在 R 中创建下三角矩阵

Question

下面是一个6x2 矩阵。

 1.178214   0.1723376
 1.121873   0.1982651
 1.120752   0.2470480
 1.121873   0.3284342
 1.165325   0.5079993
 1.128625   1.0000000

我正在尝试使用上述6x2 矩阵生成两个单独的下三角矩阵。

预期的第一个下三角矩阵低于

w1         w2         w3         w4         w5          w6
0.1723376  0          0          0          0           0 
0.1982651  0.1723376  0          0          0           0
0.2470480  0.1982651  0.1723376  0          0           0
0.3284342  0.2470480  0.1982651  0.1723376  0           0
0.5079993  0.3284342  0.2470480  0.1982651  0.1723376   0
1.0000000  0.5079993  0.3284342  0.2470480  0.1982651   0.1723376  

逻辑是

中的值

  row1 column w1,(R1:W1 0.1723376) 
  row2 column w2 (R2:W2 0.1723376)
  row3 column w3 (R3:W3 0.1723376)
  row4 column w4 (R4:W4 0.1723376)
  row5 column w5 (R5:W5 0.1723376)
  row6 column w6 (R6:W6 0.1723376)

  row2 column w1,(R2:W1 0.1982651) 
  row3 column w2 (R3:W2 0.1982651)
  row4 column w3 (R4:W3 0.1982651)
  row5 column w4 (R5:W4 0.1982651)
  row6 column w5 (R6:W5 0.1982651)

类似，其余按此模式。

第二个下三角矩阵有点复杂，涉及到初始6x2矩阵的第一列。预期矩阵低于

w1         w2         w3         w4         w5          w6
0.1723376  0          0          0          0           0 
0.1640966  0.1982651  0          0           0
0.1639326  0.1980670  0.2470480  0          0           0
0.1640966  0.1982651  0.2472952  0.2472952  0           0
0.1704523  0.2059443  0.2568733  0.2568733  0.3411550   0
0.1650842  0.1994584  0.2487835  0.2487835  0.3304109   0.4920007

逻辑如下。

column1 w1 元素计算如下

1.178214 / (1.178214 + 1.121873 + 1.120752 + 1.121873 + 1.165325 + 1.128625)
1.121873 / (1.178214 + 1.121873 + 1.120752 + 1.121873 + 1.165325 + 1.128625)
1.120752 / (1.178214 + 1.121873 + 1.120752 + 1.121873 + 1.165325 + 1.128625)
1.121873 / (1.178214 + 1.121873 + 1.120752 + 1.121873 + 1.165325 + 1.128625)
1.165325 / (1.178214 + 1.121873 + 1.120752 + 1.121873 + 1.165325 + 1.128625)
1.128625 / (1.178214 + 1.121873 + 1.120752 + 1.121873 + 1.165325 + 1.128625)

column2 w2 元素估计如下

0.1982651 = 1.121873 / (1.121873 + 1.120752 + 1.121873 + 1.165325 + 1.128625)

0.198067 = 1.120752 / (1.121873 + 1.120752 + 1.121873 + 1.165325 + 1.128625)

0.1982651 = 1.121873 / (1.121873 + 1.120752 + 1.121873 + 1.165325 + 1.128625)

0.2059443 = 1.165325 / (1.121873 + 1.120752 + 1.121873 + 1.165325 + 1.128625)

0.1994584 = 1.128625 / (1.121873 + 1.120752 + 1.121873 + 1.165325 + 1.128625)

column3 w3 元素估计如下

0.247048 = 1.120752 / (1.120752 + 1.121873 + 1.165325 + 1.128625)

0.2472952 = 1.121873 / (1.120752 + 1.121873 + 1.165325 + 1.128625)

0.2568733 = 1.165325 / (1.120752 + 1.121873 + 1.165325 + 1.128625)

0.2487835 = 1.128625 / (1.120752 + 1.121873 + 1.165325 + 1.128625)

需要帮助在 r 中生成这些矩阵。

Answer 1

也许你可以试试下面的代码

M1 <- matrix(0,nrow = 6,ncol = 6)
M1[lower.tri(M1,diag = TRUE)] <- unlist(sapply(6:1,function(k) head(m[,2],k)))

M2 <- matrix(0,nrow = 6,ncol = 6)
M2[lower.tri(M2,diag = TRUE)] <- unlist(sapply(6:1, function(k) tail(m[,1],k)))
M2 <- t(t(M2)/colSums(M2))

这样

> M1
          [,1]      [,2]      [,3]      [,4]      [,5]      [,6]
[1,] 0.1723376 0.0000000 0.0000000 0.0000000 0.0000000 0.0000000
[2,] 0.1982651 0.1723376 0.0000000 0.0000000 0.0000000 0.0000000
[3,] 0.2470480 0.1982651 0.1723376 0.0000000 0.0000000 0.0000000
[4,] 0.3284342 0.2470480 0.1982651 0.1723376 0.0000000 0.0000000
[5,] 0.5079993 0.3284342 0.2470480 0.1982651 0.1723376 0.0000000
[6,] 1.0000000 0.5079993 0.3284342 0.2470480 0.1982651 0.1723376

> M2
          [,1]      [,2]      [,3]      [,4]      [,5] [,6]
[1,] 0.1723376 0.0000000 0.0000000 0.0000000 0.0000000    0
[2,] 0.1640966 0.1982651 0.0000000 0.0000000 0.0000000    0
[3,] 0.1639326 0.1980670 0.2470480 0.0000000 0.0000000    0
[4,] 0.1640966 0.1982651 0.2472952 0.3284342 0.0000000    0
[5,] 0.1704523 0.2059443 0.2568733 0.3411550 0.5079993    0
[6,] 0.1650842 0.1994584 0.2487835 0.3304109 0.4920007    1

数据

m <- structure(c(1.178214, 1.121873, 1.120752, 1.121873, 1.165325, 
1.128625, 0.1723376, 0.1982651, 0.247048, 0.3284342, 0.5079993, 
1), .Dim = c(6L, 2L))