在 R 中，结合单个字数和字典字数

Question

我需要计算文档中的字数。在某些情况下，我需要计算特定单词的数量（例如“新鲜”），在其他情况下我需要计算一组单词的总数（“philadelphia”，“aunt”）。

我知道如何分两步执行此操作（请参阅下面的代码），但如何同时执行此操作？

下面的代码计算特定的单词。

library("quanteda")
txt <- "In west Philadelphia born and raised On the playground was where I spent most of my days Chillin' out maxin' relaxin' all cool And all shootin some b-ball outside of the school When a couple of guys who were up to no good Started making trouble in my neighborhood I got in one little fight and my mom got scared."
tokens(txt) %>% tokens_select(c("trouble", "fight")) %>% dfm()

输出为：

trouble, fight
1, 1

下面的代码计算字典单词并将总计数写入一列。

mydict <- dictionary(list(all_terms = c("chillin", "relaxin", "shootin")))
count <-dfm(txt,dictionary = mydict)

输出为：

all_terms
3

如何将两者结合起来？

我想要这样的东西：（代码是假设的，不起作用）

tokens(txt) %>% tokens_select(c("trouble", "fight"), mydict) %>% dfm()

或

tokens(txt) %>% tokens_select(c("trouble", "fight"), all_terms=c("chillin","relaxin","shootin")) %>% dfm()

期望的输出：

trouble, fight, all_terms
1, 1, 3

Answer 1

简洁重要吗，即将所有内容都放在一行中？如果没有，一个解决方案是从 dfm 对象中提取数据，然后组合成您想要的形式 - 矩阵，data.frame，tibble。

library("quanteda")
library(magritte) # for the pipe
txt <- "In west Philadelphia born and raised On the playground was where I spent most of my days Chillin' out maxin' relaxin' all cool And all shootin some b-ball outside of the school When a couple of guys who were up to no good Started making trouble in     my neighborhood I got in one little fight and my mom got scared."
mydict <- dictionary(list(all_terms = c("chillin", "relaxin", "shootin")))

first <-  dfm(tokens_select(tokens(txt), c("trouble", "fight")))
second <- dfm(txt,dictionary = mydict)

# These are the outputs you're after
first@Dimnames$features
first@x

second@Dimnames$features
second@x

# Combine into a matrix
 matrix(c(first@Dimnames$features, second@Dimnames$features), ncol = 3) %>% 
   rbind(c(first@x, second@x))

# Or make two vectors for use elsewhere
  paste(c(first@Dimnames$features, second@Dimnames$features), collapse = ", ")
  paste(c(first@x, second@x), collapse = ", ")

Answer 2

有几种方法，这可能是最简单的一种。定义一个字典，其中键等于每个特定单词的单词值，以及一组单词的组键——在您的示例中，“all_terms”。

library("quanteda")
## Package version: 2.1.2

txt <- "In west Philadelphia born and raised On the playground was where I spent most of my days Chillin' out maxin' relaxin' all cool And all shootin some b-ball outside of the school When a couple of guys who were up to no good Started making trouble in my neighborhood I got in one little fight and my mom got scared."

dict <- dictionary(list(
  trouble = "trouble",
  fight = "fight",
  all_terms = c("chillin", "relaxin", "shootin")
))

现在当你编译dfm时，你会得到你想要的。

dfmat <- dfm(txt, dictionary = dict)
dfmat
## Document-feature matrix of: 1 document, 3 features (0.0% sparse).
##        features
## docs    trouble fight all_terms
##   text1       1     1         3

要将其强制转换为更简单的对象，包括您列出的输出，您可以这样做：

# as a named numeric vector
structure(as.vector(dfmat), names = featnames(dfmat))
##   trouble     fight all_terms 
##         1         1         3

# per your output
cat(
  paste(featnames(dfmat), collapse = ", "), "\n",
  paste(as.vector(dfmat), collapse = ", ")
)
## trouble, fight, all_terms 
##  1, 1, 3

请注意，直接访问对象内部结构不是一个好主意（如另一个答案）。请改用 featnames() 等提取函数。

已添加：

另一种不创建项目命名列表的方法：

dict <- dictionary(list(all_terms = c("chillin", "relaxin", "shootin")))
single_words <- c("trouble", "fight")

tokens(txt) %>%
  tokens_lookup(dictionary = dict, exclusive = FALSE) %>%
  tokens_keep(pattern = c(names(dict), single_words)) %>%
  dfm()
## Document-feature matrix of: 1 document, 3 features (0.0% sparse).
##        features
## docs    all_terms trouble fight
##   text1         3       1     1

Answer 3

这是我在评论中建议的。

> library("quanteda")
> txt <- "In west Philadelphia born and raised On the playground was where I spent most of my days Chillin' out maxin' relaxin' all cool And all shootin some b-ball outside of the school When a couple of guys who were up to no good Started making trouble in     my neighborhood I got in one little fight and my mom got scared."
> dict <- dictionary(list(all_terms = c("chillin", "relaxin", "shootin")))
> dfmt <- dfm(txt)
> dfmt_dict <- dfm_lookup(dfmt, dict, exclusive = FALSE, cap = FALSE)
> topfeatures(dfmt_dict)
       in       and        of        my all_terms         '       the         i 
        3         3         3         3         3         3         2         2 
      all       got 
        2         2

在 R 中，结合单个字数和字典字数

In R, combing individual word count and dictionary word count

r

nlp

quanteda