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我有一个非常奇怪的问题......我正在使用to.weeklyandto.period函数将每日xts对象转换为每周数据。在大多数情况下，我将周末日期设为星期五（day.of.week函数将返回 5）（例如"2010-01-08", "2011-02-11"），但在少数情况下，我得到的不是星期五（星期六/星期日/星期四/等）

我已经尝试过to.weekly，to.period(x, period = 'weeks')并且都返回了同样的问题。

为什么会这样？有解决方法吗？

谢谢！！

[编辑：以下示例]

test.dates <- as.Date(c("2010-04-27","2010-04-28","2010-04-29","2010-04-30","2010-05-03","2010-05-04","2010-05-05","2010-05-06","2010-05-07","2010-05-10","2010-05-11","2010-05-12","2010-05-13","2010-05-14","2010-05-17","2010-05-18","2010-05-19","2010-05-20","2010-05-21","2010-05-22","2010-05-24","2010-05-25","2010-05-26","2010-05-27","2010-05-28","2010-06-01","2010-06-02","2010-06-03","2010-06-04"))

test.data <- rnorm(length(test.dates),mean=1,sd=2)

test.xts <- xts(x=test.data,order.by=test.dates)

#Function that takes in a vector of zoo/xts objects (e.g. "2010-01-08") and returns the day of the week for each
dayofweek <- function(x) {
 placeholder <- vector("list",length=length(x))
 names(placeholder) <- x

 for(i in 1:length(x)) {placeholder[[i]] <- month.day.year(x[i])}
 placeholder2 <- rep(NA,times=length(x))

 for(i in 1:length(x)) {placeholder2[i] <- day.of.week(placeholder[[i]][[1]],placeholder[[i]][[2]],placeholder[[i]][[3]])}
 return(placeholder2)}

这将返回不是星期五的日期：time(to.weekly(test.xts))[dayofweek(time(to.weekly(test.xts))) != 5]

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1 回答 1

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您的示例有两个问题：

  1. 您的dayofweek函数有点麻烦，而且结果可能不正确。
  2. 您的示例日期缺少一些日期，例如 05-23-2010。

这是您的代码的清理版本：

library(xts)
test.dates <- as.Date(c("2010-04-27","2010-04-28","2010-04-29","2010-04-30","2010-05-03","2010-05-04","2010-05-05","2010-05-06","2010-05-07","2010-05-10","2010-05-11","2010-05-12","2010-05-13","2010-05-14","2010-05-17","2010-05-18","2010-05-19","2010-05-20","2010-05-21","2010-05-22","2010-05-24","2010-05-25","2010-05-26","2010-05-27","2010-05-28","2010-06-01","2010-06-02","2010-06-03","2010-06-04"))
test.data <- rnorm(length(test.dates),mean=1,sd=2)
test.xts <- xts(x=test.data,order.by=test.dates)
test.weekly <- to.weekly(test.xts)

library(lubridate)
test.weekly[wday(test.weekly, label = TRUE, abbr = TRUE) != "Fri"]

这个函数的唯一结果是

           test.xts.Open test.xts.High test.xts.Low test.xts.Close
2010-05-22     -1.705749      1.273982    -2.084203      -1.502611

当然，问题是本周结束于05-23-2010，但时间序列中不存在该日期。因此，to.weekly使用下一个最接近的日期作为结束点，即05-22-2010. 这是你问题的根源。

这是一个更好的示例，它表明该功能没有问题to.weekly。

library(lubridate); library(xts)   
test.dates <- seq(as.Date("1900-01-01"),as.Date("2011-10-01"),by='days')
test.dates <- test.dates[wday(test.dates)!=1 & wday(test.dates)!=7] #Remove weekends
test.data <- rnorm(length(test.dates),mean=1,sd=2)
test.xts <- xts(x=test.data,order.by=test.dates)
test.weekly <- to.weekly(test.xts)
test.weekly[wday(test.weekly, label = TRUE, abbr = TRUE) != "Fri"]
于 2011-09-27T16:22:40.200 回答