Here is an example of the data.frame that I have, df<-data.frame("case"=rep(1:5,each=9),"id"=rep(1:9,times=5),"x"=round(runif(length(rep(1:5,each=9))))) "case" represents the cases, "id" the persons, and "x" is the binary state. I would like to know in how many cases any two persons a. both have "1", b. the first has "0" - the second has "1", c. the first has "0" - the second has "0", d. both have "0". There will be choose(9,2) sums, 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

library(gtools) apply(combinations(9,2), 1, function(x) with(df[df$id %in% x, ], table(x, id))) ------------------------------------------------------------------- Jacques VESLOT CNRS UMR 8090 I.B.L (2?me ?tage) 1 rue du Professeur Calmette B.P. 245 59019 Lille Cedex Tel : 33 (0)3.20.87.10.44 Fax : 33 (0)3.20.87.10.31 http://www-good.ibl.fr ------------------------------------------------------------------- Serguei Kaniovski a ?crit :> Here is an example of the data.frame that I have, > > df<-data.frame("case"=rep(1:5,each=9),"id"=rep(1:9,times=5),"x"=round(runif(length(rep(1:5,each=9))))) > > "case" represents the cases, > "id" the persons, and > "x" is the binary state. > > I would like to know in how many cases any two persons > > a. both have "1", > b. the first has "0" - the second has "1", > c. the first has "0" - the second has "0", > d. both have "0". > > There will be choose(9,2) sums, 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

sorry, answered to quickly... actually it''s easier using paste(): df <- df[order(df$case),] apply(combinations(9,2), 1, function(y) table(factor(do.call(paste, c(with(df[df$id %in% y, ], split(x, id)), sep="")), levels=c("00","01","10","11")))) ------------------------------------------------------------------- Jacques VESLOT CNRS UMR 8090 I.B.L (2?me ?tage) 1 rue du Professeur Calmette B.P. 245 59019 Lille Cedex Tel : 33 (0)3.20.87.10.44 Fax : 33 (0)3.20.87.10.31 http://www-good.ibl.fr ------------------------------------------------------------------- Serguei Kaniovski a ?crit :> Here is an example of the data.frame that I have, > > df<-data.frame("case"=rep(1:5,each=9),"id"=rep(1:9,times=5),"x"=round(runif(length(rep(1:5,each=9))))) > > "case" represents the cases, > "id" the persons, and > "x" is the binary state. > > I would like to know in how many cases any two persons > > a. both have "1", > b. the first has "0" - the second has "1", > c. the first has "0" - the second has "0", > d. both have "0". > > There will be choose(9,2) sums, 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