R - 带有逗号分隔文本列条目的文档术语矩阵

R - Document Term Matrix with comma separated text column entries

我有一个数据框,其中有一列由字符串 (project_skills) 组成,表示某项工作 (job_id) 提供的技能。我想为每个工作拆分这个字符串以获得工作提供的技能向量,然后创建一个文档术语矩阵来表示某个工作提供的技能(在所有可能的技能中)。

我得到了以下数据框:

job_id           project_skills
107182           CSS,HTML,Joomla,PHP
108169           XTCommerce,Magento,Prestashop,VirtueMart,osCommerce
112969           Google Search Console,Google Analytics,Google Webmaster Central,C++,Java,C#
114660           Marketing,Email Marketing
118686           PHP

结果应该看起来像这样(基本上是一个用逗号分隔的短语的文档术语矩阵:

        project_skills
job_id  CSS   HTML   PHP   Google Search Console   Google Analytics   Java ...
107182  1     0       0 ...         
108169  0     0       0     0                       0         
112969  0     0       0     1                       1 ...         
114660  0     0       0 ...            
118686  0     0       1 ...

我尝试了以下方法:

df <- data.frame(job_id = c(107182, 108169, 112969, 114660, 118686), project_skills = c("CSS,HTML,Joomla,PHP", "XTCommerce,Magento,Prestashop,VirtueMart,osCommerce", "Google Search Console,Google Analytics,Google Webmaster Central,C++,Java,C#", "Marketing,Email Marketing", "PHP"))

corpus <- Corpus(VectorSource(df$project_skills))
corpus <- tm_map(corpus, function(x) {
PlainTextDocument(
    strsplit(x,"\,")[[1]], 
    id=ID(x)
)
})
inspect(corpus)
dtm <- DocumentTermMatrix(corpus)
as.matrix(dtm)

但不幸的是,这会拆分所有单词而不是逗号(例如 Google Search Console 应在 DTM 中被视为一个术语)。

tm(或其他一些文本挖掘包)按单词(空格)拆分,如果您不检查,则倾向于删除 + 和 # 等标点符号。最简单的选择就是使用 strsplit。我在下面使用 tidyr 和 dplyr 显示了一个选项。首先按 job_id 分组,然后拆分列。这将创建一个嵌套,当未嵌套时创建一个长 data.frame。在这里,我为每个条目添加了值 1,它在文档术语矩阵中的作用类似于 1。然后展开成宽格式以获得预期的输出。如果您查看生成的结构,列名就是您所期望的,没有显示波浪号 (~)。

library(tidyr)
library(dplyr)

outcome <- df1 %>%
  group_by(job_id) %>% 
  mutate(project_skills = strsplit(project_skills, ",")) %>% 
  unnest() %>% 
  mutate(value = 1) %>% # add 1 for every value
  spread(key = project_skills, value = value) # use fill = 0 if you don't want NA's

head(outcome)
# A tibble: 5 x 18
# Groups:   job_id [5]
  job_id  `C#` `C++`   CSS `Email Marketin~ `Google Analyti~ `Google Search ~ `Google Webmast~  HTML  Java Joomla Magento Marketing
   <int> <dbl> <dbl> <dbl>            <dbl>            <dbl>            <dbl>            <dbl> <dbl> <dbl>  <dbl>   <dbl>     <dbl>
1 107182    NA    NA     1               NA               NA               NA               NA     1    NA      1      NA        NA
2 108169    NA    NA    NA               NA               NA               NA               NA    NA    NA     NA       1        NA
3 112969     1     1    NA               NA                1                1                1    NA     1     NA      NA        NA
4 114660    NA    NA    NA                1               NA               NA               NA    NA    NA     NA      NA         1
5 118686    NA    NA    NA               NA               NA               NA               NA    NA    NA     NA      NA        NA
# ... with 5 more variables: osCommerce <dbl>, PHP <dbl>, Prestashop <dbl>, VirtueMart <dbl>, XTCommerce <dbl>

对此有很多解决方案,但 strsplit 是您的朋友。这正是以下代码所做的:

library(udpipe)
df <- data.frame(job_id = c(107182, 108169, 112969, 114660, 118686), project_skills = c("CSS,HTML,Joomla,PHP", "XTCommerce,Magento,Prestashop,VirtueMart,osCommerce", "Google Search Console,Google Analytics,Google Webmaster Central,C++,Java,C#", "Marketing,Email Marketing", "PHP"), 
                 stringsAsFactors = FALSE)
dtm <- document_term_frequencies(x = df$project_skills, document = df$job_id, split = ",")
dtm <- document_term_matrix(dtm)
colnames(dtm)
 [1] "C#"                       "C++"                      "CSS"                      "Email Marketing"         
 [5] "Google Analytics"         "Google Search Console"    "Google Webmaster Central" "HTML"                    
 [9] "Java"                     "Joomla"                   "Magento"                  "Marketing"               
[13] "osCommerce"               "PHP"                      "Prestashop"               "VirtueMart"              
[17] "XTCommerce"              
rownames(dtm)
[1] "107182" "108169" "112969" "114660" "118686"
dim(dtm)
[1]  5 17