如何在 R 观察中提取一个单词（字符串）和几个单词？

Question

这是我的df：

Phrase               
  <chr>                 
1 I am hungry               
2 I am going to school     
3 It's raining              
4 I like talking

我想为每个观察提取一些关键词，例如（“Hungry”、“School”、“I”和“I am”），如下所示：

  Phrase               Hungry School     I `I am`
  <chr>                 <dbl>  <dbl> <dbl>  <dbl>
1 I am hungry               1      0     1      1
2 I am going to school      0      1     1      1
3 It's raining              0      0     0      0
4 I like talking            0      0     1      0

在上面的示例中，如果列名在短语内，则它可能具有数字 1，如果不在短语内，则它可能具有数字 0（或者可能是另一种提取方式）。

我尝试阅读一些类似这样的文章： https://www.r-bloggers.com/an-overview-of-keyword-extraction-techniques/ ，但我没有找到任何信息来帮助我提取这些关键词。

Answer 1

我们可以用sapply单词向量，应用grepl得到一个逻辑向量，然后用+转换为二进制

v1 <- c("Hungry", "School", "I", "I am")
cbind(df1, +(sapply(v1, function(v) grepl(paste0("\\b", toupper(v), 
       "\\b"), toupper(df1$Phrase)))))
#               Phrase Hungry School I I am
#1          I am hungry      1      0 1    1
#2 I am going to school      0      1 1    1
#3         It's raining      0      0 0    0
#4       I like talking      0      0 1    0

数据

df1 <- structure(list(Phrase = c("I am hungry", "I am going to school", 
"It's raining", "I like talking")), class = "data.frame", row.names = c("1", 
"2", "3", "4"))

Answer 2

您可以在每个单词上使用grepl ，但需要特别注意，因为您可能会有误报。

Phrase <- c("I am hungry","I am going to school","It's raining ","I like talking")

data.frame(phrase=Phrase,
    hungry = grepl("hungry",tolower(Phrase))*1,
    school = grepl("school",tolower(Phrase))*1,
    i = grepl("i\\s|\\si",tolower(Phrase))*1,
    iam = grepl("i am",tolower(Phrase))*1)

                phrase hungry scholl i iam
1          I am hungry      1      0 1   1
2 I am going to school      0      1 1   1
3        It's raining       0      0 0   0
4       I like talking      0      0 1   0

Answer 3

基础 R 解决方案：

search_words <- c("Hungry", "School", "I", "I am")
cbind(df1, data.frame(+(Vectorize(grepl)(search_words, df1, ignore.case = TRUE))))

数据：@akrun 谢谢。

df1 <- structure(list(Phrase = c("I am hungry", "I am going to school", 
"It's raining", "I like talking")), class = "data.frame", row.names = c("1", 
"2", "3", "4"))