transform data frame string variable names

Question

I have a data frame that contains dates and id's. I need to add multiple columns to this data frame based on each date. I use ddply to do this as follows:

ddply(df, "dt", transform, new_column1 = myfun(column_name_1))

However,I have a bunch of column names and would like to add multiple new columns. Is there a way that I can pass a string to transform instead of new_column1? For example I tried:

ddply(df, "dt", transform, get("some_column_name")=myfun(column_name_1))

but this does not work. Additionally, if I pass the column_name_1 to myfun as a string, can I just use get("column_name_1") within myfun to refer to the column?

UPDATE: NOT SURE HOW TO FORMAT THIS BETTER

input:
id    date    val
id1   d1      1
id2   d1      2
id3   d1      3
id4   d1      4
id1   d2      10
id2   d2      20
id3   d2      30
id4   d2      40

out (for 2 buckets for example)

id    date    val     bucket
id1   d1      1         1
id2   d1      2         1
id3   d1      3         2
id4   d1      4         2
id1   d2      10        1
id2   d2      20        1
id3   d2      30        2
id4   d2      40        2

Answer 1

Doing it with transform is slick, but why not something more basic like

tmpf <- function(x) {
   x[[new_column_name_1]] <- myfun(x[[column_name_1]])
   x[[new_column_name_2]] <- myfun(x[[column_name_2]])
   ...
   x
}
ddply(df,"dt",tmpf)

Or you can have a vector of column names to modify, or do it on the fly:

tmpf <- function(x,cols=c("column_name_1","column_name_2")) {
   newcols <- paste("new",cols,sep="_")
   for (i in seq_along(cols)) {
      x[[newcols[i]]] <- myfun(x[[cols[i]]])
   }
}

There's probably something even cleverer with assign in the appropriate environment.

If I had a reproducible example I could test this.