combination of expand.grid and mapply?

问题

I am trying to come up with a variant of mapply (call it xapply for now) that combines the functionality (sort of) of expand.grid and mapply. That is, for a function FUN and a list of arguments L1, L2, L3, ... of unknown length, it should produce a list of length n1*n2*n3 (where ni is the length of list i) which is the result of applying FUN to all combinations of the elements of the list.

If expand.grid worked to generate lists of lists rather than data frames, one might be able to use it, but I have in mind that the lists may be lists of things that won't necessarily fit into a data frame nicely.

This function works OK if there are exactly three lists to expand, but I am curious about a more generic solution. (FLATTEN is unused, but I can imagine that FLATTEN=FALSE would generate nested lists rather than a single list ...)

xapply3 <- function(FUN,L1,L2,L3,FLATTEN=TRUE,MoreArgs=NULL) {
  retlist <- list()
  count <- 1
  for (i in seq_along(L1)) {
    for (j in seq_along(L2)) {
      for (k in seq_along(L3)) {
        retlist[[count]] <- do.call(FUN,c(list(L1[[i]],L2[[j]],L3[[k]]),MoreArgs))
        count <- count+1
      }
    }
  }
  retlist
}

edit: forgot to return the result. One might be able to solve this by making a list of the indices with combn and going from there ...

回答1:

I think I have a solution to my own question, but perhaps someone can do better (and I haven't implemented FLATTEN=FALSE ...)

xapply <- function(FUN,...,FLATTEN=TRUE,MoreArgs=NULL) {
  L <- list(...)
  inds <- do.call(expand.grid,lapply(L,seq_along)) ## Marek's suggestion
  retlist <- list()
  for (i in 1:nrow(inds)) {
    arglist <- mapply(function(x,j) x[[j]],L,as.list(inds[i,]),SIMPLIFY=FALSE)
    if (FLATTEN) {
      retlist[[i]] <- do.call(FUN,c(arglist,MoreArgs))
    }
  }
  retlist
}

edit: I tried @baptiste's suggestion, but it's not easy (or wasn't for me). The closest I got was

xapply2 <- function(FUN,...,FLATTEN=TRUE,MoreArgs=NULL) {
  L <- list(...)
  xx <- do.call(expand.grid,L)
  f <- function(...) {
    do.call(FUN,lapply(list(...),"[[",1))
  }
  mlply(xx,f)
}

which still doesn't work. expand.grid is indeed more flexible than I thought (although it creates a weird data frame that can't be printed), but enough magic is happening inside mlply that I can't quite make it work.

Here is a test case:

L1 <- list(data.frame(x=1:10,y=1:10),
           data.frame(x=runif(10),y=runif(10)),
           data.frame(x=rnorm(10),y=rnorm(10)))

L2 <- list(y~1,y~x,y~poly(x,2))          
z <- xapply(lm,L2,L1)
xapply(lm,L2,L1)

来源：https://stackoverflow.com/questions/6515134/combination-of-expand-grid-and-mapply

标签

lapply