CODEWITHSUNDEEP
rdataframe
costebk08
costebk08

Reputation: 1359

Return only columns containing NA in R

I have the following data-frame:

i3<-c(1,1,1,1,2,2)
i2<-c(NA,1,1,1,2,2)
i1<-c(1,NA,2,4,5,3)
newdat1<-data.frame(i3,i2,i1)
print(newdat1)
  i3 i2 i1
1  1 NA  1
2  1  1 NA
3  1  1  2
4  1  1  4
5  2  2  5
6  2  2  3

I realize the solution for this is quite simple, but I am trying to return all the columns that any NA so that the final result looks like:

  i2 i1
1 NA  1
2  1 NA
3  1  2
4  1  4
5  2  5
6  2  3

I have found the following code which does the opposite:

newdat1<-newdat1[, sapply(newdat1, Negate(anyNA)), drop = FALSE]

But I cannot find exactly what I am looking for. Thank you.

Upvotes: 2

Views: 166

Answers (4)

David Arenburg
David Arenburg

Reputation: 92292

So I just want to bring your attention that OPs solution is actually the best one (as I expected) because apply and colSums convert the whole data.frame to a matrix, while the other solution transposes the whole data set.

OPs own sapply solution works on vectors without transforming the whole data set while implementing a Primitive function, here are some benchmarks on a bigger data set

set.seed(123)
bidData <- as.data.frame(replicate(1e4, sample(c(NA, 1:3), 1e4, replace = TRUE)))

library(microbenchmark)
microbenchmark(
  mpalanco=bidData[,!complete.cases(t(bidData)), drop = FALSE],
  mikechir=bidData[,is.na(colSums(bidData)), drop = FALSE],
  sabddem =bidData[,!apply(bidData, 2, function(x) sum(is.na(x)) == 0 ), drop = FALSE],
  OP = bidData[, sapply(bidData, anyNA), drop = FALSE])

# Unit: milliseconds
#     expr       min         lq       mean     median         uq        max neval
# mpalanco 2347.0316 2401.32940 2434.24480 2421.22703 2449.32975 2972.82020   100
# mikechir  352.8597  363.01980  425.11366  403.58777  477.06792  799.15855   100
#  sabddem 1869.2324 2025.22459 2591.11786 2812.56430 2853.55268 3655.91325   100
#       OP   17.5455   18.25625   18.99749   18.65456   19.54728   25.36552   100

Upvotes: 2

MichaelChirico
MichaelChirico

Reputation: 34703

Using base R and colSums:

newdat1[,is.na(colSums(newdat1))]

  i2 i1
1 NA  1
2  1 NA
3  1  2
4  1  4
5  2  5
6  2  3

Upvotes: 1

mpalanco
mpalanco

Reputation: 13570

newdat1[!complete.cases(t(newdat1))]

Output:

  i2 i1
1 NA  1
2  1 NA
3  1  2
4  1  4
5  2  5
6  2  3

Upvotes: 5

SabDeM
SabDeM

Reputation: 7190

A solution with apply and subsetting:

ind <- apply(newdat1, 2, function(x) sum(is.na(x)) == 0 )

newdat1[!ind]
  i2 i1
1 NA  1
2  1 NA
3  1  2
4  1  4
5  2  5
6  2  3

Upvotes: 1

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