这是我关于SO的第一个问题，请告诉我是否可以改进。我正在研究R中的自然语言处理项目，并且正在尝试构建包含测试用例的data.table。在这里，我构建了一个简化的例子：
texts.dt <- data.table(string = c("one",
"two words",
"three words here",
"four useless words here",
"five useless meaningless words here",
"six useless meaningless words here just",
"seven useless meaningless words here just to",
"eigth useless meaningless words here just to fill",
"nine useless meaningless words here just to fill up",
"ten useless meaningless words here just to fill up space"),
word.count = 1:10,
stop.at.word = c(0, 1, 2, 2, 4, 3, 3, 6, 7, 5))
这将返回我们将要处理的data.table：
string word.count stop.at.word
1: one 1 0
2: two words 2 1
3: three words here 3 2
4: four useless words here 4 2
5: five useless meaningless words here 5 4
6: six useless meaningless words here just 6 3
7: seven useless meaningless words here just to 7 3
8: eigth useless meaningless words here just to fill 8 6
9: nine useless meaningless words here just to fill up 9 7
10: ten useless meaningless words here just to fill up space 10 5
在实际应用程序中，stop.at.word列中的值是随机确定的（上限= word.count - 1）。此外，字符串不按长度排序，但不应有所不同。
代码应添加两列input和output，其中input包含从位置1到stop.at.word的子字符串，output包含跟随（单个单词），如下：
>desired_result
string word.count stop.at.word input
1: one 1 0
2: two words 2 1 two
3: three words here 3 2 three words
4: four useless words here 4 2 four useless
5: five useless meaningless words here 5 4 five useless meaningless words
6: six useless meaningless words here just 6 2 six useless
7: seven useless meaningless words here just to 7 3 seven useless meaningless
8: eigth useless meaningless words here just to fill 8 6 eigth useless meaningless words here just
9: nine useless meaningless words here just to fill up 9 7 nine useless meaningless words here just to
10: ten useless meaningless words here just to fill up space 10 5 ten useless meaningless words here
output
1:
2: words
3: here
4: words
5: here
6: meaningless
7: words
8: to
9: fill
10: just
不幸的是，我得到的是：
string word.count stop.at.word input output
1: one 1 0
2: two words 2 1 NA NA
3: three words here 3 2 NA NA
4: four useless words here 4 2 NA NA
5: five useless meaningless words here 5 4 NA NA
6: six useless meaningless words here just 6 3 NA NA
7: seven useless meaningless words here just to 7 3 NA NA
8: eigth useless meaningless words here just to fill 8 6 NA NA
9: nine useless meaningless words here just to fill up 9 7 NA NA
10: ten useless meaningless words here just to fill up space 10 5 ten NA
注意结果不一致，第1行为空字符串，第10行返回“10”。
以下是我正在使用的代码：
texts.dt[, c("input", "output") := .(
substr(string,
1,
sapply(gregexpr(" ", string),"[", stop.at.word) - 1),
substr(string,
sapply(gregexpr(" ", string),"[", stop.at.word),
sapply(gregexpr(" ", string),"[", stop.at.word + 1) - 1)
)]
我运行了很多测试，当我在控制台中尝试单个字符串时，substr指令运行良好，但在应用于data.table时失败。
我怀疑我遗漏了与data.table中的作用域相关的东西，但是我没有长时间使用这个包，所以我很困惑。
我非常感谢一些帮助。 提前谢谢！
答案 0 :(得分：5)
我可能会这样做
texts.dt[stop.at.word > 0, c("input","output") := {
sp = strsplit(string, " ")
list(
mapply(function(p,n) paste(p[seq_len(n)], collapse = " "), sp, stop.at.word),
mapply(`[`, sp, stop.at.word+1L)
)
}]
# partial result
head(texts.dt, 4)
string word.count stop.at.word input output
1: one 1 0 NA NA
2: two words 2 1 two words
3: three words here 3 2 three words here
4: four useless words here 4 2 four useless words
可替换地：
library(stringi)
texts.dt[stop.at.word > 0, c("input","output") := {
patt = paste0("((\\w+ ){", stop.at.word-1, "}\\w+) (.*)")
m = stri_match(string, regex = patt)
list(m[, 2], m[, 4])
}]
