IMTS length 2000 Sent from my iPhone
> On Feb 1, 2021, at 9:16 PM, David Winsemius <dwinsem...@comcast.net> wrote: > > Cc’ed the list as should always be your practice. > > Here’s one way (untested): > > W <- +(z>4| z<2) # assume z is of length 20 > > — > David > > Sent from my iPhone > >>> On Feb 1, 2021, at 7:08 PM, Shaami <nzsh...@gmail.com> wrote: >>> >> >> Hi Prof. David >> >> In the following state >> >> W = (1:2000)[z >4|z<2) >> >> Could you please guide how I can assign zero if condition is not satisfied? >> >> Best Regards >> >> Shaami >> >>> On Mon, 1 Feb 2021, 11:01 am David Winsemius, <dwinsem...@comcast.net> >>> wrote: >>> >>> On 1/31/21 1:26 PM, Berry, Charles wrote: >>> > >>> >> On Jan 30, 2021, at 9:32 PM, Shaami <nzsh...@gmail.com> wrote: >>> >> >>> >> Hi >>> >> I have made the sample code again. Could you please guide how to use >>> >> vectorization for variables whose next value depends on the previous one? >>> >> >>> >>> I agree with Charles that I suspect your results are not what you >>> expect. You should try using cat or print to output intermediate results >>> to the console. I would suggest you limit your examination to a more >>> manageable length, say the first 10 results while you are working out >>> your logic. After you have the logic debugged, you can move on to long >>> sequences. >>> >>> >>> This is my suggestion for a more compact solution (at least for the >>> inner loop calculation): >>> >>> set.seed(123) >>> >>> x <- rnorm(2000) >>> >>> z <- Reduce( function(x,y) { sum(y+5*x) }, x, accumulate=TRUE) >>> >>> w<- numeric(2000) >>> >>> w <- (1:2000)[ z >4 | z < 1 ] # In your version the w values get >>> overwritten and end up all being 2000 >>> >>> >>> I would also advise making a natural language statement of the problem >>> and goals. I'm thinking that you may be missing certain aspects of the >>> underying problem. >>> >>> -- >>> >>> David. >>> >>> > >>> > Glad to help. >>> > >>> > First, it could help you to trace your code. I suspect that the results >>> > are not at all what you want and tracing would help you see that. >>> > >>> > I suggest running this revision and printing out x, z, and w. >>> > >>> > #+begin_src R >>> > w = NULL >>> > for(j in 1:2) >>> > { >>> > z = NULL >>> > x = rnorm(10) >>> > z[1] = x[1] >>> > for(i in 2:10) >>> > { >>> > z[i] = x[i]+5*z[i-1] >>> > if(z[i]>4 | z[i]<1) { >>> > w[j]=i >>> > } else { >>> > w[j] = 0 >>> > } >>> > } >>> > } >>> > #+end_src >>> > >>> > >>> > You should be able to see that the value of w can easily be obtained >>> > outside of the `i' loop. >>> > [[alternative HTML version deleted]] ______________________________________________ R-help@r-project.org 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.