How to pass column variables to apply function?

Question

I have this data.frame :

  id  |  amount1  | amount2  |  day1  |  day2
 ---------------------------------------------
  A   |    10     |    32    |   0    |   34
  B   |    54     |    44    |   8    |   43
  C   |    45     |    66    |   16   |   99    

df <- data.frame(id=c('A','B','C'), amount1=c(10,54,45), amount2=c(32,44,66),  day1=c(0,8,16), day2=c(34,43,99))

on which I would like to apply a function

df$res <-  apply(df, 1, myfunc)

where

myfunc <- function(x,y) sum(x) * mean(y)

only I'd like to pass the column variables as argument to the function, so that it basically should read

 apply(df, 1, myfunc, c(amount1, amount2), c(day1, day2))

for the first row this is

myfunc(c(10,32),c(0,34))
# [1] 714

Can this be done?

Answer 1

The data.table solution.

require(data.table)
dt <- data.table(df) # don't depend on `id` column as it may not be unique
# instead use 1:nrow(dt) in `by` argument
dt[, res := myfunc(c(amount1,amount2), c(day1, day2)), by=1:nrow(dt)]
> dt
#    id amount1 amount2 day1 day2    res
# 1:  A      10      32    0   34  714.0
# 2:  B      54      44    8   43 2499.0
# 3:  C      45      66   16   99 6382.5

When you have a lot of days columns that you'd want to take the mean of and multiply with the sum of amount1 and amount2 , then I'd do it in this manner, without using myfunc . But it should be straightforward to implement one if you REALLY need a function.

# dummy example
set.seed(45)
df <- data.frame(matrix(sample(1:100, 200, replace=T), ncol=10))
names(df) <- c(paste0("amount", 1:2), paste0("day", 1:8))
df$idx <- 1:nrow(df) # idx column for uniqueness

# create a data.table
require(data.table)
calc_res <- function(df) {
    dt <- data.table(df)
    # first get the mean
    id1 <- setdiff(names(dt), grep("day", names(dt), value=TRUE))
    dt[, res := rowMeans(.SD), by=id1]
    # now product of sum(amounts) and current res
    id2 <- setdiff(names(dt), names(dt)[1:2])
    dt[, res := sum(.SD) * res, by=id2]
}
dt.fin <- calc_res(df)

Answer 2

Like this:

df$res <- apply(df, 1, function(x) myfunc(as.numeric(x[c("amount1", "amount2")]),
                                          as.numeric(x[c("day1", "day2")])))

but consider plyr::adply as an alternative:

library(plyr)
adply(df, 1, transform, res = myfunc(c(amount1, amount2), c(day1, day2)))
#   id amount1 amount2 day1 day2    res
# 1  A      10      32    0   34  714.0
# 2  B      54      44    8   43 2499.0
# 3  C      45      66   16   99 6382.5

Answer 3

This works for your example. Perhaps the same technique can be used for the real problem:

> apply(df[-1], 1, function(x) myfunc(x[1:2], x[3:4]))
## [1]  714.0 2499.0 6382.5

As flodel indicates, it is best to use the names for one of the subsetting operations, to ensure that only these columns are used for apply. A subset is necessary to prevent the vector passed by apply from being converted to character, and specifying the columns explicitly means that additional columns in the data frame will not cause this problem.

apply(df[c("amount1", "amount2", "day1", "day2")], 1, 
      function(x) myfunc(x[1:2], x[3:4])
     )

In practice, I would be more likely to code something like this:

amount <- c("amount1", "amount2")
day    <- c("day1", "day2")

df$res <- apply(df[c(amount, day)], 1, function(x) myfunc(x[amount], x[day]))