r - 使用 group_by tidyverse 自定义函数

Question

我在理解 group_by 如何在 tidyverse 中工作时遗漏了一些东西。示例将阐明:

我创建了以下函数，它需要很少的参数并计算 tibble 内的最佳权重(可能不是最漂亮但似乎有效):

library(lpSolveAPI)
library(tidyverse)

weights_fun <- function(data_tbl, objective, constraint){
  cols <- c("objective", "constraint")
  linear.dt <- data_tbl %>% select_(.dots = cols)
  lp.mod <- make.lp(0, NROW(linear.dt))
  set.objfn(lp.mod, linear.dt$amount)
  lp.control(lp.mod,sense="max")
  add.constraint(lp.mod, linear.dt$duration, "=", 6)
  add.constraint(lp.mod, rep(1, nrow(linear.dt)), "=", 1)
  set.bounds(lp.mod, upper = rep(0.4, nrow(linear.dt)))
  set.bounds(lp.mod, lower = rep(0.10, nrow(linear.dt)))
  solve(lp.mod)
  weights <- round(get.variables(lp.mod), 4)
  return(weights)
}

当我在 tibble 中只有一组时，此功能运行良好。我创建函数的方法是尝试通过对一个函数进行测试来使其工作，并希望在我稍后对数据进行切片时它会起作用。

weights_fun(one_group, "amount", "duration")
one_group$weights <- weights_fun(one_group, "amount", "duration")


  # A tibble: 5 x 6
        date country bucket   amount duration weights
      <date>   <chr>  <chr>    <dbl>    <dbl>   <dbl>
1 2006-01-31      AT     B1 4844.500  1.48475  0.1000
2 2006-01-31      AT     B2 8601.000  3.67500  0.1911
3 2006-01-31      AT     B3 8518.400  5.39900  0.4000
4 2006-01-31      AT     B4 6469.550  6.99950  0.1000
5 2006-01-31      AT     B5 7804.533 10.96133  0.2089

然后我希望我可以使用 mutate 为我的多个组创建新的权重列，如下所示，但我收到错误:

three_groups %>% 
  group_by(date, country) %>% 
  mutate(weights = weights_fun(., "amount", "duration"))

Adding missing grouping variables: `date`, `country`
Error in mutate_impl(.data, dots) : 
  Column `weights` must be length 5 (the group size) or one, not 15

那么我错过了什么？为什么我的函数为每组返回 15 而不是 5？

数据:

one_group <- structure(list(date = structure(c(13179, 13179, 13179, 13179, 
13179), class = "Date"), country = c("AT", "AT", "AT", "AT", 
"AT"), bucket = c("B1", "B2", "B3", "B4", "B5"), amount = c(4844.5, 
8601, 8518.4, 6469.55, 7804.53333333333), duration = c(1.48475, 
3.675, 5.399, 6.9995, 10.9613333333333)), .Names = c("date", 
"country", "bucket", "amount", "duration"), row.names = c(NA, 
-5L), class = c("tbl_df", "tbl", "data.frame"))

three_groups <- structure(list(date = structure(c(13179, 13179, 13179, 13179, 
13179, 13179, 13179, 13179, 13179, 13179, 13179, 13179, 13179, 
13179, 13179), class = "Date"), country = c("AT", "AT", "AT", 
"AT", "AT", "AU", "AU", "AU", "AU", "AU", "BE", "BE", "BE", "BE", 
"BE"), bucket = c("B1", "B2", "B3", "B4", "B5", "B1", "B2", "B3", 
"B4", "B5", "B1", "B2", "B3", "B4", "B5"), amount = c(4844.5, 
8601, 8518.4, 6469.55, 7804.53333333333, 4650.4, 5355.25, 5796.7, 
4899.25, 4995, 10151.38, 14484.8666666667, 9910.06666666667, 
10507.35, 9644.2), duration = c(1.48475, 3.675, 5.399, 6.9995, 
10.9613333333333, 1.8655, 3.493, 4.552, 6.3235, 7.884, 1.8558, 
3.55, 5.32466666666667, 7.01975, 12.6736666666667)), class = c("tbl_df", 
"tbl", "data.frame"), row.names = c(NA, -15L), .Names = c("date", 
"country", "bucket", "amount", "duration"))

