如何找出十大常用词汇？

Question

我在数据集中有一个单词列表

我使用R基找到前10个常用单词

排序(表(单词)，递减=真)1:10

我的问题是:如何使用字符串函数获得十大常用单词？

str_sort(表(单词)，递减=真)1:10

上面的代码没有给我提供预期的结果。有什么想法吗？

Answer 1

顾名思义，str_sort用于对字符元素进行排序。它被记录在?str_sort中。

命令或排序字符向量。

table的输出是数字输出的命名vector。如果我们正在寻找tidyverse版本，请获取count，然后使用slice_max获取前10次频率计数。

library(dplyr)
tibble(words) %>%
     count(words) %>%
     slice_max(n = 10, order_by = n)

主要的区别在于，sorting与字符元素不同，而数值对应项，即在进行排序之前，str_sort将数字输出强制到character class，这可能有不同的输出。

> str_sort(c(10, 4, 5))
[1] "10" "4"  "5" 
> sort(c(10, 4, 5))
[1]  4  5 10

包stringr不是用来排序数字输入的。字符串手册本身的标题表明

通用字符串操作的

简单一致包装器

在forcats中有一个选项可以这样做

library(forcats)
fct_count(words, sort = TRUE)$n[1:10]

Answer 2

更新：

在本例中，我们使用了sort函数的count参数：

#example:
mytext = c("This","is","a","test","for","count","of","the","words","The","words","have","been","written","very","randomly","so","that","the","test","can","be","for","checking","the","count")

library(tibble)
library(dplyr)

tibble(mytext) %>% 
  group_by(mytext) %>% 
  count(sort = TRUE) %>%
  ungroup() %>% 
  slice_max(n, n=10)

  mytext       n
   <chr>    <int>
 1 the          3
 2 count        2
 3 for          2
 4 test         2
 5 words        2
 6 a            1
 7 be           1
 8 been         1
 9 can          1
10 checking     1
11 have         1
12 is           1
13 of           1
14 randomly     1
15 so           1
16 that         1
17 The          1
18 This         1
19 very         1
20 written      1

我们可以从str_count包中使用stringr：参见这里的一个示例：

library(tidyverse)
tibble(words) %>% 
  mutate(n = str_count(words)) %>% 
  slice_max(n = 10, order_by = n)

产出：

   words           n
   <chr>       <int>
 1 appropriate    11
 2 environment    11
 3 opportunity    11
 4 responsible    11
 5 department     10
 6 difference     10
 7 experience     10
 8 individual     10
 9 particular     10
10 photograph     10
11 television     10
12 understand     10
13 university     10