使用 xts/zoo R 绘制一天中一小时的值 [英] Plot value over hour of day with xts/zoo R

问题描述

我确实有一个像这样的时间序列(分钟值):

I do have a time-series looking like this (minute values):

   "timestamp", value 
    "2012-04-09 05:03:00",2
    "2012-04-09 05:04:00",4
    "2012-04-09 05:05:00",5
    "2012-04-09 05:06:00",0
    "2012-04-09 05:07:00",0
    "2012-04-09 05:08:00",3
    "2012-04-09 05:09:00",0
    "2012-04-09 05:10:00",1

是否有一种简单的方法可以在一天中的一小时内绘制这些值:从 1 到 24 小时(或 0 和 23)的 X 轴.所以 - 第 5 小时 5:00 和 5:59 之间的所有值等等.它不取决于日期，我只对当天的小时数感兴趣.谢谢！

Is there an easy way to plot these values over the hour of the day: X-Axis from 1 to 24 hours (or 0 and 23). So - all values between 5:00 and 5:59 over the 5th hour etc. It doesn't depend wich date, I am just interested in the houers of the day. Thank you!

附加问题:我可以将其绘制为箱线图吗?现在是动物园格式的 plot(df$hh,df$coredata.._data..).

Additional Question: Can I plot this as a boxplot? Right now it's plot(df$hh,df$coredata.._data..)in zoo format.

推荐答案

使用 xts 包，你可以使用 .indexhour 泛型函数来格式化一个 xts 对象的索引数小时.

Using xts package, you can use .indexhour Generic functions to format the index of an xts object to hours.

比如我读取数据，

dat <- read.zoo(text='timestamp, value 
    "2012-04-09 05:03:00",2
    "2012-04-09 05:04:00",4
    "2012-04-09 05:05:00",5
    "2012-04-09 05:06:00",0
    "2012-04-09 05:07:00",0
    "2012-04-09 05:08:00",3
    "2012-04-09 05:09:00",0
    "2012-04-09 05:10:00",1',header=TRUE,tz='',index=0:1,sep=',')

transform(dat, hh = .indexhour(dat))
                    coredata.._data.. hh
2012-04-09 05:03:00                 2  5
2012-04-09 05:04:00                 4  5
2012-04-09 05:05:00                 5  5
2012-04-09 05:06:00                 0  5
2012-04-09 05:07:00                 0  5
2012-04-09 05:08:00                 3  5
2012-04-09 05:09:00                 0  5
2012-04-09 05:10:00                 1  5

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