similar to: Where Can I Find the Function letterTrevor?

Hey all, Has anyone ever altered an R package for image analysis to do optical mark recognition? I'm trying to find a way to semi-automate data entry of several thousand paper health surveys that are predominantly composed of check boxes. All the boxes are uniform size and shape, so it seems as if it should be possible to alter a package to recognize the boxes and output a 0 or 1 to

Hello, I have a problem with using replace() to convert a vector of dates from yyyy-mm-dd to julian date. For example, I type replace(x,2004-05-14,134) and I receive an error: Error in `[<-.data.frame`(`*tmp*`, list, value = 134) : new columns would leave holes after existing columns If I can successfully convert, I have a script that will convert all of the dates in

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,

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

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

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 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) {

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

Hi Caitlin and Ben, Thanks for your responses! My issue is that I'd like to create one continuous line, rather than 3 lines overlayed. The code I've attached works for a population of 400 and samples 100 times. I'd like to extend this to 300 samples and 3 populations. So, the x-axis would range from 0-300 samples. What I'm having trouble with is finding a way to change the

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

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

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

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

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 =

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 =

Kirsten, this may not be the most elegant approach, but I think I would create dummy variables on the fly for each disease code you have, then use the "with" function to get means. Give this a try: # setup some fake data as an example icd9 <- c(rep("dis1",4), rep("dis2",5), rep("dis3",3)) n <- length(icd9) exposure <- rnorm(n) working <-

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 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

Does anyone have any performance tuning tips when working with datasets that are extremely wide (e.g. 20,000 columns)? In particular, I am trying to perform a merge like below: merged_data <- merge(data1, data2, by.x="date",by.y="date",all=TRUE,sort=TRUE); This statement takes about 8 hours to execute on a pretty fast machine. The dataset data1 contains daily data going

Hello. I am looking for a function that will allow me to paste rows together without regard for the numbers of columns in the datasets to be joined. The only columns where it matters if they are aligned correctly are at the beginning - the rest of the columns represent differing numbers of ICD9 (disease) codes reported by each person(record) at a health visit. They are in no particular order.