Call apply-like function on each row of dataframe with multiple arguments from each row

Question

You can apply apply to a subset of the original data.

 dat <- data.frame(x=c(1,2), y=c(3,4), z=c(5,6))
 apply(dat[,c('x','z')], 1, function(x) sum(x) )

or if your function is just sum use the vectorized version:

rowSums(dat[,c('x','z')])
[1] 6 8

If you want to use testFunc

 testFunc <- function(a, b) a + b
 apply(dat[,c('x','z')], 1, function(x) testFunc(x[1],x[2]))

EDIT To access columns by name and not index you can do something like this:

 testFunc <- function(a, b) a + b
 apply(dat[,c('x','z')], 1, function(y) testFunc(y['z'],y['x']))

A data.frame is a list, so ...

For vectorized functions do.call is usually a good bet. But the names of arguments come into play. Here your testFunc is called with args x and y in place of a and b. The ... allows irrelevant args to be passed without causing an error:

do.call( function(x,z,...) testFunc(x,z), df )

For non-vectorized functions, mapply will work, but you need to match the ordering of the args or explicitly name them:

mapply(testFunc, df$x, df$z)

Sometimes apply will work - as when all args are of the same type so coercing the data.frame to a matrix does not cause problems by changing data types. Your example was of this sort.

If your function is to be called within another function into which the arguments are all passed, there is a much slicker method than these. Study the first lines of the body of lm() if you want to go that route.

Use mapply

> df <- data.frame(x=c(1,2), y=c(3,4), z=c(5,6))
> df
  x y z
1 1 3 5
2 2 4 6
> mapply(function(x,y) x+y, df$x, df$z)
[1] 6 8

> cbind(df,f = mapply(function(x,y) x+y, df$x, df$z) )
  x y z f
1 1 3 5 6
2 2 4 6 8

New answer with dplyr package

If the function that you want to apply is vectorized, then you could use the mutate function from the dplyr package:

> library(dplyr)
> myf <- function(tens, ones) { 10 * tens + ones }
> x <- data.frame(hundreds = 7:9, tens = 1:3, ones = 4:6)
> mutate(x, value = myf(tens, ones))
  hundreds tens ones value
1        7    1    4    14
2        8    2    5    25
3        9    3    6    36

Old answer with plyr package

In my humble opinion, the tool best suited to the task is mdply from the plyr package.

Example:

> library(plyr)
> x <- data.frame(tens = 1:3, ones = 4:6)
> mdply(x, function(tens, ones) { 10 * tens + ones })
  tens ones V1
1    1    4 14
2    2    5 25
3    3    6 36

Unfortunately, as Bertjan Broeksema pointed out, this approach fails if you don't use all the columns of the data frame in the mdply call. For example,

> library(plyr)
> x <- data.frame(hundreds = 7:9, tens = 1:3, ones = 4:6)
> mdply(x, function(tens, ones) { 10 * tens + ones })
Error in (function (tens, ones)  : unused argument (hundreds = 7)

Others have correctly pointed out that mapply is made for this purpose, but (for the sake of completeness) a conceptually simpler method is just to use a for loop.

for (row in 1:nrow(df)) { 
    df$newvar[row] <- testFunc(df$x[row], df$z[row]) 
}