CONFUSSING WITH select[!miss] <- 1:sum(!miss)
R is interactive so you can print the intermediate results:
ph <- data.frame(M1=c(1,NA,3,4,5), X1=1:5, X2=c(1,2,NA,4,5), X3=1:5,
Y=c(11,12,13,14,NA), row.names=paste0("R",1:5))
ph
M1 X1 X2 X3 Y R1 1 1 1 1 11 R2 NA 2 2 2 12 R3 3 3 NA 3 13 R4 4 4 4 4 14 R5 5 5 5 5 NA
miss <- apply(is.na(ph[,c("M1","X1","X2","X3")]),1, any)
miss
R1 R2 R3 R4 R5 FALSE TRUE TRUE FALSE FALSE
select <- integer(nrow(ph)) select
[1] 0 0 0 0 0
sum(!miss)
[1] 3
select[!miss] <- 1:sum(!miss) select
[1] 1 0 0 2 3 Then you can look in the Introduction to R document or ask about the steps that confuse you. Bill Dunlap TIBCO Software wdunlap tibco.com
On Tue, Dec 6, 2016 at 10:18 AM, greg holly <mak.hholly at gmail.com> wrote:
Dear All;
I am very new in R and try to understand the logic for a program has been
run sucessfully. Here select[!miss] <- 1:sum(!miss) par is confussing me. I
need to understandand the logic behind this commend line.
Thanks in advance for your help,
Greg
miss <- apply(is.na(ph[,c("M1","X1","X2","X3")]),1, any)
select <- integer(nrow(ph))
select[!miss] <- 1:sum(!miss)
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