将小时数据汇总成日常汇总 [英] Aggregating hourly data into daily aggregates

问题描述

我有一个小时的天气数据，格式如下：

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

我需要的是每日总计的最大值， min，ave这样：

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？ >

How to do this in R?

推荐答案

1）可以使用动物园紧凑地完成：

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)

这表示：

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

2）这里是一个只使用核心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

编辑：（1）由于这首先出现 text = 参数为 read.zoo 已添加到动物园包中。
（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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