两次时间观测之间的平均秒数 [英] Average number of seconds between two time observations

问题描述

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:

I have a irregular time index from an xts object. I need to find the average number of seconds between two time observations. This is the my sample data:

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.

根据答案，我制作了以下代码.该代码用于计算 A 和 B 每天的平均秒数.

Based on the answers, I have made the following code. The code is meant to calculate mean number of seconds for A and B every day.

但是代码取的是tt的索引而不是A或B，所以A和B的结果是一样的.

But the code takes the index of tt instead of A or B and so the results of A and B is same.

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")))

推荐答案

首先创建一些超过一天的测试数据，tt2.我们使用 tt 来形成它，其中 tt 来自问题.定义两个函数:

First create some test data that has more than one day, tt2. We use tt to form it where tt is from the question. Define two functions:

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

最后使用aggregate.zoo，我们按日期对每个日期组应用mean_diff_time进行聚合.

Finally using aggregate.zoo we aggregate it by date applying mean_diff_time to each date group.

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.

Update

Have revised answer in light of new features.

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