r - R 从索引中搜索特定时间

Question

我使用一个xts对象。对象的索引如下。一年中的每一小时都有一个。

"2011-01-02 18:59:00 EST"
"2011-01-02 19:58:00 EST"
"2011-01-02 20:59:00 EST"

列中是与每个索引条目关联的值。我想要做的是计算18:59全年所有星期一的值的标准偏差。一年应该有 52 个值。

我可以使用该weekdays()功能搜索星期几，但我的问题是搜索时间，例如18:59:00或任何其他时间。

Answer 1

You can do this by using interaction to create a factor from the combination of weekdays and .indexhour, then use split to select the relevant observations from your xts object.

set.seed(21)
x <- .xts(rnorm(1e4), seq(1, by=60*60, length.out=1e4))
groups <- interaction(weekdays(index(x)), .indexhour(x))
output <- lapply(split(x, groups), function(x) c(count=length(x), sd=sd(x)))
output <- do.call(rbind, output)
head(output)
#            count        sd
# Friday.0      60 1.0301030
# Monday.0      59 0.9204670
# Saturday.0    60 0.9842125
# Sunday.0      60 0.9500347
# Thursday.0    60 0.9506620
# Tuesday.0     59 0.8972697

Answer 2

您可以使用.index*函数系列（不要忘记 'index' 前面的 '.'！）：

fxts[.indexmon(fxts)==0]   # its zero-based (!) and gives you all the January values
fxts[.indexmday(fxts)==1]  # beginning of month
fxts[.indexwday(SPY)==1]   # Mondays

require(quantmod)

> fxts
                value
2011-01-02 19:58:00     1
2011-01-02 20:59:00     2
2011-01-03 18:59:00     3
2011-01-09 19:58:00     4
2011-01-09 20:59:00     5
2011-01-10 18:59:00     6
2011-01-16 18:59:00     7
2011-01-16 19:58:00     8
2011-01-16 20:59:00     9`

fxts[.indexwday(fxts)==1]  #this gives you all the Mondays

用于子集您使用的时间

fxts["T19:30/T20:00"] # this will give you the time period you are looking for

在这里你结合了工作日和时间段

fxts["T18:30/T20:00"] & fxts[.indexwday(fxts)==1] # to get a logical vector or
fxts["T18:30/T21:00"][.indexwday(fxts["T18:30/T21:00"])==1] # to get the values

>                   value
2011-01-03 18:58:00     3
2011-01-10 18:59:00     6