similar to: "Centered" dummy variables; non zero/one coding

Hi, I am new to R. My problem is with the ordered logistic model. Here is my question: Generate an order discrete variable using the variable wrwage1 = wages in first full calendar quarter after benefit application in the following way: * wage*1*Ordered *= 1 *if*0 *· wrwage*1 *< *1000 2 *if*1000 *· wrwage*1 *< *2000 3 *if*2000 *· wrwage*1 *< *3000 4 *if*3000 *· wrwage*1 *<

Hi List, I have spent more than 30 minutes, but failed to read in this file using the read.table() function. I could not figure out how to fix the following error. Error in scan(file, what, nmax, sep, dec, quote, skip, nlines, na.strings, : line 1 did not have 6 elements Any help would be be appreciated. Thanks, Pradip Muhuri ####### below is the reproducible example xd1 <-

Muchas gracias, pero claro en una muestra de 50 datos se ejecuta, en la muestra original de 1000 registros me tira error :( 2013/10/30 daniel <daniel319@gmail.com> > Amalia, > > No obtengo tus resultados. Corrí tus formulas y datos y el resultado es > x <- structure(list(ï..psraid = c(202517L, 202518L, 202520L, 202523L, > + 202527L, 202537L, 202543L, 202544L, 202551L,

Hello--I'm doing a simple crosstab using dcast: rawfreq <- dcast(nh11brfs, race3~CHCCOPD, length) with the results race3 Yes No NA 1 White non-Hispanic 446 5473 21 2 Other non-Hispanic 29 211 0 3 Hispanic 6 81 1 4 <NA> 10 83 1 How would I modify this call to exclude all missing values; that is, to obtain race3

Using the survey package I find it is convenient and easy to get estimated proportions using svymean, and their corresponding estimated standard errors. But is there any elegant/simple way to calculate corresponding confidence intervals for those proportions? Of course +/- 1.96 s.e. is a reasonable approximation for a 95% CI, but (incorrectly) assumes symmetrical distribution for a proportion.

Dear R Users: I have race-ethnicity groups identified in the factor variable Ethnic_G. I need to collapse Ethnic_G into a new variable with only two factors, 1 (White, non-Hispanic) and 2 (Minority). As seen in the code and output below, the recoded race-ethnicity variable is put into logical format, not factor format. I've used library(car) and the package was updated. Any ideas on

HI, Dear R community, I am using the xtable to create the table, but how can I see the table? The following is the codes I used: > data(tli) > tli.table <- xtable(tli[1:10, ]) > digits(tli.table)[c(2, 6)] <- 0 > print(tli.table, floating = FALSE) % latex table generated in R 2.11.0 by xtable 1.5-6 package % Thu Jul 1 20:43:43 2010 \begin{tabular}{rrlllr} \hline &

Newbie here. Many apologies in advance for using the incorrect lingo. I'm new to statistics and VERY new to R. I'm attempting to "group" or "bin" data together in order to analyze them as a combined group rather than as discrete set. I'll provide a simple example of the data for illustrative purposes. Patient ID | Charges | Age | Race 1 |

Hi all, When I try the following with pkg Survey it returns the error below: ftable(svyby(~INCOME, ~AGECL+RACECL, svymean, design=q50), rownames=list(AGECL=c("<35", "35-44", "45-54", "55-64", "65-74", ">=75"), RACECL=c("white non hispanic", "non white or

I have a function (see below) that does some bootstrapping (I am happy to expand offline why I could use existing functions.) I put my results into and empty matrix and add a row of results with each iteration. My problem is i am a new user to R and I don't understand data frames, matrices, elements, and vectors well. What I would like is to have a data frame I can manipulate outside of the

Hi R-ers: I'm drawing a plot and have used different line types (lty) for different race/ethnicity groups. I want a legend that explains what line types correspond to the different race/ethnicity groups. I used the following code: legend( 1992 , 42 , c("Hispanic" , "non-Hispanic white (NHW)" , "non-Hispanic black" , "AI/AN" , "Asian" ) ,

Dear R-Users, I have a problem on extracting T-Stat and P-Value. I have written R-code below library("Matching") data("lalonde") attach(lalonde) names(lalonde) Y <- lalonde$re78 Tr <- lalonde$treat glm1 <- glm(Tr~age+educ+black+hisp+married+nodegr+re74+re75,family=binomial,data=lalonde) pscore.predicted <- predict(glm1) rr1 <-

try resetting your factor levels and re-run? q50 <- update( q50 , INCOME = factor( INCOME ) , AGECL = factor( AGECL ) , RACECL = factor( RACECL ) ) On Sun, Jul 9, 2017 at 2:59 PM, Orsola Costantini via R-help < r-help at r-project.org> wrote: > Hi all, > > When I try the following with pkg Survey it returns the error below: > > ftable(svyby(~INCOME, ~AGECL+RACECL,

Saludo gente, antes que nada gracias por la ayuda que puedan aportarme, soy iniciante en R, estoy usando el paquete Mice para realizar imputaciones múltiples sobre variables en su mayoría categóricas. El problema está que cuando expresó este comando imp <- mice(dataset,method="polr",maxit=1) donde el dataset es un data.frame me tirá este error : iter imp variable 1 1 pial1a

I am trying to use tree to partition a data set. The data set has 3924 observations. Partitioning seems to work for small subsets of the data, but when I use the entire data set, no partitioning occurs. The variables are: RESP respondent to a survey (0 = not a respondent, 1 = respondent) AGE_P Age (continuous) ORIGIN_I Hispanic Ethnicity (1 = Hispanic, 2 = non-Hispanic) RACRECI2 Race

Hello, I'm trying to run a loop that will subset my data into specific sets by regions and by race/ethnicity. I'm trying to do this fairly compactly, and I cannot get this to work. A "simple" version of the code that I am trying to run is: names <- c("white", "black", "asian", "hispanic") for(j in names){ for(i in 1:9){

i want to ask 2 questions. 1) can R do Random-effects GLS regression which i can get from Stata? the following result is frome Stata.can I get the alike result from R? xtreg lwage educ black hisp exper expersq married union, re Random-effects GLS regression Number of obs = 4360 Group variable (i) : nr Number of groups = 545 R-sq:

Buenas tardes, Quiero aplicar la función rbind y necesito tener los mismos nombres de columnas. Como tengo unas 195 variables en cada dataframe, necesito hacerlo de una forma rápida. Tengo 9 bases de datos y tengo que fusionar todas. ¿Como puedo comprobar que los nombres de las variables son los mismos? Y de lo contrario, ¿como detecto las diferencias? He probado con

Amalia, No obtengo tus resultados. Corrí tus formulas y datos y el resultado es x <- structure(list(ï..psraid = c(202517L, 202518L, 202520L, 202523L, + 202527L, 202537L, 202543L, 202544L, 202551L, 202566L, 202570L, + 202571L, 202606L, 202619L, 202624L, 202629L, 202631L, 202632L, + 202633L, 202648L, 202657L, 202663L, 202676L, 202683L, 202685L, + 202706L, 202708L, 202709L, 202710L, 202734L,

Hi List, My goal is to force R not to print in scientific notation in the sixth column (rel_diff - for the p-value) of my data frame (not a matrix). I have used the format.pval () and printCoefmat () functions on the data frame. The R script is appended below. This issue is that use of the format.pval () and printCoefmat () functions on the data frame gives me the desired results, but coerces