Re: [R] expand.grid without expanding

From: Ray Brownrigg <ray_at_mcs.vuw.ac.nz>
Date: Thu 09 Feb 2006 - 08:34:07 EST


> From: =?iso-8859-1?q?Lu=EDs_Torgo?= <ltorgo@liacc.up.pt>
> Date: Wed, 8 Feb 2006 18:08:40 +0000
>
> Dear list,
> I've recently came across a problem that I think I've solved and that I wanted
> to share with you for two reasons:
> - Maybe others come across the same problem.
> - Maybe someone has a much simpler solution that wants to share with me ;-)
>
> The problem is as follows: expand.grid() allows you to generate a data.frame
> with all combinations of a set of values, e.g.:
> > expand.grid(par1=-1:1,par2=c('a','b'))
> par1 par2
> 1 -1 a
> 2 0 a
> 3 1 a
> 4 -1 b
> 5 0 b
> 6 1 b
>
> There is nothing wrong with this nice function except when you have too many
> combinations to fit in your computer memory, and that was my problem: I
> wanted to do something for each combination of a set of variants, but this
> set was to large for storing in memory in a data.frame generated by
> expand.grid. A possible solution would be to have a set of nested for()
> cycles but I preferred a solution that involved a single for() cycle going
> from 1 to the number of combinations and then at each iteration having some
> form of generating the combination "i". And this was the "real problem": how
> to generate a function that picks the same style of arguments as
> expand.grid() and provides me with the values corresponding to line "i" of
> the data frame that would have been created bu expand.grid(). For instance,
> if I wanted the line 4 of the above call to expand.grid() I should get the
> same as doing:
> > expand.grid(par1=-1:1,par2=c('a','b'))[4,]
> par1 par2
> 4 -1 b
>
> but obviously without having to use expand.grid() as that involves generating
> a data frame that in my case wouldn't fit in the memory of my computer.
>
> Now, the function I've created was the following:
> --------------------------------------------
> getVariant <- function(id,vars) {
> if (!is.list(vars)) stop('vars needs to be a list!')
> nv <- length(vars)
> lims <- sapply(vars,length)
> if (id > prod(lims)) stop('id above the number of combinations!')
> res <- vector("list",nv)
> for(i in nv:2) {
> f <- prod(lims[1:(i-1)])
> res[[i]] <- vars[[i]][ceiling(id / f)]
> id <- id - (ceiling(id/f)-1)*f
> }
> res[[1]] <- vars[[1]][id]
> names(res) <- names(vars)
> res
> }
> --------------------------------------
> > expand.grid(par1=-1:1,par2=c('a','b'))[4,]
> par1 par2
> 4 -1 b
> > getVariant(4,list(par1=-1:1,par2=c('a','b')))
> $par1
> [1] -1
>
> $par2
> [1] "b"
>
> I would be glad to know if somebody came across the same problem and has a
> better suggestion on how to solve this.
>
A few minor improvements:

1) let id be a vector of indices
2) use %% and %/% instead of ceiling (perhaps debateable)
3) return a data frame as does expand.grid

So your function now looks like:

getVariant <- function(id, vars) {
  if (!is.list(vars)) stop('vars needs to be a list!')   nv <- length(vars)
  lims <- sapply(vars, length)
  if (any(id > prod(lims))) stop('id above the number of combinations!')   res <- vector("list", nv)
  for(i in nv:2) {
    f <- prod(lims[1:(i-1)])
    res[[i]] <- vars[[i]][(id - 1)%/%f + 1]     id <- (id - 1)%%f + 1
  }
  res[[1]] <- vars[[1]][id]
  names(res) <- names(vars)
  return(as.data.frame(res))
}

Now, for example, you get:

> expand.grid(par1=-1:1,par2=c('a','b'),par3=c('w','x','y','z'))[12:15,]

   par1 par2 par3
12 1 b x
13 -1 a y
14 0 a y
15 1 a y
> getVariant(12:15,list(par1=-1:1,par2=c('a','b'), par3=c('w','x','y','z')))
  par1 par2 par3
1 1 b x
2 -1 a y
3 0 a y
4 1 a y
>

Note that you will run into trouble when the product of the lengths is greater than the largest representable integer on your system.

Hope this helps,
Ray Brownrigg



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