替换嵌套 ifelse() 语句链的最快方法

Question

我有这一系列的嵌套语句

data$Country == 1,"Brazil",
  ifelse(data$Country == 2, "Canada",ifelse(
    data$Country == 3, "China",ifelse(
      data$Country == 4, "Ecuador",ifelse(
        data$Country == 5, "France",ifelse(
          data$Country == 6, "Germany",ifelse(
            data$Country == 7, "India",ifelse(
              data$Country == 8, "Italy",ifelse(
                data$Country == 9, "Mexico",ifelse(
                  data$Country == 10, "Nigeria",ifelse(
                    data$Country == 11, "Poland",ifelse(
                      data$Country == 12, "Russia",ifelse(
                        data$Country == 13, "South Africa",
                        ifelse(
                          data$Country == 14, "South Korea",ifelse(
                            data$Country == 15, "Singapore",
                            ifelse(
                              data$Country == 16, "Spain",
                              ifelse(
                                data$Country == 17, "Sweden",ifelse(
                                  data$Country == 18, "United Kingdom",ifelse(
                                    data$Country == 19, "United States","l"
))))))))))))))))))))

我一直在寻找将任何编码变量转换为相应国家/地区名称的最快方法。你觉得有什么办法可以应对这种操作吗？

非常感谢

Answer 1

我不确定所需的用途。但也许您可以尝试使用命名向量。这不是最优雅的解决方案，尽管它解决了 ifelse 混乱问题；)

4 个国家的示例。中国 = "4"

countrys <- c("Brazil", "Canada",
              "China",
              "Ecuador")
names(countrys) <- c(2:5)

# Test data.frame
data <- data.frame(country = 4)

# Now we can get the country directly from the data$country:
# Careful! 4 is not '4'
unname(countrys[as.character(data$country)])

Answer 2

有2个选项：

1: case_when 来自 dplyr

library(dplyr)
data.frame(info = letters[1:5],
           country_id = 1:5) %>% 
  mutate(country_name = case_when(country_id == 1 ~ "Brazil",
                                  country_id == 2 ~ "Canada",
                                  country_id == 3 ~ "China",
                                  country_id == 4 ~ "Ecuador",
                                  country_id == 5 ~ "France",
                                  TRUE ~ "Unknown"))

  info country_id country_name
1    a          1       Brazil
2    b          2       Canada
3    c          3        China
4    d          4      Ecuador
5    e          5       France

2: 合并或加入一个国家的信息 table:

# country table
countries <- data.frame(country_id = 1:5, 
                        country_name = c("Brazil", "Canada", "China", "Ecuador", "France"))

data.frame(info = letters[1:5],
           country_id = 1:5) %>% 
  left_join(countries, by = "country_id")

  info country_id country_name
1    a          1       Brazil
2    b          2       Canada
3    c          3        China
4    d          4      Ecuador
5    e          5       France

我的偏好是 2，更少的编码和更少的错误机会。您可以将国家/地区 table 保存在您的数据库或某个文件中并进行维护，而无需更改代码。

Answer 3

这是 switch 语句的一个很好的案例，在我看来，它比 dplyr::case_when 或一系列 ifelse 的代码更具可读性，并且易于扩展，例如，如果还有其他条件，例如地区、城市等

get_country <- Vectorize(function(x){
  switch(as.character(x),
         "1" = "Brazil", "2" = "Canada", "3" = "China", "4" = "Ecuador",
         "5" = "France", "6" = "Germany", "7" = "India", "8" = "Italy", 
         "9" = "Mexico", "10" = "Nigeria", "11" = "Poland", "12" = "Russia",
         "13" = "South Africa", "14" = "South Korea", "15" = "Singapore",
         "16" = "Spain", "17" = "Sweden", "18" = "United Kingdom", "19" = "United States",
         NA)
})

data.frame(info = letters[1:5],
           country_id = 1:5) %>%
  mutate(country = get_country(country_id))

  info country_id  country
1    a          1  Brazil
2    b          2  Canada
3    c          3   China
4    d          4 Ecuador
5    e          5  France

但是像这样的长语句需要大量的打字工作。或者，一种更动态的方法，我们可以使用采用输入向量的构造函数创建 switch 语句。这里我使用 maps 包中的 ISO3166 数据集来创建 269 个国家/地区的表达式。

constructor <- function(ids, names){
  purrr::imap_chr(as.character(ids), ~paste(paste0("\"", .x ,"\""),
                                            paste0("\"", names[.y], "\""),
                                            sep = "=")) %>%
    paste0(collapse = ", ") %>%
    paste0("Vectorize(function(x) switch(as.character(x), ", ., ", NA))", collapse = "") %>%
    str2expression()
}
get_country <- eval(constructor(1:149, trimws(rworldmap::countryExData$Country)))

set.seed(1)
data.frame(info = sample(letters, size = 5, replace = T),
           country_id = sample.int(149, 5, replace = T)) %>%
  mutate(country = get_country(country_id))

  info country_id           country
1    y        122      Sierra Leone
2    h         39           Algeria
3    l         42           Eritrea
4    y        134 Trinidad & Tobago
5    w         24             Chile

为了展示力量 - 让我们用另外 2 行代码为约 20000 个城市创建一个控制流

CITIES <- maps::world.cities %>% filter(pop > 10000) %>% arrange(desc(pop))
get_city <- eval(constructor(1:nrow(CITIES), trimws(CITIES$name)))

data.frame(city_id = sample.int(23255, size = 100, replace = T),
           country_id = sample.int(269, 100, replace = T)) %>%
  mutate(country = get_country(country_id),
         city = get_city(city_id))

这种方法的一个好处是您可以轻松地优化您的控制流，方法是确保构造函数的输入向量得到优化，即最先发生的情况，然后可能制作支持嵌套方法的函数，例如 get_continent(get_country(get_city))).

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