答案 1 :(得分：5)
@ Frank mapply解决方案的替代方案是将by = 1:nrow(texts.dt)与strsplit和paste一起使用：
library(data.table)
texts.dt[, `:=` (input = paste(strsplit(string, ' ')[[1]][1:stop.at.word][stop.at.word>0],
collapse = " "),
output = strsplit(string, ' ')[[1]][stop.at.word + 1]),
by = 1:nrow(texts.dt)]
给出：
> texts.dt
string word.count stop.at.word input output
1: one 1 0 one
2: two words 2 1 two words
3: three words here 3 2 three words here
4: four useless words here 4 2 four useless words
5: five useless meaningless words here 5 4 five useless meaningless words here
6: six useless meaningless words here just 6 3 six useless meaningless words
7: seven useless meaningless words here just to 7 3 seven useless meaningless words
8: eigth useless meaningless words here just to fill 8 6 eigth useless meaningless words here just to
9: nine useless meaningless words here just to fill up 9 7 nine useless meaningless words here just to fill
10: ten useless meaningless words here just to fill up space 10 5 ten useless meaningless words here just
您可以将[[1]]包裹在strsplit中，而不是使用unlist，如下所示：unlist(strsplit(string, ' '))（而不是strsplit(string, ' ')[[1]]）。这将给你相同的结果。
另外两个选择：
使用 stringi 包1）：
library(stringi)
texts.dt[, `:=`(input = paste(stri_extract_all_words(string[stop.at.word>0],
simplify = TRUE)[1:stop.at.word],
collapse = " "),
output = stri_extract_all_words(string[stop.at.word>0],
simplify = TRUE)[stop.at.word+1]),
1:nrow(texts.dt)]
2）或来自this answer的改编：
texts.dt[stop.at.word>0,
c('input','output') := tstrsplit(string,
split = paste0("(?=(?>\\s+\\S*){",
word.count - stop.at.word,
"}$)\\s"),
perl = TRUE)
][, output := sub('(\\w+).*','\\1',output)]
两者都给出了：
> texts.dt
string word.count stop.at.word input output
1: one 1 0 NA NA
2: two words 2 1 two words
3: three words here 3 2 three words here
4: four useless words here 4 2 four useless words
5: five useless meaningless words here 5 4 five useless meaningless words here
6: six useless meaningless words here just 6 3 six useless meaningless words
7: seven useless meaningless words here just to 7 3 seven useless meaningless words
8: eigth useless meaningless words here just to fill 8 6 eigth useless meaningless words here just to
9: nine useless meaningless words here just to fill up 9 7 nine useless meaningless words here just to fill
10: ten useless meaningless words here just to fill up space 10 5 ten useless meaningless words here just
答案 2 :(得分：5)
System.out.format(fmt, value);
我不确定我理解dt[, `:=`(input = sub(paste0('((\\s*\\w+){', stop.at.word, '}).*'), '\\1', string),
output = sub(paste0('(\\s*\\w+){', stop.at.word, '}\\s*(\\w+).*'), '\\2', string))
, by = stop.at.word][]
# string word.count stop.at.word
# 1: one 1 0
# 2: two words 2 1
# 3: three words here 3 2
# 4: four useless words here 4 2
# 5: five useless meaningless words here 5 4
# 6: six useless meaningless words here just 6 3
# 7: seven useless meaningless words here just to 7 3
# 8: eigth useless meaningless words here just to fill 8 6
# 9: nine useless meaningless words here just to fill up 9 7
#10: ten useless meaningless words here just to fill up space 10 5
# input output
# 1: one
# 2: two words
# 3: three words here
# 4: four useless words
# 5: five useless meaningless words here
# 6: six useless meaningless words
# 7: seven useless meaningless words
# 8: eigth useless meaningless words here just to
# 9: nine useless meaningless words here just to fill
#10: ten useless meaningless words here just
对于第一行没有任何意义的逻辑，但是如果确实需要的话，那么微不足道的修复留给了OP。