R矩阵包: Demean sparse matrix

Question

是否有一种简单的方法可以按列贬低稀疏矩阵，同时将零值视为缺失(使用 Matrix 包)？

我似乎遇到两个问题:

找到合适的列意味着

空单元格被视为零而不是缺失:

M0 <- matrix(rep(1:5,4),nrow = 4)
M0[2,2] <- M0[2,3] <- 0
M <- as(M0, "sparseMatrix")
M
#[1,] 1 5 4 3 2
#[2,] 2 . . 4 3
#[3,] 3 2 1 5 4
#[4,] 4 3 2 1 5
colMeans(M)
#[1] 2.50 2.50 1.75 3.25 3.50

正确的结果应该是:

colMeans_correct <- colSums(M) / c(4,3,3,4,4)
colMeans_correct
#[1] 2.500000 3.333333 2.333333 3.250000 3.500000

减去列平均值

对缺失的单元格也进行减法:

sweep(M, 2, colMeans_correct)
#4 x 5 Matrix of class "dgeMatrix"
#     [,1]       [,2]       [,3]  [,4] [,5]
#[1,] -1.5  1.6666667  1.6666667 -0.25 -1.5
#[2,] -0.5 -3.3333333 -2.3333333  0.75 -0.5
#[3,]  0.5 -1.3333333 -1.3333333  1.75  0.5
#[4,]  1.5 -0.3333333 -0.3333333 -2.25  1.5

附注希望发布由两个问题组成的问题不是问题。它们连接到相同的任务，似乎反射(reflect)了相同的问题 - 区分缺失值和实际零值。

Answer 1

一种选择是将 colSums 除以非零逻辑矩阵的 colSums

colSums(M)/colSums(M!=0)
#[1] 2.500000 3.333333 2.333333 3.250000 3.500000

或者另一种选择是将 0 替换为 NA 并使用 na.rm = TRUE 参数获取 colMeans

colMeans(M*NA^!M, na.rm = TRUE)
#[1] 2.500000 3.333333 2.333333 3.250000 3.500000

<小时/>

或者@user20650评论

colSums(M) / diff(M@p)
#[1] 2.500000 3.333333 2.333333 3.250000 3.500000

其中“p”是?sparseMatrix中提到的指针

In typical usage, p is missing, i and j are vectors of positive integers and x is a numeric vector. These three vectors, which must have the same length, form the triplet representation of the sparse matrix.

If i or j is missing then p must be a non-decreasing integer vector whose first element is zero. It provides the compressed, or “pointer” representation of the row or column indices, whichever is missing. The expanded form of p, rep(seq_along(dp),dp) where dp <- diff(p), is used as the (1-based) row or column indices.