Convert R list to dataframe with missing/NULL elements

Question

Given a list:

alist = list(
  list(name="Foo",age=22),
  list(name="Bar"),
  list(name="Baz",age=NULL)
 )

what's the best way to convert this into a dataframe with name and age columns, with missing values (I'll accept NA or "" in that order of preference)?

Simple methods using ldply fail because it tries to convert each list element into a data frame, but the one with the NULL barfs because the lengths don't match. Best I have at the moment is:

> ldply(alist,function(s){t(data.frame(unlist(s)))})
  name  age
1  Foo   22
2  Bar <NA>
3  Baz <NA>

but that's pretty icky and the numeric variable becomes a factor...

Kind of annoying that check.rows=FALSE still gives an error about rows having different numbers of observations... — Joshua Ulrich
– Joshua Ulrich, Commented Apr 3, 2013 at 17:26

flodel · Accepted Answer · 2013-04-03 17:41:33Z

18

Step1: remove NULL items

non.null.list <- lapply(alist, Filter, f = Negate(is.null))

Step2: stack everything together:

library(plyr)
rbind.fill(lapply(non.null.list, as.data.frame))
#   name age
# 1  Foo  22
# 2  Bar  NA
# 3  Baz  NA

Edit: In case you had a variable that is NULL for all your list items, it would not show up in your final output. If instead, you'd like a column filled with NA, the first step should not remove NULLs but replace them with NAs:

Step 1 alternative: replace NULL with NA:

non.null.list <- lapply(alist, lapply, function(x)ifelse(is.null(x), NA, x))

edited Apr 3, 2013 at 17:41

answered Apr 3, 2013 at 17:22

flodel

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2 Comments

flodel Over a year ago

That second step was the subject of this question: stackoverflow.com/questions/15753091/…, for which Josh O'Brien had a longer but more scalable answer.

Spacedman Over a year ago

I'm holding out for a solution that only has one loop/apply step... Otherwise this seems to crack it.

Community · Accepted Answer · 2017-05-23 12:02:03Z

11

A comment mentioned wanting only a single loop, which can be achieved with @flodel's answer just by putting the body of the two loops together:

rbind.fill(lapply(alist, function(f) {
  as.data.frame(Filter(Negate(is.null), f))
}))

giving

  name age
1  Foo  22
2  Bar  NA
3  Baz  NA

edited May 23, 2017 at 12:02

CommunityBot

11 silver badge

answered Apr 3, 2013 at 21:06

Brian Diggs

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