3

我有一个大型数据集，我希望从中获得一列的汇总估计值（平均值、中位数、计数等），当它们按另外两列分组时。

非常努力地弄清楚如何使用purrr- 希望让这个工作流程能够点击未来的项目......但非常卡住。

作为一个可重现的示例，这适用于按am和分组vs，并估计mpg

library(tidyverse)
library(rlang)

mtcars %>%
  group_by(am, vs) %>%
  summarise(mean_mpg = mean(mpg),
            median_mpg = median(mpg),
            count = n())

然而，为了扩展这个例子，假设我想为amand分组vs；然后am和gear; 然后am和carb。直觉上，这似乎是map应该处理的事情。

group_vars <- c("vs", "gear", "carb")
group_syms <- rlang::syms(group_vars)
sym_am <- rlang::sym("am")

mtcars %>%
  map_df(~group_by(!!sym_am, !!!group_syms) %>%
           summarise(mean_mpg = mean(mpg),
           summarise(median_mpg = median(mpg),
           summarise(count = n())
  )

#Error in !sym_am : invalid argument type
4

2 回答 2

1

我们可以使用map2from来使用多个符号作为参数，purrr然后在输出中对其进行评估group_bysummarise

library(tidyverse)
map2_df(list(sym_am), group_syms, ~ mtcars %>%
         group_by(!!.x, !!.y) %>% 
         summarise(mean_mgp = mean(mpg), median_mpg = median(mpg),count = n()))
于 2017-12-18T16:52:08.190 回答
1

这是一种方法

library(tidyverse)

variable_grp <- c("vs", "gear", "carb")
constant_grp <- c("am")
group_vars <- lapply(variable_grp, function(i) c(constant_grp, i))

map(group_vars, ~group_by_at(mtcars, .x) %>% 
                summarise(  mean_mgp = mean(mpg),
                        median_mpg = median(mpg),
                        count = n()))

这将生成每个组的汇总统计信息列表。与您的问题一起使用map_df的问题是每个组的列名不同（第一组：am，vs；第二组：am，gear ...）。variable_column因此，如果您正在使用，则需要重命名map_df

map_df(group_vars, ~group_by_at(mtcars, .x) %>% 
                summarise(  mean_mgp = mean(mpg),
                        median_mpg = median(mpg),
                        count = n()) %>%
                setNames(c("am", "variable_column", "mean_mpg", "median_mpg", "count")))

# A tibble: 17 x 5
# Groups:   am [2]
      # am variable_column mean_mpg median_mpg count
   # <dbl>           <dbl>    <dbl>      <dbl> <int>
 # 1     0               0 15.05000      15.20    12
 # 2     0               1 20.74286      21.40     7
 # 3     1               0 19.75000      20.35     6
 # 4     1               1 28.37143      30.40     7
 # 5     0               3 16.10667      15.50    15
 # 6     0               4 21.05000      21.00     4
 # 7     1               4 26.27500      25.05     8
 # 8     1               5 21.38000      19.70     5
 # 9     0               1 20.33333      21.40     3
# 10     0               2 19.30000      18.95     6
# 11     0               3 16.30000      16.40     3
# 12     0               4 14.30000      14.30     7
# 13     1               1 29.10000      29.85     4
# 14     1               2 27.05000      28.20     4
# 15     1               4 19.26667      21.00     3
# 16     1               6 19.70000      19.70     1
# 17     1               8 15.00000      15.00     1

您可以variable_column使用.id参数map_df和 post-map_df保存名称mutate

map_df(group_vars, ~group_by_at(mtcars, .x) %>% 
                summarise(  mean_mgp = mean(mpg),
                        median_mpg = median(mpg),
                        count = n()) %>%
                setNames(c("am", "variable_column", "mean_mpg", "median_mpg", "count")),
            .id="variable_col_name") %>%
            mutate(variable_col_name = variable_grp[as.numeric(variable_col_name)])

# A tibble: 17 x 6
# Groups:   am [2]
   # variable_col_name    am variable_column mean_mpg median_mpg count
               # <chr> <dbl>           <dbl>    <dbl>      <dbl> <int>
 # 1                vs     0               0 15.05000      15.20    12
 # 2                vs     0               1 20.74286      21.40     7
 # 3                vs     1               0 19.75000      20.35     6
 # 4                vs     1               1 28.37143      30.40     7
 # 5              gear     0               3 16.10667      15.50    15
 # 6              gear     0               4 21.05000      21.00     4
 # 7              gear     1               4 26.27500      25.05     8
 # 8              gear     1               5 21.38000      19.70     5
 # 9              carb     0               1 20.33333      21.40     3
# 10              carb     0               2 19.30000      18.95     6
# 11              carb     0               3 16.30000      16.40     3
# 12              carb     0               4 14.30000      14.30     7
# 13              carb     1               1 29.10000      29.85     4
# 14              carb     1               2 27.05000      28.20     4
# 15              carb     1               4 19.26667      21.00     3
# 16              carb     1               6 19.70000      19.70     1
# 17              carb     1               8 15.00000      15.00     1
于 2017-12-18T16:45:32.660 回答