Question

如果我计算factorial(100)，那么我会得到[1] 9.332622e+157的答案，但当我尝试计算更大的阶乘时，请说factorial(1000)我会得到[1] Inf的答案

有没有办法在计算阶乘时使用任意精度，以便我可以计算说factorial(1000000)？

Answer 1

对于任意精度，您可以使用gmp或Rmpfr。对于具体的因子gmp商品factorialZ和Rmpfr商品factorialMpfr。所以你可以运行类似下面的内容

> Rmpfr::factorialMpfr(200)
1 'mpfr' number of precision  1246   bits 
[1] 788657867364790503552363213932185062295135977687173263294742533244359449963403342920304284011984623904177212138919638830257642790242637105061926624952829931113462857270763317237396988943922445621451664240254033291864131227428294853277524242407573903240321257405579568660226031904170324062351700858796178922222789623703897374720000000000000000000000000000000000000000000000000
> gmp::factorialZ(200)
Big Integer ('bigz') :
[1] 788657867364790503552363213932185062295135977687173263294742533244359449963403342920304284011984623904177212138919638830257642790242637105061926624952829931113462857270763317237396988943922445621451664240254033291864131227428294853277524242407573903240321257405579568660226031904170324062351700858796178922222789623703897374720000000000000000000000000000000000000000000000000

HTH

Answer 2

我写了一个网络刮刀; @Khashaa的答案可能更快，但我经历了概念验证并磨练我的新生rvest技能：

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

另一个可能的优势是使用Wolfram的计算能力而不是你自己（我不知道包选项的效率如何，我想他们只是使用渐近近似，所以这可能不是关注，只是以为我提到了它）

是否有可用于计算R中的大因子的包或技术？

2 个答案: