DayMatchQ
更多信息和选项
- 可以使用下列格式指定日期:
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DateObject[…] 日期对象 {y,m,d} 年,月,日 {y,m} 指定月份的第一天 {y} y 年的 1 月1 日 "string" 作为字符串给出的日期("Jan. 1, 2000") {y,m,d,h,m,s} 精确的时间 time 绝对时间 - 可用的日期类型有:
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All 任何日历中的日期(默认) "Weekday" 被视为工作日的日期 "Weekend" 被视为周末的日期 "Holiday" 由 HolidayCalendar 规定的假期 "NonHoliday" 不是假期的日期 "BusinessDay" 非假期工作日 "NonBusinessDay" 假期或周末 "WeekdayHoliday" 发生在工作日的假期 "WeekendHoliday" 发生在周末的假期 "WeekendNonHoliday" 非假期的周末 Monday, Tuesday, … 一周的某一天 "EndOfMonth" 给定月份的最后一天 "BeginningOfMonth" 给定月份的第一天 - 在犹太历和伊斯兰历中,工作日是 Sunday 到 Thursday.
- DayMatchQ 使用下列选项:
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CalendarType Automatic 指定日期的日历 HolidayCalendar {"UnitedStates","Default"} 用于工作日和假期计算的假期安排和子指定
范例
打开所有单元 关闭所有单元基本范例 (5)
DayMatchQ[Today, "Weekend"]DayMatchQ[Today, "BusinessDay"]DayMatchQ[Yesterday, "Holiday"]DayMatchQ[DateObject[{2022, 7, 4}, "Day", "Gregorian", -6.], "Holiday", HolidayCalendar -> "Brazil"]DayMatchQ[DateObject[{2022, 7, 4}, "Day"], "Holiday", HolidayCalendar -> "UnitedStates"]范围 (3)
DayMatchQ["aug. 4, 1980", "Holiday"]Length[Select[DayRange[{2000}, {2050}, "Holiday"], DayMatchQ[#, Monday]&]] / DayCount[{2000}, {2050}, "Holiday"]//NDayMatchQ[Wednesday][DateObject[{2025, 4, 23}, "Day"]]DayMatchQ["WeekendHoliday"][DateObject[{2025, 1, 1}, "Day"]]选项 (3)
CalendarType (1)
HolidayCalendar (2)
DayMatchQ[{2026, 4, 6}, "Holiday"]DayMatchQ[{2026, 4, 6}, "Holiday", HolidayCalendar -> "HongKong"]CalendarData["UnitedStates"]DayMatchQ[DateObject[{2022, 4, 15}], "Holiday", CalendarType -> "Gregorian", HolidayCalendar -> {"UnitedStates", "NYSE"}]DayMatchQ[DateObject[{2022, 4, 15}], "Holiday", CalendarType -> "Gregorian", HolidayCalendar -> "UnitedStates"]应用 (1)
属性和关系 (2)
DayRange 的日期类型指定如果不是 All,则等价于 Select 陈述:
DayRange[{2022}, {2023}, "Weekend"] === Select[DayRange[{2022}, {2023}], DayMatchQ[#, "Weekend"]&]DayMatchQ[date,"BusinessDay"] 等价于 BusinessDayQ[date]:
DayMatchQ[Today, "BusinessDay"]BusinessDayQ[Today]可能存在的问题 (2)
DayMatchQ[{2022, 7, 30}, "Weekend"]DayMatchQ[{2022, 7, 30}, "Holiday", HolidayCalendar -> {"Gregorian" -> {{True&, # == 7&, # == 30&}}}]sundayGregorian = DateObject[{2022, 7, 31}]DayMatchQ[sundayGregorian, "Holiday", HolidayCalendar -> {"Gregorian" -> {{True&, # == 7&, # == 31&}}}]但是,在犹太历中,星期天不是周末的一部分,因此可以作为假日:
sundayJewish = CalendarConvert[sundayGregorian, "Jewish"]DayMatchQ[sundayJewish, "Holiday", HolidayCalendar -> {"Gregorian" -> {{True&, # == 7&, # == 31&}}}]相关指南
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- 日期和时间
文本
Wolfram Research (2012),DayMatchQ,Wolfram 语言函数,https://reference.wolfram.com/language/ref/DayMatchQ.html (更新于 2025 年).
CMS
Wolfram 语言. 2012. "DayMatchQ." Wolfram 语言与系统参考资料中心. Wolfram Research. 最新版本 2025. https://reference.wolfram.com/language/ref/DayMatchQ.html.
APA
Wolfram 语言. (2012). DayMatchQ. Wolfram 语言与系统参考资料中心. 追溯自 https://reference.wolfram.com/language/ref/DayMatchQ.html 年
BibTeX
@misc{reference.wolfram_2026_daymatchq, author="Wolfram Research", title="{DayMatchQ}", year="2025", howpublished="\url{https://reference.wolfram.com/language/ref/DayMatchQ.html}", note=[Accessed: 07-September-2026]}
BibLaTeX
@online{reference.wolfram_2026_daymatchq, organization={Wolfram Research}, title={DayMatchQ}, year={2025}, url={https://reference.wolfram.com/language/ref/DayMatchQ.html}, note=[Accessed: 07-September-2026]}