CODEWITHSUNDEEP
r
ifett
ifett

Reputation: 275

How to subset a list and remove empty vectors?

require(plyr)

list <- list(a=c(1:3),
             b=c(1:5),
             c=c(1:8),
             d=c(1:10))

llply(list,function(x)(subset(x,subset=(x>5))))

The above returns:

$a
integer(0)

$b
integer(0)

$c
[1] 6 7 8

$d
[1]  6  7  8  9 10

How to return a list only with the existing values, here $c and $d?

Upvotes: 0

Views: 540

Answers (3)

Emilio Herrero
Emilio Herrero

Reputation: 11

You can use 'lengths()':

# Your original list
your_list1 <- list(a=c(1:3),
                   b=c(1:5),
                   c=c(1:8),
                   d=c(1:10))

# Your subseted list
your_list2 <- lapply(your_list1,function(x) x[ x > 5 ])

# Your 'non_empty_elements' list    
your_final_list <- your_list2[(lengths(your_list2) > 0)]


> your_final_list
$c
[1] 6 7 8

$d
[1]  6  7  8  9 10

Pdt: Names assigned to objets that you create must make sense to you, but I agree Stephen Henderson that it's not a good practice to use a functions' name for objects. I think it is not necessary either, to use such long names as I have, except for pedagogical reasons.

Upvotes: 0

Thomas
Thomas

Reputation: 44527

You don't need plyr for this:

out <- lapply(list, function(x) x[ x > 5 ] )
out[ sapply(out, length) > 0 ]

Result:

$c
[1] 6 7 8

$d
[1]  6  7  8  9 10

Upvotes: 1

Stephen Henderson
Stephen Henderson

Reputation: 6522

You really shouldn't call a list variable "list" as it's a function name. If it's called L then:

L[lapply(L,length)>0]

this should give a nonempty list

Upvotes: 0

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