similar to: subset function

Dear R-Users, How can I put a pre-defined regression model into to an object of class lm in order to use the predict.lm function. A simplified example: I would normally run a regression analysis on a dataset, > germany<-lm(RENT~AGE1, in.mi01) > summary(germany) Call: lm(formula = RENT ~ AGE1, data = in.mi01) Residuals: Min 1Q Median 3Q Max

This is probably a simple problem but I don't see a solution. I have a data.frame with a number of columns where I would like 0 <- NA thus I have df1[,144:157] <- NA if df1[, 144: 157] ==0 and df1[, 190:198] <- NA if df1[, 190:198] ==0 but I cannot figure out a way do this. cata <- c( 1,1,6,1,1,NA) catb <- c( 1,2,3,4,5,6) doga <- c(3,5,3,6,4, 0) dogb <- c(2,4,6,8,10,

Dear R-Users, I used the box.tidwell () function of the car Package. When I used the following formula: semi.sub.in.mi1.boxtidwell_h<-box.tidwell(RENT_LG ~ I(age+1)+I(age2+1)+X06A + I(X08B+1) + I(X22+1) + I(X24+1) + X31A, ~B_YEAR + C_X01 + C_X14 + C_X19 + C_X29A +C_X21 + C_X23 + D_X12 + D_X17 + D_X18 + D_X25 + D_X27 + D_X30 + D_X32 + D_X35, data = semi.sub.in.mi1) everything is

Dear all here is my data called "rglp" structure(list(vzorek = structure(1:17, .Label = c("179/1/1", "179/2/1", "180/1", "181/1", "182/1", "183/1", "184/1", "185/1", "186/1", "187/1", "188/1", "189/1", "190/1", "191/1", "192/1",

Hi all, I'm having some trouble with rbind - this may be a bug or it may be my misunderstanding. If I do fileName <- paste(tempdir(),"test.txt",sep="/") file.create(fileName) x <- read.table(fileName, col.names=c("one","two","three")) I get a data frame with no rows, as documented. If I then try to rbind this with another data frame

Hello, I want to create a new variable which includes 4 age categories in this way: if (age>=12 && age<32) age1==1 if (age>=32 && age<52) age1==2 if (age>=52 && age<72) age1==3 if (age>=72 && age<100) age1==4 but I get the results only for the first observation. how can I apply this condition to all observations? Thanks in advance,

Hi I have two processes which take with a certain probability (p1 and p2) x number of years to complete (age1 and age2). As soon as thge first process is completed, the second one begins. I want to calculate the time it takes for the both processes to be completed. I have the following script which gives me the answer, butI think there must be a more elegant way of doing the calculations

Dear list I'm using the reshape command and want to reshape a wide data set to a long one e.g. I have the variables y1,y2,y3,age1,age2,age3,sex,ethnic I want my new long data set to consist of the variables y (which has been created from y1,y2,y3), age (which has been created from age1,age2,age3), sex and ethnic I have tried to use the command:

Hi there, I have a rather large data set and perform the following cox model: test1 <- list(tstart,tstop,death1,chemo1,radio1,horm1) out1<-coxph( Surv(tstart,tstop, death1) ~ chemo1+chemo1:log(tstop+1)+horm1+horm1:log(tstop+1)+age1+grade1+grade1:log(tstop+1)+positive1+positive1:log(tstop+1)+size1+size1:log(tstop+1), test1) out1 Up to here everything works fine (with each covariate

Hi, Is there a simple command for computing the standard error of a combination of parameter estimates in a GLM? For example: riskdata$age1 <- riskdata$age riskdata$age2 <- ifelse(riskdata$age<67,0,riskdata$age-67) model <- glm(death~age1+age2+ldl, data=riskdata,family=binomial(link=logit)) And we want to find the standard error of the partial linear predictor for

Dear all, I would like to know whether it is possible to unlist elements and keep the original format of the data. To make it more clear, let me give an exemple: I have a list l of dataframes that I created with apply but which looks like this: x1=data.frame(Name=LETTERS[1:2],Age=1:2) x2=data.frame(Name=LETTERS[3:4],Age=3:4) l=list(x1,x2) l [[1]] Name Age 1 A 1 2 B 2 [[2]] Name

Hi, I am an R newbie and one thing I am having trouble with binding variables that I have created within one data frame into a new data frame when using na.omit(). To illustrate this problem I will give the example I am working on and the approah I have been using:- data.frame1<-filepath.... attach(data.frame1) #create a new variable using a function new.variable<-rep(1,length(weight3))

Hi all, I am running regressions with many covariates, most of which remain the same each time (control variables). Instead of writing 30 demographic variables every regression, is there a way I could call them all at once using a variable called, perhaps "demog"? I have tried: > demog <- list(age1, age2, age3) but I get an error when I try to call a list in a regression. I also

Hi all, I am hoping someone can help me out with this: If I have dataframe of years and ages and the first column and first row are filled with leading values: Df<- age1 age2 age3 Yr1 1 0.4 0.16 Yr2 1.5 0 0 Yr3 0.9 0 0 Yr4 1 0 0 Yr5 1.2 0 0 Yr6 1.4 0 0 Yr7 0.8 0 0 Yr8 0.6 0 0 Yr9 1.1 0 0 Now the rest of the cells need to be filled according to the previous year and age

Hi everyone: I have been struggling with this repeated data type for whole afternoon,I sent two emails to server for help,many people kindly responded , hereby thank you so much,but since I dont want to write to much in email,so I divide the problem in parts,so far this seem did not work out very well,so this is my whole problem~ first I have example of data here:

Hi all, The last e-mails about beginners gave me the courage to post a question; from a beginner's perspective, there are a lot of questions that I'm tempted to ask. But I'm trying to find the answers either in the documentation, either in the about 15 free books I have, either in the help archives (I often found many similar questions posted in the past). Being an (still actual)

Hello all, I apologize for the simplistic question, but I have been having some trouble learning how to do mediation analysis in R. Ideally, I would like to use Preacher's Bootstrapping test for mediation (Preacher & Hayes, 2004). I have attempted to use the mediate package to set this up, using code that looks basically like this: model.m <- lm(data$outcome ~ data$mediator +

Hello I have the following data frame which I read from an EXCEL file, and when i try to call one of its columns with a space in their names I am not being able to. For example if I do EURODOLLAR$ED1.Comdty Date I obtain the following error: Error: inesperado símbolo en "EURODOLLAR$ED1.Comdty Date" I have also tried using . or _ instead of the space and have obtained no succes. How do I

Hello, I need to import a data set. I have never imported data files with R. I have always worked on simulated data. I have looked at R Data Import/Export manual. It is a bit peculiar because my data base is already an R object called "japan". I guess it is not yet a data set, and I don't know how to manipulate variables from it. When I type "japan", here is an extract

Simple example of 5 groups of 4 replicates. >set.seed(5) >tmp <- rnorm(20) >gp <- as.factor(rep(1:5,each=4)) >summary(glm(tmp ~ -1 + gp, data=data.frame(tmp, gp)))$coefficients Estimate Std. Error t value Pr(>|t|)gp1 -0.1604613084 0.4899868 -0.3274809061 0.7478301gp2 0.0002487984 0.4899868 0.0005077655 0.9996016gp3 0.0695463698 0.4899868