similar to: Kaplan-Meier function from survfit

Dear R users: When I create a table without Major Column headings, my *regular* column headings appear correct in the typeset latex file. The major row heading and row groups are as they should. w <- latex(mytab,title="",file="tab/my.tex",ctable=TRUE,caption="Descriptive statistics by

Hi All, I am fitting a tree to censored survival data using the rpart package and wanted to better understand the results. I am trying to interpret the output from the tree. I am interested in understanding what "yval" is for a survival tree. I see in the output of summary, the phrase "estimated rate". The estimated rate is 1 for the entire tree, and more of less for each

I am trying to use lm for a simple linear fit with weights. The results I get from IDL (which I am more familiar with) seem correct and intuitive, but the "lm" function in R gives outputs that seem strange to me. Unweighted case: > x<-1:4 > y<-(1:4)^2 > summary(lm(y~x)) Call: lm(formula = y ~ x) Residuals: 1 2 3 4 1 -1 -1 1 Coefficients:

Hi, I installed the package Hmisc with the command install.packages("Hmisc") without errors. When I try to load the library with command library(Hmisc) I get the error > library(Hmisc) Error in library(Hmisc) : there is no package called 'Hmisc' > version _ platform i386-pc-linux-gnu arch i386 os linux-gnu system i386,

Hi there, we are beginners in R and we are trying to fit the following time series using ar(2): > x <- c(1.89, 2.46, 3.23, 3.95, 4.56, 5.07, 5.62, 6.16, 6.26, 6.56, 6.98, > 7.36, 7.53, 7.84, 8.09) The reason of choosing the present time series is that the we have previously calculated analitically the autoregressive coefficients using the direct inversion method as 1.1, 0.765, 0.1173.

I have a data set like this: if I want to less than 200000 obs from this data set. How can I do these? > str(p1982) 'data.frame': 465979 obs. of 6 variables: $ p : Factor w/ 1982 levels "154l_aa","1A0P_aa",..: 1 1 1 1 1 1 1 1 1 1 ... $ aa : Factor w/ 19 levels "ALA","ARG","ASN",..: 2 16 4 5 18 3 19 3 2 9 ... $ as : num 152.0

Full_Name: Richard Huggins Version: 1.7.1 OS: windows 2000 Submission from: (NULL) (131.172.4.44) > x<-rnorm(100,2,1) > mean(x) [1] 1.73299 > summary(fivenum(x)) Min. 1st Qu. Median Mean 3rd Qu. Max. -0.3655 1.1070 1.7430 1.7320 2.3840 3.7910 > summary(x) Min. 1st Qu. Median Mean 3rd Qu. Max. -0.3655 1.1070 1.7430 1.7330 2.3830 3.7910 >

I have a data set like this. I want to do glm, but I get this error: Error in model.matrix.default(mt, mf, contrasts) : cannot allocate vector of length 932889958 I am wondering if my data set is too large or I did something wrong. Is there some limitation for data size for R? thanks, Aimin > p1982<- read.csv("p_1982_aa.csv") > names(p1982) [1] "p"

Greetings. I'm trying to recreate in R some regression models I've done in SAS, but I'm not getting the same results. My advisor suspects this may be due to differences in precision between R and SAS. Does anyone know where I can find specifications for R's type double? (It doesn't seem to be in the R Language Definition.) Thanks in advance for any help anyone can

Dear R users: I'm trying to generate a output file with fixed format using function "sprintf" in R. However, the execution time in R is very long even the toy data (smaller size df) seems to work fine. The syntax that I used is as follows: df.fmt <- sprintf("%2s%2s%2.4f", df$v1, df$v2, df$v3) write.table(df.fmt, output.name,...) The actual dataset is a df with the

Hi all there I am enjoying R since 2 weeks and I come to my first deadlock, il am trying to use predict.Arima in the ts package. I get a "Error in cbind(...) : cannot create a matrix from these types" -- Start R session ----------------------------------------------------- > fitdiv <- arima(data, c(2, 0, 3), xreg = y ) ; print(fitdiv) Call: arima(x = data, order = c(2, 0, 3),

Dear R-tisans, I am trying to calculate the 12th root of a transition (square) matrix, but can't seem to obtain an accurate result. I realize that this post is laced with intimations of quantitative finance, but the question is both R-related and broadly mathematical. That said, I'm happy to post this to R-SIG-Finance if I've erred in posting this to the general list. I've

Hi, I need a help. I am new in R and I need to run a nested anova with fixed and random factors (Mixed Model). I have a design with three factors: Day, Area and Plot and the dependent variable is density. The factors Day and Area are fixed while Plot is random, factor Area is nested in factor Day, and factor Plot is nested in Area. I can do it using aov by: mod1<-aov(density~ day +

Hola a todos! perdón por molestarlos nuevamente con los contrastes, pero estoy trantando de entender que es lo que está haciendo R y de donde vienen los valores que informa pero  no lo logro. Creí haberlo entendido pero a la hora de usar mis datos los resultados no dan como deberían. Tengo dos variables explicativas que son factores con 3 niveles cada uno. Esta es la tabla de medias de la

Dear all, I'm using R 2.8.1 under Windows Vista on a dual core 2,4 GhZ with 4 GB of RAM. I'm trying to reproduce a result out of "Analysis of Financial Time Series" by Ruey Tsay. In R I'm using the fGarch library. After fitting a ar(3)-garch(1,1)-model > model<-garchFit(~arma(3,0)+garch(1,1), analyse) I'm saving the results via > result<-model

Dear R users, I want to plot a one variable continuous function f(x) vs x, x=[0,1]. Say for example: f(x)= x^2. Now, using the command plot(f~x) I will get a curve where the range of x-axis is [0,1] with all equispaced label. But, I need something else, and that is: my curve will be such that 80% on x-axis the range would be [0,0.5] and the rest 20% would contain [0.5,1]. Let me draw informally

Dear R-listers, I am giving part of my R code : ########################################################### n=15 m=1 library("partitions") library("gregmisc") library("combinat") x = t(restrictedparts(n-m,m)) l = length(x[,1]) for(u in 1:l){ A= unique(matrix( unlist(permn(x[u,])), ncol=m, byrow=TRUE )) }

Hello, Or maybe logical_idx <- max_usage_hours_per_region$Region %in% input$Region Another option is ?match Hope this helps, Rui Barradas ?s 15:41 de 07/11/20, Jeff Newmiller escreveu: > This looks odd... > > max_usage_hours_per_region[input$Region,] > > This would only work if you had rownames on that data frame corresponding to the names of the Regions. This is a

Hi, I am running a simple linear model with (say) 5 independent variables. Is there a simple way of getting the variance-covariance matrix of the coeffcient estimates? None of the values of the lm() seem to provide this. Thanks in advance, Ritwik Sinha rsinha@darwin.cwru.edu Grad Student Case Western Reserve University [[alternative HTML version deleted]]

Dear Sir/madam, I'm getting a problem with a R-code which calculate Fisher Information Matrix for Hybrid Censored Weibull Distribution. My problem is that: when I take weibull(scale=1,shape=2) { i.e shape>1} I got my desired result but when I take weibull(scale=1,shape=0.5) { i.e shape<1} it gives error : Error in integrate(int2, lower = 0, upper = t) : the integral is probably