Question

更新：

我有以下数据框：

df <- data.frame(clause = c("Hello world my dearest","Hello world my dearest","Hello world my dearest","Hello world my dearest","Hello world my dearest","Hello world my dearest"),
                 word = c("Hello", "Hello", "world", "my", "dearest", "dearest"),
                 syllable = c("He", "lo", "world", "my", "dea", "rest"),
                 phrase_ID = c("1", "1", "1", "2", "2", "2"))

这基本上显示了“ hello world，我最亲爱的”子句的语法，包括 2个短语，4个单词和6个音节。这些词组只用一个短语ID表示。

我进行了以下转换，以获取音节的位置和总数在单词和短语中。

setDT(df)[, word_ID := rleid(word, phrase_ID)]
df[, poss_syll_in_word := sequence(.N), by = word_ID]

我现在要做的是对位置和数量进行相同的转换短语中的单词和条款中的单词：

df$poss_word_in_phrase <- c("1", "1", "2", "1", "2", "2")
df$n_word_in_phrase <- c("2", "2", "2", "2", "2", "2")

我找不到解决方法。有什么想法吗？

Answer 1

编辑

基于更新的数据，我想我们可以使用data.table::rleid和n_distinct来获得预期的输出。

在data.table

df[, c("poss_word_in_phrase", "n_word_in_phrase") := 
            list(rleid(word_ID), uniqueN(word_ID)), phrase_ID]

或与dplyr

df %>%
  group_by(phrase_ID) %>%
  mutate(poss_word_in_phrase = data.table::rleid(word_ID), 
         n_word_in_phrase = n_distinct(word_ID))

使用tidyverse，我们可以在每个单词处拆分字符串以找出其在word中的位置，并使用str_count来计算单词数。

library(tidyverse)

df %>%
  mutate(poss_word_in_phrase = map2_dbl(str_split(phrase, "\\s+"), word, 
                               ~match(.y, .x)), 
         n_word_in_phrase = str_count(phrase, "\\w+"), 
         poss_word_in_clause = map2_dbl(str_split(clause, "\\s+"), word, 
                               ~match(.y, .x)),
         n_word_in_clause =  str_count(clause, "\\w+"))

#                     clause      phrase    word syllable poss_word_in_phrase
#1 Hello world my dearest Hello world   Hello       He                   1
#2 Hello world my dearest Hello world   Hello       lo                   1
#3 Hello world my dearest Hello world   world    world                   2
#4 Hello world my dearest  my dearest      my       my                   1
#5 Hello world my dearest  my dearest dearest      dea                   2
#6 Hello world my dearest  my dearest dearest     rest                   2

#  n_word_in_phrase poss_word_in_clause n_word_in_clause
#1                2                   1                4
#2                2                   1                4
#3                2                   2                4
#4                2                   3                4
#5                2                   4                4
#6                2                   4                4

数据

通过在clause列中假设您的意思是"world"而不是"word"，我稍微纠正了您的输入数据。

df <- data.frame(clause = c("Hello world my dearest","Hello world my dearest",
    "Hello world my dearest","Hello world my dearest",
    "Hello world my dearest","Hello world my dearest"),
    phrase = c("Hello world", "Hello world", "Hello world", "my dearest", 
     "my dearest", "my dearest"),
    word =  c("Hello", "Hello", "world", "my", "dearest", "dearest"),
    syllable = c("He", "lo", "world", "my", "dea", "rest"))

R：组内唯一值的计数

1 个答案: