Question

我有一个非常大的数据集，其中DateTime列包含POSIXct-Values。我需要根据DateTime列确定季节（冬季 - 夏季）。我已经创建了一个在小数据集上工作正常的函数，但是当我在大数据集上使用它时会崩溃。任何人都可以看到我的错误吗？

我创建了4个函数：

3个子功能，以便我可以进行逻辑比较和选择使用* apply functions
1确定季节的功能

以下是功能：

require(lubridate)

# function for logical comparison (to be used in *apply)
greaterOrEqual <- function(x,y){
  ifelse(x >= y,T,F)
}

# function for logical comparison (to be used in *apply)
less <- function(x,y){
  ifelse(x < y,T,F)
}

# function for logical comparison (to be used in *apply)
selFromLogic <- function(VecLogic,VecValue){
  VecValue[VecLogic]
}

# Main Function to determine the season
getTwoSeasons <- function(input.date) {
  Winter1Start <- as.POSIXct("2000-01-01 00:00:00", tz = "UTC")
  Winter1End <- as.POSIXct("2000-04-15 23:59:59", tz = "UTC")

  SummerStart <- Winter1End + 1
  SummerEnd <- as.POSIXct("2000-10-15 23:59:59", tz = "UTC")

  Winter2Start <- SummerEnd + 1
  Winter2End <- as.POSIXct("2000-12-31 00:00:00", tz = "UTC")

  year(input.date) <- year(Winter1Start)
  attr(input.date, "tzone") <- attr(Winter1Start, "tzone")

  SeasonStart <- c(Winter1Start,SummerStart,Winter2Start)
  SeasonsEnd <- c(Winter1End,SummerEnd,Winter2End)
  Season_names <- as.factor(c("WinterHalfYear","SummerHalfYear","WinterHalfYear"))

  Season_select <- sapply(SeasonStart, greaterOrEqual, x = input.date) & sapply(SeasonsEnd, less, x = input.date)
  Season_return <- apply(Season_select,MARGIN = 1,selFromLogic,VecValue = Season_names)

  return(Season_return)
}

这是一种测试功能的方法：

dates <- Sys.time() + seq(0,10000,10)
getTwoSeasons(dates)

我会感激任何帮助，这让我发疯了！

Answer 1

我将@Lars Arne Jordanger的优雅方法打包成一个功能：

getTwoSeasons <- function(input.date){
  numeric.date <- 100*month(input.date)+day(input.date)
  ## input Seasons upper limits in the form MMDD in the "break =" option:
  cuts <- base::cut(numeric.date, breaks = c(0,415,1015,1231)) 
  # rename the resulting groups (could've been done within cut(...levels=) if "Winter" wasn't double
  levels(cuts) <- c("Winter", "Summer","Winter")
  return(cuts)
}

对某些样本数据进行测试似乎工作正常：

getTwoSeasons(as.POSIXct("2016-01-01 12:00:00")+(0:365)*(60*60*24))

Answer 2

如果您对恢复四个季节感兴趣，可以使用以下代码：

library(lubridate)
getSeason <- function(input.date){
  numeric.date <- 100*month(input.date)+day(input.date)
  ## input Seasons upper limits in the form MMDD in the "break =" option:
  cuts <- base::cut(numeric.date, breaks = c(0,319,0620,0921,1220,1231)) 
  # rename the resulting groups (could've been done within cut(...levels=) if "Winter" wasn't double
  levels(cuts) <- c("Winter","Spring","Summer","Fall","Winter")
  return(cuts)
}

单元测试：

getSeason(as.POSIXct("2016-01-01 12:00:00")+(0:365)*(60*60*24))

Answer 3

为了完整起见，值得注意的是lubridate现在有四分之一（和一个学期）的功能。 quarter将年份分为四分之一，将semester分成两半：

quarter(x, with_year = FALSE, fiscal_start = 1)
semester(x, with_year = FALSE)

