Question

note: this is a direct follow up to this previous question

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I have very long dataframe consisting of two columns that I am using as arguments for a function that will find the value of a third column using mapply as so:

df$3rd <- mapply(myfunction, A=df$1st, B=df$2nd)

where myfunction has arguments A and B. While this works great for small datasets, it stalls for large datasets so I was thinking a good way to approach the problem would be to apply this function using ddply. I don’t know if ddply is the best approach for this problem but I am also having some trouble with syntax. So suggestions for either would be appreciated.

This is what I am trying:

> df$3rd <- ddply(df, .(1st), function(x) x$3rd <-
> mapply(myfunction, A=x$1st, B=df$second))

and this is the error I am getting:

Error in `$<-.data.frame`(`*tmp*`, "n", value = c(1L, 1L, 1L, 1L, 1L,  : 
  replacement has 112 rows, data has 16

EDIT:

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In light of the answer and comments I I am posting a small reproducible example below – it is one of the answers from the previous question. However as the commenters below note, ddply is probably not the way to go. I am trying Ramnath’s solution right now.

library(reshape2)
foo <- data.frame(x = c('a', 'a', 'a', 'b', 'b', 'b'), 
                  y = c('ab', 'ac', 'ad', 'ae', 'fx', 'fy'))
bar <- data.frame(x = c('c', 'c', 'c', 'd', 'd', 'd'), 
                  y = c('ab', 'xy', 'xz', 'xy', 'fx', 'xz'))

nShared <- function(A, B) {
    length(intersect(with(foo, y[x==A]), with(bar, y[x==B])))
}

# Enumerate all combinations of groups in foo and bar
(combos <- expand.grid(foo.x=unique(foo$x), bar.x=unique(bar$x)))

# Find number of elements in common among all pairs of groups
combos$n <- mapply(nShared, A=combos$foo.x, B=combos$bar.x)

# Reshape results into matrix form
dcast(combos, foo.x ~ bar.x)
#   foo.x c d
# 1     a 1 0
# 2     b 0 1

Answer 1

ddply isn’t what you’re after here, ddply(df,.(1st), FUNCTION) is more like:

for each val in unique(df$1st)
    outdf[nrow(outdf)+1,] = FUNCTION( df[df$1st==val] )

That is, it makes outdf consisting of FUNCTION applied to subsets of df determined by column 1st.

In any case, I think your error might be because you have df instead of x in function(x) x$3rd<-mapply(myfunction,A=x$1st, B=df$second) (the B argument)? Although it is hard to tell without a working example.

What exactly does myfunction do? I think your best bet is to vectorise myfunction so that you can just do df$third <- myfunction( A=df$first, B=df$second ).

For example, if myfunction <- function(A,B) { A+B }, instead of doing mapply(myfunction,df$first,df$second) you could equivalently do myfunction(df$first,df$second) and not even need mapply at all.