stringr::str_sub输出是意外的

Question

考虑一下折叠式data.frame：

df <- structure(list(sufix = c("atizado", "atoria", "atório", "auta", 
                         "áutico", "ável"), min_stem_len = c(4, 5, 3, 5, 4, 2), replacement = c("", 
                                                                                                "", "", "", "", ""), exceptions = list(character(0), character(0), 
                                                                                                                                       character(0), character(0), character(0), c("afável", "razoável", 
                                                                                                                                                                                   "potável", "vulnerável"))), .Names = c("sufix", "min_stem_len", 
                                                                                                                                                                                                                          "replacement", "exceptions"), row.names = 21:26, class = c("tbl_df", 
                                                                                                                                                                                                                                                                                    "tbl", "data.frame"))

我在这个sufix的变量data.frame中有一个字符串列表。现在我有了一个单词word <- "amável"，我想得到这个单词的sufix，长度与df$sufix的每个字相同。

我用的是折页码：

library(stringr)
word <- "amável"
str_sub(word, start = -stringr::str_length(df$sufix))

但这会产生这样的结果：

> str_sub(word, start = -stringr::str_length(df$sufix))
[1] "amável" "mável"  "mável"  "vel"    "mável"  "vel"   
> df$sufix
[1] "atizado" "atoria"  "atório"  "auta"    "áutico"  "ável"

我原以为得到的向量的最后一个元素是"ável“，因为：

> str_length("ável")
[1] 4
> str_sub(word, start = -4)
[1] "ável"

这里有一个更简单、可重复的例子：

set.seed(100)
a <- sample(1:10, 10000, replace = T)
res <- rep("ábc", 10000) %>% str_sub(start = -a)
sum(ifelse(a > 3, 3, a) != str_length(res))
[1] 2504

Answer 1

这已在stringi的开发分支中得到修正，请参见https://github.com/gagolews/stringi/issues/227 (因为stringr的str_sub依赖于stringi中的stri_sub )。一旦对CRAN进行了更新，任何来自“一般公众”的人都应该复制正确的行为：

str_sub(word, start = -stringr::str_length(df$sufix))
## [1] "amável" "amável" "amável" "ável"   "amável" "ável"  

Answer 2

如果你注意到，所有的结果都是错误的(除了第一个)。

他们应该是

[1] "amável" "amável" "amável" "ável"   "amável" "ável" 

这件事可以很容易地通过

library(stringi)
stri_sub(rep(word, 6), from = -stri_length(df$suffix))

我敢打赌，您同样可以重用您的stringr代码。

###编辑以添加###

我现在明白你的意思了。当然，有一种奇怪的行为，很可能是对á这个特殊角色的认识。见下面的例子：

df <- data.frame(suffix = c("Lorem","ipsum","dolor","sit","amet","consectetur","adipiscing", "elit","Donec","arcu")) 
df$len <- stri_length(df$suffix)

然后看看结果的第7元素中的奇怪行为：

stri_sub("amavel", from = -df$len)
##  [1] "mavel"  "mavel"  "mavel"  "vel"    "avel"   "amavel" "amavel" "avel"  
##  [9] "mavel"  "avel" 

# Compared to
stri_sub("amável", from = -df$len)
##  [1] "mável"  "mável"  "mável"  "vel"    "ável"   "amável" "mável"  "ável"  
##  [9] "mável"  "ável"

奇怪的是，在最后一种情况下，如果使用rep，则会更正结果：

stri_sub(rep("amável", 10), from = -df$len)
## [1] "mável"  "mável"  "mável"  "vel"    "ável"   "amável" "amável" "ável"  
## [9] "mável"  "ável"

# note how the 7th element is now correct.

因此，尽管这有点麻烦，但--上面提供的解决方案--应该可以运行。

我试着查看stri_sub的代码，它引用了C_stri_sub，但对我来说，这是一个死胡同。也许一个更熟悉C和/或字符串操作的人会来帮忙呢？

###第二次编辑###

在我看来，问题在于对stri_sub调用中字符串的重复。查看您在编辑中添加的代码的替代代码：

set.seed(100)
a <- sample(1:10, 10000, replace = TRUE)
res <- stri_sub(rep("ábc", 10000), from = -a)
sum(ifelse(a > 3, 3, a) != stri_length(res))
## [1] 0