similar to: Arima contents

Hi, there Since glm cannot handle factors very well. I try to use bigglm like this: logit_model <- bigglm(responser~var1+var2+var3, data, chunksize=1000, family=binomial(), weights=~trial, sandwich=FALSE) fitted <- predict(logit_model, data) only var2 is factor, var1 and var3 are numeric. I expect fitted should be a vector of value falls in (0,1) However, I get something like this:

Hi all Suppose I do the following: set.seed(1000) library(xtable) x <- runif( 10 ) y <- 1 + 2*x + rnorm( length(x) ) test.lm <- lm( y ~ x ) summary( test.lm ) xtable ( summary( test.lm ) ) The final xtable output follows: % latex table generated in R 1.8.1 by xtable 1.2-2 package % Tue Jun 22 09:56:36 2004 \begin{table}[ht] \begin{center} \begin{tabular}{rrrrr} \hline &

anova.lm() gives the sequential tests: > set.seed(1) > dat <- data.frame(y=rnorm(10), x1=runif(10), x2=runif(10)) > anova(lm(y ~ x1 + x2, dat)) Analysis of Variance Table Response: y Df Sum Sq Mean Sq F value Pr(>F) x1 1 1.1483 1.1483 2.0943 0.1911 x2 1 0.4972 0.4972 0.9068 0.3727 Residuals 7 3.8383 0.5483 > anova(lm(y ~ x2 + x1,

Hello Lets say as an example I have a dataframe with the following attributes: rownum(1:405), colnum(1:287), year(2000:2009), daily(rownum x colnum x year) and foragePotential (0:1, by 0.01). The data is actually stored in a netcdf file and I'm trying to provide a conceptual version of the data. Ok. I need to calculate a moving mean and a moving variance for each cell on the following

Dear All, I'm trying to use the ols function in the Design library (version 2.1.1) of R to estimate parameters of a linear model, and then use the contrast function in the same library to test various contrasts. As a simple example, suppose I have three factors: feature (3 levels), group (2 levels), and patient (3 levels). Patient is coded as a non-unique identifier and is

Dear list and Martin, I'm testing different approaches to fit an electricity demand time series and come upon the fracdiff package (v 1.3-1) for fitting fractional ARIMA models. The following questions are motivated by this package. 1. Despite having a help page, the residuals and fitted functions don't seem to have implementation, or did i miss something obvious? Alternatively, having a

Hello Helpers, I have some problems with fitting the model for my data... -->my Literatur says (crawley testbook)= Non-normality of errors-->I get a banana shape Q-Q plot with opening of banana downwards Structure of data: origin wt pes gender 1 wild 5.35 147.0 male 2 wild 5.90 148.0 male 3 wild 6.00 156.0 male 4 wild 7.50 157.0 male 5 wild 5.90

Dear R users, I have a response variable in a csv file called "y" and a matrix of predictor variables in a csv file called "mat". I have used the function "nls" I have specified the nonlinear relation between these variable.The code I have witten is called Rprog which begins with the phrase: L.minor.m1<-nls(Y~a ....etc.. The program when I execute the program, I

Hi, this is not a new doubt, but is a doubt that I cant find a good response. Look this output: > m.lme <- lme(Yvar~Xvar,random=~1|Plot1/Plot2/Plot3) > anova(m.lme) numDF denDF F-value p-value (Intercept) 1 860 210.2457 <.0001 Xvar 1 2 1.2352 0.3821 > summary(m.lme) Linear mixed-effects model fit by REML Data: NULL AIC BIC

Hello R users, I have been using R for a while now for basic stats but I'm now trying to get my head around looping scripts and in some places I am failing! I have a data set with c. 1200 data points on 98 individual animals with data on each row representing a daily measure and I am asking the question "what variables affect the animal's behaviour?" the dataset includes

Hi. I have a Shuttle box with an AMD Athlon XP 2200+ and 1GB of RAM. I''m normally running it with Debian sarge/sid and kernel 2.6.10-1-k7, as built by Debian. I want to use Xen on it. I built a xen0 kernel which is as close to the Debian kernel as I can (no power management, no HPET timers, broken ISA drivers disabled), disabled /lib/tls, and booted with the new kernel. Everything works.

Hi List, I have spent more than 30 minutes, but failed to read in this file using the read.table() function. I could not figure out how to fix the following error. Error in scan(file, what, nmax, sep, dec, quote, skip, nlines, na.strings, : line 1 did not have 6 elements Any help would be be appreciated. Thanks, Pradip Muhuri ####### below is the reproducible example xd1 <-

Hi all, Using: R version 2.8.1 Patched (2009-03-07 r48068) on OSX (10.5.6) with survival version: Version: 2.35-3 Date: 2009-02-10 I get the following using the first example in ?summary.survfit: > summary( survfit( Surv(futime, fustat)~1, data=ovarian)) Call: survfit(formula = Surv(futime, fustat) ~ 1, data = ovarian) time n.risk n.event survival

On 04/26/2013 05:08 AM, tanmx_star wrote: > Hi all, Hi, thanks for the update and sorry for the delay in reviewing. I just had a look at your proposal. > I have updated my GSoS proposal: "FastPolly: Reducing LLVM-Polly Compiling overhead" (https://gist.github.com/tanstar/5441808). I think the pass ordering problem you discussed early can be also investigated in this project!

Hi Experts, I am new to R, and was wondering how to do 3D linear regression in R. In other words, I need to Fit a 3-Dimensional Line to Data Points (input). I googled before posting this, and found that it is possible in Matlab and other commercial packages. For example, see the Matlab link:

Hi, I am trying use lm() on some data, the code works fine but I would like to use a more efficient way to do this. The data looks like this (the data is very sparse with a few 1s, -1s and the rest 0s): > head(adj0708) MARGIN Poss P235 P247 P703 P218 P430 P489 P83 P307 P337.... 1 64.28571 29 0 0 0 0 0 0 0 0 0 0 0 0 0 2 -100.00000 6 0 0

Hi, I ran the following in R (on item means): aov(VAR ~(a*b)+Error(item/(a*b)), data = item) I got this result: Error: item Df Sum Sq Mean Sq F value Pr(>F) a 1 7.7249e+13 7.7249e+13 11.329 0.003934 ** Residuals 16 1.0910e+14 6.8187e+12 --- Signif. codes: 0 ?***? 0.001 ?**? 0.01 ?*? 0.05 ?.? 0.1 ? ? 1 Error: item:a Df Sum Sq

Hi all, Can anyone clear my doubts about what conclusions to take with the following what puts of some time series tests: > adf.test(melbmax) Augmented Dickey-Fuller Test data: melbmax Dickey-Fuller = -5.4075, Lag order = 15, p-value = 0.01 alternative hypothesis: stationary Warning message: p-value smaller than printed p-value in: adf.test(melbmax)

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

Hello all, I tried to extract the p-value from adf.test in tseries; however, I got the error message such as > ht=adf.test(list.var$aa) > ht$p-value Error in ht$p - value : non-numeric argument to binary operator > ht Augmented Dickey-Fuller Test data: list.var$aa Dickey-Fuller = -2.3147, Lag order = 4, p-value = 0.4461 alternative hypothesis: stationary > ht$data [1]