r - 使用 r 根据模式列表中的精确匹配拆分文本

Question

我有文字和图案。

text <- "It is only a very poor quality car that can give big problems with automatic gearbox" 
patterns <- c("very poor","big problems")

拆分文本

unlist(strsplit(text, "(\\s+)|(?!')(?=[[:punct:]])", perl = TRUE))

输出:

[1] "It"        "is"        "only"      "a"         "very"      "poor"      "quality"   "car"       "that"      "can"      
[11] "give"      "big"       "problems"  "with"      "automatic" "gearbox"

我需要的是匹配句子中的模式列表而不是“非常”“差”变成“非常差”与“大问题”相同。

示例输出:

[1] "It"     "is"     "only"    "a"    "very poor"   "quality"   "car"  "that"   "can"      
[10] "give"   "big problems"  "with"   "automatic"   "gearbox"

我应该怎么做？

Answer 1

这是一种方法:

library(stringr)
text <- "It is only a very poor quality car that can give big problems with automatic gearbox" 
patterns <- c("very poor","big problems")
patterns_ns <- setNames(str_replace_all(patterns, " ", "&&"), patterns)
text_ns <- str_replace_all(text, patterns_ns)
text_split <- str_replace_all(unlist(str_split(text_ns, "\\s")), "&&", " ")
text_split

我假设 "&&" 是一个字符串，它实际上并没有出现在您的源文本中，并且您想在空格处拆分。