使用 R 和术语文档矩阵创建频率 table

Question

我创建了以下由一些电子邮件主题行组成的数据框。

 df <- data.frame(subject=c('Free ! Free! Free ! Clear Cover with New Phone',
                            'Offer ! Buy New phone and get earphone at 1000. Limited Offer!'))

我创建了一个从上述数据框派生的常用词列表。我已将这些关键字添加到数据框并将它们虚拟编码为 0

 most_freq_words <- c('Free', 'New', 'Limited', 'Offer')



Subject                                               Free New Limited Offer                                                    

 'Free Free Free! Clear Cover with New Phone',          0   0     0      0
 'Offer ! Buy New phone and get earphone at             0   0     0      0
 1000. Limited Offer!'

我想获取邮件主题中单词出现的频率。输出应该如下

  Subject                                             Free New Limited Offer                                                    

 'Free Free Free!  Clear Cover with New Phone',         3   1     0      0
 'Offer ! Buy New phone and get earphone at             0   1     1      2
 1000. Limited Offer!'

我试过下面的代码

for (i in 1:length(most_freq_words)){
df[[most_freq_words[i]]] <- as.numeric(grepl(tolower(most_freq_words[i]), 
tolower(df$subject)))}

然而，这表明句子中是否存在该词。我需要上面给出的输出。我请求某人帮助我

Answer 1

将 grepl 替换为 gregexpr，然后检查 1st 列表项的 length。此外，for-loop 也应该在 df 的每一行上运行。保持 OP 的 for-loop 意图，修改后的代码将如下所示：

for (i in 1:length(most_freq_words)){
  for(j in 1:nrow(df)){
    df[j,most_freq_words[i]] <- ifelse(gregexpr(tolower(most_freq_words[i]),
       tolower(df$subject[j]))[[1]][[1]] >0,
    length(gregexpr(tolower(most_freq_words[i]), tolower(df$subject[j]))[[1]]), 0)
  }
}  


> df
                                                         subject Free New Limited Offer
1                Free ! Free! Free ! Clear Cover with New  Phone    3   1       0     0
2 Offer ! Buy New phone and get earphone at 1000. Limited Offer!    0   1       1     2

Answer 2

我用 tidytext 包处理了这个任务。首先，我在数据集中添加了一个分组变量。然后，我使用 unnest_token() 分隔单词。除了 most_freq_words 中的单词，我删除了所有单词。然后，我统计每个单词在每个句子中出现了多少次。最后，我将 long-format 数据转换为 wide-format 数据。如果您仍然想要原始句子，可以轻松地将其添加到输出中（例如，在 spread() 行之后添加 cbind(subject = df$subject)）

library(dplyr)
library(tidytext)

df <- data.frame(subject=c('Free ! Free! Free ! Clear Cover with New Phone',
                           'Offer ! Buy New phone and get earphone at 1000. Limited Offer!'),
                 stringsAsFactors = FALSE)

most_freq_words <- c('Free', 'New', 'Limited', 'Offer')

mutate(df, group = 1:n()) %>%
unnest_tokens(input = subject, output = word, token = "words", to_lower = FALSE) %>%
filter(word %in% most_freq_words) %>%
count(group, word) %>%
spread(key = word, value = n, fill = 0)

  group  Free Limited   New Offer
  <int> <dbl>   <dbl> <dbl> <dbl>
1     1  3.00    0     1.00  0   
2     2  0       1.00  1.00  2.00

Answer 3

这里是 tidyverse 的另一个选项。我们使用 map 遍历 'most_freq_words'，用 str_count 从 'df' 的 'subject' 列获取它的计数，转换为 tibble，设置'most_freq_words' 中的列名称并将列与原始数据集绑定 'df'

library(tidyverse)
most_freq_words %>% 
      map(~ str_count(df$subject, .x) %>%
                    as_tibble %>% 
                    set_names(.x)) %>% 
      bind_cols(df, .)
#                                                         subject Free New Limited Offer
#1                 Free ! Free! Free ! Clear Cover with New Phone    3   1       0     0
#2 Offer ! Buy New phone and get earphone at 1000. Limited Offer!    0   1       1     2

使用 R 和术语文档矩阵创建频率 table

Create Frequency table using R and Term document Matrix

r

frequency

text-mining

grepl

term-document-matrix