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Meissner, Tony (DFW)

2010-Jul-15 04:52 UTC

[R] taking daily means from hourly data

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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Gabor Grothendieck

2010-Jul-15 05:33 UTC

head link

[R] taking daily means from hourly data

On Thu, Jul 15, 2010 at 12:52 AM, Meissner, Tony (DFW)
<Tony.Meissner at sa.gov.au> 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
> : ? ? ? ? : ? ? ? ? ? ? ? : ? ? ? : ? ? ? ? ? ? ? : ? ? ? ? ?: ? ? : ? ? ?
?: ? ? ? ? :
> : ? ? ? ? : ? ? ? ? ? ? ? : ? ? ? : ? ? ? ? ? ? ? : ? ? ? ? ?: ? ? : ? ? ?
?: ? ? ? ? :
>
There are many possibilities.  Here are three.

#1 can be done with only the core of R.

#2 produces a zoo series which seems to be the logical representation
since it is, in fact, a series so its now already in the form for
other series operations.  See the 3 vignettes that come with zoo.

with #3 its easy to take different functions (avg, count, etc.) of
different columns and if you already know SQL its particularly
convenient.  See http://sqldf.googlecode.com

# DF2 is used in #1 and #3
DF2 <- data.frame(DF, Day = as.Date(format(DF$Time)))

# 1 - aggregate
aggregate(cbind(flow, direction, velocity) ~ Day, DF2, mean)

# 2 - zoo
library(zoo)
z <- read.zoo(DF, header = TRUE, tz = "GMT")
aggregate(z, as.Date, mean)

# 3 - sqldf
library(sqldf)
sqldf("select
   Day, avg(flow) Flow, avg(direction) Direction, avg(velocity) Velocity
   from DF2
   group by Day")

Allan Engelhardt

2010-Jul-15 05:38 UTC

head link

[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
>
>
>
>
> 	[[alternative HTML version deleted]]
>
>    
>
>
> ______________________________________________
> R-help@r-project.org mailing list
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide
http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.
>    
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