将稀疏矩阵索引列表转换为R中的矩阵

Question

I have this list of strings:

dat <- list(V1=c("1:23","4:12"),V2=c("1:3","2:12","6:3"))

the list elements V1 and V2 are the columns. 1:23 means "the first entry in this column has value 23". All other entries should be zero. The dimension of the matrix is indicated by the highest entry, in this case we have 2 columns (V1 and V2) and the highest row number is a 6, so it would result in a 2x6 matrix like this:

matrix(c(23,3,
     0,12,
     0,0,
     12,0,
     0,0,
     0,3),nrow=6,ncol=2,byrow=T)

how can this convertion be achieved?

Answer 1

Solution:

dat <- list(V1=c("1:23","4:12"),V2=c("1:3","2:12","6:3"))
y <- inverse.rle(list(values = 1:length(dat),lengths = sapply(dat,length)))

x <-  as.numeric(unlist(sapply(dat,function(y)sapply(strsplit(y,":"),function(x)x[1]))))
val <- as.numeric(unlist(sapply(dat,function(y)sapply(strsplit(y,":"),function(x)x[2]))))

num_row <- max(x)
num_col <- max(y) 
m = matrix(0, nrow = num_row, ncol = num_col)
m[cbind(x,y)] <- val
m

Answer 2

You may also try

library(dplyr)
library(tidyr)
library(Matrix)

 d1 <- unnest(dat,col) %>% 
           separate(x, into=c('row', 'val'), ':', convert=TRUE)  %>% 
           extract(col, into='col', '\\D+(\\d+)', convert=TRUE)

 as.matrix(with(d1, sparseMatrix(row, col, x=val)))
 #     [,1] [,2]
 #[1,]   23    3
 #[2,]    0   12
 #[3,]    0    0
 #[4,]   12    0
 #[5,]    0    0
 #[6,]    0    3