Create column with minimum (first) level of factors by group r - r

Question

I have a data frame with a mixture of patient and appointment information. Each patient may have attended multiple appointments. Some patient information is collected at each appointment, with the result that some patient information is conflicting and some is missing.
I want to fill in the missing patient information for appointments where it wasn't recorded based upon data recorded at other appointments. And (this is where I'm stuck) I want to take the 'minimum' level (earliest in order of levels) recorded for factors that have conflicting information for the same patient. (In the example below the factor levels are in alphabetical order , but this isn't always the case).
This is similar to this question, but I am using factors instead of characters, and I have multiple factors that I want the minimum for so can't filter by row.
eg. I have
df.have <- data.frame(
grp_id = rep(1:3, each = 2),
grpvar1 = factor(c("B", "A", "B", "C", NA, "A")),
grpvar2 = factor(c("a", "b", "c", NA, NA, "x")),
appt_id = 1:6)
I want
grp_id grpvar1 grpvar2 appt_id
1 A a 1
1 A a 2
2 B c 3
2 B c 4
3 A x 5
3 A x 6
or at least
grp_id grpvar1 grpvar1.1
1 A a
2 B c
3 A x

Answer 1

We can try with summarise_each. As we need the first level of 'grpvar' variables, we need to make sure that the unused levels are dropped before we do that (using droplevels).
df.have %>%
group_by(grp_id) %>%
summarise_each(funs(first(levels(droplevels(.)))), grpvar1:grpvar2)
# grp_id grpvar1 grpvar2
# <int> <chr> <chr>
#1 1 A a
#2 2 B c
#3 3 A x
Or if we use mutate_each, we get the first output
df.have %>%
group_by(grp_id) %>%
mutate_each(funs(levels(droplevels(.))[1]), grpvar1:grpvar2)
# grp_id grpvar1 grpvar2 appt_id
# <int> <chr> <chr> <int>
#1 1 A a 1
#2 1 A a 2
#3 2 B c 3
#4 2 B c 4
#5 3 A x 5
#6 3 A x 6
if we need the output as factor columns for 'grpvar's.
df.have %>%
group_by(grp_id) %>%
mutate_each(funs(factor(levels(droplevels(.))[1])), grpvar1:grpvar2)
# grp_id grpvar1 grpvar2 appt_id
# <int> <fctr> <fctr> <int>
#1 1 A a 1
#2 1 A a 2
#3 2 B c 3
#4 2 B c 4
#5 3 A x 5
#6 3 A x 6
Or using data.table
library(data.table)
setDT(df.have)[, lapply(.SD, function(x) levels(droplevels(x))[1]) ,
grp_id, .SDcols = grpvar1:grpvar2]

Answer 2

While revisiting this for a different project with a larger dataset I realised that it was much more efficient (though required more keystrokes) to convert the factors to numbers, take the minimum, then reconvert back to factors.
library(data.table)
library(dplyr)
set.seed(1)
n <- 100L
dat <- data.table(
grp_id = rep(1:n/10, each = 10),
grpvar1 = factor(sample(c(LETTERS, NA), n, replace = TRUE), levels = LETTERS),
grpvar2 = factor(sample(c(letters, NA), n, replace = TRUE), levels = letters),
appt_id = 1:n)
cols <- c("grpvar1","grpvar2")
dplyr_fct <- function(data, cols) {
data %>%
group_by(grp_id) %>%
mutate_each(funs(factor(levels(droplevels(.))[1])), one_of(cols))
}
dt_fct <- function(data, cols) {
data[, lapply(.SD, function(x) levels(droplevels(x))[1]), grp_id, .SDcols = cols]}
dt_nmbr <- function(data, cols) {
dat_out <- copy(data)
v_lvl = lapply(dat_out[, .SD, .SDcols = cols], levels)
# Convert factors to numeric
for(col in cols) set(dat_out, j = col, value = as.numeric(dat_out[[col]]))
# Select highest value
dat_out[, (cols):= lapply(.SD, min, na.rm = TRUE), by = grp_id, .SDcols = cols]
# Convert back to factor
for(col in cols) set(dat_out, j = col,
value = factor(dat_out[[col]], levels = 1:length(v_lvl[[col]]), labels = v_lvl[[col]]))
assign("dat_out", dat_out, envir = .GlobalEnv)
}
mbm <- microbenchmark::microbenchmark(
mbm_dplyr_fct = dplyr_fct(dat, cols),
mbm_dt_fct = dt_fct(dat, cols),
mbm_dt_nmbr = dt_nmbr(dat, cols)
)
mbm
Unit: milliseconds
expr min lq mean median uq max neval cld
mbm_dplyr_fct 84.487484 85.829834 90.988740 87.015878 91.159178 120.22171 100 c
mbm_dt_fct 56.768529 58.007094 60.988083 58.831850 60.269427 87.11799 100 b
mbm_dt_nmbr 4.181538 4.406392 4.540248 4.557948 4.619757 6.04197 100 a