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...,1,2) How can I replace the values in Y with the index (!) of the corresponding values in X. So 2 appears in X in the first coordinate, so all 2’s in Y should be replaced by 1, etc. Thank you for your help, Serguei ________________________________________ Austrian Institute of Economic Research (WIFO) P.O.Box 91 Tel.: +43-1-7982601-231 1103 Vienna, Austria Fax: +43-1-7989386 Mail: Serguei.Kaniovski@wifo.ac.at http://www.wifo.ac.at/Serguei.Kaniovski [[alternative HTML version deleted]]

...0 0 0 0 0 0 0 0 0 0 0 0 0 1 is displayed with [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] in rows, and the same in columns. The matrix was generated using "apply" Serguei ___________________________________________________________________ Austrian Institute of Economic Research (WIFO) Name: Serguei Kaniovski P.O.Box 91 Tel.: +43-1-7982601-231 Arsenal Objekt 20 Fax: +43-1-7989386 1103 Vienna, Austria Mail: Serguei.Kaniovski@wifo.ac.at A-1030 Wien http://www.wifo.ac.at/Serguei.Kaniovski [[alternative HTML version deleted]]

...rmat*x does not produce permat*temp1 due to the way the element-by-element multiplication works in R. Can you help to simplify the above code? Thank you in advance, Serguei Kaniovski -- ___________________________________________________________________ Austrian Institute of Economic Research (WIFO) Name: Serguei Kaniovski P.O.Box 91 Tel.: +43-1-7982601-243 Arsenal Objekt 20 Fax: +43-1-7989386 1103 Vienna, Austria Mail: Serguei.Kaniovski at wifo.ac.at http://www.wifo.ac.at/Serguei.Kaniovski

...br;29;1 bul;10;2 cze;12;2 grc;12;2 hun;12;2 prt;12;2 rom14;2 aut;10;3 bel;12;3 The first group has the largest sum of weights, the second, with the second largest, and so on. Thanks for your help, Serguei Kaniovski ________________________________________ Austrian Institute of Economic Research (WIFO) P.O.Box 91 Tel.: +43-1-7982601-231 1103 Vienna, Austria Fax: +43-1-7989386 Mail: Serguei.Kaniovski@wifo.ac.at http://www.wifo.ac.at/Serguei.Kaniovski [[alternative HTML version deleted]]

...aniovski The data looks like: person;gr;x mike;gr1;1 jane;gr1;0 bill;gr1;0 jack;gr2;1 mike;gr2;1 jane;gr2;0 bill;gr2;0 alex;gr2;1 james;gr2;1 mike;gr3;0 bill;gr3;1 jane;gr3;1 ___________________________________________________________________ Austrian Institute of Economic Research (WIFO) Name: Serguei Kaniovski P.O.Box 91 Tel.: +43-1-7982601-231 Arsenal Objekt 20 Fax: +43-1-7989386 1103 Vienna, Austria Mail: Serguei.Kaniovski@wifo.ac.at A-1030 Wien http://www.wifo.ac.at/Serguei.Kaniovski [[alternative HTML version deleted]]

...mat))) or mat[which(duplicated(rownames(mat))),] return only half of the required indices. How can I find the remaining ones, ie the matches, so that I can add them up? Thanks, Serguei ___________________________________________________________________ Austrian Institute of Economic Research (WIFO) Name: Serguei Kaniovski P.O.Box 91 Tel.: +43-1-7982601-231 Arsenal Objekt 20 Fax: +43-1-7989386 1103 Vienna, Austria Mail: Serguei.Kaniovski@wifo.ac.at A-1030 Wien http://www.wifo.ac.at/Serguei.Kaniovski [[alternative HTML version deleted]]

...=1 in a univariate model lm (y~x). Entering lm((y-x)~1) does not help since anova test requires the same dependent variable. What is the right way to proceed? Thank you for your help and marry xmas, Serguei Kaniovski ________________________________________ Austrian Institute of Economic Research (WIFO) P.O.Box 91                          Tel.: +43-1-7982601-231 1103 Vienna, Austria        Fax: +43-1-7989386 Mail: Serguei.Kaniovski@wifo.ac.at http://www.wifo.ac.at/Serguei.Kaniovski [[alternative HTML version deleted]]

...matrix(rnorm(nobs, mean = 1, sd = 0.3), ncol = vars)) colnames(dat) <- paste("var",1:4) (cl <- kmeans(dat, k)) schwarz <- sum(cl$withinss)+ vars*k*log(nobs) Thanks for your help, Serguei ________________________________________ Austrian Institute of Economic Research (WIFO) P.O.Box 91 Tel.: +43-1-7982601-231 1103 Vienna, Austria Fax: +43-1-7989386 Mail: Serguei.Kaniovski@wifo.ac.at http://www.wifo.ac.at/Serguei.Kaniovski [[alternative HTML version deleted]]

...t I would like to have the original format, ie John.Mike John.Jim John.Steve Mike.Jim Mike.Steve Jim.Steve with two columns containing the p-values of the tests. Thanks a lot, Serguei -- ___________________________________________________________________ Austrian Institute of Economic Research (WIFO) Name: Serguei Kaniovski P.O.Box 91 Tel.: +43-1-7982601-231 Arsenal Objekt 20 Fax: +43-1-7989386 1103 Vienna, Austria Mail: Serguei.Kaniovski at wifo.ac.at http://www.wifo.ac.at/Serguei.Kaniovski

