There are many ways to get the output you want, so your question is ill defined. But it is easy to see where your code goes wrong. And it should be easy for you to fix it.

If you subset a vector with the '[' operator, this is like putting a vector of indices "into" the square brackets. So, to debug such code, just print out your indices. And understand how the bracketing and operator priorities work:

i <- 2; n <- 7

1:n+(i-1)*(n+1) # 1:7 gets produced first, then (i-1)*(n+1) == (1)*(8) is added to that vector

[1] 9 10 11 12 13 14 15

But consider:

i <- 2; n <- 7

1:(n+(i-1)*(n+1)) # 7+(1)*(8)

[1] 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

You can probably figure it out now.

B.

> On Aug 18, 2017, at 7:31 AM, Moohwan Kim <

[hidden email]> wrote:

>

> Dear R users,

>

> I have the following codes:

> zeta <- rep(1,8)

> n <- 7

> for (i in 1:2){

> beta <- zeta[1:n+(i-1)*(n+1)]

> print(beta)

> parm <- zeta[i*(n+1)]

> print(parm)

> }

> ###################

> The output is as follows:

> [1] 1 1 1 1 1 1 1

> [1] 1

> [1] NA NA NA NA NA NA NA

> [1] NA

> #######################

> The outcome I want to get is:

> [1] 1 1 1 1 1 1 1

> [1] 1

> [1] 1 1 1 1 1 1 1

> [1] 1

>

> How could I get the desired outcome?

>

> best,

> kmh

>

> [[alternative HTML version deleted]]

>

> ______________________________________________

>

[hidden email] mailing list -- To UNSUBSCRIBE and more, see

>

https://stat.ethz.ch/mailman/listinfo/r-help> PLEASE do read the posting guide

http://www.R-project.org/posting-guide.html> and provide commented, minimal, self-contained, reproducible code.

______________________________________________

[hidden email] mailing list -- To UNSUBSCRIBE and more, see

https://stat.ethz.ch/mailman/listinfo/r-helpPLEASE do read the posting guide

http://www.R-project.org/posting-guide.htmland provide commented, minimal, self-contained, reproducible code.