Keep list index number unchanged after removing NULL and character[0] in R -

is there way in r keep index number same before, while removing null , character[0] element list. example:

ld: [[1]] [1]"d2hgdh"  [[2]] character(0)  [[3]] character(0)  [[4]] [1] "tnf"   "il6"   "cxcl1"  [[5]] character(0)  [[6]] character(0)  [[7]] [1] "icam1"  [[8]] null 

than removing null , character[0] using following code.

#remove character 0 element rmchar <- lapply(seq(ld) , function(x) ld[[x]][!ld[[x]] == "character(0)" ])  # remove null is.nullob <- function(x) if(!(is.function(x))) is.null(x) | all(sapply(x,      is.null)) else false rmnullobs <- function(x) { x <- filter(negate(is.nullob), x) lapply(x, function(x) if (is.list(x)) rmnullobs(x) else x) }  rmnull <- rmnullobs(rmchar) 

so after running index changed like:

[[1]] [1] "d2hgdh"  [[2]] [1] "tnf" "il6" "cxcl1"  [[3]] [1] "icam1" 

which don't want. want index number same in list ld. desired output looking like:

[[1]] [1] "d2hgdh"  [[4]] [1] "tnf" "il6" "cxcl1"  [[7]] [1] "icam1" 

any way this? if yes how?

i appreciated help. thank you.

screenshot of output.

similar @richards's comments here solution case

ld = list(c("d2hgdh"), character(length = 0), character(length = 0),        c("tnf","il6","cxcl1"),character(length = 0),        character(length = 0), c("icam1"), null)  # convert list names  ld <- setnames(as.list(ld), c(1:length(ld))) 

lines below copied r remove null elements list of lists

#remove null , character values  is.nullob <- function(x) is.null(x) | all(sapply(x, is.null))  ## recursively step down list, removing such objects       rmnullobs <- function(x) {      x <- filter(negate(is.nullob), x)      lapply(x, function(x) if (is.list(x)) rmnullobs(x) else x)  }  newlist <- rmnullobs(ld) 

to indexes of list left use

names(newlist)