Count how many values in some cells of a row are not NA (in R)

Question

I've got a data frame with lots of columns. For each row of the data frame, I'd like to get a count of how many columns are NA. The problem is that I'm only interested in a few of the columns, and want to (efficiently) call those columns out.

Using mutate the way I do in the fake sample below gives me the right answer.

library(stringr)

df  <- data_frame(
         id = 1:10
       , name = fruit[1:10]
       , word1 = c(words[1:5],NA,words[7:10])
       , word2 = words[11:20]
       , word3 = c(NA,NA,NA,words[25],NA,NA,words[32],NA,NA,words[65])
    ) %>%
    mutate(
        n_words = 
            as.numeric(!is.na(word1)) + 
            as.numeric(!is.na(word2)) + 
            as.numeric(!is.na(word3)) 
    )

However, it's painful to type and read even for a toy example like this--when I have more than 3 columns to count, it's largely useless. Is there a more R/dplyr-y way to write this, maybe using select() style syntax (eg. n_words = !count_blank(word1:word3))?

I considered using summarize() sans grouping, however, I need the data in the columns I'm counting and if I add them to group_by, I'm in the same boat of calling out nearly all columns again.

Answer 1

You can use is.na() over the selected columns, then rowSums() the result:

library(stringr)
df <- data_frame(
  id = 1:10
  , name = fruit[1:10]
  , word1 = c(words[1:5],NA,words[7:10])
  , word2 = words[11:20]
  , word3 = c(NA,NA,NA,words[25],NA,NA,words[32],NA,NA,words[65]))

df$word_count <- rowSums( !is.na( df [,3:5]))

df
      id         name    word1     word2   word3 n_words
   <int>        <chr>    <chr>     <chr>   <chr>   <dbl>
1      1        apple        a    actual    <NA>       2
2      2      apricot     able       add    <NA>       2
3      3      avocado    about   address    <NA>       2
4      4       banana absolute     admit   agree       3
5      5  bell pepper   accept advertise    <NA>       2
6      6     bilberry     <NA>    affect    <NA>       1
7      7   blackberry  achieve    afford alright       3
8      8 blackcurrant   across     after    <NA>       2
9      9 blood orange      act afternoon    <NA>       2
10    10    blueberry   active     again   awful       3

Edit

Using dplyr you could do this:

df %>% 
    select(3:5) %>% 
    is.na %>% 
    `!` %>% 
    rowSums

Answer 2

library(dplyr)
library(stringr)

df  <- data_frame(
  id = 1:10
  , name = fruit[1:10]
  , word1 = c(words[1:5],NA,words[7:10])
  , word2 = words[11:20]
  , word3 = c(NA,NA,NA,words[25],NA,NA,words[32],NA,NA,words[65])
) 

library(purrr)
# Rowwise sum of NAs
df %>% by_row(~ sum(is.na(.)), .collate = 'cols')

# Rowwise sum of non-NAs for word columns
df %>% 
  select(starts_with('word')) %>% 
  by_row(~ sum(!is.na(.)), .collate = 'cols')

Answer 3

Another dplyr solution:

library(stringr)

## define count function

count_na <- function(x) sum(!is.na(x))

df$count_na <- df %>%

  select(starts_with("word")) %>%

    apply(., 1, count_na)

## A tibble: 10 × 6
      id         name    word1     word2   word3 n_words
   <int>        <chr>    <chr>     <chr>   <chr>   <int>
1      1        apple        a    actual    <NA>       2
2      2      apricot     able       add    <NA>       2
3      3      avocado    about   address    <NA>       2
4      4       banana absolute     admit   agree       3
5      5  bell pepper   accept advertise    <NA>       2
6      6     bilberry     <NA>    affect    <NA>       1
7      7   blackberry  achieve    afford alright       3
8      8 blackcurrant   across     after    <NA>       2
9      9 blood orange      act afternoon    <NA>       2
10    10    blueberry   active     again   awful       3