Re: [R] in continuation with the earlier R puzzle
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 126 189 255 201 252 216 263
297 333 146 ...
Thanks
Raghu
On Tue, Jul 13, 2010 at 12:01 PM, Petr PIKAL wrote:
> Hi
>
> r-help-boun...@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@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'
[[alternative HTML version deleted]]
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Re: [R] in continuation with the earlier R puzzle
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 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 126 189 255 201 252 216 263
> 297 333 146 ...
>
>
> Thanks
> Raghu
>
> On Tue, Jul 13, 2010 at 12:01 PM, Petr PIKAL
wrote:
> Hi
>
> r-help-boun...@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@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'
__
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Re: [R] in continuation with the earlier R puzzle
I wanted to point out one thing that Ted said, about initializing the
vectors ('s' in your example). This can make a dramatic speed
difference if you are using a for loop (the difference is neglible
with vectorized computations).
Also, a lot of benchmarks have been flying around, each from a
different system and using random numbers without identical seeds. So
to provide an overall comparison of all the methods I saw here plus
demonstrate the speed difference for initializing a vector (if you
know its desired length in advance), I ran these benchmarks.
Notes:
I did not want to interfere with your objects so I used different
names. The equivalencies are: news1o = x; s2o = y; s = z.
system.time() automatically calculates the time difference from
proc.time() between start and finish .
> ##R version info
> sessionInfo()
R version 2.11.1 (2010-05-31)
x86_64-pc-mingw32
#snipped
>
> ##Some Sample Data
> set.seed(10)
> x <- rnorm(10^6)
> set.seed(15)
> y <- rnorm(10^6)
>
> ##Benchmark 1
> z.1 <- NULL
> system.time(for(i in 1:length(x)) {
+ if(x[i] > y[i]) {
+ z.1[i] <- 1
+ } else {
+ z.1[i] <- -1}
+ }
+ )
user system elapsed
1303.83 174.24 1483.74
>
> ##Benchmark 2
> #initialize 'z' at length
> z.2 <- vector("numeric", length = 10^6)
> system.time(for(i in 1:length(x)) {
+ if(x[i] > y[i]) {
+ z.2[i] <- 1
+ } else {
+ z.2[i] <- -1}
+ }
+ )
user system elapsed
3.770.003.77
>
> ##Benchmark 3
>
> z.3 <- NULL
> system.time(z.3 <- ifelse(x > y, 1, -1))
user system elapsed
0.380.000.38
>
> ##Benchmark 4
>
> z.4 <- vector("numeric", length = 10^6)
> system.time(z.4 <- ifelse(x > y, 1, -1))
user system elapsed
0.310.000.31
>
> ##Benchmark 5
>
> system.time(z.5 <- 2*(x > y) - 1)
user system elapsed
0.010.000.01
>
> ##Benchmark 6
>
> system.time(z.6 <- numeric(length(x))-1)
user system elapsed
0 0 0
> system.time(z.6[x > y] <- 1)
user system elapsed
0.030.000.03
>
> ##Show that all results are identical
>
> identical(z.1, z.2)
[1] TRUE
> identical(z.1, z.3)
[1] TRUE
> identical(z.1, z.4)
[1] TRUE
> identical(z.1, z.5)
[1] TRUE
> identical(z.1, z.6)
[1] TRUE
I have not replicated these on other system, but tentatively, it
appears that loops are significantly slower than ifelse(), which in
turn is slower than options 5 and 6. However, when using the same
test data and the same system, I did not find an appreciable
difference between options 5 and 6 speed wise.
Cheers,
Josh
On Mon, Jul 12, 2010 at 7:09 AM, Raghu wrote:
> 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
> + }
>
> --
> 'Raghu'
>
> [[alternative HTML version deleted]]
>
> __
> R-help@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.
>
--
Joshua Wiley
Ph.D. Student, Health Psychology
University of California, Los Angeles
http://www.joshuawiley.com/
__
R-help@r-project.org mailing list
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
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Re: [R] in continuation with the earlier R puzzle
Thanks to you all. I stand corrected Ted and Manuela:) I am just an end user
and trying to pick up from such forums. Many thanks sirs.
On Mon, Jul 12, 2010 at 5:45 PM, Huso, Manuela wrote:
> Using Ted Harding's example:
>
> news1o <- runif(100)
> s2o<- runif(100)
>
> pt1 <- proc.time()
> s <- numeric(length(news1o))-1 # Set all of s to -1
> s[news1o>s2o] <-1# Change to 1 only those values of s
> # for which news1o>s2o
> pt2<- proc.time()
> pt2-pt1 # Takes even less time...
