R - создать ID предложения из столбца слов

Question

У меня есть датафрейм (тиббл) слов, который выглядит следующим образом.

   text      confidence type          start_time end_time
   <chr>     <chr>      <chr>         <chr>      <chr>   
 1 Angela    0.7482     pronunciation 0.04       0.32    
 2 very      1.0        pronunciation 0.32       0.59    
 3 powerful  1.0        pronunciation 0.59       1.29    
 4 .         0.0        punctuation   NA         NA      
 5 And       1.0        pronunciation 1.3        1.65    
 6 with      1.0        pronunciation 1.65       1.87    
 7 every     1.0        pronunciation 1.88       2.24    
 8 hurricane 1.0        pronunciation 2.24       2.75    
 9 there's   0.8826     pronunciation 2.75       2.96    
10 that      1.0        pronunciation 2.96       3.22    
11 one's     0.6438     pronunciation 3.22       3.73    
12 own       0.748      pronunciation 3.73       4.02    
13 .         0.0        punctuation   NA         NA      
14 It's      0.9278     pronunciation 4.02       4.19    
15 usually   0.851      pronunciation 4.19       4.51    

Я пытаюсь создать значение идентификатора предложения, чтобы я мог сгруппировать слова в предложения. Я бы хотел, чтобы идентификаторы начинались / заканчивались на type = punctuation.

   text      confidence type          start_time end_time sentence_id
   <chr>     <chr>      <chr>         <chr>      <chr>          <dbl>
 1 Angela    0.7482     pronunciation 0.04       0.32               1
 2 very      1.0        pronunciation 0.32       0.59               1
 3 powerful  1.0        pronunciation 0.59       1.29               1
 4 .         0.0        punctuation   NA         NA                 1
 5 And       1.0        pronunciation 1.3        1.65               2
 6 with      1.0        pronunciation 1.65       1.87               2
 7 every     1.0        pronunciation 1.88       2.24               2
 8 hurricane 1.0        pronunciation 2.24       2.75               2
 9 there's   0.8826     pronunciation 2.75       2.96               2
10 that      1.0        pronunciation 2.96       3.22               2
11 one's     0.6438     pronunciation 3.22       3.73               2
12 own       0.748      pronunciation 3.73       4.02               2
13 .         0.0        punctuation   NA         NA                 2
14 It's      0.9278     pronunciation 4.02       4.19               3
15 usually   0.851      pronunciation 4.19       4.51               3

Я уверен, что есть относительно простой способ сделать это, но я не совсем понимаю. У кого-нибудь есть предложения? Если это поможет, вот dput:

structure(list(text = c("Angela", "very", "powerful", ".", "And", 
"with", "every", "hurricane", "there's", "that", "one's", "own", 
".", "It's", "usually"), confidence = c("0.7482", "1.0", "1.0", 
"0.0", "1.0", "1.0", "1.0", "1.0", "0.8826", "1.0", "0.6438", 
"0.748", "0.0", "0.9278", "0.851"), type = c("pronunciation", 
"pronunciation", "pronunciation", "punctuation", "pronunciation", 
"pronunciation", "pronunciation", "pronunciation", "pronunciation", 
"pronunciation", "pronunciation", "pronunciation", "punctuation", 
"pronunciation", "pronunciation"), start_time = c("0.04", "0.32", 
"0.59", NA, "1.3", "1.65", "1.88", "2.24", "2.75", "2.96", "3.22", 
"3.73", NA, "4.02", "4.19"), end_time = c("0.32", "0.59", "1.29", 
NA, "1.65", "1.87", "2.24", "2.75", "2.96", "3.22", "3.73", "4.02", 
NA, "4.19", "4.51")), row.names = c(NA, -15L), class = c("tbl_df", 
"tbl", "data.frame"))

Answer 1

Один dplyr вариант может быть:

df %>%
 mutate(sentence_id = rev(cumsum(rev(type) == "punctuation")),
        sentence_id = max(sentence_id) - sentence_id + 1)

   text      confidence type          start_time end_time sentence_id
   <chr>     <chr>      <chr>         <chr>      <chr>          <dbl>
 1 Angela    0.7482     pronunciation 0.04       0.32               1
 2 very      1.0        pronunciation 0.32       0.59               1
 3 powerful  1.0        pronunciation 0.59       1.29               1
 4 .         0.0        punctuation   <NA>       <NA>               1
 5 And       1.0        pronunciation 1.3        1.65               2
 6 with      1.0        pronunciation 1.65       1.87               2
 7 every     1.0        pronunciation 1.88       2.24               2
 8 hurricane 1.0        pronunciation 2.24       2.75               2
 9 there's   0.8826     pronunciation 2.75       2.96               2
10 that      1.0        pronunciation 2.96       3.22               2
11 one's     0.6438     pronunciation 3.22       3.73               2
12 own       0.748      pronunciation 3.73       4.02               2
13 .         0.0        punctuation   <NA>       <NA>               2
14 It's      0.9278     pronunciation 4.02       4.19               3
15 usually   0.851      pronunciation 4.19       4.51               3

Answer 2

мы можем использовать cumsum

   df$id <-  cumsum(df$type == "punctuation")+1

# A tibble: 15 x 6
   text      confidence type          start_time end_time    id
   <chr>     <chr>      <chr>         <chr>      <chr>    <dbl>
 1 Angela    0.7482     pronunciation 0.04       0.32         1
 2 very      1.0        pronunciation 0.32       0.59         1
 3 powerful  1.0        pronunciation 0.59       1.29         1
 4 .         0.0        punctuation   <NA>       <NA>         2
 5 And       1.0        pronunciation 1.3        1.65         2
 6 with      1.0        pronunciation 1.65       1.87         2
 7 every     1.0        pronunciation 1.88       2.24         2
 8 hurricane 1.0        pronunciation 2.24       2.75         2
 9 there's   0.8826     pronunciation 2.75       2.96         2
10 that      1.0        pronunciation 2.96       3.22         2
11 one's     0.6438     pronunciation 3.22       3.73         2
12 own       0.748      pronunciation 3.73       4.02         2
13 .         0.0        punctuation   <NA>       <NA>         3
14 It's      0.9278     pronunciation 4.02       4.19         3
15 usually   0.851      pronunciation 4.19       4.51         3

Answer 3

Функциональный код, разделенный на несколько столбцов, чтобы вы могли понять мой процесс мыслей 3,0)