将 R 中的时间数据绘制为各种分辨率(分钟、小时、秒等) [英] Plot time data in R to various resolutions (to the minute, to the hour, to the second, etc.)

问题描述

我有一些 CSV 格式的数据，例如:

时间戳"，计数"《2009-07-20 16:30:45》，10《2009-07-20 16:30:45》，15"2009-07-20 16:30:46", 8"2009-07-20 16:30:46", 6"2009-07-20 16:30:46", 8《2009-07-20 16:30:47》，20

我可以使用 read.cvs 将其读入 R.我想绘制:

1. 每秒条目数，因此:<前>"2009-07-20 16:30:45", 2"2009-07-20 16:30:46", 3《2009-07-20 16:30:47》, 1

2. 每秒平均值:<前>《2009-07-20 16:30:45》，12.52009-07-20 16:30:46"，7.333《2009-07-20 16:30:47》，20

3. 同 1 &2 然后按分钟，然后按小时.

在 R 中有什么方法可以做到这一点(按秒/分钟/等& 情节收集)?

解决方案

读取您的数据，并将其转换为 动物园 对象:

R>X <- read.csv("/tmp/so.csv")R＞X <- zoo(X$Count, order.by=as.POSIXct(as.character(X[,1])))

请注意，由于时间戳不唯一，这将显示警告.

任务 1 使用 aggregate 和 length 进行计数:

R>聚合(X，力，长度)2009-07-20 16:30:45 2009-07-20 16:30:46 2009-07-20 16:30:472 3 1

任务 2 使用 aggregate:

R>聚合(X，力，平均值)2009-07-20 16:30:45 2009-07-20 16:30:46 2009-07-20 16:30:4712.500 7.333 20.000

任务 3 可以通过聚合高阶索引来完成.您可以对聚合的结果调用 plot:

plot(aggregate(X, force, mean))

I have some data in CSV like:

"Timestamp", "Count"
"2009-07-20 16:30:45", 10
"2009-07-20 16:30:45", 15
"2009-07-20 16:30:46", 8
"2009-07-20 16:30:46", 6
"2009-07-20 16:30:46", 8
"2009-07-20 16:30:47", 20

I can read it into R using read.cvs. I'd like to plot:

1. Number of entries per second, so:

   "2009-07-20 16:30:45", 2
   "2009-07-20 16:30:46", 3
   "2009-07-20 16:30:47", 1
   

2. Average value per second:

   "2009-07-20 16:30:45", 12.5
   "2009-07-20 16:30:46", 7.333
   "2009-07-20 16:30:47", 20
   

3. Same as 1 & 2 but then by Minute and then by Hour.

Is there some way to do this (collect by second/min/etc & plot) in R?

解决方案

Read your data, and convert it into a zoo object:

R> X <- read.csv("/tmp/so.csv")
R> X <- zoo(X$Count, order.by=as.POSIXct(as.character(X[,1])))

Note that this will show warnings because of non-unique timestamps.

Task 1 using aggregate with length to count:

R> aggregate(X, force, length)
2009-07-20 16:30:45 2009-07-20 16:30:46 2009-07-20 16:30:47 
                  2                   3                   1 

Task 2 using aggregate:

R> aggregate(X, force, mean)
2009-07-20 16:30:45 2009-07-20 16:30:46 2009-07-20 16:30:47 
             12.500               7.333              20.000 

Task 3 can be done the same way by aggregating up to higher-order indices. You can call plot on the result from aggregate:

plot(aggregate(X, force, mean))

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