CODEWITHSUNDEEP
rlistmatrixsparse-matrix
spore234
spore234

Reputation: 3640

convert list of sparse matrix indices to matrix in R

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?

Upvotes: 5

Views: 351

Answers (2)

akrun
akrun

Reputation: 887138

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

Upvotes: 4

Eric Brooks
Eric Brooks

Reputation: 667

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

Upvotes: 4

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