date:20110420

Dear Baptiste,

thank you for your help. I tried to built in your suggestions into the splom. 
Below is the result. I decided to create the plotmath-expressions outside the 
function splom2, this will allow me later to horizontally shift (via phantom,
for example) the table entries so that they will be vertically aligned 
according 
to the = signs. Although an array is allowed to contain expressions, I could 
not manage to create it properly. Do you know a solution?

Cheers,

Marius

library(lattice) 
library(grid)
library(gridExtra)

## splom with customized lower.panel
## x: data
## expr.arr: array of expressions [(i,j,) entry contains expressions which are
##   plotted in a grid table in the lower panel (i,j)]
splom2 - function(x, expr.arr){
## function for creating table 
table.fun - function(vec){ # single values for one panel
grid.table(vec,
   parse=TRUE, # parse labels as expressions
   theme=theme.list(
   gpar.corefill=gpar(fill=NA, col=NA), # make bg transparent
   core.just=left) # justification of labels
   ) 
}
## splom
splom(x, varname.cex=1.4,
  superpanel=function(z, ...){
  panel.pairs(z, upper.panel=panel.splom, lower.panel=function(i,j){
  table.fun(expr.arr[i,j,])
  }, ...)
  })
}

## create data and array of expressions
d - 4
x - matrix(runif(d*1000), ncol=d)
expr.arr - array(, dim=c(d,d,3), dimnames=c(i,j,val)) # d x d x 3 
elements
for(i in 1:d){
for(j in 1:d){
# expr.arr[i,j,] - expression(italic(a)==0, italic(bbb)==0, 
italic(c)==0) # does no work
expr.arr[i,j,] - c(bquote(italic(a)==.(0)), bquote(italic(bbb)==.(0)), 
bquote(italic(c)==.(0))) # same here
}
}

## plot
splom2(x, expr.arr)


On 2011-04-20, at 01:33 , baptiste auguie wrote:

 Hi,
 
 You may want to wait advice from someone who actually understands (the
 labyrinth that is) lattice's help for splom, but the following might
 be a start. I didn't understand what values you actually wanted
 displayed in the lower triangle panels, so I made up some random ones
 in a 3x3 matrix of 3-vectors.
 
 library(lattice)
 library(gridExtra)
 
 info - function(x){
  grid.table(c(bquote(italic(a)==.(x[1])),
   bquote(italic(b)==.(x[2])),
   bquote(italic(c)==.(x[3]))),
 core.just=left,
 parse=TRUE)
 }
 
 U - matrix(runif(3000), ncol=3)
 
 splom(U,
 superpanel=function(z, ...){
   ## dummy 3x3 matrix of 3 values to diplay
   values - replicate(9, round(rnorm(3), 3), simplify=FALSE)
   dummy - matrix(values, ncol=3, byrow=F)
   panel.pairs(z, upper.panel=panel.splom,
   lower.panel=function(i, j, ...){
 print(paste(i,j)) # current panel indices
 info(dummy[i,j] [[1]]) # access the list elements
   }, ...)
   })
 
 HTH,
 
 baptiste
 
 
 On 20 April 2011 10:17, Marius Hofert m_hof...@web.de wrote:
 Dear Baptiste,
 
 there is one tricky part left: how can I create a matrix with the 
 grid.table()
 objects as output? Is this possible? If not, maybe one can try to work with
 panel.splom (which can address single panels and thus call info() for each
 row-column index pair (i,j)), but I'm not sure if this will work.
 
 Cheers,
 
 Marius
 
 library(lattice)
 library(gridExtra)
 
 splom2 - function(x, a, b, c){
## function for the additional information
info - function(a., b., c.){ # single values for one panel
grid.table(c(bquote(italic(a)==.(a.)),
 bquote(italic(b)==.(b.)),
 bquote(italic(c)==.(c.))
 ),
   parse=TRUE, # parse labels as expressions
   theme=theme.list(
   gpar.corefill=gpar(fill=NA, col=NA), # make bg transparent
   core.just=right) # justification of labels
   )
}
labs - matrix(, nrow=ncol(x), ncol=ncol(x)) # should be a matrix of 
 grid.table() objects
for(i in 1:ncol(x)) for(j in 1:ncol(x)) labs[i,j] - info(a.=a[i,j], 
 b.=b[i,j], c.=c[i,j])
## splom
splom(x, superpanel=function(z,...){
  df=data.frame(rows=as.vector(row(a)),
  columns=as.vector(col(a)), labs=as.vector(labs))
  df=subset(df,columnsrows) # subset for lower left triangle
  with(df,{
  panel.text(x=rows, y=columns, labels=labs)
  })
  panel.pairs(z, upper.panel=panel.splom, 
 lower.panel=function(...){}, ...)
  })
 }
 
 ## generate data
 U - matrix(runif(3000), ncol=3)
 
 ## build information
 a - cor(U, method=kendall)
 b - diag(ncol=3, nrow=3)
 c - diag(ncol=3, nrow=3)
 
 ## plot with information
 splom2(U, a, b, c)
 
 
 

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[R] Error Installing or Updating Packages (Maybe because of a proxy)

2011-04-20 Thread Majid Einian

Dear R Helpers,
(I am using Ubuntu lucid and R 2.13.0
When I try to update packages I get this error:

 update.packages()
--- Please select a CRAN mirror for use in this session ---
Loading Tcl/Tk interface ... done
Error in ret[i, ] - c(pkgs[i], lib, desc) :
  number of items to replace is not a multiple of replacement length

I had no problem before (case 1) but now (case 2) I cannot get it to work,
googleing did not help:
case 1:
 * connecting directly without any proxy setting (at my university)
 * using R 2.12.2
case 2:
 * connecting through proxy setting (at my workplace)
 * using R 2.13.0

I set the proxy in terminal too but it does not help (echo $http_proxy gives
me http://192.168.0.1:8080/)

-- 
Majid Einian
PhD Candidate in Economics
Graduate School of Management and Economics
Sharif University of Technology

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Re: [R] Axes Alignment Problem for Multiple Plots

2011-04-20 Thread Matthieu Stigler


Guido

You missed William's e-mails, which solved the problem:
use

 yaxs=i


in second call. Also, William made the good point that you can rather 
just use lines() in second call.


Good William!

Mat



Le 20/04/2011 08:00, Barbaglia, Guido (ESA) a écrit :

Dear Matthieu,

thanks for your clarification! Basically, what I need to do is to plot 
different series on the same chart using different types of plot (lines, 
barplot, ...) and it is fundamental that the various charts have the same 
reference system. I hope that someone in the list is able to fix my problem!


Best Regards
Guido Barbaglia


From: mat [matthieu.stig...@gmail.com]
Sent: 19 April 2011 22:58
To: John Kane
Cc: r-help@r-project.org; Barbaglia, Guido (ESA)
Subject: Re: [R] Axes Alignment Problem for Multiple Plots

Ok, I can replicate your problem, with following code:

dat- 1:10
 barplot(dat, beside=TRUE,ylim=c(0,100));
 par()$usr;
 par(new=T);
 plot(dat, ylim=c(0,100), type=l);
 par()$usr;

So it looks like even if you specify yourself ylim, the resulting
effective ylim (usr[3:4] ) will be different! More surprinsingly,
setting the usr parameter before is not effective:
 par(new=T, usr=c(0,11,0,100));

this will not preventpar()$usr; to be different than fixed :-(

No idea, but hopefully someone else on the list will be able to provide
advice!

