为什么我的自定义函数在改变 tibble 时不起作用？

Question

我创建了一个自定义函数来将工作日添加到日期。该函数依赖于以下包：

library(tidyverse)
library(lubridate)
library(tidyquant)

这是我创建的函数：

add_workingdays <- function(start_date, number_of_days, switch_count_weekendsholidays = TRUE,  remove_weekends = TRUE, holidays = NULL){
  start_date <- start_date %>% as.Date()
  if (!is.Date(start_date)) stop("add_workingdays(): start_date must be a date.", call. = FALSE)
  target_date <- start_date + number_of_days
  if(switch_count_weekendsholidays){
    target_date_lenght <- tidyquant::WORKDAY_SEQUENCE(start_date, target_date, remove_weekends, holidays = holidays) %>% length()
    while(target_date_lenght != number_of_days) {
      target_date <- target_date + 1
      target_date_lenght <- tidyquant::WORKDAY_SEQUENCE(start_date, target_date, remove_weekends, holidays = holidays) %>% length()
    }
  }
  target_date %>% return()
}

当我在以下场景中运行该函数时，它可以正常工作。

add_workingdays(start_date = '2022-04-08' %>% as.Date(), number_of_days = 5)
[1] "2022-04-14"
'2022-04-08' %>% as.Date() %>% add_workingdays(number_of_days = 5)
[1] "2022-04-14"

但是当我尝试在 tibble 的 mutate 函数中使用它时，我收到了我不理解的错误消息。

我使用以下代码，最后给出错误：

tibble(
+   dates = rep('2022-04-08' %>% as.Date()), #) seq.Date(from = '2022-04-08' %>% as.Date(), by = 'days', length.out = 5),
+   days_to_add = rep(10:5)
+ ) %>% 
+   print() %>% 
+   mutate(
+     target_date = add_workingdays(start_date = dates, number_of_days = days_to_add)
+   )
# A tibble: 6 x 2
  dates      days_to_add
  <date>           <int>
1 2022-04-08          10
2 2022-04-08           9
3 2022-04-08           8
4 2022-04-08           7
5 2022-04-08           6
6 2022-04-08           5
Error in `mutate()`:
! Problem while computing `target_date =
  add_workingdays(start_date = dates, number_of_days =
  days_to_add)`.
Caused by error in `seq.Date()`:
! 'from' must be of length 1
Run `rlang::last_error()` to see where the error occurred.

任何人都可以向我解释在 mutate 函数中使用此自定义函数时我做错了什么吗？

Answer 1

该错误与 mutate 无关。它来自于 tidyquant::WORKDAY_SEQUENCE 函数中的参数输入不正确。如果您通过将其打印到控制台来检查此函数，您会得到以下信息：

tidyquant::WORKDAY_SEQUENCE #with no parens

#function (start_date, end_date, remove_weekends = TRUE, holidays = NULL) 
#{
#    day_sequence <- DATE_SEQUENCE(start_date, end_date, by = "day")
#    ret_tbl <- tibble::tibble(day_sequence = day_sequence) %>% 
#    ...#and more

这表明此函数使用 DATE_SEQUENCE 函数。同样，如果您检查此函数，您会发现它使用 seq.Date 函数，该函数需要 from 参数为单个日期。例如：

seq.Date(from = as.Date("2020-01-01"), to = as.Date("2020-01-03"), by = 'days')
#[1] "2020-01-01" "2020-01-02" "2020-01-03"
WORKDAY_SEQUENCE("2020-01-01", "2020-01-03")
#[1] "2020-01-01" "2020-01-02" "2020-01-03"

如果您向 from 输入多个日期数据，您会收到相同的错误：

seq.Date(from = c(as.Date("2020-01-01"), as.Date("2020-01-02")), to = as.Date("2020-01-03"), by = 'days')
#Error in seq.Date(c(as.Date("2020-01-01"), as.Date("2020-01-02")), #as.Date("2020-01-03"),  : 
#  'from' must be of length 1

WORKDAY_SEQUENCE(c("2020-01-01", "2020-01-02"), "2020-01-03")
#Error in seq.Date(from = AS_DATE(start_date), to = AS_DATE(end_date),  : 
#  'from' must be of length 1

为避免此错误，当您想要输入 start_date 的列时，您必须对 WORKDAY_SEQUENCE 函数进行矢量化，这意味着您将该函数应用于列中的每个日期。正如@danloo所示，您可以使用Vectorize。您还可以使用*apply 系列。

情况 1：start_date 为多个日期，end_date 为单个日期。
在这种情况下，您可以使用 lapply 或 sapply。

lapply(c("2020-01-01", "2020-01-02"), WORKDAY_SEQUENCE, end_date = "2020-01-03")
#[[1]]
#[1] "2020-01-01" "2020-01-02" "2020-01-03"

#[[2]]
#[1] "2020-01-02" "2020-01-03"

您还需要对 length() 函数进行矢量化。由于 lapply 返回一个列表，因此如果您需要返回一个向量以便可以使用它在数据框中创建列，则可以使用 unlist()：

c("2020-01-01", "2020-01-02") %>% 
      lapply(WORKDAY_SEQUENCE, end_date = "2020-01-03") %>% 
      lapply(length) %>% 
      unlist()
#[1] 3 2

因此，如果您想遵循上面的 lapply 方法，则应将函数中的相关行修改为

target_date_lenght <- tidyquant::WORKDAY_SEQUENCE(start_date,
                      target_date, remove_weekends, 
                      holidays = holidays) %>% 
                      length()

target_date_length <- start_date %>% 
                       lapply(tidyquant::WORKDAY_SEQUENCE, 
                              end_date = target_date,
                              remove_weekends = remove_weekends, 
                              holidays = holidays) %>% 
                       lapply(length) %>% 
                       unlist()

情况 2：start_date 都有多个日期和结束日期
在这种情况下，您可以使用 Map 或 mapply

target_date_length <- mapply(tidyquant::WORKDAY_SEQUENCE, 
                             start_date = start_date, 
                             end_date = end_date, 
                             remove_weekends = remove_weekends, 
                             holidays = holidays)%>% 
                      lapply(length) %>% 
                      unlist()

Answer 2

您的函数必须使用 Vectorize 进行矢量化，或者使用 purrr::map 或 lapply 单独应用于每个元素：

library(tidyverse)
library(lubridate)
#> 
#> Attaching package: 'lubridate'
#> The following objects are masked from 'package:base':
#> 
#>     date, intersect, setdiff, union
library(tidyquant)
#> Loading required package: PerformanceAnalytics
#> Loading required package: xts
#> Loading required package: zoo
#> 
#> Attaching package: 'zoo'
#> The following objects are masked from 'package:base':
#> 
#>     as.Date, as.Date.numeric
#> 
#> Attaching package: 'xts'
#> The following objects are masked from 'package:dplyr':
#> 
#>     first, last
#> 
#> Attaching package: 'PerformanceAnalytics'
#> The following object is masked from 'package:graphics':
#> 
#>     legend
#> Loading required package: quantmod
#> Loading required package: TTR
#> Registered S3 method overwritten by 'quantmod':
#>   method            from
#>   as.zoo.data.frame zoo
#> ══ Need to Learn tidyquant? ════════════════════════════════════════════════════
#> Business Science offers a 1-hour course - Learning Lab #9: Performance Analysis & Portfolio Optimization with tidyquant!
#> </> Learn more at: https://university.business-science.io/p/learning-labs-pro </>

add_workingdays <- function(start_date, number_of_days, switch_count_weekendsholidays = TRUE, remove_weekends = TRUE, holidays = NULL) {
  start_date <- start_date %>% as.Date()
  if (!is.Date(start_date)) stop("add_workingdays(): start_date must be a date.", call. = FALSE)
  target_date <- start_date + number_of_days
  if (switch_count_weekendsholidays) {
    target_date_lenght <- tidyquant::WORKDAY_SEQUENCE(start_date, target_date, remove_weekends, holidays = holidays) %>% length()
    while (target_date_lenght != number_of_days) {
      target_date <- target_date + 1
      target_date_lenght <- tidyquant::WORKDAY_SEQUENCE(start_date, target_date, remove_weekends, holidays = holidays) %>% length()
    }
  }
  target_date %>% return()
}

add_workingdays(start_date = "2022-04-08" %>% as.Date(), number_of_days = 5)
#> [1] "2022-04-14"


c("2022-04-08", "2022-04-09") %>%
  as.Date() %>%
  add_workingdays(number_of_days = 5)
#> Error in seq.Date(from = AS_DATE(start_date), to = AS_DATE(end_date), : 'from' must be of length 1

add_workingdays <- Vectorize(add_workingdays)
c("2022-04-08", "2022-04-09") %>%
  as.Date() %>%
  add_workingdays(number_of_days = 5)
#> [1] 19096 19097

tibble(
  dates = rep("2022-04-08" %>% as.Date()), # ) seq.Date(from = '2022-04-08' %>% as.Date(), by = 'days', length.out = 5),
  days_to_add = rep(10:5)
) %>%
  mutate(
    target_date = add_workingdays(start_date = dates, number_of_days = days_to_add)
  )
#> # A tibble: 6 × 3
#>   dates      days_to_add target_date
#>   <date>           <int>       <dbl>
#> 1 2022-04-08          10       19103
#> 2 2022-04-08           9       19102
#> 3 2022-04-08           8       19101
#> 4 2022-04-08           7       19100
#> 5 2022-04-08           6       19097
#> 6 2022-04-08           5       19096

^{创建于 2022 年- 04-13 由 reprex 包 (v2.0.0)}