Calculate value based on multiple other values - r

Question

I would like to make a some new variables in R based on multiple (>100) other variables.
My dataset looks like this
sub_id diag_1_ais diag_2_ais diag_3_ais diag_4_ais diag_5_ais diag_1_br diag_2_br diag_3_br diag_4_br diag_5_br
1 1 1 1 2 2 1 6 0 1 6 1
2 2 2 3 2 5 1 3
3 3 0 0 <NA> 4 1 0 0 <NA> 2 2
4 4 NA 1 2 2 NA 1 1 4
5 5 NA 4 2 3 5 NA 4 3 4 3
The variables diag_x_ais can take integers from 0-6, and diag_x_br can take integers between 1-6.
I would like to make 6 new variables corresponding to the 6 possible diag_x_br values, i.e. the new variables would be called br_1, br_2 ... br_6. These new variables shall then be filled with the maximum value of the corresponding diag_x_ais variables, i.e.
if diag_1_br, diag_2_br, and diag_4_br are all 3, then br_3 should take the maximum value of diag_1_ais, diag_2_ais, and diag_4_ais.
Please also see the example dataset below:
sub_id diag_1_ais diag_2_ais diag_3_ais diag_4_ais diag_5_ais diag_1_br diag_2_br diag_3_br diag_4_br diag_5_br br_1 br_2 br_3 br_4 br_5 br_6
1 1 1 1 2 2 1 6 0 1 6 1 2 NA NA NA NA 2
2 2 2 1 4 3 5 5 2 2 1 3 3 4 5 NA 2 NA
3 3 0 0 NA 4 1 0 0 NA 2 2 NA 4 NA NA NA NA
4 4 NA 1 2 2 NA 1 1 4 2 NA NA 2 NA NA
5 5 NA 4 2 3 5 NA 4 3 4 3 NA NA 5 4 NA NA
Hereafter, I would like a final variable which calculates the sum of the up to three largest br_x variables, example displayed below:
sub_id diag_1_ais diag_2_ais diag_3_ais diag_4_ais diag_5_ais diag_1_br diag_2_br diag_3_br diag_4_br diag_5_br br_1 br_2 br_3 br_4 br_5 br_6 sum3
1 1 1 1 2 2 1 6 0 1 6 1 2 NA NA NA NA 2 4
2 2 2 1 4 3 5 5 2 2 1 3 3 4 5 NA 2 NA 12
3 3 0 0 NA 4 1 0 0 NA 2 2 NA 4 NA NA NA NA 4
4 4 NA 1 2 2 NA 1 1 4 2 NA NA 2 NA NA 4
5 5 NA 4 2 3 5 NA 4 3 4 3 NA NA 5 4 NA NA 9
My actual dataset has 60 diag_x_ais variables and 60 diag_x_br variables and 4000 rows.
I hope that someone can help me do this in R. Thank you!

Answer 1

I think you could use the following solution. I made a slight modification so that we only sum the first 3 max values:
library(dplyr)
library(purrr)
df %>%
bind_cols(as.data.frame(t(map_dfr(1:6, function(a) pmap_dfc(df, ~ {x <- c(...)[grepl("br", names(df))]
inds <- which(x == a)
if(length(inds) != 0) {
y <- c(...)[grepl("ais", names(df))]
max(y[inds])
} else {
NA
}})))) %>%
setNames(paste0("br", 1:6))) %>%
rowwise() %>%
mutate(sum = sum(sort(as.numeric(c_across(starts_with("br"))), decreasing = TRUE)[1:3], na.rm = TRUE)) %>%
select(starts_with("br"), sum)
Resulting output
# A tibble: 5 x 7
# Rowwise:
br1 br2 br3 br4 br5 br6 sum
<chr> <chr> <chr> <chr> <chr> <chr> <dbl>
1 2 NA NA NA NA 2 4
2 3 4 5 NA 2 NA 12
3 NA 4 NA NA NA NA 4
4 2 NA NA 2 NA NA 4
5 NA NA 5 4 NA NA 9

