如何在没有时间戳的情况下将小时值汇总为24小时平均值 [英] How to aggregate hourly values into 24h-average means without timestamp
问题描述
我有3个站的"mydata_hourly"(实际上更多)，并且它们在一年中的小时温度值.这使我一年可以进行8760次每小时测量.现在，我希望具有相同的结构，但(365)24h平均意味着'mydata_daily'.
I have 'mydata_hourly' with 3 station (actually more) and their hourly temperature values over one year. This gives me 8760 hourly measurements in one year. Now I want to have the same structure but with the (365) 24h-average means 'mydata_daily'.
我尝试了一些for循环操作，但没有成功.我听说过有关聚合函数的内容.我发现有时间戳的东西是我不幸的.
I have tried something with a for loop, this didn't work out. I have heard something about an aggregate function. I found something with a timestamp, what I don't have unfortunately.
.
my_data_hourly <- structure(c(8.29, 7.96, 8.14, 7.27, 7.37, 7.3, 7.23, 7.53,
7.98, 10.2, 12.39, 14.34, 14.87, 14.39, 12.54, 11.84, 10.3, 10.62,
10.65, 10.56, 10.43, 10.35, 9.85, 9.12, 8.95, 8.82, 8.92, 9.33,
9.44, 9.3, 9.15, 9.37, 9.54, 10.24, 12.13, 12.43, 12.65, 13,
13.18, 13.58, 13.64, 13.75, 13.85, 13.94, 13.79, 13.84, 13.94,
14.26, 24.93, 24.64, 23.67, 21.46, 21.33, 20.83, 21.12, 21.1,
23.75, 25.39, 30.72, 30.71, 30.81, 30.92, 32.61, 32.37, 32.49,
30.68, 30.23, 30.45, 28.1, 26.9, 25.09, 25.07, 24.59, 24.22,
23.05, 22.21, 22.07, 21.6, 21.24, 21.22, 21.85, 24.87, 28.85,
29.42, 30.82, 30.97, 31.32, 30.81, 30.83, 29.9, 30.01, 30.31,
30, 27.91, 25.78, 25.88, 8.78, 8.47, 8.49, 7.65, 8.63, 9.02,
9.02, 8.11, 7.63, 9.19, 11.25, 12.24, 13.62, 12.09, 10.6, 11.1,
10.16, 10.44, 9.58, 10.04, 10.01, 10.23, 9.51, 9.2, 9.34, 9.6,
9.4, 9.45, 9.36, 9.26, 9.3, 9.46, 9.58, 9.89, 10.6, 11.04, 12.1,
12.61, 13.12, 13.47, 13.55, 13.51, 13.63, 13.84, 13.93, 14.17,
13.97, 13.86), .Dim = c(48L, 3L), .Dimnames = list(NULL, c("station1",
"station2", "station3")))
.
