similar to: tune svm

Dear experts, I just noticed that I get different results conducting type III anova using drop1 or the Anova command from the car package. I suppose I made a mistake and hope you can offer me some help. I have no idea where I got wrong and would be very grateful for explaination as R is new terrain for me. If I run the commands in line, they produce the same results. But if I run them in

Hi, I have a transition matrix T for which I want to find the steady state matrix for. This could be approximated by taking T^n , for large n. T= [ 0.8797 0.0382 0.0527 0.0008 0.0212 0.8002 0.0041 0.0143 0.0981 0.0273 0.8802 0.0527 0.0010 0.1343 0.0630 0.9322] According to a text book I have T^200 should have reached the steady state L L

Dear R-listers Is there anyone who knows why I get different eigenvectors when I run MatLab and R? I run both programs in Windows Me. Can I make R to produce the same vectors as MatLab? #R Matrix PA9900<-c(11/24 ,10/53 ,0/1 ,0/1 ,29/43 ,1/24 ,27/53 ,0/1 ,0/1 ,13/43 ,14/24 ,178/53 ,146/244 ,17/23 ,15/43 ,2/24 ,4/53 ,0/1 ,2/23 ,2/43 ,4/24 ,58/53 ,26/244 ,0/1 ,5/43) #R-syntax

Hi I am working with this data: my data summary is: > summary(spi) open high low close volume Min. :4315 Min. :4365 Min. :4301 Min. :4352 Min. : 0 1st Qu.:4480 1st Qu.:4497 1st Qu.:4458 1st Qu.:4475 1st Qu.:11135 Median :4609 Median :4631 Median :4594 Median :4614 Median :14439 Mean :4620

Dear R-Helpers, I want to compare the results of outputs from glmmPQL and lmer analyses. I could do this if I could extract the coefficients and standard errors from the summaries of the lmer models. This is easy to do for the glmmPQL summaries, using > glmm.fit <- try(glmmPQL(score ~ x*type, random = ~ 1 | subject, data = df, family = binomial), TRUE) > summary(glmmPQL.fit)$tTable

Hello, I have an input file: http://r.789695.n4.nabble.com/file/n3700031/testOut.txt testOut.txt where col 1 is chromosome, column2 is start of region, column 3 is end of region, column 4 and 5 is base position, column 6 is total reads, column 7 is methylation data, and column 8 is the strand. I would like a summary output file such as:

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My name is Giovanna and I am a PhD student in Norway. I am a beginner with statistics and R, hence my ignorance. Apologies from now..... I have been collecting data on time performances of 5 subjects using a 1:3 scale tower yarder. The task was consisting in yarding 5 small logs placed on permanently marked course. Four subjects had different previous experiences (None, Some) and the fifth was a

Hello all R-users _question 1_ I need to make a statistical model and respective ANOVA table but I get distinct results for the T-test (in summary(lm.object) function) and the F-test (in anova(lm.object) ) shouldn't this two approach give me the same result, i.e to indicate the same significants terms in both tests??????? obs. The system has two restrictions: 1) sum( x_i ) = 1 2) sum(

I am attempting to use package boot to summarize and compare the performance of three models. I'm using R 2.13.0 in a Win32 environment. My statistic function returns a vector of 6 values, 3 of which are error rates for different models, and 3 are pairwise differences between those error rates. It looks like: multiEst<-function(dat,i) { .... c(E1,E2,E3,E2-E1,E3-E1,E3-E2); }

Hello, I probably have a syntax error in trying to generate an expected survival curve from a weighted cox model, but I can't see it. I used the help sample code to generate a weighted model, with the addition of a "weights=albumin" argument (I only chose albumin because it had no missing values, not because of any real relevance). Below are my code with the resulting error

This nightly tester is now using an llvm-g++ that produces the new ODR linkage types. This means that many more functions are being considered by the inter-procedural optimization passes (for example, "linkonce" functions defined in a header). The result seems to be pretty huge swings (both good and bad) in the C++ tests in the testsuite, see below. Note that this tester is often

before I accuse somebody to "overbill" I would like you to calculate with me: Rate: 0.0189 for calling Taiwan via NuFone Duration: 930 seconds Lets vote for the answers: 0.7269 or 0.2929 ??? bye Ronald Wiplinger

Hello , I have estimated the following model, a sarima: p=9 d=1 q=2 P=0 D=1 Q=1 S=12 In R 2.12.2 Call: arima(x = xdata, order = c(p, d, q), seasonal = list(order = c(P, D, Q), period = S), optim.control = list(reltol = tol)) Coefficients: ar1 ar2 ar3 ar4 ar5 ar6 ar7 ar8 ar9 0.3152 0.8762 -0.4413 0.0152 0.1500 0.0001 -0.0413 -0.1811

Hi all R-users, I?m trying to assign colors on those p-value in my table output that fall above a certain critical value, let?s say a p-value >0.05. My table looks like this: Assets ADF-Level P-Value ADF-First D P-Value ADF-Second D P-Value [1,] Liabilities -2.3109 0.1988 -3.162 0.025 -6.0281

Hi, I have just tagged 3.7.1-rc1, so it is ready for testing. As a reminder, when doing regression testing, use the 3.7.0 release as your baseline. Thanks, Tom

Dear list, I would like to simulate individual survival times from a model that has been fitted using the survreg procedure (library survival). Output shown below. My plan is to extract the shape and scale arguments for use with rweibull() since my error terms are assumed to be Weibull, but it does not make any sense. The mean survival time is easy to predict, but I would like to simulate

Hi, I am trying to fit a model with lmer in R and proc glimmix in SAS. I have simplified my code but I am surprised to see I get different results from the two softwares. My R code is : lmer(y~age_cat + (1|cat),data=fic,family=binomial(link = "logit"), NaGQ=1) My SAS code is : ods output Glimmix.Glimmix.ParameterEstimates=t_estimates; proc glimmix data=tab_psi method=laplace;

Hello, I'm new here, but will try to be as specific and complete as possible. I'm trying to use “lm“ to first estimate parameter values from a set of calibration measurements, and then later to use those estimates to calculate another set of values with “predict.lm”. First I have a calibration dataset of absorbance values measured from standard solutions with known concentration of

Hello , I have estimated the following model, a sarima: p=9 d=1 q=2 P=0 D=1 Q=1 S=12 In R 2.12.2 Call: arima(x = xdata, order = c(p, d, q), seasonal = list(order = c(P, D, Q), period = S), optim.control = list(reltol = tol)) Coefficients: ar1 ar2 ar3 ar4 ar5 ar6 ar7 ar8 ar9 0.3152 0.8762 -0.4413 0.0152 0.1500 0.0001 -0.0413 -0.1811