仅将匹配的模式保留在 R 中的句子列表中答案

【问题标题】：Keep the matched patterns only in a list of sentences in R仅将匹配的模式保留在 R 中的句子列表中
【发布时间】：2017-09-30 16:20:34
【问题描述】：

我有句子列表和单词列表，我想更新每个句子以仅保留单词列表中的单词。

比如我有以下几个字

"美国","英国","德国","澳大利亚","意大利","in","to"

以及以下句子：

“我在德国生活了 2 年”、“我从意大利搬到美国”、“美国、英国和澳大利亚的人说英语”

我想删除单词列表中未出现的句子中的所有单词所以预期的输出是以下句子： “在德国”、“意大利到美国”、“在美国英国澳大利亚”

如何使用应用函数来做到这一点

mywords=data.frame(words=c("USA","UK","Germany","Australia","Italy","in","to"),
                   stringsAsFactors = F)
mysentences=data.frame(sentences=c("I lived in Germany 2 years",
                                   "I moved from Italy to USA",
                                   "people in USA, UK and Australia speak English"),
                   stringsAsFactors = F)

【问题讨论】：

我第一次看错了；这里有一个非常相似的问题与接受的答案 - stackoverflow.com/questions/28891130/…
@neilfws - 可以很容易地适应 - 例如sapply(strsplit(sentence, "[[:space:]|[:punct:]]"), intersect, vect)。

标签： r

【解决方案1】：

如果将此文本转换为整洁的数据格式，则可以使用连接来查找匹配的单词。然后你可以使用purrr::map_chr() 回到你需要的字符串。

library(tidyverse)
library(tidytext)

mywords <- data_frame(word = c("USA","UK","Germany","Australia","Italy","in","to"))

mysentences <- data_frame(sentences = c("I lived in Germany 2 years",
                                        "I moved from Italy to USA",
                                        "people in USA, UK and Australia speak English"))

mysentences %>% 
    mutate(id = row_number()) %>% 
    unnest_tokens(word, sentences, to_lower = FALSE) %>% 
    inner_join(mywords) %>% 
    nest(-id) %>%
    mutate(sentences = map(data, unlist),
           sentences = map_chr(sentences, paste, collapse = " ")) %>%
    select(-data)

#> Joining, by = "word"
#> # A tibble: 3 × 2
#>      id           sentences
#>   <int>               <chr>
#> 1     1          in Germany
#> 2     2        Italy to USA
#> 3     3 in USA UK Australia

【讨论】：

【解决方案2】：

您也可以使用 stringr。我很抱歉发了两次。搞错了。

vect <- c("USA","UK","Germany","Australia","Italy","in","to")
sentence <- c("I lived in Germany 2 years", "I moved from Italy to USA", "people in USA, UK and Australia speak English")

library(stringr)
li <- str_extract_all(sentence,paste0(vect,collapse="|"))
d <- list()
for(i in 1:length(li){
  d[i] <- paste(li[[i]],collapse=" ")
}

unlist(d)

输出：

 > unlist(d)
[1] "in Germany"         
[2] "Italy to USA"       
[3] "in USA UK Australia"

【讨论】：

【解决方案3】：

这适用于较短的单词列表

library(stringr)
mywords_regex <- paste0(mywords$word, collapse = "|")
sapply(str_extract_all(mysentences$sentences, mywords_regex), paste, collapse = " ")

[1] "in Germany"          "Italy to USA"        "in USA UK Australia"

【讨论】：

【解决方案4】：

这里有两种方法。第一个将单词列表折叠成正则表达式，然后使用str_detect 将单词与正则表达式匹配：

library(tidyverse)
library(glue)

mywords=data_frame(words=c("USA","UK","Germany","Australia","Italy","in","to"))
mysentences=data_frame(sentences=c("This is a sentence with no words of word list",
                                   "I lived in Germany 2 years",
                                   "I moved from Italy to USA",
                                   "people in USA, UK and Australia speak English"))
mysentences %>% 
  filter(sentences %>% 
           str_detect(mywords$words %>% collapse(sep = "|") %>% regex(ignore_case = T)))
#> # A tibble: 3 × 1
#>                                       sentences
#>                                           <chr>
#> 1                    I lived in Germany 2 years
#> 2                     I moved from Italy to USA
#> 3 people in USA, UK and Australia speak English

第二种方法使用fuzzyjoin 的regex_semi_join（在幕后使用str_detect 并为您完成上述工作）

library(fuzzyjoin)
mysentences %>%
  regex_semi_join(mywords, by= c(sentences = "words"))
#> # A tibble: 3 × 1
#>                                       sentences
#>                                           <chr>
#> 1                    I lived in Germany 2 years
#> 2                     I moved from Italy to USA
#> 3 people in USA, UK and Australia speak English

【讨论】：

【解决方案5】：

谢谢大家，

我通过使用 intersect 函数的 answer 启发的以下代码解决了这个问题

vect <- data.frame( c("USA","UK","Germany","Australia","Italy","in","to"),stringsAsFactors = F)
sentence <- data.frame(c("I lived in Germany 2 years", "I moved from Italy to USA",
                         "people in USA     UK and    Australia speak English"),stringsAsFactors = F)

sentence[,1]=gsub("[^[:alnum:] ]", "", sentence[,1]) #remove special characters
sentence[,1]=sapply(sentence[,1], FUN =  function(x){ paste(intersect(strsplit(x, "\\s")[[1]], vect[,1]), collapse=" ")})

【讨论】：