Question

我正试图找到一个简单的方法来分离动态变化列表（被动对象）的元素。问题是当“列表”只有一个元素时，我在字符串前面有一个不需要的“和”。所以，我想要一个简单的代码，当我有一个元素时不要添加“和”。

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

我们可以将sub与paste

一起使用

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

还有另一种方式：

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

Answer 3

我认为你需要一些逻辑来检测你的向量是否有多个元素，所以像：

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

这给出了

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

Answer 4

我以前使用过一个（其他）现有函数，但不记得是哪个函数，但是这很好用，我总是使用 knitr！希望我以后记得。

knitr::combine_words(c("something"))

<块引用>

东西

knitr::combine_words(c("something","something"))

<块引用>

某事和某事

knitr::combine_words(c("something","something","something"))

<块引用>

某事，某事，某事

用逗号分隔列表，用“和”分隔最后一个元素

4 个答案: