使用R中的稀疏矩阵从矢量中提取元素，而不转换为密集矩阵

Posted 2021-03-29

我想从矢量x1中提取所有元素，其中第i列存在于稀疏矩阵中。我需要删除所有稀疏元素，但结果应该在他们自己的对象/列表/矩阵中逐行存在。

鉴于：

> x1
 [1]  1  2  3  4  5  6  7  8  9 10
> sparse_mat
8 x 10 sparse Matrix of class "ngCMatrix"

[1,] | | | . . . . . . .
[2,] . | | | . . . . . .
[3,] . . | | | . . . . .
[4,] . . . | | | . . . .
[5,] . . . . | | | . . .
[6,] . . . . . | | | . .
[7,] . . . . . . | | | .
[8,] . . . . . . . | | |

期望的结果：

     [,1] [,2] [,3]
[1,]    1    2    3
[2,]    2    3    4
[3,]    3    4    5
[4,]    4    5    6
[5,]    5    6    7
[6,]    6    7    8
[7,]    7    8    9
[8,]    8    9   10

---

更完整的例子与评论

library(Matrix)
library(purrr)

x1 <- 1:10

create_seq_sparse <- function(n, len) {
  bandSparse(m = n, n = n - len + 1L, k = seq_len(len) - 1L)
}

sparse_mat <- create_seq_sparse(10, 3)
sparse_mat
#> 8 x 10 sparse Matrix of class "ngCMatrix"
#>                         
#> [1,] | | | . . . . . . .
#> [2,] . | | | . . . . . .
#> [3,] . . | | | . . . . .
#> [4,] . . . | | | . . . .
#> [5,] . . . . | | | . . .
#> [6,] . . . . . | | | . .
#> [7,] . . . . . . | | | .
#> [8,] . . . . . . . | | |

# If there's a better way to do this, please advise?
mat_x1_mult_sparse <- t(t(sparse_mat) * x1) 
mat_x1_mult_sparse
#> 8 x 10 sparse Matrix of class "dgCMatrix"
#>                          
#> [1,] 1 2 3 . . . . . .  .
#> [2,] . 2 3 4 . . . . .  .
#> [3,] . . 3 4 5 . . . .  .
#> [4,] . . . 4 5 6 . . .  .
#> [5,] . . . . 5 6 7 . .  .
#> [6,] . . . . . 6 7 8 .  .
#> [7,] . . . . . . 7 8 9  .
#> [8,] . . . . . . . 8 9 10

# This is nice, but can't use in conjunction with keep?
# mat_x1_mult_sparse[1, , drop = FALSE] 

# Desired results, but this approach I think I lose the advantages of the sparse matrix?
mat_x1_mult_sparse[1, ] %>% keep(~ .x != 0)
#> [1] 1 2 3
mat_x1_mult_sparse[2, ] %>% keep(~ .x != 0)
#> [1] 2 3 4
# etc...
mat_x1_mult_sparse[8, ] %>% keep(~ .x != 0)
#> [1]  8  9 10

答案

一种选择是利用summary方法来获得非稀疏元素的索引

library(Matrix)
i1 <- summary(sparse_mat)
i2 <- as.matrix(i1[order(i1[,1]),]) # order by the row index
# multiply the sparse matrix by the replicated 'x1', extract elements
# with i2 index and convert it to n column matrix
matrix((sparse_mat * x1[col(sparse_mat)])[i2], ncol = 3, byrow = TRUE)
#.     [,1] [,2] [,3]
#[1,]    1    2    3
#[2,]    2    3    4
#[3,]    3    4    5
#[4,]    4    5    6
#[5,]    5    6    7
#[6,]    6    7    8
#[7,]    7    8    9
#[8,]    8    9   10

另一答案

一旦我注意到你不想让你的矩阵稀疏，就删除了之前的回答;仍然，想法是利用矩阵的i槽：

# convert to dgCMatrix since ngCMatrix can only be on/off
out = as(sparse_mat, 'dgCMatrix')

# subset to the "on" elements of sparse_mat, 
#   and replace with the column number. The column number is
#   not stored directly so we have to make it ourselves, basically
#   by looking for when the value in @i stays the same or goes down
out[sparse_mat] = c(1L, cumsum(diff(sparse_mat@i) <= 0) + 1L)
out
# 8 x 10 sparse Matrix of class "dgCMatrix"
#                          
# [1,] 1 2 3 . . . . . .  .
# [2,] . 2 3 4 . . . . .  .
# [3,] . . 3 4 5 . . . .  .
# [4,] . . . 4 5 6 . . .  .
# [5,] . . . . 5 6 7 . .  .
# [6,] . . . . . 6 7 8 .  .
# [7,] . . . . . . 7 8 9  .
# [8,] . . . . . . . 8 9 10

这应该是非常有效的，因为[的dgCMatrix方法应该是聪明的，你的替换正是所需的长度（没有浪费的元素）。