：Matlab函數(shù)-創(chuàng)新互聯(lián)
瀏覽路徑：http://redsoil1982.com.cn/article24/dissje.html

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