在 R 中，用值“A”填充前 1000 个 df 行，用值“B”填充接下来的 1000 个 df 行，等等

Question

我的数据如下所示：

  Keyword                    AdGroup index
  <chr>                      <chr>   <int>
1 memes 2019                 ""          1
2 new movies coming out      ""          2
3 black friday tv deals 2019 ""          3
4 wheel of time tv series    ""          4
5 nitro cold brew            ""          5
6 new tv shows               ""          6

我尝试了以下代码：

adgroup<- function(){
  browser()
  mytop<-1
  mybot<-1000
  i = 1
  mygroup <- paste0("Adgroup ",i)
  
  while (mytop < 15001153){
    myoutput$AdGroup[myoutput$index >= mytop & myoutput$index <= mybot] <- mygroup
    i<-i+1
    mytop <- mytop + 1000
    mybot <- mybot + 1000
    mygroup <- paste0("Adgroup ",i)
    print(paste0(i," ",mytop))
  }
    
}

当我逐步完成代码时，第一组填充了“Adgroup 1”。 但是，当我在代码完成后查看第一组时，它是空白的。

谢谢。

Answer 1

你想要像下面这样的东西吗？

addgroup <- function(x, size = 1000, group = "Adgroup "){
  n <- nrow(x)
  passes <- n %/% size
  remaining <- n %% size
  inx <- seq_len(size) - size
  if(passes){
    for(i in seq_len(passes)){
      inx <- inx + size
      x$AdGroup[inx] <- paste0(group, i)
    }
  }
  if(remaining){
    i <- passes + 1L
    inx <- (inx + size)[seq_len(remaining)]
    x$AdGroup[inx] <- paste0(group, i)
  }
  x
}

addgroup(myoutput, size = 10)

---

测试数据创建代码

set.seed(2021)
n <- 34
myoutput <- data.frame(Keyword = sample(letters, n, TRUE), AdGroup = "")
myoutput$index <- seq_len(n)

Answer 2

df[["group"]] <- (seq_len(nrow(df)) - 1) %/% 1000