R语言开发捷克地区下一个最近工作日查询函数的技术问询

Got it, let's fix your next business day calculation! The core problem with your current code is that next_business_dt simply adds 1 day without accounting for weekends, holidays, or even consecutive stretches of non-working days. Here's a robust approach that handles all these edge cases:

Step 1: Define Helper Functions #

First, let's create a helper function to check if a given date is a valid business day (non-weekend + non-holiday):

library(dplyr)
library(lubridate)

# Your holiday list
business_holidays <- c("01-01", "04-19", "04-22", "05-01", "05-08", "07-05", "07-06", "09-28", "10-28", "11-17", "12-24", "12-25", "12-26")

# Helper function: Check if a date is a business day
is_business_day <- function(date) {
  # Rule 1: Exclude weekends (wday() returns 1=Sunday, 7=Saturday)
  if (wday(date) %in% c(1, 7)) {
    return(FALSE)
  }
  # Rule 2: Exclude non-weekend holidays
  if (format(date, "%m-%d") %in% business_holidays) {
    return(FALSE)
  }
  # If neither, it's a business day
  return(TRUE)
}

Next, create the main function to find the next business day starting from the day after your input date:

# Main function: Get the next nearest business day
get_next_business_day <- function(start_date) {
  next_day <- start_date + days(1)
  
  # Keep incrementing the day until we hit a valid business day
  while (!is_business_day(next_day)) {
    next_day <- next_day + days(1)
  }
  
  return(next_day)
}

Step 2: Apply to Your Dataset #

Now you can integrate this into your existing dataframe to calculate next_business_dt correctly:

df <- data.frame( date = seq(as.Date("2019-01-01"), as.Date("2019-12-31"), "days") ) %>%
  mutate(
    weekend = if_else(wday(date) %in% c(7, 1), 1, 0),
    business_holiday = if_else(format(date, "%m-%d") %in% business_holidays & weekend == 0, 1, 0),
    # Use our function to get the correct next business day
    next_business_dt = purrr::map(date, get_next_business_day) %>% as.Date()
  )

Step 3: Test Edge Cases #

Let's verify this works for your examples and tricky scenarios:

• For 2019-07-09: get_next_business_day(as.Date("2019-07-09")) returns 2019-07-10 (correct, since it's a weekday and not a holiday)
• For 2019-07-04: The next day is 2019-07-05 (holiday), followed by 2019-07-06 (another holiday) — the function skips both and returns 2019-07-08 (the next valid Monday)
• For 2019-12-23: The function skips 2019-12-24, 2019-12-25, 2019-12-26 (all holidays) and returns 2019-12-27 (the next Friday)

This approach handles any number of consecutive non-working days, whether they're weekends, holidays, or a mix of both.

内容的提问来源于stack exchange，提问作者user4394417