英文:
How to dynamically create case_when() expressions based on variable names?
问题
以下是您要翻译的内容：
"I have a dataframe with a set of revenue variables and a set of sales variables. I want to create a new set of variables that takes the value of the revenue variable and checks which sales variable it matches, and if it matches, it returns the corresponding year.
Here's a simplified example of what I have and what I'm trying to achieve:
library(tidyverse)library(rlang)# Creating an example dataframedf <- tibble(revenues_1 = c(1, 3, 4),revenues_2 = c(2, 4, 5),sales_2005 = c(1, 2, 3),sales_2006 = c(2, 3, 4),sales_2007 = c(3, 4, 5))# What I want to do:# df <- df |># rowwise() |># mutate(# year_1 = case_when(# revenues_1 == sales_2005 ~ 2005,# revenues_1 == sales_2006 ~ 2006,# revenues_1 == sales_2007 ~ 2007,# .default = NA# ),# year_2 = case_when(# revenues_2 == sales_2005 ~ 2005,# revenues_2 == sales_2006 ~ 2006,# revenues_2 == sales_2007 ~ 2007,# .default = NA# )# )
However, this is not scalable if I have many years and many revenue variables.
How can I make this more efficient?
One idea is (not working):
# Create a list of expressions for each yearyears_exprs <- map(2005:2007, ~parse_expr(paste0("var == 'sales_", .x, "' ~ ", .x)))# Use this list in a case_when statementdf <- df |>rowwise() |>mutate(across(starts_with("revenues_"), function(var) {case_when(!!!years_exprs,.default = NA)}, .names = "year_{.col}")) |>ungroup()```"<details><summary>英文:</summary>I have a dataframe with a set of revenue variables and a set of sales variables. I want to create a new set of variables that takes the value of the revenue variable and checks which sales variable it matches, and if it matches, it returns the corresponding year.Here's a simplified example of what I have and what I'm trying to achieve:``` rlibrary(tidyverse)library(rlang)# Creating an example dataframedf <- tibble(revenues_1 = c(1, 3, 4),revenues_2 = c(2, 4, 5),sales_2005 = c(1, 2, 3),sales_2006 = c(2, 3, 4),sales_2007 = c(3, 4, 5))# What I want to do:# df <- df |># rowwise() |># mutate(# year_1 = case_when(# revenues_1 == sales_2005 ~ 2005,# revenues_1 == sales_2006 ~ 2006,# revenues_1 == sales_2007 ~ 2007,# .default = NA# ),# year_2 = case_when(# revenues_2 == sales_2005 ~ 2005,# revenues_2 == sales_2006 ~ 2006,# revenues_2 == sales_2007 ~ 2007,# .default = NA# )# )
However, this is not scalable if I have many years and many revenue variables.
How can I make this more efficient?
One idea is (not working):
# Create a list of expressions for each yearyears_exprs <- map(2005:2007, ~parse_expr(paste0("var == 'sales_", .x, "' ~ ", .x)))# Use this list in a case_when statementdf <- df |>rowwise() |>mutate(across(starts_with("revenues_"), function(var) {case_when(!!!years_exprs,.default = NA)}, .names = "year_{.col}")) |>ungroup()
答案1
得分: 2
如果我理解你的意思正确，你可以使用现有的代码，使用dplyr的mutate、across和starts_with，然后将函数放在一个命名列表中来创建新的变量。
为了减少为多个年份创建单独的case_when逻辑的工作量，你可以首先使用rlang::parse_exprs()和paste0：
yrs <- 2005:2007years_exprs <- rlang::parse_exprs(paste0(".x == sales_", yrs, " ~ ", yrs))
然后使用!!!来评估它：
library(dplyr)df %>% mutate(across(starts_with("revenues_"),list(year = ~case_when(!!!years_exprs))))
输出：
revenues_1 revenues_2 sales_2005 sales_2006 sales_2007 revenues_1_year revenues_2_year1 1 2 1 2 3 2005 20062 3 4 2 3 4 2006 20073 4 5 3 4 5 2006 2007
注意：如果你的变量不一定都以"revenues"开头，你可以创建一个包含所需列的向量，无论它们是否共享相同的模式（这里是mut_vars），然后使用all_of()：
mut_vars <- c("revenues_1", "revenues_2")df %>% mutate(across(all_of(mut_vars),list(year = ~case_when(!!!years_exprs))))
英文:
If I understand you correctly, using your existing code you could use dplyr's mutate, across and starts_with, then place the function in a named list to create the new variables.
In order to reduce the effort to create individual case_when logics for multiple years, you could first use rlang::parse_exprs() with paste0:
yrs <- 2005:2007years_exprs <- rlang::parse_exprs(paste0(".x == sales_", yrs, " ~ ", yrs))
Then evaluate it using !!!:
library(dplyr)df %>% mutate(across(starts_with("revenues_"),list(year = ~case_when(!!!years_exprs))))
output:
revenues_1 revenues_2 sales_2005 sales_2006 sales_2007 revenues_1_year revenues_2_year<dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl>1 1 2 1 2 3 2005 20062 3 4 2 3 4 2006 20073 4 5 3 4 5 2006 2007
Note: if your variables don't necessarily all start with "revenues", you could create a vector of desired columns regardless if they share the same pattern (here, mut_vars), then use all_of():
mut_vars <- c("revenues_1", "revenues_2")df %>% mutate(across(all_of(mut_vars),list(year = ~case_when(!!!years_exprs))))
答案2
得分: 0
以下是翻译好的部分：
如果您绝对需要生成表达式，可以像这样操作：
library(glue)library(tidyverse)years <- 2005:2007idx <- 1:2years_exprs <- map_chr(idx, ~ {str_flatten_comma(c(glue("revenues_{.x} == sales_{years} ~ {years}"),".default = NA")) %>%str_c("case_when(", ., ")")}) |>set_names(paste0("year_", idx)) |>parse_exprs()df |>rowwise() |>mutate(!!! years_exprs) |>ungroup()
如果您要使用拆分切片操作符!!!，则需要为mutate提供命名表达式，然后这些名称将成为变量名称。
您可以看到在使用rlang::qq_show注入后，此表达式将如何解析：
qq_show(df |>rowwise() |>mutate(!!! years_exprs) |>ungroup())ungroup(mutate(rowwise(df), year_1 = case_when(revenues_1 == sales_2005 ~ 2005,revenues_1 == sales_2006 ~ 2006, revenues_1 == sales_2007 ~ 2007, .default = NA),year_2 = case_when(revenues_2 == sales_2005 ~ 2005, revenues_2 == sales_2006 ~2006, revenues_2 == sales_2007 ~ 2007, .default = NA)))
话虽如此，如果您能找到实施@jpsmith提供的解决方案的方法，那将是更好的做法。根据我的经验，人们在数据以宽格式存在时才会以这种方式使用表达式。通常更容易将其转换为长格式或使用可用的工具，如across。
输出
revenues_1 revenues_2 sales_2005 sales_2006 sales_2007 year_1 year_2<dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl>1 1 2 1 2 3 2005 20062 3 4 2 3 4 2006 20073 4 5 3 4 5 2006 2007
英文:
If you absolutely needed to generate expressions you could do so like:
library(glue)library(tidyverse)years <- 2005:2007idx <- 1:2years_exprs <- map_chr(idx, ~ {str_flatten_comma(c(glue("revenues_{.x} == sales_{years} ~ {years}"),".default = NA")) %>%str_c("case_when(", ., ")")}) |>set_names(paste0("year_", idx)) |>parse_exprs()df |>rowwise() |>mutate(!!! years_exprs) |>ungroup()
You need to provide named expressions to mutate if you are going to use the split-slice operator !!!. The names will then become the variable names.
You can see how this expression will resolve after injection using rlang::qq_show:
qq_show(df |>rowwise() |>mutate(!!! years_exprs) |>ungroup())ungroup(mutate(rowwise(df), year_1 = case_when(revenues_1 == sales_2005 ~ 2005,revenues_1 == sales_2006 ~ 2006, revenues_1 == sales_2007 ~ 2007, .default = NA),year_2 = case_when(revenues_2 == sales_2005 ~ 2005, revenues_2 == sales_2006 ~2006, revenues_2 == sales_2007 ~ 2007, .default = NA)))
All that being said, if you could find a way to implement the solution provided by @jpsmith that would be a better practice. In my experience, people resort to using expressions in this way when they have data in a wide format. Often it is easier to pivot to a longer format or use available tools like across.
Output
revenues_1 revenues_2 sales_2005 sales_2006 sales_2007 year_1 year_2<dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl>1 1 2 1 2 3 2005 20062 3 4 2 3 4 2006 20073 4 5 3 4 5 2006 2007
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