Remove rows where all variables are NA using dplyr

Question

I'm having some issues with a seemingly simple task: to remove all rows where all variables are NA using dplyr. I know it can be done using base R ( Remove rows in R matrix where all data is NA and Removing empty rows of a data file in R ), but I'm curious to know if there is a simple way of doing it using dplyr.

Example:

library(tidyverse)
dat <- tibble(a = c(1, 2, NA), b = c(1, NA, NA), c = c(2, NA, NA))
filter(dat, !is.na(a) | !is.na(b) | !is.na(c))

The filter call above does what I want but it's infeasible in the situation I'm facing (as there is a large number of variables). I guess one could do it by using filter_ and first creating a string with the (long) logical statement, but it seems like there should be a simpler way.

Another way is to use rowwise() and do() :

na <- dat %>% 
  rowwise() %>% 
  do(tibble(na = !all(is.na(.)))) %>% 
  .$na
filter(dat, na)

but that does not look too nice, although it gets the job done. Other ideas?

Answer 1

Since dplyr 0.7.0 new, scoped filtering verbs exists. Using filter_any you can easily filter rows with at least one non-missing column:

# dplyr 0.7.0
dat %>% filter_all(any_vars(!is.na(.)))

Using @hejseb benchmarking algorithm it appears that this solution is as efficient as f4.

UPDATE:

Since dplyr 1.0.0 the above scoped verbs are superseded. Instead the across function family was introduced, which allows to perform a function on multiple (or all) columns. Filtering rows with at least one column being not NA looks now like this:

# dplyr 1.0.0
dat %>% filter(if_any(everything(), ~ !is.na(.)))

Answer 2

I would suggest to use the wonderful janitor package here. Janitor is very user-friendly:

janitor::remove_empty(dat, which = "rows")

Answer 3

Benchmarking

@DavidArenburg suggested a number of alternatives. Here's a simple benchmarking of them.

library(tidyverse)
library(microbenchmark)

n <- 100
dat <- tibble(a = rep(c(1, 2, NA), n), b = rep(c(1, 1, NA), n))

f1 <- function(dat) {
  na <- dat %>% 
    rowwise() %>% 
    do(tibble(na = !all(is.na(.)))) %>% 
    .$na
  filter(dat, na)
}

f2 <- function(dat) {
  dat %>% filter(rowSums(is.na(.)) != ncol(.))
}

f3 <- function(dat) {
  dat %>% filter(rowMeans(is.na(.)) < 1)
}

f4 <- function(dat) {
  dat %>% filter(Reduce(`+`, lapply(., is.na)) != ncol(.))
}

f5 <- function(dat) {
  dat %>% mutate(indx = row_number()) %>% gather(var, val, -indx) %>% group_by(indx) %>% filter(sum(is.na(val)) != n()) %>% spread(var, val) 
}

# f1 is too slow to be included!
microbenchmark(f2 = f2(dat), f3 = f3(dat), f4 = f4(dat), f5 = f5(dat))

Using Reduce and lapply appears to be the fastest:

> microbenchmark(f2 = f2(dat), f3 = f3(dat), f4 = f4(dat), f5 = f5(dat))
Unit: microseconds
 expr        min          lq       mean      median         uq        max neval
   f2    909.495    986.4680   2948.913   1154.4510   1434.725 131159.384   100
   f3    946.321   1036.2745   1908.857   1221.1615   1805.405   7604.069   100
   f4    706.647    809.2785   1318.694    960.0555   1089.099  13819.295   100
   f5 640392.269 664101.2895 692349.519 679580.6435 709054.821 901386.187   100

Using a larger data set 107,880 x 40 :

dat <- diamonds
# Let every third row be NA
dat[seq(1, nrow(diamonds), 3), ]  <- NA
# Add some extra NA to first column so na.omit() wouldn't work
dat[seq(2, nrow(diamonds), 3), 1] <- NA
# Increase size
dat <- dat %>% 
  bind_rows(., .) %>%
  bind_cols(., .) %>%
  bind_cols(., .)
# Make names unique
names(dat) <- 1:ncol(dat)
microbenchmark(f2 = f2(dat), f3 = f3(dat), f4 = f4(dat))

f5 is too slow so it is also excluded. f4 seems to do relatively better than before.

