从列中提取国家名称（或其他实体）答案

【问题标题】：extract country names (or other entity) from column从列中提取国家名称（或其他实体）
【发布时间】：2019-11-05 10:08:48
【问题描述】：

我在location 列中有一个包含国家和城市的data.frame，我想通过匹配来自library(maps)（或任何其他国家名称集合）的world.cities$country.etc 数据框来提取前者。

考虑这个例子：

df <- data.frame(location = c("Aarup, Denmark",
                              "Switzerland",
                              "Estonia: Aaspere"),
                 other_col = c(2,3,4))

我尝试使用此代码

df %>% extract(location,
               into = c("country", "rest_location"),
               remove = FALSE,
               function(x) x[which x %in% world.cities$country.etc])

但我没有成功；我期待这样的事情：

          location other_col     country rest_location
1   Aarup, Denmark         2     Denmark       Aarup, 
2      Switzerland         3 Switzerland              
3 Estonia: Aaspere         4     Estonia     : Aaspere

【问题讨论】：

标签： r dataframe

【解决方案1】：

我们可以通过将所有国家名称粘贴在一起来创建一个模式，并使用str_extract_all 来获取与location 中的模式匹配的所有国家名称，并删除与国家名称匹配的单词以获得rest_location。

library(maps)
library(stringr)

all_countries <- str_c(unique(world.cities$country.etc), collapse = "|")
df$country <- sapply(str_extract_all(df$location, all_countries), toString)
df$rest_location <- str_remove_all(df$location, all_countries)
#OR can also do
#df$rest_location <- str_remove_all(df$location, df$country)

df
#          location other_col     country rest_location
#1   Aarup, Denmark         2     Denmark       Aarup, 
#2      Switzerland         3 Switzerland              
#3 Estonia: Aaspere         4     Estonia     : Aaspere

将sapply 和toString 用于country，因为如果location 中有多个国家/地区名称，它们都将连接在一个字符串中。

【讨论】：

【解决方案2】：

你可以试试这个作为起点

library(tidyverse)
df %>% 
  rownames_to_column() %>% 
  separate_rows(location) %>% 
  mutate(gr = location %in% world.cities$country.etc) %>% 
  mutate(gr = ifelse(gr, "country", "rest_location")) %>% 
  spread(gr, location) %>% 
  right_join(df %>% 
              rownames_to_column(), 
              by = c("rowname", "other_col")) %>% 
  select(location, other_col, country, rest_location)
          location other_col     country rest_location
1   Aarup, Denmark         2     Denmark         Aarup
2      Switzerland         3 Switzerland          <NA>
3 Estonia: Aaspere         4     Estonia       Aaspere

值得注意的是，这仅适用于位置列中只有两个“单词”的情况。如有必要，您必须指定一个合适的单独例如sep=",|:"

【讨论】：

【解决方案3】：

Base R（不包括地图包）：

# Import the library: 

library(maps)

# Split the string on the spaces: 

country_city_vec <- strsplit(df$location, "\\s+")

# Replicate the other col's rows by the split string vec: 

rolled_out_df <- data.frame(other_col = rep(df$other_col, sapply(country_city_vec, length)), 

                            location = gsub("[[:punct:]]", "", unlist(country_city_vec)), stringsAsFactors = F)

# Match with the world df: 

matched_with_world_df <- merge(df,

                               setNames(rolled_out_df[rolled_out_df$location %in% world.cities$country.etc,],
                                        c("other_col", "country")),

                               by = "other_col", all.x = T)

# Extract the city/location drilldown: 

matched_with_world_df$rest_location <- trimws(gsub("[[:punct:]]",
                                                   "",
                                                   gsub(paste0(matched_with_world_df$country,
                                                               collapse = "|"),
                                           "", matched_with_world_df$location)), "both")

【讨论】：