search for: cutpoints

Below is a diff to vcut.c and its vcut.1 manpage, which - rewrites the manpage in the mdoc(7) markup as offered in my previous message, and while there: - drops the [] around cutpoint as it is _not_ optional - adds the [] around [+] as it _is_ optional - drops the argument name difference between "cutpoint" and "cuttime" - adds the usage sentence about "." output

Dear List, Does anybody no how to compare mean survival times for two (more) groups in R? What test statistics should I use? Thank you very much! Joe [[alternative HTML version deleted]]

Hola a todos, Al ejecutar el código que veis más abajo:  library(OptimalCutpoints)prediccion<-c(0.49165923,0.52759793,0.30213400,0.33468349,0.14979703,0.47401846,0.52216404,0.42018794,0.92168073,0.76893929,0.83362668,0.38251162,0.70803701,0.49165923,0.94462558) real<-c(0,1,0,0,0,0,1,1,1,1,1,0,1,0,1)datos_OPTIMO<-cbind(prediccion,real) cutpoint1 <- optimal.cutpoints(X...

.... ### They claim that one can recode an n-chotomy into n-1 dichotomous ### indicators. Each observation in the original dataset begets n-1 ### lines in the augmented version. After creating the dichotomous ### indicator, one uses an ordinary dichotomous logit model to ### estimate parameters and cutpoints for an ordinal logit ### model. Cole, et al., are very clear. ### There is an additional benefit to the augmented data approach. ### One can explicitly test the proportional odds assumption by checking ### for interactions between the included independent variables and the ### "level" of...

Dear R users, I still feel new to R so please apologize if I am doing something stupid here. My use of the polygon() function produces a result that I cannot comprehend: In a plot, I would like to shade the area below a normal distribution. However, I do not want the entire area to be shaded, but just the area on the right side of a vertical line that I draw through the distribution (in

..., since the documentation is insufficient in this case. According to him Jarrod (forums), the a posteriori means of the coefficients of the covariates are the probit scale. According to my study, these coefficients are the scores of standard normal distribution. More scores should not correspond to cutpoints? In this case, we would have j (response variable categories) -1 cutpoints, ie, two cutpoints. The output shows me only one cutpoint. How can then calculate the probabilities with only one cutpoint? According to the documentation (Vignettes, page 22), if P (y = k) = F (yk | l (vlatente), 1) - F (yk...

Dear Mrs Rachel Pearce, I am looking for a function "cutpt-coxph" in R - like you did some years ago. How have you solved the problem? Have you found it or a similar function? thank you, Sincerely, Friederike "The title says it all really; I am looking for a function along the lines of cutpt.coxph as described in "Survival Analysis Using S" (Tableman and Kim), Chapter

...to "likfit" to ensure convergence. Although not the most elegant solution, the code works ... I just want to re-code it to avoid using the "for" loop. Any help would be greatly appreciated. This is what I've got .... RootDev <- 600 splits <- NULL for (cutpoint in cutpoints) { LRGdata <- split(gdata, gdata[,4] <= cutpoint) vgrmL <- variog(as.geodata(LRGdata$"TRUE"), covar.col=covcol)) vgrmR <- variog(as.geodata(LRGdata$"FALSE"), covar.col=covcol)) mlL<- likfit(as.geodata(LRGdata$"TRUE"), ini=expand.grid(seq(0...

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How does one use the vcut from Vorbis-tools package? I tried to enter samples, miliseconds, seconds, h:mm:ss as cut point, but nothing worked (I even tried to enter 1 as cutpoint, but I always got "Cutpoint not within stream." message)... I'm doing this on Windows, if it matters... -- Jernej Simoncic, jernej.simoncic@guest.arnes.si http://www2.arnes.si/~sopjsimo/ ICQ: 26266467

Dear all, I'm trying to manage with user defined split function in rpart (file rpart\tests\usersplits.R in http://cran.r-project.org/src/contrib/rpart_3.1-34.tar.gz - see bottom of the email). Suppose to have the following data.frame (note that x's values are already sorted) > D y x 1 7 0.428 2 3 0.876 3 1 1.467 4 6 1.492 5 3 1.703 6 4 2.406 7 8 2.628 8 6 2.879 9 5 3.025 10 3 3.494