> # user system elapsed
> # 0.040.000.05
>
> >::<>::<>::<>::<>::<>::<>::<>::<>::<>::<>::<>::<
> Please note: I will be out of the office and out
> of email contact from 7/11-7/25/2010
> >::<>::<>::<>::<>::<>::<>::<>::<>::<>::<>::<>::<
> Manuela Huso
> Consulting Statistician
> 201H Richardson Hall
> Department of Forest Ecosystems and Society
> Oregon State University
> Corvallis, OR 97331
> ph: 541-737-6232
> fx: 541-737-1393
>
>
> -Original Message-
> From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org]
> On Behalf Of Ted Harding
> Sent: Monday, July 12, 2010 9:36 AM
> To: r-help@r-project.org
> Cc: Raghu
> Subject: Re: [R] in continuation with the earlier R puzzle
>
> On 12-Jul-10 14:09:30, Raghu wrote:
> > 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
> > + }
> >
> > --
> > 'Raghu'
>
> Many operations over the whole length of vectors can be done
> in "vectorised" form, in which an entire vector is changed
> in one operation based on the values of the separate elemnts
> of other vectors, also all take into account in a single
> operation. What happens "behind to scenes" is that the single
> element by element operations are performed by a function
> in a precompiled (usually from C) library. Hence R already
> does what you are suggesting as a "might as well" alternative!
>
> Below is an example, using long vectors. The first case is a
> copy of your R loop above (with some additional initialisation
> of the vectors). The second achieves the same result in the
> "vectorised" form.
>
> news1o <- runif(100)
> s2o<- runif(100)
> s <- numeric(length(news1o))
>
> proc.time()
> #user system elapsed
> # 1.728 0.680 450.257
> for(i in 1:length(news1o)){ ### Using a loop
>if(news1o[i]>s2o[i])
>s[i]<- 1
>else
>s[i]<- (-1)
> }
> proc.time()
> #user system elapsed
> # 11.184 0.756 460.340
> s2 <- 2*(news1o > s2o) - 1 ### Vectorised
> proc.time()
> #user system elapsed
> # 11.348 0.852 460.663
>
> sum(s2 != s)
> # [1] 0 ### Results identical
>
> Result: The loop took (11.184 - 1.728) = 9.456 seconds,
> Vectorised, it took (11.348 - 11.184) = 0.164 seconds.
>
> Loop/Vector = (11.184 - 1.728)/(11.348 - 11.184) = 57.65854
>
> i.e. nearly 60 times as long.
>
> Ted.
>
>
> E-Mail: (Ted Harding)
> Fax-to-email: +44 (0)870 094 0861
> Date: 12-Jul-10 Time: 17:36:07
> -- XFMail --
>
> __
> R-help@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'
[[alternative HTML version deleted]]
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Re: [R] in continuation with the earlier R puzzle
Using Ted Harding's example:
news1o <- runif(100)
s2o<- runif(100)
pt1 <- proc.time()
s <- numeric(length(news1o))-1 # Set all of s to -1
s[news1o>s2o] <-1# Change to 1 only those values of s
# for which news1o>s2o
pt2<- proc.time()
pt2-pt1 # Takes even less time...
# user system elapsed
# 0.040.000.05
>::<>::<>::<>::<>::<>::<>::<>::<>::<>::<>::<>::<
Please note: I will be out of the office and out
of email contact from 7/11-7/25/2010
>::<>::<>::<>::<>::<>::<>::<>::<>::<>::<>::<>::<
Manuela Huso
Consulting Statistician
201H Richardson Hall
Department of Forest Ecosystems and Society
Oregon State University
Corvallis, OR 97331
ph: 541-737-6232
fx: 541-737-1393
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
Behalf Of Ted Harding
Sent: Monday, July 12, 2010 9:36 AM
To: r-help@r-project.org
Cc: Raghu
Subject: Re: [R] in continuation with the earlier R puzzle
On 12-Jul-10 14:09:30, Raghu wrote:
> 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
> + }
>
> --
> 'Raghu'
Many operations over the whole length of vectors can be done
in "vectorised" form, in which an entire vector is changed
in one operation based on the values of the separate elemnts
of other vectors, also all take into account in a single
operation. What happens "behind to scenes" is that the single
element by element operations are performed by a function
in a precompiled (usually from C) library. Hence R already
does what you are suggesting as a "might as well" alternative!
Below is an example, using long vectors. The first case is a
copy of your R loop above (with some additional initialisation
of the vectors). The second achieves the same result in the
"vectorised" form.
news1o <- runif(100)
s2o<- runif(100)
s <- numeric(length(news1o))
proc.time()
#user system elapsed
# 1.728 0.680 450.257
for(i in 1:length(news1o)){ ### Using a loop
if(news1o[i]>s2o[i])
s[i]<- 1
else
s[i]<- (-1)
}
proc.time()
#user system elapsed
# 11.184 0.756 460.340
s2 <- 2*(news1o > s2o) - 1 ### Vectorised
proc.time()
#user system elapsed
# 11.348 0.852 460.663
sum(s2 != s)
# [1] 0 ### Results identical
Result: The loop took (11.184 - 1.728) = 9.456 seconds,
Vectorised, it took (11.348 - 11.184) = 0.164 seconds.
Loop/Vector = (11.184 - 1.728)/(11.348 - 11.184) = 57.65854
i.e. nearly 60 times as long.
Ted.
E-Mail: (Ted Harding)
Fax-to-email: +44 (0)870 094 0861
Date: 12-Jul-10 Time: 17:36:07
-- XFMail --
__
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and provide commented, minimal, self-contained, reproducible code.
Re: [R] in continuation with the earlier R puzzle
On 12-Jul-10 14:09:30, Raghu wrote:
> 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
> + }
>
> --
> 'Raghu'
Many operations over the whole length of vectors can be done
in "vectorised" form, in which an entire vector is changed
in one operation based on the values of the separate elemnts
of other vectors, also all take into account in a single
operation. What happens "behind to scenes" is that the single
element by element operations are performed by a function
in a precompiled (usually from C) library. Hence R already
does what you are suggesting as a "might as well" alternative!
Below is an example, using long vectors. The first case is a
copy of your R loop above (with some additional initialisation
of the vectors). The second achieves the same result in the
"vectorised" form.
news1o <- runif(100)
s2o<- runif(100)
s <- numeric(length(news1o))
proc.time()
#user system elapsed
# 1.728 0.680 450.257
for(i in 1:length(news1o)){ ### Using a loop
if(news1o[i]>s2o[i])
s[i]<- 1
else
s[i]<- (-1)
}
proc.time()
#user system elapsed
# 11.184 0.756 460.340
s2 <- 2*(news1o > s2o) - 1 ### Vectorised
proc.time()
#user system elapsed
# 11.348 0.852 460.663
sum(s2 != s)
# [1] 0 ### Results identical
Result: The loop took (11.184 - 1.728) = 9.456 seconds,
Vectorised, it took (11.348 - 11.184) = 0.164 seconds.
Loop/Vector = (11.184 - 1.728)/(11.348 - 11.184) = 57.65854
i.e. nearly 60 times as long.
Ted.
E-Mail: (Ted Harding)
Fax-to-email: +44 (0)870 094 0861
Date: 12-Jul-10 Time: 17:36:07
-- XFMail --
__
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Re: [R] in continuation with the earlier R puzzle
On Jul 12, 2010, at 10:09 AM, Raghu wrote:
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
+ }
Perhaps:
s <- 2*( news1o > s2o[1:length(news1o)] ) - 1
...which I think will throw errors under pretty much the same
conditions that would cause errors in that loop.
--
David Winsemius, MD
West Hartford, CT
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Re: [R] in continuation with the earlier R puzzle
> 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
> + }
You might give ifelse() a shot here.
s <- ifelse(news1o > s2o, 1, -1)
Learning to think in vectors is important in R, just like thinking in sets
is important for SQL, or thinking in rows and steps is important in SAS.
cur
--
Curt Seeliger, Data Ranger
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Re: [R] in continuation with the earlier R puzzle
I don't know what is wrong with your code but I believe you should use
ifelse instead of a for loop:
s <- ifelse(news1o > s2o, 1 , -1 )
Alain
On 12-Jul-10 16:09, Raghu wrote:
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
+ }
--
Alain Guillet
Statistician and Computer Scientist
SMCS - IMMAQ - Université catholique de Louvain
Bureau c.316
Voie du Roman Pays, 20
B-1348 Louvain-la-Neuve
Belgium
tel: +32 10 47 30 50
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