R: summerise verb with varying length - r

Question

[this question is now thoroughly rewritten. I hope this would clarify things]
I have a dataset describing several tests with multiple-answer questions. Each line contains the raw answers of one participant, and the score that participant was awarded for each question. Each test has a different answer key:
df <- data.frame(id =c(1, 2, 3, 4, 5, 6, 7, 8, 9,10), # participant's id
test =c(1, 1, 1, 1, 1, 2, 2, 2, 2, 2), # id of question set
ans01=c(1, 2, 3, 4, 1, 1, 2, 3, 3, 4), # raw answers
ans02=c(2, 2,NA, 3, 4, 4, 3, 1, 1, 2),
bin01=c(1, 0, 0, 0, 1, 0, 1, 0, 0, 0), # item scores
bin02=c(1, 1, 0, 0, 0, 1, 0, 0, 0, 0))
My problem is that the answer key is missing, and I need to recreate it from the dataset.
Currently, my solution is simple and creates a separate answer key for each test:
library(dplyr)
key_data <- df %>%
group_by(test) %>%
summarise(key01 = mean(ans01[bin01 == 1], na.rm=TRUE),
key02 = mean(ans02[bin02 == 1], na.rm=TRUE))
However, while this is ok for a short tests, it is not so ok for longer tests, containing dozens of questions.
Also, I want to be able to do so for future sets of tests, so flexibility is needed for the number of items.
Therefore, the question is whether there is a way to do so without writing a line for each item key.
Maybe loop through all variables, or passing string vectors as variable names?
[I answered my own question with a not-very-elegant solution. I'm sure this can be achieved in a much better way]

Answer 1

Hi you could use summarise_at and define the scope at the first argument.
See this exmaple using mtcars dataset.:
library(dplyr)
selection_a<-c("mpg","cyl","vs")
selection_b<-c("mpg","cyl","vs","qsec","carb")
# Use first selection (A)
mtcars %>%
summarise_at(selection_a , ~ mean(.x, na.rm = TRUE))
# Use second Selection (B)
mtcars %>%
summarise_at(selection_b , ~ mean(.x, na.rm = TRUE))
# combine selections (A+C)
selection_c<-c("gear","carb")
mtcars %>%
summarise_at( c(selection_a,selection_c), ~ mean(.x, na.rm = TRUE))

Answer 2

As a side note, I'm answering my own question with a-not-very-elegant-solution that I'm currently using. I'm sure there are shorter and much better solutions..
This required key table can be acquired with a loop, as follows:
MaxItems <- 2
pad0 <- function(x, n = 2) {
n0_pad <- n - nchar(x)
return(paste0(strrep("0",n0_pad), x))
}
library(dplyr)
## create the structure for the key table
keys <- df %>%
group_by(test) %>%
summarise(a01 = mean(ans01[bin01 == 1], na.rm=TRUE))
## add items to key table
for (i in 2:MaxItems) {
keysTemp <- df %>%
rename("tmpAns" = paste0("ans",pad0(i)),
"tmpBin" = paste0("bin",pad0(i))) %>%
group_by(test) %>%
summarise(tmpKey = mean(tmpAns[tmpBin == 1], na.rm=TRUE))
colnames(keysTemp)[2] <- paste0("a",pad0(i))
keys <- keys %>%
left_join(keysTemp, by = c("test"))
}