Marta Miguel
2012-Sep-18 11:17 UTC
[R] Data frame divison by another data frame with common groups and different length
Dear all, I have two different data frames, that have two common variables: date and sample. Here is a small extract of both of them> head(traffic)datet sessiont samplet buddleiat 1 07-08-2012 1 1 1 2 07-08-2012 1 1 1 3 07-08-2012 1 1 1 4 07-08-2012 1 2 3 5 07-08-2012 1 2 1 6 07-08-2012 1 2 2 7 07-08-2012 2 3 4 8 07-08-2012 2 3 5 9 07-08-2012 2 3 5 10 07-08-2012 2 4 8 11 07-08-2012 2 4 4 12 07-08-2012 2 4 6 13 08-08-2012 1 1 9 14 08-08-2012 1 1 12 15 08-08-2012 1 1 7 (...)> head(encounters)datec samplec Bladen 1 07-08-2012 1 9 2 07-08-2012 1 6 3 07-08-2012 1 8 4 07-08-2012 1 8 5 07-08-2012 1 5 6 07-08-2012 1 4 7 07-08-2012 1 7 8 07-08-2012 1 6 9 07-08-2012 1 4 10 07-08-2012 1 2 11 07-08-2012 1 7 12 07-08-2012 1 8 (...) They don't have the same length. Using the function tapply, I managed to calculate the mean of the column buddleiat (from traffic) by sample and date. tapply(buddleiat, list(samplet,datet), mean) Now I want to divide each different value of Bladen by the value that has the same number of sample and date of the buddleiat mean calculated. Being that, different values from Bladen can be divided by the same value of buddleiat mean. I tried using this code:> budt=(Bladen/tapply(buddleiat, list(samplet,datet), mean))Error: dims [product 68] do not match the length of object [1360] But they don't have the same length, and I am not sure that it will do what I am asking. Do you have any suggestions? Is there a function I could use or will it be easier to do it "manually" by creating a third data frame with both variables and repeated values of the buddleiat means? I hope I made myself understand. Thanks in advance, Marta Miguel [[alternative HTML version deleted]]
R. Michael Weylandt
2012-Sep-18 14:52 UTC
[R] Data frame divison by another data frame with common groups and different length
On Tue, Sep 18, 2012 at 12:17 PM, Marta Miguel <martamiguel12 at gmail.com> wrote:> Dear all, > > > I have two different data frames, that have two common variables: date and > sample. Here is a small extract of both of them > >> head(traffic) > datet sessiont samplet buddleiat > 1 07-08-2012 1 1 1 > 2 07-08-2012 1 1 1 > 3 07-08-2012 1 1 1 > 4 07-08-2012 1 2 3 > 5 07-08-2012 1 2 1 > 6 07-08-2012 1 2 2 > 7 07-08-2012 2 3 4 > 8 07-08-2012 2 3 5 > 9 07-08-2012 2 3 5 > 10 07-08-2012 2 4 8 > 11 07-08-2012 2 4 4 > 12 07-08-2012 2 4 6 > 13 08-08-2012 1 1 9 > 14 08-08-2012 1 1 12 > 15 08-08-2012 1 1 7 > (...) > >> head(encounters) > datec samplec Bladen > 1 07-08-2012 1 9 > 2 07-08-2012 1 6 > 3 07-08-2012 1 8 > 4 07-08-2012 1 8 > 5 07-08-2012 1 5 > 6 07-08-2012 1 4 > 7 07-08-2012 1 7 > 8 07-08-2012 1 6 > 9 07-08-2012 1 4 > 10 07-08-2012 1 2 > 11 07-08-2012 1 7 > 12 07-08-2012 1 8 > (...) > > They don't have the same length. > > Using the function tapply, I managed to calculate the mean of the column > buddleiat (from traffic) by sample and date. > > tapply(buddleiat, list(samplet,datet), mean) > > > Now I want to divide each different value of Bladen by the value that has > the same number of sample and date of the buddleiat mean calculated. Being > that, different values from Bladen can be divided by the same value of > buddleiat mean. > > > I tried using this code: > >> budt=(Bladen/tapply(buddleiat, list(samplet,datet), mean)) > Error: dims [product 68] do not match the length of object [1360] > > But they don't have the same length, and I am not sure that it will do what > I am asking. > > > Do you have any suggestions? Is there a function I could use or will it be > easier to do it "manually" by creating a third data frame with both > variables and repeated values of the buddleiat means? > >I believe changing tapply() --> ave() will be a good start. Cheers, Michael> I hope I made myself understand. > > > Thanks in advance, > > Marta Miguel > > [[alternative HTML version deleted]] > > ______________________________________________ > R-help at 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.
Rui Barradas
2012-Sep-18 18:34 UTC
[R] Data frame divison by another data frame with common groups and different length
Hello,
Try the following.
agg <- aggregate(buddleiat ~ samplet + datet, data = traffic, FUN = mean)
mrg <- merge(encounters, agg,
by.x = c("samplec", "datec"),
by.y = c("samplet", "datet"))
mrg$Div <- with(mrg, Bladen/buddleiat)
Hope this helps,
Rui Barradas
Em 18-09-2012 12:17, Marta Miguel escreveu:> Dear all,
>
>
> I have two different data frames, that have two common variables: date and
> sample. Here is a small extract of both of them
>
>> head(traffic)
> datet sessiont samplet buddleiat
> 1 07-08-2012 1 1 1
> 2 07-08-2012 1 1 1
> 3 07-08-2012 1 1 1
> 4 07-08-2012 1 2 3
> 5 07-08-2012 1 2 1
> 6 07-08-2012 1 2 2
> 7 07-08-2012 2 3 4
> 8 07-08-2012 2 3 5
> 9 07-08-2012 2 3 5
> 10 07-08-2012 2 4 8
> 11 07-08-2012 2 4 4
> 12 07-08-2012 2 4 6
> 13 08-08-2012 1 1 9
> 14 08-08-2012 1 1 12
> 15 08-08-2012 1 1 7
> (...)
>
>> head(encounters)
> datec samplec Bladen
> 1 07-08-2012 1 9
> 2 07-08-2012 1 6
> 3 07-08-2012 1 8
> 4 07-08-2012 1 8
> 5 07-08-2012 1 5
> 6 07-08-2012 1 4
> 7 07-08-2012 1 7
> 8 07-08-2012 1 6
> 9 07-08-2012 1 4
> 10 07-08-2012 1 2
> 11 07-08-2012 1 7
> 12 07-08-2012 1 8
> (...)
>
> They don't have the same length.
>
> Using the function tapply, I managed to calculate the mean of the column
> buddleiat (from traffic) by sample and date.
>
> tapply(buddleiat, list(samplet,datet), mean)
>
>
> Now I want to divide each different value of Bladen by the value that has
> the same number of sample and date of the buddleiat mean calculated. Being
> that, different values from Bladen can be divided by the same value of
> buddleiat mean.
>
>
> I tried using this code:
>
>> budt=(Bladen/tapply(buddleiat, list(samplet,datet), mean))
> Error: dims [product 68] do not match the length of object [1360]
>
> But they don't have the same length, and I am not sure that it will do
what
> I am asking.
>
>
> Do you have any suggestions? Is there a function I could use or will it be
> easier to do it "manually" by creating a third data frame with
both
> variables and repeated values of the buddleiat means?
>
>
> I hope I made myself understand.
>
>
> Thanks in advance,
>
> Marta Miguel
>
> [[alternative HTML version deleted]]
>
> ______________________________________________
> R-help at 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.