similar to: Error in Zero inflated model (ziP) with bam

Do you have the mgcv package installed (I think it's part of the standard distro, though) /loaded? ziP is there, not in BAM. Other than that, sorry, no clue. Cheers, Bert Bert Gunter "The trouble with having an open mind is that people keep coming along and sticking things into it." -- Opus (aka Berkeley Breathed in his "Bloom County" comic strip ) On Mon, Nov 6,

Hi all: I've got a data of ".bam" which is created from my partner under linux sysyem. My system is window xp, and I wanna know how to read the .bam file. Many thanks! My best [[alternative HTML version deleted]]

Hello, I'm trying to use the bam() function in the R mgcv package for a large set of grouped binary data. However, I have found that this function does not take data in the format of cbind(numerator, denominator) on the left hand side of the formula. As an example, consider the following dat1 <- data.frame(id=rep(1:6, each=3), num=rbinom(18, size=10, prob=0.8), den=rbinom(18, size=5,

Hello Jim, Thanks again. I am getting the two-headed arrow but I cannot seem to get the coordinates right for the arrow to appear beneath the map. These coordinates puts the arrow on the left hand side. Thanks again! Sincerely, Milu On Tue, Apr 12, 2016 at 1:15 PM, Jim Lemon <drjimlemon at gmail.com> wrote: > Hi Milu, > There is a two-headed arrow on the image you sent, and it

Dear useRs, I reran an analysis with bam (mgcv, version 1.7-17) originally conducted using an older version of bam (mgcv, version 1.7-11) and this resulted in the same estimates, but much lower standard errors (in some cases 20 times as low) and lower p-values. This obviously results in a larger set of significant predictors. Is this result expected given the improvements in the new version? Or

Dear David, The device was the issue. The quartz() device works fine but pdf() does not. Now I just need to figure out the limits for map for Europe. Thanks for all your help and patience. Sincerely, Milu On Sun, Apr 10, 2016 at 7:10 PM, David Winsemius <dwinsemius at comcast.net> wrote: > > > On Apr 10, 2016, at 4:12 AM, Miluji Sb <milujisb at gmail.com> wrote: > >

Hi Milu, My fault here. As I don't have the data to make the map and try out my suggestions I mixed up the x and y coordinates. Try this: par(xpd=TRUE) arrows(-19.75966,53,33.60000,53,code=3) par(xpd=FALSE) Jim On Tue, Apr 12, 2016 at 10:11 PM, Miluji Sb <milujisb at gmail.com> wrote: > Hello Jim, > > Thanks again. I am getting the two-headed arrow but I cannot seem to get

Dear Jim, Thanks again! I do want the arrows at the bottom (beneath the map). This is what I am doing: # Draw the map eps_europe <- mapCountryData(n, nameColumnToPlot="eps_score", mapTitle="EPS Score - Europe",colourPalette=colourPalette, catMethod="fixedWidth", missingCountryCol = "white", mapRegion="Europe", addLegend=FALSE) # ISO3 codes on

This is what I was looking for. Thank you everyone! Sincerely, Milu <https://www.avast.com/sig-email?utm_medium=email&utm_source=link&utm_campaign=sig-email&utm_content=webmail> Mail priva di virus. www.avast.com <https://www.avast.com/sig-email?utm_medium=email&utm_source=link&utm_campaign=sig-email&utm_content=webmail>

Hello everyone. I am doing a logistic gam (package mgcv) on a pretty large dataframe (130.000 cases with 100 variables). Because of that, the gam is fitted on a random subset of 10000. Now when I want to predict the values for the rest of the data, I get the following error: > gam.basis_alleakti.1.pr=predict(gam.basis_alleakti.1, +

Hello David, This is exactly what I want but I still can't get the arrows. R and R studio is updated. Thanks again! Sincerely, Milu On Sat, Apr 9, 2016 at 10:29 PM, David Winsemius <dwinsemius at comcast.net> wrote: > > > On Apr 9, 2016, at 1:27 PM, David Winsemius <dwinsemius at comcast.net> > wrote: > > > > > >> On Apr 9, 2016, at 11:18 AM,

Dear useRs, This is the first time I post to this list and I would appreciate any help available. I've used the excellent mgcv package for a while now to investigate geographical patterns of language variation, and it has has always worked without any problems for me. The problem below occurs using R 2.14.0 (both 32 and 64 bit versions in Windows and the 64 bit version in Unix) and mgcv (both

> On Apr 10, 2016, at 1:45 PM, Miluji Sb <milujisb at gmail.com> wrote: > > Dear David, > > The device was the issue. The quartz() device works fine but pdf() does not. Now I just need to figure out the limits for map for Europe. Thanks for all your help and patience. After plotting a map of Europe with base graphics the coordinates of the lower-left and upper-right

Ista, et. al: efficiency? (Note: I needed to correct my previous post: do.call() is required for pmax() over the data frame) > x <- data.frame(matrix(runif(12e6), ncol=12)) > system.time(r1 <- do.call(pmax,x)) user system elapsed 0.049 0.000 0.049 > identical(r1,r2) [1] FALSE > system.time(r2 <- apply(x,1,max)) user system elapsed 2.162 0.045 2.207 ##

Dear all, I am running a panel regression with time and location fixed effects: ### reg1 <- lm(lny ~ factor(id) + factor(year) + x1+ I(x1)^2 + x2+ I(x2)^2 , data=mydata, na.action="na.omit") ### My goal is to use the estimation for prediction. However, I have 8,500 IDs, which is resulting in very slow computation. Ideally, I would like to do the following: ### reg2 <-

> On Apr 10, 2016, at 4:12 AM, Miluji Sb <milujisb at gmail.com> wrote: > > Hello David, > > This is exactly what I want but I still can't get the arrows. R and R studio is updated. Thanks again! I didn't try it in Rstudio until just now (and I don't remember that you ever mentioned RStudio as a possible issue.) The plotting I see in the default graphics Rstudio

Hi R users, I'm using the 'bam' function in mgcv to examine trends in a remotely sensed vegetation index. I have one random effect variable, 'cons' (with six levels) which identifies different subjects within this analysis. My model is specified as follows: rm4<-bam(trend~factor(zone)+s(cons,bs="re")+s(year,

or in base R : ?xtabs ?? as in: xtabs(~previous_location + current_location,data=x) (You can convert the 0s to NA's if you like) Cheers, Bert Bert Gunter "The trouble with having an open mind is that people keep coming along and sticking things into it." -- Opus (aka Berkeley Breathed in his "Bloom County" comic strip ) On Tue, May 8, 2018 at 9:21 AM, Huzefa

On Tue, Feb 20, 2018 at 11:58 AM, Bert Gunter <bgunter.4567 at gmail.com> wrote: > Ista, et. al: efficiency? > (Note: I needed to correct my previous post: do.call() is required for > pmax() over the data frame) > > > x <- data.frame(matrix(runif(12e6), ncol=12)) > > > system.time(r1 <- do.call(pmax,x)) > user system elapsed > 0.049 0.000

xtabs does this automatically if your cross classifying variables are factors with levels all the cities (sorted, if you like): > x <- sample(letters[1:5],8, rep=TRUE) > y <- sample(letters[1:5],8,rep=TRUE) > xtabs(~ x + y) y x c d e a 1 0 0 b 0 0 1 c 1 0 0 d 1 1 1 e 1 1 0 > lvls <- sort(union(x,y)) > x <- factor(x, levels = lvls) > y <-