Matthieu




Le 19. 04. 11 20:47, John Kane a écrit :

What is Coredata(Z0)?
It would be very useful. as the posting guidelines suggest to supply working 
code and sample data.


--- On Tue, 4/19/11, Barbaglia, Guido (ESA)guido.barbag...@fao.org   wrote:


From: Barbaglia, Guido (ESA)guido.barbag...@fao.org
Subject: [R] Axes Alignment Problem for Multiple Plots
To: r-help@r-project.orgr-help@r-project.org
Cc: Stigler, Matthieu (EST)matthieu.stig...@fao.org
Received: Tuesday, April 19, 2011, 11:42 AM
Dear all,

 I'm trying to plot, in the same window,
two different series, using barplot() for the first one and
plot() for the second. What happens is that the second chart
has a different axes origin, therefore the final plot is
wrong. This piece of code shows the differences between the
values of par()$usr:

 barplot(coredata(Z0), beside=TRUE,
ylim=c(0,100));
 par()$usr;
 par(new=T);
 plot(coredata(Z0), ylim=c(0,100));
 par()$usr;

I would like to know how it is possible to edit the values
of par()$usr[3:4] of the second chart in order to be the
same of the first one or, alternatively, how can I plot
together bar and line series within the same reference
system.


Best Regards
Guido Barbaglia
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Re: [R] What to do with positive likelihoods

2011-04-20 Thread Ingmar Visser

Michael,

It's not entirely clear what you are doing from the text (and the code is
incompplete), so just a few remarks.
I assume you are after the log likelihood rather than the likelihood
(otherwise the issue of positive/negative would not show up, dnorm always
returns positive numbers).

Taking the sum of log's is not the same as taking the log of a sum:

 sum(log(1:10))
[1] 15.10441
 log(sum(1:10))
[1] 4.007333


Positive log likelihoods are not problematic; as you note, when variances
are small, log likelihoods may turn positive; this does not change the use
of the log likelihood (eg using it for estimating parameters by maximizing
it, or for testing models).

hth, Ingmar

On Wed, Apr 20, 2011 at 4:13 AM, Turchin, Michael 
michael.turc...@childrens.harvard.edu wrote:

 Hi all,

 I'll preface this with saying I've gone through the archives, and am still
 in need of some help.

 I've been using this likelihood model with mean = 0 and s.d. = sqrt( (c + (
 1 / N1 ) + ( 1 / N2 ) ) * x * ( 1 - x )), where c is a genetic drift
 parameter (usually very small, like between .005 - .001), N1 and N2 are my
 population sizes (~200), and x is a value between 0 and 1. The values I'm
 testing are usually between -.25 to .25, so my command looks like

 dnorm(.1, 0, sqrt( (c + ( 1 / N1 ) + ( 1 / N2 ) ) * x * ( 1 - x )))

 Originally, I was doing this over multiple data points at once, summing up
 the values I was going to test and their variances, and just running the
 likelihood on these summed values once (getting one final likelihood in
 return). I've recently switched this over to running the likelihood on each
 data point and its associated variance one at a time, and summing the
 likelihoods afterwards. However, upon doing this, I'm now getting positive
 likelihoods since the individual variances are so small (.01 to .09, for
 instance). I'm not sure what to do, because I think these small variances
 are messing up the behavior of my final data -- the patterns I'm getting are
 not what I expected, whereas my previous method of summing multiple data
 points and just taking one likelihood value did return what I expected.

 I'm not sure if getting positive likelihoods somewhat implies that the
 behavior of the model / results are off. Should I be using a different
 function than dnorm, now that my variances are so small? Using pnorm instead
 returns my data to what's expected, but my understanding is that pnorm gets
 me a probability now, not a likelihood. Could I use the output from pnorm in
 a likelihood ratio test (which was my original plan)?

 Thanks for any help,
 ~Michael Turchin
 Children's Hospital Boston
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[R] Allocation of memory to a data frame

2011-04-20 Thread santosh

Hello gRoup,

A while back I was advised that overallocating an object and filling
it up is better than rbind.

I am trying to optimize some code. I have the following object (empty
which I know in advance).
Is there an easy to create an empty data frame with 100,000 empty rows
based on the structure below?

 dput(d)
structure(list(A = character(0), B = character(0), C = numeric(0),
D = character(0), E = character(0), F = character(0), G =
character(0),
H = structure(numeric(0), class = Date), I = numeric(0),
J = integer(0)), .Names = c(A, B, C, D, E, F,
G, H, I, J), row.names = integer(0), class = data.frame)

I can use dataFrame in R.utils and do it explicitly but wondering if
there an easier way since I have the structure with me.

Thank you.

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[R] Sweave stops when opening X11 device fails

2011-04-20 Thread Andreas Borg


Hi all,

I've run into the following problem with Sweave: I frequently run Sweave 
from a xterm console within an X session owned by a different user, i.e. 
my colleague is logged in on this computer and I do su with my 
username and start R and Sweave afterwards. Now, when Sweave comes to a 
figure chunk, it sees that there is an X server running and tries to 
show whatever I plot in that chunk on the screen, additionally to 
writing a pdf file. The problem is that I am not logged into the X 
session myself, and the script fails with a message saying someting like 
could not open device X11 (I have German messages, so I do not know 
what the exact phrase would be in English). When I log in with putty, 
where there is no X11 at all, everything works fine. Is there a way to 
prevent Sweave from failing in the former case? Would this even account 
as a bug? I think it would be nice if the script just kept on running 
without plotting on the screen if opening X11 fails, just like it does 
when no X11 is available at all.


Best regards,

Andreas

--
Andreas Borg
Medizinische Informatik

UNIVERSITÄTSMEDIZIN
der Johannes Gutenberg-Universität
Institut für Medizinische Biometrie, Epidemiologie und Informatik
Obere Zahlbacher Straße 69, 55131 Mainz
www.imbei.uni-mainz.de

Telefon +49 (0) 6131 175062
E-Mail: b...@imbei.uni-mainz.de

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Re: [R] Saving run time in loop

2011-04-20 Thread Dennis Murphy

Hi:

Perhaps rollapply() in the zoo package might be helpful.

Dennis

On Tue, Apr 19, 2011 at 11:30 PM, vincent.deluard
vincent.delu...@trimtabs.com wrote:
 Hi r users,

 I am trying to compute the moving variance of a large matrix. I now use a
 loop but I am looking for a faster solution. Here is a sample of the code.

 Source= matrix(rnorm(400),ncol=100)
 variances= matrix(rep(NA,4*100),ncol=100)

 for (i in 1:80) {variances[,i]=apply(Source[,i:(i+80)],1,var)}

 any idea? Many thanks in advance.


 Vincent.

 --
 View this message in context: 
 http://r.789695.n4.nabble.com/Saving-run-time-in-loop-tp3462228p3462228.html
 Sent from the R help mailing list archive at Nabble.com.

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[R] BMA, logistic regression, odds ratio, model reduction etc

2011-04-20 Thread khosoda

Hi everybody,
I apologize for long mail in advance.

