把唯一的/本能的单词数到新的列中

Question

我有一个具有列的数据集，其中包含一行中的一组国家。有时国家不止一次被重复，我想在我的数据集的每一行中计算出独特的国家数如下：

> class(address_countries2$address_countries)
[1] "character"

> head(address_countries2)
                    address_countries
1                         China China
2                   China China China
3                         China China
4                         China China
5 China China China China China China
6                China China Uk China

所需的输出将是如下所示的新列：

                    address_countries n_countries
1                         China China           1
2                   China China China           1
3                         China China           1
4                         China China           1
5 China China China China China China           1
6                China China Uk China           2

这段代码给出了每一行的字数：

address_countries2 <- address_countries2 %>% 
  select(address_countries) %>% 
  mutate(n_countries = str_count(address_countries, boundary("word")))


> head(address_countries2)
                    address_countries n_countries
1                         China China           2
2                   China China China           3
3                         China China           2
4                         China China           2
5 China China China China China China           6
6                China China Uk China           4

我尝试过用n_distinct()以及str_count()添加unique()和distinct()，但是我得到了以下错误：

Error in mutate_impl(.data, dots) : 
  Column `n_countries` must be length 34760 (the number of rows) or one, not 39

有什么建议吗？

Answer 1

试试这个：

你的data.frame

address_countries2<-data.frame(address_countries=c("Chian","China China","China UK"))

国家数目：

list_country<-strsplit(as.character(address_countries2$address_countries)," ")
list_country
[[1]]
[1] "Chian"

[[2]]
[1] "China" "China"

[[3]]
[1] "China" "UK"  

加入"n_countries“栏

address_countries2$n_countries<-unlist(lapply(lapply(list_country, unique),length))

输出量

address_countries2
        address_countries n_countries
        1             Chian           1
        2       China China           1
        3          China UK           2

Answer 2

您可以将address_countries拆分为一个列表，然后使用n_distinct。

library(purrr)
library(dplyr)
library(stringr)

df %>% 
  mutate(n_countries = map_int(address_countries, ~
                                 .x %>% 
                                 str_trim %>% 
                                 str_split(" ") %>% 
                                 unlist() %>% 
                                 n_distinct))

map_int在comme之后将函数应用于address_countries的每个元素，并输出一个整数

str_trim移除向量开头和结尾处的空白

str_split拆分向量，使用" "作为分裂模式。

unlist将str_split的结果转化为向量

n_distinct计算结果向量的唯一元素。

数据

df <- tibble(address_countries = c("China China", "China China China", "China China", 
                                   "China China", "China China China China China China", 
                                   "China China Uk China"))

Answer 3

这应该给你你想要的：

ac$n_countries <- lengths(lapply(strsplit(ac$countries, split = ' '), unique))

结果：

> ac
    countries n_countries
1       Chian           1
2 China China           1
3    China UK           2

数据：

ac <- data.frame(countries = c("Chian","China China","China UK"), stringsAsFactors = FALSE)