R 在表中工作-用于处理边距和和单个值的函数

这是我的小数据集

set.seed(123)
X1 <- rep(1:3, each = 4)
X2 <- c(1:4, 1:4, 1:4)
Y <- rnorm (12, 4, 20)
myd <- data.frame(X1, X2, Y)

# "unmelten" form of the dataframe 
aggregate(Y ~ X1, myd, 'c')
  X1         Y.1         Y.2         Y.3         Y.4
1  1  -7.2095129  -0.6035498  35.1741663   5.4101678
2  2   6.5857547  38.3012997  13.2183241 -21.3012247
3  3  -9.7370570  -4.9132394  28.4816359  11.1962765

另外两个例子来说明我的观点

For X2, Y1.1 cell
     (X2,Y.1) - sum(X2 row) - sum(Y.1 column) + grand total (= sum(Y))

For X3, Y.3 cell  
     (X3,Y.3) - sum(X3 row) - sum(Y.3 column) + grand total (= sum(Y))

同样，这将为矩阵中的每个单元格输出一个大小为3 x 4的矩阵

我正试图写一个函数来实现这一点，但我能得到任何想法。请帮帮我就这么做

z1 <- aggregate(Y ~ X1, myd, 'c')[,-1]
ans <- z1 - rowSums(z1) - colSums(z1) + sum(z1)

         [,1]      [,2]      [,3]      [,4]
[1,]  64.98307  65.92300  20.13181  34.45743
[2,]  31.60013 106.46100  75.71199 -40.37646
[3,] -17.03576  31.87768 108.41788  85.46648

z1
z1 <- aggregate(Y ~ X1, myd, 'c')[,-1]
ans <- z1 - rowSums(z1) - colSums(z1) + sum(z1)

         [,1]      [,2]      [,3]      [,4]
[1,]  64.98307  65.92300  20.13181  34.45743
[2,]  31.60013 106.46100  75.71199 -40.37646
[3,] -17.03576  31.87768 108.41788  85.46648