similar to: aggregate() function, strange behavior for augmented data

All, I have data across 5 time points that I am graphing via xyplot, along with error bars. For one of the variables I have missing data for two of the time points. The code below is okay but I can't seem to get the lines to connect across the missing time points. Does anyone now how to rectify this? Cheers, David Afshartous library(lattice) ## the data junk = data.frame( Visit =

Hi R users,I have a question about augmented prediction plot (?augPred). The covariate of my data set is c(0, 0.01, 0.1, 1, 10, 100, 1000) and I have fitted a nonlinear mixed effects model.I use plot(augPred(out.nlme....)) to get the augmented prediction plot. However, because the scale of the covariate is too large thus I am not able to see the detail difference at c(0,0.01, 0.1, and 1). Could

All, I've been using aggregate() to compute means and standard deviations at time/treatment combinations for a longitudinal dataset, using na.rm = TRUE for missing data. This was working fine before, but now when I re-run some old code it isn't. I've backtracked my steps and can't seem to find out why it was working before but not now. In any event, below is a reproducible

I am trying to define a general R function that has a function as the output that depends on the user's input arguments (this may make more sense by looking at the toy example below). My real use for this type of code is to allow a user to choose from many parameterizations of the same general model. My "issue" is that when I compile a package with this type of code in it I get a

Dear R-experts: I am using a function called AUC whose arguments are data, time, id, and dv. data is the name of the dataframe, time is the independent variable column name, id is the subject id and dv is the dependent variable. The function computes area under the curve by trapezoidal rule, for each subject id. I would like to embed this in aggregate to further subset by each

Hi All: I have been looking for an elegant way to do the following, but haven't found it, I have never had a good understanding of any of the "apply" functions. A simplified idea is I have an array, say: junk(5, 10, 3) where (5, 10, 3) give the dimension sizes, and I want to reverse the second dimension, so I could do: junk1 <- junk[, rev(seq_len(10), ] but what I am

All, How does one extract the level-1 variance from a model fit via lmer()? In the code below the level-2 variance component may be obtained via subscripting, but what about the level-1 variance, viz., the 3.215072 term? (actually this term squared) Didn't see anything in the archives on this. Cheers, David > fm <- lmer( dv ~ time.num*drug + (1 | Patient.new), data=dat.new )

Classification: UNCLASSIFIED Caveats: NONE Dear R users, How can I retrieve an evaluated gradient value from "deriv" function provided below? I want to retrieve b0 value. Appreciate your help. Kyong junk1<-deriv(~(1/b1)*k-b0/b1,"b0",c("k","b0","b1"),formal=T) junk1(k=0,b0=-14.0236,b1=2.44031) [1] 5.746647 attr(, "gradient"):

Thanks to all for responses/. There was a question of exactly what was wanted. It is the generalization of the obvious example I gave, >>> junk1 <- junk[, rev(seq_len(10), ] so that junk[1,1,1 ] = junk1[1,10,1] junk[1,2,1] = junk1[1,9,1] etc. The genesis of this is the program is downloading data from a variety of sources on (time, altitude, lat, lon) coordinates, but all

How about this: f <- function(a,wh){ ## a is the array; wh is the index to be reversed l<- lapply(dim(a),seq_len) l[[wh]]<- rev(l[[wh]]) do.call(`[`,c(list(a),l)) } ## test z <- array(1:120,dim=2:5) ## I omit the printouts f(z,2) f(z,3) Cheers, Bert Bert Gunter "The trouble with having an open mind is that people keep coming along and sticking things into

On the off chance that anyone is still interested, here is the corrected function using aperm(): z <- array(1:120,dim=2:5) f2 <- function(a, wh) { idx <- seq_len(length(dim(a))) dims <- setdiff(idx, wh) idx <- append(idx[-1], idx[1], wh-1) aperm(apply(a, dims, rev), idx) } all.equal(f(z, 1), f2(z, 1)) # [1] TRUE all.equal(f(z, 2), f2(z, 2)) # [1] TRUE

> On 1 Jun 2017, at 22:42, Roy Mendelssohn - NOAA Federal <roy.mendelssohn at noaa.gov> wrote: > > Thanks to all for responses/. There was a question of exactly what was wanted. It is the generalization of the obvious example I gave, > >>>> junk1 <- junk[, rev(seq_len(10), ] > > > so that > > junk[1,1,1 ] = junk1[1,10,1] > junk[1,2,1] =

Thanks again. I am going to try the different versions. But I probably won't be able to get to it till next week. This is probably at the point where anything further should be sent to me privately. -Roy > On Jun 1, 2017, at 1:56 PM, David L Carlson <dcarlson at tamu.edu> wrote: > > On the off chance that anyone is still interested, here is the corrected function using

My error. Clearly I did not do enough testing. z <- array(1:24,dim=2:4) > all.equal(f(z,1),f2(z,1)) [1] TRUE > all.equal(f(z,2),f2(z,2)) [1] TRUE > all.equal(f(z,3),f2(z,3)) [1] "Attributes: < Component ?dim?: Mean relative difference: 0.4444444 >" [2] "Mean relative difference: 0.6109091" # Your earlier example > z <- array(1:120, dim=2:5) >

Here is an alternative approach using apply(). Note that with apply() you are reversing rows or columns not indices of rows or columns so apply(junk, 2, rev) reverses the values in each column not the column indices. We actually need to use rev() on everything but the index we are interested in reversing: f2 <- function(a, wh) { dims <- seq_len(length(dim(a))) dims <-

?? > z <- array(1:24,dim=2:4) > all.equal(f(z,3),f2(z,3)) [1] "Attributes: < Component ?dim?: Mean relative difference: 0.4444444 >" [2] "Mean relative difference: 0.6109091" In fact, > dim(f(z,3)) [1] 2 3 4 > dim(f2(z,3)) [1] 3 4 2 Have I made some sort of stupid error here? Or have I misunderstood what was wanted? Cheers, Bert Bert Gunter

All, For data consisting of serial measurements on subjects, one may use the aggregate function to say compute the peak response for each subject for each design condition. Is there a way to alter this or another one-liner to also retain the time at which the peak occurred and thus avoid writing a doing this via a loop? I suppose one could attempt to employ the split function but that's

On my debian box dovecot does not start properly any more, and mails don't get delivered or sent. I've googled around without finding any useful leads. Hopefully someone is willing to educate me. My guess is it must be a config problem after an upgrade, since I didn't change anything myself directly. Not sure when it started, since I this mail server is not our default machine (yet).

All, Is there an efficient way to apply say "mean" or "median" to a dataframe according to say all combinations of two variables in the dataframe? Below is a simple example and the outline of a "manual" solution that will work but is not very efficient (could also generalize this to a function). Searched the archives and docs but didn't see anything close to

Dear Users, please help with the following DF test: ===== library(tseries) library(timeSeries) Y=c(3519,3803,4332,4251,4661,4811,4448,4451,4343,4067,4001,3934,3652,3768 ,4082,4101,4628,4898,4476,4728,4458,4004,4095,4056,3641,3966,4417,4367 ,4821,5190,4638,4904,4528,4383,4339,4327,3856,4072,4563,4561,4984,5316 ,4843,5383,4889,4681,4466,4463,4217,4322,4779,4988,5383,5591,5322,5404