基于另一列的综合因子

Question

该示例显示了不同工厂的生产输出的测量值， 其中第一列表示工厂 最后一列是生产量。

factory <- c("A","A","B","B","B","B","B","C","D")
production <- c(15, 2, 1, 1, 2, 1, 2,20,5)
df <- data.frame(factory, production)
df
  factory production
1       A         15
2       A          2
3       B          1
4       B          1
5       B          2
6       B          1
7       B          2
8       C         20
9       D          5

现在我想根据数据集中的总产量将工厂归为更少的级别。

使用普通的 forcats::fct_lump，我可以按你出现的行数来汇总它们，例如制作 3 个关卡：

library(tidyverse)    
df %>% mutate(factory=fct_lump(factory,2))
      factory production
    1       A         15
    2       A          2
    3       B          1
    4       B          1
    5       B          2
    6       B          1
    7       B          2
    8   Other         20
    9   Other          5

但我想根据总和（生产）对它们进行汇总，保留前 n=2 个工厂（按总产量）并将其余工厂汇总。期望的结果：

1       A         15
2       A          2
3   Other          1
4   Other          1
5   Other          2
6   Other          1
7   Other          2
8       C         20
9   Other          5

有什么建议吗？

谢谢！

Answer 1

这里的关键是应用特定的理念，以便根据生产总和将工厂组合在一起。请注意，这种理念与您在（真实）数据集中拥有的实际值有关。

选项 1

以下示例将总产量等于或小于 15 的工厂组合在一起。如果您想要另一个分组，您可以修改阈值（例如，使用 18 而不是 15）

factory <- c("A","A","B","B","B","B","B","C","D")
production <- c(15, 2, 1, 1, 2, 1, 2,20,5)
df <- data.frame(factory, production, stringsAsFactors = F)

library(dplyr)

df %>%
  group_by(factory) %>%
  mutate(factory_new = ifelse(sum(production) > 15, factory, "Other")) %>%
  ungroup()

# # A tibble: 9 x 3
#   factory production factory_new
#   <chr>        <dbl> <chr>      
# 1 A               15 A          
# 2 A                2 A          
# 3 B                1 Other      
# 4 B                1 Other      
# 5 B                2 Other      
# 6 B                1 Other      
# 7 B                2 Other      
# 8 C               20 C          
# 9 D                5 Other 

我正在创建 factory_new 而不删除（原始）factory 列。

选项 2

这是一个示例，您可以根据工厂的产量对工厂进行排名/排序，然后您可以选择一些顶级工厂保持原样并将其余工厂分组

factory <- c("A","A","B","B","B","B","B","C","D")
production <- c(15, 2, 1, 1, 2, 1, 2,20,5)
df <- data.frame(factory, production, stringsAsFactors = F)

library(dplyr)

# get ranked factories based on sum production
df %>%
  group_by(factory) %>%
  summarise(SumProd = sum(production)) %>%
  arrange(desc(SumProd)) %>%
  pull(factory) -> vec_top_factories

# input how many top factories you want to keep
# rest will be grouped together
n = 2

# apply the grouping based on n provided
df %>%
  group_by(factory) %>%
  mutate(factory_new = ifelse(factory %in% vec_top_factories[1:n], factory, "Other")) %>%
  ungroup()

# # A tibble: 9 x 3
#   factory production factory_new
#   <chr>        <dbl> <chr>      
# 1 A               15 A          
# 2 A                2 A          
# 3 B                1 Other      
# 4 B                1 Other      
# 5 B                2 Other      
# 6 B                1 Other      
# 7 B                2 Other      
# 8 C               20 C          
# 9 D                5 Other 

Answer 2

只需指定权重参数w：

> df %>% 
+   mutate(factory = fct_lump_n(factory, 2, w = production))
  factory production
1       A         15
2       A          2
3   Other          1
4   Other          1
5   Other          2
6   Other          1
7   Other          2
8       C         20
9   Other          5

注意：使用forcats::fct_lump_n，因为不再推荐通用的fct_lump。

Answer 3

我们也可以通过使用ave 创建一个逻辑条件来使用base R

df$factory_new <- "Other"
i1 <- with(df, ave(production, factory, FUN = sum) > 15)
df$factory_new[i1] <- df$factory[i1]