[R] CONFUSSING WITH select[!miss] <- 1:sum(!miss)
William Dunlap
wdunlap at tibco.com
Tue Dec 6 20:04:52 CET 2016
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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>
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