如何将函数应用于列表中的向量

Question

I have a list of vectors, I wish to apply a function I created to the list of vectors that should only return one integer. The list of vectors are: (3,3);(3,2,1);(3,1,1,1).

alpha.6.3 <- list(c(3, 3), c(3, 2, 1), c(3, 1, 1, 1))

The function determines the combination of the vector elements and their sum(n) minus the previous element, and returns the product between them. Ex: (3,3) => 6C3.3C3 = 20; (3,2,1) => (6C3)(3C2)(1C1) = 60 and so on.

Wen I apply this to the entire list, using sapply, the output is undesirable.

> list 
[[1]]
[1] 3 3

[[2]]
[1] 1 2 3

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

combVecFunc <-  function(n,x){
  ans = c()

  for(j in 1:length(x)){
    ans[j] = comb(n,x[j])
    n = n-x[j]
  }
  print(prod(ans))
}

prodVecFunc <- function(n,list){

  ans <- sapply(list, function(i){
    x <- c()
      for(i in 1:length(list)){
        x[i]<- combVecFunc(n,list[[i]])
      }
      return(x)
    })
  print(ans)
}

> prodVecFunc(n=6,list)
[1] 20
[1] 60
[1] 120
[1] 20
[1] 60
[1] 120
[1] 20
[1] 60
[1] 120
     [,1] [,2] [,3]
[1,]   20   20   20
[2,]   60   60   60
[3,]  120  120  120

Thus I should have a list of integers after the function is applied namely (20,60,120). Rather than the vector with multiple answers.

Answer 1

There is no need for two functions, just combVecFunc will do what is asked for after corrected.

1. The base R function to compute the number of combinations of n elements taken k at a time is choose , not comb (that doesn't even exist and was throwing an error).
2. The function should take only one input argument, the sum n can be computed by the function itself.

So the code would become the following.

combVecFunc <-  function(x){
  n <- sum(x)
  ans = numeric(length(x))

  for(j in seq_along(x)){
    ans[j] = choose(n, x[j])
    n = n - x[j]
  }
  prod(ans)
}

alpha.6.3 <- list(c(3, 3), c(3, 2, 1), c(3, 1, 1, 1))

sapply(alpha.6.3, combVecFunc)
#[1]  20  60 120

Edit.

The comment by user @Cole can make the function much simpler.

combVecFunc2 <-  function(x){
  n <- sum(x)
  ans <- choose(n - c(0, cumsum(x[-length(x)])), x)
  prod(ans)
}

sapply(alpha.6.3, combVecFunc2)
#[1]  20  60 120