[R] expand.grid game

[baptiste auguie]

Wow!

system.time({
all = blockparts(rep(9,8),17)
print( dim(all[,all[1,]!=0])[2] ) # remove leading 0s
})

## 229713
   user  system elapsed
  0.160   0.068   0.228

In some ways I think this is close to Hadley's suggestion, though I
didn't know how to implement it.

Thanks a lot to everybody who participated, I have learned interesting
things from a seemingly innocuous question.

Best regards,

baptiste


2009/12/21 Robin Hankin <rksh1 at cam.ac.uk>:
> Hi
>
> library(partitions)
> jj <- blockparts(rep(9,8),17)
> dim(jj)
>
> gives 318648
>
>
> HTH
>
> rksh
>
>
>
> baptiste auguie wrote:
>>
>> Dear list,
>>
>> In a little numbers game, I've hit a performance snag and I'm not sure
>> how to code this in C.
>>
>> The game is the following: how many 8-digit numbers have the sum of
>> their digits equal to 17?
>> The brute-force answer could be:
>>
>> maxi <- 9 # digits from 0 to 9
>> N <- 5 # 8 is too large
>> test <- 17 # for example's sake
>>
>> sum(rowSums(do.call(expand.grid, c(list(1:maxi), rep(list(0:maxi),
>> N-1)))) == test)
>> ## 3675
>>
>> Now, if I make N = 8, R stalls for some time and finally gives up with:
>> Error: cannot allocate vector of size 343.3 Mb
>>
>> I thought I could get around this using Reduce() to recursively apply
>> rowSum to intermediate results, but it doesn't seem to help,
>>
>>
>> a=list(1:maxi)
>> b=rep(list(0:maxi), N-1)
>>
>> foo <- function(a, b, fun="sum", ...){
>>  switch(fun,
>>         'sum' =  rowSums(do.call(expand.grid, c(a, b))),
>>         'mean' =  rowMeans(do.call(expand.grid, c(a, b))),
>>         apply(do.call(expand.grid, c(a, b)), 1, fun, ...)) # generic case
>> }
>>
>> sum(Reduce(foo, list(b), init=a) == test)
>> ## 3675 # OK
>>
>> Same problem with N=8.
>>
>> Now, if N was fixed I could write a little C code to do this
>> calculation term-by-term, something along those lines,
>>
>> test = 0;
>>
>> for (i1=1, i1=9, i1++) {
>>  for (i2=0, i2=9, i2++) {
>>
>>  [... other nested loops ]
>>
>>  test = test + (i1 + i2 + [...] == 17);
>>
>>  } [...]
>> }
>>
>> but here the number of for loops, N, should remain a variable.
>>
>> In despair I coded this in R as a wicked eval(parse()) construct, and
>> it does produce the expected result after an awfully long time.
>>
>> makeNestedLoops <- function(N=3){
>>
>>  startLoop <- function(ii, start=1, end=9){
>>    paste("for (i", ii, " in seq(",start,", ",end,")) {\n", sep="")
>>  }
>>
>>  first <- startLoop(1)
>>  inner.start <- lapply(seq(2, N), startLoop, start=0)
>>  calculation <- paste("test <- test + (", paste("i", seq(1, N),
>> sep="", collapse="+"), "==17 )\n")
>>  end <- replicate(N, "}\n")
>>  code.to.run <- do.call(paste, c(list(first), inner.start, calculation,
>> end))
>>  cat(code.to.run)
>>  invisible(code.to.run)
>> }
>>
>> test <- 0
>> eval(parse(text = makeNestedLoops(8)) )
>> ## 229713
>>
>> I hope I have missed a better way to do this in R. Otherwise, I
>> believe what I'm after is some kind of C or C++ macro expansion,
>> because the number of loops should not be hard coded.
>>
>> Thanks for any tips you may have!
>>
>> Best regards,
>>
>> baptiste
>>
>> ______________________________________________
>> R-help at r-project.org mailing list
>> https://stat.ethz.ch/mailman/listinfo/r-help
>> PLEASE do read the posting guide
>> http://www.R-project.org/posting-guide.html
>> and provide commented, minimal, self-contained, reproducible code.
>>
>
>
> --
> Robin K. S. Hankin
> Uncertainty Analyst
> University of Cambridge
> 19 Silver Street
> Cambridge CB3 9EP
> 01223-764877
>
>