search for: dat3new

...5?? c?? e??? 12 #or you have cases like these: dat3<- read.table(text=" id1?? id2?? value a????? b?????? 10 c????? d??????? 11 b???? a???????? 10 a????? b??????? 10 e????? c???????? 12 c????? e???????? 12 c????? d???????? 11 ",sep="",header=TRUE,stringsAsFactors=FALSE) ?dat3New<-dat3[apply(dat3[,-3],1,function(x) {x1<- order(x); x1[1]<x1[2]}),] dat3New[!duplicated(dat3New$value),] #? id1 id2 value #1?? a?? b??? 10 #2?? c?? d??? 11 #6?? c?? e??? 12 A.K. >Hi everybody, > >I was hoping that someone could help me with this problem. I have a table with...

Hi everybody, I would like to know whether it is possible to compare to tables for certain parameters. I have these two tables: gene table name chr start end str accession Length gen1 4 646752 646838 + MI0005806 86 gen12 2L 243035 243141 - MI0005821 106 gen3 2L 159838 159928 + MI0005813 90 gen7 2L

Hi, I have a table in which one column has the name of the objects as shown below. Name Budlamp-Woodcutter Complex - 15 to 60% slope (60/25/15) Budlamp-Woodcutter Complex - 15 to 60% slope (60/25/15) Terrarossa-Blacktail-Pyeatt Complex - 1 to 40% slope (40/35/15/10) Terrarossa-Blacktail-Pyeatt Complex - 1 to 40% slope (40/35/15/10) How can I split the single column into three columns

HI Elisa, You could use ?cut() vec1<-c(33,18,13,47,30,10,6,21,39,25,40,29,14,16,44,1,41,4,15,20,46,32,38,5,31,12,48,27,36,24,34,2,35,11,42,9,8,7,26,22,43,17,19,28,23,3,49,37,50,45) label1<-unlist(lapply(mapply(c,lapply(seq(0,45,5),function(x) x),lapply(seq(5,50,5),function(x) x),SIMPLIFY=FALSE),function(i) paste(i[1],"<x<=",i[2],sep="")))