I have data of 104 patients, which consists of 15 explanatory variables
and one binary outcome (poor/good). The outcome consists of 25 poor
results and 79 good results. I tried to analyze the data with logistic
regression. However, the 15 variables and 25 events means events per
variable (EPV) is much less than 10 (rule of thumb). Therefore, I used R
package, BMA to perform logistic regression with BMA to avoid this
problem.

model 1 (full model):
x1, x2, x3, x4 are continuous variables and others are binary data.

 x16.bic.glm - bic.glm(outcome ~ ., data=x16.df,
glm.family=binomial, OR20, strict=FALSE)
 summary(x16.bic.glm)
(The output below has been cut off at the right edge to save space)

  62  models were selected
 Best  5  models (cumulative posterior probability =  0.3606 ):

 p!=0EV SDmodel 1model2
Intercept100-5.1348545  1.652424-4.4688  -5.15
-5.1536
age3.3   0.0001634  0.007258  .
sex4.0
   .M   -0.0243145  0.220314  .
side  10.8
.R   0.0811227  0.301233  .
procedure 46.9  -0.5356894  0.685148  .  -1.163
symptom3.8  -0.0099438  0.129690  .  .
stenosis   3.4  -0.0003343  0.005254  .
x13.7  -0.0061451  0.144084  .
x2   100.0   3.1707661  0.892034 3.2221 3.11
x351.3  -0.4577885  0.551466-0.9154 .
HT 4.6
  .positive  0.0199299  0.161769  .  .
DM 3.3
  .positive -0.0019986  0.105910  .  .
IHD3.5
   .positive 0.0077626  0.122593  .  .
smoking9.1
   .positive 0.0611779  0.258402  .  .
hyperlipidemia16.0
  .positive  0.1784293  0.512058  .  .
x4 8.2   0.0607398  0.267501  .  .


nVar   2  2
 1  3  3
BIC   -376.9082
-376.5588  -376.3094  -375.8468  -374.5582
post prob0.104
0.087  0.077  0.061  0.032

[Question 1]
Is it O.K to calculate odds ratio and its 95% confidence interval from
EV (posterior distribution mean) and“SD”(posterior distribution
standard deviation)?
For example, 95%CI of EV of x2 can be calculated as;
 exp(3.1707661)
[1] 23.82573 - odds ratio
 exp(3.1707661+1.96*0.892034)
[1] 136.8866
 exp(3.1707661-1.96*0.892034)
[1] 4.146976
-- 95%CI (4.1 to 136.9)
Is this O.K.?

[Question 2]
Is it permissible to delete variables with small value of p!=0 and
EV, such as age (3.3% and 0.0001634) to reduce the number of
explanatory variables and reconstruct new model without those variables
for new session of BMA?

model 2 (reduced model):
I used R package, pvclust, to reduce the model. The result suggested
x1, x2 and x4 belonged to the same cluster, so I picked up only x2.
Based on the subject knowledge, I made a simple unweighted sum, by
counting the number of clinical features. For 9 features (sex, side,
HT2, hyperlipidemia, DM, IHD, smoking, symptom, age), the sum ranges
from 0 to 9. This score was defined as ClinicalScore. Consequently, I
made up new data set (x6.df), which consists of 5 variables (stenosis,
x2, x3, procedure, and ClinicalScore) and one binary outcome
(poor/good). Then, for alternative BMA session...

 BMAx6.glm - bic.glm(postopDWI_HI ~ ., data=x6.df,
glm.family=binomial, OR=20, strict=FALSE)
 summary(BMAx6.glm)
(The output below has been cut off at the right edge to save space)
Call:
bic.glm.formula(f = postopDWI_HI ~ ., data = x6.df, glm.family =
binomial, strict = FALSE, OR = 20)


  13  models were selected
 Best  5  models (cumulative posterior probability =  0.7626 ):

p!=0EV SD   model 1model 2
Intercept   100-5.6918362  1.81220-4.4688-6.3166
stenosis  8.1  -0.0008417  0.00815  .  .
x2  100.0   3.0606165  0.87765 3.2221 3.1154
x3   46.5  -0.3998864  0.52688-0.9154  .
procedure   49.3   0.5747013  0.70164  . 1.1631
ClinicalScore   27.1   0.0966633  0.19645  .  .


nVar 2  2  1
 3  3
BIC -376.9082  -376.5588
-376.3094  -375.8468  -375.5025
post prob  0.208  0.175
0.154  0.122  0.103

[Question 3]
Am I doing it correctly or not?
I mean this kind of model

Re: [R] splom, plotmath: how to add three lines of information with alignment?

2011-04-20 Thread baptiste auguie

On 20 April 2011 19:23, Marius Hofert m_hof...@web.de wrote:
 Dear Baptiste,

 thank you for your help. I tried to built in your suggestions into the splom.
 Below is the result. I decided to create the plotmath-expressions outside the
 function splom2, this will allow me later to horizontally shift (via phantom,
 for example) the table entries so that they will be vertically aligned 
 according
 to the = signs. Although an array is allowed to contain expressions, I could
 not manage to create it properly. Do you know a solution?

 Cheers,

 Marius

 library(lattice)
 library(grid)
 library(gridExtra)

 ## splom with customized lower.panel
 ## x: data
 ## expr.arr: array of expressions [(i,j,) entry contains expressions which are
 ##           plotted in a grid table in the lower panel (i,j)]
 splom2 - function(x, expr.arr){
    ## function for creating table
    table.fun - function(vec){ # single values for one panel
        grid.table(vec,
                   parse=TRUE, # parse labels as expressions
                   theme=theme.list(
                   gpar.corefill=gpar(fill=NA, col=NA), # make bg transparent
                   core.just=left) # justification of labels
                   )
    }
    ## splom
    splom(x, varname.cex=1.4,
          superpanel=function(z, ...){
              panel.pairs(z, upper.panel=panel.splom, 
 lower.panel=function(i,j){
                  table.fun(expr.arr[i,j,])
              }, ...)
          })
 }

 ## create data and array of expressions
 d - 4
 x - matrix(runif(d*1000), ncol=d)
 expr.arr - array(, dim=c(d,d,3), dimnames=c(i,j,val)) # d x d x 3 
 elements
 for(i in 1:d){
    for(j in 1:d){
        # expr.arr[i,j,] - expression(italic(a)==0, italic(bbb)==0, 
 italic(c)==0) # does no work
        expr.arr[i,j,] - c(bquote(italic(a)==.(0)), 
 bquote(italic(bbb)==.(0)), bquote(italic(c)==.(0))) # same here
    }
 }

 ## plot
 splom2(x, expr.arr)


Initializing your array with a list seems to work for whatever reason
that's well above my head,

expr.arr - array(list(NA,NA,NA), dim=c(d,d,3), dimnames=c(i,j,val))

As for the alignment, there should be a better way using multiple
columns in grid.table, but parse is struggling to interpret the = sign
on its own. Again, I'm not sure why, this is deep magic to me.


library(gridExtra)
d = matrix(c(italic(a)==phantom(''), round(pi,4),
  italic(b)==phantom(), round(pi,6)), ncol=2, byrow=T)

grid.table(d, parse=T,theme=theme.list(
  gpar.corefill=gpar(fill=NA, col=NA),
  core.just=left, padding.h = unit(0, mm) ))

HTH,

baptiste


 On 2011-04-20, at 01:33 , baptiste auguie wrote:

 Hi,

 You may want to wait advice from someone who actually understands (the
 labyrinth that is) lattice's help for splom, but the following might
 be a start. I didn't understand what values you actually wanted
 displayed in the lower triangle panels, so I made up some random ones
 in a 3x3 matrix of 3-vectors.

 library(lattice)
 library(gridExtra)

 info - function(x){
  grid.table(c(bquote(italic(a)==.(x[1])),
               bquote(italic(b)==.(x[2])),
               bquote(italic(c)==.(x[3]))),
             core.just=left,
             parse=TRUE)
 }

 U - matrix(runif(3000), ncol=3)

 splom(U,
     superpanel=function(z, ...){
       ## dummy 3x3 matrix of 3 values to diplay
       values - replicate(9, round(rnorm(3), 3), simplify=FALSE)
       dummy - matrix(values, ncol=3, byrow=F)
       panel.pairs(z, upper.panel=panel.splom,
                   lower.panel=function(i, j, ...){
                     print(paste(i,j)) # current panel indices
                     info(dummy[i,j] [[1]]) # access the list elements
                   }, ...)
       })

 HTH,

 baptiste


 On 20 April 2011 10:17, Marius Hofert m_hof...@web.de wrote:
 Dear Baptiste,

 there is one tricky part left: how can I create a matrix with the 
 grid.table()
 objects as output? Is this possible? If not, maybe one can try to work with
 panel.splom (which can address single panels and thus call info() for each
 row-column index pair (i,j)), but I'm not sure if this will work.

