Hi Guys,
I have three character vectors (final, final2 and final3) of same length. I would like to concatenate final2 and final3 onto final omitting all the NAs in final2 and final3 and get the result as shown below.
I tried something like this, but won't get what I need:
idx2 <- which(!is.na(final2))
idx3<-which(!is.na(final3))
paste(final, final2[idx2], final[idx3], sep=":")
final
[1] "chr999:9992053031:9992053031-C(34),G()" "chr999:99933999574:99933999574-C(22),A(2)" "chr999:115389448:115389448-C(25),A()"
[4] "chr999:120094925:120094925-A(30),C()" "chr999:120833042:120833042-C(48),T()" "chr999:124348684:124348684-C(60),T()"
[7] "chr999:126686718:126686718-G(18),A()" "chr999:126691538:126691538-C(9),T()" "chr999:129913632:129913632-G(28),T()"
final2
[1] NA "G(2)" NA NA NA NA
final3
[1] NA NA NA "T(34)" NA NA
result
[1] "chr999:9992053031:9992053031-C(34),G()" "chr999:99933999574:99933999574-C(22),A(2),G(2)" "chr999:115389448:115389448-C(25),A()"
[4] "chr999:120094925:120094925-A(30),C(),T(34)" "chr999:120833042:120833042-C(48),T()" "chr999:124348684:124348684-C(60),T()"
Here is my own answer:
paste2 <- function(...,sep=",") {
L <- list(...)
L <- lapply(L,function(x) {x[is.na(x)] <- ""; x})
gsub(paste0("(^",sep,"|",sep,"$)"),"",
gsub(paste0(sep,sep),sep,
do.call(paste,c(L,list(sep=sep)))))
}
final.output<-paste2(final,final2,final3)
Thanks for sharing the solution, although for this type of queries StackExchange is more appropriate.
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I guess this belongs on SO.
I got it figured out, thank you!
Hello MAPK!
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