[R] expand.grid game
Robin Hankin
rksh1 at cam.ac.uk
Mon Dec 21 09:37:38 CET 2009
Hello again everybody.
I fired off my reply before reading the correspondence about
the leading zeros.
You can also assume that there is at least one block
at the leading position, [so that position can take 0,1,2,...,8 additional
blocks] and distribute the remaining 16 blocks amongst all 8
places:
system.time(dim(blockparts(c(8,rep(9,7)),16)))
user system elapsed
0.056 0.016 0.069
>
this is faster! :-)
best wishes
rksh
baptiste auguie wrote:
> 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
>>
>>
>>
--
Robin K. S. Hankin
Uncertainty Analyst
University of Cambridge
19 Silver Street
Cambridge CB3 9EP
01223-764877
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