R 将表达式传递给聚合密钥

我很容易做到：

library(tsibble)
tourism %>% 
  aggregate_key(Purpose * (State / Region), Trips = sum(Trips))

假设我有

string <- 'Purpose * (State / Region)'

一些函数可以是什么

---

感谢您的帮助

我们可以使用

parse\u expr

aggregate_key(tourism, .spec = !!rlang::parse_expr(string), Trips = sum(Trips))
# A tsibble: 34,000 x 5 [1Q]
# Key:       Purpose, State, Region [425]
#   Quarter Purpose      State        Region        Trips
#     <qtr> <chr*>       <chr*>       <chr*>        <dbl>
# 1 1998 Q1 <aggregated> <aggregated> <aggregated> 23182.
# 2 1998 Q2 <aggregated> <aggregated> <aggregated> 20323.
# 3 1998 Q3 <aggregated> <aggregated> <aggregated> 19827.
# 4 1998 Q4 <aggregated> <aggregated> <aggregated> 20830.
# 5 1999 Q1 <aggregated> <aggregated> <aggregated> 22087.
# 6 1999 Q2 <aggregated> <aggregated> <aggregated> 21458.
# 7 1999 Q3 <aggregated> <aggregated> <aggregated> 19914.
# 8 1999 Q4 <aggregated> <aggregated> <aggregated> 20028.
# 9 2000 Q1 <aggregated> <aggregated> <aggregated> 22339.
#10 2000 Q2 <aggregated> <aggregated> <aggregated> 19941.
# … with 33,990 more rows

---

或者使用

glue

和

parse

eval(parse(text = glue::glue('tourism %>% aggregate_key({string}, Trips = sum(Trips))')))
# A tsibble: 34,000 x 5 [1Q]
# Key:       Purpose, State, Region [425]
   Quarter Purpose      State        Region        Trips
     <qtr> <chr*>       <chr*>       <chr*>        <dbl>
 1 1998 Q1 <aggregated> <aggregated> <aggregated> 23182.
 2 1998 Q2 <aggregated> <aggregated> <aggregated> 20323.
 3 1998 Q3 <aggregated> <aggregated> <aggregated> 19827.
 4 1998 Q4 <aggregated> <aggregated> <aggregated> 20830.
 5 1999 Q1 <aggregated> <aggregated> <aggregated> 22087.
 6 1999 Q2 <aggregated> <aggregated> <aggregated> 21458.
 7 1999 Q3 <aggregated> <aggregated> <aggregated> 19914.
 8 1999 Q4 <aggregated> <aggregated> <aggregated> 20028.
 9 2000 Q1 <aggregated> <aggregated> <aggregated> 22339.
10 2000 Q2 <aggregated> <aggregated> <aggregated> 19941.
# … with 33,990 more rows

eval（解析（text=glue:：glue（'tourism%>%aggregate_key（{string}，Trips=sum（Trips））））
#一个tsibble:34000x5[1Q]
#关键词：目的、州、地区[425]
四分之一目的州地区旅行
1 1998年第一季度23182。
2 1998年第2季度20323。
3 1998年第3季度19827。
1998年第4季度20830。
5 1999年第一季度22087。
6 1999年第2季度21458。
一九九九年第七季一九九四年第三季。
8 1999年4月28日。
9 2000年第一季度22339。
10 2000年第2季度19941。
#…还有33990行

它似乎不起作用``R>library（tsibble）R>string tourism%%>%+aggregate\u key（as.formula（string），Trips=sum（Trips））错误：在

group\u by（）中添加计算列时出现问题。

。

mutate（）

column

as.formula（string）

存在x问题。ℹ

as.formula（string）=agg_vec（

as.formula（string）

）

。找不到x对象“as.formula（string）”，请运行

rlang:：last_error（）

查看错误发生的位置```fabletools：：聚合_key@Andrea我想知道更新是否有助于您感谢您的帮助，但是

eval parse

应该是一个极端的选择。我正在寻找一种在标准评估范围内与fable合作的方法frame@Andrea好的，我添加了

parse\u expr

。希望它能起作用

tourism %>%
    aggregate_key( .spec = !!rlang::parse_expr(string), Trips = sum(Trips))

eval(parse(text = glue::glue('tourism %>% aggregate_key({string}, Trips = sum(Trips))')))
# A tsibble: 34,000 x 5 [1Q]
# Key:       Purpose, State, Region [425]
   Quarter Purpose      State        Region        Trips
     <qtr> <chr*>       <chr*>       <chr*>        <dbl>
 1 1998 Q1 <aggregated> <aggregated> <aggregated> 23182.
 2 1998 Q2 <aggregated> <aggregated> <aggregated> 20323.
 3 1998 Q3 <aggregated> <aggregated> <aggregated> 19827.
 4 1998 Q4 <aggregated> <aggregated> <aggregated> 20830.
 5 1999 Q1 <aggregated> <aggregated> <aggregated> 22087.
 6 1999 Q2 <aggregated> <aggregated> <aggregated> 21458.
 7 1999 Q3 <aggregated> <aggregated> <aggregated> 19914.
 8 1999 Q4 <aggregated> <aggregated> <aggregated> 20028.
 9 2000 Q1 <aggregated> <aggregated> <aggregated> 22339.
10 2000 Q2 <aggregated> <aggregated> <aggregated> 19941.
# … with 33,990 more rows