similar to: String variables in GLM

Buenas noches, mi pregunta es tal vez sencilla. tengo esta libreria y estos datos library(nlme) data(MathAchieve) Infoest = MathAchieve en esta base quiero contar la cantidad de casos en la que la columna Sex, toma el valor Female, como puedo hacerle el filtro para ver solo los Females. y como puedo dentro del mismo conjunto de datos agregar la variable numerica que me cuente 1 si es Female y 0

I currently have a data set in which gender is inputed as "Male" and "Female" , and I'm trying to convert this into "1" and "0". I found a website which reccomended using two commands: data$scode[data$sex=="M"] <- "1" data$scode[data$sex=="F"] <- "2" to convert to numbers, and: data$scode <-

Hi A.K Regarding my question on comparing normal/ obese/overweight with blood pressure change, I did finally as per the first suggestion of stacking the data and creating a normal category . This only gives me a obese not obese 14, but when I did with the wide format hoping to get a obese14,normal14,overweight 14 Vs hibp 21, i could not complete any of the models. This time I classified obese=1

Dear people, I have got a question concerning the underlying numerical codes when reading an SPSS file into R. I used the package foreign and when I look at a variable I get the verbal codes. I would like to know how it is possible to get the underlying numerical codes as output, which are the same as in my SPSS file. Thank you very much in advance for your help! Marion [[alternative HTML

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Hi i want to recode Male =1 and female =2 in R what is the syntax for same ? thanks Sameer [[alternative HTML version deleted]]

Hi, To be able to match cases with a benchmark I need to have a data.frame with a character id variable. however, I am surprised why this seems to be so hard. In fact I was unable to succeed. Here is what I tried: >test1 <-expand.grid(ID = 1:2, sex = c("male","female")) >is(test1[,2]) [1] "factor" "oldClass" >test2 <-expand.grid(ID =

I'm trying to develop some graphic methods for glm objects, but they only apply for models where all predictors are discrete factors. How can I test for this in a function, given the glm model object? That is, I want something that will serve as an equivalent of is.discrete.glm() in the following context: myplot.glm <- function(model, ...) { if (!inherits(model,"glm"))

Hi R-helpers, I am trying to use this ifelse statement to recode a variable: > data$SOCIAL_STATUS<-ifelse(data$SOCIAL_STATUS=="B" & data$MALE>4, "C", "B") (i.e., if social status is B and there are more than 4 males, then recode social status to C; otherwise, leave it B) But, it's not working. See the below R output. Notice that there were 71

Hello R-users/experts, I am new to R- I have a simple question: Let say I have a data set as follows temp:[file attached] the data structure is a follows: sex age female 28 female 53 female 53 female 36 male 42 male 29 male 43 male 36 male 41 Here we are grouping all male value into male and all female value in to female

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:

Dear list, I have following data set, which I want to plot the "Scale" variable on the x-axis and "Mean"´on the y-axis for each Ageclass and for each sex. The Mean value of each Ageclass for each sex would be connected by a line. Totally, there should be 6 lines, from which three present the Mean values of each Ageclass for respective sex. Are there any easy ways to do

Hello, I'm trying to do bar plot where 'sex' will be the category axis and 'occupation' will represent the bars and the clusters will represent the mean 'income'. sex occupation income 1 female j 12 2 male b 34 3 male j 22 4 female j 54 5 male b 33 6

I'm working on a quick tutorial for my students, and was planning on using Mann-Whitney U as one of the tests. I have the following (fake) data grade <- c("MVG", "VG", "VG", "G", "MVG", "G", "VG", "G", "VG") sex <- c( "male", "male", "female", "male",

Hi all, I have data like this: >sample <- read.csv(file="sample.csv",sep=",",header=TRUE) > sample stdate Domain sex age Login 1 01/11/09 xxx FeMale 25 2 2 01/11/09 xxx FeMale 35 4 3 01/11/09 xxx Male 18 30 4 01/11/09 xxx Male 31 3 5 02/11/09 xxx Male 32 11 6 02/11/09 xxx Male 31 1 7 02/11/09

Hi, I don't know how to make prediction with a factor in the linear model. Say yd=c(1,2,3,4,5) > sl=c(2,3,4,5,6) > sex=c("male","male","female","female","male") > sex=factor(sex) > m=lm(sl~yd+sex+sex:yd) How to make a prediction with new data like yd=c(4,5,6,7,8) for male and female separately ? Thank you very

dear£º in the example£¨nomogram£©£¬I don't understand the meanings of the program which have been marked by red line.And how to compile the program(L <- .4*(sex=='male') + .045*(age-50) + (log(cholesterol - 10)-5.2)*(-2*(sex=='female') + 2*(sex=='male'))). n <- 1000 # define sample size set.seed(17) # so can reproduce the results age <- rnorm(n, 50, 10)

I have the following problem. It is not of earthshaking importance, but still I have spent a considerable amount of time thinking about it. PROBLEM: Is there any way I can have a single textfile that contains both a) data b) programcode The program should act on the data, if the textfile is source()'ed into R. BOUNDARY CONDITION: I want the data written in the textfile in exactly

I am trying to add confidence (error) bars to lattice barcharts (and dotplots, and xyplots). I found this helpful message from Deepayan Sarkar and based teh code below on it: http://finzi.psych.upenn.edu/R/Rhelp02a/archive/50299.html However, I can't get it to work with groups, as illustrated. I am sure I am missing something elementary, but I am unsure what. Using R 2.1.1 on various

Dear R-helpers, Suppose I have dataset like this below: data(HairEyeColor) dfHEC <- as.data.frame(as.table(HairEyeColor)) my.dfHEC <- data.frame(Hair=rep(dfHEC$Hair,dfHEC$Freq), Eye=rep(dfHEC$Eye,dfHEC$Freq), Sex=rep(dfHEC$Sex,dfHEC$Freq)) my.dfHEC my.dfHEC$Hair my.dfHEC$Eye my.dfHEC$Sex and I know all levels for Hair, Eye and Sex. In my case, I