有关详情，请参阅：https://www.rdocumentation.org/packages/lubridate/versions/1.7.4/topics/quarter

Answer 4

经过几个小时的调试后，我发现了自己的错误，实在是太荒谬了：

如果找不到DateTimeValue的季节，apply返回list - 对象而不是vector（当DateTime值等于2000-12-31 00:00:00时就是这种情况）。返回列表创建了计算时间和所描述的崩溃的过度增加。这是固定代码：

# input date and return 2 season
getTwoSeasons <- function(input.date) {
  Winter1Start <- as.POSIXct("2000-01-01 00:00:00", tz = "UTC")
  Winter1End <- as.POSIXct("2000-04-15 23:59:59", tz = "UTC")

  SummerStart <- Winter1End + 1
  SummerEnd <- as.POSIXct("2000-10-15 23:59:59", tz = "UTC")

  Winter2Start <- SummerEnd + 1
  Winter2End <- as.POSIXct("2001-01-01 00:00:01", tz = "UTC")

  SeasonStart <- c(Winter1Start,SummerStart,Winter2Start)
  SeasonsEnd <- c(Winter1End,SummerEnd,Winter2End)
  Season_names <- factor(c("WinterHalf","SummerHalf","WinterHalf"))

  year(input.date) <- year(Winter1Start)
  attr(input.date, "tzone") <- attr(Winter1Start, "tzone")

  Season_selectStart <- vapply(X = SeasonStart,function(x,y){x <= input.date},FUN.VALUE = logical(length(input.date)),y = input.date)
  Season_selectEnd   <- vapply(X = SeasonsEnd,function(x,y){x > input.date},FUN.VALUE = logical(length(input.date)),y = input.date)
  Season_selectBoth  <- Season_selectStart & Season_selectEnd
  Season_return <- apply(Season_selectBoth,MARGIN = 1,function(x,y){y[x]}, y = Season_names)
  return(Season_return)
}

“sub”功能现已集成在main函数中，两个sapply函数已替换为vapply。

PS：时区仍然存在问题，因为c（）将时区剥离。我修复它时会更新代码。

Answer 5

也可以使用以下策略：基本观察是 substr可以提取我们需要的月和日信息决定是夏天还是冬天。然后想法将其转换为 month.date 形式的数字，以及夏天的测试归结为数字大于4.15但小于10.16。

下面的示例显示了如何在日期向量中完成此操作首先转换为所描述的替代演示上面，然后是一个向量，告诉它是夏天“真”或冬天将基于此创建“FALSE”。

DateTime <- as.POSIXct(x  = "2000-01-01 00:00:00",
                       tz = "UTC") +
    (0:1000)*(60*60*24)

DateTime_2 <- as.numeric(paste(
    substr(x = DateTime,
           start = 6,
           stop = 7),
    substr(x = DateTime,
           start = 9,
           stop = 10),
    sep = "."))

.season <- (DateTime_2 > 4.15) & (DateTime_2 < 10.16)

Answer 6

使用POSXlt代替POSXct。

我根据自己所用季节的定义制定了自己的功能。我为非s年创建了名为 normal 的矢量，为leap年创建了名为 leap 的矢量，每个季节名称均重复编号。从1月1日开始出现。并创建了以下功能。

SEASON <- function(datee){
  
  datee <- as.POSIXlt(datee)
  season <- vector()
  normal <- rep(c("Winter","Spring","Summer","Monsoon","Autumn","Winter"), c(46,44,91,77,76,31))
  leap <- rep(c("Winter","Spring","Summer","Monsoon","Autumn","Winter"), c(46,45,91,77,76,31))

  
  if(leap_year(year(datee)) == FALSE){
    season <- normal[datee$yday+1]
  } else {
    season <- leap[datee$yday+1]
  }
  return(season)
}

让我们测试一下某些数据集。

Dates <- seq(as.POSIXct("2000-01-01"), as.POSIXct("2010-01-01"), by= "day")
sapply(Dates, SEASON)

有效。

使用R

6 个答案: