从非零索引MATLAB处的向量减去行和列



假设我在matlab中有以下矩阵:

A =[0    0    4    0; 
    0    5    0    3; 
    1    2    0    0];

给定以下向量:

b1 = [1 2 3];
b2 = [2 3 4 5];

输出应该看起来像这样:

C1 =[0    0    3    0; 
     0    3    0    1; 
    -2   -1    0    0];
C2 =[0    0    0    0; 
     0    2    0   -2; 
    -2   -1    0    0];

c1和c2是从> non-Zero Elements 的向量中的原始矩阵A的列和行的缩写。注意实际上是稀疏矩阵。显然,不使用循环的答案将不胜感激!谢谢

这可能会更有效地记忆:

A =[0    0    4    0; 
    0    5    0    3; 
    1    2    0    0];
b1 = [1 2 3].';   % transpose so it's a column vector
b2 = [2 3 4 5].';
[Arows Acols Avals] = find(A);
C1 = sparse([Arows;Arows], [Acols;Acols], [Avals;-b1(Arows)]);
C2 = sparse([Arows;Arows], [Acols;Acols], [Avals;-b2(Acols)]);

结果:

>> full(C1)
ans =
   0   0   3   0
   0   3   0   1
  -2  -1   0   0
>> full(C2)
ans =
   0   0   0   0
   0   2   0  -2
  -1  -1   0   0

这利用了sparse添加为重复下标给出的值的事实。A可能是稀疏的或满的。

无需使用循环。首先执行减法,然后替换应保持0的元素。

C1 = A - repmat(b1.',1,size(A,2));
C2 = A - repmat(b2,size(A,1),1);
C1(A==0)=0;
C2(A==0)=0;
C1 =
     0     0     3     0
     0     3     0     1
    -2    -1     0     0
C2 =
     0     0     0     0
     0     2     0    -2
    -1    -1     0     0

在稀疏矩阵上测试

您还可以确认这将在稀疏Matirces

上起作用
A = sparse(10,10);
A(5:6,5:6)=rand(2);
b1 = rand(10,1);
b2 = rand(1,10);
B1 = A - repmat(b1,1,size(A,2));
B2 = A - repmat(b2,size(A,1),1);
B1(A==0)=0;
B2(A==0)=0;
C1 = A ~= 0; // save none zero elements of A
b1 = b1.';   // transpose b1
b1 = [b1, b1, b1, b1];  // create matrix of same size as A
C1 = C1.*b1;            
C1 = A-C1;

C1:

 0     0     3     0
 0     3     0     1
-2    -1     0     0

接下来是C2

 C2 = A ~= 0;
 k = [b2; b2; b2];
 C2 = C2.*k;
 C2 = A-C2;

C2:

 0     0     0     0
 0     2     0    -2
-1    -1     0     0

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