Answer 2

You could use some heavy data.transforming most likely not very efficient on large datasets. There are some empty values, NA and 0 in your dataset. I didn't handle them (and replaced the empty values by NA to make importing easier).
library(tidyr)
library(dplyr)
data %>%
pivot_longer(-sub_id,
names_to = c("name", "cat"),
names_pattern = ".*_(\\d+)_(.*)") %>%
pivot_wider(names_from = "cat") %>%
group_by(sub_id, br) %>%
summarise(value = max(ais), .groups = "drop") %>%
filter(br %in% 1:6) %>%
group_by(sub_id) %>%
mutate(sum = sum(tail(sort(value), 3))) %>%
pivot_wider(names_from = br,
names_glue = "br_{br}") %>%
select(sub_id, paste0("br_", 1:6), sum)
This returns
# A tibble: 5 x 8
sub_id br_1 br_2 br_3 br_4 br_5 br_6 sum
<dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl>
1 1 2 NA NA NA NA 2 4
2 2 3 4 5 NA 2 NA 12
3 3 NA 4 NA NA NA NA 4
4 4 2 NA NA 2 NA NA 4
5 5 NA NA 5 4 NA NA 9
Piping an addtional right_join(data, by = "sub_id") gives you your example output (minus the order of your columns).
I took an idea from this answer.
Data
data <- structure(list(sub_id = c(1, 2, 3, 4, 5), diag_1_ais = c(1, 2,
0, NA, NA), diag_2_ais = c(1, 1, 0, NA, 4), diag_3_ais = c(2,
4, NA, 1, 2), diag_4_ais = c(2, 3, 4, 2, 3), diag_5_ais = c(1,
5, 1, 2, 5), diag_1_br = c(6, 5, 0, NA, NA), diag_2_br = c(0,
2, 0, NA, 4), diag_3_br = c(1, 2, NA, 1, 3), diag_4_br = c(6,
1, 2, 1, 4), diag_5_br = c(1, 3, 2, 4, 3)), row.names = c(NA,
-5L), class = c("tbl_df", "tbl", "data.frame"))

Answer 3

For the first part:
data <- data.frame(sub_id = c(1,2,3,4,5),
diag_1_ais = c(1,2,0,NA,NA),
diag_2_ais = c(1,1,0,NA,4),
diag_3_ais = c(2,4,NA,1,2),
diag_4_ais = c(2,3,4,2,3),
diag_5_ais = c(1,5,1,2,5),
diag_1_br = c(6,5,0,NA,NA),
diag_2_br = c(0,2,0,NA,4),
diag_3_br = c(1,2,NA,1,3),
diag_4_br = c(6,1,2,1,4),
diag_5_br = c(1,3,2,4,3))
calc_br <- function(data, value, firstBr, lastBr) {
br <- c()
for (i in 1:nrow(data)){
if (length(which(data[i,c(firstBr:lastBr)] %in% value))!=0){
br <- c(br, c(max(data[i,which(data[i,c(firstBr:lastBr)] %in% value)+1])))
}
else {
br <- c(br, c(NA))
}
}
result <- br
}
firstBr = 7
lastBr = 11
data$br_1 <- calc_br(data,1,firstBr,lastBr)
data$br_2 <- calc_br(data,2,firstBr,lastBr)
data$br_3 <- calc_br(data,3,firstBr,lastBr)
data$br_4 <- calc_br(data,4,firstBr,lastBr)
data$br_5 <- calc_br(data,5,firstBr,lastBr)
data$br_6 <- calc_br(data,6,firstBr,lastBr)
This should yield the same results as in your example. You should only have to exchange lastBr and firstBr (to 62 and 122 i would guess).
For the second part this should do the trick:
br_sum <- c()
for (i in 1:nrow(data)){
br_sum <- c(br_sum, sum(data[i,lastBr+tail(order(data[i,c((lastBr+1):(lastBr+6))], na.last = NA), 3)]))
}
data$br_sum <- br_sum
For completness here my results:
sub_id diag_1_ais diag_2_ais diag_3_ais diag_4_ais diag_5_ais diag_1_br
1 1 1 1 2 2 1 6
2 2 2 1 4 3 5 5
3 3 0 0 NA 4 1 0
4 4 NA NA 1 2 2 NA
5 5 NA 4 2 3 5 NA
diag_2_br diag_3_br diag_4_br diag_5_br br_1 br_2 br_3 br_4 br_5 br_6 br_sum
1 0 1 6 1 2 NA NA NA NA 2 4
2 2 2 1 3 3 4 5 NA 2 NA 12
3 0 NA 2 2 NA 4 NA NA NA NA 4
4 NA 1 1 4 2 NA NA 2 NA NA 4
5 4 3 4 3 NA NA 5 4 NA NA 9