search for: 0.198

Hi folks, attached you can find a tiny patch for klibc-0.198 to get rid of the symbolic link pointing to the kernel sources, and to get rid of some obsolete rebuilds during incremental. Feel free to include it. Regards Harri -------------- next part -------------- diff -ur old/klibc-0.198/MCONFIG klibc-0.198/MCONFIG --- old/klibc-0.198/MCONFIG 2004-10-14 06:32:24.000000000 +0200 +++

Hi, In the following results I interpret exp(coef) as the factor that multiplies the base hazard rate if the corresponding variable is TRUE. For example, when the bucket is ks008 and fidelity <= 3, then the rate, compared to the base rate h_0(t), is h(t) = 0.200 h_0(t). My question is then, to what case does the base hazard rate correspond to? I would expect the reference to be the first

Hola, ¿qué tal? Tienes que escribir a la dirección r-help-es en r-project.org, no a las que has usado (que son de administración de la lista). Un saludo, Carlos J. Gil Bellosta http://www.datanalytics.com El día 3 de septiembre de 2013 12:39, Jose Betancourt B. <betanster en gmail.com> escribió: > Quisiera saber en el paquete Epiestim > > Como lograr concatenar dos vectores,

I just tried the rsync version of R-1.1.0 on one of my alphas: It compiles without problems (gcc/g77 2.95.2, system is DU4.0E) but make check stops in base-Ex.R at > X <- cbind(1, 1:7) > str(s <- svd(X)); D <- diag(s$d) List of 3 $ d: num [1:2] 12.07 1.16 $ u: num [1:7, 1:2] -0.0976 -0.1788 -0.2601 -0.3413 -0.4225 ... $ v: num [1:2, 1:2] -0.198 -0.980 -0.980 0.198 >

Dear list-members I am new to R and a statistics beginner. I really like the ease with which I can extract and manipulate data in R, and would like to use it primarily. I've been learning by checking analyses that have already been run in SAS. In an experiment with Y being a response variable, and group a 2-level between-subject factor, and time a 5-level within-subject factor. 2

Dear R helpers, I have been using the loadings function from the multiv library and I get the typical output (see below). When I try to export these results to a file using a write.table() I get the following error message "Error in as.data.frame.default(x[[i]], optional = TRUE) : can't coerce loadings into a data.frame" Any idea why write.table is doing that and any

I have a data frame, lets call it dat, with 3 columns ( mc, yr, ret) which represent market cap, year, and return. mc is a factor, mc, and ret are real numbers. I want to add a column to the data calculated as follows. For each year, I want to split the data by mc decile, then calculate the mean ret within that mc decile, and finally subtract that year's decile mean from the raw return. Then

Dear R-friends, For the following data: > xy x y i 1 731 0.313 2 2 739 0.340 2 3 790 0.373 2 4 855 0.451 2 5 980 0.608 2 6 575 0.156 1 7 608 0.207 1 8 630 0.249 1 9 670 0.332 1 10 838 0.377 1 11 964 0.466 1 > coplot(y ~ x|i, data=xy) coplot gives 3 panels, rather than 2, namely one for i=1 and one for i=2. Futhermore, when I extand data fram xy to have i=3 as follows:

I am using rpart in classification mode and am confused about this particular model's predictions. > predict(fit, train[8,]) -1 1 8 0.5974089 0.4025911 > predict(fit, train[8,], type="class") 1 Levels: -1 1 So, it seems like there is a 60% change of being class -1 according the the "prob" output (which is the default for classification) but gives

Dear R-listers, I am an MD and clinical epidemiologist developing a measure of comorbidity severity for patients with liver disease. Having developed my comorbidity score as the linear predictor from a Cox regression model I want to compare the discriminative ability of my comorbidity measure with the "old" comorbidity measure, Charlson's Comorbidity Index. I have nearly 10,000

I am reposting my message from April 8th because I never received a response to the original post: Dear R-listers, I am an MD and clinical epidemiologist developing a measure of comorbidity severity for patients with liver disease. Having developed my comorbidity score as the linear predictor from a Cox regression model I want to compare the discriminative ability of my comorbidity measure with

Hi I've managed to compile klibc 0.198 and run the hello world program. How can I compile my own program? Can someone give an example or provide a simple wrapper program for doing this? like uclibc did with their earlier versions. I looked at the make process and did a bit reading and figure out this much: gcc -Wp,[preprocessor options] -MT,[target_name_I_think] tests/hello.o,

I am a total newbie at R but experienced with computers. If this is not the right forum for this question, please let me know one that is. I searched in the R manuals in the Help section, the Nabble help and Rseek but no luck. I am trying to install the package gcl on my recently installed R 2.8.1 I searched on cran for the windows version and found a zip file for the package In RGui, I ran

Hi r-users,   I would like to extract the data that match.  Attached is my data: I'm interested in matchind the value in column 'intg' with value in column 'rand_no' > cbind(z=z,intg=dd,rand_no = rr)             z  intg rand_no    [1,]  0.00 0.000   0.001    [2,]  0.01 0.000   0.002    [3,]  0.02 0.000   0.002    [4,]  0.03 0.000   0.003    [5,]  0.04 0.000   0.003    [6,] 

Dear R community members, I think I am asking a very simple question, but I really looked up in the faqs and manuals and found nothing helpful. I am trying to plot a data frame with the following structure (this is just a small extract): glo conc odor line series X1 X2 X3 X4 X5 X6 X7 X8 X9 X10 X11 X12 X13 1 0 AIR LN1 UP -0.488

Dear R users I have a big matrix like 6021 1188 790 290 1174 1015 1990 6613 6288 100714 6021 1 0.658 0.688 0.474 0.262 0.163 0.137 0.32 0.252 0.206 1188 0.658 1 0.917 0.245 0.331 0.122 0.148 0.194 0.168 0.171 790 0.688 0.917 1 0.243 0.31 0.122 0.15 0.19 0.171 0.174 290 0.474

Hola Jose, si CONCATENAR significa APILAR, es decir, concantenar verticalmente, por decirlo de algun modo, podrias hacerlo con rbind(): nuevovector <- rbind(vector1,vector2) Si ademas quieres que cada valor de los vectores originales sea identificado en el nuevovector, puedes usar: nuevovector <- stack(vector1,vector2) en este ultimo caso se agrega una columna adicional tipo factor, con

I have these questions: (1) Use Poisson regression to estimate the main effects of car, age, and dist (each treated as categorical and modelled using indicator variables) and interaction terms. (2) It was determined by one study that all the interactions were unimportant and decided that age and car could be treated as though they were continuous variables. Fit a model incorporating these

Hello everybody, I am sorry if my questions are too simple or not easily understandable. I’m not a native English speaker and this is my first analysis using this function. I have a problem with a cross correlation function and I would like to understand how I have to perform it in R. I have yearly data of an independent variable (x) from 1982 to 2010, and I also have yearly data of a variable

Hi R users: What is the difference between in the computation of the diag of the "hat" matrix in: "lm.influence" and the matrix operations with "solve()" and "t()"? I mean, this is my X matrix x1 x2 x3 x4 x5 [1,] 0.297 0.310 0.290 0.220 0.1560 [2,] 0.360 0.390 0.369 0.297 0.2050 [3,] 0.075 0.058 0.047 0.034 0.0230 [4,] 0.114 0.100