R creating a nested list from multiple vectors

Question

I am trying to create a nested list where the value from one vector becomes the name and the process repeats until the last vector which stores an actual value. I was thinking that the code I have below would work, but it isn't

chips[[toString(aCor[i])]]=list(toString(bCor[i])=list(toString(cCor[i])=list(toString(dCor[i])=eCor[i])))

I was expecting something like this if aCor=c(1,2,2,1), bCor=c(4,5,6,4), cCor=c(3,3,2,3), dCor=c(1,4,5,1), eCor=c(1,3,4,7)

Resulting list ["1"=["4"=["3"=["1"= 7]]], "2"=["5"=["3"=["4"=3]],"6"=["2"=["5"=4]]]]

$1
$1$4
$1$4$3
$1$4$3$1
[1] 7

$2
$2$5
$2$5$3
$1$4$3$4
[1] 3

$2
$2$6
$2$6$2
$1$6$2$5
[1] 4

Sorry if the expected list is formatted poorly. I wasnt sure the best way to do it. If there is a better way of doing this than a list I am open to suggestions, I would have used a dictionary in python and this was the closest I could find that would replicate it. I am currently getting this error

Error in parse(text = script) : parse error in text argument: unexpected '=' in function argument before

@akrun I am currently getting this error so I cant give a reproducible example Error in parse(text = script) : parse error in text argument: unexpected '=' in function argument before — Nikita Belooussov
– Nikita Belooussov, Commented Aug 6, 2020 at 6:37
It is just to understand what your input and expected would be — akrun
– akrun, Commented Aug 6, 2020 at 6:38
Would this be helpful? stackoverflow.com/questions/52872581/… — Ronak Shah
– Ronak Shah, Commented Aug 6, 2020 at 7:31

akrun · Accepted Answer · 2020-08-06 08:05:07Z

2

We could use lst from purrr or dplyr (or need setNames) which would work with assignment (:=)

nm1 <- 'a'
nm2 <- 'b'
dplyr::lst(!! nm1 := lst(!! nm2 := 5))
#$a
#$a$b
#[1] 5

If we need to create a nested list, use a for loop

lst1 <- list()

for(i in seq_along(aCor)) {
      an <- as.character(aCor[i])
      
      bn <- as.character(bCor[i])
      cn <- as.character(cCor[i])
      dn <- as.character(dCor[i])
      lst1[[an]][[bn]][[cn]][[dn]] <- eCor[[i]]
      

}

-output

lst1
#$`1`
#$`1`$`4`
#$`1`$`4`$`3`
#$`1`$`4`$`3`$`1`
#[1] 7




#$`2`
#$`2`$`5`
#$`2`$`5`$`3`
#$`2`$`5`$`3`$`4`
#[1] 3



#$`2`$`6`
#$`2`$`6`$`2`
#$`2`$`6`$`2`$`5`
#[1] 4

data

aCor <- c(1,2,2,1)
bCor <- c(4,5,6,4)
cCor <- c(3,3,2,3)
dCor <- c(1,4,5,1)
eCor <- c(1,3,4,7)

edited Aug 6, 2020 at 8:05

answered Aug 6, 2020 at 6:42

akrun

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13 Comments

Nikita Belooussov Over a year ago

why are you using := instead of = and why the !!

akrun Over a year ago

@NikitaBelooussov The = won't evaluate an object it would give list(nm1 = 5)

Nikita Belooussov Over a year ago

and the !! does what?

akrun Over a year ago

It evaluates the object

Nikita Belooussov Over a year ago

Thanks for your help. I am not sure why your code didnt work, I put up the code that seems like it is going to work in my case

|

Nikita Belooussov · Accepted Answer · 2020-08-06 08:46:09Z

This is the code I ended up making with the help of akrun in this question and another code that I found on this page written by IRTFM.

require(dplyr)

aCor <- c(1,2,2,1)
bCor <- c(4,5,6,4)
cCor <- c(3,3,2,3) 
dCor <- c(1,4,5,1)
eCor <- c(1,3,4,7)

appendList <- function (x, val) 
{
   stopifnot(is.list(x), is.list(val))
   xnames <- names(x)
   for (v in names(val)) {
      x[[v]] <- if (v %in% xnames && is.list(x[[v]]) && is.list(val[[v]])) 
        appendList(x[[v]], val[[v]])
        else c(val[[v]])
    }
   x
}


chips <- list()

for(i in seq_along(aCor)) {
   an <- as.character(aCor[i]) 
   bn <- as.character(bCor[i])
   cn <- as.character(cCor[i])
   dn <- as.character(dCor[i])
   #chips[[an]]=lst(!! bn := lst(!! cn := lst(!! dn := eCor[i])))
   list1=lst(!! an := lst(!! bn := lst(!! cn  := lst(!! dn:= eCor[i]))))
   chips=appendList(chips,list1)
}

chips

Output:

$1
$1$4
$1$4$3
$1$4$3$1
[1] 7

$`2`
$`2`$`5`
$`2`$`5`$`3`
$`2`$`5`$`3`$`4`
[1] 3

$`2`$`6`
$`2`$`6`$`2`
$`2`$`6`$`2`$`5`
[1] 4

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