search for: 1.0000

dears R-users, I'm interested in model selection problem, and i have faced some problems that i would like to ask for help. well, this is a very small example with 4 variable (just one var. is the response - z) with 100 individuals i would like to do a stepwise search, for the "best" model, and a use BIC criteria. I know when I have a lot of variables, let's say 120, I know,

All, Given the following commands: > ag.m35<-read.table("m35.txt",header=TRUE,sep=",") > ag.m35.lp<-subset(ag.m35, race=="lp") > aov.m35.lp=aov(year~time,data=ag.m35.lp) > anova.m35.lp=anova(aov.m35.lp) > anova.m35.lp Analysis of Variance Table Response: year Df Sum Sq Mean Sq F value Pr(>F) time 1 11.56 11.5649 1.7282

Hello everyone, I am using *mlogit* to analyse my choice experiment data. I have *3 alternatives* for each individual and for each individual I have *9 questions*. I have a response from *516 individuals*. So it is a panel of 9*516 observations. I have arranged the data in long format (it contains 100 columns indicating different variables and identifiers). In mlogit I tried the following

[this message needed manual improvement by the mailing list administrator since it was `HTMLified' .. ``please do not''] Apologies for bothering you about a fairly trivial matter. I have been getting some inconsistencies with the display digits in R V1.5. I have been using the hypergeometric distribution function, and have found that when printing out the results from this

Hello, My name is Rodrigo, I am using R program and I have a trouble. I am trying to do a dendrogram with genetics information. Let me explain... The Similarity Matrix was already did, and with this matrix I want to construct a dendrogram. So, the distance is done. I need to transform this matrix (that I have) in a dendrogram, I woud be very grateful if someone could help me. PS: I am sending

Dear Forum, I have a question about interaction estimate in the Cox model: why the hazard ratio of the interaction is not produced in the summary of the model? (Instead, the estimate of the coefficient is given in the print of the model.) # Example: modINT <-cph( Surv(T_BASE, T_FIN,STATUS)~ NYHA + ASINI + RFP + FE_REC + XX_PR*XX_DISF) print(modINT) coef se(coef) z

Hello, I am attempting to train a random forest model using the randomForest package on 500,000 rows and 8 columns (7 predictors, 1 response). The data set is the first block of data from the UCI Machine Learning Repo dataset "Record Linkage Comparison Patterns" with the slight modification that I dropped two columns with lots of NA's and I used knn imputation to fill in other gaps.

Hi R users, Could anyone let me know how to export a paired t-test output table (see below) to an excel file? Jason, with(score2,pairwise.t.test(values,ind, + p.adjust.method="holm", paired=T)) Pairwise comparisons using paired t tests data: values and ind test1 test2 test3 test4 test5 test6 test1 1.0000 - - -

Hi R-helpers, I need to read some file with different lines (I don't know the number of lines to skip) and I would like to find a way to start reading the data.frame from the word "source". ex: djhsafk asdfhkjash shdfjkash asfhjkash #those lines contain numbers and words, I want to skip then but they have different sizes asdfhjkash asdfhjksa source tret 2 res 3 Can

Hi list, One of my co-workers found this problem with 'cor' in his code and I confirm it too (see below). He's using R 2.2.1 under Win 2K and I'm using R 2.3.0 under Win XP. =========================================== > R.Version() $platform [1] "i386-pc-mingw32" $arch [1] "i386" $os [1] "mingw32" $system [1] "i386, mingw32" $status

My data is called xc and has more than 15 variables. When I used summary(xc) it gave me the detail description of each variable. Summary(xc) Y1 x1 x2 x3 .. Min. :0.0000 Min. : 1.000 Min. : 1.000 Min. : 1.000 1st Qu. :0.0000 1st Qu.: 1.000 1st Qu.: 1.000 1st Qu.: 2.000 Median :1.0000 Median : 1.000

Hi, I downloaded both gee and geepack, and I am trying to understand the differences between the two libraries. I used the same data and estimated the same model, with a correlation structure autoregressive of order 1. Surprisingly for me, I found very different results. Coefficients are slightly different in value but sometimes opposite in sign. Moreover, the estimate of rho (correlation

I was unsure of what suitable goodness-of-fit tests existed in R for logistic regression. After searching the R-help archive I found that using the Design models and resid, could be used to calculate this as follows: d <- datadist(mydataframe) options(datadist = 'd') fit <- lrm(response ~ predictor1 + predictor2..., data=mydataframe, x =T, y=T) resid(fit, 'gof'). I set up a

Okay, I deleted most of this thread, so I was waiting for another message to respond to, so unfortunately this will be out of place in the thread. This is in response to Miroslav's idea about variable block sizes. I may be a bit out of my league here as I'm just starting to look at how the actual encoding gets done. But it seems to me that you could make a decent guess about when

Dear All: I run a logistic regression (using lrm in the Design package), and after that, I use the command "summary" to get the marginal effects of each variable. But one strange thing happens on my binary dependent variable: The marginal effect of it jumping from 0 to 1 is 1.77. I believe the marginal effect of binary variable x1 has interpretation should be P(Y=1|x1=1,

Hi I am trying to read selected fields from a xml file with R using xml package. So far I have learned the basics of this package by going through the manual, examples, tutorial, and so on (www.omegahat.org/RSXML) . The problem is that I am getting stuck when it comes down to more complex xml files. I am a novice in R and xml, and was wondering if someone could help me out with here.

Hello, I would like to perform a multivariate regression analysis to model the relationship between m responses Y1, ... Ym and a single set of predictor variables X1, ..., Xr. Each response is assumed to follow its own regression model, and the error terms in each model can be correlated. Based on my readings of the R help archives and R documentation, the function lm() should be able to

Hey guys, I'd really appreciate any help. I have a multivariate analysis done, the output of which is: > GraphData <-read.table("eigen.coa") > GraphData V1 V2 V3 V4 1 1 0.371970 0.8552 0.8552 2 2 0.061785 0.1420 0.9972 3 3 0.001211 0.0028 1.0000 4 4 0.000000 0.0000 1.0000 > summary(GraphData) V1 V2 V3

Dear Sirs, I am Professor at Indira Gandhi Krishi Vishwavidyalaya, Raipur, Chhattisgarh, India. While taking classes, I found the *by() *function producing following error when I use FUN=mean or median and some other functions, however, FUN=summary works. Given below is the output of the example I used on a built-in dataset "mtcars", along with error message reproduced herewith: >

--l4Siqd0eqV Content-Type: text/plain; charset=us-ascii Content-Transfer-Encoding: 7bit Hello, I run the following job. Please, compare the results of summary and table concerning berufl. From similar SPSS/PSPP runs, the result of table is correct. Did I misunderstand anything or is there a bug? What does the difference come from? What does '(other)' mean? What about the strange