使用“to.weekly”错误的周末结束日期“xts”中的函数包裹
我有一个非常奇怪的问题...我正在使用 to.weekly
和 to.period
函数将每日 xts
对象转换为每周数据。在大多数情况下,我得到的周末日期为星期五(day.of.week
函数将返回 5)(例如 "2010-01-08"
,< code>“2011-02-11”),但在某些情况下我会得到周五以外的时间(周六/周日/周四/等),
我尝试过 to.weekly
和to.period(x, period = 'weeks')
两者都返回相同的问题。
为什么会发生这种情况?有解决方法吗?
谢谢!!
[编辑:下面的示例]
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)}
这将返回非星期五的日期:time(to.weekly(test.xts))[dayofweek(time(to.weekly(test. xts))) != 5]
I have a really odd issue... I am using the to.weekly
and to.period
function to convert a daily xts
object to weekly data. In most instances, I get the week-ending date as a Friday (day.of.week
function will return 5) (e.g. "2010-01-08"
, "2011-02-11"
), but there are a few cases where I get something other than Friday (Saturday/Sunday/Thursday/etc.)
I have tried to.weekly
and to.period(x, period = 'weeks')
and both return the same problem.
Why is this happening? Is there a work-around for this??
Thanks!!
[EDIT: EXAMPLE BELOW]
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)}
This returns the date(s) that are not Friday: time(to.weekly(test.xts))[dayofweek(time(to.weekly(test.xts))) != 5]
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您的示例有两个问题:
dayofweek
函数有点麻烦,而且结果可能不正确。这是代码的清理版本:
该函数的唯一结果是
当然,问题是本周于
05-23-2010
结束,但该日期不存在于时间序列。因此,to.weekly
使用下一个最接近的日期作为结束点,即05-22-2010
。这就是你的问题的根源。这是一个更好的示例,它表明
to.weekly
函数没有问题。You have 2 problems with your example:
dayofweek
function is a bit cumbersome, and probably incorrect in its results.Here is a cleaned-up version of your code:
The only result of this function is
The problem of course, is that this week ends on
05-23-2010
, but that date is not present in the time series. Therefore,to.weekly
uses the next closest date as the end point, which is05-22-2010
. This is the source of your problem.Here is a better example, which reveals no issue with the
to.weekly
function.