用逗號和最後一個元素與「和」分隔列表

Question

我想要一個簡短的方法來分隔動態變化列表（反應對象）的元素。問題是當「列表」只有一個元素時，我在字符串之前有一個不需要的「and」。所以，我想要一個簡單的代碼，認爲當我有一個元素不添加「和」時。

list<-c("something") 
list<-c("something","something") 
list<-c("something","something","something") 

paste0(c(paste(head(list, n=length(list) -1), collapse = ", ") , 
     "and",paste(tail(list, n=1))), collapse= " ") 

[1] " and something" # not what is expected 
[1] "something and something" # ok 
[1] "something, something and something" #ok 

Answer 1

我們可以使用sub與paste

fPaste <- function(vec) sub(",\\s+([^,]+)$", " and \\1", toString(vec)) 


fPaste("something") 
#[1] "something" 

fPaste(c("something","something")) 
#[1] "something and something" 

fPaste(c("something","something","something")) 
#[1] "something, something and something" 

fPaste(c("something","something","something", "something")) 
#[1] "something, something, something and something" 

Answer 2

我想你需要一些邏輯來檢測你的載體有一個以上的元素，所以是這樣的：

## Note I've change list to x since its 
## confusing, as it's actually a vector 
if(length(x) > 1) " and " else "" 

這給

paste0(c(
    paste(x[-length(x)], collapse = ", ") , 
     if(length(x) > 1) " and " else "", 
     x[length(x)] 
     ), collapse= "") 

Answer 3

而又另一種方式：

i <- length(list) 
if (i <= 1) { 
    as.character(list) 
} else { 
    paste0(paste0(list[1:(i - 1)], collapse = ", "), " and ", list[i]) 
}