filter for complete cases in data.frame using dplyr (case-wise deletion)

问题

Is it possible to filter a data.frame for complete cases using dplyr? complete.cases with a list of all variables works, of course. But that is a) verbose when there are a lot of variables and b) impossible when the variable names are not known (e.g. in a function that processes any data.frame).

library(dplyr)
df = data.frame(
    x1 = c(1,2,3,NA),
    x2 = c(1,2,NA,5)
)

df %.%
  filter(complete.cases(x1,x2))

回答1:

Try this:

df %>% na.omit

or this:

df %>% filter(complete.cases(.))

or this:

library(tidyr)
df %>% drop_na

If you want to filter based on one variable's missingness, use a conditional:

df %>% filter(!is.na(x1))

df %>% drop_na(x1)

Other answers indicate that of the solutions above na.omit is much slower but that has to be balanced against the fact that it returns row indices of the omitted rows in the na.action attribute whereas the other solutions above do not.

str(df %>% na.omit)
## 'data.frame':   2 obs. of  2 variables:
##  $ x1: num  1 2
##  $ x2: num  1 2
##  - attr(*, "na.action")= 'omit' Named int  3 4
##    ..- attr(*, "names")= chr  "3" "4"

ADDED Have updated to reflect latest version of dplyr and comments.

ADDED Have updated to reflect latest version of tidyr and comments.

回答2:

This works for me:

df %>%
  filter(complete.cases(df))

Or a little more general:

library(dplyr) # 0.4
df %>% filter(complete.cases(.))

This would have the advantage that the data could have been modified in the chain before passing it to the filter.

Another benchmark with more columns:

set.seed(123)
x <- sample(1e5,1e5*26, replace = TRUE)
x[sample(seq_along(x), 1e3)] <- NA
df <- as.data.frame(matrix(x, ncol = 26))
library(microbenchmark)
microbenchmark(
  na.omit = {df %>% na.omit},
  filter.anonymous = {df %>% (function(x) filter(x, complete.cases(x)))},
  rowSums = {df %>% filter(rowSums(is.na(.)) == 0L)},
  filter = {df %>% filter(complete.cases(.))},
  times = 20L,
  unit = "relative")

#Unit: relative
#             expr       min        lq    median         uq       max neval
 #         na.omit 12.252048 11.248707 11.327005 11.0623422 12.823233    20
 #filter.anonymous  1.149305  1.022891  1.013779  0.9948659  4.668691    20
 #         rowSums  2.281002  2.377807  2.420615  2.3467519  5.223077    20
 #          filter  1.000000  1.000000  1.000000  1.0000000  1.000000    20

回答3:

Here are some benchmark results for Grothendieck's reply. na.omit() takes 20x as much time as the other two solutions. I think it would be nice if dplyr had a function for this maybe as part of filter.

library('rbenchmark')
library('dplyr')

n = 5e6
n.na = 100000
df = data.frame(
    x1 = sample(1:10, n, replace=TRUE),
    x2 = sample(1:10, n, replace=TRUE)
)
df$x1[sample(1:n, n.na)] = NA
df$x2[sample(1:n, n.na)] = NA


benchmark(
    df %>% filter(complete.cases(x1,x2)),
    df %>% na.omit(),
    df %>% (function(x) filter(x, complete.cases(x)))()
    , replications=50)

#                                                  test replications elapsed relative
# 3 df %.% (function(x) filter(x, complete.cases(x)))()           50   5.422    1.000
# 1               df %.% filter(complete.cases(x1, x2))           50   6.262    1.155
# 2                                    df %.% na.omit()           50 109.618   20.217

回答4:

This is a short function which lets you specify columns (basically everything which dplyr::select can understand) which should not have any NA values (modeled after pandas df.dropna()):

drop_na <- function(data, ...){
    if (missing(...)){
        f = complete.cases(data)
    } else {
        f <- complete.cases(select_(data, .dots = lazyeval::lazy_dots(...)))
    }
    filter(data, f)
}

[drop_na is now part of tidyr: the above can be replaced by library("tidyr")]

Examples:

library("dplyr")
df <- data.frame(a=c(1,2,3,4,NA), b=c(NA,1,2,3,4), ac=c(1,2,NA,3,4))
df %>% drop_na(a,b)
df %>% drop_na(starts_with("a"))
df %>% drop_na() # drops all rows with NAs

回答5:

try this

df[complete.cases(df),] #output to console

OR even this

df.complete <- df[complete.cases(df),] #assign to a new data.frame

The above commands take care of checking for completeness for all the columns (variable) in your data.frame.

回答6:

Just for the sake of completeness, dplyr::filter can be avoided altogether but still be able to compose chains just by using magrittr:extract (an alias of [):

library(magrittr)
df = data.frame(
  x1 = c(1,2,3,NA),
  x2 = c(1,2,NA,5))

df %>%
  extract(complete.cases(.), )

The additional bonus is speed, this is the fastest method among the filter and na.omit variants (tested using @Miha Trošt microbenchmarks).

来源：https://stackoverflow.com/questions/22353633/filter-for-complete-cases-in-data-frame-using-dplyr-case-wise-deletion

标签

dplyr

magrittr