Take Symmetrical Mean of a tibble (ignoring the NAs) - r

Question

I have a tibble where the rows and columns are the same IDs and I would like to take the mean (ignoring the NAs) to make the df symmetrical. I am struggling to see how.
data <- tibble(group = LETTERS[1:4],
A = c(NA, 10, 20, NA),
B = c(15, NA, 25, 30),
C = c(20, NA, NA, 10),
D = c(10, 12, 15, NA)
)
I would normally do
A <- as.matrix(data[-1])
(A + t(A))/2
But this does not work because of the NAs.
Edit: below is the expected output.
output <- tibble(group = LETTERS[1:4],
A = c(NA, 12.5, 20, 10),
B = c(12.5, NA, 25, 21),
C = c(20, 25, NA, 12.5),
D = c(10, 21, 12.5, NA))

Answer 1

Here is a suggestion using tidyverse code.
library(tidyverse)
data <- tibble(group = LETTERS[1:4],
A = c(NA, 10, 20, NA),
B = c(15, NA, 25, 30),
C = c(20, NA, NA, 10),
D = c(10, 12, 15, NA)
)
A <- data %>%
pivot_longer(-group, values_to = "x")
B <- t(data) %>%
as.data.frame() %>%
setNames(LETTERS[1:4]) %>%
rownames_to_column("group") %>%
pivot_longer(-group, values_to = "y") %>%
left_join(A, by = c("group", "name")) %>%
mutate(
mean = if_else(!(is.na(x) | is.na(y)), (x + y)/2, x),
mean = if_else(is.na(mean) & !is.na(y), y, mean)
) %>%
select(-x, -y) %>%
pivot_wider(names_from = name, values_from = mean)
B
## A tibble: 4 x 5
# group A B C D
# <chr> <dbl> <dbl> <dbl> <dbl>
#1 A NA 12.5 20 10
#2 B 12.5 NA 25 21
#3 C 20 25 NA 12.5
#4 D 10 21 12.5 NA

Answer 2

Okay so this is how I ended up doing this. I would have preferred if I didnt use a for loop because the actual data I have is much bigger but beggars cant be choosers!
A <- as.matrix(data[-1])
for (i in 1:nrow(A)){
for (j in 1:ncol(A)){
if(is.na(A[i,j])){
A[i,j] <- A[j, i]
}
}
}
output <- (A + t(A))/2
output %>%
as_tibble() %>%
mutate(group = data$group) %>%
select(group, everything())
# A tibble: 4 x 5
group A B C D
<chr> <dbl> <dbl> <dbl> <dbl>
1 A NA 12.5 20 10
2 B 12.5 NA 25 21
3 C 20 25 NA 12.5
4 D 10 21 12.5 NA