Question

我想返回数字向量v中元素的字符串向量，重复的次数与元素的值相同，我必须使用嵌套for循环。

这是我的代码：

ReturnValueTimes <- function(v) {
  emptyString <- ""
  for(i in 1:length(v)){
    for(j in 1:v[i]){
      emptyString <- c(emptyString, v[i])
      repeatedNumbers <- paste(emptyString, collapse =  "")
    }
  }
  return(repeatedNumbers)
}

测试案例：

> ReturnValueTimes(c(2,4,3))
[1] "224444333"
However, I want this to be something like:
> ReturnValueTimes(c(2, 4, 3))
[1] "22"   "4444" "333" 
> ReturnValueTimes(c(7, 1))
[1] "7777777" "1"

我应该在代码中添加哪些条件（语句）来分隔字符串？或者改变一些东西以获得理想的结果？

Answer 1

好吧，我不明白为什么你应该为此写一个R函数，因为基础r中已有一个

strrep(x<-c(5,2,3),x)
[1] "55555" "22"    "333"  
strrep(x<-c(7,4,9),x)
[1] "7777777"   "4444"      "999999999"

虽然你仍然可以写你的：

ReturnValueTimes=function(x)strrep(x,x)
ReturnValueTimes(c(5,2,3))
[1] "55555" "22"    "333"

现在，您可以编写代码：

ReturnValueTimes3 <- function(v) {
  store_here=character(length(v))# A character vector which will store
  for(i in 1:length(v)){
    emptyString <- ""#This is defined inside the first loop in order to rewrite its value after the inner loop is done
    for(j in 1:v[i]){
      emptyString <- c(emptyString, v[i])
      repeatedNumbers <- paste(emptyString, collapse =  "")
    }
   store_here[i] <- repeatedNumbers
  }
  return(store_here)
}

ReturnValueTimes3(c(3,4,5,6))
[1] "333"    "4444"   "55555"  "666666"

或者你可以这样做：

sapply(c(3,4,5,6),function(x)paste0(rep(x,x),collapse=""))
[1] "333"    "4444"   "55555"  "666666"

如果你想写你的，但要避免明确的for loop（包括*申请家庭）

ReturnValueTimes2=function(x){
pattern=paste0("(\\d{",x,"})",collapse="")
substitution=paste0("\\",1:length(x),collapse=" ")
vect=paste0(rep(x,x),collapse = "")
strsplit(sub(pattern,substitution,vect)," ")[[1]]
}
 ReturnValueTimes2(c(5,3,4,6))
[1] "55555"  "333"    "4444"   "666666"

Answer 2

使用purrr中的地图功能

library(tidyverse)
ReturnValueTimes <- function (vec) map_chr(vec, function(x) str_c(rep(x, x), collapse = ""))
ReturnValueTimes(c(2, 3, 4))
#> [1] "22"   "333"  "4444"

由reprex package创建于2018-02-03（v0.1.1.9000）。

关于R中循环中字符串的分离

2 个答案: