R将字典迭代为多个值的函数

Question

我正在尝试创建一个循环（或最有效的方法）来迭代 R 中的一系列日历（或 Python！）到 post 所有假期（理想情况下是所有工作日，但似乎我可以为此设计两个部分——因为我希望标记周末）。目标是拥有一个如下所示的数据框：

Country | ISO Code (if available) | Dates
United States of America| US| 12.24.2020
United States of America| US| 12.25.2020
United States of America| US| 01.01.2021
United Kingdom| UK| 12.24.2020
United Kingdom| UK| 12.25.2020
United Kingdom| UK| 01.01.2021

我目前拥有的：

    require("lattice")
    require("reticulate")
    require("RcppQuantuccia")
    require("tidyverse")
    require("tidytable")

fun_Holidays <- function(cal) {
    setCalendar(cal)
    getHolidays(as.Date("2019-01-01"), as.Date("2030-12-31"))
}
cal_dic <- data.table(calendar=calendars)
as.list(cal_dic)

cal_dic 是 RcppQuantuccia 上所有可用日历的列表，但如果我运行:

fun_Holidays(cal_dic)

我得到的只是错误（因为它不是迭代的）：

ERROR: Error in setCalendar(cal): Expecting a single string value: [type=list; extent=1].

我也在 Python 中使用 Holidays 包对此进行了尝试，并取得了进一步的进展，但 ISO 代码没有正确附加：

all_holidays = []
country_list = ['Angola','Argentina','Aruba','Australia','Austria','Bangladesh','Belarus','Belgium','Botswana','Brazil',
                'Bulgaria','Burundi','Canada','Chile','China','Colombia','Croatia','Curacao','Czechia','Denmark','Djibouti','DominicanRepublic',
                'Egypt','England','Estonia','Finland','France','Georgia','Germany','Greece','Honduras','HongKong','Hungary','Iceland','India','Ireland','IsleOfMan',
                'Israel','Italy','Jamaica','Japan','Kenya','Korea','Latvia','Lesotho','Lithuania','Luxembourg','Malaysia','Malawi','Mexico','Morocco','Mozambique','Netherlands',
                'Namibia','NewZealand','Nicaragua','Nigeria','NorthernIreland','Norway','Paraguay','Peru','Poland','Portugal','PortugalExt','Romania','Russia','SaudiArabia','Scotland',
                'Serbia','Singapore','Slovakia','Slovenia','SouthAfrica','Spain','Swaziland','Sweden','Switzerland','Turkey','Ukraine','UnitedArabEmirates','UnitedKingdom',
                'UnitedStates','Venezuela','Vietnam','Wales','Zambia','Zimbabwe']
    

for country in country_list:
    for holiday in holidays.CountryHoliday(country, years = np.arange(2018,2030,1)).items():
        all_holidays.append({'date' : holiday[0], 'holiday' : holiday[1], 'country': country, 'code': code})
all_holidays = pd.DataFrame(all_holidays)
all_holidays

    date    holiday country code
0   2018-09-17  Dia do Herói Nacional   Angola  NZ
1   2018-01-01  Ano novo    Angola  NZ
2   2018-03-30  Sexta-feira Santa   Angola  NZ
3   2018-02-13  Carnaval    Angola  NZ
4   2018-02-04  Dia do Início da Luta Armada    Angola  NZ
... ... ... ... ...
14386   2029-08-15  Zimbabwe Heroes' Day    Zimbabwe    NZ
14387   2029-08-13  Defense Forces Day  Zimbabwe    NZ
14388   2029-12-22  Unity Day   Zimbabwe    NZ
14389   2029-12-25  Christmas Day   Zimbabwe    NZ
14390   2029-12-26  Boxing Day  Zimbabwe    NZ
14391 rows × 4 columns

我觉得很奇怪，在 csv 中没有按日期按国家/地区列出的假期主列表或类似的列表来帮助处理时间序列 - 但也许只有我这样！ :)

谢谢！

编辑：我也一直在看：https://workalendar.github.io/workalendar/

因为它有最大的国家列表，但它比假期更难处理 - 但如果有人有解决方案从 workaldendar 中获取“主日历”，那将是惊人的！

Answer 1

使用 lapply 获取 calendars 中每个值的日期列表。

library(RcppQuantuccia)

fun_Holidays <- function(cal) {
  setCalendar(cal)
  getHolidays(as.Date("2019-01-01"), as.Date("2030-12-31"))
}


lapply(calendars, fun_Holidays)

要创建包含国家名称和日期的单个数据框，您可以使用 -

do.call(rbind, lapply(calendars, function(x) {
  dates <- fun_Holidays(x)
  if(length(dates))
    data.frame(country = x, dates)
})) -> result

head(result)

#  country      dates
#1  TARGET 2019-01-01
#2  TARGET 2019-04-19
#3  TARGET 2019-04-22
#4  TARGET 2019-05-01
#5  TARGET 2019-12-25
#6  TARGET 2019-12-26

或者用 purrr -

purrr::map_df(calendars, function(x) {
  dates <- fun_Holidays(x)
  if(length(dates))
    data.frame(country = x, dates)
}) -> result

R将字典迭代为多个值的函数

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