Re: [R] taking daily means from hourly data

Wed, 14 Jul 2010 22:59:50 -0700 

Thanks Alan and Gabor.  Alan’s code appears to be the simplest to run and 
Gabor provided some further insight.


Tschüß
Tony Meissner
Principal Scientist (Monitoring)
Resources Monitoring Group
Science, Monitoring and Information Division
Department for Water
"Imagine" ©
•(ph) (08) 8595 2209
•(mob) 0401 124 971
Ê(fax) (08) 8595 2232
• 28 Vaughan Terrace, Berri SA 5343
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From: Allan Engelhardt [mailto:all...@cybaea.com]
Sent: Thursday, 15 July 2010 3:09 PM
To: Meissner, Tony (DFW)
Cc: r-help@r-project.org
Subject: Re: [R] taking daily means from hourly data

This is one way:

df <- data.frame(Time=as.POSIXct("2009-01-01", format="%Y-%m-%d") + seq(0, 
60*60*24*365-1, 60*60),

                 lev.morgan=3+runif(24*365),

                 lev.lock2=3+runif(24*365),

                 flow=1000+rnorm(24*365, 200),

                 direction=runif(24*365, 0, 360),

                 velocity=runif(24*365, 0, 10))

(df2 <- aggregate(df[ , -1], list(date=as.Date(df$Time)), FUN=mean, na.rm=TRUE))


Hope this helps

Allan


On 15/07/10 05:52, Meissner, Tony (DFW) wrote:

I have a data frame (morgan) of hourly river flow, river levels and wind 
direction and speed thus:

         Time           hour lev.morgan lev.lock2 lev.lock1 flow   direction  
velocity

1  2009-07-06 15:00:00   15      3.266     3.274     3.240 1710.6   180.282    
4.352

2  2009-07-06 16:00:00   16      3.268     3.272     3.240 1441.8   192.338    
5.496

3  2009-07-06 17:00:00   17      3.268     3.271     3.240 1300.1   202.294    
2.695

4  2009-07-06 18:00:00   18      3.267     3.274     3.241 1099.1   237.161    
2.035

5  2009-07-06 19:00:00   19      3.265     3.277     3.243  986.6   237.576    
0.896

6  2009-07-06 20:00:00   20      3.266     3.281     3.242 1237.6   205.686    
1.257

7  2009-07-06 21:00:00   21      3.267     3.280     3.242 1513.3    26.080    
0.664

8  2009-07-06 22:00:00   22      3.267     3.281     3.242 1819.5   264.280    
0.646

9  2009-07-06 23:00:00   23      3.267     3.281     3.242 1954.4   337.137    
0.952

10 2009-07-07 00:00:00    0      3.267     3.281     3.242 1518.9   260.006    
0.562

11 2009-07-07 01:00:00    1      3.267     3.281     3.242 1082.6   252.172    
0.673

12 2009-07-07 02:00:00    2      3.267     3.280     3.243 1215.9   190.007    
1.286

13 2009-07-07 03:00:00    3      3.267     3.279     3.244 1093.5   260.415    
1.206

:         :               :       :               :          :     :        :   
      :

:         :               :       :               :          :     :        :   
      :



Time is of class POSIXct

I wish to take daily means of the flow, levels, and wind parameters and put 
them into a new dataframe.  I envisage doing this with the following example 
code:



morgan$fTime <- factor(substr(as.character(morgan$Time),1,10))

dflow <- tapply(morgan[,"flow"], morgan$fTime, mean)

day <- tapply(morgan[,"Time"], morgan$fTime, mean)

  :

  :



daily <- as.data.frame(cbind(day,dflow, dlev.morg,dlev.lock2, ...))

daily$day <- with(daily, as.POSIXct("1970-01-01", "%Y-%m-%d", 
tz="Australia/Adelaide") + day)

rownames(daily) <- NULL



Is there a more efficient way of doing this?  I am running R-2.11.0 under 
Windows XP



Tschüß

Tony Meissner

Principal Scientist (Monitoring)

Resources Monitoring Group

Science, Monitoring and Information Division

Department for Water

"Imagine" ©

*(ph) (08) 8595 2209

*(mob) 0401 124 971

*(fax) (08) 8595 2232

* 28 Vaughan Terrace, Berri SA 5343

        PO Box 240, Berri SA 5343

        DX 51103

***The information in this e-mail may be confidential and/or legally 
privileged.  Use or disclosure of the information by anyone other than the 
intended recipient is prohibited and may be unlawful.  If you have received 
this e-mail in error, please advise by return e-mail or by telephoning +61 8 
8595 2209









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