计算“行”标记中的单词

时间:2017-05-08 20:14:26

标签: r tidyr tidytext

我在R中是全新的,所以这个问题似乎很明显。但是,我没有管理,也没有找到解决方案

如何计算我的代币中的单词数量(实际上是评论)? 因此,有一个数据集的评论(reviewText)与产品ID(asin)相关联

amazonr_tidy_sent = amazonr_tidy_sent%>%unnest_tokens(word, reviewText, token = "lines") amazonr_tidy_sent = amazonr_tidy_sent %>% anti_join(stop_words)%>%ungroup()

我尝试以下列方式

wordcounts <- amazonr_tidy_sent %>% group_by(word, asin)%>% summarize(word = n())

但这不合适。我假设,没有办法计算,因为作为令牌的行不能“分开”

非常感谢

2 个答案:

答案 0 :(得分:0)

如果适合您的分析,您可以多次使用unnest_tokens()

首先,您可以使用unnest_tokens()来获取所需的行。请注意,我正在添加一列来跟踪每一行的id;你可以随意调用它,但重要的是要有一个列,它会记下你在哪一行。

library(tidytext)
library(dplyr)
library(janeaustenr)


d <- data_frame(txt = prideprejudice)

d_lines <- d %>%
    unnest_tokens(line, txt, token = "lines") %>%
    mutate(id = row_number())

d_lines

#> # A tibble: 10,721 × 2
#>                                                                        line
#>                                                                       <chr>
#>  1                                                      pride and prejudice
#>  2                                                           by jane austen
#>  3                                                                chapter 1
#>  4  it is a truth universally acknowledged, that a single man in possession
#>  5                            of a good fortune, must be in want of a wife.
#>  6   however little known the feelings or views of such a man may be on his
#>  7 first entering a neighbourhood, this truth is so well fixed in the minds
#>  8 of the surrounding families, that he is considered the rightful property
#>  9                                 of some one or other of their daughters.
#> 10 "my dear mr. bennet," said his lady to him one day, "have you heard that
#> # ... with 10,711 more rows, and 1 more variables: id <int>

现在您可以再次使用unnest_tokens() ,但这次使用words,这样您就可以获得每个单词的一行。请注意,您仍然知道每个单词来自哪一行。

d_words <- d_lines %>%
    unnest_tokens(word, line, token = "words")

d_words
#> # A tibble: 122,204 × 2
#>       id      word
#>    <int>     <chr>
#>  1     1     pride
#>  2     1       and
#>  3     1 prejudice
#>  4     2        by
#>  5     2      jane
#>  6     2    austen
#>  7     3   chapter
#>  8     3         1
#>  9     4        it
#> 10     4        is
#> # ... with 122,194 more rows

现在你可以做任何你想要的计数,例如,也许你想知道每行有多少个单词?

d_words %>%
    count(id)

#> # A tibble: 10,715 × 2
#>       id     n
#>    <int> <int>
#>  1     1     3
#>  2     2     3
#>  3     3     2
#>  4     4    12
#>  5     5    11
#>  6     6    15
#>  7     7    13
#>  8     8    11
#>  9     9     8
#> 10    10    15
#> # ... with 10,705 more rows

答案 1 :(得分:0)

通过使用str_split分割每一行,我们可以计算每行的单词数。

一些示例数据(包含换行符和停用词):

library(dplyr)
library(tidytext)
d = data_frame(reviewText = c('1 2 3 4 5 able', '1 2\n3 4 5\n6\n7\n8\n9 10 above', '1!2', '1',
                          '!', '', '\n', '1', 'able able', 'above above', 'able', 'above'),
           asin = rep(letters, each = 2, length.out = length(reviewText)))

计算单词数:

by_line %>%
    group_by(asin) %>%
    summarize(word = sum(sapply(strsplit(word, '\\s'), length)))

   asin  word
  <chr> <int>
1     a    17
2     b     2
3     c     1
4     d     1
5     e     4
原始代码中的

注意:,因为您按行拆分数据,因此不会删除大多数停用词。只会删除完全是单个停用词的行。

要从wordcount中排除停用词,请使用以下命令:

by_line %>%
    group_by(asin) %>%
    summarize(word = word %>% strsplit('\\s') %>%
                  lapply(setdiff, y = stop_words$word) %>% sapply(length) %>% sum)

   asin  word
  <chr> <int>
1     a    15
2     b     2
3     c     1
4     d     1
5     e     0
6     f     0