This programme
for(T in 1:3){
for(j in 1:(5-1)){
for(k in (j+1):5){
for(l in (j+2):5){
print(paste("1 JKL:", j,k,l,sep=" "))
}
}
}
}
Prints out (among other things)
[1] "1 JKL: 4 5 6"
That is for(l in (j+2):5) sets l to 6 one more than the upper limit.
cheers
Worik
[[alternative HTML version deleted]]
On Aug 31, 2009, at 8:31 PM, Worik R wrote:
> This programme
>
> for(T in 1:3){
> for(j in 1:(5-1)){
> for(k in (j+1):5){
> for(l in (j+2):5){
> print(paste("1 JKL:", j,k,l,sep=" "))
> }
> }
> }
> }
>
> Prints out (among other things)
> [1] "1 JKL: 4 5 6"
>
> That is for(l in (j+2):5) sets l to 6 one more than the upper limit.
Except try this at the console and see if enlightenment occurs:
6:5
The second entry in n:m may not be an upper limit.
David Winsemius, MD
Heritage Laboratories
West Hartford, CT
(Continue reading)
Answer: No.
On Tue, 1 Sep 2009, Worik R wrote:
> This programme
>
> for(T in 1:3){
> for(j in 1:(5-1)){
> for(k in (j+1):5){
> for(l in (j+2):5){
> print(paste("1 JKL:", j,k,l,sep=" "))
> }
> }
> }
> }
>
> Prints out (among other things)
> [1] "1 JKL: 4 5 6"
>
> That is for(l in (j+2):5) sets l to 6 one more than the upper limit.
Check out help(":") which explains that in from:to from is the starting
value and to is the end value. Note that it does not impose that
from <= to or that to is the upper limit.
For j = 4, you get l to be in 6:5.
Z
> cheers
> Worik
(Continue reading)
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