 Cheers,

 Marius

 library(lattice)
 library(gridExtra)

 splom2 - function(x, a, b, c){
    ## function for the additional information
    info - function(a., b., c.){ # single values for one panel
        grid.table(c(bquote(italic(a)==.(a.)),
                     bquote(italic(b)==.(b.)),
                     bquote(italic(c)==.(c.))
                     ),
                   parse=TRUE, # parse labels as expressions
                   theme=theme.list(
                   gpar.corefill=gpar(fill=NA, col=NA), # make bg transparent
                   core.just=right) # justification of labels
                   )
    }
    labs - matrix(, nrow=ncol(x), ncol=ncol(x)) # should be a matrix of 
 grid.table() objects
    for(i in 1:ncol(x)) for(j in 1:ncol(x)) labs[i,j] - info(a.=a[i,j], 
 b.=b[i,j], c.=c[i,j])
    ## splom
    splom(x, superpanel=function(z,...){

Re: [R] Allocation of memory to a data frame




On 20.04.2011 09:46, santosh wrote:

Hello gRoup,

A while back I was advised that overallocating an object and filling
it up is better than rbind.

I am trying to optimize some code. I have the following object (empty
which I know in advance).
Is there an easy to create an empty data frame with 100,000 empty rows
based on the structure below?


dput(d)

structure(list(A = character(0), B = character(0), C = numeric(0),
 D = character(0), E = character(0), F = character(0), G =
character(0),
 H = structure(numeric(0), class = Date), I = numeric(0),
 J = integer(0)), .Names = c(A, B, C, D, E, F,
G, H, I, J), row.names = integer(0), class = data.frame)



Replace the zeros by your number of rows?

Uwe Ligges




I can use dataFrame in R.utils and do it explicitly but wondering if
there an easier way since I have the structure with me.

Thank you.

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Re: [R] Axes Alignment Problem for Multiple Plots

2011-04-20 Thread Barbaglia, Guido (ESA)

I've missed William's e-mail indeed, I've tested it and works great, thanks a 
lot!!!

Many Thanks
Guido

From: Matthieu Stigler [matthieu.stig...@gmail.com]
Sent: 20 April 2011 09:30
To: Barbaglia, Guido (ESA)
Cc: wdun...@tibco.com; r-help@r-project.org
Subject: Re: [R] Axes Alignment Problem for Multiple Plots

Guido

You missed William's e-mails, which solved the problem:
use

  yaxs=i

in second call. Also, William made the good point that you can rather
just use lines() in second call.

Good William!

Mat

Le 20/04/2011 08:00, Barbaglia, Guido (ESA) a écrit :
 Dear Matthieu,

 thanks for your clarification! Basically, what I need to do is to plot 
 different series on the same chart using different types of plot (lines, 
 barplot, ...) and it is fundamental that the various charts have the same 
 reference system. I hope that someone in the list is able to fix my problem!

 Best Regards
 Guido Barbaglia

 From: mat [matthieu.stig...@gmail.com]
 Sent: 19 April 2011 22:58
 To: John Kane
 Cc: r-help@r-project.org; Barbaglia, Guido (ESA)
 Subject: Re: [R] Axes Alignment Problem for Multiple Plots

 Ok, I can replicate your problem, with following code:

 dat- 1:10
  barplot(dat, beside=TRUE,ylim=c(0,100));
  par()$usr;
  par(new=T);
  plot(dat, ylim=c(0,100), type=l);
  par()$usr;

 So it looks like even if you specify yourself ylim, the resulting
 effective ylim (usr[3:4] ) will be different! More surprinsingly,
 setting the usr parameter before is not effective:
  par(new=T, usr=c(0,11,0,100));

 this will not preventpar()$usr; to be different than fixed :-(

 No idea, but hopefully someone else on the list will be able to provide
 advice!

 Matthieu

 Le 19. 04. 11 20:47, John Kane a écrit :
 What is Coredata(Z0)?
 It would be very useful. as the posting guidelines suggest to supply working 
 code and sample data.

 --- On Tue, 4/19/11, Barbaglia, Guido (ESA)guido.barbag...@fao.org   wrote:

 From: Barbaglia, Guido (ESA)guido.barbag...@fao.org
 Subject: [R] Axes Alignment Problem for Multiple Plots
 To: r-help@r-project.orgr-help@r-project.org
 Cc: Stigler, Matthieu (EST)matthieu.stig...@fao.org
 Received: Tuesday, April 19, 2011, 11:42 AM
 Dear all,

  I'm trying to plot, in the same window,
 two different series, using barplot() for the first one and
 plot() for the second. What happens is that the second chart
 has a different axes origin, therefore the final plot is
 wrong. This piece of code shows the differences between the
 values of par()$usr:

  barplot(coredata(Z0), beside=TRUE,
 ylim=c(0,100));
  par()$usr;
  par(new=T);
  plot(coredata(Z0), ylim=c(0,100));
  par()$usr;

 I would like to know how it is possible to edit the values
 of par()$usr[3:4] of the second chart in order to be the
 same of the first one or, alternatively, how can I plot
 together bar and line series within the same reference
 system.

 Best Regards
 Guido Barbaglia
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 PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
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 reproducible code.

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Re: [R] Error Installing or Updating Packages (Maybe because of a proxy)

If the internet connection from R works, can you please verify that you 
do not have any R base package from an old R version in a current R 
library that you may have in the .libPaths() already?


Uwe Ligges



On 20.04.2011 09:25, Majid Einian wrote:

Dear R Helpers,
(I am using Ubuntu lucid and R 2.13.0
When I try to update packages I get this error:


update.packages()

--- Please select a CRAN mirror for use in this session ---
Loading Tcl/Tk interface ... done
Error in ret[i, ]- c(pkgs[i], lib, desc) :
   number of items to replace is not a multiple of replacement length

I had no problem before (case 1) but now (case 2) I cannot get it to work,
googleing did not help:
case 1:
  * connecting directly without any proxy setting (at my university)
  * using R 2.12.2
case 2:
  * connecting through proxy setting (at my workplace)
  * using R 2.13.0

I set the proxy in terminal too but it does not help (echo $http_proxy gives
me http://192.168.0.1:8080/)



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Re: [R] How to check if a value of a variable is in a list




On 20.04.2011 05:00, Frederik Lang wrote:

Hi all,


I am working with some social network analysis in R and ran into a problem I
just cannot solve.

Each observation in my data consists of a respondent, some characteristics
and up to five friends. The problem is that all of these five friends might
no show up later as a respondent (observation). Therefore I might not have
characteristics on all the friends listed in the data and I want to restrict
my data to only those friends that I also have as respondents. The data
(without characteristics) look like this:

*resp  f1  f2  f3  f4  f5*
ID1   ID5   ID37   ID6ID2   ID53
ID2   ID1ID4ID17  NANA
...

