similar to: list of data.frames

I would like to propose adding a new plot function to the 'graphics' package. The new function is called areaplot() and I have implemented it as a generic function that supports a variety of data classes. An area plot consists of a simple line, like plot(x, y, type="l"), except the area between 0 and the line is a filled polygon. Areas can be stacked on top of each other,

Dear list members, I want to perform in R the analysis "simultaneous confidence interval for multiple proportions", as illustrated in the article of Agresti et al. (2008) "Simultaneous confidence intervals for comparing binomial parameter", Biometrics 64, 1270-1275. If I am not wrong the R function implementing the Agresti et al. method is prop.test(). I ask an help because I

Hello! Very sorry for a probably very simple question - I looked but did not find an answer in the archives. I have a table "counts" (below) that shows counts by Option within each of my 2 groups. However, my groups have different sizes (N1=255 and N2=68). Table "prop" shows the resulting proportions within each group. I would like to compare the proportions in 2 groups using

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 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, Does anyone know which method from Newcombe (1998)* is implemented in prop.test for comparing two proportions? I would guess it is the method based on the Wilson score (for single proportion), with and without continuity correction for prop.test(..., correct=FALSE) and prop.test(..., correct=TRUE). These methods would correspond to no. 10 and 11 tested in Newcombe, respectively. Can

Hi, I'm working on the codes below however every time I run them when they get to OpenBUGS I keep getting the error message: array index is greater than array upper bound for hab. Any help would be greatly appreciated, Suzie Codes: ungulate <- read.csv(file.choose ()) #ungulate ungulate <- as.matrix(ungulate);colnames(ungulate)<-NULL;rownames(ungulate)<-NULL

Dear R-helpers: I am an entry-level R user and have a question related to overlaying a barchart and and a xyplot using latticeExtra. My problem is that when I overlay them I fail to align their x-axes. I show my problem below through an example. #the example data frame is provided below vec <-c(1,5.056656,0.5977967,0.06126587,0.08557778, 2,4.601049,0.5995989,0.05002188,0.11410027,

Dear all, Does someone know an R function implementing the method of Sison and Glaz (1995) (see full ref below) for computing Simultaneous confidence intervals for multinomial proportions? As alternative method, I think to boostrap the mean of each proportion and get in that way confidence interval of the mean. I observed 21 times a response that could be one out of 8 categories

Dear Josh, Thanks for your help! Does your answer mean, that you agree the two methods should do the same, and what I was guessing, despite the small differences? What I prefer about ci.pd is, that the help clearly says which method is implemented, which is not the case for prop.test. But I do not know who has programmed the function. Best wishes Steffi Stefanie von Felten, PhD Statistician

Picking up an ancient thread (from Oct 2007), I have a somewhat more complex problem than given in Simon Wood's example below. My full model has more than two smooths as well as factor variables as in this simplified example: b <- gam(y~fv1+s(x1)+s(x2)+s(x3)) Judging from Simon's example, my guess is to fit reduced models to get components of deviance as follows: b1 <-

While building a package, I see the following: * checking R code for possible problems ... NOTE cheat.fit: no visible binding for global variable 'Zobs' plot.jml: no visible binding for global variable 'Var1' I see the issue has come up before, but I'm having a hard time discerning how solutions applied elsewhere would apply here. The entire code for both functions is below,

Hi r-users,   I would like to use prop.test code and I also calculate manually to test the proportions for 2 groups.  The problem is the answer for the p-value calculated manually are different from prop.test.  Here are the results:   ## Manually   z value: z= (phat-p)/sqrt(pq/n) = (.084-.081)/sqrt(.081(1-.081)/691)=0.289, pvalue=0.7718   ## Using prop.test code > low <- c(56,58) > tot

Dear R-Helpers, I've looked high and low for a function that provides frequencies, proportions and cumulative proportions side-by-side. Below is the table I need. Is there a function that already does it? Thanks, Bob > # Generate some test scores > myValues <- c(70:95) > Score <- ( sample( myValues, size=1000, replace=TRUE) ) > head(Score) [1] 77 71 81 88 83 93 > >

Full_Name: Robert W. Baer, Ph.D. Version: 1.6.2 OS: Windows 2000 Submission from: (NULL) (198.209.172.106) Problem: prop.test() does not seem to produce appropriate confidence intervals for the case where the vector length of x and n is one. (I am not certain about higher vector lengths.) As an example, I include x=6 and n=42 which has a mean proportion of 0.115. When I calculate the 95% CI

Hi, I have a seemingly simple proportional test. ?here is the question I am trying to answer: ? There is a test running each day in the lab, the test comes out as either positive or negative. So at the end of each month, we can calculate a positive rate in that month as the proportion of positive test results. The data look like: ? Month??? ??# positive?????? # total tests??? positive rate

When one has raw data it is easy to create a table of one variable against another and then calculate proportions For example a.nice.table<-table(a,b) prop.table(a.nice.table,1) However, I looked at several papers and created a data frame of the aggregate data. That means I acually created a table except it is a data frame. The first column lists the name of the first author and the year. I

Dear R users, I read your methods of extracting the variance explained by each predictor in different places. My question is: using the method you suggested, the sum of the deviance explained by all terms is not equal to the deviance explained by the full model. Could you tell me what caused such problem? > set.seed(0) > n<-400 > x1 <- runif(n, 0, 1) > ## to see problem

Following is my code, can some one help on the error at the bottom? > mh<-function(iterations,alpha,beta){ + data<-read.table("epidemic.txt",header = TRUE) + attach(data, warn.conflicts = F) + k<-97 + d <- (sqrt((x-x[k])^2 + (y-y[k])^2)) + p <- 1-exp(-alpha*d^(-beta)) + p.alpha<-1 - exp(-3*d^(-beta)) + p.beta <- 1 - exp(alpha*d^(-2)) +

Hi, I often need to take columns from two data.frames and 'splice' them together (alternately taking a column from the first data frame, then from the second). For example: x <- table(sample(letters[1:9], 100, replace=TRUE), sample(letters[1:4], 100, replace=TRUE)) y <- prop.table(x) splice <- function (x, y) { z <- matrix(rep(NA, (ncol(x) * 2) * nrow(x)), nrow