在所有其他列中的行均为NA的条件下过滤一列的行，并重复n列

Question

Apologies if this has been asked before; I wasn't quite sure how to phrase the question, which might have prevented other questions from showing up in my search.

My situation is that I have a data set like this:

toy <- 
  data.frame(
    Serves_1 = c("yes", NA, "yes", "no", "yes", "no"),
    Serves_2 = c(NA, NA, "no", "no", "no", "yes"),
    Serves_3 = c(NA, "no", "yes", "no", NA, "no"),
    Serves_4 = c(NA, "yes", "yes", "no", "yes", "no")
  )
toy

I'm trying to determine how many rows have a non-NA for one column and NAs for all other columns. So take for example, column Serves_1:

toy %>%
  filter(
    !is.na(Serves_1) &
      is.na(Serves_2) &
      is.na(Serves_3) &
      is.na(Serves_4)
  ) %>%
  nrow

There is one row where Serves_1 has a non-NA value and, simultaneously, all other columns have NA for that row.

This code works fine, but I need to repeat this procedure for each column. I could just move the exclamation mark down the line for each column. But in my real dataset, I have to do this for over 20 columns.

Is there a more efficient way to do this (preferably using dplyr)?

Answer 1

You can use rowSums :

library(dplyr)
toy <- 
  data.frame(
    Serves_1 = c("yes", NA, "yes", "no", "yes", "no"),
    Serves_2 = c(NA, NA, "no", "no", "no", "yes"),
    Serves_3 = c(NA, "no", "yes", "no", NA, "no"),
    Serves_4 = c(NA, "yes", "yes", "no", "yes", "no")
  ) %>% 
  mutate(na_sum = rowSums(is.na(.)))

This gives you:

  Serves_1 Serves_2 Serves_3 Serves_4 na_sum
1      yes     <NA>     <NA>     <NA>      3
2     <NA>     <NA>       no      yes      2
3      yes       no      yes      yes      0
4       no       no       no       no      0
5      yes       no     <NA>      yes      1
6       no      yes       no       no      0

You can then filter rows where na_sum == 3 to get all rows where one value is not NA and the rest are:

toy %>% 
  filter(na_sum ==3)

Which gives us:

  Serves_1 Serves_2 Serves_3 Serves_4 na_sum
1      yes     <NA>     <NA>     <NA>      3

Answer 2

additional option

sum(apply(toy, 1, function(x) (length(x) - 1 == sum(is.na(x)))))