Now, let's say that ID37 never appears as a respondent, then I want to
replace that value with a NA so that it looks like this:

*resp  f1  f2  f3  f4  f5*
ID1   ID5NAID6ID2   ID53
ID2   ID1ID4ID17  NANA


I thought I could check if for each entry, the value goes again in a list of
the respondents.

How do I do this?



See

?%in%

Uwe Ligges




Kind regards,


Frederik

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Re: [R] Loading a package shared lib works in 2.12 but not 2.11




On 13.04.2011 17:32, Von Der Hirschheydt, Juergen wrote:

Hi,

I have a problem with an external library on a previous R version.

We've created our own package containing a mixture of C++ as well as R
code which works fine under R 2.12.1. However, trying to install the
very same package ZIP file on R 2.11.1 will issue an error when loading
a library:



A zip file is a binary package that has been compiled for an almost 
fixed triple (R version, OS, hardware), where OS is Windows in 32-bit, 
64-bit or (exclusive) both of them in this case.


Therefore, you have to download a zip that was made for R-2.11.x which 
is available from yourCRANmirror/bin/windows/contrib/2.11 or more easily 
just via install.packages() that picks uop the correct repository. 
Alternatively, install from sources - and read the R Installation and 
Administration manual.


Uwe Ligges




 utils:::menuInstallLocal()
   package 'quasar' successfully unpacked and MD5 sums checked
 require(quasar)
   Loading required package: quasar
   Error in library.dynam(lib, package, package.lib) :
 shared library 'quasar' not found
   In addition: Warning message:
   package 'quasar' was built under R version 2.12.1

(quasar being the name of the package). Does anyone have some pointers
why it works under 2.12.1 but not 2.11.1 ? I've looked over the
changelog but couldn't find any clue there.

Has the mechanism for useDynLib(quasar) in the NAMESPACE file changed
?

Thanks for your time,
Cheers,

Juergen

===
Please access the attached hyperlink for an important el...{{dropped:4}}

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Re: [R] having trouble with R CMD INSTALL


Not sure what you are really doing, but

R --vanilla CMD INSTALL something

runs in a vanilla style. I just tested with a local .Rprofile under with 
R-2.13.0 under Windows.
Perhaps it is specific to the package your are talking about? Perhaps 
you can make it available for further inspection of this problem?


Uwe Ligges




On 19.04.2011 01:55, Gene Leynes wrote:

Hello,

I was having trouble passing in command line options when doing an package
install earlier.


From An Introduction in R



In addition, you can 
use25http://cran.r-project.org/doc/manuals/R-intro.html#fn-25options
--arch=, --no-environ, --no-init-file, --no-site-file and --vanillabetween
R and CMD: these affect any R processes run by the tools. (Here --vanillais 
equivalent to --no-environ
--no-site-file --no-init-file.)


R CMD cmd arg


When I tried doing this, two unexpected things happened:

1. --vanilla doesn't have an effect when used with R CMD  (R --vanilla
CMD INSTALL rj_0.5.2-1.tar.gz)
2. When loading my functions from Rprofile.site that seems to interfere
with the INSTALL command.  I temporarily deleted the Rprofile.site contents,
and the install worked as expected.  Maybe it's because I'm also parsing the
arguments after the --args flag?

The second item is probably user error on my part.
The first item seems like a possible R bug, but I'm not ruling out user
error.  See ?user.error for a sample error message


I finally got the package installed, but wanted to report this if it really
is a bug.



The lines below come from a DOS prompt:

c:\Users\gene.leynes\Downloadsdir /B
gdata_2.8.1.zip
rj_0.5.2-1.tar.gz
c:\Users\gene.leynes\DownloadsR --vanilla

R version 2.13.0 (2011-04-13)
Copyright (C) 2011 The R Foundation for Statistical Computing
ISBN 3-900051-07-0
Platform: i386-pc-mingw32/i386 (32-bit)

R is free software and comes with ABSOLUTELY NO WARRANTY.
You are welcome to redistribute it under certain conditions.
Type 'license()' or 'licence()' for distribution details.

   Natural language support but running in an English locale

R is a collaborative project with many contributors.
Type 'contributors()' for more information and
'citation()' on how to cite R or R packages in publications.

Type 'demo()' for some demos, 'help()' for on-line help, or
'help.start()' for an HTML browser interface to help.
Type 'q()' to quit R.


R.Version()

$platform
[1] i386-pc-mingw32

$arch
[1] i386

$os
[1] mingw32

$system
[1] i386, mingw32

$status
[1] 

$major
[1] 2

$minor
[1] 13.0

$year
[1] 2011

$month
[1] 04

$day
[1] 13

$`svn rev`
[1] 55427

$language
[1] R

$version.string
[1] R version 2.13.0 (2011-04-13)


q()


c:\Users\gene.leynes\DownloadsR --vanilla CMD INSTALL rj_0.5.2-1.tar.gz
Function_II.R :
Function_IM.R :
Function_NAsummary.R :
Function_YahooAdjCloseAsZoo.R :
Function_addbg.R :
Function_bgfun.R :
Function_boxplot2.R :
Function_clipdir.R :
Function_clipper.R :
Function_clipstar.R :
Function_comma.R :
Function_count.na.R :
Function_dftab.R :
Function_enclose.R :
Function_eye.R :
Function_factors.R :
Function_finder.R :
Function_headstr.R :
Function_inin.R :
Function_lll.R :
Function_loader.R :
Function_ma.R :
Function_makebm.R :
Function_makegm.R :
Function_makegroups.R :
Function_plot100colors.R :
Function_plotRunningSd.R :
Function_saver.R :
Function_sourcedir.R :
Function_stacker.R :
Function_subzoo.R :
Function_svr.R :
Function_tictoc.R :
Function_writexls.R :
Function_wtt_3.0.R :
heatplot.R :
loadfuns.R :
Opening: nextArgrj_0.5.2-1.tar.gz
Warning in file(file, r, encoding = encoding) :
   cannot open file 'nextArgrj_0.5.2-1.tar.gz': No such file or directory
Error in file(file, r, encoding = encoding) :
   cannot open the connection
Calls: source -  file
Execution halted

c:\Users\gene.leynes\Downloads

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[R] coercion

2011-04-20 Thread Stat Consult

Dear ALL

I don't know why I can see this error in run this sentences.

In paste(V, 1L:cols, sep = ) : NAs introduced by coercion



DATA-read.delim (D:\\DATA\\GeneExpression.txt,header=FALSE)

I will be glad if you help me.

Best Regards,
Stat Consult

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[R] odfWeave \ odfTable and encoding

2011-04-20 Thread kristell.desseaux

Dear All,

I usually use the package Sweave and Latex, but I want to explore the 
opportunities available with odfWeave. So I started exploring this new 
package.

I find myself hang while attempting to insert tables in the document 
odt. I built a data.frame to present the results of statistical 
analysis, more complex than a conventional contingency table.

Now when I run the command odfWeave, I get the following error that 
appears.

Post-processing the contents

Input is not proper UTF-8, indicate encoding !

Bytes: 0xE9 0x6D 0x69 0x6E

Erreur : 1: Input is not proper UTF-8, indicate encoding !

Bytes: 0xE9 0x6D 0x69 0x6E

It seems to be an encoding problem since I managed to turn the function 
odfWeave with another type of table less complex no accent, no % ...

