If I calculate factorial(100) then I get an answer of [1] 9.332622e+157 but when I try to calculate a larger factorial, say factorial(1000) I get an answer of [1] Inf
Is there a way to use arbitrary precision when calculating factorials such that I can calculate say factorial(1000000)?
For arbitrary precision you can use either gmp or Rmpfr. For specifically factorial gmp offers factorialZ and Rmpfr has factorialMpfr. So you can run something like below
> Rmpfr::factorialMpfr(200)
1 'mpfr' number of precision 1246 bits
[1] 788657867364790503552363213932185062295135977687173263294742533244359449963403342920304284011984623904177212138919638830257642790242637105061926624952829931113462857270763317237396988943922445621451664240254033291864131227428294853277524242407573903240321257405579568660226031904170324062351700858796178922222789623703897374720000000000000000000000000000000000000000000000000
> gmp::factorialZ(200)
Big Integer ('bigz') :
[1] 788657867364790503552363213932185062295135977687173263294742533244359449963403342920304284011984623904177212138919638830257642790242637105061926624952829931113462857270763317237396988943922445621451664240254033291864131227428294853277524242407573903240321257405579568660226031904170324062351700858796178922222789623703897374720000000000000000000000000000000000000000000000000
HTH
I wrote a web scraper; @Khashaa's answer is probably faster, but I went through for proof of concept and to hone my nascent rvest skills:
library(rvest)
Factorial<-function(n){
x<-strsplit(strsplit((html(paste0(
#%21 is URL speak for !
"http://www.wolframalpha.com/input/?i=",n,"%21")) %>%
#to understand this part, I recommend going to the site
# and doing "Inspect Element" on the decimal representation
html_nodes("area") %>% html_attr("href")),
split="[=&]")[[1]][2],split="\\+")[[1]]
cat(paste0(substr(x[1],1,8), #8 here can be changed to the precision you'd like;
# could also make it match printing options...
"e+",gsub(".*E","",x[3])))
}
> Factorial(10000)
2.846259e+35659
Another possible advantage is using Wolfram's computing power instead of your own (I don't know how efficient the package options are, I imagine they just use asymptotic approximations so this probably isn't a concern, just thought I'd mention it)
来源:https://stackoverflow.com/questions/32123499/is-there-a-package-or-technique-availabe-for-calculating-large-factorials-in-r