R help - Sep 2012 - average environmental data if AnimalID and Time is duplicated

Hello,
I tried for about three hours now to solve this problem but I can't figure
it out. I am sure someone knows how do it. At least I hope so. 

I have a data frame somewhat like this: 

myframe <- data.frame (ID=c("Ernie", "Ernie",
"Bert", "Bert"),
Timestamp=c("24.09.2012 09:00", "24.09.2012 09:00",
"24.09.2012 10:00",
"25.09.2012 10:00"),
Hunger=c("1","5","2","2"),
Temp=c("25","30","27","28")
)
myframe

As you can see for Ernie I do have different data for 24.09.2012 9:00. Now I
would like to average the Hunger and Temp value for this timestamp to get a
data frame without duplicated Times and the respective average Temp and
Hunger. 

I tried something like 
Meanframe<-  by(myframe[, 3:4], duplicated(myframe$ID,
Zusatzdaten3$Timestamp) == TRUE, mean)
but it doesn't work and I guess that it is also totally crap ;-)

Tagmarie




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On 09/26/2012 08:53 PM, Tagmarie wrote:
> Hello,
> I tried for about three hours now to solve this problem but I can't
figure
> it out. I am sure someone knows how do it. At least I hope so.
>
> I have a data frame somewhat like this:
>
> myframe<- data.frame (ID=c("Ernie", "Ernie",
"Bert", "Bert"),
> Timestamp=c("24.09.2012 09:00", "24.09.2012 09:00",
"24.09.2012 10:00",
> "25.09.2012 10:00"),
Hunger=c("1","5","2","2"),
Temp=c("25","30","27","28")
> )
> myframe
>
> As you can see for Ernie I do have different data for 24.09.2012 9:00. Now
I
> would like to average the Hunger and Temp value for this timestamp to get a
> data frame without duplicated Times and the respective average Temp and
> Hunger.
>
> I tried something like
> Meanframe<-  by(myframe[, 3:4], duplicated(myframe$ID,
> Zusatzdaten3$Timestamp) == TRUE, mean)
> but it doesn't work and I guess that it is also totally crap ;-)
>
Hi Tagmarie,
Your problem is that both Hunger and Temp are read in as factors. If you 
try it like this:

by(as.numeric(as.character(myframe[,3])),myframe[,"ID"],mean)
by(as.numeric(as.character(myframe[,4])),myframe[,"ID"],mean)

You might get what you want. The "as.character" call is necessary, 
otherwise you will get the wrong mean values.

Jim

Hello,

Why do you have Hunger and Temp recorded as characters? Between double 
quotes?


myframe <- data.frame (ID=c("Ernie", "Ernie",
"Bert", "Bert"),
Timestamp=c("24.09.2012 09:00", "24.09.2012 09:00",
"24.09.2012 10:00",
"25.09.2012 10:00"),
Hunger=c("1","5","2","2"),
Temp=c("25","30","27","28")
)
str(myframe)

myframe$Hunger <- as.numeric(levels(myframe$Hunger)[myframe$Hunger])
myframe$Temp <- as.numeric(levels(myframe$Temp)[myframe$Temp])

aggregate(cbind(Hunger, Temp) ~ ID, data = myframe, FUN = mean)

Hope this helps,

Rui Barradas

Em 26-09-2012 11:53, Tagmarie escreveu:
> Hello,
> I tried for about three hours now to solve this problem but I can't
figure
> it out. I am sure someone knows how do it. At least I hope so.
>
> I have a data frame somewhat like this:
>
> myframe <- data.frame (ID=c("Ernie", "Ernie",
"Bert", "Bert"),
> Timestamp=c("24.09.2012 09:00", "24.09.2012 09:00",
"24.09.2012 10:00",
> "25.09.2012 10:00"),
Hunger=c("1","5","2","2"),
Temp=c("25","30","27","28")
> )
> myframe
>
> As you can see for Ernie I do have different data for 24.09.2012 9:00. Now
I
> would like to average the Hunger and Temp value for this timestamp to get a
> data frame without duplicated Times and the respective average Temp and
> Hunger.
>
> I tried something like
> Meanframe<-  by(myframe[, 3:4], duplicated(myframe$ID,
> Zusatzdaten3$Timestamp) == TRUE, mean)
> but it doesn't work and I guess that it is also totally crap ;-)
>
> Tagmarie
>
>
>
>
> --
> View this message in context:
http://r.789695.n4.nabble.com/average-environmental-data-if-AnimalID-and-Time-is-duplicated-tp4644218.html
> Sent from the R help mailing list archive at Nabble.com.
>
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