如何向 R 中的函数提交“分布向量”？

Question

I want to write an R-function, say f , which has inputs x and n , where x is some kind of "list of distributions" and f is supposed to draw n samples from each distribution in x .

What is a good way to implement this in R?

My current idea is

f = function(x,n){
  
  out = list()
  
  for(i in 1:length(x)){
    
    name = sub("\\(.*", "",x[i])
    size = ifelse(name=="sample",paste("size=",n),paste0("n=",n))
    args = paste(size,gsub("[\\(\\)]", "", regmatches(x[i], gregexpr("\\(.*?\\)", x[i]))[[1]]),sep=",")
    out[[i]] = eval(parse(text=paste0(name,"(",args,")")))
    
  }
  
  return(out)
  
}

f(x = c("rnorm(mean=1,sd=2)","sample(0:1,replace=TRUE)","rbinom(size=10,prob=0.1)"), n = 10)

I don't like this implementation, because

1. n is not always the input name for the sample size (eg in sample it is size ),
2. the code will crash if not all inputs for the distributions are properly defined.

Can I improve the implementation, for example with x of class alist?

Answer 1

You could change your input and create a list of function names and arguments. For each distribution we set the n / size -value to 1 .

ls_func <- list("rnorm" = list(mean = 0, sd = 1, n = 1),
                "sample" = list(x = 0:1, replace = TRUE, size = 1),
                "rbinom" = list(size = 10, prob = 0.1, n = 1))

Your function takes those distributions and replicates them n -times:

g <- function(ls_func, n) {
  out = list()
  
  for(i in seq_along(ls_func)){
    out[[i]] <- replicate(do.call(names(ls_func)[i], ls_func[[i]]), n = n)
  }
  
  return(out)
}

so

set.seed(4096)
g(ls_func, 10)

returns

[[1]]
 [1]  0.1894398 -0.1622468  0.5327100 -1.5747229 -0.6884024 -0.3092226 -0.0879258 -0.4320240 -0.7799596  0.4525895

[[2]]
 [1] 0 1 0 0 0 1 1 1 0 0

[[3]]
 [1] 0 0 1 1 0 1 1 1 1 0

or. Basically it's not a good approach to use eval(parse(text=...)) to execute functions. Use do.call instead.

---

You can remove the for -loop:

g <- function(ls_func, n) {
  out = list()
 
  out <- lapply(seq_along(ls_func), function(i) replicate(do.call(names(ls_func)[i], ls_func[[i]]), n = n))
  
  return(out)
}

Note: This code also crashes, if your distributions aren't defined properly. To avoid this, you need some error handling. Look for try and stop functions.

Answer 2

I've been putting together an R package -- distionary -- that can help with this.

First make a list of input distributions:

library(distionary)
x <- list(
  dst_norm(1, 2^2),
  dst_empirical(0:1),
  dst_binom(10, 0.1)
)

The function for drawing from a distribution is realize() , which fits nicely with lapply() (or purrr's map() ):

set.seed(123)
lapply(x, realize, n = 10)
#> [[1]]
#>  [1] -0.1209513  0.5396450  4.1174166  1.1410168  1.2585755  4.4301300
#>  [7]  1.9218324 -1.5301225 -0.3737057  0.1086761
#> 
#> [[2]]
#>  [1] 0 0 0 0 0 0 0 0 1 1
#> 
#> [[3]]
#>  [1] 3 2 1 2 0 1 2 0 0 0

Putting this code in a function is then straightforward:

f <- function(x, n) {
  lapply(x, realize, n = n)
}
set.seed(123)
f(x, n = 10)
#> [[1]]
#>  [1] -0.1209513  0.5396450  4.1174166  1.1410168  1.2585755  4.4301300
#>  [7]  1.9218324 -1.5301225 -0.3737057  0.1086761
#> 
#> [[2]]
#>  [1] 0 0 0 0 0 0 0 0 1 1
#> 
#> [[3]]
#>  [1] 3 2 1 2 0 1 2 0 0 0