how to use is.na() with dplyr, when column name is stored as string?

Question

I'm new to tidy evaluation- although some things make sense, I usually just try different combinations of {{}}, !! + quo, sym() etc.

But I can't get is.na() to work? here's an example:

column <- "Sepal.Length"
iris_na <- rbind(iris, rep(NA,5))
iris_na %>%
  subset(is.na(column))

The reason your attempts didn't work is that `subset` isn't a ‘dplyr’ function and performs no tidy evaluation. — Konrad Rudolph, Dec 01 '22 at 15:59
With `dplyr`, use `iris_na %>% filter(is.na(.data[[column]]))`. `filter` is the dplyr version of the base R function `subset` — MrFlick, Dec 01 '22 at 16:00
you can try with this syntax `iris_na[column] %>% subset(is.na(.))` — rral, Dec 01 '22 at 16:02

score 0 · Accepted Answer · answered Dec 01 '22 at 15:59

You can simple use get

 iris_na %>%
    subset(is.na(get(column)))
    Sepal.Length Sepal.Width Petal.Length Petal.Width Species
151           NA          NA           NA          NA    <NA>

Also you can use filter instead of subset and get the same result

how to use is.na() with dplyr, when column name is stored as string?

1 Answers1