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

Question

Is there any way in R to keep the index number same as before, while removing NULL and character[0] element from list. For 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 and character[0] using the 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 this the index is changed like:

[[1]]
[1] "D2HGDH"

[[2]]
[1] "TNF" "IL6" "CXCL1"

[[3]]
[1] "ICAM1"

which i don't want. I want the index number same as in list ld. So basically my desired output is looking something like:

[[1]]
[1] "D2HGDH"

[[4]]
[1] "TNF" "IL6" "CXCL1"

[[7]]
[1] "ICAM1"

Any way to do this? If yes than How?

I appreciated any help. Thank You.

Screenshot of my output.

The point is: if you remove some elements, indexes are changed (since index is the position of an element in the list)... so basically you can't...maybe you should think to use names instead of indexes... — digEmAll, Apr 19 '16 at 09:46
use named list elements instead. `names(myList) <- paste0("X", 1:length(mylist))` These names will be fixed — Richard Telford, Apr 19 '16 at 09:49

score 1 · Answer 1 · edited May 23 '17 at 11:52

1

Similar to @Richards's comments here is a solution for your 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 the list to names 
ld <- setNames(as.list(ld), c(1:length(ld)))

Lines below copied from R Remove NULL elements from list of lists

#Remove null and character values 
is.NullOb <- function(x) is.null(x) | all(sapply(x, is.null))

## Recursively step down into list, removing all 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 get the indexes of list that were left use

names(newList)

edited May 23 '17 at 11:52

Community

1
1

answered Apr 19 '16 at 09:57

discipulus

2,665
3
34
51

But total I have 1900 list, how can i write the names in each list.? – priyanka nimavat Apr 19 '16 at 09:59
Look at my solution, the names are automatically generated, not typed by hand, if what is required is numeric from 1 to 1900. – discipulus Apr 19 '16 at 10:00
Still the index numbers are changed. Thanks – priyanka nimavat Apr 19 '16 at 11:30
Did you use my code to remove null and character values? – discipulus Apr 19 '16 at 11:33
Could you run the above code and show a screenshot here? – discipulus Apr 19 '16 at 11:51
Yap surely i will show a screenshot here. Please wait. – priyanka nimavat Apr 19 '16 at 11:57
Let us [continue this discussion in chat](http://chat.stackoverflow.com/rooms/109554/discussion-between-lovedynasty-and-priyanka-nimavat). – discipulus Apr 19 '16 at 12:55
Instead of `c(1:length(ld))`, we'd usually write `seq_along(ld)`, fyi. Regarding filtering, in the OP's case, `ld[ lengths(ld) > 0L ]` seems to work fine. – Frank Apr 19 '16 at 12:58

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

1 Answers1