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> c2v2_safe<-list('12345_at'=strsplit(df[1,2], "\t")[[1]]) and it works. > > The question is how to replace 12345_at with the df[i, 1] variables. for this particular case: names(c2v2_safe) = df[1, 1] ? b

try the following ( i haven't tested it) result = lapply(df[,2], function(x) x) names(result) <- df[,1] HTH, weiwei On 7/24/07, Li, Aiguo (NIH/NCI) [C] <liai at="" mail.nih.gov=""> wrote: > Hi all, > > > > I need to make a list as the following format: > > $`1555385_at` > > [1] "00604" "01031" > > > > $`224869_s_at` > > [1] NA > > > > $`210477_x_at` > > [1] "04010" "04310" "04510" "04620" "04664" "04910" "04912" "04920" > "04930" "05120" "05210" > > > > $`212807_s_at` > > [1] NA > > > > Now I have the following data frame. > > > df[1,1] > > [1] "1007_s_at" > > > > >df[1,2] > > [[1]] > > [1] "244" "501" "578" "579" "604" "611" "844" "877" "909" > "951" "954" "1181" "1305" "1312" > > > > How can I use a for loop to make a list object as the above format? > > > > I tried the following: > > c2v2_safe<-list('12345_at'=strsplit(df[1,2], "\t")[[1]]) and it works. > > > > The question is how to replace 12345_at with the df[i, 1] variables. > > > > Thanks, > > > > AG Lee > > > > > > > [[alternative HTML version deleted]] > > _______________________________________________ > Bioconductor mailing list > Bioconductor at stat.math.ethz.ch > https://stat.ethz.ch/mailman/listinfo/bioconductor > Search the archives: http://news.gmane.org/gmane.science.biology.informatics.conductor > -- Weiwei Shi, Ph.D Research Scientist GeneGO, Inc. "Did you always know?" "No, I did not. But I believed..." ---Matrix III