Here is my current configuration, on Windows 7, SessionInfo ()

R version 2.10.1 (2009-12-14)

i386-pc-mingw32

locale:

[1] 
LC_COLLATE=French_France.1252LC_CTYPE=French_France.1252LC_MONETARY=French_France.1252
 
LC_NUMERIC=CLC_TIME=French_France.1252

attached base packages:

[1] gridsplinesstatsgraphicsgrDevices utilsdatasetsmethodsbase

other attached packages:

[1] MASS_7.3-4survival_2.35-7 
chron_2.3-38xtable_1.5-6gdata_2.8.0odfWeave_0.7.11 XML_3.2-0lattice_0.17-26

loaded via a namespace (and not attached):

[1] gtools_2.6.2 tools_2.10.1

I also found this function that allows me to know the encoding in place 
l10n_info R ().

$MBCS

[1] FALSE

$`UTF-8`

[1] FALSE

$`Latin-1`

[1] TRUE

$codepage

[1] 1252

I wanted to know if you knew how to change the encoding of reference in R?

Thank you very much.

Kristell

-- 
Kristell DESSEAUX

Biostatisticienne

Département de Biostatistique et Informatique Médicale
Hôpital Saint Louis, 1 avenue Claude Vellefaux
75475 PARIS Cedex 10

Tel : +33 (0)1 42 49 97 48
Sec : +33 (0)1 42 49 97 42
Fax : +33 (0)1 42 49 97 45


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[R] How to get R plots with FastRweb

2011-04-20 Thread MLSC MANIPAL

Dear friends,

I am working in a web service project which uses integration of Java with R.
I have used RJava to connect with Java and that is working fine. As R
produces more interactive plots, I would also like to pipe plots generate
from R on web page. I came to know that FastRWeb, R2HTML, brew and
WebGraphics, Cairo together can be used to do that. I have installed all
those successfully. I am developing my web service project with NetBeans.
For testing purpose I have used the excersie (kmeans) given in the FastRWeb
document.I have put the kmeans.png.R program in my netbeansProject/R
directory where other java code runs properly.

student@mlscubl30:~/NetBeansProjects/R$ ls
build  build.xml  dist  kmeans.png.R  nbproject  src  test  web
student@mlscubl30:~/NetBeansProjects/R$

When i try to execute it with http://localhost:8080/R/kmeans.png in browser
it does not execute it. Hence can you please tell me what exactly i have to
do in order to make it up and run?

thanks in advance.

Regards,
Devi
Manipal Life Sciences Center

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[R] user input

2011-04-20 Thread Ivan Calandra


Dear users,

I have looked on different sources and found different functions to 
prompt the user to provide input. However, I couldn't find one that does 
exactly what I'm looking for.


select.list() and menu() are nice because a graphic window appears to 
prompt the user. However, the user can only choose from a predefined 
list of choices. readline() and scan() are more free in the input but 
prompt the user in the console.


Examples:
a - select.list(choices=c(0.0,0.1,0.2), title=select number)
a## I cannot choose 0.15
[1] 0.1

b - readline(prompt=select number )
select number 0.2## in the console, not really visible
b
[1] 0.2


I would like both free input and a pop-up window. Is there a function 
for this?


Thanks in advance,
Ivan

--
Ivan CALANDRA
PhD Student
University of Hamburg
Biozentrum Grindel und Zoologisches Museum
Abt. Säugetiere
Martin-Luther-King-Platz 3
D-20146 Hamburg, GERMANY
+49(0)40 42838 6231
ivan.calan...@uni-hamburg.de

**
http://www.for771.uni-bonn.de
http://webapp5.rrz.uni-hamburg.de/mammals/eng/1525_8_1.php

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[R] grid.table + splom: how to nicely align panel entries

Dear expeRts,

is there a way to get the entries in each panel correctly aligned according to 
the 
equality signs?

Here is the wish-list:
(1) the equality signs in each panel should be vertically aligned  
(2) the numbers should be aligned on the decimal point

One could adjust the phantom()-arguments by hand to achieve (1), but is there a 
simpler solution? For (2) I have no idea.

Cheers,

Marius


library(lattice) 
library(grid)
library(gridExtra)

## splom with customized lower.panel
## x: data
## arr: array of containing expressions which are plotted in a grid table in 
the 
##  lower panel (i,j)]
splom2 - function(x, arr){
## function for creating table 
table.fun - function(vec){ # vector containing lines for table for *one* 
panel
grid.table(matrix(vec, ncol=2, byrow=TRUE),
   parse=TRUE, # parse labels as expressions
   theme=theme.list(
   gpar.corefill=gpar(fill=NA, col=NA), # make bg transparent
   core.just=left, padding.h=unit(0,mm)) # justification of 
labels
   ) 
}
## splom
splom(x, varname.cex=1.4,
  superpanel=function(z, ...){
  panel.pairs(z, upper.panel=panel.splom, lower.panel=function(i,j){
  table.fun(arr[i,j,])
  }, ...)
  })
}

## create data and array of expressions
d - 4
x - matrix(runif(d*1000), ncol=d) # data to be plotted with splom
arr - array(list(rep(NA, 3*2)), dim=c(d,d,3*2), dimnames=c(i,j,val)) # 
array containing the table entries per panel
f - function(i,j) (i+j)*10+0.1 # dummy function
for(i in 1:d){
for(j in 1:d){
arr[i,j,] - c(alpha==phantom(), round(pi,4),
   italic(bbb)==phantom(), round(pi,6),
   gamma==phantom(), f(i,j))
}
}

## plot
splom2(x, arr)

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Re: [R] splom, plotmath: how to add three lines of information with alignment?

Dear Baptiste, 

many thanks, that worked :-)

For the alignment, I made a new minimal example and posted it under the subject 
grid.table + splom: how to nicely align panel entries. 

Cheers,

Marius

On 2011-04-20, at 10:22 , baptiste auguie wrote:

 On 20 April 2011 19:23, Marius Hofert m_hof...@web.de wrote:
 Dear Baptiste,
 
 thank you for your help. I tried to built in your suggestions into the splom.
 Below is the result. I decided to create the plotmath-expressions outside the
 function splom2, this will allow me later to horizontally shift (via phantom,
 for example) the table entries so that they will be vertically aligned 
 according
 to the = signs. Although an array is allowed to contain expressions, I 
 could
 not manage to create it properly. Do you know a solution?
 
 Cheers,
 
 Marius
 
 library(lattice)
 library(grid)
 library(gridExtra)
 
 ## splom with customized lower.panel
 ## x: data
 ## expr.arr: array of expressions [(i,j,) entry contains expressions which 
 are
 ##   plotted in a grid table in the lower panel (i,j)]
 splom2 - function(x, expr.arr){
## function for creating table
table.fun - function(vec){ # single values for one panel
grid.table(vec,
   parse=TRUE, # parse labels as expressions
   theme=theme.list(
   gpar.corefill=gpar(fill=NA, col=NA), # make bg transparent
   core.just=left) # justification of labels
   )
}
## splom
splom(x, varname.cex=1.4,
  superpanel=function(z, ...){
  panel.pairs(z, upper.panel=panel.splom, 
 lower.panel=function(i,j){
  table.fun(expr.arr[i,j,])
  }, ...)
  })
 }
 
 ## create data and array of expressions
 d - 4
 x - matrix(runif(d*1000), ncol=d)
 expr.arr - array(, dim=c(d,d,3), dimnames=c(i,j,val)) # d x d x 3 
 elements
 for(i in 1:d){
for(j in 1:d){
# expr.arr[i,j,] - expression(italic(a)==0, italic(bbb)==0, 
 italic(c)==0) # does no work
expr.arr[i,j,] - c(bquote(italic(a)==.(0)), 
 bquote(italic(bbb)==.(0)), bquote(italic(c)==.(0))) # same here
}
 }
 
 ## plot
 splom2(x, expr.arr)
 
 
 Initializing your array with a list seems to work for whatever reason
 that's well above my head,
 
 expr.arr - array(list(NA,NA,NA), dim=c(d,d,3), dimnames=c(i,j,val))
 
 As for the alignment, there should be a better way using multiple
 columns in grid.table, but parse is struggling to interpret the = sign
 on its own. Again, I'm not sure why, this is deep magic to me.
 
 
 library(gridExtra)
 d = matrix(c(italic(a)==phantom(''), round(pi,4),
  italic(b)==phantom(), round(pi,6)), ncol=2, byrow=T)
 
 grid.table(d, parse=T,theme=theme.list(
  gpar.corefill=gpar(fill=NA, col=NA),
  core.just=left, padding.h = unit(0, mm) ))
 
 HTH,
 
 baptiste
 
 
 On 2011-04-20, at 01:33 , baptiste auguie wrote:
 
 Hi,
 
 You may want to wait advice from someone who actually understands (the
 labyrinth that is) lattice's help for splom, but the following might
 be a start. I didn't understand what values you actually wanted
 displayed in the lower triangle panels, so I made up some random ones
 in a 3x3 matrix of 3-vectors.
 
 library(lattice)
 library(gridExtra)
 
 info - function(x){
  grid.table(c(bquote(italic(a)==.(x[1])),
   bquote(italic(b)==.(x[2])),
   bquote(italic(c)==.(x[3]))),
 core.just=left,
 parse=TRUE)
 }
 
 U - matrix(runif(3000), ncol=3)
 
 splom(U,
 superpanel=function(z, ...){
   ## dummy 3x3 matrix of 3 values to diplay
   values - replicate(9, round(rnorm(3), 3), simplify=FALSE)
   dummy - matrix(values, ncol=3, byrow=F)
   panel.pairs(z, upper.panel=panel.splom,
   lower.panel=function(i, j, ...){
 print(paste(i,j)) # current panel indices
 info(dummy[i,j] [[1]]) # access the list elements
   }, ...)
   })
 
 HTH,
 
 baptiste
 
 
 On 20 April 2011 10:17, Marius Hofert m_hof...@web.de wrote:
 Dear Baptiste,
 
 there is one tricky part left: how can I create a matrix with the 
 grid.table()
 objects as output? Is this possible? If not, maybe one can try to work with
 panel.splom (which can address single panels and thus call info() for each
 row-column index pair (i,j)), but I'm not sure if this will work.
 
 Cheers,
 
 Marius
 
 library(lattice)
 library(gridExtra)
 
 splom2 - function(x, a, b, c){
## function for the additional information
info - function(a., b., c.){ # single values for one panel
grid.table(c(bquote(italic(a)==.(a.)),
 bquote(italic(b)==.(b.)),
 bquote(italic(c)==.(c.))
 ),
   parse=TRUE, # parse labels as expressions
   theme=theme.list(
   gpar.corefill=gpar(fill=NA, col=NA), # make bg 
 transparent
   core.just=right) #

Re: [R] coercion


See the body of all messages to R-help:

PLEASE do read the posting guide 
http://www.R-project.org/posting-guide.html and provide commented, 
minimal, self-contained, reproducible code.


and we do not know what D:\\DATA\\GeneExpression.txt is.

Uwe Ligges




On 20.04.2011 08:56, Stat Consult wrote:

Dear ALL

I don't know why I can see this error in run this sentences.

In paste(V, 1L:cols, sep = ) : NAs introduced by coercion



DATA-read.delim (D:\\DATA\\GeneExpression.txt,header=FALSE)

I will be glad if you help me.

Best Regards,
Stat Consult

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Re: [R] user input

2011-04-20 Thread Ivan Calandra


Thanks for your answer.
But I don't understand anything, I would say it's far beyond my current 
knowledge.

Could you please explain me in more details how this works?

Ivan

Le 4/20/2011 11:40, Juan Carlos Borrás a écrit :

An alternative is to run your R programs as scripts (#!/usr/bin/env
Rscript) and pass user input as command line parameters.
After all you can already set runtime params if you're developing with
you IDE of choice, right?

Cheers,
jcb!
___
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On Wed, Apr 20, 2011 at 12:00 PM, Ivan Calandra
ivan.calan...@uni-hamburg.de  wrote:

Dear users,

I have looked on different sources and found different functions to prompt
the user to provide input. However, I couldn't find one that does exactly
what I'm looking for.


--
Ivan CALANDRA
PhD Student
University of Hamburg
Biozentrum Grindel und Zoologisches Museum
Abt. Säugetiere
Martin-Luther-King-Platz 3
D-20146 Hamburg, GERMANY
+49(0)40 42838 6231
ivan.calan...@uni-hamburg.de

**
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Re: [R] user input

2011-04-20 Thread Andreas Borg


Hi Ivan,

there are also data.entry() and edit(), which are more tailored to data 
frames and might look awkward for entering a single value. You could 
also take a look at CRAN, maybe there is a GUI package that does what 
you want.


Andreas

Ivan Calandra schrieb:

Dear users,

I have looked on different sources and found different functions to 
prompt the user to provide input. However, I couldn't find one that 
does exactly what I'm looking for.


select.list() and menu() are nice because a graphic window appears to 
prompt the user. However, the user can only choose from a predefined 
list of choices. readline() and scan() are more free in the input but 
prompt the user in the console.


Examples:
a - select.list(choices=c(0.0,0.1,0.2), title=select number)
a## I cannot choose 0.15
[1] 0.1

b - readline(prompt=select number )
select number 0.2## in the console, not really visible
b
[1] 0.2


I would like both free input and a pop-up window. Is there a 
function for this?


Thanks in advance,
Ivan




--
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Medizinische Informatik

UNIVERSITÄTSMEDIZIN
der Johannes Gutenberg-Universität
Institut für Medizinische Biometrie, Epidemiologie und Informatik
Obere Zahlbacher Straße 69, 55131 Mainz
www.imbei.uni-mainz.de

Telefon +49 (0) 6131 175062
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Re: [R] grid.table + splom: how to nicely align panel entries

2011-04-20 Thread baptiste auguie

On 20 April 2011 21:16, Marius Hofert m_hof...@web.de wrote:
 Dear expeRts,

 is there a way to get the entries in each panel correctly aligned according 
 to the
 equality signs?

 Here is the wish-list:
 (1) the equality signs in each panel should be vertically aligned

You can put the equal signs in their own column,

library(gridExtra)
d = matrix(c(italic(a), phantom()==phantom(), round(pi,4),
italic(b), phantom()==phantom(), round(pi,6)), ncol=3, byrow=T)
grid.table(d, parse=T,theme=theme.list(core.just=left))

 (2) the numbers should be aligned on the decimal point

You could place some phantom()s to do this,

align.digits = function(l)
{

sp - strsplit(as.character(l), \\.)
chars - sapply(sp, function(x) nchar(x)[1])
n = max(chars) - chars
l0 = sapply(n, function(x) paste(rep(0, x), collapse=))
labels = sapply(seq_along(sp), function(i) {
  as.expression(bquote(phantom(.(l0[i])) * .(sp[[i]][1])*.*.(sp[[i]][2])))})

return(labels)
}

library(gridExtra)

d - align.digits(l = c(125.3, 1.2344))
grid.table(d, parse=T,core.just=left)

HTH,

baptiste

 One could adjust the phantom()-arguments by hand to achieve (1), but is there 
 a
 simpler solution? For (2) I have no idea.

