[R] in continuation with the earlier R puzzle

[Petr PIKAL]

Hi

I do not use any of mentioned libraries so I can not directly answer it. I 
would try to use debug(expr.frame) to see at what time the error is 
thrown.

I have no idea why did you obtain error. Try to evaluate code in peaces 
e.g. what is result of

list(MAf=quote(SMA(Last, 20)), MAs=quote(SMA(Last, 50)))

and look for differences between results got from spx data and nifty data.

Regards
Petr



Raghu <r.raghuraman at gmail.com> napsal dne 13.07.2010 13:17:42:

> Many many thanks to all of you. The beer cleared the air of doubts!
> Pls look at the following lines of code. This is taken from the example 
of 
> tradesys documentation. When I run the given example using the 
data.frame spx 
> it works just very fine but while I use some other data.frame (here 
nifty) it 
> crashes. Now I can intuit that the total rows in the column named "Last" 
are 
> 3637 and if i do a 20d MA and a 50d MA the respective rows for each of 
them 
> are 3618 and 3588. Why does expr.frame crash for one data.frame and not 
for 
> the other? I have given str() for both below for youe kind perusal.
> 
> library(tradesys)
> > library(TTR)
> > x=nifty[,c("Open","Last")]
> > d <- expr.frame(x, list(MAf=quote(SMA(Last, 20)), MAs=quote(SMA(Last, 
50))))
> Error in data.frame(c(1000, 1001.53, 987.17, 976.28, 960.32, 951.93, 
949.29,  : 
>   arguments imply differing number of rows: 3637, 3618, 3588
> 
> 
> str(nifty)
> 'data.frame':   3637 obs. of  6 variables:
>  $ Date..GMT.: Factor w/ 3637 levels "01/01/1996","01/01/1997",..: 321 
687 807
> 929 1052 1172 1537 1650 1764 1886 ...
>  $ Open      : num  1000 1002 987 976 960 ...
>  $ High      : num  1000 1002 987 976 960 ...
>  $ Low       : num  1000 989 977 963 952 ...
>  $ Last      : num  1000 989 978 964 953 ...
>  $ Date      : num  321 687 807 929 1052 ...
> > str(spx)
> 'data.frame':   14940 obs. of  5 variables:
>  $ Open  : num  16.7 16.9 16.9 17 17.1 ...
>  $ High  : num  16.7 16.9 16.9 17 17.1 ...
>  $ Low   : num  16.7 16.9 16.9 17 17.1 ...
>  $ Close : num  16.7 16.9 16.9 17 17.1 ...
>  $ Volume: num  1260000 1890000 2550000 2010000 2520000 2160000 2630000 
> 2970000 3330000 1460000 ...
> 
> 
> Thanks 
> Raghu
> 

> On Tue, Jul 13, 2010 at 12:01 PM, Petr PIKAL <petr.pikal at precheza.cz> 
wrote:
> Hi
> 
> r-help-bounces at r-project.org napsal dne 12.07.2010 16:09:30:
> 
> > When I just run a for loop it works. But if I am going to run a for 
loop
> > every time for large vectors I might as well use C or any other
> language.
> > The reason R is powerful is becasue it can handle large vectors 
without
> each
> > element being manipulated? Please let me know where I am wrong.
> >
> > for(i in 1:length(news1o)){
> > + if(news1o[i]>s2o[i])
> > + s[i]<-1
> > + else
> > + s[i]<--1
> > + }

> Think in R not in C. Why using loops when you can use whole object
> directly. It is like drinking beer from snifters. It is possible but 
using
> pints is preferable and more convenient.
> 
> news1o>s2o
> 
> gives you a logical vector the same length
> 
> and you can use it directly for further selection or computation. You 
can
> consider FALSE as 0 and TRUE as 1 and use it as numeric vector
> so
> 
> x<-runif(10)
> y<-runif(10)
> 
> c(-1,1)[(x>y)+1]
> 
> selects -1 when FALSE and 1 when TRUE.
> 
> or you can use it in mathematical operation directly
> 
> (x>y)*2-1
> 
> Regards
> Petr
> 
> >
> > --
> > 'Raghu'
> >
> >    [[alternative HTML version deleted]]
> >
> > ______________________________________________
> > R-help at r-project.org mailing list
> > 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.

> 
> 
> 
> -- 
> 'Raghu'