...ns for manual or pre-made solutions. Example: set.seed(2) x <- matrix(rnorm(1000),ncol=10) #obviously no real clusters here... row.hclust <- hclust(dist(x)) row.dendro <- as.dendrogram(row.hclust) heatmap(x, Rowv=row.dendro) row.cut <- cutree(row.hclust,3)[row.hclust$order] cutpoints <- which(row.cut[-1]!=row.cut[-length(row.cut)]) ymin <- par("usr")[3] #in general incorrect ymax <- par("usr")[4] #in general incorrect for (i in cutpoints){ thisy <- ymin + (ymax-ymin)*(i-1)/nrow(x) abline(h=thisy,lw=3) }

R 2.1.1 Win 2k Would someone suggest a method (or methods) that can be used to determine ntile cutpoints of a vector, i.e. to determine values that can be used to divide a vector into thirds such as 0-33 centile, 34-66 centile, 67-100 centile. If for example I had a vector: 1,2,3,4,5,6,7,8,9 and wanted to divide the vector into thirds I would have cut-points of 3, and 6. Thanks, John John Sorkin M.D...

Dear List: I have searched the archives and my R books and cannot find a method to transform a continuous variable into a binary variable. For example, I have test score data along a continuous scale. I want to create a new variable in my dataset that is 1=above a cutpoint (or passed the test) and 0=otherwise. My instinct tells me that this will require a combination of the transform

I am new to R and I am trying to do a monte carlo simulation where I generate data and interject error then test various cut points; however, my output was garbage (at x equal zero, I did not get .50) I am basically testing the performance of classifiers. Here is the code: n <- 1000; # Sample size fitglm <- function(sigma,tau){ x <- rnorm(n,0,sigma) intercept <- 0 beta

Dear forumites As a newbie I try to figure out whether R can do a certain job quicker than other programs and it seems so, but I don't find a solution to a seemingly simple problem: I have built a matrix of distance with as.matrix(dist()) with several hundred rows and columns 1 2 3 4 1 0 1.3 1.6 2.1 2 1.3 0

Dear all, I have two variables, y and x. It seems that the relationship between them is Piecewise Linear Functions. The cutpoint is 20. That is, when x<20, there is a linear relationship between y and x; while x>=20, there is another different linear relationship between them. How can i specify their relationships in R correctly? # glm(y~I(x<20)+I(x>=20),family = binomial, data =

I am running my code in a loop and it does not work but when I run it outside the loop I get the values I want. n <- 1000; # Sample size fitglm <- function(sigma,tau){ x <- rnorm(n,0,sigma) intercept <- 0 beta <- 0 ystar <- intercept+beta*x z <- rbinom(n,1,plogis(ystar)) xerr <- x + rnorm(n,0,tau) model<-glm(z ~ xerr, family=binomial(logit))

...an attribute on NULL and > attach( rt ) Error in attach(rt) : attach only works for lists and data frames the code that I use to build the matrix is : # break up the classes into cats and test for departure from zero signif <- symnum(fit3sls[[2]]$residuals, corr = FALSE, na = FALSE, cutpoints = c(-5,.1,.2,.3,.4, 5), symbols = c("A","B","C","D","E")) rt <- noquote( cbind( fit3sls[[2]]$residuals, signif ) ) colnames(table) <- c("resid","group") boxplot( resid ~ signif, data = rt ) which doesn't work. W...

...e")[unclass(iris$Species)], lower.panel=panel.cor, splitvar=iris$Species) ------------ #here the code I tried to implement to show the significance levels test <- cor.test(x, y, ,use="complete.obs") Signif <- symnum(test$p.value, corr = FALSE, na = FALSE, cutpoints = c(0, 0.001, 0.01, 0.05, 0.1, 1), symbols = c("***", "**", "*", ".", " ")) -- View this message in context: http://www.nabble.com/display-p-values--and-significance-levels-tp21769056p21769056.html Sent from the R help mailing lis...