使字符向量的重复元素唯一，但不像 make.unique()

Question

当多个参考文献具有相同的作者（或多个作者）和出版年份时，通常的做法是在年份之后包含一个小写字母。我正在为此寻找一个优雅的功能：

# what I have
have <- c("Dawkins (2008)",
          "Dawkins (2008)",
          "Stephenson (2008)")

# what I want
want <- c("Dawkins (2008a)",
          "Dawkins (2008b)",
          "Stephenson (2008)")

# this would do the job, but is not really what I want
make.unique(have)
#> [1] "Dawkins (2008)"    "Dawkins (2008).1"  "Stephenson (2008)"

^{由 reprex package (v2.0.1)}

于 2022-02-24 创建

编辑：解决方案基于下面@akrun 的回答

library(dplyr)
library(stringr)

have <- c("Dawkins (2008)",
          "Dawkins (2008)",
          "Stephenson (2008)")

f <- function(x){
  v1 <- ave(x, x, FUN = function(x) if(length(x) > 1) letters[seq_along(x)] else "")
  stringr::str_replace(x, "\)", stringr::str_c(v1, ")"))
}

data.frame(ha = have) %>% 
  mutate(want = f(ha))
#>                  ha              want
#> 1    Dawkins (2008)   Dawkins (2008a)
#> 2    Dawkins (2008)   Dawkins (2008b)
#> 3 Stephenson (2008) Stephenson (2008)

^{由 reprex package (v2.0.1)}

于 2022-02-24 创建

Answer 1

我们可能会根据重复的length提取出letters（假设重复的长度不会大于26），然后使用str_replace插入字母收盘前 )

library(stringr)
v1 <- ave(have, have, FUN = function(x) 
  if(length(x) > 1) letters[seq_along(x)] else "")
str_replace(have, "\)", str_c(v1, ")"))
[1] "Dawkins (2008a)"   "Dawkins (2008b)"   "Stephenson (2008)"

使字符向量的重复元素唯一，但不像 make.unique()

Makes duplicate elements of a character vector unique, but not like make.unique()

r

duplicates

unique

编辑：解决方案基于下面@akrun 的回答