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I get the following error when I run qplot() qplot(grade, read,data = hhm.long.m, geom = c("point", "smooth")) Error in smooth.construct.cr.smooth.spec(object, data, knots) : x has insufficient unique values to support 10 knots: reduce k. I am not sure how to tackle this problem. When I take a subsample (< 1000) than I am able to run that function but with my sample

.... How might I do this in R? Can I do it with the survival library? If possible, I'd also like to control for sex. Thanks, Chris -- Christopher David Desjardins PhD student, Quantitative Methods in Education MS student, Statistics University of Minnesota 192 Education Sciences Building http://cddesjardins.wordpress.com

Hi, Let's say I have the following data: > a=matrix(c(1,2,4,4,2,1,1,2,4),nrow=3,byrow=T) > a [,1] [,2] [,3] [1,] 1 2 4 [2,] 4 2 1 [3,] 1 2 4 What syntax should I use to get R to tell me the column that corresponds to the maximum value for each row? For my example, I would like to get a vector that says 3, 1, 3 because the maximum value for row 1 is

Hi, I am trying to merge data similar to the example data below > dat0 id var1 var2 var3 2 1 0 1 3 1 0 1 4 0 1 1 5 0 1 1 > dat1 id var4 var5 var6 2 1 0 1 3 1 0 1 6 0 1 1 7 0 1 1 > dat2 id

...l using priors to overcome the lack of a mixed response? Also, please cc me on an email as I am a digest subscriber. Thanks, Chris -- Christopher David Desjardins PhD student, Quantitative Methods in Education MS student, Statistics University of Minnesota 192 Education Sciences Building http://cddesjardins.wordpress.com

Hi, I am having trouble with syntax for a for loop. Here is what I am trying to do. class=c(rep(1,3),rep(2,3),rep(3,3)) out1=rnorm(length(class)) out2=rnorm(length(class)) out3=rnorm(length(class)) data=data.frame(class,out1,out2,out3) dat.split=split(data,data$class) for(i in 1:3){ sub[i]=dat.split[i] } However, the for loop doesn't work. I want to assign each split to a different

Hi, I am trying to write a loop to recode my data from -999 to NA in R. What's the most efficient way to do this? Below is what I'm presently doing, which is inefficient. Thanks, Chris dat0 <- read.table("time1.dat") colnames(dat0) <- c("e1dq", "e1arcp", "e1dev", "s1prcp", "s1nrcp", "s1ints",

Hi, I have 45 models that I have named: 1, 2, 3, ... , 45 and I am trying to plot them in order of ascending BIC values. I am however unclear as to how I can get the models to line up on the x-axis by BIC and not by numeric order. For example, if model 5 has a lower BIC than 1, I want it to be the first point on the left hand side of the curve. This seems to work in plot: plot(1:45,

Hi I have two questions about using ggplot2. First, I have multiple columns of data that I would like to combine into one histogram where each column of data would correspond to one bar in the histogram. Each column has 0 or 1s and I want my bars in the histogram to correspond to the sum of the 1s in each column. Does that make sense? Second, is there a way to completely turn off the legend?

Hi, I have 45 models that I have named: 1, 2, 3, ... , 45 and I am trying to plot them in order of ascending BIC values. I am however unclear as to how I can get the models to line up on the x-axis by BIC and not by numeric order. For example, if model 5 has a lower BIC than 1, I want it to be the first point on the left hand side of the curve. This seems to work in plot: plot(1:45,

Hi, I want to submit a package to CRAN and I am getting the following Note: * checking CRAN incoming feasibility ... NOTE New submission How can I take care of this? And/or is it a big deal? Thanks and sorry if this is something that I easily overlooked I have googled this topic for a while and found nothing. Chris [[alternative HTML version deleted]]

Hi, I am using ggplot2 to with the following code: gmathk2 <- qplot(time,math,colour=Kids,data=kids.ach.lm.k5,geom="smooth",method="lm",formula=y~ns(x,1)) + opts(title="Smoother Plot: Math K-5") + xlab("Time") + ylab("Math") + scale_colour_brewer(pal="Set1"); gmathk2 This plots all the smoother for all the x values. What I'd like

Hi, I have my data the following way: y A B C 0 1 1 2 0 1 2 1 1 1 1 2 0 1 1 2 1 1 1 2 1 1 2 1 0 1 2 2 . . . And so on. How can I make my data look like the following: y A B C 2 1 1 2 1 1 2 1 0 1 2 2 . . . In other words how can I change my ungrouped cases into grouped cases? Thanks! Chris [[alternative

Hi, I am simulating data from a zero-inflated negative binomial model. I have 4 different conditions and I've saved various information about the models in zinbsumbstat-jklm.rdata (as shown below). What I want to do is print information from each of these *.rdata and store them in one large object (a matrix or a data frame). However, within each data the objects all have the same name. So I

I have a data frame in long format and I would like to randomly divide this data frame in half. The data frame consists of 39622 rows and I initially tried ... randomsample1 <- data[sample(nrow(data),19811), ] Where allows me to randomly select half of the rows and assign them to randomsample1 but then I couldn't figure out how to select those rows that were not selected and assign

...Surv numeric call 3 -none- call ties 1 -none- character -- Christopher David Desjardins PhD student, Quantitative Methods in Education MS student, Statistics University of Minnesota 192 Education Sciences Building http://cddesjardins.wordpress.com

Hi, I am trying to recreate the right graph on page 524 of Gelman's 2006 paper "Prior distributions for variance parameters in hierarchical models" in Bayesian Analysis, 3, 515-533. I am only interested, however, in recreating the portion of the graph for the overlain prior density for the half-Cauchy with scale 25 and not the posterior distribution. However, when I try:

I have 3 vectors: p1, p2, and p3. I would like each vector to be any possible value between 0 and 1 and p1 + p2 + p3 = 1. I want to graph these and I've thought about using scatterplot3d(). Here's what I have so far. library(scatterplot3d) p1 <- c(1,0,0,.5,.5,0,.5,.25,.25,.34,.33,.33,.8,.1,.1,.9,.05,.05) p2 <- c(0,1,0,.5,0,.5,.25,.5,.25,.33,.34,.33,.1,.8,.1,.05,.9,.05) p3 <-

Hi, I have the following data: > data[1:20,c(1,2,20)] idr schyear year 1 8 0 1 9 1 1 10 NA 2 4 NA 2 5 -1 2 6 0 2 7 1 2 8 2 2 9 3 2 10 4 2 11 NA 2 12 6 3 4 NA 3 5 -2 3 6 -1 3 7 0 3 8 1 3 9 2 3 10 3 3 11 NA What I want to do is

Hi, I have longitudinal school suspension data on students. I would like to figure out how many students (id_r) have no suspensions (sus), i.e. have a code of '0'. My data is in long format and the first 20 records look like the following: > suslm[1:20,c(1,7)] id_r sus 11 0 15 10 16 0 18 0 19 0 19 0 20 0 21 0 21 0 22 0 24 0 24