我在理解group_by如何在tidyverse中工作时遗漏了一些东西。示例将澄清：
我创建了以下函数，它需要很少的参数并计算tibble中的最佳权重（可能不是很漂亮但似乎有效）：
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)
}
当我只有一个组在tibble时，这个功能很好用。我创建函数的方法是尝试通过在一个函数上进行测试来使其工作，并希望在以后对数据进行切片时它将起作用。
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，但是我收到错误：
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
那我错过了什么？为什么我的函数每组返回15而不是5？
DATA：
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"))
EXTRA：正如Jimbou所示，分组工作正在进行，但我的功能在某种程度上被破坏了。对某些变量进行硬编码会使这个变得有效，所以我只需要找出正确的变量引用吗？
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
答案 0 :(得分：0)
将回答我自己的问题，但这似乎是解决问题，而且我的部分知识很差。感谢Jimbou。欢迎提供更好的答案。
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