Remove Columns from Dataframe Where Some of Values Are Na

Here's a convenient way to do it using the dplyr function select_if(). Combine not (!), any() and is.na(), which is equivalent to selecting all columns that don't contain any NA values.

library(dplyr)
Itun %>%
    select_if(~ !any(is.na(.)))

You can use transpose twice:

newdf <- t(na.omit(t(df)))

Alternatively, select(where(~FUNCTION)) can be used:

library(dplyr)

(df <- data.frame(x = letters[1:5], y = NA, z = c(1:4, NA)))
#>   x  y  z
#> 1 a NA  1
#> 2 b NA  2
#> 3 c NA  3
#> 4 d NA  4
#> 5 e NA NA

# Remove columns where all values are NA
df %>% 
  select(where(~!all(is.na(.))))
#>   x  z
#> 1 a  1
#> 2 b  2
#> 3 c  3
#> 4 d  4
#> 5 e NA
  
# Remove columns with at least one NA  
df %>% 
  select(where(~!any(is.na(.))))
#>   x
#> 1 a
#> 2 b
#> 3 c
#> 4 d
#> 5 e

data[,!apply(is.na(data), 2, any)]

A base R method related to the apply answers is

Itun[!unlist(vapply(Itun, anyNA, logical(1)))]
  v1
1  1
2  1
3  2
4  1
5  2
6  1

Here, vapply is used as we are operating on a list, and, apply, it does not coerce the object into a matrix. Also, since we know that the output will be logical vector of length 1, we can feed this to vapply and potentially get a little speed boost. For the same reason, I used anyNA instead of any(is.na()).

Another alternative with the dplyr package would be to make use of the Filter function

Filter(function(x) !any(is.na(x)), Itun)

with data.table would be a little more cumbersome

setDT(Itun)[,.SD,.SDcols=setdiff((1:ncol(Itun)),
                                which(colSums(is.na(Itun))>0))]