Hi everyone,Good news, I think... I ended up being able to do the rolling
maximum with a for loop using the rollMax function in package Zoo suggested by
Gabor. Here's what I did:
library(zoo)for (i in 1:136)Pmax[,i] <- rollMax(Pmat[,i], n=7, trim=FALSE,
na.rm=FALSE)
I think, well, I'm pretty certain, that some of you will tell me that the for
loop is not the most efficient way of accomplishing this. I tried to do a
sapply function but I never got it to work. Here's my attempt and the error I
got:
library(zoo)> Pmax <- Pmat[,sapply(Pmat, function(x) {rollMax((x), n=7,
trim=FALSE, na.rm=FALSE)})]
Error in `[.data.frame`(Pmat, , sapply(Pmat, function(x) { :
undefined columns selected
So any comments on the reasons for my failure would be highly appreciated.
Thanks and have a great day!
Rita ________________________________________ "If you think education is
expensive, try ignorance"--Derek Bok
> From: [email protected]
> Date: Thu, 7 Apr 2011 15:41:30 -0400
> Subject: Re: [R] df with max function applied to 6 lags of a variable?!?
> To: [email protected]
> CC: [email protected]
>
> On Thu, Apr 7, 2011 at 2:29 PM, Rita Carreira <[email protected]>
> wrote:
> >
> > Hello everyone!
> > I have a data frame of 136 variables with 270 observations. I would like to
> > create a new data frame such that each element of that data frame contains
> > the maximum value of the 6 prior lags of the initial data frame. So for
> > example, if my original data frame, A, were
> > A1=c(7.72, 7.94, 7.56, 7.54, 0.93, 0.59, 7.21, 8.00, 7.19, 7.57)A2=c(4.27,
> > 3.70, 3.80, 3.67, 3.83, 3.95, 4.02, 2.06, 3.28, 2.17)A3=c(4.16, 2.91, 3.89,
> > 3.73, 4.00, 3.81, 3.86, 1.88, 3.74, 3.75)A4=c(3.56, 3.27, 3.42, 3.32, 3.48,
> > 3.54, 3.60, 2.04, 3.38, 3.37)A5=c(8.24, 8.67, 8.49, 0.00, 9.39, 0.00, 8.49,
> > 8.94, 0.00, 0.00)A6=c(4.28, 4.18, 4.20, 4.14, 4.40, 4.20, 4.20, 4.15, 4.17,
> > 4.16)
> > A <- cbind(A1, A2, A3, A4, A5, A6)
> > My new data frame, Amax, where each element is amax(i,j)=max[a(i-1,j),
> > a(i-2,j), a(i-3,j), a(i-4,j), a(1-5,j), a(i-6,j)] would be:NA NA NA
> > NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA
> > NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA
> > NA NA 7.94
> > 4.27
> > 4.16
> > 3.56
> > 9.39
> > 4.40 7.94
> > 4.02
> > 4.00
> > 3.60
> > 9.39
> > 4.40 8.00
> > 4.02
> > 4.00
> > 3.60
> > 9.39
> > 4.40 8.00
> > 4.02
> > 4.00
> > 3.60
> > 9.39
> > 4.40 8.00
> > 4.02
> > 4.00
> > 3.60
> > 9.39
> > 4.40
> > So the question is: what is the most efficient way of doing this in R?
> > Thanks!Rita ________________________________________ "If you think
> > education is expensive, try ignorance"--Derek Bok
>
>
> See rollmax in the zoo package and see ?rollmax for the various
> options. e.g. using the built in BOD data frame:
>
> > library(zoo)
> > as.data.frame(rollmax(BOD, 3, na.pad = TRUE, align = "right"))
> Time demand
> 1 NA NA
> 2 NA NA
> 3 3 19.0
> 4 4 19.0
> 5 5 19.0
> 6 7 19.8
>
>
> --
> Statistics & Software Consulting
> GKX Group, GKX Associates Inc.
> tel: 1-877-GKX-GROUP
> email: ggrothendieck at gmail.com
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