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。这是我的示例数据：

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"

非常感谢任何帮助。

编辑：

根据答案，我制作了以下代码。 该代码旨在计算 A 和 B 每天的平均秒数。

但代码采用 tt 的索引而不是 A 或 B，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，tt2。我们使用 tt 在 tt 来自问题的地方形成它。定义两个函数：

• mean_diff_time 从其参数中删除 NA，然后将其转换为数值，取其平均差异。
• 将其参数转换为 Date 类的日期

最后使用 aggregate.zoo 我们按日期聚合它，将 mean_diff_time 应用于每个日期组。

library(xts)

# create test input tt2 having >1 day (tt is from question)
tt2 <- tt
time(tt2) <- time(tt) + seq(1, 24*60*60, length = 60)

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

aggregate(tt2, dates, mean_diff_time, coredata = FALSE)
##                       A        B
##     2016-04-01 3029.006 1648.939
##     2016-04-02 5416.096 1632.957

更新

根据新功能修改了答案。

Answer 2

为了补充 G.Grothendieck 的答案，您还可以使用 mean(diff(index(tt))) 将结果作为 difftime 对象返回：

> mean(diff(index(tt)))
Time difference of 5.576271 secs

或简单地mean(diff(.index(tt))) 以获取数字结果：

> mean(diff(.index(tt)))
[1] 5.576271

编辑：

> lapply(tt, function(x){mean(diff(.index(x[!is.na(x)])))})
$A
[1] 14.95455

$B
[1] 6.211538