similar to: memory and step()

Dear R-ers, I have a data frame data with predictors x1 through x5 and the response variable y. I am running a simple regression: reg<-lm(y~x1, data=data) I would like to loop through all predictors. Something like: predictors<-c("x1","x2",... "x10) for(i in predictors){ reg<-lm(y~i) etc. } But it's not working. I am getting an error: Error in

Hi R users, Is there an easy way in R to generate the results table below using table 1 and the formula (simplified version of the real problem)? It would be easy if I knew the R equivalent of SAS's retain function, but could not find one. Thanks in Advance for any help! table1: ID X2 X3 1.00 1.00 0 2.00 0.00 3.00 1.00 4.00 3058 5.00 0.00 6.00 6.00 Formula: X3 = x2 + (.24 *

Full_Name: Jerome Asselin Version: 1.3.1 OS: MacOS 9.2 Submission from: (NULL) (142.103.173.46) The step() function attempts to calculate the deviance of fitted models even if does not really need it. As a consequence, the step() function gives an error when it is used with coxph(). (There is currently no method to calculate the deviance of coxph() fits.) The code below gives an example of how

Dear R-users I applied vegan's varpart function to partition the effects of explanatory matrices. Adj. R square for the unique fraction [a] is 0.25. Does anyone know why the decomposition by hand using rda gives me a different result for [a] (constrained proportion is 0.32)? I used cbind() for the conditional fractions, but it should be similar to condition()? Thanks very much

Hello All, I need help with my dataframe, it is big but here I am using a small table as an example. My dataframe df looks like: X1 X2 X3 1 2011-02 0.00 96.00 2 2011-02 0.00 2.11 3 2011-02 2.00 3.08 4 2011-02 0.06 2.79 5 2011-02 0.00 96.00 6 2011-02 0.00 97.00 7 2011-02 0.08 2.23 I want values in columns X2 and X3 to be checked if they are greater than

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,

Hello all, I am trying to do factorial regression using lm() like this (example): model<-lm(y ~ x1 + x2 + x3 + x4 + x1*x2*x3*x4) The final term 'x1*x2*x3*x4' adds all possible interactions between explanatory variables to the model. i.e. x1:x2, x1:x2:x3, etc, etc. Now, the issue is that some of the interactions are significant and some are not. I can manually remove

Hello Lady or Sir, I want to do model selection with 'step' function. But there is a trouble to me. For example, there are 5 independent variables, X0,X1,X2,X3 and X4; and 1 dependent variable y. I want to run stepwise regression y ~ X0 + X1 + X2 + X3 + X4, and I want to keep X0 always in the model. Every time when I run command 'step' with 'keep', error information

Hi, I am trying to fit a logistic regression using glm, but my explanatory variables are of mixed type: some are numeric, some are ordinal, some are categorical, say If x1 is numeric, x2 is ordinal, x3 is categorical, is the following formula OK? *model <- glm(y~x1+x2+x3, family=binomial(link="logit"), na.action=na.pass)* * * *Thanks,* * * *-Jack* [[alternative HTML version

hello, I am looking for a step() function for GAM's. In the book Statistical Computing by Crawley and a removal of predictors has been done "by hand" model <- gam(y ~s(x1) +s(x2) + s(x3)) summary(model) model2 <- gam(y ~s(x2) + s(x3)) # removal of the unsignificant variable #then comparing these two models if an significant increase occurs. anova(model, model2,

Dear R-users I would like to apply classification and regression tree(CART) to the following data. I have some question on using 'tree' package. The data contains one response variable Y and five explanatory variables. The explanatory variable "x2" is categorical and not ordinal. But, the result obtained after running following R code has indicated that x2 is regard as

If I have a model line = lm(y~x1) and I want to use a for loop to change the number of explanatory variables, how would I do this? So for example I want to store the model objects in a list. model1 = lm(y~x1) model2 = lm(y~x1+x2) model3 = lm(y~x1+x2+x3) model4 = lm(y~x1+x2+x3+x4) model5 = lm(y~x1+x2+x3+x4+x5)... model10. model_function = function(x){ for(i in 1:x) { } If x =1, then the list

Dear R users, I have a data where I desire to subset according to certain conditions. However, the script is very messy as there are about 30 distinct conditions. (i.e. same script but with different conditions) I would like to make a user defined function so that I can input the desired conditions and just get the results accordingly. Below is an arbitrary data set & sample statements

Thanks for your reply. I use mvrnorm from the *MASS* package and lmrob from the *robustbase* package. To further explain my data generating process, the idea is as follows. The explanatory variables are generated my a multivariate normal distribution where the covariance matrix of the variables is defined by Sigma in my code, with ones on the diagonal and rho = 0.15 on the non-diagonal. Then y

What is 'd'? What is 'n'? On Sun, Mar 4, 2018 at 12:14 PM, Christien Kerbert < christienkerbert at gmail.com> wrote: > Thanks for your reply. > > I use mvrnorm from the *MASS* package and lmrob from the *robustbase* > package. > > To further explain my data generating process, the idea is as follows. The > explanatory variables are generated my a

The partial R-square (or coefficient of partial determination, or squared partial correlation coefficients) measures the marginal contribution of one explanatory variable when all others are already included in multiple linear regression model. The following link has very clear explanations on partial and semi-partial correlation: http://www.psy.jhu.edu/~ashelton/courses/stats315/week2.pdf In

Given an x-vector with, say, 3 elements, I would like to compute the following vector of permutated products (1-x1)*(1-x2)*(1-x3) (1-x1)*(1-x2)*x3 (1-x1)*x2*(1-x3) x1*(1-x2)*(1-x3) (1-x1)*x2*x3 x1*(1-x2)*x3 x1*x2*(1-x3) x1*x2*x3 Now, I already have the correctly sorted matrix of permutations! So, the input looks something like: #input x<-c(0.3,0.1,0.2) Nx<-length(x) Ncomb<-2^Nx

Dear list members, I want to perform an MM-regression. This seems an easy task using the function lmrob(), however, this function provides me with NA coefficients. My data generating process is as follows: rho <- 0.15 # low interdependency Sigma <- matrix(rho, d, d); diag(Sigma) <- 1 x.clean <- mvrnorm(n, rep(0,d), Sigma) beta <- c(1.0, 2.0, 3.0, 4.0) error <- rnorm(n = n,

d is the number of observed variables (d = 3 in this example). n is the number of observations. 2018-03-04 11:30 GMT+01:00 Eric Berger <ericjberger at gmail.com>: > What is 'd'? What is 'n'? > > > On Sun, Mar 4, 2018 at 12:14 PM, Christien Kerbert < > christienkerbert at gmail.com> wrote: > >> Thanks for your reply. >> >> I use

He, I used the "coxph" function, with four covariates. Let's say something like that > model.1 <- coxph(Surv(Time,Event)~X1+X2+X3+X4,data=DATA) So I obtain the 4 coefficients B1,B2,B3,B4 such that h(t) = h0(t) exp(B1*X1+ B2*X2 + B3*X3 + B4*X4). When I use the function on the same data > predict.coxph(model.1,type="lp") how it works in making the prediction?