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r - 将 15 分钟的时间序列数据汇总到每日

转载 作者:行者123 更新时间:2023-12-04 10:59:51 26 4
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这是我的文本文件中的数据:(我已经显示了 10,000 行中的 10 行)Index 是行名,temp 是时间序列,m 是以毫米为单位的值。

     "Index" "temp" "m"
1 "2012-02-07 18:15:13" "4297"
2 "2012-02-07 18:30:04" "4296"
3 "2012-02-07 18:45:10" "4297"
4 "2012-02-07 19:00:01" "4297"
5 "2012-02-07 19:15:07" "4298"
6 "2012-02-07 19:30:13" "4299"
7 "2012-02-07 19:45:04" "4299"
8 "2012-02-07 20:00:10" "4299"
9 "2012-02-07 20:15:01" "4300"
10 "2012-02-07 20:30:07" "4301"

我使用这个在 r 中导入:

    x2=read.table("data.txt", header=TRUE)

我尝试使用以下代码将时间序列聚合到每日数据:

   c=aggregate(ts(x2[, 2], freq = 96), 1, mean)

我已将频率设置为 96,因为对于 15 分钟的数据,24 小时将包含在 96 个值中。

它返回给我:

    Time Series:
Start = 1
End = 5
Frequency = 1
[1] 5366.698 5325.115 5311.969 5288.542 5331.115

但我想要与原始数据相同的格式,即我还想要值旁边的时间序列。我需要帮助才能实现这一目标。

最佳答案

将数据转换为 xts 对象后,使用 xts 包中的 apply.daily:

像这样的东西应该可以工作:

x2 = read.table(header=TRUE, text='     "Index" "temp" "m"
1 "2012-02-07 18:15:13" "4297"
2 "2012-02-07 18:30:04" "4296"
3 "2012-02-07 18:45:10" "4297"
4 "2012-02-07 19:00:01" "4297"
5 "2012-02-07 19:15:07" "4298"
6 "2012-02-07 19:30:13" "4299"
7 "2012-02-07 19:45:04" "4299"
8 "2012-02-07 20:00:10" "4299"
9 "2012-02-07 20:15:01" "4300"
10 "2012-02-07 20:30:07" "4301"')

x2$temp = as.POSIXct(strptime(x2$temp, "%Y-%m-%d %H:%M:%S"))
require(xts)
x2 = xts(x = x2$m, order.by = x2$temp)
apply.daily(x2, mean)
## [,1]
## 2012-02-07 20:30:07 4298.3

更新:您的问题以可重现的格式出现(带有假数据)

我们并不总是需要实际的数据集来帮助解决问题....

set.seed(1) # So you can get the same numbers as I do
x = data.frame(datetime = seq(ISOdatetime(1970, 1, 1, 0, 0, 0),
length = 384, by = 900),
m = sample(2000:4000, 384, replace = TRUE))
head(x)
# datetime m
# 1 1970-01-01 00:00:00 2531
# 2 1970-01-01 00:15:00 2744
# 3 1970-01-01 00:30:00 3146
# 4 1970-01-01 00:45:00 3817
# 5 1970-01-01 01:00:00 2403
# 6 1970-01-01 01:15:00 3797
require(xts)
x2 = xts(x$m, x$datetime)
head(x2)
# [,1]
# 1970-01-01 00:00:00 2531
# 1970-01-01 00:15:00 2744
# 1970-01-01 00:30:00 3146
# 1970-01-01 00:45:00 3817
# 1970-01-01 01:00:00 2403
# 1970-01-01 01:15:00 3797
apply.daily(x2, mean)
# [,1]
# 1970-01-01 23:45:00 3031.302
# 1970-01-02 23:45:00 3043.250
# 1970-01-03 23:45:00 2896.771
# 1970-01-04 23:45:00 2996.479

更新 2:替代方案

(使用我在上述更新中提供的虚假数据。)

data.frame(time = x[seq(96, nrow(x), by=96), 1],
mean = aggregate(ts(x[, 2], freq = 96), 1, mean))
# time mean
# 1 1970-01-01 23:45 3031.302
# 2 1970-01-02 23:45 3043.250
# 3 1970-01-03 23:45 2896.771
# 4 1970-01-04 23:45 2996.479

关于r - 将 15 分钟的时间序列数据汇总到每日,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/11184085/

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