在 Dplyr 中排除 Dot 中的周末

Question

这是此答案的后续问题：https://stackoverflow.com/a/45254762/5893585

我正在使用 prophet 包中的 dplyr 中的 do 函数。尝试此操作时，我想制作一个不包括周末的未来数据框。以下是我当前的代码：

当前数据框：

dataset
          ds     group     y
  2021-12-15         A     5
  2021-12-16         A     6
  2021-12-15         B    10
  2021-12-16         B     7
         etc       etc   etc

预测

library(dplyr)
library(prophet)

data = dataset %>%  
group_by(group) %>%
do(predict(prophet(., daily.seasonality = TRUE, yearly.seasonality = TRUE), 
make_future_dataframe(prophet(.,daily.seasonality = TRUE, yearly.seasonality = TRUE), periods = 14))) %>%
select(ds, group, yhat)

如何重写上面的代码来过滤make_future_dataframe 数据集没有周末？

我希望它看起来像这样，但这不起作用：

data = dataset %>%  
  group_by(group) %>%
  do(predict(prophet(., daily.seasonality = TRUE, yearly.seasonality = TRUE), 
  make_future_dataframe(prophet(.[which(weekdays(.$ds) != 'Saturday' | weekdays(.$ds) != 'Sunday'),],daily.seasonality = TRUE, yearly.seasonality = TRUE), periods = 14))) %>%
  select(ds, group, yhat)

Answer 1

我们可以在预测之前删除周末：

df %>% 
  group_by(group) %>% 
  mutate(weekdays = weekdays(ds)) %>% 
  filter(weekdays != "Saturday" & weekdays != "Sunday") %>% 
  do(predict(prophet(., daily.seasonality = TRUE, yearly.seasonality = TRUE), 
             make_future_dataframe(prophet(.[which(weekdays(.$ds) != 'Saturday' | weekdays(.$ds) != 'Sunday'),],daily.seasonality = TRUE, yearly.seasonality = TRUE), periods = 14))) %>%
  select(ds, group, yhat)