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Dear R-List, I would like to recode my data according to quantile breaks, i.e. all data within the range of 0%-25% should get a 1, >25%-50% a 2 etc. Is there a nice way to do this with all columns in a dataframe. e.g. df<-

Dear R-List, I have a dataframe with one grouping variable (x) and three response variables (y,z,w). df<-data.frame(x=c(rep(1,3),rep(2,4),rep(3,5)),y=rnorm(12),z=c(3,4,5,NA,NA,NA,NA,1,2,1,2,1),w=c(1,2,3,3,4,3,5,NA,5,NA,7,8)) >df      x            y            z     w      1      0.29306106  3      1      1      0.54797780  4      2      1     -1.38365548  5      3      2     -0.20407986

Dear R-List, (I am not sure whether this list is the right place for my question...) I have a dataframe df.cfa

Dear R-List, I have carried out a factor analysis using fa (psych) with nfactors=2, rotation="oblimin" and fm="pa". Now I have to report both pattern AND structure matrix. As I have understood R-Documentation, this can be obtained by setting the "oblique.scores" argument TRUE (structure matrix) or FALSE (pattern matrix), respectively. However both produced the same

Dear R-List,   I would like to have a large number of stratified random subsamples drawn from my dataframe and automatically test for correlation differences in every subsample.   Let this be my dataframe   df<-data.frame(group=c(rep(1,5),rep(2,5),rep(3,5)),a=c(3,4,5,6,3,4,5,4,5,4,1,2,1,2,1),b=c(1,2,3,4,5,3,4,3,4,5,6,5,6,2,3),c=c(2,2,3,3,5,1,1,6,6,5,6,1,1,2,1))   Then I would like to have n

Dear R-List, I apologize for not posting a reproducible example - the reason is that I actually do not succeed in reproducing my specific problem with generated data. Let me still describe the problem: I want to impute missing data using the "pan" package. a) It works, when I use a fraction of my data (e.g. 200 out of 44000 cases) b) It works, when I generate a dataset of equal

Hi Eik, thank you so much - it works perfectly! Thank you and best wishes Alain   Eik Vettorazzi <E.Vettorazzi@uke.de> hat am 6. Februar 2013 um 17:01 geschrieben: > Hi Alain, > here is a one-liner for a df without the rowSum column > >

Dear R-List, I tried to fit a partial credit model using the "pcmdat" from eRm-package comparing the results of mirt, eRm, ltm and winsteps. The results where quite different, though. I cannot figure out what went wrong and I do not know which result I can rely on. This is what I did in R library(mirt) #load(file="u3.RData")

Dear R-List, I have a dataframe area<-c(rep(1,10),rep(2,10),rep(3,10),rep(4,10),rep(5,10)) type<-c(rep(1:10,5)) a<-rnorm(50) b<-rnorm(50) c<-rnorm(50) d<-rnorm(50) df<-cbind(area,type,a,b,c,d) df area type a b c d [1,] 1 1 0.45608192 0.240378547 2.05208079 -1.18827462 [2,] 1 2 -0.12119506 -0.028078577

Dear R-List, I have a dataframe with one column "name.of.report" containing character values, e.g. >df$name.of.report "jeff_2001_teamx" "teamy_jeff_2002" "robert_2002_teamz" "mary_2002_teamz" "2003_mary_teamy" ... (i.e. the bit of interest is not always at same position) Now I want to recode the column "name.of.report"

Dear R-List, I am sure there must be a very simple way to do this - I just do not know how... This is what I want to do: #my dataframe df<-data.frame(id=c("x01","x02","x03","x04","x05","x06"),a=c(1,2,NA,4,5,6),b=c(2,4,6,8,10,NA),c=c(NA,3,9,12,NA,NA),sum=c(3,9,15,24,15,6))    id    a     b     c   sum 1 x01  1     2    NA   3 2

Dear R-List, I would like to recode categorial variables into binary data, so that all values above median are coded 1 and all values below 0, separating each var into two equally large groups (e.g. good performers = 0 vs. bad performers =1). I have not succeeded so far in finding a nice solution to do that in R. I thought there might be a better way than ordering each column and recoding the

Dear R-List, I would like to recode categorial variables into binary data, so that all values above median are coded 1 and all values below 0, separating each var into two equally large groups (e.g. good performers = 0 vs. bad performers =1). I have not succeeded so far in finding a nice solution to do that in R. I thought there might be a better way than ordering each column and recoding the