R help - Jun 2010 - partial solutions to: if else statement problem

Thank you to all for your help!

I received
two equal alternative solutions that bypassed elegantly the problem
from peter.l.e.konings at gmail.com
and
rafael.bjork at gmail.com

rr.dia2.corr <- rr.dia.2m
rr.dia2.corr[which(med.hyper == 1)] <- rr.dia.2m[which(med.hyper == 1)] - 5

 From pburns at pburns.seanet.com
I received a hint to understand the problem why my code does not work.
p 13 on
http://lib.stat.cmu.edu/S/Spoetry/Tutor/R_inferno.pdf

I have to digest this yet

Peter

Hi

r-help-bounces at r-project.org napsal dne 08.06.2010 14:21:10:
> Thank you to all for your help!
> 
> I received
> two equal alternative solutions that bypassed elegantly the problem
> from peter.l.e.konings at gmail.com
> and
> rafael.bjork at gmail.com
> 
> rr.dia2.corr <- rr.dia.2m
> rr.dia2.corr[which(med.hyper == 1)] <- rr.dia.2m[which(med.hyper == 1)] 
- 5

Try to compare it with my suggestion

my.result <- rr.dia.2m - (med.hyper == 1)*5
all.eqal(rr.dia2.corr, my.result)

which by my opinion will be same.

Regards
Petr

> 
>  From pburns at pburns.seanet.com
> I received a hint to understand the problem why my code does not work.
> p 13 on
> http://lib.stat.cmu.edu/S/Spoetry/Tutor/R_inferno.pdf
> 
> I have to digest this yet
> 
> Peter
> 
> 
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