Question

我正在使用一堆气候模型输出（CMIP5 models，具体而言）。这些是时间戳温度，风等的网络。

他们都使用UTC中的days since YYYY-mm-dd 00:00:00约定。我一直在使用lubridate转换为更简单的日期（而非日期时间）对象：

library(tidyverse)
input$date.utc =
  ymd_hms('0001-01-01 00:00:00', tz = 'UTC') +
  days(floor(input$time))

我遇到了两个问题。一个是每个模型都有不同的时代。这很容易修复。另一个更难的问题是并非所有模型都使用格里高利历。有些人使用365天的变化，没有闰年。

我认为没有办法在lubridate函数中指定非公历日历。这可能吗？

Answer 1

我无法在Customer c = Customer.retrieve(this.customer); Map<String, Object> params = new HashMap<String, Object>(); params.put("source", above_token); Card card = customer.getSources().create(params);中找到此功能，因此我编写了一个函数来至少计算日期向量和给定时期的每个元素之间的闰天数：

lubridate

有了这个，我可以通过添加缺少的闰天数将365天日历转换为格里高利：

# count_leap_days: returns an integer for the number of leap days between a
# supplied vector of dates and a fixed epoch
count_leap_days <- function(dates, epoch = ymd('1850-01-01'), proleptic = FALSE)
{
  require(lubridate)

  # check input
  if (!is(epoch, 'Date') | !is(dates, 'Date'))
  {
    stop('count_leap_days: both arguments must be Date objects.')
  }
  if (any(dates <= epoch))
  {
    stop('count_leap_days: dates should all be later than epoch.')
  }
  if (proleptic = FALSE & epoch < ymd('1582-10-15'))
  {
    message('count_leap_days: ',
      'no leap days before 1582-10-15 unless proleptic = TRUE.')
    epoch = ymd('1582-10-15')
  }

  # get the year range
  # exclude start (end) years if they begin after (start before) feb 29
  y.epoch = year(epoch) +
    ifelse(epoch >= ymd(paste0(year(epoch), '-03-01')), 1, 0)
  y.dates = year(dates) -
    ifelse(dates <= ymd(paste0(year(dates), '-02-28')), 1, 0)
  span = y.dates - y.epoch + 1

  # a year is a leap year if it's:
  #   - divisble by 4. but
  #   - NOT if it's also divisible by 100, unless
  #   - it's also divisible by 400.
  # all years div. by 4 are also div. by 100, and
  # all years div. by 100 are also div. by 400.
  # hence, total days = (div. by 4) - (div. by 100) + (div. by 400)
  div4 = span %/% 4 +
    ifelse(
      (y.epoch %% 4) %in% c(0, (4 - (span %% 4) - 1):(4 - 1)) &
      (y.dates %% 4) %in% 0:(span %% 4 - 1),
    1, 0)
  div100 = span %/% 100 +
    ifelse(
      (y.epoch %% 100) %in% c(0, (100 - (span %% 100) - 1):(100 - 1)) &
      (y.dates %% 100) %in% 0:(span %% 100 - 1),
    1, 0)
  div400 = span %/% 400 +
    ifelse(
      (y.epoch %% 400) %in% c(0, (400 - (span %% 400) - 1):(400 - 1)) &
      (y.dates %% 400) %in% 0:(span %% 400 - 1),
    1, 0)
  return(div4 - div100 + div400) 
}

使用非Gregorian日历从时期转换

1 个答案: