I have an example dataframe s1

s1=data.frame(c1=c("red","green","blue","yellow","orange","black","white"),col1=c("car1","car2","car3","car4","car5","car6","car7"))
s1=s1 %>% remove_rownames %>% column_to_rownames(var="c1")

There is only one column-col1 and the row names are red,green, blue and so on.

       col1
red    car1
green  car2
blue   car3
yellow car4
orange car5
black  car6
white  car7

I also have a matrix containg only 1' and 0's

m1= matrix(c(1,0,0,1,0),nrow =7, ncol =3, byrow = TRUE) 

     [,1] [,2] [,3]
[1,]    1    0    0
[2,]    1    0    1
[3,]    0    0    1
[4,]    0    1    0
[5,]    0    1    0
[6,]    1    0    0
[7,]    1    0    1

I want to create a list such that each element of the list contains the row names of s1, only if the corresponding element of the matrix is 1. I need to iterate this over all columns of the matrix.

the output should looks something like this

l1=list(c("red","green","black","white"),c("yellow","orange"),c("green","blue","white"))

I tried using this code but i was unable to apply it to every column

row.names(s1)[which(m1[,1]==1)]

Note that both the actual dataframe and matrix are much bigger. Thank you!

CodePudding user response：

Split by columns of 'm1' with asplit on a logical matrix and use the index for subsetting the rownames of 's1'

lapply(asplit(m1 == 1, 2), function(x) row.names(s1)[x])

-output

[[1]]
[1] "red"   "green" "black" "white"

[[2]]
[1] "yellow" "orange"

[[3]]
[1] "green" "blue"  "white"

Or use apply with MARGIN=2 to loop over the columns and subset

apply(m1==1, 2, function(x) row.names(s1)[x])

CodePudding user response：

A more primitive and involved solution:

l1 <- ifelse(m1==1, rownames(s1), '') %>% 
  t() %>%
  split(seq(nrow(.))) %>%
  lapply(\(x){x[x!='']})

l1

$`1`
[1] "red"   "green" "black" "white"

$`2`
[1] "yellow" "orange"

$`3`
[1] "green" "blue"  "white"