问题描述
我正在处理来自家庭调查的数据，我想计算对各种问题的回答频率表（每个受访者可能有多个回答）。为了加快调查数据的分析，我编写了一些本地函数，同时试图保持 tidyverse 逻辑。我已经成功编写了一个函数，该函数允许我在使用分组变量的同时计算频率。
library(tibble)
library(dplyr)
my_df <- tibble(id = c(1,1,2,3,4,5,7,8,8),country = c("A","A","B","B"),region = c("ax","ax","ay","bx","by","by"),district = c("ax1","ax1","ax2","ay1","ay2","bx1","by1","by1"),question = c("answer1","answer2","answer1","answer2"))
freq <- function(df,var,id_var,...) {
n <- df %>%
group_by(...) %>%
summarise(n = NROW(unique({{id_var}})),.groups = "drop") %>%
left_join(distinct(df,{{var}}),by = character(),.)
df %>%
group_by(...,{{var}}) %>%
summarise(cases = n(),.groups = "drop") %>%
left_join(n) %>%
mutate(freq_answer = cases/n*100) %>%
ungroup()
}
my_df %>%
freq(question,id,country,region)
# A tibble: 7 x 6
country region question cases n freq_answer
<chr> <chr> <chr> <int> <int> <dbl>
1 A ax answer1 2 2 100
2 A ax answer2 1 2 50
3 A ay answer1 1 2 50
4 A ay answer2 1 2 50
5 B bx answer1 2 2 100
6 B by answer1 1 2 50
7 B by answer2 2 2 100
出于报告目的，在某些情况下，我想计算每个分组级别的频率并将结果合并到单个数据框中。我找到了一个解决方案，可以让我对固定数量的分组变量执行此操作，并且能够获得所需的结果。显然，如果我使用更多或更少的分组变量，我将不得不指定额外的函数。
freq_sum <- function(df,group1,group2) {
df0 <- freq({{df}},{{var}},{{id_var}}) %>%
add_column({{group1}} :="Total",.before = 1) %>%
add_column({{group2}} :="Total",.after = 1)
df1 <- freq({{df}},{{id_var}},{{group1}}) %>%
add_column({{group2}} :="Total",.after = 1)
df2 <- freq({{df}},{{group1}},{{group2}})
rbind(df2,df1,df0)
}
my_df %>%
freq_sum(question,region)
country region question cases n freq_answer
<chr> <chr> <chr> <int> <int> <dbl>
1 A ax answer1 2 2 100
2 A ax answer2 1 2 50
3 A ay answer1 1 2 50
4 A ay answer2 1 2 50
5 B bx answer1 2 2 100
6 B by answer1 1 2 50
7 B by answer2 2 2 100
8 A Total answer1 3 3 100
9 A Total answer2 2 3 66.7
10 B Total answer1 3 4 75
11 B Total answer2 2 4 50
12 Total Total answer1 6 7 85.7
13 Total Total answer2 4 7 57.1
我的问题：有没有人对如何使 freq_sum 函数更通用/优雅而不需要预先指定分组变量的数量有什么建议？
我对如何实现这一目标有了一些初步想法，但不确定如何实施它们，或者它们一开始是否可行。
freq_sum <- function(df,...) {
df0 <- df %>%
freq({{var}},{{id}},...)
grouping_vars <- df0 %>%
select(1:{{var}}) %>%
select(-last_col()) %>%
names()
# From grouping_vars create a list with vectors that contain increasingy less grouping variables.
[1] "country" "region" "district"
[2] "country" "region"
[3] "country"
# Use the elements of the list as input in the freq() function.
# Add the missing grouping variables to the resulting data frames.
# Combine all dataframes in a single data frame.
}
解决方法
如果有人遇到类似问题：借助以下两个问题的答案，我找到了一个涉及按预期工作的 for 循环的解决方案，让我可以自由选择要汇总的分组变量的数量。
R: Create empty tibble/data frame with column names coming from a vector
Adding column if it does not exist
freq_sum <- function(df,var,id_var,...) {
var_names <- names(select(df,...))
df_total <- bind_rows(setNames(rep("Total",length(var_names)),var_names))
df_final <- df %>% freq({{var}},{{id_var}},...)
for (i in 1:length(var_names)-1) {
v <- var_names[1:i]
df_final <- df %>%
freq({{var}},across(v)) %>%
add_column(!!!df_total[!names(df_total) %in% names(.)]) %>%
rbind(df_final,.) %>%
distinct()
}
df %>%
freq({{var}},{{id_var}}) %>%
add_column(!!!df_total[!names(df_total) %in% names(.)]) %>%
rbind(df_final,.)
}
my_df %>%
freq_sum(question,id,country,region)
# A tibble: 13 x 6
country region question cases n freq_answer
<chr> <chr> <chr> <int> <int> <dbl>
1 A ax answer1 2 2 100
2 A ax answer2 1 2 50
3 A ay answer1 1 2 50
4 A ay answer2 1 2 50
5 B bx answer1 2 2 100
6 B by answer1 1 2 50
7 B by answer2 2 2 100
8 A Total answer1 3 3 100
9 A Total answer2 2 3 66.7
10 B Total answer1 3 4 75
11 B Total answer2 2 4 50
12 Total Total answer1 6 7 85.7
13 Total Total answer2 4 7 57.1