Question

假设我有一个字符串，例如以下内容。

x <- 'The world is at end. What do you think?   I am going crazy!    These people are too calm.'

我只需要在标点符号!?.上进行分割并跟随空格并保留标点符号。

这会删除标点并在分割部分留下前导空格

vec <- strsplit(x, '[!?.][:space:]*')

如何分割留下标点符号的句子？

Answer 1

您可以使用perl=TRUE启用PCRE并使用lookbehind断言。

strsplit(x, '(?<![^!?.])\\s+', perl=TRUE)

正则表达式：

(?<!          look behind to see if there is not:
 [^!?.]       any character except: '!', '?', '.'
)             end of look-behind
\s+           whitespace (\n, \r, \t, \f, and " ") (1 or more times)

Live Demo

Answer 2

qdap package中的sentSplit函数仅为此任务创建：

library(qdap)
sentSplit(data.frame(text = x), "text")

##   tot                       text
## 1 1.1       The world is at end.
## 2 2.2         What do you think?
## 3 3.3          I am going crazy!
## 4 4.4 These people are too calm.

Answer 3

看看this question。像[:space:]这样的字符类是在括号表达式中定义的，因此您需要将它括在一组括号中。尝试：

vec <- strsplit(x, '[!?.][[:space:]]*')
vec
# [[1]]
# [1] "The world is at end"       "What do you think"        
# [3] "I am going crazy"          "These people are too calm"

这摆脱了领先的空间。要保持标点符号，请使用perl = TRUE：

使用正面的lookbehind断言

vec <- strsplit(x, '(?<=[!?.])[[:space:]]*', perl = TRUE)
vec
# [[1]]
# [1] "The world is at end."       "What do you think?"        
# [3] "I am going crazy!"          "These people are too calm."

Answer 4

您可以使用字符串替换标点符号后面的空格，例如zzzzz，然后在该字符串上拆分。

x <- gsub("([!?.])[[:space:]]*","\\1zzzzz","The world is at end. What do you think?   I am going crazy!    These people are too calm.")
strsplit(x, "zzzzz")

替换字符串中的\1指的是模式的带括号的子表达式。

Answer 5

从qdap version 1.1.0开始，您可以使用sent_detect函数，如下所示：

library(qdap)
sent_detect(x)

## [1] "The world is at end."       "What do you think?"        
## [3] "I am going crazy!"          "These people are too calm."

R string删除分割时的标点符号

5 个答案: