similar to: Objects in Views

Hi, Does anyone know a way to override .blank? method that rails provides. I want to add additional custom checks (specific to my application) to this method. Thanks, Pratik

Hello all,will_paginate don''t work on jruby platform, the Model.paginate find works ok, accepts all kinds of params, this is config.gem config.gem ''will_paginate'', :version => ''2.3.11'', :source => ''http://gemcutter.org'', but <%= will_paginate @collection %> doesn''t work it always produce nil(so nothing on page).

I have application that i send email normal with native ruby. But when i execute the sample application with jruby, it dont send email and dont get error. What can be wrong? -- Atenciosamente, Paulo Coutinho. Blog: www.prsolucoes.com/blog Site: www.prsolucoes.com Msn: paulo-QE/7f1ia5mR0ubjbjo6WXg@public.gmane.org -- You received this message because you are subscribed to the Google Groups

Dear R users, i'm using a custom function to fit ancova models to a dataset. The data are divided into 12 groups, with one dependent variable and one covariate. When plotting the data, i'd like to add separate regression lines for each group (so, 12 lines, each with their respective individual slopes). My 'model1' uses the group*covariate interaction term, and so the coefficients

I was just wondering about locales and .yml files. Is it better to store the multilanguage strings in .yml files than in databases? And if yes, why? I was also wondering how rails are loading this files (for example, I have 4 languages in my web app, each has her own .yml file, will my rails app loads all the files in ram and then it will call each variable inside my web app? Or something else?)

Hello R-Help, ----------------------------------------------------------------------------------------------------------------------------------------- Issues (there are 2): 1) Possible bug when using lmtest::encomptest() with a linear model created using nlme::lmList() 2) Possible modification to lmtest::encomptest() to fix confusing fail when models provided are, in fact, nested. I have

Full_Name: Menelaos Stavrinides Version: 1.3. 1 OS: Windows 98 Submission from: (NULL) (193.129.76.90) When model simplification is used in glm (binomial errors) and anova is used two compare two competitive models one can use either an "F" or a "Chi" test. R always performs an F test (Although when test="Chi" the test is labeled as Chi, there isn't any

API-> hello_message_api.rb. class HelloMessageApi < ActionWebService::API::Base api_method :hello_message, :expects => [{:firstname=>:string}, {:lastname=>:string}], :returns => [:string] end controller -> class HelloMessageController < ApplicationController web_service_api HelloMessageApi web_service_dispatching_mode :direct wsdl_service_name

Dear r-helpers, Spencer Graves and Manual Morales proposed the following methods to simulate p-values in lme4: ************preliminary************ require(lme4) require(MASS) summary(glm(y ~ lbase*trt + lage + V4, family = poisson, data = epil), cor = FALSE) epil2 <- epil[epil$period == 1, ] epil2["period"] <- rep(0, 59); epil2["y"] <- epil2["base"]

Hello all, Actually i am having a small problem.. googled it but did''nt find any clue.. So what i want to do is that i want to insert ruby variable in css style property.. as like .. <div class="abc" style="left: <%= @var * 6 %> px;"> But this is not working.. I am doing it because i want to keep the left property dynamic .. :) Is there any possible way to

Hi, I'm trying to compare models, one of which has all parameters fixed using offsets. The log-likelihoods seem reasonble in all cases except the model in which there are no free parameters (model3 in the toy example below). Any help would be appreciated. Cheers, Jarrod x<-rnorm(100) y<-rnorm(100, 1+x) model1<-lm(y~x) logLik(model1) sum(dnorm(y, predict(model1),

I am using anova.lm to compare 3 linear models. Model 1 has 1 variable, model 2 has 2 variables and model 3 has 3 variables. All models are fitted to the same data set. anova.lm(model1,model2) gives me: Res.Df RSS Df Sum of Sq F Pr(>F) 1 135 245.38 2 134 184.36 1 61.022 44.354 6.467e-10 *** anova.lm(model1,model2,model3) gives

Hello, I have a problem concerning logistic regressions. When I add a quadratic term to my linear model, I cannot draw the line through my scatterplot anymore, which is no problem without the quadratic term. In this example my binary response variable is "incidence", the explanatory variable is "sun": > model0<-glm(incidence~1,binomial) >

Hello, I want to understand how to tell if a model is significant. For example I have vectX1 and vectY1. I seek first what model is best suited for my vectors and then I want to know if my result is significant. I'am doing like this: model1 <- lm(vectY1 ~ vectX1, data= d), model2 <- nls(vectY1 ~ a*(1-exp(-vectX1/b)) + c, data= d, start = list(a=1, b=3, c=0)) aic1 <- AIC(model1)

This has got to be incredibly simple but I nevertheless can't figure it out as I am apparently brain dead. I just want to convert the elements of a character vector to variable names, so as to then assign formulas to them, e.g: z = c("model1","model2"); I want to assign formulas, such as lm(y~x[,1]) and lm(y~x[,2]), to the variables "model1" and

Hello, I have a simple theorical question about regresion... Let's suppose I have this: Model 1: Y = B0 + B1*X1 + B2*X2 + B3*X3 and Model 2: Y = B0 + B2*X2 + B3*X3 I.E. Model1 = lm(Y~X1+X2+X3) Model2 = lm(Y~X2+X3) The Ajusted R-Square for Model1 is 0.9 and the Ajusted R-Square for Model2 is 0.99, among many other significant improvements. And I want to do the anova test to choose the best

Dear All I am attempting some power analyses, based on simulated data. My experimental set up is thus: Bleach: main effect, three levels (control, med, high), Fixed. Temp: main effect, two levels (cold, hot), Fixed. Main effect interactions, six levels (fixed) For each main-effect combination I have three replicates. Within each replicate I can take varying numbers of measurements (response

Dear all, I am analysing growth data - response variable - using GAM and GAMM models, and 4 covariates: mean size, mean capture year, growth interval, having tumors vs. not The models work fine, and fit the data well, however when I try to compare models using AIC I cannot get an AIC value. This is the code for the gam model:

Dear community, I've check it while working, but just to reassure myself. Let's say we have 2 models: model1 <- lm(vdep ~ log(v1) + v2 + v3 + I(v4^2) , data = mydata) model2 <- lm(vdep ~ log(v1) + v2 + v3 + v4^2, data = mydata) So in model1 you really square v4; and in model2, v4*^2 *doesn't do anything, does it? Model2 could be rewritten: model2b <- lm(vdep ~

Hello everyone, Sorry if my question is not clear, my first language is not English, but Portuguese. I am building a model for my data, using non-binomial error. I am having a bit of a problem when updating the model to remove parameters that I no do no autocorrelate with other variables (I have used a autocorrelation function for this). So my first model looks like this: