基于另一列的块因子

Question

示例显示了不同工厂的产量测量值，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

现在我想根据他们在这个数据集中的总产量将工厂集中到更少的级别。

对于正常的 forcats::fct_lump，我可以根据它们出现的行数来合并它们，例如制作 3 个级别：

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（产量）将它们合并，保留前 n=2 家工厂（按总产量）并将其余工厂合并。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

这里的关键是应用一种特定的理念，以便根据生产总和将工厂分组在一起。请注意，这种理念与您在（真实）数据集中的实际值有关。

选项 1

下面的示例将总产量等于或小于 15 的工厂组合在一起。如果你想要另一个分组，你可以修改阈值（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

我们也可以通过使用 ave

创建逻辑条件来使用 base R

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

Answer 3

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 fct_lump。

基于另一列的块因子

lump factor based on another column

r

tidyverse

forcats