similar to: Creating Better Table in R

Hi, I would really appreciate all the help I can get. Unfortunately, I am really new to statistics! I hope you guys don't mind this. I am trying to find significance levels, beta, R, R squared, adjusted R squared, standard error and t test. FILE http://r.789695.n4.nabble.com/file/n4541923/datpat.csv datpat.csv Variables (written exactly as in the Excel file) I am trying to examine are :

Hi, this is related to academic research I am trying to conduct. Please pardon my lack of socialization for this forum. For my project, I had to combine two different datasets, Democracy dataset from Pippa Norris and World Bank Patents dataset. My issue arrises from just loading the file into R. My colleagues proficient in R have been stumped as well. Often times the file would seem to load

I'm using the fixed argument in arima. Shouldn't ar4, ar5, and ar6 display as zero in the output? Call: arima(x = window(log(hhprice), start = c(1990, 1), end = c(2003, 3)), order = c(7, 1, 0), xreg = window(ts.union(exa1 = lag(exa, -1), exa12 = lag(exa, -12), exb1 = lag(exb, -1), exc1 = lag(exc, -1), exc12 = lag(exc, -12)), start = c(1990, 1), end = c(2003, 3)),

Dear R-List users, Can anyone explain exactly the difference between Weights options in lm glm and gls? I try the following codes, but the results are different. > lm1 Call: lm(formula = y ~ x) Coefficients: (Intercept) x 0.1183 7.3075 > lm2 Call: lm(formula = y ~ x, weights = W) Coefficients: (Intercept) x 0.04193 7.30660 > lm3 Call:

Dear All, Hans Borchers and I have been trying to compute "exact" derivatives in R using the idea of complex-step derivatives that Hans has proposed. This is a really, really cool idea. It gives "exact" derivatives with only a minimal effort (same as that involved in computing first-order forward-difference derivative). Unfortunately, we cannot implement this in R as the

Hi, Sorry for the delay in response. I measured the code size change and noticed small changes in both directions for individual programs. I found a 30k binary size growth for the entire testsuite + SPEC. I attached an updated performance report that includes both compile time and performance measurements. Thanks, Nadav On Jul 14, 2013, at 10:55 PM, Nadav Rotem <nrotem at apple.com>

Dear Mr. and Mrs. I'm very grateful for these software and this list. My question is: when a use linear multiple regression (lm()) for my data, abundance ichthyoplankton ~ salinity + temperature + month of the year(f1 is a factor: 1 for january, 2 for february, ..., 12 for december), the summary() of results is ... Coefficients: Estimate Std. Error t value Pr(>|t|)

Suppose I have a square matrix P P <- matrix(c(.3,.7, .7, .3), ncol=2) I know that > P * P Returns the element by element product, whereas > P%*%P Returns the matrix product. Now, P^2 also returns the element by element product. But, is there a slick way to write P %*% P %*% P Obviously, P^3 does not return the result I expect. Thanks, Harold [[alternative HTML version

Greetings, I'd be grateful if a good Samaritan helps me to approach this problem.... with my data, I've created the following model lm(formula = OUTCOME ~ VAR1 + VAR2) summary(model) Call: lm(formula = OUTCOME ~ VAR1 + VAR2) Residuals: Min 1Q Median 3Q Max -1.4341 -0.3621 0.1879 0.4994 0.7696 Coefficients: Estimate Std. Error t value Pr(>|t|) (Intercept) 1.89020

Hi, all. I want to write some functions like glm() so i studied it. In glm.fit(), it calls a fortran subroutine named "dqrfit" to compute least squares solutions to the system x * b = y To learn how "dqrfit" works, I just follow how glm() calls "dqrfit" by my own example, my codes are given below: > qr <- >

Hi, the following strange error appears when I use qda: > qda1 <- qda(as.data.frame(mfilters[cvtrain,]),as.factor(traingroups)) Error: function is not a closure That's also strange: > qda1 <- qda(mfilters[cvtrain,],as.factor(traingroups)) Error in qda.default(mfilters[cvtrain, ], as.factor(traingroups)) : length of dimnames must match that of dims Some backgroud: >

Greetings, Following the example in help(predict.lm): x <- rnorm(15) y <- x + rnorm(15) new <- data.frame(x = seq(-3, 3, 0.5)) predict(lm(y ~ x), new) predicts the response elements corresponding to new$x as can be viewed by: plot(x,y) lines(new$x,predict(lm(y ~ x), new)) I am trying to extend this fitting and prediction over a variety of factors as

Hello: I am doing forward stepwise analysis on the glm model. I am trying to use model operator in the "upper" scope formula, for example, scope=list(lower=~1,upper=~ .^2) but the upper bound of the scope seems to be ignored and add1 is not performed at all, while if the terms are explicitly listed in the formula, the step function seems to work

This command cdmoutcome<- glm(log(value)~factor(year) > +log(gdppcpppconst)+log(gdppcpppconstAII) > +log(co2eemisspc)+log(co2eemisspcAII) > +log(dist) > +fdiboth > +odapartnertohost > +corrupt > +log(infraindex) > +litrate > +africa >

On Jul 14, 2013, at 9:52 PM, Chris Lattner <clattner at apple.com> wrote: > > On Jul 13, 2013, at 11:30 PM, Nadav Rotem <nrotem at apple.com> wrote: > >> Hi, >> >> LLVM’s SLP-vectorizer is a new pass that combines similar independent instructions in a straight-line code. It is currently not enabled by default, and people who want to experiment with it

Greetings all The help file for GAMM in mgcv indicates that the log likelihood for a GAMM reported using summary(my.gamm$lme) (as an example) is not correct. However, in a past R-help post (included below), there is some indication that the likelihood ratio test in anova.lme(mygamm$lme, mygamm1$lme) is valid. How can I tell if anova.lme results are meaningful (are AIC, BIC, and logLik

Hi, Is there a function for raising a matrix to a power? For example if you like to compute A%*%A%*%A, is there an abbreviation similar to A^3? Atte Tenkanen > A=rbind(c(1,1),c(-1,-2)) > A [,1] [,2] [1,] 1 1 [2,] -1 -2 > A^3 [,1] [,2] [1,] 1 1 [2,] -1 -8 But: > A%*%A%*%A [,1] [,2] [1,] 1 2 [2,] -2 -5

Hello, I need help with a partial least square regression in R. I have read both the vignette and the post on R bloggers but it is hard to figure out how to do it. Here is the script I wrote: library(pls) plsrcue<- plsr(cue~fb+cn+n+ph+fung+bact+resp, data = cue, ncomp=7, na.action = NULL, method = "kernelpls", scale=FALSE, validation = "LOO", model = TRUE, x = FALSE, y =

Hello All, I have posted to this list before regarding the same issue so I apologize for the multiple e-mails. I am still struggling with this issue so I thought I'd give it another try. This time I have included reproducible code and a subset of the data I am analyzing. I am running an ANOVA with three factors: GROUP (5 levels), FEATURE (2 levels), and PATIENT (2 levels), where

I have two variables which show a typical quantile relation I would like to fit quantile regression models based on both logarithm and polynomial of second order functions within quantreg. Any help appreciated... Cheers Duccio Herewith the values: ------ var1 var2 0.96429 0.00138 1 0.02316 1.03145 0.09323 1.24088 0.77128 1.39869 0.86732 1.33728 0.63674 1.48299 0.96194