B=[1 2 3;4 5 6;7 8 9;10 11 12]
d=[-3 0 2];
A=spdiags(B,d,4,7); %主对角线之下的对角线溢出时候，取最后一个值
B=spdiags(B,d,7,4); %主对角线之上的对角线溢出时候，取最后一个值
C=spdiags(B,d,4,4);
full(A)
full(B)
full(C)
Why would I get
ans =

2 0 3 0 0 0 0
0 5 0 6 0 0 0
0 0 8 0 9 0 0
10 0 0 11 0 12 0

ans =

2 0 9 0
0 5 0 12
0 0 8 0
1 0 0 11
0 4 0 0
0 0 7 0
0 0 0 10

ans =

0 0 8 0
0 5 0 0
0 0 0 0
2 0 0 0
Why is C so different from the previous two? In terms of having the numbers not shown on main diagnol and also having the numbers arranged from bottom to top？B=[1 2 3;4 5 6;7 8 9;10 11 12]
d=[-3 0 2];
A=spdiags(B,d,4,7); %主对角线之下的对角线溢出时候，取最后一个值
B=spdiags(B,d,7,4); %主对角线之上的对角线溢出时候，取最后一个值
C=spdiags(B,d,4,4);
full(A)
full(B)
full(C)
Why would I get
ans =

2 0 3 0 0 0 0
0 5 0 6 0 0 0
0 0 8 0 9 0 0
10 0 0 11 0 12 0

ans =

2 0 9 0
0 5 0 12
0 0 8 0
1 0 0 11
0 4 0 0
0 0 7 0
0 0 0 10

ans =

0 0 8 0
0 5 0 0
0 0 0 0
2 0 0 0
Why is C so different from the previous two? In terms of having the numbers not shown on main diagnol and also having the numbers arranged from bottom to top？ B=[1 2 3;4 5 6;7 8 9;10 11 12]
d=[-3 0 2];
A=spdiags(B,d,4,7); %主对角线之下的对角线溢出时候，取最后一个值
B=spdiags(B,d,7,4); %主对角线之上的对角线溢出时候，取最后一个值
C=spdiags(B,d,4,4);
full(A)
full(B)
full(C)
Why would I get
ans =

2 0 3 0 0 0 0
0 5 0 6 0 0 0
0 0 8 0 9 0 0
10 0 0 11 0 12 0

ans =

2 0 9 0
0 5 0 12
0 0 8 0
1 0 0 11
0 4 0 0
0 0 7 0
0 0 0 10

ans =

0 0 8 0
0 5 0 0
0 0 0 0
2 0 0 0
Why is C so different from the previous two? In terms of having the numbers not shown on main diagnol and also having the numbers arranged from bottom to top？ matlab MATLAB Answers — New Questions

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