Question

我在尝试按小时计算平均温度时遇到了麻烦。

我有一个日期，时间（hh：mm：ss p.m./a.m。）和温度的数据框。我需要的是按小时提取平均温度，以绘制每日温度变化。

我是R的新手，但尝试了解我所知道的：我首先尝试将小时数转换为数字，然后提取前两个字符，然后计算均值，但效果不佳。此外，我有很多文件可以分析说，与我找到的“解决方案”相比，拥有更加自动化和清洁的东西要好得多。

我认为在R中计算平均值必须是一个更好的方法，所以我一直在寻找其他帖子中的答案。不幸的是，我找不到从时间数据中提取统计数据的明确答案。

我的数据看起来像这样

          date     hour temperature
1   28/12/2013 13:03:01      41.572
2   28/12/2013 13:08:01      46.059
3   28/12/2013 13:13:01       48.55
4   28/12/2013 13:18:01      49.546
5   28/12/2013 13:23:01      49.546
6   28/12/2013 13:28:01      49.546
7   28/12/2013 13:33:01      50.044
8   28/12/2013 13:38:01      50.542
9   28/12/2013 13:43:01      50.542
10  28/12/2013 13:48:01       51.04
11  28/12/2013 13:53:01      51.538
12  28/12/2013 13:58:01      51.538
13  28/12/2013 14:03:01      50.542
14  28/12/2013 14:08:01       51.04
15  28/12/2013 14:13:01       51.04
16  28/12/2013 14:18:01      52.534
17  28/12/2013 14:23:01      53.031
18  28/12/2013 14:28:01      53.031
19  28/12/2013 14:33:01      53.031
20  28/12/2013 14:38:01      51.538
21  28/12/2013 14:43:01      53.031
22  28/12/2013 14:48:01      53.529
etc (24hs data)

我希望R计算每小时的平均值（不考虑分钟或秒的差异，只需按小时计算）

有什么建议吗？非常感谢你提前！

此致玛利亚

Answer 1

将日期和小时列合并到POSIXct列中，并按小时休息时间cut()合并：

df <- read.table(header=TRUE, stringsAsFactors=FALSE, text="
date hour temperature
28/12/2013 13:03:01  41.572
28/12/2013 13:08:01  46.059
28/12/2013 13:13:01  48.55
28/12/2013 13:18:01  49.546
28/12/2013 13:23:01  49.546
28/12/2013 13:28:01  49.546
28/12/2013 13:33:01  50.044
28/12/2013 13:38:01  50.542
28/12/2013 13:43:01  50.542
28/12/2013 13:48:01  51.04
28/12/2013 13:53:01  51.538
28/12/2013 13:58:01  51.538
28/12/2013 14:03:01  50.542
28/12/2013 14:08:01  51.04
28/12/2013 14:13:01  51.04
28/12/2013 14:18:01  52.534
28/12/2013 14:23:01  53.031
28/12/2013 14:28:01  53.031
28/12/2013 14:33:01  53.031
28/12/2013 14:38:01  51.538
28/12/2013 14:43:01  53.031
28/12/2013 14:48:01  53.529
28/12/2013 15:01:01  50.77")

df$datehour <- cut(as.POSIXct(paste(df$date, df$hour),
   format="%d/%m/%Y %H:%M:%S"), breaks="hour") 
head(df)
        date     hour temperature            datehour
1 28/12/2013 13:03:01      41.572 2013-12-28 13:00:00
2 28/12/2013 13:08:01      46.059 2013-12-28 13:00:00
3 28/12/2013 13:13:01      48.550 2013-12-28 13:00:00
4 28/12/2013 13:18:01      49.546 2013-12-28 13:00:00
5 28/12/2013 13:23:01      49.546 2013-12-28 13:00:00
6 28/12/2013 13:28:01      49.546 2013-12-28 13:00:00

现在按小时列汇总：

means <- aggregate(temperature ~ datehour, df, mean)
head(means)
             datehour temperature
1 2013-12-28 13:00:00    49.17192
2 2013-12-28 14:00:00    52.23470
3 2013-12-28 15:00:00    50.77000

plot(as.POSIXct(means$datehour), means$temperature, type="l", las=1,
     main="Hourly Avg Temperatures", xlab="Hour", ylab="")

但是，对于时间序列数据，我喜欢使用包xts：

require(xts)
df.xts <- xts(df$temperature, as.POSIXct(paste(df$date, df$hour),
      format="%d/%m/%Y %H:%M:%S"))
head(df.xts)
                      [,1]
2013-12-28 13:03:01 41.572
2013-12-28 13:08:01 46.059
2013-12-28 13:13:01 48.550
2013-12-28 13:18:01 49.546
2013-12-28 13:23:01 49.546
2013-12-28 13:28:01 49.546

means <- period.apply(df.xts, endpoints(df.xts, "hours"), mean)
head(means)
                        [,1]
2013-12-28 13:58:01 49.17192
2013-12-28 14:48:01 52.23470
2013-12-28 15:01:01 50.77000

注意时间戳是每小时的最后一个条目。我们可以使用此函数将时间戳（向下）与小时的开头对齐：

align.time.down = function(x,n){ index(x) = index(x)-n; align.time(x,n) }
means.rounded <- align.time.down(means, 60*60)  
         # 2nd argument is the number of seconds to adjust/round to,
         # just like function align.time()

head(means.rounded)
                        [,1]
2013-12-28 13:00:00 49.17192
2013-12-28 14:00:00 52.23470
2013-12-28 15:00:00 50.77000

plot(means.rounded, las=1, main="Hourly Avg Temperatures")

Answer 2

如果在问题中给出样本数据和预期输出，则总是更容易。

Data.table包的解决方案

require(data.table)
data <- fread('temp.csv',sep=',') #Assuming your data is in temp.csv
#if above step not executed, convert the data frame to data.table 
data <- data.table(data)
> str(data)
Classes ‘data.table’ and 'data.frame':  12 obs. of  3 variables:
$ date       : chr  "28/12/2013" "28/12/2013" "28/12/2013" "28/12/2013" ...
$ hour       : chr  "13:03:01" "13:08:01" "13:13:01" "13:18:01" ...
$ temperature: num  41.6 46.1 48.5 49.5 49.5 ...

> data
      date     hour    temperature      avg
1: 27/12/2013 13:00:00       42.99 35.78455
2: 27/12/2013 14:00:00       65.97 35.78455
3: 27/12/2013 15:00:00       63.57 35.78455 

  data[,list(avg=mean(temperature)),by=hour] #dataset is sorted by hour
    hour   avg
1: 13:00:00 42.99
2: 14:00:00 65.97
3: 15:00:00 63.57
  data[,list(avg=mean(temperature)),by="date,hour"] #data set is grouped by date,then hour
        date     hour   avg
1: 27/12/2013 13:00:00 42.99
2: 27/12/2013 14:00:00 65.97
3: 27/12/2013 15:00:00 63.57

data[,list(avg=mean(temperature)),by=list(date,hour(as.POSIXct(data$hour, format = "%H:%M:%S")))] # to group by hour only 
     date     hour    avg
1: 27/12/2013    1 29.530
2: 27/12/2013    4 65.970

如何在R中按小时计算变量的平均值

2 个答案: