Re: [R] Factors.

2004-07-07 Thread Uwe Ligges
[last week,] [EMAIL PROTECTED] wrote:
Hello,
I'm new with R. I need some help; I have a matrix of data to wich i want to 
apply the function dudi.acm to perform multiple correspondence analysis. 
However to use it all variables must be factors, so how can i turn each column 
of the matrix into a factor? I've tried as.factor. It works isolated for each 
column, but when I form the matrix of all factors it doesnt work.
Please help me!

[Looks like there was no answer until now.]
a) Please tell us which package the functions you are using are in (I 
know, it is ade4).
b) Please follow the posting guide which tells you some more things 
(which are not *that* important here).
c) Please read the help page! ?dudi.acm tells you
df, df1, df2: data frames containing only factors
   ^^^
You cannot have a matrix of factors, because the required attributes 
cannot be set for each column separately. Instead, use a data.frame.

Uwe Ligges
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[R] Daily time series

2004-07-07 Thread Vito Ricci
Hi, 

I'm dealing with time series with 1 observaton for day
(data sampled daily). I will create a ts object using
that time series and the function ts(). 
In ts() help is written:

The value of argument 'frequency' is used when the
series is sampled an integral number of times in each
unit time interval. For example, one could use a value
of '7' for 'frequency' when the data are sampled
daily, and the natural time period is a week, or '12'
when the data are sampled monthly and the natural time
period
is a year.  Values of '4' and '12' are assumed in
(e.g.) 'print' methods to imply a quarterly and
monthly series respectively.

But what value should assume start in ts function?

Here is a time series:

1/1 10
2/1 20
3/1 30
4/1 40
5/1 50
6/1 60

x-c(10,20,30,40,50,60) ## observation
serie-ts(dati, start=c(1,1),frequency=7) ##creating
ts object

serie  ## printing ts output

Time Series:
Start = c(1, 1) 
End = c(1, 6) 
Frequency = 7 
[1] 10 20 30 40 50 60

Could someone help me?

Thanks in advance.

Sincerely.
Vito Ricci


=
Diventare costruttori di soluzioni

Visitate il portale http://www.modugno.it/
e in particolare la sezione su Palese http://www.modugno.it/archivio/cat_palese.shtml

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[R] boxplot a list of objects

2004-07-07 Thread Lefebure Tristan
Hi list,

#Imagine we have vectors of different length (in practice 100 vectors):
a-c(1:10)
b-c(1:20)
c-c(1:30)

#then we got a list of the names of those objects:

list-c(a,b,c)

#I don't find how to boxplot them using a less stupid way than :

boxplot(get(list[1]),get(list[2]),get(list[3]))


Thanks for any advice !


-- 

Tristan LEFEBURE
Laboratoire d'écologie des hydrosystèmes fluviaux (UMR 5023)
Université Lyon I - Campus de la Doua
6 rue Dubois 69622 Villeurbanne - France

Phone: (33) (0)4 72 43 29 45
Fax: (33) (0)4 72 43 15 23

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[R] Howto debug R on Windows XP?

2004-07-07 Thread Brueckner-Keutmann-GbR
Hello,

I start working with R and I have tried to debug R on a Windows XP system.
Unfortunately I am not able to set a breakpoint in the package SJava, which
I am interested in.
So far I succeed to compile R with the DEBUG=T option, and followed the
hints given in the manual/FAQs about debugging.
After starting the gdb, I also succeed with

 break WinMain
 run

so that the program stops there. The debugger is also
able to find und list the function R_ReadConsole, but the command

 break R_ReadConsole

is replied:

Cannot access memory at address 0x1e6a0

Can anybody give a hint how to continue?

My motivation is to use the SJava package and to continue the work Jens
Oehlschlaegel and Ingo von Otte started with.

Thanks for your help!

Herbert Brückner

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[R] AW: Howto debug R on Windows XP?

2004-07-07 Thread Brueckner-Keutmann-GbR


-Ursprüngliche Nachricht-
Von: Brueckner-Keutmann-GbR [mailto:[EMAIL PROTECTED]
Gesendet: Wednesday, July 07, 2004 9:48 AM
An: R_Help Mailing List
Betreff: Howto debug R on Windows XP?


Hello,

I start working with R and I have tried to debug R on a Windows XP system.
Unfortunately I am not able to set a breakpoint in the package SJava, which
I am interested in.
So far I succeed to compile R with the DEBUG=T option, and followed the
hints given in the manual/FAQs about debugging.
After starting the gdb, I also succeed with

 break WinMain
 run

so that the program stops there. The debugger is also
able to find und list the function R_ReadConsole, but the command

 break R_ReadConsole

is replied:

Cannot access memory at address 0x1e6a0

Can anybody give a hint how to continue?

My motivation is to use the SJava package and to continue the work Jens
Oehlschlaegel and Ingo von Otte started with.

Thanks for your help!

Herbert Brückner

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Re: [R] Random intercept model with time-dependent covariates, results different from SAS

2004-07-07 Thread Dan Bebber
Hello,
I have been struggling with a similar problem, i.e. fitting an LME model to
unbalanced repeated measures data.
I found Linear Mixed Models by John Fox
(http://socserv2.socsci.mcmaster.ca/jfox/Books/Companion/appendix-mixed-mode
ls.pdf)
quite helpful.
Fox gives examples which are unbalanced, so I guess that balance is not a
requirement (assuming Fox is correct). However, the sample sizes are large
compared to yours (and mine), which may make a difference.

Dan Bebber


Dr. Daniel P. Bebber
Department of Plant Sciences
University of Oxford
South Parks Road
Oxford
OX1 3RB
Tel. 01865 275060
Web. http://www.forestecology.co.uk/

Data, data, data! he cried impatiently. I can't make bricks without
clay
- Sherlock Holmes, The Adventure of the Copper Beeches, 1892



 Message: 24
 Date: Sun,  4 Jul 2004 19:21:32 +1000
 From: Keith Wong [EMAIL PROTECTED]
 Subject: Re: [R] Random intercept model with time-dependent covariates,
results different from AS
 To: Prof Brian Ripley [EMAIL PROTECTED]
 Cc: [EMAIL PROTECTED]
 Message-ID: [EMAIL PROTECTED]
 Content-Type: text/plain; charset=ISO-8859-1

 Thank you for the very prompt response. I only included a small
 part of the
 output to make the message brief. I'm sorry it did not provide
 enough detail to
 answer my question. I have appended the summary() and anova()
 outputs to the
 two models I fitted in R.

 Quoting Prof Brian Ripley [EMAIL PROTECTED]:

  Looking at the significance of a main effect (group) in the
 presence of an
  interaction (time:group) is hard to interpret, and in your case
 is I think
  not even interesting.  (The `main effect' probably represents difference
  in intercept for the time effect, that is the group difference
 at the last
  time.  But see the next para.)  Note that the two systems are returning
  different denominator dfs.


 I take your point that the main effect is probably not interesting in the
 presence of an interaction. I was checking the results for
 consistency to see
 if I was doing the right thing. I was not 100% sure that the SAS
 code was in
 itself correct.

  At this point you have not told us enough.  My guess is that you have
  complete balance with the same number of subjects in each
 group.  In that
  case the `group' effect is in the between-subjects stratum (as
 defined for
  the use of Error in aov, which you could also do), and thus R's 11 df
  would be right (rather than 44, without W and Z).  Without balance Type
  III tests get much harder to interpret and the `group' effect
 would appear
  in two strata and there is no simple F test in the classical theory.  So
  further guessing, SAS may have failed to detect balance and so used the
  wrong test.

 I had not appreciated the need for balance: in actual fact, one
 group has 5
 subjects and the other 7. Will this be a problem? Would the R
 analysis still be
 valid in that case?


  The time-dependent covariates muddy the issue more, and I
 looked mainly at
  the analyses without them.  Again, a crucial fact is not here: do the
  covariates depend on the subjects as well?

 Yes the covariates are measures of blood pressure and pulse, and
 they depend on
 the subjects as well.

  The good news is that the results _are_ similar.  You do have different
  time behaviour in the two groups.  So stop worrying about tests of
  uninteresting hypotheses and concentrate of summarizing that difference.
 
  --
  Brian D. Ripley,  [EMAIL PROTECTED]
  Professor of Applied Statistics,  http://www.stats.ox.ac.uk/~ripley/
  University of Oxford, Tel:  +44 1865 272861 (self)
  1 South Parks Road, +44 1865 272866 (PA)
  Oxford OX1 3TG, UKFax:  +44 1865 272595


 Thank you. I was concerned that one or both methods were
 incorrect given the
 results were inconsistent. Perhaps reassuringly, the parameter
 estimates for
 the fixed effects in both SAS and R were the same.

 Is the model specification OK for the model with just time, group
 and their
 interaction?
 Is the model specification with the 2 time dependent covariates
 appropriate?

 Once again, I'm very grateful for the time you've taken to answer
 my questions.

 Keith

 [Output from the 2 models fitted in R follows]

  g1 = lme(Y ~ time + group + time:group, random = ~ 1 | id, data
 = datamod)

  anova(g1)
 numDF denDF   F-value p-value
 (Intercept) 144  3.387117  0.0725
 time444 10.620547  .0001
 group   111  0.508092  0.4908
 time:group  444  3.961726  0.0079
  summary(g1)
 Linear mixed-effects model fit by REML
  Data: datamod
AIC  BIClogLik
   372.4328 396.5208 -174.2164

 Random effects:
  Formula: ~1 | id
 (Intercept) Residual
 StdDev:11.05975 3.228684

 Fixed effects: Y ~ time + group + time:group
   Value Std.Error DF   t-value p-value
 (Intercept)   8.250  4.073428 44  2.025321  0.0489
 time1-0.250  1.614342 44 

Re: [R] boxplot a list of objects

2004-07-07 Thread Lefebure Tristan
thanks a lot !

an other simple solution proposed by Stefano Guazzetti is  :
boxplot(list(a, b, c))

(ok I will never use again a function name for an object name)

On Wednesday 07 July 2004 10:48, Unternährer Thomas, uth wrote:
 One possibility ist

 boxplot(sapply(ListOfNames, get, env = .GlobalEnv))

 Hope that this helps

 Thomas


 -Ursprüngliche Nachricht-
 Von: Lefebure Tristan [mailto:[EMAIL PROTECTED]
 Gesendet: Mittwoch, 7. Juli 2004 10:34
 An: [EMAIL PROTECTED]
 Betreff: [R] boxplot a list of objects


 Hi list,

 #Imagine we have vectors of different length (in practice 100 vectors):
 a-c(1:10)
 b-c(1:20)
 c-c(1:30)

 #then we got a list of the names of those objects:

 list-c(a,b,c)

 #I don't find how to boxplot them using a less stupid way than :

 boxplot(get(list[1]),get(list[2]),get(list[3]))


 Thanks for any advice !

-- 

Tristan LEFEBURE
Laboratoire d'écologie des hydrosystèmes fluviaux (UMR 5023)
Université Lyon I - Campus de la Doua
6 rue Dubois 69622 Villeurbanne - France

Phone: (33) (0)4 72 43 29 45
Fax: (33) (0)4 72 43 15 23

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Re: [R] AW: Howto debug R on Windows XP?

2004-07-07 Thread Duncan Murdoch
On Wed, 7 Jul 2004 10:50:18 +0200, Brueckner-Keutmann-GbR
[EMAIL PROTECTED] wrote:


-Ursprüngliche Nachricht-
Von: Brueckner-Keutmann-GbR [mailto:[EMAIL PROTECTED]
Gesendet: Wednesday, July 07, 2004 9:48 AM
An: R_Help Mailing List
Betreff: Howto debug R on Windows XP?


Hello,

I start working with R and I have tried to debug R on a Windows XP system.
Unfortunately I am not able to set a breakpoint in the package SJava, which
I am interested in.
So far I succeed to compile R with the DEBUG=T option, and followed the
hints given in the manual/FAQs about debugging.
After starting the gdb, I also succeed with

 break WinMain
 run

so that the program stops there. The debugger is also
able to find und list the function R_ReadConsole, but the command

 break R_ReadConsole

is replied:

Cannot access memory at address 0x1e6a0

Can anybody give a hint how to continue?

My motivation is to use the SJava package and to continue the work Jens
Oehlschlaegel and Ingo von Otte started with.

I've written some web pages about debugging in R that are mostly
oriented towards Windows users.  Go to
http://www.stats.uwo.ca/faculty/murdoch/software/debuggingR  I'm not
sure what's going wrong for you, but you're not doing things the way I
would.

These web pages are quite new; please let me know what is missing or
unclear.

Duncan Murdoch

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[R] Rmetrics Documentation Update

2004-07-07 Thread Diethelm Wuertz
I like to announce that some of the Rmetrics
Documents have been updated to Version
R 191.10057
   Rmetrics Flyer: 
http://www.itp.phys.ethz.ch/econophysics/R/pdf/DocRmetrics.pdf
   Rmetrics Fact Sheet: 
http://www.itp.phys.ethz.ch/econophysics/R/pdf/DocFactsheet.pdf
   Rmetrics Reference Card: 
http://www.itp.phys.ethz.ch/econophysics/R/pdf/DocRefcard.pdf

Unfortunately, the User Guides are still behind, having Version No 1.8.1.
They will be updated in the near future.
  
Best Regards

Diethelm
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[R] Using permax with Data Frame Containing Missing Values

2004-07-07 Thread Brian Lane
I'm new to this site so I hope this isn't too naive a problem. I'm trying 
to use the permax function with a data frame containing gene expression 
measurements taken from 79 microarray experiments with 3000 genes per 
array. The data contains missing values and every time I use permax with 
the data frame I get the error:

NA/NaN/Inf in foreign function call (arg 1)
Could anyone suggest how I might get round this problem?
Regards,
Brian Lane
Dept of Haematology
University of Liverpool
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[R] vectorizing sapply() code (Modified by Aaron J. Mackey)

2004-07-07 Thread Aaron J. Mackey
[ Not sure why, but the first time I sent this it never seemed to go 
through; apologies if you're seeing this twice ... ]

I have some fully functional code that I'm guessing can be done 
better/quicker with some savvy R vector tricks; any help to make this 
run a bit faster would be greatly appreciated; I'm particularly stuck 
on how to calculate using row-wise vectors without iterating 
explicitly over the dataframe or table ...

library(stats4);
d - data.frame( ix=c(0,1,2,3,4,5,6,7),
 ct=c(253987,  9596, 18680,  2630,  8224,  3590,  5534, 
18937),
 A=c(  0, 1, 0, 1, 0, 1, 0, 
1),
 B=c(  0, 0, 1, 1, 0, 0, 1, 
1),
 C=c(  0, 0, 0, 0, 1, 1, 1, 
1)
   );
ct - round(logb(length(d$ix), 2))
ll - function( th=0.5,
a1=log(0.5), a2=log(0.5), a3=log(0.5),
b1=log(0.5), b2=log(0.5), b3=log(0.5)
  ) {
  a - exp(sapply(1:ct, function (x) { get(paste(a, x, sep=)) }));
  b - exp(sapply(1:ct, function (x) { get(paste(b, x, sep=)) }));
  -sum( d$ct * log( sapply( d$ix,
function (ix, th, a, b) {
  x - d[ix+1,3:(ct+2)]
  (th * prod((b ^ (1-x)) * ((1-b) ^ x   
 ))) +
  ((1-th) * prod((a ^ x) * ((1-a) ^ 
(1-x
},
th, a, b
  )
  )
  );
}

ml - mle(ll,
  lower=c(0+1e-5, rep(log(0+1e-8), 2*ct)),
  upper=c(1-1e-5, rep(log(1-1e-8), 2*ct)),
  method=L-BFGS-B
 );
For those interested in the math, this is the MLE procedure to estimate 
the false positive/false negative rates (a and b) of three diagnostic 
(A, B and C) tests that have the observed performance recapitulated in 
dataframe d, but no gold standard (sometimes called latent class 
analysis, or LCA).

Thanks for any help,
-Aaron
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Re: [R] Using permax with Data Frame Containing Missing Values

2004-07-07 Thread Robert Gentleman
There is currently no handling of NAs in permax. Your only simple
option is to drop those rows with NA's in them, or to perform some
sort of imputation.

I will mention it to the package's author, 
 Robert

On Wed, Jul 07, 2004 at 12:47:05PM +0100, Brian Lane wrote:
 I'm new to this site so I hope this isn't too naive a problem. I'm trying 
 to use the permax function with a data frame containing gene expression 
 measurements taken from 79 microarray experiments with 3000 genes per 
 array. The data contains missing values and every time I use permax with 
 the data frame I get the error:
 
 NA/NaN/Inf in foreign function call (arg 1)
 
 Could anyone suggest how I might get round this problem?
 
 Regards,
 Brian Lane
 Dept of Haematology
 University of Liverpool
 
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-- 
+---+
| Robert Gentleman phone : (617) 632-5250   |
| Associate Professor  fax:   (617)  632-2444   |
| Department of Biostatistics  office: M1B20|
| Harvard School of Public Health  email: [EMAIL PROTECTED]|
+---+

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[R] Mapinfo mid and mif files

2004-07-07 Thread Andrew McCulloch

Hi ,

Has anyone any experience of converting Mapinfo mid and mif
files into a format that can be used with the R spatial packages. 
Thanks.

yours sincerely
Andrew McCulloch
Department of Urban Studies
University of Glasgow
G12 8RS

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Re: [R] Creating Binary Outcomes from a continuous variable

2004-07-07 Thread Roger Bivand
On Wed, 7 Jul 2004, Doran, Harold wrote:

 Dear List:
 

?cut

  
 
 I have searched the archives and my R books and cannot find a method to
 transform a continuous variable into a binary variable. For example, I
 have test score data along a continuous scale. I want to create a new
 variable in my dataset that is 1=above a cutpoint (or passed the test)
 and 0=otherwise.
 
  
 
 My instinct tells me that this will require a combination of the
 transform command along with a conditional selection. Any help is much
 appreciated.
 
  
 
 Thanks,
 
 Harold
 
 
   [[alternative HTML version deleted]]
 
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-- 
Roger Bivand
Economic Geography Section, Department of Economics, Norwegian School of
Economics and Business Administration, Breiviksveien 40, N-5045 Bergen,
Norway. voice: +47 55 95 93 55; fax +47 55 95 93 93
e-mail: [EMAIL PROTECTED]

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Re: [R] Mapinfo mid and mif files

2004-07-07 Thread F. Tusell
Not sure if it will work straight with .mid or .mif files, but it works 
for me on .tab files, which were
supposed to be read with the Excel map tool package and beleive confro 
to the MapInfo format.

I asked Roger Bivand at the recent useR!2004 meeting and he made several 
suggestions. Following the
leads he gave me, I found a small program (ogr2ogr) which translates my 
.tab files into .shp files,
which I can then read into R. The MapInfo format is explicitly supported.

I use Debian Linux (sarge), and ogr2ogr  is part of the package 
gdal-bin. A batch
file like

for i in *.tab ; do
ogr2ogr ${i%.*}.shp $i ;
done
converted a bunch of files in a snap. Hope it may be of help to you.
ft.
--
Fernando TUSELLe-mail:
Departamento de Econometría y Estadística   [EMAIL PROTECTED] 
Facultad de CC.EE. y Empresariales Tel:   (+34)94.601.3733
Avenida Lendakari Aguirre, 83  Fax:   (+34)94.601.3754
E-48015 BILBAO  (Spain)Secr:  (+34)94.601.3740

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Re: [R] Creating Binary Outcomes from a continuous variable

2004-07-07 Thread Marc Schwartz
On Wed, 2004-07-07 at 07:57, Doran, Harold wrote:
 Dear List:
 
 I have searched the archives and my R books and cannot find a method to
 transform a continuous variable into a binary variable. For example, I
 have test score data along a continuous scale. I want to create a new
 variable in my dataset that is 1=above a cutpoint (or passed the test)
 and 0=otherwise.

 My instinct tells me that this will require a combination of the
 transform command along with a conditional selection. Any help is much
 appreciated.

Example:

 a - rnorm(20)
 b - ifelse(a  0, 0, 1)

 a
 [1] -1.0735800 -0.6788456  1.9979801 -0.4026760  0.1781791 -1.1540434
 [7] -1.0842728  1.6042602 -0.7950492 -0.1194323  0.4450296  1.9269333
[13] -0.4456181 -0.8374677 -1.1898772  1.7353067  1.8619422 -0.1679996
[19] -0.2656138 -1.5529884
 b
 [1] 0 0 1 0 1 0 0 1 0 0 1 1 0 0 0 1 1 0 0 0

HTH,

Marc Schwartz

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R: [R] Creating Binary Outcomes from a continuous variable

2004-07-07 Thread Guazzetti Stefano
consider a cutpoint of 20

x-runif(100, min=1, max=50)
as.integer(x  20)

Stefano

 -Messaggio originale-
 Da: Doran, Harold [mailto:[EMAIL PROTECTED]
 Inviato: mercoledì 7 luglio 2004 14.57
 A: [EMAIL PROTECTED]
 Oggetto: [R] Creating Binary Outcomes from a continuous variable
 
 
 Dear List:
 
  
 
 I have searched the archives and my R books and cannot find a 
 method to
 transform a continuous variable into a binary variable. For example, I
 have test score data along a continuous scale. I want to create a new
 variable in my dataset that is 1=above a cutpoint (or passed the test)
 and 0=otherwise.
 
  
 
 My instinct tells me that this will require a combination of the
 transform command along with a conditional selection. Any help is much
 appreciated.
 
  
 
 Thanks,
 
 Harold
 
 
   [[alternative HTML version deleted]]
 
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[R] Creating Binary Outcomes from a continuous variable

2004-07-07 Thread Doran, Harold
Dear List:

 

I have searched the archives and my R books and cannot find a method to
transform a continuous variable into a binary variable. For example, I
have test score data along a continuous scale. I want to create a new
variable in my dataset that is 1=above a cutpoint (or passed the test)
and 0=otherwise.

 

My instinct tells me that this will require a combination of the
transform command along with a conditional selection. Any help is much
appreciated.

 

Thanks,

Harold


[[alternative HTML version deleted]]

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[R] lme with poly(x,2) terms

2004-07-07 Thread Andrew Beckerman
Hi there.
Mac OSX 3.3.4 R 1.9.1
I am analysing a data set with the following model
m4- 
lme(fixed=sr~time*poly(energy,2)*poly(dist,2),random=~time|pot,data=deh)

where time is one of six months, pot is a jar in which the repeated  
measures of species number (sr) was made.  energy and dist  
(disturbance) are fixed experimental treatments. We are trying to test  
the hypothesis that there is an interaction between energy and  
disturbance that varies through time, with the expectation that sr  
varies quadratically with energy and with disturbance.  Our difficulty  
is interpreting the various outputs from the model, assuming it is  
specified correctly - sorry if this is more a stats question than a R  
mechanics question.

summary(m1) and anova(m1) produce the tables below the .
Q1) Am i correct to assume that the anova table is sequential?
Q2) How does one interpret the fixed effects/coefficients table?  Do  
the insignificant terms for poly(dist)2 all the way down (Up) to its  
main effect suggest that a quadratic function in dist is not  
significant?
Q3) If we remove the quadratic term in dist and compare it to the model  
with poly(dist,2), the anova says the polynomial is significant

 anova(update(m2,~.,method=ML),update(m4,~.,method=ML))
  Model df  AIC  BIClogLik
Test L.Ratio p-value
update(m2, ~., method = ML) 1 16 2781.683 2858.271 -1374.841
update(m4, ~., method = ML) 2 22 2771.380 2876.688 -1363.690 1 vs  
2  22.303  0.0011

despite only the main effect of poly(dist,2) being significant in the  
terms. Is the best approach to use the anova test or the coefficients?   
How does one justify the insignificance of every term with poly(dist)2  
in it?

Many thanks in advance
andrew
-
summary(m1)
Linear mixed-effects model fit by REML
 Data: deh
   AIC  BIClogLik
  2687.974 2792.830 -1321.987
Random effects:
 Formula: ~time | pot
 Structure: General positive-definite, Log-Cholesky parametrization
StdDevCorr
(Intercept) 1.5503393 (Intr)
time0.1858609 -0.862
Residual0.9234853
Fixed effects: sr ~ time * poly(energy, 2) * poly(dist, 2)
Value Std.Error  DF   t-value  
p-value
(Intercept)8.2424   0.14576 721  56.54737   
0.
time  -1.1447   0.02376 721 -48.16926   
0.
poly(energy, 2)1  18.2052   4.34118 721   4.19361   
0.
poly(energy, 2)2 -43.8133   4.34213 721 -10.09028   
0.
poly(dist, 2)1-9.9600   4.34169 721  -2.29403   
0.0221
poly(dist, 2)2   -10.6639   4.34198 721  -2.45599   
0.0143
time:poly(energy, 2)1  1.7320   0.70705 721   2.44961   
0.0145
time:poly(energy, 2)2  5.6245   0.70695 721   7.95608   
0.
time:poly(dist, 2)1   -0.6569   0.70701 721  -0.92908   
0.3532
time:poly(dist, 2)20.0400   0.70697 721   0.05657   
0.9549
poly(energy, 2)1:poly(dist, 2)1  356.6786 128.77967 721   2.76968   
0.0058
poly(energy, 2)2:poly(dist, 2)1  -99.7288 128.60505 721  -0.77547   
0.4383
poly(energy, 2)1:poly(dist, 2)2  -11.4295 129.65263 721  -0.08816   
0.9298
poly(energy, 2)2:poly(dist, 2)2  149.5420 129.80979 721   1.15201   
0.2497
time:poly(energy, 2)1:poly(dist, 2)1 -79.3803  20.96606 721  -3.78613   
0.0002
time:poly(energy, 2)2:poly(dist, 2)1  59.4570  20.93577 721   2.83997   
0.0046
time:poly(energy, 2)1:poly(dist, 2)2 -20.6131  21.10723 721  -0.97659   
0.3291
time:poly(energy, 2)2:poly(dist, 2)2 -22.3304  21.13159 721  -1.05673   
0.2910

 anova(m4)
   numDF denDF   F-value p-value
(Intercept)1   721  888.6686  .0001
time   1   721 2321.2473  .0001
poly(energy, 2)2   721   77.1328  .0001
poly(dist, 2)  2   721   22.9940  .0001
time:poly(energy, 2)   2   721   34.6873  .0001
time:poly(dist, 2) 2   7210.4551  0.6345
poly(energy, 2):poly(dist, 2)  4   7212.5824  0.0361
time:poly(energy, 2):poly(dist, 2) 4   7216.1290  0.0001
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[R] KalmanSmooth problem

2004-07-07 Thread Hazenberg21, Pieter
Hello,
In R I am trying to use Kalman filtering to find a solution for an hydrological 
problem. With Kalman Filtering I want to estimate the discharge comming from three 
storage bassins. I have programmed a function in R which can run KalmanSmooth. When 
I'm asking for the function and putting in values, R detects the following error: 
Error in as.vector(data) : Argument S1 is missing, with no default.
I have try to find a solution for this error in the R help file, and in different 
manuals, but I can't find it. Please help me find a solution.
Question: What does R mean with S1 and what am I doing wrong?
Here is the way I have programmed the hydrological problem in R. 
 
 discharge=read.table(file=C:/Program Files/R/rw1090/discharge.txt,header=T)
 deb=discharge[,1]
 deb
  [1] 11.545313  8.045465  5.670868  4.044584  2.919311  2.306668  2.940956
  [8]  4.238159  5.017374  3.818236  2.928805  2.262183  1.757765  1.633945
 [15]  2.295130  3.454054  4.035224  3.193967  2.533181  2.012406  1.600836
 [22]  1.652155  2.428678  3.642827  4.019545  3.209473  2.563617  2.048347
 [29]  1.637041  1.828952  2.757842  4.050821  4.147013  3.316503  2.652490
 [36]  2.121535  1.696934  2.027763  3.107366  4.429670  4.160178  3.327950
 [43]  2.662237  2.129710  1.703717  2.158095  3.337039  4.582359  3.905901
 [50]  3.124690  2.499732  1.999772  1.599810  2.130893  3.302622  4.336081
 [57]  3.468857  2.775081  2.220062  1.776048  1.560859  2.169537  3.348081
 [64]  4.170552  3.336440  2.669151  2.135320  1.708256  1.648859  2.374217
 [71]  3.624091  4.248563  3.398850  2.719080  2.175264  1.740211  1.826122
 [78]  2.704749  4.056438  4.437309  3.549847  2.839878  2.271902  1.817522
 [85]  2.053994  3.107875  4.548436  4.600601  3.680481  2.944385  2.355508
 [92]  1.884406  2.273248  3.490148  4.949898  4.584409  3.667527  2.934022
 [99]  2.347217  1.84
 Kalm = function(x,O1,O2,O3,T1,T2,T3,T4,T5,t,ga){
+ t=array(c(1+ga*O1+t/O1*(-(1/T2)-(1/T3)-(1/T1)),t/O1*(1/T2),t/O1*(1/T3),
+ t/O2*(1/T2),1+ga*O2+t/O2*(-(1/T2)-(1/T4)),t/O2*(1/T4),
+ t/O3*(1/T3),t/O3*(1/T4),1+ga*O3+t/O3*(-(1/T3)-(1/T4)-(1/T5))),dim=c(3,3));
+ h=0.5;
+ r=array(c(1,0,0,0,1,0,0,0,1),dim=c(3,3));
+ q=1;
+ v=r*q*t(r);
+ a=10.14286;
+ z=array(c((1/T1),0,0,0,0,0,0,0,(1/T5)), dim=c(3,3));
+ p=array(c(1,0,0,0,1,0,0,0,1),dim=c(3,3));
+ pn=array(c(1,0,0,0,1,0,0,0,1),dim=c(3,3));
+ kal=KalmanSmooth(x, list(T=t,Z=z,h=h,V=v,a=a,P=p,Pn=pn), nit=0)
+ kal}
 
 Kalm(deb,4,.5,5,.7,.1,2,3,4,1,0.65)
Error in as.vector(data) : Argument S1 is missing, with no default
 
First I thought I had to make a timeserie of deb. But this doesn't change the problem.
Lot's of thanks trying to help me. 
Best regards,
Pieter Hazenberg
Student Hydrology and Watermanagement
Wageningen University
The Netherlands

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Re: [R] Daily time series

2004-07-07 Thread Vincent Goulet
Hi Vito,

Short answer: argument 'start' can take any value you want. For monthly 
observations (the case 'ts()' handles most nicely, with quarterly 
observations), 'start' will be used to specify the year and month (or 
quarter) of the first observation.

From what I can gather from the help page, something like, say, 

ts(x, freq=7, start=c(35, 1))

would mean that the first observation is on the first day of week 35.

Bear in mind that, to the best of my knowledge, the value of 'start' has 
absolutely no impact on calculations. It is merely there for labeling 
purposes.

Hope this helps!

On Wednesday 07 July 2004 03:30, Vito Ricci wrote:
 Hi,

 I'm dealing with time series with 1 observaton for day
 (data sampled daily). I will create a ts object using
 that time series and the function ts().
 In ts() help is written:

 The value of argument 'frequency' is used when the
 series is sampled an integral number of times in each
 unit time interval. For example, one could use a value
 of '7' for 'frequency' when the data are sampled
 daily, and the natural time period is a week, or '12'
 when the data are sampled monthly and the natural time
 period
 is a year.  Values of '4' and '12' are assumed in
 (e.g.) 'print' methods to imply a quarterly and
 monthly series respectively.

 But what value should assume start in ts function?

 Here is a time series:

 1/1 10
 2/1 20
 3/1 30
 4/1 40
 5/1 50
 6/1 60

 x-c(10,20,30,40,50,60) ## observation
 serie-ts(dati, start=c(1,1),frequency=7) ##creating
 ts object

 serie  ## printing ts output

 Time Series:
 Start = c(1, 1)
 End = c(1, 6)
 Frequency = 7
 [1] 10 20 30 40 50 60

 Could someone help me?

 Thanks in advance.

 Sincerely.
 Vito Ricci


 =
 Diventare costruttori di soluzioni

 Visitate il portale http://www.modugno.it/
 e in particolare la sezione su Palese
 http://www.modugno.it/archivio/cat_palese.shtml

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-- 
  Vincent Goulet, Associate Professor
  École d'actuariat
  Université Laval, Québec 
  [EMAIL PROTECTED]   http://vgoulet.act.ulaval.ca

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Re: [R] fast NA elimination ?

2004-07-07 Thread Roger D. Peng
I find complete.cases() to be very useful for this kind of stuff (and 
very fast).  As in,

 d - data.frame(x = c(1,2,3,NA,5), y = c(1,NA,3,4,5))
 d
   x  y
1  1  1
2  2 NA
3  3  3
4 NA  4
5  5  5
 complete.cases(d)
[1]  TRUE FALSE  TRUE FALSE  TRUE
 use - complete.cases(d)
 d[use, ]
  x y
1 1 1
3 3 3
5 5 5

-roger
ivo welch wrote:
dear R wizards: an operation I execute often is the deletion of all 
observations (in a matrix or data set) that have at least one NA. (I now 
need this operation for kde2d, because its internal quantile call 
complains;  could this be considered a buglet?)   usually, my data sets 
are small enough for speed not to matter, and there I do not care 
whether my method is pretty inefficient (ok, I admit it: I use the sum() 
function and test whether the result is NA)---but now I have some bigger 
data sets. Is there a recommended method of doing NA elimination most 
efficiently? sincerely, /iaw
---
ivo welch
professor of finance and economics
brown / nber / yale

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--
Roger D. Peng
http://www.biostat.jhsph.edu/~rpeng/
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Re: [R] fast NA elimination ?

2004-07-07 Thread Marc Schwartz
On Wed, 2004-07-07 at 09:35, ivo welch wrote:
 dear R wizards: an operation I execute often is the deletion of all 
 observations (in a matrix or data set) that have at least one NA. (I 
 now need this operation for kde2d, because its internal quantile call 
 complains;  could this be considered a buglet?)   usually, my data sets 
 are small enough for speed not to matter, and there I do not care 
 whether my method is pretty inefficient (ok, I admit it: I use the 
 sum() function and test whether the result is NA)---but now I have some 
 bigger data sets. Is there a recommended method of doing NA elimination 
 most efficiently? sincerely, /iaw
 ---
 ivo welch
 professor of finance and economics
 brown / nber / yale


Take a look at ?complete.cases

HTH,

Marc Schwartz

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Re: [R] Code density functions

2004-07-07 Thread F Z
Thanks to Andy Liaw, James Holtman and Uwe Ligges;  I downloaded and looked 
at the source and found what I need!!

Francisco

From: Uwe Ligges [EMAIL PROTECTED]
To: F Z [EMAIL PROTECTED]
CC: [EMAIL PROTECTED], [EMAIL PROTECTED]
Subject: Re: [R] Code density functions
Date: Wed, 07 Jul 2004 07:51:35 +0200
F Z wrote:
Dear Andy
Thanks for your reply.  I don't seem to find the file that you suggested. 
I tried:

file.show('C:/Program Files/R/rw1091/src/nmath/dnorm.c.')
You have installed a binary distribution.
You need to get the source tarball (directly accessible via the main CRAN 
page) that includes the file.

Uwe Ligges

NULL
Warning message:
file.show(): file C:/Program Files/R/rw1091/src/nmath/dnorm.c. does not 
exist

Then I looked at the directory and tried a file with similar name:
file.show('C:/Program Files/R/rw1091/src/include/Rmath.h')

But this file does not show the actual code used to calculate the 
densities, only the declarations of the procedures.

What am I doing wrong?
Thanks again!
Francisco Zagmutt :)

From: Liaw, Andy [EMAIL PROTECTED]
To: 'F Z' [EMAIL PROTECTED],[EMAIL PROTECTED]
Subject: RE: [R] Code density functions
Date: Tue, 6 Jul 2004 18:35:24 -0400
Dear insert your name here:
You need to look at the C-level source codes, in R-1.9.1/src/nmath/d*.c.
Andy
 From: F Z

 Hello

 I would like to see the algorithm that R uses to generate
 density functions
 for several distributions (i.e. Normal,Weibull, etc).  I tried:

 dnorm
 function (x, mean = 0, sd = 1, log = FALSE)
 .Internal(dnorm(x, mean, sd, log))
 environment: namespace:stats

 How can I see the code used for densities?

 Thanks!
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Re: [R] fast NA elimination ?

2004-07-07 Thread F Z
Hi Ivo
Try ?na.omit
Example :
d - data.frame(x = c(1:5,NA), y = c(NA,3:7)) d
  x  y
1  1 NA
2  2  3
3  3  4
4  4  5
5  5  6
6 NA  7
do-na.omit(d)
do
 x y
2 2 3
3 3 4
4 4 5
5 5 6
I usually pass na.omit within the data argument of a function i.e.  
m-lm(x~y,data=na.omit(d)). In this way you don't have to store 2 datasets.

I hopw that this helps
Francisco
From: Marc Schwartz [EMAIL PROTECTED]
Reply-To: [EMAIL PROTECTED]
To: ivo welch [EMAIL PROTECTED]
CC: R-Help [EMAIL PROTECTED]
Subject: Re: [R] fast NA elimination ?
Date: Wed, 07 Jul 2004 09:41:39 -0500
On Wed, 2004-07-07 at 09:35, ivo welch wrote:
 dear R wizards: an operation I execute often is the deletion of all
 observations (in a matrix or data set) that have at least one NA. (I
 now need this operation for kde2d, because its internal quantile call
 complains;  could this be considered a buglet?)   usually, my data sets
 are small enough for speed not to matter, and there I do not care
 whether my method is pretty inefficient (ok, I admit it: I use the
 sum() function and test whether the result is NA)---but now I have some
 bigger data sets. Is there a recommended method of doing NA elimination
 most efficiently? sincerely, /iaw
 ---
 ivo welch
 professor of finance and economics
 brown / nber / yale
Take a look at ?complete.cases
HTH,
Marc Schwartz
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Technology 101. http://special.msn.com/tech/technology101.armx
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[R] Win32 C code

2004-07-07 Thread Simon Cullen
Hi,
I'm trying to get C code working with R. This is my first time writing C  
on Windows and I'm making a mess of it. Help!

I'm following the example in Roger Peng's An Introduction to the .C  
interface to R. The C code is:

#include R.h
void hello(int *n){
int i;
for(i=0; i  *n; i++) {
Rprintf(Hello, world!\n);
}
}
I seem to be unable to make Windows pay attention to additions to the PATH  
variable so I stuck the code (test.c) into the $R_HOME\bin directory. I  
copied into the same directory mingw32-make.exe and renamed it make.exe  
(as the perl script SHLIB seems to want a make.exe).

When I type Rcmd SHLIB test.c at a command prompt I get the following:
C:\Program Files\R\rw1091\binRcmd SHLIB test.c
C:/PROGRA~1/R/rw1091/src/gnuwin32/MkRules:110: warning: overriding  
commands for target `.c.d'
C:/PROGRA~1/R/rw1091/src/gnuwin32/MkRules:98: warning: ignoring old  
commands for target `.c.d'
C:/PROGRA~1/R/rw1091/src/gnuwin32/MkRules:126: warning: overriding  
commands for target `.c.o'
C:/PROGRA~1/R/rw1091/src/gnuwin32/MkRules:114: warning: ignoring old  
commands for target `.c.o'
MkRules:110: warning: overriding commands for target `.c.d'
MkRules:98: warning: ignoring old commands for target `.c.d'
MkRules:126: warning: overriding commands for target `.c.o'
MkRules:114: warning: ignoring old commands for target `.c.o'
MkRules:110: warning: overriding commands for target `.c.d'
MkRules:98: warning: ignoring old commands for target `.c.d'
MkRules:126: warning: overriding commands for target `.c.o'
MkRules:114: warning: ignoring old commands for target `.c.o'
make: *** No rule to make target `'test.c'', needed by `makeMakedeps'.   
Stop.

I'm obviously an idiot but any help offered would be much appreciated.
--
SC
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Re: [R] Enumeration in R

2004-07-07 Thread Paul Roebuck
On Tue, 7 Jul 2004, Peter Dalgaard wrote:

 Paul Roebuck [EMAIL PROTECTED] writes:

  I want the equivalent of this 'C' declaration.
  enum StoplightColor {
  green = 3,
  yellow = 5,
  red = 7
  };

 I think you *dis*abled it by specifying an initializer which doesn't
 check the validity:

Thanks. It all seems obvious once it's pointed out to you.
I have a couple related questions:

1) When I was reading Chambers book (pg 288), my initial
impression was that there was a way I could have done this
class without specifying the representation in 'setClass'
- I just couldn't figure out how to assign values that
way. Could this class have been defined 'slotless'? If so,
how?

2) How do you define a class and instantiate some, yet
prevent more from being created after that. Possibly better
stated, from the package's API view, I would like these to
be instance variables of opaque types. So I would like to
create my 'global' constants in an initialization routine
then prevent use of 'new' to create any more. I tried the
following with no success. Possible?

setMethod(new,
  stoplightColor,
  function(Class, ...) stop(can't make any more))

3) This seems kind of painful for trivial stuff. My idea
was to move some of the validation error checking out of my
project by converting certain function arguments into classes
that could be validated upon creation, improving the clarity
of project routines. What is the canonical style used in R
package authoring?




- stoplightColor.R 
setClass(stoplightColor,
 representation(value = integer),
 prototype = integer(1))
stoplightColor - function(value) {
new(stoplightColor, value)
}
valid.stoplightColor - function(object) {
valid - switch(as([EMAIL PROTECTED], character),
3 = TRUE,
5 = TRUE,
7 = TRUE,
FALSE)
if (valid == FALSE)
return('Invalid value - must be [3|5|7]');
return(TRUE);
}
setValidity(stoplightColor, valid.stoplightColor)
initialize.stoplightColor - function(.Object, value) {
if (missing(value) || is.na(value))
stop('Argument value is missing or NA')
[EMAIL PROTECTED] - as.integer(value)
validObject(.Object)
.Object
}
setMethod(initialize,
  signature(.Object = stoplightColor),
  initialize.stoplightColor)
stoplightColor.as.integer - function(from) {
return([EMAIL PROTECTED])
}
setAs(stoplightColor, integer, stoplightColor.as.integer)

green - stoplightColor(3)
yellow - stoplightColor(5)
red - stoplightColor(7)

--
SIGSIG -- signature too long (core dumped)

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[R] Histograms, density, and relative frequencies

2004-07-07 Thread Bret Collier
R-users,
I have been using R for about 1 year, and I have run across a 
couple of graphics problem that I am not quite sure how to address.  I have 
read up on the email threads regarding the differences between density and 
relative frequencies (count/sum(count) on the R list, and I am hoping that 
someone could provide me with some advice/comments concerning my 
approach.  I will admit that some of the underlying mathematics of the 
density discussion are beyond my current understanding, but I am looking 
into it.

I have a data set (600,000 obs) used to parameterize a probabilistic causal 
model where each obs is a population response for one of 2 classes (either 
regs1 and regs2).  I have been attempting to create 1 marginal probability 
plot with 2 lines (one for each class).  Using my rather rough code, I 
created a plot that seems to adhere to the commonly used (although from 
what I can understand wrong) relative frequency histogram approach.

My rough code looks like this:
bk - c(0, .05, .1, .15, .2, .25,.3, .35, 1)
par(mfrow=c(1, 1))
fawn1 - hist(MFAWNRESID[regs1], plot=F, breaks=bk)
fawn2 - hist(MFAWNRESID[regs2], plot=F, breaks=bk)
count1 - fawn1$counts/sum(fawn1$counts)
count2 - fawn2$counts/sum(fawn2$counts)
b - c(0, .05, .1, .15, .2, .25, .3, .35)
plot(count1~b,xaxt=n, xlim=c(0, .5), ylim=c(0, .40), pch=., bty=l)
lines(spline(count1~b), lty=c(1), lwd=c(2), col=black)
lines(spline(count2~b), lty=c(2), lwd=c(2), col=black)
axis(side=1, at=c(0, .05, .1, .15, .2,  .25, .3, .35))
Using the above, I get frequency values for regs1 that look like this 
(which is the same as output for my probabilistic model):
 count1
[1] 1.213378e-01 3.454324e-01 3.365343e-01 1.580839e-01 3.342101e-02
[6] 4.698426e-03 4.488942e-04 4.322685e-05

First, count1 is the frequency of occurrence within range 0-0.05, but when 
plotted is the value at b=0 and does not really represent the range?  Are 
there any suggestions on a technique to approach this?

Next:  Using the above code, the x-axis values end at 0.35, but the axis 
continues (because bk ends at 1)?  While there is the chance of occurrence 
out past .35, it is low and I want to extend the lines to about .35 and 
clip the x-axis.  But, I have been unable to figure out how to clip  Could 
someone point me in the correct direction?

TIA,
Bret A. Collier
Arkansas Cooperative Fish and Wildlife Research Unit
Department of Biological Sciences University of Arkansas
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Re: [R] Win32 C code

2004-07-07 Thread Paul Roebuck
On Wed, 7 Jul 2004, Simon Cullen wrote:

 I'm trying to get C code working with R. This is my first time writing C
 on Windows and I'm making a mess of it. Help!
 ...
 I seem to be unable to make Windows pay attention to additions to the PATH
 variable so I stuck the code (test.c) into the $R_HOME\bin directory. I
 copied into the same directory mingw32-make.exe and renamed it make.exe
 (as the perl script SHLIB seems to want a make.exe).
 ...


I wrote a batch file to get mine working. Change directory
paths to match your setup.

WINBUILD.CMD
-
@cls
@SETLOCAL
@set PROJ=rwt
@set RBINDIR=C:\R\rw1091\bin
@set TOOLSBINDIR=C:\Rtools\bin
@set MINGWBINDIR=C:\MinGW\bin
@set PERLBINDIR=C:\Perl\bin
@set TEXBINDIR=C:\PROGRA~1\TeXLive\bin\win32
@set HCCBINDIR=C:\PROGRA~1\HTMLHE~1
    Next line split for readability 
@set PATH=%TOOLSBINDIR%;%RBINDIR%;%MINGWBINDIR%;%PERLBINDIR%;
  %TEXBINDIR%;%HCCBINDIR%;%WINDIR%\system32;%WINDIR%
@echo PATH=%PATH%
Rcmd build --binary %PROJ%
Rcmd check %PROJ%
@ENDLOCAL


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SIGSIG -- signature too long (core dumped)

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Re: [R] Win32 C code

2004-07-07 Thread Simon Cullen
On Wed, 7 Jul 2004 12:32:41 -0500 (CDT), Paul Roebuck  
[EMAIL PROTECTED] wrote:

I wrote a batch file to get mine working. Change directory
paths to match your setup.
That batch file was exactly what I needed! Thanks.
--
SC
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[R] Observational error in ARIMA

2004-07-07 Thread Guiming Wang
Hi, 

Does anyone know how to include observation errors in the arima of R, which is 
implemented with the Kalman filter.  I want to estimate observational error variance 
for noisy data in the context of ARMA model using arima of R.  I read the manual and 
tried the example codes, but did not find the solution.  From the outputs of the 
components model, it seems to me that the default setting of the arima does not 
include the observational error in the fitting. The elements of matrix H are zeros 
when executing the example codes.  Am I right on this one?  Thanks in advance.

Sincerely,

Guiming Wang
Natural Resource Ecology Lab
Colorado State University
Fort Collins, CO 80523
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[R] Importing an Excel file

2004-07-07 Thread Park, Kyong H Mr. RDECOM
Hello, R users,
I am  a very beginner of R and tried read.csv to import an excel file after
saving an excel file as csv. But it added alternating rows of fictitious NA
values after row number 16. When I applied read.delim, there were trailing
several commas at the end of each row after row number 16 instead of NA
values. Appreciate your help. 

Kyong



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[R] Sorry forgot to mention about OS system in an previous email

2004-07-07 Thread Park, Kyong H Mr. RDECOM
Sorry for not mentioning  about OS for importing an excel file in an
previous email. I'm using R 1.9.1 with Windows 2000.

Kyong

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[R] lost messages

2004-07-07 Thread Aaron J. Mackey
I've posted a message twice to this list, and never seen it appear yet 
... perhaps this one will go through ... ?

--
Aaron J. Mackey, Ph.D.
Dept. of Biology, Goddard 212
University of Pennsylvania   email:  [EMAIL PROTECTED]
415 S. University Avenue office: 215-898-1205
Philadelphia, PA  19104-6017 fax:215-746-6697
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Re: [R] Generate a matrix Q satisfying t(Q)%*%Q=Z and XQ=W

2004-07-07 Thread Stephane DRAY
thanks,
I want to create matrices for simulation purpose (in order to evaluate the 
efficiency of different methods on this simulated data set). So, I want to 
create a data matrix Q (m individuals and r variables). I want to specify 
the variance-covariance structure for this matrix (t(Q)%*%Q=Z ) but I want 
also to create another constraint due to another matrix of data. I want 
that the covariance of Q and X are equal to those given in W (XQ=W).
The example I gave is just to illustrate my problem and perhaps it has no 
solution (I cannot see it because I have no idea how to construct Q such as 
Q=Q1=Q2)


At 00:00 07/07/2004, Spencer Graves wrote:
 Is a solution even possible for the matrices in your example?
I've tried a few things that have suggested that a solution may not be 
possible.
 What can you tell us of the problem that you've translated into 
this?  I see a minimization problem subject to constraints, but I'm not 
certain which are the constraints and what is the objective function.
For example, are you trying to find Q to minimize sum((Z-X'X)^2) subject 
to XQ=W or do you want to minimize sum((XQ-W)^2) subject to Q'Q=Z or 
something else?
 If it were my problem, I think I would work for a while with the 
singular value decompositions of X, W and Z, and see if that would lead 
me to more information about Q, including conditions under which a 
solution existed, expressions for Q when multiple solutions existed, and 
a solution minimizing your chosen objective function when solutions do 
not exist.  (A google search produced many hits for singular value 
decomposition, implemented as svd in R.)
 hope this helps.  spencer graves

Stephane DRAY wrote:
Hello,
I have a question that is not directly related to R ... but I try to do 
it in R ;-) :

I would like to generate a matrix Q satisfying (for a given Z, X and W) 
the two following conditions:

t(Q)%*%Q=Z  (1)
XQ=W (2)
where:
Q is m rows and r columns
X is p rows and m columns
D is p rows and r columns
C is r rows and r columns
with mp,r
e.g:
m=6,
p=2
r=3
Z=matrix(c(1,.2,.5,.2,1,.45,.5,.45,1),3,3)
X=matrix(c(.1,.3,.5,.6,.2,.1,.8,1,.4,.2,.2,.9),2,6)
W=matrix(c(0,.8,.4,.6,.2,0),2,3)
#Create a matrix satisfying (1) is easy:
A=matrix(runif(18),6,3)
Q1=svd(A)$u%*%chol(Z)
#For the second condition (2), a solution is given by
Q2=A%*%ginv(X%*%A)%*%W


I do not know how to create a matrix Q that satisfies the two 
conditions.  I have try to construct an iterative procedure without 
success (no convergence):

eps=10
i=0
while(eps.5)
{
Q1=svd(Q2)$u%*%chol(Z)
Q2=Q1%*%ginv(X%*%Q1)%*%W
eps=sum(abs(Q1-Q2))
cat(i,:,eps,\n)
i=i+1
}
Perhaps someone could have any idea to solve the problem, or a reference 
on this kind of question or the email of another list where I should ask 
this question.

Thanks in advance,
Sincerely.
Stéphane DRAY
-- 

Département des Sciences Biologiques
Université de Montréal, C.P. 6128, succursale centre-ville
Montréal, Québec H3C 3J7, Canada
Tel : 514 343 6111 poste 1233
E-mail : [EMAIL PROTECTED]
-- 

Web
http://www.steph280.freesurf.fr/
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Stéphane DRAY
-- 

Département des Sciences Biologiques
Université de Montréal, C.P. 6128, succursale centre-ville
Montréal, Québec H3C 3J7, Canada
Tel : 514 343 6111 poste 1233
E-mail : [EMAIL PROTECTED]
-- 

Web  http://www.steph280.freesurf.fr/
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[R] vectorizing sapply() code (Modified by Aaron J. Mackey)

2004-07-07 Thread Aaron J. Mackey
[ Not sure why, but the first two times I sent this it never seemed to 
go through; apologies if you're seeing this thrice ... ]

I have some fully functional code that I'm guessing can be done 
better/quicker with some savvy R vector tricks; any help to make this 
run a bit faster would be greatly appreciated; I'm particularly stuck 
on how to calculate using row-wise vectors without iterating 
explicitly over the dataframe or table ...

library(stats4);
d - data.frame( ix=c(0,1,2,3,4,5,6,7),
 ct=c(253987,  9596, 18680,  2630,  8224,  3590,  5534, 
18937),
 A=c(  0, 1, 0, 1, 0, 1, 0, 
1),
 B=c(  0, 0, 1, 1, 0, 0, 1, 
1),
 C=c(  0, 0, 0, 0, 1, 1, 1, 
1)
   );
ct - round(logb(length(d$ix), 2))
ll - function( th=0.5,
a1=log(0.5), a2=log(0.5), a3=log(0.5),
b1=log(0.5), b2=log(0.5), b3=log(0.5)
  ) {
  a - exp(sapply(1:ct, function (x) { get(paste(a, x, sep=)) }));
  b - exp(sapply(1:ct, function (x) { get(paste(b, x, sep=)) }));
  -sum( d$ct * log( sapply( d$ix,
function (ix, th, a, b) {
  x - d[ix+1,3:(ct+2)]
  (th * prod((b ^ (1-x)) * ((1-b) ^ x   
 ))) +
  ((1-th) * prod((a ^ x) * ((1-a) ^ 
(1-x
},
th, a, b
  )
  )
  );
}

ml - mle(ll,
  lower=c(0+1e-5, rep(log(0+1e-8), 2*ct)),
  upper=c(1-1e-5, rep(log(1-1e-8), 2*ct)),
  method=L-BFGS-B
 );
For those interested in the math, this is the MLE procedure to estimate 
the false positive/false negative rates (a and b) of three diagnostic 
(A, B and C) tests that have the observed performance recapitulated in 
dataframe d, but no gold standard (sometimes called latent class 
analysis, or LCA).

Thanks for any help,
-Aaron
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Re: [R] Importing an Excel file

2004-07-07 Thread Marc Schwartz
On Wed, 2004-07-07 at 13:21, Park, Kyong H Mr. RDECOM wrote:
 Hello, R users,
 I am  a very beginner of R and tried read.csv to import an excel file after
 saving an excel file as csv. But it added alternating rows of fictitious NA
 values after row number 16. When I applied read.delim, there were trailing
 several commas at the end of each row after row number 16 instead of NA
 values. Appreciate your help. 
 
 Kyong


Yep. This is one of the behaviors that I had seen with Excel when I was
running Windows XP. Seemingly empty cells outside the data range would
get exported in the CSV file causing a data integrity problem.

It is one of the reasons that I installed OpenOffice under Windows and
used Calc to open the Excel files and then do the CSV exports before I
switched to Linux :-)

Depending upon the version of Excel you are using, you might try to
highlight and copy only the rectangular range of cells in the sheet that
actually have data to a new sheet and then export the new sheet to a CSV
file.

Do not just click on the upper left hand corner of the sheet to
highlight the entire sheet to copy it. Only highlight the range of cells
you actually need for copying.

Another option is to use the read.xls() function in the 'gregmisc'
package on CRAN or install OpenOffice.

HTH,

Marc Schwartz

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Re: [R] KalmanSmooth problem

2004-07-07 Thread Uwe Ligges
Hazenberg21, Pieter wrote:
Hello,
In R I am trying to use Kalman filtering to find a solution for an hydrological problem. With Kalman Filtering I want to estimate the discharge comming from three storage bassins. I have programmed a function in R which can run KalmanSmooth. When I'm asking for the function and putting in values, R detects the following error: Error in as.vector(data) : Argument S1 is missing, with no default.
I have try to find a solution for this error in the R help file, and in different manuals, but I can't find it. Please help me find a solution.
Question: What does R mean with S1 and what am I doing wrong?
Here is the way I have programmed the hydrological problem in R. 
 

discharge=read.table(file=C:/Program Files/R/rw1090/discharge.txt,header=T)
deb=discharge[,1]
deb
  [1] 11.545313  8.045465  5.670868  4.044584  2.919311  2.306668  2.940956
  [8]  4.238159  5.017374  3.818236  2.928805  2.262183  1.757765  1.633945
 [15]  2.295130  3.454054  4.035224  3.193967  2.533181  2.012406  1.600836
 [22]  1.652155  2.428678  3.642827  4.019545  3.209473  2.563617  2.048347
 [29]  1.637041  1.828952  2.757842  4.050821  4.147013  3.316503  2.652490
 [36]  2.121535  1.696934  2.027763  3.107366  4.429670  4.160178  3.327950
 [43]  2.662237  2.129710  1.703717  2.158095  3.337039  4.582359  3.905901
 [50]  3.124690  2.499732  1.999772  1.599810  2.130893  3.302622  4.336081
 [57]  3.468857  2.775081  2.220062  1.776048  1.560859  2.169537  3.348081
 [64]  4.170552  3.336440  2.669151  2.135320  1.708256  1.648859  2.374217
 [71]  3.624091  4.248563  3.398850  2.719080  2.175264  1.740211  1.826122
 [78]  2.704749  4.056438  4.437309  3.549847  2.839878  2.271902  1.817522
 [85]  2.053994  3.107875  4.548436  4.600601  3.680481  2.944385  2.355508
 [92]  1.884406  2.273248  3.490148  4.949898  4.584409  3.667527  2.934022
 [99]  2.347217  1.84
Kalm = function(x,O1,O2,O3,T1,T2,T3,T4,T5,t,ga){
+ t=array(c(1+ga*O1+t/O1*(-(1/T2)-(1/T3)-(1/T1)),t/O1*(1/T2),t/O1*(1/T3),
+ t/O2*(1/T2),1+ga*O2+t/O2*(-(1/T2)-(1/T4)),t/O2*(1/T4),
+ t/O3*(1/T3),t/O3*(1/T4),1+ga*O3+t/O3*(-(1/T3)-(1/T4)-(1/T5))),dim=c(3,3));
+ h=0.5;
+ r=array(c(1,0,0,0,1,0,0,0,1),dim=c(3,3));
+ q=1;
+ v=r*q*t(r);
+ a=10.14286;
+ z=array(c((1/T1),0,0,0,0,0,0,0,(1/T5)), dim=c(3,3));
+ p=array(c(1,0,0,0,1,0,0,0,1),dim=c(3,3));
+ pn=array(c(1,0,0,0,1,0,0,0,1),dim=c(3,3));
+ kal=KalmanSmooth(x, list(T=t,Z=z,h=h,V=v,a=a,P=p,Pn=pn), nit=0)
+ kal}
 

Kalm(deb,4,.5,5,.7,.1,2,3,4,1,0.65)
Error in as.vector(data) : Argument S1 is missing, with no default
 
First I thought I had to make a timeserie of deb. But this doesn't change the problem.
Lot's of thanks trying to help me. 
a) Please tell us R Version and OS (OK, implicitly done that we are 
talking about R-1.9.0 on Windows).
b) Please tell us which packages you are using (I don't know 
KalmanSmooth(), for example).
c) Please try to specify reproducible examples.

Conclusion for a-c): Please read the posting-guide.
My hint is to try to debug yourself, at least *try*, starting eith 
calling traceback() right after the error appeared, in order to get a 
guess where the error really happens. Then look at the data that is 
passed to the function -  and I'm pretty sure you will get at least an 
idea what goes wrong.

Uwe Ligges


Best regards,
Pieter Hazenberg
Student Hydrology and Watermanagement
Wageningen University
The Netherlands
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Re: [R] Importing an Excel file

2004-07-07 Thread Marc Schwartz
On Wed, 2004-07-07 at 13:44, Marc Schwartz wrote:
 On Wed, 2004-07-07 at 13:21, Park, Kyong H Mr. RDECOM wrote:
  Hello, R users,
  I am  a very beginner of R and tried read.csv to import an excel file after
  saving an excel file as csv. But it added alternating rows of fictitious NA
  values after row number 16. When I applied read.delim, there were trailing
  several commas at the end of each row after row number 16 instead of NA
  values. Appreciate your help. 
  
  Kyong


One other thing:

The default delimiting characters in read.csv() and read.delim() are NOT
the same.

The former uses a comma and the latter a TAB character. If you did not
change the defaults in Excel when you created your CSV file, that would
account for the difference behaviors upon import.

Be sure that the delimiting character in the R function you use properly
corresponds to the actual delimiting character in your CSV file.

Marc

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Re: [R] Generate a matrix Q satisfying t(Q)%*%Q=Z and XQ=W

2004-07-07 Thread Spencer Graves
 How about generating matrices (X1|X2), dim(X1) = c(n, k1), dim(X2) 
= c(n, k2), with mean 0 and covariance matrix as follows: 

(S11 | S12)
(S21 | S22),
with S12 = W, S22 = Z and S11 = whatever you want?  With X = t(X1), and 
Q = X2, we have E(XQ) = W and E(Q'Q) = Z. 

 This can be done using rmvnorm in package mvtnorm. 

 hope this helps.  spencer graves

Stephane DRAY wrote:
thanks,
I want to create matrices for simulation purpose (in order to evaluate 
the efficiency of different methods on this simulated data set). So, I 
want to create a data matrix Q (m individuals and r variables). I want 
to specify the variance-covariance structure for this matrix 
(t(Q)%*%Q=Z ) but I want also to create another constraint due to 
another matrix of data. I want that the covariance of Q and X are 
equal to those given in W (XQ=W).
The example I gave is just to illustrate my problem and perhaps it has 
no solution (I cannot see it because I have no idea how to construct Q 
such as Q=Q1=Q2)


At 00:00 07/07/2004, Spencer Graves wrote:
 Is a solution even possible for the matrices in your example?
I've tried a few things that have suggested that a solution may not 
be possible.
 What can you tell us of the problem that you've translated into 
this?  I see a minimization problem subject to constraints, but I'm 
not certain which are the constraints and what is the objective 
function.
For example, are you trying to find Q to minimize sum((Z-X'X)^2) 
subject to XQ=W or do you want to minimize sum((XQ-W)^2) subject to 
Q'Q=Z or something else?
 If it were my problem, I think I would work for a while with the 
singular value decompositions of X, W and Z, and see if that would 
lead me to more information about Q, including conditions under which 
a solution existed, expressions for Q when multiple solutions 
existed, and a solution minimizing your chosen objective function 
when solutions do not exist.  (A google search produced many hits for 
singular value decomposition, implemented as svd in R.)
 hope this helps.  spencer graves

Stephane DRAY wrote:
Hello,
I have a question that is not directly related to R ... but I try to 
do it in R ;-) :

I would like to generate a matrix Q satisfying (for a given Z, X and 
W) the two following conditions:

t(Q)%*%Q=Z  (1)
XQ=W (2)
where:
Q is m rows and r columns
X is p rows and m columns
D is p rows and r columns
C is r rows and r columns
with mp,r
e.g:
m=6,
p=2
r=3
Z=matrix(c(1,.2,.5,.2,1,.45,.5,.45,1),3,3)
X=matrix(c(.1,.3,.5,.6,.2,.1,.8,1,.4,.2,.2,.9),2,6)
W=matrix(c(0,.8,.4,.6,.2,0),2,3)
#Create a matrix satisfying (1) is easy:
A=matrix(runif(18),6,3)
Q1=svd(A)$u%*%chol(Z)
#For the second condition (2), a solution is given by
Q2=A%*%ginv(X%*%A)%*%W


I do not know how to create a matrix Q that satisfies the two 
conditions.  I have try to construct an iterative procedure without 
success (no convergence):

eps=10
i=0
while(eps.5)
{
Q1=svd(Q2)$u%*%chol(Z)
Q2=Q1%*%ginv(X%*%Q1)%*%W
eps=sum(abs(Q1-Q2))
cat(i,:,eps,\n)
i=i+1
}
Perhaps someone could have any idea to solve the problem, or a 
reference on this kind of question or the email of another list 
where I should ask this question.

Thanks in advance,
Sincerely.
Stéphane DRAY
-- 

Département des Sciences Biologiques
Université de Montréal, C.P. 6128, succursale centre-ville
Montréal, Québec H3C 3J7, Canada
Tel : 514 343 6111 poste 1233
E-mail : [EMAIL PROTECTED]
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Stéphane DRAY
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Université de Montréal, C.P. 6128, succursale centre-ville
Montréal, Québec H3C 3J7, Canada
Tel : 514 343 6111 poste 1233
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[R] lme: extract variance estimate

2004-07-07 Thread Stephen Ellner
Spencer Graves wrote: 

 Have you considered VarCorr?  I've used it with lme, and the 
documentation in package lme4 suggests it should work with GLMM, which 
might also do what you want from glmmPQL. 

Thanks for the pointer (I was not aware that nlme and lme4 had different
versions of lme), but I'm still stuck at the same place using lme4: VarCorr(fit)
 Groups   NameVariance Std.Dev.
 yeart(Intercept) 0.040896 0.20223 
 Residual 0.091125 0.30187 

The number I need to extract and store is the .20223, but all the 
components I can find in VarCorr(fit) are something else. 

u=VarCorr(fit); slotNames(u)
[1] scalereSumry  useScale
 [EMAIL PROTECTED]
[1] 0.3018693
 [EMAIL PROTECTED]
$yeart
An object of class corrmatrix
(Intercept)
(Intercept)   1
Slot stdDev:
(Intercept) 
  0.6699156 
 [EMAIL PROTECTED]
[1] TRUE

In glmmML the estimate is returned as the $sigma component
of the model, but I also need the same info from 'family=gaussian' 
models. 


Stephen P. Ellner ([EMAIL PROTECTED])
Department of Ecology and Evolutionary Biology
Corson Hall, Cornell University, Ithaca NY 14853-2701
Phone (607) 254-4221FAX (607) 255-8088

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Re: [R] Generate a matrix Q satisfying t(Q)%*%Q=Z and XQ=W

2004-07-07 Thread Stephane DRAY
Thanks but it does not solve my problem because I would like to generate Q 
(X2) for a given X(X1). X is fixed. But It seems hard to do it and perhaps 
It would be easier to change my approach.

Thanks again for your help
At 14:55 07/07/2004, Spencer Graves wrote:
 How about generating matrices (X1|X2), dim(X1) = c(n, k1), dim(X2) = 
c(n, k2), with mean 0 and covariance matrix as follows:
(S11 | S12)
(S21 | S22),

with S12 = W, S22 = Z and S11 = whatever you want?  With X = t(X1), and Q 
= X2, we have E(XQ) = W and E(Q'Q) = Z.
 This can be done using rmvnorm in package mvtnorm.
 hope this helps.  spencer graves

Stephane DRAY wrote:
thanks,
I want to create matrices for simulation purpose (in order to evaluate 
the efficiency of different methods on this simulated data set). So, I 
want to create a data matrix Q (m individuals and r variables). I want to 
specify the variance-covariance structure for this matrix (t(Q)%*%Q=Z ) 
but I want also to create another constraint due to another matrix of 
data. I want that the covariance of Q and X are equal to those given in W 
(XQ=W).
The example I gave is just to illustrate my problem and perhaps it has no 
solution (I cannot see it because I have no idea how to construct Q such 
as Q=Q1=Q2)


At 00:00 07/07/2004, Spencer Graves wrote:
 Is a solution even possible for the matrices in your example?
I've tried a few things that have suggested that a solution may not be 
possible.
 What can you tell us of the problem that you've translated into 
this?  I see a minimization problem subject to constraints, but I'm not 
certain which are the constraints and what is the objective function.
For example, are you trying to find Q to minimize sum((Z-X'X)^2) subject 
to XQ=W or do you want to minimize sum((XQ-W)^2) subject to Q'Q=Z or 
something else?
 If it were my problem, I think I would work for a while with the 
singular value decompositions of X, W and Z, and see if that would lead 
me to more information about Q, including conditions under which a 
solution existed, expressions for Q when multiple solutions existed, 
and a solution minimizing your chosen objective function when solutions 
do not exist.  (A google search produced many hits for singular value 
decomposition, implemented as svd in R.)
 hope this helps.  spencer graves

Stephane DRAY wrote:
Hello,
I have a question that is not directly related to R ... but I try to do 
it in R ;-) :

I would like to generate a matrix Q satisfying (for a given Z, X and W) 
the two following conditions:

t(Q)%*%Q=Z  (1)
XQ=W (2)
where:
Q is m rows and r columns
X is p rows and m columns
D is p rows and r columns
C is r rows and r columns
with mp,r
e.g:
m=6,
p=2
r=3
Z=matrix(c(1,.2,.5,.2,1,.45,.5,.45,1),3,3)
X=matrix(c(.1,.3,.5,.6,.2,.1,.8,1,.4,.2,.2,.9),2,6)
W=matrix(c(0,.8,.4,.6,.2,0),2,3)
#Create a matrix satisfying (1) is easy:
A=matrix(runif(18),6,3)
Q1=svd(A)$u%*%chol(Z)
#For the second condition (2), a solution is given by
Q2=A%*%ginv(X%*%A)%*%W


I do not know how to create a matrix Q that satisfies the two 
conditions.  I have try to construct an iterative procedure without 
success (no convergence):

eps=10
i=0
while(eps.5)
{
Q1=svd(Q2)$u%*%chol(Z)
Q2=Q1%*%ginv(X%*%Q1)%*%W
eps=sum(abs(Q1-Q2))
cat(i,:,eps,\n)
i=i+1
}
Perhaps someone could have any idea to solve the problem, or a 
reference on this kind of question or the email of another list where I 
should ask this question.

Thanks in advance,
Sincerely.
Stéphane DRAY
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Département des Sciences Biologiques
Université de Montréal, C.P. 6128, succursale centre-ville
Montréal, Québec H3C 3J7, Canada
Tel : 514 343 6111 poste 1233
E-mail : [EMAIL PROTECTED]
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Département des Sciences Biologiques
Université de Montréal, C.P. 6128, succursale centre-ville
Montréal, Québec H3C 3J7, Canada
Tel : 514 343 6111 poste 1233
E-mail : [EMAIL PROTECTED]
-- 

Web
http://www.steph280.freesurf.fr/
-- 

Stéphane DRAY
-- 

Département des Sciences Biologiques
Université de Montréal, C.P. 6128, succursale centre-ville
Montréal, Québec H3C 3J7, Canada
Tel : 514 343 6111 poste 1233
E-mail : [EMAIL PROTECTED]

[R] a small bug in spatstat::rmh

2004-07-07 Thread Evgueni Parilov
Time to time, rmh.default fails to simulate a lookup-type process on a 
statement:

if(all.equal(diff(r),rep(deltar,nlook-1))) {
   equisp - 1
   par - c(beta,nlook,equisp,deltar,rmax,h)
   } else {
   equisp - 0
   par - c(beta,nlook,equisp,deltar,rmax,h,r)
   }
According to the manual, all.equal should not be used in if-statement 
directly. This works:
identical(all.equal(diff(r), rep(deltar, nlook - 1)),TRUE)

Evgueni
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Re: [R] Importing an Excel file

2004-07-07 Thread Richard Mller
Am Mittwoch, 7. Juli 2004 20:21 schrieb Park, Kyong H Mr. RDECOM:
 Hello, R users,
 I am  a very beginner of R and tried read.csv to import an excel file after
 saving an excel file as csv. But it added alternating rows of fictitious NA
 values after row number 16. When I applied read.delim, there were trailing
 several commas at the end of each row after row number 16 instead of NA
 values. Appreciate your help.

I import my OpenOffice calc files as follows (OOo or Excel won't make any 
difference, the csv-format is the same):

inp - (scan(file, sep=;, dec=,, list(0,0), skip = 13, nlines = 58)
x - inp[[1]]; y - inp [[2]]

sep=;: column separator ;
dec=, decimal separator ,
list(0,0): first two columns
skip: no of lines to skip (these lines contain comments etc.)
nlines=58: 58 lines of values to plot
hth, Richard
--
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[EMAIL PROTECTED]  - www.oeko-sorpe.de

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[R] NAMESPACE and tests for unexported functions

2004-07-07 Thread Paul Roebuck
How do you get around the problem of having tests for
functions that are not exported in NAMESPACE? It seems
rather self-defeating to have to export everything so
that 'R CMD CHECK pkg' won't crash when it encounters
a test case for an internal function. I don't want
someone using the package to call the function, but the
package itself should be able to see its own contents.

--
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Re: [R] lme: extract variance estimate

2004-07-07 Thread Spencer Graves
 I just tried it in both lme4 and nlme.  I got it in nlme but not 
lme4.  I'm sure lme4 is better in many ways, but I could not figure out 
how to get what you want in lme4. 

 Specifically, I tried the following: 

DF - data.frame(x=rep(letters[1:2], 2), y=rep(1:2, 2)+0.01*(1:4))
fit - lme(y~1, random=~1|x, data=DF)
VC - VarCorr(fit)
VC
VC[1,2]
 When I did this in library(nlme), I got the following: 

 VC
x = pdLogChol(1)
   Variance StdDev   
(Intercept) 0.5101563004 0.71425227
Residual0.0001999596 0.01414071
 VC[1,2]
[1] 0.71425227

 However, when I did it in library(lme4), I got something different: 

 VC
Groups   NameVariance Std.Dev.
x(Intercept) 0.50995  0.71411
Residual 2e-040.014142
 VC[1,2]
Error in VC[1, 2] : incorrect number of dimensions
 I was concerned by the differences in the estimates in this case, 
so I ported this problem to S-Plus 6.2.  There I got the lme4 answers 
from both lme and varcomp. 

 hope this helps.  spencer graves
Stephen Ellner wrote:
Spencer Graves wrote: 

 

   Have you considered VarCorr?  I've used it with lme, and the 
documentation in package lme4 suggests it should work with GLMM, which 
might also do what you want from glmmPQL. 
   

Thanks for the pointer (I was not aware that nlme and lme4 had different
versions of lme), but I'm still stuck at the same place using lme4: VarCorr(fit)
Groups   NameVariance Std.Dev.
yeart(Intercept) 0.040896 0.20223 
Residual 0.091125 0.30187 

The number I need to extract and store is the .20223, but all the 
components I can find in VarCorr(fit) are something else. 

u=VarCorr(fit); slotNames(u)
[1] scalereSumry  useScale
 

[EMAIL PROTECTED]
   

[1] 0.3018693
 

[EMAIL PROTECTED]
   

$yeart
An object of class corrmatrix
   (Intercept)
(Intercept)   1
Slot stdDev:
(Intercept) 
 0.6699156 
 

[EMAIL PROTECTED]
   

[1] TRUE
In glmmML the estimate is returned as the $sigma component
of the model, but I also need the same info from 'family=gaussian' 
models. 

Stephen P. Ellner ([EMAIL PROTECTED])
Department of Ecology and Evolutionary Biology
Corson Hall, Cornell University, Ithaca NY 14853-2701
Phone (607) 254-4221FAX (607) 255-8088
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RE: [R] NAMESPACE and tests for unexported functions

2004-07-07 Thread Liaw, Andy
Not sure if this is recommended for testing purposes, but you can try using
`:::'; e.g., mypkg:::invisibleFunction(...).

Andy

 From: Paul Roebuck
 
 How do you get around the problem of having tests for
 functions that are not exported in NAMESPACE? It seems
 rather self-defeating to have to export everything so
 that 'R CMD CHECK pkg' won't crash when it encounters
 a test case for an internal function. I don't want
 someone using the package to call the function, but the
 package itself should be able to see its own contents.
 
 --
 SIGSIG -- signature too long (core dumped)

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Re: [R] KalmanSmooth problem

2004-07-07 Thread Uwe Ligges
Spencer Graves wrote:
Hi, Uwe:  KalmanSmooth is in the stats package in R 1.9.1.  R is 
moving so fast that it is impossible to keep current with all parts of 
it.  Best Wishes, Spencer Graves
U  thanks, Spencer. My apologies to Pieter.
Indeed, looks like I forgot to upgrade my R version on the machine at 
home. I only did so in my office and on my laptop.

Uwe

Uwe Ligges wrote:
Hazenberg21, Pieter wrote:
Hello,
In R I am trying to use Kalman filtering to find a solution for an 
hydrological problem. With Kalman Filtering I want to estimate the 
discharge comming from three storage bassins. I have programmed a 
function in R which can run KalmanSmooth. When I'm asking for the 
function and putting in values, R detects the following error: Error 
in as.vector(data) : Argument S1 is missing, with no default.
I have try to find a solution for this error in the R help file, and 
in different manuals, but I can't find it. Please help me find a 
solution.
Question: What does R mean with S1 and what am I doing wrong?
Here is the way I have programmed the hydrological problem in R. 

discharge=read.table(file=C:/Program 
Files/R/rw1090/discharge.txt,header=T)
deb=discharge[,1]
deb

  [1] 11.545313  8.045465  5.670868  4.044584  2.919311  2.306668  
2.940956
  [8]  4.238159  5.017374  3.818236  2.928805  2.262183  1.757765  
1.633945
 [15]  2.295130  3.454054  4.035224  3.193967  2.533181  2.012406  
1.600836
 [22]  1.652155  2.428678  3.642827  4.019545  3.209473  2.563617  
2.048347
 [29]  1.637041  1.828952  2.757842  4.050821  4.147013  3.316503  
2.652490
 [36]  2.121535  1.696934  2.027763  3.107366  4.429670  4.160178  
3.327950
 [43]  2.662237  2.129710  1.703717  2.158095  3.337039  4.582359  
3.905901
 [50]  3.124690  2.499732  1.999772  1.599810  2.130893  3.302622  
4.336081
 [57]  3.468857  2.775081  2.220062  1.776048  1.560859  2.169537  
3.348081
 [64]  4.170552  3.336440  2.669151  2.135320  1.708256  1.648859  
2.374217
 [71]  3.624091  4.248563  3.398850  2.719080  2.175264  1.740211  
1.826122
 [78]  2.704749  4.056438  4.437309  3.549847  2.839878  2.271902  
1.817522
 [85]  2.053994  3.107875  4.548436  4.600601  3.680481  2.944385  
2.355508
 [92]  1.884406  2.273248  3.490148  4.949898  4.584409  3.667527  
2.934022
 [99]  2.347217  1.84

Kalm = function(x,O1,O2,O3,T1,T2,T3,T4,T5,t,ga){

+ 
t=array(c(1+ga*O1+t/O1*(-(1/T2)-(1/T3)-(1/T1)),t/O1*(1/T2),t/O1*(1/T3),
+ t/O2*(1/T2),1+ga*O2+t/O2*(-(1/T2)-(1/T4)),t/O2*(1/T4),
+ 
t/O3*(1/T3),t/O3*(1/T4),1+ga*O3+t/O3*(-(1/T3)-(1/T4)-(1/T5))),dim=c(3,3)); 

+ h=0.5;
+ r=array(c(1,0,0,0,1,0,0,0,1),dim=c(3,3));
+ q=1;
+ v=r*q*t(r);
+ a=10.14286;
+ z=array(c((1/T1),0,0,0,0,0,0,0,(1/T5)), dim=c(3,3));
+ p=array(c(1,0,0,0,1,0,0,0,1),dim=c(3,3));
+ pn=array(c(1,0,0,0,1,0,0,0,1),dim=c(3,3));
+ kal=KalmanSmooth(x, list(T=t,Z=z,h=h,V=v,a=a,P=p,Pn=pn), nit=0)
+ kal}
 

Kalm(deb,4,.5,5,.7,.1,2,3,4,1,0.65)

Error in as.vector(data) : Argument S1 is missing, with no default
 
First I thought I had to make a timeserie of deb. But this doesn't 
change the problem.
Lot's of thanks trying to help me. 

a) Please tell us R Version and OS (OK, implicitly done that we are 
talking about R-1.9.0 on Windows).
b) Please tell us which packages you are using (I don't know 
KalmanSmooth(), for example).
c) Please try to specify reproducible examples.

Conclusion for a-c): Please read the posting-guide.
My hint is to try to debug yourself, at least *try*, starting eith 
calling traceback() right after the error appeared, in order to get a 
guess where the error really happens. Then look at the data that is 
passed to the function -  and I'm pretty sure you will get at least an 
idea what goes wrong.

Uwe Ligges


Best regards,
Pieter Hazenberg
Student Hydrology and Watermanagement
Wageningen University
The Netherlands
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Re: [R] a small bug in spatstat::rmh

2004-07-07 Thread Uwe Ligges
Evgueni Parilov wrote:
Time to time, rmh.default fails to simulate a lookup-type process on a 
statement:

if(all.equal(diff(r),rep(deltar,nlook-1))) {
   equisp - 1
   par - c(beta,nlook,equisp,deltar,rmax,h)
   } else {
   equisp - 0
   par - c(beta,nlook,equisp,deltar,rmax,h,r)
   }
According to the manual, all.equal should not be used in if-statement 
directly. This works:
identical(all.equal(diff(r), rep(deltar, nlook - 1)),TRUE)

Evgueni
According to library(help = spatstat), Adrian Baddeley 
[EMAIL PROTECTED] (in CC) is maintainer of the spatstat package 
- and therefore the right addressee of this message.

Uwe Ligges
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RE: [R] KalmanSmooth problem

2004-07-07 Thread Hazenberg21, Pieter
Dear Uwe and Spencer,
Thanks for replying so soon. The R version I wrote is indeed R 1.9.1, and the package 
is stats. Sorry I didn't wrote it in my first email. I've have tried to use the 
comments you gave, and simplified my function. 
It is changed to:
 
 Kalm = function(x){
+ t=array(c(-1.4248, 2.5, .1250, 20, -19.3183, .6667, .1, .0667, 1.7945), dim=c(3,3))
+ h=0.5;
+ r=array(c(1,0,0,0,1,0,0,0,1),dim=c(3,3));
+ q=1;
+ v=r*q*t(r);
+ a=10.14286;
+ z=array(c(.3574, 0, 0, 0, 0, 0, 0, 0, .2), dim=c(3,3))
+ p=array(c(1,0,0,0,1,0,0,0,1),dim=c(3,3));
+ pn=array(c(1,0,0,0,1,0,0,0,1),dim=c(3,3));
+ kal=KalmanSmooth(x,list(T=t,Z=z,h=h,V=v,a=a,P=p,Pn=pn),nit=0)
+ kal}
Kalm(deb)
 
Wonderfully it is running quite well now. Thank you for your spare time.
Best regards,
Pieter Hazenberg

-Oorspronkelijk bericht- 
Van: Uwe Ligges [mailto:[EMAIL PROTECTED] 
Verzonden: wo 7-7-2004 22:09 
Aan: Spencer Graves; Hazenberg21, Pieter 
CC: R Help Mailing List 
Onderwerp: Re: [R] KalmanSmooth problem



Spencer Graves wrote:

 Hi, Uwe:  KalmanSmooth is in the stats package in R 1.9.1.  R is
 moving so fast that it is impossible to keep current with all parts of
 it.  Best Wishes, Spencer Graves

U  thanks, Spencer. My apologies to Pieter.

Indeed, looks like I forgot to upgrade my R version on the machine at
home. I only did so in my office and on my laptop.

Uwe


 Uwe Ligges wrote:

 Hazenberg21, Pieter wrote:

 Hello,
 In R I am trying to use Kalman filtering to find a solution for an
 hydrological problem. With Kalman Filtering I want to estimate the
 discharge comming from three storage bassins. I have programmed a
 function in R which can run KalmanSmooth. When I'm asking for the
 function and putting in values, R detects the following error: Error
 in as.vector(data) : Argument S1 is missing, with no default.
 I have try to find a solution for this error in the R help file, and
 in different manuals, but I can't find it. Please help me find a
 solution.
 Question: What does R mean with S1 and what am I doing wrong?
 Here is the way I have programmed the hydrological problem in R.

 discharge=read.table(file=C:/Program
 Files/R/rw1090/discharge.txt,header=T)
 deb=discharge[,1]
 deb



   [1] 11.545313  8.045465  5.670868  4.044584  2.919311  2.306668 
 2.940956
   [8]  4.238159  5.017374  3.818236  2.928805  2.262183  1.757765 
 1.633945
  [15]  2.295130  3.454054  4.035224  3.193967  2.533181  2.012406 
 1.600836
  [22]  1.652155  2.428678  3.642827  4.019545  3.209473  2.563617 
 2.048347
  [29]  1.637041  1.828952  2.757842  4.050821  4.147013  3.316503 
 2.652490
  [36]  2.121535  1.696934  2.027763  3.107366  4.429670  4.160178 
 3.327950
  [43]  2.662237  2.129710  1.703717  2.158095  3.337039  4.582359 
 3.905901
  [50]  3.124690  2.499732  1.999772  1.599810  2.130893  3.302622 
 4.336081
  [57]  3.468857  2.775081  2.220062  1.776048  1.560859  2.169537 
 3.348081
  [64]  4.170552  3.336440  2.669151  2.135320  1.708256  1.648859 
 2.374217
  [71]  3.624091  4.248563  3.398850  2.719080  2.175264  1.740211 
 1.826122
  [78]  2.704749  4.056438  4.437309  3.549847  2.839878  2.271902 
 1.817522
  [85]  2.053994  3.107875  4.548436  4.600601  3.680481  2.944385 
 2.355508
  [92]  1.884406  2.273248  3.490148  4.949898  4.584409  3.667527 
 2.934022
  [99]  2.347217  1.84

 Kalm = function(x,O1,O2,O3,T1,T2,T3,T4,T5,t,ga){



 +
 t=array(c(1+ga*O1+t/O1*(-(1/T2)-(1/T3)-(1/T1)),t/O1*(1/T2),t/O1*(1/T3),
 + t/O2*(1/T2),1+ga*O2+t/O2*(-(1/T2)-(1/T4)),t/O2*(1/T4),
 +
 t/O3*(1/T3),t/O3*(1/T4),1+ga*O3+t/O3*(-(1/T3)-(1/T4)-(1/T5))),dim=c(3,3));

 + h=0.5;
 + r=array(c(1,0,0,0,1,0,0,0,1),dim=c(3,3));
 + q=1;
 + v=r*q*t(r);
 + a=10.14286;
 + z=array(c((1/T1),0,0,0,0,0,0,0,(1/T5)), dim=c(3,3));
 + p=array(c(1,0,0,0,1,0,0,0,1),dim=c(3,3));
 + pn=array(c(1,0,0,0,1,0,0,0,1),dim=c(3,3));
 + kal=KalmanSmooth(x, list(T=t,Z=z,h=h,V=v,a=a,P=p,Pn=pn), nit=0)
 + kal}
 

 Kalm(deb,4,.5,5,.7,.1,2,3,4,1,0.65)



 Error in as.vector(data) : Argument S1 is missing, with no default
 
 First I thought I had to make a timeserie of deb. But this doesn't
 change the problem.
 Lot's of 

[R] question about seq.dates from chron vs. as.POSIXct

2004-07-07 Thread Laura Holt
Dear R People:
Here is an interesting question:
library(chron)
xt - seq.dates(from=01/01/2004,by=days,length=5)
xt
[1] 01/01/04 01/02/04 01/03/04 01/04/04 01/05/04

#Fine so far
as.POSIXct(xt)
[1] 2003-12-31 18:00:00 Central Standard Time
[2] 2004-01-01 18:00:00 Central Standard Time
[3] 2004-01-02 18:00:00 Central Standard Time
[4] 2004-01-03 18:00:00 Central Standard Time
[5] 2004-01-04 18:00:00 Central Standard Time

Why do the dates change, please?  Presumably the as.POSIXct is taking the xt 
as midnight GMT and converting to Central Standard Time.

Is the best solution to:
as.POSIXlt(xt, CST)
[1] 2004-01-01 CST 2004-01-02 CST 2004-01-03 CST 2004-01-04 CST
[5] 2004-01-05 CST

Thanks in advance!
Sincerely,
Laura Holt
who is corrupted by dates and times
mailto: [EMAIL PROTECTED]
download! http://toolbar.msn.click-url.com/go/onm00200413ave/direct/01/
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Re: [R] lost messages

2004-07-07 Thread Aaron J. Mackey
The second two resends did go through (I checked on the web archive), 
but the first did not; additionally, the second two resends both came 
through with the exact same timestamp, even though I resent them over 
the space of 4 hours or so (and only after I had sent the lost 
messages message, which did come through to me just fine).  So there 
was something stuck, but seems to be unstuck now.

Thanks to all,
-Aaron
On Jul 7, 2004, at 4:05 PM, (Ted Harding) wrote:
On 07-Jul-04 Aaron J. Mackey wrote:
I've posted a message twice to this list, and never seen it appear yet
... perhaps this one will go through ... ?
Hi Aaron,
Yes, it did get through. If you have not received it yourself, then
possibly your subscription to R-help has got set to nomail or
equivalent. It would be worth checking.
Best wishes,
Ted.
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[R] text editor for R

2004-07-07 Thread Yi-Xiong Sean Zhou
Hi, 

 

What is the best text editor for programming in R? I am using JEdit as the
text editor, however, it does not have anything specific for R. It will be
nice to have a developing environment where the keywords are highlighted,
plus some other debugging functions. 

 

Yi-Xiong

 


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Re: [R] text editor for R

2004-07-07 Thread Sundar Dorai-Raj

Yi-Xiong Sean Zhou wrote:
Hi, 

 

What is the best text editor for programming in R? I am using JEdit as the
text editor, however, it does not have anything specific for R. It will be
nice to have a developing environment where the keywords are highlighted,
plus some other debugging functions. 

 

Yi-Xiong
 
best is subjective, but (X)Emacs with ESS seems to be the most 
popular. See the following for more:

http://cran.us.r-project.org/other-software.html
--sundar
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Re: [R] text editor for R

2004-07-07 Thread Marc Schwartz
On Wed, 2004-07-07 at 17:47, Yi-Xiong Sean Zhou wrote:
 Hi, 

 What is the best text editor for programming in R? I am using JEdit as the
 text editor, however, it does not have anything specific for R. It will be
 nice to have a developing environment where the keywords are highlighted,
 plus some other debugging functions. 
 
 Yi-Xiong

More information is available at:

http://www.sciviews.org/_rgui/

Your e-mail headers suggest that you are using Windows. Thus, perhaps
the two best choices (subject to challenge by others) would be:

1. R-WinEdt (Under IDE/Script Editors)

2. ESS for Windows

The above two tools provide for a wide variety of functionality beyond
syntax highlighting.

There is a syntax highlighting file listed at the above site for jEdit.

HTH,

Marc Schwartz

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Re: [R] text editor for R

2004-07-07 Thread Murray Jorgensen
I tried R-WinEdt a few years ago, but as I remember it interfered with 
my usual use of WinEdt which is as a front end to MiKTeX. Is there a way 
to use WinEdt both ways?

Murray Jorgensen
Marc Schwartz wrote:
On Wed, 2004-07-07 at 17:47, Yi-Xiong Sean Zhou wrote:
Hi, 

What is the best text editor for programming in R? I am using JEdit as the
text editor, however, it does not have anything specific for R. It will be
nice to have a developing environment where the keywords are highlighted,
plus some other debugging functions. 

Yi-Xiong

More information is available at:
http://www.sciviews.org/_rgui/
Your e-mail headers suggest that you are using Windows. Thus, perhaps
the two best choices (subject to challenge by others) would be:
1. R-WinEdt (Under IDE/Script Editors)
2. ESS for Windows
The above two tools provide for a wide variety of functionality beyond
syntax highlighting.
There is a syntax highlighting file listed at the above site for jEdit.
HTH,
Marc Schwartz
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--
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Department of Statistics, University of Waikato, Hamilton, New Zealand
Email: [EMAIL PROTECTED]Fax 7 838 4155
Phone  +64 7 838 4773 wk+64 7 849 6486 homeMobile 021 1395 862
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RE: [R] text editor for R

2004-07-07 Thread Liaw, Andy
Uwe would be the authority on this 8-), but my impression is that if you
keep two separate shortcuts, you should be fine.  The one for R-WinEdt has
flags that sets it up for R, which should not be used in the one for MikTeX.

Andy

 From: Murray Jorgensen
 
 I tried R-WinEdt a few years ago, but as I remember it 
 interfered with 
 my usual use of WinEdt which is as a front end to MiKTeX. Is 
 there a way 
 to use WinEdt both ways?
 
 Murray Jorgensen
 
 Marc Schwartz wrote:
 
  On Wed, 2004-07-07 at 17:47, Yi-Xiong Sean Zhou wrote:
  
 Hi, 
  
  
 What is the best text editor for programming in R? I am 
 using JEdit as the
 text editor, however, it does not have anything specific 
 for R. It will be
 nice to have a developing environment where the keywords 
 are highlighted,
 plus some other debugging functions. 
 
 Yi-Xiong
  
  
  More information is available at:
  
  http://www.sciviews.org/_rgui/
  
  Your e-mail headers suggest that you are using Windows. 
 Thus, perhaps
  the two best choices (subject to challenge by others) would be:
  
  1. R-WinEdt (Under IDE/Script Editors)
  
  2. ESS for Windows
  
  The above two tools provide for a wide variety of 
 functionality beyond
  syntax highlighting.
  
  There is a syntax highlighting file listed at the above 
 site for jEdit.
  
  HTH,
  
  Marc Schwartz
  
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 http://www.R-project.org/posting-guide.html
  
  
 
 -- 
 Dr 
 Murray Jorgensen  http://www.stats.waikato.ac.nz/Staff/maj.html
 Department of Statistics, University of Waikato, Hamilton, New Zealand
 Email: [EMAIL PROTECTED]Fax 7 838 4155
 Phone  +64 7 838 4773 wk+64 7 849 6486 homeMobile 021 1395 862
 
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[R] omit complete cases

2004-07-07 Thread ivo_welch-rstat8783
thanks everyone.   all solutions were better than what I had, and 
simple.

R is an interesting experience.  Extremely powerful and awe-inspiring 
for its elegance;  things work like I would never have believed how 
elegantly  they work.  The IQ in the subsetting alone is superbly 
clever.  And then it turns around: figuring out how to do simple things 
can take a long time---until I realize that it is somewhere somehow 
built in already.  So, R (and the answers from helpful souls on this 
list) often makes me feel quite stupid.  Yes, I first search; yes, I 
always look---but either I did not know or I had forgotten.

I used to use perl for much work, and although there is much to like 
about it, R seems to be even better for most tasks---except that there 
is one perl resource that R cannot beat:  the Perl Cookbook.  if I only 
had an R cookbook...

regards,
/ivo
---
ivo welch
professor of finance and economics
brown / nber / yale
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[R] command line interface

2004-07-07 Thread Lana Schaffer

How can plots (histograms) be implemented with the command line interface to R?
Lana Schaffer

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RE: [R] text editor for R

2004-07-07 Thread Kevin Wang
Hi,

 Uwe would be the authority on this 8-), but my impression is that if you
 keep two separate shortcuts, you should be fine.  The one for R-WinEdt has
 flags that sets it up for R, which should not be used in the one for MikTeX.

I think Andy is correct.  A few years ago (back in the dark ages -- before
I discovered Emacs/ESS), I had two short cuts, one calls R-WinEdt (i.e.
with flags...etc) and the other with just a normal WinEdt icon.

However, I *think* now you can interact R-WinEdt within R directly (I
tried the new version about 2 ~ 3 months ago just for fun, and that seemed
to be the case, but I can't quite remember *_*).

Cheers,

Kevin


Ko-Kang Kevin Wang
PhD Student
Centre for Mathematics and its Applications
Building 27, Room 1004
Mathematical Sciences Institute (MSI)
Australian National University
Canberra, ACT 0200
Australia

Homepage: http://wwwmaths.anu.edu.au/~wangk/
Ph (W): +61-2-6125-2431
Ph (H): +61-2-6125-7411
Ph (M): +61-40-451-8301

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Re: [R] text editor for R

2004-07-07 Thread David Scott
On Thu, 8 Jul 2004, Murray Jorgensen wrote:

 I tried R-WinEdt a few years ago, but as I remember it interfered with 
 my usual use of WinEdt which is as a front end to MiKTeX. Is there a way 
 to use WinEdt both ways?
 
 Murray Jorgensen
 
This problem annoyed me for a while too. My solution (which is not perhaps 
ideal) is this. You want two different incarnations of WinEdt, one for 
TeX, the other for R. On the desktop I have a shortcut to WinEdt which is 
the one for TeX stuff. I open the other one with R syntax highlighting etc 
by starting R and using library(RWinEdt). To do this you have to install 
the RWinEdt package and SWinRegistry. This is all well explained in the 
ReadMe.txt for RWinEdt.

I think with the right additions to the Target field in a shortcut to 
WinEdt you can call up the incarnation of WinEdt that is suitable for R. I 
haven't done that. You would then have two shortcuts to WinEdt, one for 
your TeX stuff, one for R.

Uwe Ligges is the guru for this though.

David Scott


David Scott Department of Statistics, Tamaki Campus
The University of Auckland, PB 92019
AucklandNEW ZEALAND
Phone: +64 9 373 7599 ext 86830 Fax: +64 9 373 7000
Email:  [EMAIL PROTECTED] 


Graduate Officer, Department of Statistics

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[R] S data library

2004-07-07 Thread Kimpel, Mark W
I am currently working my way through Statistical Models in S
(Chambers  Hastie, 1992) and it would be helpful to me if I had access
to the data library they refer to and which contains the authors sample
datasets. Are these datasets available as an R package or any other type
of download?
 
Thanks,
 
Mark W. Kimpel MD
Department of Psychiatry
Indiana University School of Medicine
 

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Re: [R] text editor for R

2004-07-07 Thread Peter Flom
I use WinEdt for both.  I simply installed it twice, and set up one
version for R and one for LaTeX, I have seperate icons on the desktop,
with different names, and it works fine.

HTH

Peter

Peter L. Flom, PhD
Assistant Director, Statistics and Data Analysis Core
Center for Drug Use and HIV Research
National Development and Research Institutes
71 W. 23rd St
www.peterflom.com
New York, NY 10010
(212) 845-4485 (voice)
(917) 438-0894 (fax)


 [EMAIL PROTECTED] 07/07/04 7:37 PM 
I tried R-WinEdt a few years ago, but as I remember it interfered with 
my usual use of WinEdt which is as a front end to MiKTeX. Is there a way

to use WinEdt both ways?

Murray Jorgensen

Marc Schwartz wrote:

 On Wed, 2004-07-07 at 17:47, Yi-Xiong Sean Zhou wrote:
 
Hi, 
 
 
What is the best text editor for programming in R? I am using JEdit as
the
text editor, however, it does not have anything specific for R. It
will be
nice to have a developing environment where the keywords are
highlighted,
plus some other debugging functions. 

Yi-Xiong
 
 
 More information is available at:
 
 http://www.sciviews.org/_rgui/
 
 Your e-mail headers suggest that you are using Windows. Thus, perhaps
 the two best choices (subject to challenge by others) would be:
 
 1. R-WinEdt (Under IDE/Script Editors)
 
 2. ESS for Windows
 
 The above two tools provide for a wide variety of functionality beyond
 syntax highlighting.
 
 There is a syntax highlighting file listed at the above site for
jEdit.
 
 HTH,
 
 Marc Schwartz
 
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 https://www.stat.math.ethz.ch/mailman/listinfo/r-help
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-- 
Dr Murray Jorgensen  http://www.stats.waikato.ac.nz/Staff/maj.html
Department of Statistics, University of Waikato, Hamilton, New Zealand
Email: [EMAIL PROTECTED]Fax 7 838 4155
Phone  +64 7 838 4773 wk+64 7 849 6486 homeMobile 021 1395 862

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Re: [R] text editor for R

2004-07-07 Thread Murray Jorgensen
Oh, yes. I think I did do something like this, but for some reason the 
two incarnations bothered me.

Murray Jorgensen
David Scott wrote:
On Thu, 8 Jul 2004, Murray Jorgensen wrote:

I tried R-WinEdt a few years ago, but as I remember it interfered with 
my usual use of WinEdt which is as a front end to MiKTeX. Is there a way 
to use WinEdt both ways?

Murray Jorgensen
This problem annoyed me for a while too. My solution (which is not perhaps 
ideal) is this. You want two different incarnations of WinEdt, one for 
TeX, the other for R. On the desktop I have a shortcut to WinEdt which is 
the one for TeX stuff. I open the other one with R syntax highlighting etc 
by starting R and using library(RWinEdt). To do this you have to install 
the RWinEdt package and SWinRegistry. This is all well explained in the 
ReadMe.txt for RWinEdt.

I think with the right additions to the Target field in a shortcut to 
WinEdt you can call up the incarnation of WinEdt that is suitable for R. I 
haven't done that. You would then have two shortcuts to WinEdt, one for 
your TeX stuff, one for R.

Uwe Ligges is the guru for this though.
David Scott

David Scott	Department of Statistics, Tamaki Campus
		The University of Auckland, PB 92019
		Auckland	NEW ZEALAND
Phone: +64 9 373 7599 ext 86830		Fax: +64 9 373 7000
Email:	[EMAIL PROTECTED] 

Graduate Officer, Department of Statistics
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Email: [EMAIL PROTECTED]Fax 7 838 4155
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[R] read.frame

2004-07-07 Thread S Peri
Hello group, 
  I am learning R and I am new to many concepts.I face
the following errors when I am trying to execute the
following. I have 4 text files with protein accession
numbers. I wanted to represent them in a venn diagram
and for that I using intersect and setdiff functions.

My data looks like this:

file1.txt (c):
NP_05
NP_20
NP_30
NP_53

file2.txt(e):
NP_05
NP_20
NP_30
NP_31
NP_53
NP_55
NP_87

file3.txt(h):
NP_05
NP_20
NP_30
NP_53
NP_55
NP_57
NP_87

file4.txt (s):
NP_05
NP_20
NP_30
NP_33
NP_53
NP_55
NP_87
NP_000168


Now I did the following FIRST time:
c=read.table(file1.txt)
e=read.table(file2.txt)
s=read.table(file4.txt)
h=read.table(file3.txt)

 class(c)
[1] data.frame
 class(s)
[1] data.frame
 CiS=intersect(c,s)
 CiS
NULL data frame with 0 rows 
# Why am I getting NULL data error. I know there
are common elements between c and S. ##

 CiS-intersect(read.matrix(c,s))
Error in unique(y[match(x, y, 0)]) : Argument y is
missing, with no default
 CiS-intersect(read.frame(c,s))
Error in unique(y[match(x, y, 0)]) : Argument y is
missing, with no default

# Why am I getting this error.



Second thing I did:

I loaded the data as data.frame instead read.table(). 
Again I never get intersection of C,E and S,H. 


Can any one please help me. 
thank you

SP

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R cookbook (Re: [R] omit complete cases)

2004-07-07 Thread Arin Basu
Hi Ivo:

You might check out Paul Jobnson's following page:

http://www.ukans.edu/~pauljohn/R/Rtips.html  

HTH,
Arin



On Thu, 08 Jul 2004 [EMAIL PROTECTED] wrote :

...I used to use perl for much work, and although there is much to like about it, R 
seems to be even better for most tasks---except that there is one perl resource that 
R cannot beat:  the Perl Cookbook.  if I only had an R cookbook...

regards,

/ivo
---


Arindam Basu MD MPH DBI
Assistant Director
Fogarty International Program on Environmental Health in India
IPGMER
244 AJC Bose Road, Kolkata 700027
India
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Re: [R] read.frame

2004-07-07 Thread S Peri
Dear Alec, 
 Thank you for your response and it worked. Further, I
have another mathematical problem. I apologize ahead
as this question is not apt for this list. I am a
biologist working at Johns Hopkins School of Medicine.

As I listed in my previous e-mail (attached below)I
mentioned I have 4 protein sets.  I am now trying to
calculate the combinatorics of these sets.  My
ultimate aim is to draw a venn diagram and find out
the proteins that are unique to set C,S,E and H. 
 I drew a venn diagram graph and I am banging my head
to deduce the combinations.  It is easy for me to
deduce the intersections - that means the protein
entries that are present in common. However, it proved
very difficult to deduce the following: 

I could calculate the following:

NP_*** present in both C and E (C intersection E)
NP_*** present in both C and H (C intersection H)
NP_*** present in both C and S (C ^ S)
NP_*** present in both E and H (E ^ H)
NP_*** present in both E and S (E ^ S)
NP_*** present in both H and S (H ^ S)

NP_*** present in C, E and H (C^E^H)
NP_*** present in C, H and S (C^H^S)
NP_*** present in E, H and S (E^H^S)
NP_*** present in E, S and C (E^S^C)


It is very difficult for me to deduce the following:

NP_ entries specific to E
NP_ entries specific to H
NP_ entries specific to S


I waster many pages but could not derive some solution
to get unique elements for sets, s,e,h,and C.

Can any one help me by suggesting some way to get
these.

Thank you and I apologise again for posting the wrong
question. 

SP


--- S Peri [EMAIL PROTECTED] wrote:
 Hello group, 
   I am learning R and I am new to many concepts.I
 face
 the following errors when I am trying to execute the
 following. I have 4 text files with protein
 accession
 numbers. I wanted to represent them in a venn
 diagram
 and for that I using intersect and setdiff
 functions.
 
 My data looks like this:
 
 file1.txt (c):
 NP_05
 NP_20
 NP_30
 NP_53
 
 file2.txt(e):
 NP_05
 NP_20
 NP_30
 NP_31
 NP_53
 NP_55
 NP_87
 
 file3.txt(h):
 NP_05
 NP_20
 NP_30
 NP_53
 NP_55
 NP_57
 NP_87
 
 file4.txt (s):
 NP_05
 NP_20
 NP_30
 NP_33
 NP_53
 NP_55
 NP_87
 NP_000168
 
 
 Now I did the following FIRST time:
 c=read.table(file1.txt)
 e=read.table(file2.txt)
 s=read.table(file4.txt)
 h=read.table(file3.txt)
 
  class(c)
 [1] data.frame
  class(s)
 [1] data.frame
  CiS=intersect(c,s)
  CiS
 NULL data frame with 0 rows 
 # Why am I getting NULL data error. I know there
 are common elements between c and S. ##
 
  CiS-intersect(read.matrix(c,s))
 Error in unique(y[match(x, y, 0)]) : Argument y is
 missing, with no default
  CiS-intersect(read.frame(c,s))
 Error in unique(y[match(x, y, 0)]) : Argument y is
 missing, with no default
 
 # Why am I getting this error.
 
 
 
 Second thing I did:
 
 I loaded the data as data.frame instead
 read.table(). 
 Again I never get intersection of C,E and S,H. 
 
 
 Can any one please help me. 
 thank you
 
 SP
 
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RE: [R] question about seq.dates from chron vs. as.POSIXct

2004-07-07 Thread Gabor Grothendieck


Yes, its assuming GMT.

I find it best to use an intermediate conversion
to character to avoid these sorts of problems.
If, as in your example, you just have dates and
no times then the following would do it (and has the
advantage that you don't have to specify your
time zone explicitly so it will still work if
someone in another timezone tries your code):

   as.POSIXct(format(as.Date(xt)))

The reason to convert it to Date first, before 
formatting is so that the format will use Date's
default format (which is accepted by as.POSIXct)
rather than chron's default format.

---

Date:   Wed, 07 Jul 2004 15:30:47 -0500
From:   Laura Holt [EMAIL PROTECTED]
To: [EMAIL PROTECTED]
Subject:[R] question about seq.dates from chron vs. as.POSIXct

Dear R People:

Here is an interesting question:

library(chron)
xt - seq.dates(from=01/01/2004,by=days,length=5)
xt
[1] 01/01/04 01/02/04 01/03/04 01/04/04 01/05/04

#Fine so far
as.POSIXct(xt)
[1] 2003-12-31 18:00:00 Central Standard Time
[2] 2004-01-01 18:00:00 Central Standard Time
[3] 2004-01-02 18:00:00 Central Standard Time
[4] 2004-01-03 18:00:00 Central Standard Time
[5] 2004-01-04 18:00:00 Central Standard Time

Why do the dates change, please? Presumably the as.POSIXct is taking the xt
as midnight GMT and converting to Central Standard Time.

Is the best solution to:
as.POSIXlt(xt, CST)
[1] 2004-01-01 CST 2004-01-02 CST 2004-01-03 CST 2004-01-04 CST
[5] 2004-01-05 CST


Thanks in advance!

Sincerely,
Laura Holt
who is corrupted by dates and times
mailto: [EMAIL PROTECTED]

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Re: [R] read.frame

2004-07-07 Thread DJNordlund
 
In a message dated 7/7/2004 9:42:03 PM Pacific Daylight Time,  
[EMAIL PROTECTED] writes:

Dear  Alec, 
Thank you for your response and it worked. Further, I
have  another mathematical problem. I apologize ahead
as this question is not apt  for this list. I am a
biologist working at Johns Hopkins School of  Medicine.

As I listed in my previous e-mail (attached  below)I
mentioned I have 4 protein sets.  I am now trying  to
calculate the combinatorics of these sets.  My
ultimate aim is  to draw a venn diagram and find out
the proteins that are unique to set  C,S,E and H. 
I drew a venn diagram graph and I am banging my head
to  deduce the combinations.  It is easy for me to
deduce the  intersections - that means the protein
entries that are present in common.  However, it proved
very difficult to deduce the following: 

I could  calculate the following:

NP_*** present in both C and E (C intersection  E)
NP_*** present in both C and H (C intersection H)
NP_*** present in  both C and S (C ^ S)
NP_*** present in both E and H (E ^ H)
NP_***  present in both E and S (E ^ S)
NP_*** present in both H and S (H ^  S)

NP_*** present in C, E and H (C^E^H)
NP_*** present in C, H and S  (C^H^S)
NP_*** present in E, H and S (E^H^S)
NP_*** present in E, S and  C (E^S^C)


It is very difficult for me to deduce the  following:

NP_ entries specific to E
NP_ entries specific to  H
NP_ entries specific to S





I am not sure about your notation, but in R terms, if you want the items  
specific to E, H, or S, how about something like:
 
specific to E  --   setdiff(e, union(c,union(h,s)))
 
specific to H  --   setdiff(h,  union(c,union(e,s)))
 
specific to S  --   setdiff(s,  union(c,union(e,h)))
 
Dan Nordlund



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