我有长格式的数据，每个个体（通过“ID”标识）有多个观察（“Visit”）。每个个体的观察次数不同。我想创建一个新列，其中包括最后一次访问，我已经实现了，以及一个包括倒数第二次访问的列。
我的数据如下所示：
```R
ID <- c(1000,1000,1000,1001,1001,1001,1001,1002,1002,1002,1002,1002)
Visit <- c("BL","V02","V03","BL","V02","V03","V04","BL","V02","V03","V04","V05")
df <- data.frame(ID,Visit)

df <- df %>%
  group_by(ID) %>%
  mutate(lastVisit = last(Visit))

ID Visit lastVisit secondlastVisit
1000 BL    V03       V02
1000 V02   V03       V02
1000 V03   V03       V02
1001 BL    V04       V03
1001 V02   V04       V03
1001 V03   V04       V03
1001 V04   V04       V03
1002 BL    V05       V04
1002 V02   V05       V04
1002 V03   V05       V04
1002 V04   V05       V04
1002 V05   V05       V04

希望这有助于您的工作！如果您需要进一步的帮助，请随时告诉我。
<details>
<summary>英文:</summary>
I have data in long format with several observations (&quot;Visit&quot;) per individual (identified by &quot;ID&quot;). The number of observations per individual varies. I would like to create a new column with the last visit, which I have accomplished, and a column with the second last Visit.
My data looks like this:

The lastVisit column is created by the following code:

The desired output is like this:

I have tried using ```secondlastVisit  = lag(Visit)```, but this does not give the desired output. A method using dplyr::mutate is preferred. 
Thanks!
</details>
# 答案1
**得分**: 2
🧙🏾‍♂️: I understand you need a translation of the code snippet you provided. Here's the translated code:
``` r
函数`dplyr::nth()`可以实现你想要的功能；负索引表示从末尾开始。
library("magrittr")
library("dplyr")
ID <- c(1000, 1000, 1000, 1001, 1001, 1001, 1001, 1002, 1002, 1002, 1002, 1002)
Visit <- c("BL", "V02", "V03", "BL", "V02", "V03", "V04", "BL", "V02", "V03", "V04", "V05")
df <- data.frame(ID, Visit)
df <- df %>%
  group_by(ID) %>%
  mutate(lastVisit = last(Visit)) %>%
  mutate(secondlastVisit = nth(Visit, -2L))
df
#> # A tibble: 12 x 4
#> # Groups:   ID [3]
#>       ID Visit lastVisit secondlastVisit
#>    <dbl> <chr> <chr>     <chr>          
#>  1  1000 BL    V03       V02            
#>  2  1000 V02   V03       V02            
#>  3  1000 V03   V03       V02            
#>  4  1001 BL    V04       V03            
#>  5  1001 V02   V04       V03            
#>  6  1001 V03   V04       V03            
#>  7  1001 V04   V04       V03            
#>  8  1002 BL    V05       V04            
#>  9  1002 V02   V05       V04            
#> 10  1002 V03   V05       V04            
#> 11  1002 V04   V05       V04            
#> 12  1002 V05   V05       V04

df <- df %>%
  按ID分组 %>%
  添加列(secondLastVisit = Visit[which(lastVisit == Visit) - 1])