如何删除 tsibble 对象聚合中的时间分量？

Question

我一直在使用 tsibble 包，但我不知道如何从聚合结果中删除时间分量的正确方法。所以在下面的数据集中，我想要按地区和州划分平均行程。是将 tsibble 转换为 tibble 的正确方法（可能是，我只是不确定）还是我缺少一些实现聚合的选项？

library(tsibble)
library(dplyr)

tourism %>% group_by(Region, State) %>% summarise(Mean_trips = mean(Trips))

# A tsibble: 6,080 x 4 [1Q]
# Key:       Region, State [76]
# Groups:    Region [76]
   Region   State           Quarter Mean_trips
   <chr>    <chr>             <qtr>      <dbl>
 1 Adelaide South Australia 1998 Q1       165.
 2 Adelaide South Australia 1998 Q2       112.
 3 Adelaide South Australia 1998 Q3       148.

## This is not what I want, this is what I want:

tourism %>% as_tibble %>% group_by(Region, State) %>% summarise(Mean_trips = mean(Trips))

# A tibble: 76 x 3
# Groups:   Region [76]
   Region                     State              Mean_trips
   <chr>                      <chr>                   <dbl>
 1 Adelaide                   South Australia        143.  
 2 Adelaide Hills             South Australia          7.18

Answer 1

如果我们在 tourism 数据上使用 select(-Quarter)，它会给出一条信息性错误消息。

library(tsibble)
library(dplyr)

tourism %>% select(-Quarter)

Error: Column Quarter (index) can't be removed. Do you need as_tibble() to work with data frame?

因此，as_tibble 是转换为 tibble 的正确方法。

tourism %>% 
    as_tibble %>% 
    group_by(Region, State) %>% 
    summarise(Mean_trips = mean(Trips))

#   Region                     State              Mean_trips
#   <chr>                      <chr>                   <dbl>
# 1 Adelaide                   South Australia        143.  
# 2 Adelaide Hills             South Australia          7.18
# 3 Alice Springs              Northern Territory      14.2 
# 4 Australia's Coral Coast    Western Australia       47.4 
#...

Answer 2

为了完整起见：来自 tsibble

的 reference manual

Column-wise verbs, including select(), transmute(), summarise(), mutate() & transmute(), keep the time context hanging around. That is, the index variable cannot be dropped for a tsibble. If any key variable is changed, it will validate whether it’s a tsibble internally. Use as_tibble() to leave off the time context.

不能删除时间分量，as_tibble() 是转换为 tibble 的正确选择。

如何删除 tsibble 对象聚合中的时间分量？

How to remove the temporal component in an aggregation of a tsibble object?

r

dplyr

tibble

tsibble