额外:正如 Jimbou 所展示的，分组正在起作用，但我的功能不知何故被破坏了。硬编码一些变量会使这个工作，所以我只需要找出正确的变量引用？

weights_fun1 <- function(objective, constraint){
  lp.mod <- make.lp(0, 5)
  set.objfn(lp.mod, objective)
  lp.control(lp.mod,sense="max")
  add.constraint(lp.mod, constraint, "=", 6)
  add.constraint(lp.mod, rep(1, 5), "=", 1)
  set.bounds(lp.mod, upper = rep(0.4, 5))
  set.bounds(lp.mod, lower = rep(0.10, 5))
  solve(lp.mod)
  weights <- round(get.variables(lp.mod), 4)
  return(weights)
}

three_groups %>% 
  group_by(date, country) %>% 
  mutate(weights = weights_fun1(amount, duration))

# A tibble: 15 x 6
# Groups:   date, country [3]
         date country bucket    amount  duration weights
       <date>   <chr>  <chr>     <dbl>     <dbl>   <dbl>
 1 2006-01-31      AT     B1  4844.500  1.484750  0.1000
 2 2006-01-31      AT     B2  8601.000  3.675000  0.1911
 3 2006-01-31      AT     B3  8518.400  5.399000  0.4000
 4 2006-01-31      AT     B4  6469.550  6.999500  0.1000
 5 2006-01-31      AT     B5  7804.533 10.961333  0.2089
 6 2006-01-31      AU     B1  4650.400  1.865500  0.1000
 7 2006-01-31      AU     B2  5355.250  3.493000  0.1000
 8 2006-01-31      AU     B3  5796.700  4.552000  0.1235
 9 2006-01-31      AU     B4  4899.250  6.323500  0.2765
10 2006-01-31      AU     B5  4995.000  7.884000  0.4000
11 2006-01-31      BE     B1 10151.380  1.855800  0.1000
12 2006-01-31      BE     B2 14484.867  3.550000  0.4000
13 2006-01-31      BE     B3  9910.067  5.324667  0.1000
14 2006-01-31      BE     B4 10507.350  7.019750  0.2136
15 2006-01-31      BE     B5  9644.200 12.673667  0.1864

Answer 1

将回答我自己的问题，但这似乎是解决方法，而我的 tidyverse 知识很少。感谢金宝。欢迎更好的答案。

修改功能:

weights_fun1 <- function(objective, constraint, rows){
  lp.mod <- make.lp(0, rows[1])
  set.objfn(lp.mod, objective)
  lp.control(lp.mod,sense="max")
  add.constraint(lp.mod, constraint, "=", 6)
  add.constraint(lp.mod, rep(1, rows[1]), "=", 1)
  set.bounds(lp.mod, upper = rep(0.4, rows[1]))
  set.bounds(lp.mod, lower = rep(0.10, rows[1]))
  solve(lp.mod)
  weights <- round(get.variables(lp.mod), 4)
  return(weights)
}

three_groups %>% 
  group_by(date, country) %>% 
  mutate(rows = n()) %>% #create helper column, as couldn't figure out other way now
  mutate(weights = weights_fun1(amount, duration, rows))


# A tibble: 15 x 7
# Groups:   date, country [3]
         date country bucket    amount  duration  rows weights
       <date>   <chr>  <chr>     <dbl>     <dbl> <int>   <dbl>
 1 2006-01-31      AT     B1  4844.500  1.484750     5  0.1000
 2 2006-01-31      AT     B2  8601.000  3.675000     5  0.1911
 3 2006-01-31      AT     B3  8518.400  5.399000     5  0.4000
 4 2006-01-31      AT     B4  6469.550  6.999500     5  0.1000
 5 2006-01-31      AT     B5  7804.533 10.961333     5  0.2089
 6 2006-01-31      AU     B1  4650.400  1.865500     5  0.1000
 7 2006-01-31      AU     B2  5355.250  3.493000     5  0.1000
 8 2006-01-31      AU     B3  5796.700  4.552000     5  0.1235
 9 2006-01-31      AU     B4  4899.250  6.323500     5  0.2765
10 2006-01-31      AU     B5  4995.000  7.884000     5  0.4000
11 2006-01-31      BE     B1 10151.380  1.855800     5  0.1000
12 2006-01-31      BE     B2 14484.867  3.550000     5  0.4000
13 2006-01-31      BE     B3  9910.067  5.324667     5  0.1000
14 2006-01-31      BE     B4 10507.350  7.019750     5  0.2136
15 2006-01-31      BE     B5  9644.200 12.673667     5  0.1864