两次观察之间的平均秒数

Question

我有一个来自xts对象的不规则时间索引.我需要找到两次观察之间的平均秒数.这是我的样本数据:

dput(tt)
structure(c(1371.25, NA, 1373.95, NA, NA, 1373, NA, 1373.95, 
1373.9, NA, NA, 1374, 1374.15, NA, 1374, 1373.85, 1372.55, 1374.05, 
1374.15, 1374.75, NA, NA, 1375.9, 1374.05, NA, NA, NA, NA, NA, 
NA, NA, 1375, NA, NA, NA, NA, NA, 1376.35, NA, NA, NA, NA, NA, 
NA, NA, NA, NA, NA, 1376.25, NA, 1378, 1376.5, NA, NA, NA, 1378, 
1378, NA, NA, 1378.8, 231.9, 231.85, NA, 231.9, 231.85, 231.9, 
231.8, 231.9, 232.6, 231.95, 232.35, 232, 232.1, 232.05, 232.05, 
232.05, 231.5, 231.3, NA, NA, 231.1, 231.1, 231.1, 231, 231, 
230.95, 230.6, 230.6, 230.7, 230.6, 231, NA, 231, 231, 231.45, 
231.65, 231.4, 231.7, 231.3, 231.25, 231.25, 231.4, 231.4, 231.85, 
231.75, 231.5, 231.55, 231.35, NA, 231.5, 231.5, NA, 231.5, 231.25, 
231.15, 231, 231, 231, 231.05, NA), .Dim = c(60L, 2L), .indexCLASS = c("POSIXct", 
"POSIXt"), tclass = c("POSIXct", "POSIXt"), .indexTZ = "Asia/Calcutta", tzone = "Asia/Calcutta", index = structure(c(1459482299, 
1459482301, 1459482302, 1459482303, 1459482304, 1459482305, 1459482306, 
1459482307, 1459482309, 1459482310, 1459482311, 1459482312, 1459482314, 
1459482315, 1459482316, 1459482317, 1459482318, 1459482319, 1459482320, 
1459482321, 1459482322, 1459482323, 1459482324, 1459482326, 1459482328, 
1459482329, 1459482330, 1459482331, 1459482332, 1459482336, 1459482337, 
1459482338, 1459482339, 1459482342, 1459482344, 1459482346, 1459482347, 
1459482348, 1459482349, 1459482590, 1459482591, 1459482594, 1459482595, 
1459482596, 1459482597, 1459482598, 1459482599, 1459482602, 1459482603, 
1459482604, 1459482609, 1459482610, 1459482611, 1459482612, 1459482613, 
1459482618, 1459482619, 1459482620, 1459482622, 1459482628), tzone = "Asia/Calcutta", tclass = c("POSIXct", 
"POSIXt")), .Dimnames = list(NULL, c("A", "B")), class = c("xts", 
"zoo"))

这是我的尝试:

difftime(index(tt),index(lag.xts(tt, k=1)), units=c("auto"))
Time differences in secs
 [1] 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
attr(,"tclass")
[1] "POSIXct" "POSIXt"

任何帮助都非常感谢.

编辑:

根据答案,我制作了以下代码.该代码旨在计算每天A和B的平均秒数.

但是代码采用tt而不是A或B的索引,因此A和B的结果是相同的.

fun.time= function(x) mean(diff(time(x)))
df.time<-do.call(rbind, lapply(split(tt, "days"), FUN=function (x) {do.call(cbind, lapply(as.list(x), fun.time))})) 


dput(df.time)
structure(c(5.57627118644068, 5.57627118644068), .Dim = 1:2, .Dimnames = list(
    NULL, c("A", "B")))

Answer 1

首先创建一些超过一天的测试数据.使用tt我们创建的问题tt2.

剩下的代码在删除该列的NA值后按列和按天计算连续时间的差异.我们lapply在列和列内我们aggregate按日期.为此,我们删除当前列的NA,然后构造一个zoo对象,其值为数字秒,其索引是格式化的时间.数值是这样的,我们可以避免处理difftime对象及其不可预测的单位,同时format导致as.Date相对于tzone数据属性的日期; 否则,as.Date将取相对于GMT的日期而不是所需的日期(基于下面海报的评论).这将返回一个zoo对象,但as.xts如果xts输出很重要,则很容易应用于结果.

library(xts)

# test input
tt2 <- tt
time(tt2) <- time(tt) + seq(1, 24*60*60, length = 60)

do.call(cbind, lapply(tt2, function(x) {
  times <- time(na.omit(x))
  aggregate(zoo(as.numeric(times), format(times)), as.Date, function(x) mean(diff(x)))
}))

给出以下动物园系列:

                A      B
2016-04-01 3029.0 1648.9
2016-04-02 5416.1 1633.0

注1:如果我们只按列而不是按日期需要平均差异,那么我们可以将其aggregate简化为下面的单行代码.在这里,我们将恢复使用tt作为测试输入,因为这足以说明这种情况.请注意,我们再次将索引转换为数字,以避免difftime输出单位的不可预测性.

sapply(tt, function(x) mean(diff(as.numeric(time(na.omit(x))))))

给这个命名的数字向量:

      A       B 
14.9545  6.2115 

注2: 随着动物园的开发版本,这可以简化.在该版本的动物园中存在一个coredata参数aggregate.zoo,如果设置为FALSE整个动物园对象将被发送到该函数而不仅仅是该coredata部分.在下面的代码中,我们定义函数以获取NA删除后的平均差异并将索引转换为字符,然后将Date转换为具有使用tzone输入属性为其时区(或本地时区)的效果.然后我们aggregate.zoo按日期对每列进行应用,cbind并将结果列表重新组合在一起:

library(xts)

mean_diff_time <- function(x) mean(diff(as.numeric(time(na.omit(x)))))
dates <- function(x) as.Date(format(x))

do.call("cbind", lapply(as.zoo(tt2), aggregate, dates, mean_diff_time, coredata = FALSE))

更新:重新安排了演示文稿.