将每小时数据聚合为每日聚合 [英] Aggregating hourly data into daily aggregates

问题描述

我有以下格式的每小时天气数据:

I have an hourly weather data in the following format:

Date,DBT
01/01/2000 01:00,30
01/01/2000 02:00,31
01/01/2000 03:00,33
...
...
12/31/2000 23:00,25

我需要的是每天的最大值、最小值、平均值，如下所示:

What I need is a daily aggregate of max, min, ave like this:

Date,MaxDBT,MinDBT,AveDBT
01/01/2000,36,23,28
01/02/2000,34,22,29
01/03/2000,32,25,30
...
...
12/31/2000,35,9,20

如何在 R 中做到这一点?

How to do this in R?

推荐答案

1) 这可以使用 zoo 紧凑地完成:

1) This can be done compactly using zoo:

L <- "Date,DBT
01/01/2000 01:00,30
01/01/2000 02:00,31
01/01/2000 03:00,33
12/31/2000 23:00,25"

library(zoo)
stat <- function(x) c(min = min(x), max = max(x), mean = mean(x))
z <- read.zoo(text = L, header = TRUE, sep = ",", format = "%m/%d/%Y", aggregate = stat)

这给出:

> z
           min max     mean
2000-01-01  30  33 31.33333
2000-12-31  25  25 25.00000

2) 这是一个仅使用核心 R 的解决方案:

2) here is a solution that only uses core R:

DF <- read.csv(text = L)
DF$Date <- as.Date(DF$Date, "%m/%d/%Y")
ag <- aggregate(DBT ~ Date, DF, stat) # same stat as in zoo solution 

最后一行给出:

> ag
        Date  DBT.min  DBT.max DBT.mean
1 2000-01-01 30.00000 33.00000 31.33333
2 2000-12-31 25.00000 25.00000 25.00000

(1)自从首次出现以来，read.zoo 的text= 参数被添加到zoo 包中.(2) 小改进.

(1) Since this first appeared the text= argument to read.zoo was added in the zoo package. (2) minor improvements.

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