similar to: select cases

bonjour, Je voudrais faire une analyse des correspondances multiples avec R. avec les repr?sentation graphiques correspondantes avec R. je ne sais pas comment proc?der .. en vour remerciant par avance Faouzi

Hi, poids taille fumeur sexe sport etat 85 184 oui homme 1 malade 65 175 oui homme 1 malade 74 180 oui homme 2 gueri 79 175 oui homme 2 malade 71 165 non homme 3 gueri 80 185 non homme 3 gueri 75 180 non homme 4 malade 69 155 non homme 4 malade 74 168 oui

Hi everybody! I'm sure that I overlook something and feel quite stupid to ask, but I have not found an easy solution to the following problem: Take e.g. the Orthodont data from the nlme package: > head(Orthodont) Grouped Data: distance ~ age | Subject distance age Subject Sex 1 26.0 8 M01 Male 2 25.0 10 M01 Male 3 29.0 12 M01 Male 4 31.0 14 M01 Male

Hi, I have imported some data to R from stata and my factor variables have an Indeterminate level which I don't really want. For example the variable sex has the levels Male, Female and Indeterminate. There are no 'Indeterminate' values in the data. Can somebody tell me how to get rid of this level as it restricting my cox ph model. Thanks Neil

The data is in the attachment. What I wanna get is: , , Sex = Male Eye Hair Brown Blue Hazel Green Black 32 11 10 3 Brown 53 50 25 15 Red 10 10 7 7 Blond 3 30 5 8 , , Sex = Female Eye Hair Brown Blue Hazel Green Black 36 9 5 2 Brown 66 34 29 14 Red 16 7 7 7 Blond 4

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

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

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",

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

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 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, 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

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 =

Hi dear listers Do you know how to made a mean by one or more categorical variables? I can do that quite easy on SAS by in R I just can't do it E.g. Some fishes Sex Lenght male 15 fema 20 fema 17 fema 19 male 18 So the idea is mean(Lenght) by sex, in order to have sex: Male mean=XX Sex: Fema mean=YY In this case is quite easy to do it by a loop, but I have a huge DB so is not an

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 hope this formatting is correct as it is my first time. I am trying to do comparisons of values in a data frame that has some factor variables. One instance is > train$sex[2] [1] Male Levels: Female Male So the value is Male but a comparison like "Male" == train$sex[2] will always return FALSE because of the level information included. Another problem this creates is

Hi R users, I am trying to extract some attributes (age, sex, area) from dataframe "AB" that has 101,269 observations of 28 variables to dataframe "t2" that has 47 observations of 6 variables. They share a column called "id", which is a factor with 47 levels. I want to end up with a dataframe that has 47 observations of 9 variables (the original 6 variables of t2,

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

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

Dear people, I am including an example of a dataframe: mydataframe<-data.frame(X=c(1:4),total_bill=c(16.99,10.34,21.01,23.68),tip=c(1.01,1.66,3.50,3.31),sex=c("Male","Male","Male","Female")) When I use the sapply function getting the information about the factors works: sapply(mydataframe,function(x)is.factor(x)) X total_bill tip