...", d. both have "x=0", The difficulty is that the number of "names" and their identity changes from case to case. Thanks a lot for you help, Serguei Kaniovski -- ___________________________________________________________________ Austrian Institute of Economic Research (WIFO) Name: Serguei Kaniovski P.O.Box 91 Tel.: +43-1-7982601-231 Arsenal Objekt 20 Fax: +43-1-7989386 1103 Vienna, Austria Mail: Serguei.Kaniovski at wifo.ac.at http://www.wifo.ac.at/Serguei.Kaniovski

...93/0473;0 093/0473;1 093/0473;1 093/0473;1 093/0473;1 093/0473;1 093/0473;1 093/0473;1 093/0499;0 093/0499;0 093/0499;1 093/0499;1 093/0499;1 093/0499;1 093/0499;1 093/0499;1 093/0499;1 -- ___________________________________________________________________ Austrian Institute of Economic Research (WIFO) Name: Serguei Kaniovski P.O.Box 91 Tel.: +43-1-7982601-231 Arsenal Objekt 20 Fax: +43-1-7989386 1103 Vienna, Austria Mail: Serguei.Kaniovski at wifo.ac.at http://www.wifo.ac.at/Serguei.Kaniovski

...trivial example) for(i in 1:5) { x<-i^2 y<-i^3 } How can I create a data.frame and a 3D plot of (i,x(i),y(i)), i.e. for each iteration Thanks, Serguei Kaniovski -- ___________________________________________________________________ ??sterreichisches Institut f??r Wirtschaftsforschung (WIFO) Name: Serguei Kaniovski Postadresse: Postfach 91 Tel.: +43-1-7982601-231 A-1103 Wien Fax : +43-1-7989386 Standort: Arsenal Objekt 20 Mail: Serguei.Kaniovski at wifo.ac.at A-1030 Wien http://www.wifo.ac.at/

..."combn"-function in "combinat" package: library(combinat) combn(c(0.1,0.2,0.3,0.4),2) , but to no avail... Thank you for your help, Serguei Kaniovski -- ___________________________________________________________________ ??sterreichisches Institut f??r Wirtschaftsforschung (WIFO) Name: Serguei Kaniovski Postadresse: Postfach 91 Tel.: +43-1-7982601-231 A-1103 Wien Fax : +43-1-7989386 Standort: Arsenal Objekt 20 Mail: Serguei.Kaniovski at wifo.ac.at A-1030 Wien http://www.wifo.ac.at/

Dear All, I have several socio-economic and geographic variables for the 27 EU countries. I would to use these data to derive a correlation matrix between groups of countries (for a different application). I thought of using kmeans to cluster the groups, and then calibrate between group correlations using distances between the centroids, and within group correlations using distances in a cluster

...I have looked at several packages, including Gaussian mixture models in Mclust, but I am not sure what is the best way, or the best package to use for this task. Greatly appreciate any suggestions! Serguei Kaniovski ________________________________________ Austrian Institute of Economic Research (WIFO) P.O.Box 91 Tel.: +43-1-7982601-231 1103 Vienna, Austria Fax: +43-1-7989386 Mail: Serguei.Kaniovski@wifo.ac.at http://www.wifo.ac.at/Serguei.Kaniovski [[alternative HTML version deleted]]

...ums, denoted s_ij for 1<=i<j<=9, where i and j are running indices for an "id"-pair. Please help, this is way beyond my knowledge of R! Thank you, Serguei Kaniovski -- ___________________________________________________________________ Austrian Institute of Economic Research (WIFO) Name: Serguei Kaniovski P.O.Box 91 Tel.: +43-1-7982601-231 Arsenal Objekt 20 Fax: +43-1-7989386 1103 Vienna, Austria Mail: Serguei.Kaniovski at wifo.ac.at http://www.wifo.ac.at/Serguei.Kaniovski

...,0,(fr[4]-(1-x[1])*(1-x[2])-x[3]*sqrt(x[1]*(1-x[1])*x[2]*(1-x[2])))^2/fr[4]) } sval=rep(0.1,3) fr<-c(0,0.1,0.2,0.3) optim(sval,obj, method="BFGS")$par Thank you, Serguei -- ___________________________________________________________________ Austrian Institute of Economic Research (WIFO) Name: Serguei Kaniovski P.O.Box 91 Tel.: +43-1-7982601-231 Arsenal Objekt 20 Fax: +43-1-7989386 1103 Vienna, Austria Mail: Serguei.Kaniovski at wifo.ac.at http://www.wifo.ac.at/Serguei.Kaniovski

Hello All, I have a panel date - here a small-scale example: df <- data.frame(cbind(rep(c("AUT","BEL","DEN","GER"),4),cbind(rep(c(1999,2000,2001,2002),4)),sample(10,16,replace=T))) names(df) <- c("country","year","x") SORT <- c("GER","BEL","DEN","AUT") I need to compute the

Hallo All, I would like to apply a function to all permutations of variables in a dataframe (except the first). What is the best way to achieve this? I produce the permutations using: nvar <- ncol(dat) - 1 perms <- as.matrix( expand.grid(rep( list(1:0) , nvar ))[ , nvar:1] ) Thanks in advance Serguei Test-dataframe, comma-delimited: code,wav,w,area,gdp,def,pop,coast,milspend,agr

...create a dummy (indicator) variable for use in regression (using factor?), such that it takes the value of 1 if the country is in the pair (i.e. EITHER an i-country OR an j-country). Thank you for your help, Serguei ________________________________________ Austrian Institute of Economic Research (WIFO) P.O.Box 91                          Tel.: +43-1-7982601-231 1103 Vienna, Austria        Fax: +43-1-7989386 Mail: Serguei.Kaniovski@wifo.ac.at http://www.wifo.ac.at/Serguei.Kaniovski [[alternative HTML version deleted]]