不包括Dot in Dplyr的周末 [英] Exclude Weekends in Dot in Dplyr

问题描述

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

我在prophet包的dplyr中使用do函数。尝试此操作时，我想创建一个不包括周末的未来数据帧。以下是我当前的代码：

当前数据帧：

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)

如何重写上述代码以筛选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)

推荐答案

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), 
  filter(make_future_dataframe(prophet(., daily.seasonality = TRUE, yearly.seasonality = TRUE), periods = 14), weekdays(ds) != "Saturday" & weekdays(ds) != "Sunday"))) %>%
  select(ds, group, yhat)

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