Question

如何改进此功能？

powerset = function(s){

    len = length(s)
    l = vector(mode="list",length=2^len) ; l[[1]]=numeric()
    counter = 1L
    for(x in 1L:length(s)){
        for(subset in 1L:counter){
            counter=counter+1L
            l[[counter]] = c(l[[subset]],s[x])
        }
    }
    return(l)
}

测试向量

x <- 1:3
powerset(x)

输出

[[1]]
numeric(0)

[[2]]
[1] 1

[[3]]
[1] 2

[[4]]
[1] 1 2

[[5]]
[1] 3

[[6]]
[1] 1 3

[[7]]
[1] 2 3

[[8]]
[1] 1 2 3

Answer 1

试试这个：

    powerset<-function(s) {
      n<-length(s)
      do.call(c,lapply(0:n,function(x) combn(s,x,simplify=FALSE)))
    }
    s<-1:3
    powerset(s)

Answer 2

如果你的矢量非常大，那么来自combnPrim的gRbase会更有效率。与@nicola功能相同，但将combn替换为combnPrim

 library(gRbase)
 powersetPrim <- function(s) {
  n<-length(s)
   do.call(c,lapply(0:n,function(x) combnPrim(s,x,simplify=FALSE)))
  }
 s<-1:3
 powersetPrim(s)

基准

较小的样本

s <- 1:25
microbenchmark(powersetPrim(), powerset(), unit='relative', times=10L)
#Unit: relative
#          expr      min      lq     mean   median       uq      max neval cld
#powersetPrim() 1.000000 1.00000 1.000000 1.000000 1.000000 1.000000    10  a 
#   powerset()  5.254663 4.21187 3.644953 3.386665 2.874453 3.844787    10  b

更有效地计算大矢量（2 ^ 2000）的功率集

2 个答案:

基准