hourly_measure Station1 Station2 Station3
[1,] 8.29 24.93 8.78
[2,] 7.96 24.64 8.47
[3,] 8.14 23.67 8.49
[4,] 7.27 21.46 7.65
[5,] 7.37 21.33 8.63
[6,] 7.30 20.83 9.02
[7,] 7.23 21.12 9.02
[8,] 7.53 21.10 8.11
[9,] 7.98 23.75 7.63
[10,] 10.20 25.39 9.19
[11,] 12.39 30.72 11.25
[12,] 14.34 30.71 12.24
[13,] 14.87 30.81 13.62
[14,] 14.39 30.92 12.09
[15,] 12.54 32.61 10.60
[16,] 11.84 32.37 11.10
[17,] 10.30 32.49 10.16
[18,] 10.62 30.68 10.44
[19,] 10.65 30.23 9.58
[20,] 10.56 30.45 10.04
[21,] 10.43 28.10 10.01
[22,] 10.35 26.90 10.23
[23,] 9.85 25.09 9.51
[24,] 9.12 25.07 9.20
[25,] 8.95 24.59 9.34
[26,] 8.82 24.22 9.60
[27,] 8.92 23.05 9.40
[28,] 9.33 22.21 9.45
[29,] 9.44 22.07 9.36
[30,] 9.30 21.60 9.26
[31,] 9.15 21.24 9.30
[32,] 9.37 21.22 9.46
[33,] 9.54 21.85 9.58
[34,] 10.24 24.87 9.89
[35,] 12.13 28.85 10.60
[36,] 12.43 29.42 11.04
[37,] 12.65 30.82 12.10
[38,] 13.00 30.97 12.61
[39,] 13.18 31.32 13.12
[40,] 13.58 30.81 13.47
[41,] 13.64 30.83 13.55
[42,] 13.75 29.90 13.51
[43,] 13.85 30.01 13.63
[44,] 13.94 30.31 13.84
[45,] 13.79 30.00 13.93
[46,] 13.84 27.91 14.17
[47,] 13.94 25.78 13.97
[48,] 14.26 25.88 13.86
因此，从理论上讲，我希望在my_data_daily [1,1]中包含mydata_hourly [1:24,1] 和mydata_daily [2,1]中的mydata_hourly [25:48,1]
So in theory I want to have mydata_hourly[1:24,1] in my_data_daily[1,1] and mydata_hourly[25:48,1] in mydata_daily[2,1]
推荐答案
这些是时间序列，最好为它们使用时间序列表示，这将有助于绘图和其他时间序列处理.
These are time series and it is probably best to use time series representations for them which will facilitate plotting and other time series processing.
I)ts 假设您的数据是最后注释中可重复显示的矩阵m.将其转换为频率为24的ts时间序列，然后将其汇总，如图所示.不使用任何软件包.
I) ts Suppose your data is the matrix m shown reproducibly in the Note at the end. Convert that to a ts time series with frequency 24 and then aggregate it as shown. No packages are used.
tt <- ts(m, frequency = 24)
aggregate(tt, 1, mean)
给予:
Time Series:
Start = 1
End = 2
Frequency = 1
Station1 Station2 Station3
1 10.06333 26.89042 9.794167
2 11.71000 25.40542 11.585000
2)zooreg :一种替代方法是使用zoo软件包创建zooreg对象.
2) zooreg An alternative is to create zooreg objects using the zoo package.
library(zoo)
z <- zooreg(m, frequency = 24)
aggregate(z, as.integer, mean)
给予:
Station1 Station2 Station3
1 10.06333 26.89042 9.794167
2 11.71000 25.40542 11.585000
注意
Lines <- "
Station1 Station2 Station3
8.29 24.93 8.78
7.96 24.64 8.47
8.14 23.67 8.49
7.27 21.46 7.65
7.37 21.33 8.63
7.30 20.83 9.02
7.23 21.12 9.02
7.53 21.10 8.11
7.98 23.75 7.63
10.20 25.39 9.19
12.39 30.72 11.25
14.34 30.71 12.24
14.87 30.81 13.62
14.39 30.92 12.09
12.54 32.61 10.60
11.84 32.37 11.10
10.30 32.49 10.16
10.62 30.68 10.44
10.65 30.23 9.58
10.56 30.45 10.04
10.43 28.10 10.01
10.35 26.90 10.23
9.85 25.09 9.51
9.12 25.07 9.20
8.95 24.59 9.34
8.82 24.22 9.60
8.92 23.05 9.40
9.33 22.21 9.45
9.44 22.07 9.36
9.30 21.60 9.26
9.15 21.24 9.30
9.37 21.22 9.46
9.54 21.85 9.58
10.24 24.87 9.89
12.13 28.85 10.60
12.43 29.42 11.04
12.65 30.82 12.10
13.00 0.97 12.61
13.18 31.32 13.12
13.58 30.81 13.47
13.64 30.83 13.55
13.75 29.90 13.51
13.85 30.01 13.63
13.94 30.31 13.84
13.79 30.00 13.93
13.84 27.91 14.17
13.94 25.78 13.97
14.26 25.88 13.86"
m <- as.matrix(read.table(text = Lines, header = TRUE))
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