lubridate - 选择每周的第一个非星期一。

Question

有一些财务数据，我想通过只选择每周的第一个非星期一来过滤它。通常是星期二，但如果星期二是假期，有时可能是星期三。

这是我在大多数情况下都有效的代码

XLF <- quantmod::getSymbols("XLF", from = "2000-01-01", auto.assign = FALSE)

library(tibble)
library(lubridate)
library(dplyr)
xlf <- as_tibble(XLF) %>% rownames_to_column(var = "date") %>% 
         select(date, XLF.Adjusted)  
xlf$date <- ymd(xlf$date)

# We create Month, Week number and Days of the week columns
# Then we remove all the Mondays
xlf <- xlf %>% mutate(Year = year(date), Month = month(date), 
                      IsoWeek = isoweek(date), WDay = wday(date)) %>% 
               filter(WDay != 2)

# Creating another tibble just for ease of comparison
xlf2 <- xlf %>% 
          group_by(Year, IsoWeek) %>% 
          filter(row_number() == 1) %>% 
          ungroup()

也就是说，到目前为止，我还无法解决一些问题。

问题在于它跳过了星期二的“2002-12-31”，因为它被视为 2003 年第一周 ISO 的一部分。 有几个类似的问题。
我的问题是，我如何在没有此类问题的情况下选择每周的第一个非星期一，同时留在 tidyverse（即不必使用 xts / zoo 类）？

Answer 1

您可以自己创建一个持续增加的周数。也许不是最优雅的解决方案，但对我来说效果很好。

as_tibble(XLF) %>% 
  rownames_to_column(var = "date")%>% 
  select(date, XLF.Adjusted)%>%
  mutate(date = ymd(date),
         Year = year(date),
         Month = month(date),
         WDay = wday(date),
         WDay_label = wday(date, label = T))%>% 
  # if the weekday number is higher in the line above or 
  # if the date in the previous line is more than 6 days ago
  # the week number should be incremented
  mutate(week_increment  = (WDay < lag(WDay) | difftime(date, lag(date), unit = 'days') > 6))%>%
  # the previous line causes the first element to be NA due to 
  # the fact that the lag function can't find a line above
  # we correct this here by setting the first element to TRUE
  mutate(week_increment = ifelse(row_number() == 1,
                                 TRUE,
                                 week_increment))%>%
  # we can sum the boolean elements in a cumulative way to get a week number
  mutate(week_number = cumsum(week_increment))%>%
  filter(WDay != 2)%>%
  group_by(Year, week_number) %>% 
  filter(row_number() == 1)