similar to: replace() error: new columns would leave holes after existing columns

Dear R experts, I want to plot a line chart with another secondary axis placed right to the standard secondary axis which one can access with the axis command, so that the data lines are seen in the same plot. Is there any way to do this in R? Many thanks, Kirsten.

Hi Jim, Thanks for your help, I really appreciate it. Perhaps I'm misunderstanding, but does this formula run different ajustment values for this function? logit(p = doc$value, adjust = 0.025) I'm looking to plot the p-values of different adjustment values. Thanks so much, Kirsten On Wed, Jul 12, 2017 at 8:49 PM, Jim Lemon <drjimlemon at gmail.com> wrote: > Hi Kirsten,

Hello Speex, I think I've noticed a bug in version 1.1.6. In the source file "speexdec.c", lines 558 to 560 say wav_format = strlen(outFile)>=4 && ( strcmp(outFile+strlen(outFile)-4,".wav")==0 || strcmp(inFile+strlen(inFile)-4,".WAV")==0); . I think each

Hello everyone, I have a .csv file with the following format: uniqueID SubjectID Distance_miles Tag 1 1001 5.5 3 2 1001 7 1 3 1001 6.5 1 4 1001 5 1 5 1002

Hi Ben, That's exactly right! Except for each set it's the sample population that is 400, 800 or 300. I want to take 3 samples, each of 100, where only the population differs. I can do this separately, but I'm having trouble putting them all on the same graph. I'd like to have sample on the x axis (1-300) and estimate on the y axis. I want to show how population affects the

Hi Kirsten, Perhaps this will help: set.seed(3) kmdf<-data.frame(group=rep(1:4,each=20), prop=c(runif(20,0.25,1),runif(20,0.2,0.92), runif(20,0.15,0.84),runif(20,0.1,0.77))) km.glm<-glm(prop~group,kmdf,family=quasibinomial(link="logit")) summary(km.glm) pval<-0.00845 padjs<-NA npadj<-1 # assume you have five comparisons in this family for(method in p.adjust.methods) {

I have a researcher who is consistently get the warning message: In max(i) : no non-missing arguments to max; returning -Inf Best as I can tell the code is working properly and the output is as expected. I would like some help in understanding why he is getting this error message and what its implications are. I have his code. Sincerely, Kirsten Miles Support Specialist Research Computing Lab

Hi Kirsten, I can run your example code but I can't quite follow your division of sampling. Can you restate the the task? Below is what I think you are asking for, but I have the feeling I may be off the mark. Set A: 400 samples, draw 100 in range of 5 to 15 Set B: 800 samples, draw 100 in range of 5 to 15 Set C: 300 samples, draw 100 in range of 5 to 15 Ben > On Aug 5, 2017, at

Hello all, I have a data set where the response variable is the percent cover of a specific plant (represented in cover classes 0,1,2,3,4,5, or 6). This data set has a lot of zeros (plots where the plant was not present). I am trying to model cover class of the plant as a function of both total nitrogen and shrub cover. After quite a bit of research I have come across a conditional approach

Hi All, I'm trying to create two side-by-side contour plots with one legend by modifying the code found here: http://wiki.cbr.washington.edu/qerm/sites/qerm/images/b/bb/Example_4-panel_v1a.R I've been able to set up the other two functions called in the code, but I can't find reference to print.letterTrevor.R anywhere other than the page linked above. I've searched all the help

Not a reproducible example, so a bit of guessing here, but a) don't try to assign results to variables inside the ifelse. That is, remove all the single-equals signs and "test" variables. If you really need to conditionally assign variables then use "if"... but chances are good you don't need that. b) "closure" is effectively another word

Hi everyone, I'm having some trouble with my ifelse statements. I'm trying to put 12 conditions within 3 groups. Here is the code I have so far: dat$cond <- ifelse(test = dat$cond == "cond1" | dat$cond == "cond2" | dat$cond == "cond3" dat$cond == "cond4" yes = "Uniform" no = ifelse(test =

Hello! I have measurements (length and volume) of fish collected in two years. I want to know if the the relationship between length and volume is the same for both years. The number of fish measured is different for each year. I don't know whether lm or aov is more appropriate to use. Here are the two output options: Call: lm(formula = Volume ~ Length * Year) Residuals: Min 1Q

Hey everyone, I'm working on a contour plot depicting asymptomatic prevalence at varying durations of infectiousness and force of infection. I've been able to work everything out except for this one - my legend title keeps getting cut off. Here's what I have: filled.contour(x=seq(2,30,length.out=nrow(asym_matrix)), y=seq(1,2,length.out=ncol(asym_matrix)), asym_matrix, color =

Hello R-help! I am a systems administrator for the University of Southern California. I take care of it's general-purpose research cluster, and have recently been asked to provide access to a parallelized R platform. I do not have any previous experience using R, and have only ever had to do anything more then 'yum -y install R'. We have a myrinet MPI network, and use mpich1 as our

Hi! Thanks for taking the time to read this. The code below creates a graph that takes 100 samples that are between 5% and 15% of the population (400). What I'd like to do, however, is add two other sections to the graph. It would look something like this: from 1-100 samples take 100 samples that are between 5% and 15% of the population (400). From 101-200 take 100 samples that are between

I downloaded version 0.3.12 of TZInfo from [1], and observed that it has a very robust set of timezones including, for example, ''America/ Anchorage'': irb(main):001:0> require ''rubygems'' => false irb(main):002:0> require ''tzinfo'' => true irb(main):003:0> TZInfo::Timezone.get(''America/Anchorage'') =>

I have a list of ICD9 (disease) codes with various formats - 3 digit, 4 digit, 5 digit. The first three digits of these codes are what I am most interested in. I would like to either add zeros to the 3 and 4 digit codes to make them 5 digit codes or add decimal points to put them all in the format ###.##. I did not see a function that allows me to do this in the formatting command. This seems

Is anyone else here from Alaska interested in a ruby users group to meet in Anchorage once a month? It might be a long shot but I thought i''d put the possibility of such a group out there if others are interested. If so, we''ll coordinate some times that work. Benjamin -------------- next part -------------- An HTML attachment was scrubbed... URL:

To anyone who can help: I have two brief questions concerning sapply. Following below is the code for my example. The two problems are described at the end of the code: site <- rep(2:6, each = 12) tillage <- rep(c(1,-1), each = 6, times = 5) carbon <- c(18.23, 16.06, 17.81, 16.07, 17.26, 17.08, 14.92, 15.88, 12.11, 14.23, 16.99, 13.57, 20.34, 20.3,