从 dataframe 中删除所有值为 NA 的列

Question

I have a data frame where some of the columns contain NA values.

How can I remove columns where all rows contain NA values?

Answer 1

尝试这个：

df <- df[,colSums(is.na(df))<nrow(df)]

Answer 2

The two approaches offered thus far fail with large data sets as (amongst other memory issues) they create is.na(df) , which will be an object the same size as df .

Here are two approaches that are more memory and time efficient

An approach using Filter

Filter(function(x)!all(is.na(x)), df)

and an approach using data.table (for general time and memory efficiency)

library(data.table)
DT <- as.data.table(df)
DT[,which(unlist(lapply(DT, function(x)!all(is.na(x))))),with=F]

examples using large data (30 columns, 1e6 rows)

big_data <- replicate(10, data.frame(rep(NA, 1e6), sample(c(1:8,NA),1e6,T), sample(250,1e6,T)),simplify=F)
bd <- do.call(data.frame,big_data)
names(bd) <- paste0('X',seq_len(30))
DT <- as.data.table(bd)

system.time({df1 <- bd[,colSums(is.na(bd) < nrow(bd))]})
# error -- can't allocate vector of size ...
system.time({df2 <- bd[, !apply(is.na(bd), 2, all)]})
# error -- can't allocate vector of size ...
system.time({df3 <- Filter(function(x)!all(is.na(x)), bd)})
## user  system elapsed 
## 0.26    0.03    0.29 
system.time({DT1 <- DT[,which(unlist(lapply(DT, function(x)!all(is.na(x))))),with=F]})
## user  system elapsed 
## 0.14    0.03    0.18 

Answer 3

Update

You can now use select with the where selection helper. select_if is superceded, but still functional as of dplyr 1.0.2. (thanks to @mcstrother for bringing this to attention).

library(dplyr)
temp <- data.frame(x = 1:5, y = c(1,2,NA,4, 5), z = rep(NA, 5))
not_all_na <- function(x) any(!is.na(x))
not_any_na <- function(x) all(!is.na(x))

> temp
  x  y  z
1 1  1 NA
2 2  2 NA
3 3 NA NA
4 4  4 NA
5 5  5 NA

> temp %>% select(where(not_all_na))
  x  y
1 1  1
2 2  2
3 3 NA
4 4  4
5 5  5

> temp %>% select(where(not_any_na))
  x
1 1
2 2
3 3
4 4
5 5

Old Answer

dplyr now has a select_if verb that may be helpful here:

> temp
  x  y  z
1 1  1 NA
2 2  2 NA
3 3 NA NA
4 4  4 NA
5 5  5 NA

> temp %>% select_if(not_all_na)
  x  y
1 1  1
2 2  2
3 3 NA
4 4  4
5 5  5

> temp %>% select_if(not_any_na)
  x
1 1
2 2
3 3
4 4
5 5

Answer 4

Another way would be to use the apply() function.

If you have the data.frame

df <- data.frame (var1 = c(1:7,NA),
                  var2 = c(1,2,1,3,4,NA,NA,9),
                  var3 = c(NA)
                  )

then you can use apply() to see which columns fulfill your condition and so you can simply do the same subsetting as in the answer by Musa, only with an apply approach.

> !apply (is.na(df), 2, all)
 var1  var2  var3 
 TRUE  TRUE FALSE 

> df[, !apply(is.na(df), 2, all)]
  var1 var2
1    1    1
2    2    2
3    3    1
4    4    3
5    5    4
6    6   NA
7    7   NA
8   NA    9

Answer 5

Late to the game but you can also use the janitor package. This function will remove columns which are all NA, and can be changed to remove rows that are all NA as well.

df <- janitor::remove_empty(df, which = "cols")

Answer 6

df[sapply(df, function(x) all(is.na(x)))] <- NULL

Answer 7

Another options with purrr package:

library(dplyr)

df <- data.frame(a = NA,
                 b = seq(1:5), 
                 c = c(rep(1, 4), NA))

df %>% purrr::discard(~all(is.na(.)))
df %>% purrr::keep(~!all(is.na(.)))

Answer 8

The accepted answer does not work with non-numeric columns. From this answer , the following works with columns containing different data types

Filter(function(x) !all(is.na(x)), df)

Answer 9

You can use Janitor package remove_empty

library(janitor)

df %>%
  remove_empty(c("rows", "cols")) #select either row or cols or both

Also, Another dplyr approach

 library(dplyr) 
 df %>% select_if(~all(!is.na(.)))

OR

df %>% select_if(colSums(!is.na(.)) == nrow(df))

this is also useful if you want to only exclude / keep column with certain number of missing values eg

 df %>% select_if(colSums(!is.na(.))>500)

Answer 10

I hope this may also help. It could be made into a single command, but I found it easier for me to read by dividing it in two commands. I made a function with the following instruction and worked lightning fast.

naColsRemoval = function (DataTable) { na.cols = DataTable [ , .( which ( apply ( is.na ( .SD ) , 2 , all ) ) )] DataTable [ , unlist (na.cols) := NULL , with = F] }

.SD will allow to limit the verification to part of the table, if you wish, but it will take the whole table as

Answer 11

一个方便的base R选项可能是colMeans() ：

df[, colMeans(is.na(df)) != 1]

Answer 12

From my experience of having trouble applying previous answers, I have found that I needed to modify their approach in order to achieve what the question here is:

How to get rid of columns where for ALL rows the value is NA?

First note that my solution will only work if you do not have duplicate columns (that issue is dealt with here (on stack overflow)

Second, it uses dplyr .

Instead of

df <- df %>% select_if(~all(!is.na(.)))

I find that what works is

df <- df %>% select_if(~!all(is.na(.)))

The point is that the "not" symbol "!" needs to be on the outside of the universal quantifier. Ie the select_if operator acts on columns. In this case, it selects only those that do not satisfy the criterion

every element is equal to "NA"

Answer 13

An old question, but I think we can update @mnel's nice answer with a simpler data.table solution:

DT[, .SD, .SDcols = \\(x) !all(is.na(x))]

(I'm using the new \\(x) lambda function syntax available in R>=4.1, but really the key thing is to pass the logical subsetting through .SDcols .

Speed is equivalent.

microbenchmark::microbenchmark(
  which_unlist  = DT[,which(unlist(lapply(DT, \(x) !all(is.na(x))))),with=F],
  sdcols = DT[, .SD, .SDcols = \(x) !all(is.na(x))],
  times = 2
)
#> Unit: milliseconds
#>          expr      min       lq     mean   median       uq      max neval cld
#>  which_unlist 51.32227 51.32227 56.78501 56.78501 62.24776 62.24776     2   a
#>        sdcols 43.14361 43.14361 49.33491 49.33491 55.52621 55.52621     2   a

Answer 14

看门人::remove_constant() 做得很好。