Question

考虑下面的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"))

我在此data.frame的变量sufix中有一个字符串列表。现在我有一个单词word <- "amável"，我希望得到这个单词的sufix与df$sufix的每个单词的长度相同。

我正在使用以下代码：

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

但这会输出：

> 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"

我期待结果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"

这是一个更简单的可重复示例：

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

如果您注意到，所有结果都是错误的（第一个除外）。

他们应该是

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

这可以通过

轻松解决

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

我打赌你可以重复使用stringr代码。

###编辑添加###

我现在明白你的意思了。绝对存在一种奇怪的行为，很可能是特殊角色á。请参阅以下示例：

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个元素中的奇怪行为：

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"

很奇怪，如果使用rep，结果会在最后一种情况下得到纠正：

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.

所以即使它有点hacky，上面提供的解决方案应该可行。

我尝试查看stri_sub的代码，它引用了C_stri_sub，但这对我来说是一个死胡同。也许有人更了解C和/或字符串操作可以来帮忙吗？

### SECOND EDIT ###

在我看来问题是在stri_sub 的调用中重复了字符串。查看您在编辑中添加的替代代码：

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

Answer 2

已在TaskCanceledException的开发分支中修复此问题，请参阅https://github.com/gagolews/stringi/issues/227（来自stringi的{{1}}依赖于str_sub stringr ）。一旦CRAN上有可用的更新，任何来自“普通公众”的人都可以复制正确的行为，而不是：

stri_sub

stringr :: str_sub输出是意外的

2 个答案: