6

这是数据：

library(tidyverse)

data <- tibble::tribble(
  ~var1, ~var2, ~var3,  ~var4,    ~var5,
    "a",   "d",   "g",  "hello",    1L,
    "a",   "d",   "h",  "hello",    2L,
    "b",   "e",   "h",  "k",        4L,
    "b",   "e",   "h",  "k",        7L,
    "c",   "f",   "i",  "hello",    3L,
    "c",   "f",   "i",  "hello",    4L
  )

和向量，我想使用：

filter_var <- c("hello")
groupby_vars1 <- c("var1", "var2", "var3")
groupby_vars2 <- c("var1", "var2")
joinby_vars1 <- c("var1", "var2")
joinby_vars2 <- c("var1", "var2", "var3")

2nd & 5th 和 3rd & 4th 向量相同，但请假设它们不同并将它们保留为不同的向量。

现在我想创建一个通用函数，我可以在其中获取数据和这些向量来获得结果。 

my_fun <- function(data, filter_var, groupby_vars1,groupby_vars2, joinby_vars1, joinby_vars2) {

  data2 <- data %>% filter(var4 == filter_var) 

  data3 <- data2 %>%
    group_by(groupby_vars1) %>% 
    summarise(var6 = sum(var5))

  data4 <- data3 %>%
    ungroup() %>%
    group_by(groupby_vars2) %>% 
    summarise(avg = mean(var6,na.rm = T))

  data5 <- data3 %>% left_join(data4, by = joinby_vars1)

  data6 <- data %>% left_join(data5, by = joinby_vars2)
}

问题是向函数提供多个变量的多个向量以用作主体中的 dplyr 参数。我尝试查看http://dplyr.tidyverse.org/articles/programming.html，但无法解决上述问题。

4

1 回答 1

8

group_by不能将groupby_vars...字符串作为输入。您需要使用rlang::syms()将字符串向量转换为变量，然后使用取消引用!!!它们，以便可以在内部评估它们group_by 

library(tidyverse)
library(rlang)

data <- tibble::tribble(
  ~var1, ~var2, ~var3,  ~var4,    ~var5,
  "a",   "d",   "g",  "hello",    1L,
  "a",   "d",   "h",  "hello",    2L,
  "b",   "e",   "h",  "k",        4L,
  "b",   "e",   "h",  "k",        7L,
  "c",   "f",   "i",  "hello",    3L,
  "c",   "f",   "i",  "hello",    4L
)

filter_var <- c("hello")
groupby_vars1 <- c("var1", "var2", "var3")
groupby_vars2 <- c("var1", "var2")
joinby_vars1  <- c("var1", "var2")
joinby_vars2  <- c("var1", "var2", "var3")

my_fun <- function(data, filter_var, 
                   groupby_vars1, groupby_vars2, 
                   joinby_vars1,  joinby_vars2) {

  groupby_vars1 <- syms(groupby_vars1)
  groupby_vars2 <- syms(groupby_vars2)

  data2 <- data %>% 
    filter(var4 == filter_var) 

  data3 <- data2 %>%
    group_by(!!! groupby_vars1) %>% 
    summarise(var6 = sum(var5))

  data4 <- data3 %>%
    ungroup() %>%
    group_by(!!! groupby_vars2) %>% 
    summarise(avg = mean(var6, na.rm = TRUE))

  data5 <- data3 %>% 
    left_join(data4, by = joinby_vars1)

  data6 <- data %>% 
    left_join(data5, by = joinby_vars2)

  return(data6)
}

my_fun(data, filter_var, 
       groupby_vars1, groupby_vars2, 
       joinby_vars1,  joinby_vars2)

#> # A tibble: 6 x 7
#>   var1  var2  var3  var4   var5  var6   avg
#>   <chr> <chr> <chr> <chr> <int> <int> <dbl>
#> 1 a     d     g     hello     1     1   1.5
#> 2 a     d     h     hello     2     2   1.5
#> 3 b     e     h     k         4    NA  NA  
#> 4 b     e     h     k         7    NA  NA  
#> 5 c     f     i     hello     3     7   7  
#> 6 c     f     i     hello     4     7   7

另一种方法：使用外部解析字符串向量，parse_exprs然后在函数内部取消引用它们。另请参阅

my_fun2 <- function(data, filter_var, 
                   groupby_vars1, groupby_vars2, 
                   joinby_vars1,  joinby_vars2) {

  data2 <- data %>% 
    filter(var4 == filter_var) 

  data3 <- data2 %>%
    group_by(!!! groupby_vars1) %>% 
    summarise(var6 = sum(var5))

  data4 <- data3 %>%
    ungroup() %>%
    group_by(!!! groupby_vars2) %>% 
    summarise(avg = mean(var6, na.rm = TRUE))

  data5 <- data3 %>% 
    left_join(data4, by = joinby_vars1)

  data6 <- data %>% 
    left_join(data5, by = joinby_vars2)

  return(data6)
}

my_fun2(data, filter_var, 
        parse_exprs(groupby_vars1), parse_exprs(groupby_vars2), 
        joinby_vars1,  joinby_vars2) 

identical(my_fun(data, filter_var, 
                 groupby_vars1, groupby_vars2, 
                 joinby_vars1,  joinby_vars2),
          my_fun2(data, filter_var, 
                  parse_exprs(groupby_vars1), parse_exprs(groupby_vars2), 
                  joinby_vars1,  joinby_vars2))

[1] TRUE

由reprex 包（v0.2.0）于 2018 年 4 月 24 日创建。

于 2018-04-24T23:44:13.343 回答