> microbenchmark(f2 = f2(dat), f3 = f3(dat), f4 = f4(dat))
Unit: milliseconds
 expr      min       lq      mean    median       uq      max neval
   f2 34.60212 42.09918 114.65140 143.56056 148.8913 181.4218   100
   f3 35.50890 44.94387 119.73744 144.75561 148.8678 254.5315   100
   f4 27.68628 31.80557  73.63191  35.36144 137.2445 152.4686   100

Answer 4

Starting with dyplr 1.0, the colwise vignette gives a similar case as an example:

filter(across(everything(), ~ !is.na(.x))) #Remove rows with *any* NA

We can see it uses the same implicit "& logic" filter uses with multiple expressions. So the following minor adjustment selects all NA rows:

filter(across(everything(), ~ is.na(.x))) #Remove rows with *any* non-NA

But the question asks for the inverse set: Remove rows with all NA.

1. We can do a simple setdiff using the previous, or
2. we can use the fact that across returns a logical tibble and filter effectively does a row-wise all() (ie &).

Eg:

rowAny = function(x) apply(x, 1, any)
anyVar = function(fcn) rowAny(across(everything(), fcn)) #make it readable
df %<>% filter(anyVar(~ !is.na(.x))) #Remove rows with *all* NA

Or:

filterout = function(df, ...) setdiff(df, filter(df, ...))
df %<>% filterout(across(everything(), is.na)) #Remove rows with *all* NA

Or even combinine the above 2 to express the first example more directly:

df %<>% filterout(anyVar(~ is.na(.x))) #Remove rows with *any* NA

---

In my opinion, the tidyverse filter function would benefit from a parameter describing the 'aggregation logic'. It could default to "all" and preserve behavior, or allow "any" so we wouldn't need to write anyVar -like helper functions.

Answer 5

使用 dplyr 1.0 的解决方案很简单，不需要辅助函数，您只需要在正确的位置添加一个否定。

dat %>% filter(!across(everything(), is.na))

Answer 6

dplyr 1.0.4 introduced the if_any() and if_all() functions:

dat %>% filter(if_any(everything(), ~!is.na(.)))

or, more verbose:

dat %>% filter(if_any(everything(), purrr::negate(is.na)))

"Take dat and keep all rows where any entry is non-NA"

Answer 7

Here's another solution that uses purrr::map_lgl() and tidyr::nest() :

library(tidyverse)

dat <- tibble(a = c(1, 2, NA), b = c(1, NA, NA), c = c(2, NA, NA))

any_not_na <- function(x) {
  !all(map_lgl(x, is.na))
}


dat_cleaned <- dat %>%
  rownames_to_column("ID") %>%
  group_by(ID) %>%
  nest() %>%
  filter(map_lgl(data, any_not_na)) %>%
  unnest() %>%
  select(-ID)
## Warning: package 'bindrcpp' was built under R version 3.4.2

dat_cleaned
## # A tibble: 2 x 3
##       a     b     c
##   <dbl> <dbl> <dbl>
## 1    1.    1.    2.
## 2    2.   NA    NA

I doubt this approach will be able to compete with the benchmarks in @hejseb's answer, but I think it does a pretty good job at showing how the nest %>% map %>% unnest pattern works and users can run through it line-by-line to figure out what's going on.

Answer 8

You can use the function complete.cases from dplyr using the dot (.) for specify the previous dataframe on the chain.

library(dplyr)
df = data.frame(
    x1 = c(1,2,3,NA),
    x2 = c(1,2,NA,5),
    x3 = c(NA,2,3,5)
)
df %>%
   filter(complete.cases(.))

  x1 x2 x3
1  2  2  2

Answer 9

I a neat solution what works in dplyr 1.0.1 is to use rowwise()

dat %>%
  rowwise() %>%
  filter(!all(is.na(across(everything())))) %>%
  ungroup()

very similar to @Callum Savage 's comment on the top post but I missed it on the first pass, and without the sum()

Answer 10

(tidyverse 1.3.1)

data%>%rowwise()%>%
filter(!all(is.na(c_across(is.numeric))))

data%>%rowwise()%>%
filter(!all(is.na(c_across(starts_with("***")))))