 Cheers,

 Marius


 library(lattice)
 library(grid)
 library(gridExtra)

 ## splom with customized lower.panel
 ## x: data
 ## arr: array of containing expressions which are plotted in a grid table in 
 the
 ##      lower panel (i,j)]
 splom2 - function(x, arr){
    ## function for creating table
    table.fun - function(vec){ # vector containing lines for table for *one* 
 panel
        grid.table(matrix(vec, ncol=2, byrow=TRUE),
                   parse=TRUE, # parse labels as expressions
                   theme=theme.list(
                   gpar.corefill=gpar(fill=NA, col=NA), # make bg transparent
                   core.just=left, padding.h=unit(0,mm)) # justification 
 of labels
                   )
    }
    ## splom
    splom(x, varname.cex=1.4,
          superpanel=function(z, ...){
              panel.pairs(z, upper.panel=panel.splom, 
 lower.panel=function(i,j){
                  table.fun(arr[i,j,])
              }, ...)
          })
 }

 ## create data and array of expressions
 d - 4
 x - matrix(runif(d*1000), ncol=d) # data to be plotted with splom
 arr - array(list(rep(NA, 3*2)), dim=c(d,d,3*2), dimnames=c(i,j,val)) # 
 array containing the table entries per panel
 f - function(i,j) (i+j)*10+0.1 # dummy function
 for(i in 1:d){
    for(j in 1:d){
        arr[i,j,] - c(alpha==phantom(), round(pi,4),
                       italic(bbb)==phantom(), round(pi,6),
                       gamma==phantom(), f(i,j))
    }
 }

 ## plot
 splom2(x, arr)

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Re: [R] coercion

2011-04-20 Thread Petr Savicky

On Wed, Apr 20, 2011 at 11:26:19AM +0430, Stat Consult wrote:
 Dear ALL
 
 I don't know why I can see this error in run this sentences.
 
 In paste(V, 1L:cols, sep = ) : NAs introduced by coercion

Check, what cols is. This warning may be obtained, if cols
is a character value. For example

  cols - a
  paste(V, 1L:cols, sep = )

  Error in 1L:cols : NA/NaN argument
  In addition: Warning message:
  In paste(V, 1L:cols, sep = ) : NAs introduced by coercion

 DATA-read.delim (D:\\DATA\\GeneExpression.txt,header=FALSE)

What is the relationship of this to the command above?

If you include a reproducible example, you have a better chance
to get a useful answer.

Petr Savicky.

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Re: [R] odfWeave \ odfTable and encoding

2011-04-20 Thread Wincent

This has been a long-lasting issue without good solution for windows
because windows does not support UTF-8 encoding.

If you are using Linux, you can set UTF-8 locale through Sys.setlocale.

Regards

On 20 April 2011 16:51, kristell.desseaux
kristell.desse...@univ-paris-diderot.fr wrote:
 Dear All,

 I usually use the package Sweave and Latex, but I want to explore the
 opportunities available with odfWeave. So I started exploring this new
 package.

 I find myself hang while attempting to insert tables in the document
 odt. I built a data.frame to present the results of statistical
 analysis, more complex than a conventional contingency table.

 Now when I run the command odfWeave, I get the following error that
 appears.

 Post-processing the contents

 Input is not proper UTF-8, indicate encoding !

 Bytes: 0xE9 0x6D 0x69 0x6E

 Erreur : 1: Input is not proper UTF-8, indicate encoding !

 Bytes: 0xE9 0x6D 0x69 0x6E

 It seems to be an encoding problem since I managed to turn the function
 odfWeave with another type of table less complex no accent, no % ...

 Here is my current configuration, on Windows 7, SessionInfo ()

 R version 2.10.1 (2009-12-14)

 i386-pc-mingw32

 locale:

 [1]
 LC_COLLATE=French_France.1252LC_CTYPE=French_France.1252LC_MONETARY=French_France.1252
 LC_NUMERIC=CLC_TIME=French_France.1252

 attached base packages:

 [1] gridsplinesstatsgraphicsgrDevices utilsdatasetsmethodsbase

 other attached packages:

 [1] MASS_7.3-4survival_2.35-7
 chron_2.3-38xtable_1.5-6gdata_2.8.0odfWeave_0.7.11 XML_3.2-0lattice_0.17-26

 loaded via a namespace (and not attached):

 [1] gtools_2.6.2 tools_2.10.1

 I also found this function that allows me to know the encoding in place
 l10n_info R ().

 $MBCS

 [1] FALSE

 $`UTF-8`

 [1] FALSE

 $`Latin-1`

 [1] TRUE

 $codepage

 [1] 1252

 I wanted to know if you knew how to change the encoding of reference in R?

 Thank you very much.

 Kristell

 --
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 Biostatisticienne

 Département de Biostatistique et Informatique Médicale
 Hôpital Saint Louis, 1 avenue Claude Vellefaux
 75475 PARIS Cedex 10

 Tel : +33 (0)1 42 49 97 48
 Sec : +33 (0)1 42 49 97 42
 Fax : +33 (0)1 42 49 97 45


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-- 
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Re: [R] Whole genome searching of 100bp D sequence

2011-04-20 Thread zoolium

Thanks, I've done so.

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[R] Fibonacci

2011-04-20 Thread Georgina Imberger

Hi!

I am trying to work out the code to get a Fibonacci sequence, using the
while() loop and only one variable. And I can't figure it out.

Fibonacci-c(1,1)
while (max(Fibonacci)500){
Fibonacci-c(Fibonacci, (max(Fibonacci) + ?(Fibanacci)))
}


How can I tell R to take the value one before the max value? (Without
defining another variable)

(Probably super easy... I am a beginner...)

Thanks,
Georgie

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[R] Integrate na.rm in own defined functions

2011-04-20 Thread Mauro

It`s probably an easy question, but couldn`t figure it out.

I`ve defined a function like: 

rmse-function (x){
dquared-x^2
sum1-sum(x^2)
rmse-sqrt((1/length(x))*sum1)
rmse}

My problem is, that I have NA Values in x and the above function returns NA.

I`m looking for a way to use na.rm=TRUE like in for instance
mean(x,na.rm=TRUE).

Does anybody know how to do this?


Thank you very much.

Mauro

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[R] 'Record' row values every time the binary value in a collumn changes

2011-04-20 Thread baboon2010

My question is twofold.

Part 1:
My data looks like this:

(example set, real data has 2*10^6 rows)
binary-c(1,1,1,0,0,0,1,1,1,0,0)
Chromosome-c(1,1,1,1,1,1,2,2,2,2,2)
start-c(12,17,18,20,25,36,12,15,16,17,19)
Table-cbind(Chromosome,start,binary)
  Chromosome start binary
 [1,]  112  1
 [2,]  117  1
 [3,]  118  1
 [4,]  120  0
 [5,]  125  0
 [6,]  136  0
 [7,]  212  1
 [8,]  215  1
 [9,]  216  1
[10,]  217  0
[11,]  219  0

As output I need a shortlist for each binary block: giving me the starting
and ending position of each block.
Which for these example would look like this:
 Chromosome2 position_start position_end binary2
[1,]   1 12   18   1
[2,]   1 20   36   0
[3,]   2 12   16   1
[4,]   2 17   19   0

Part 2:
Based on the output of part 1, I need to assign the binary to rows of
another data set. If the position value in this second data set falls in one
of the blocks defined in the shortlist made in part1,the binary value of the
shortlist should be assigned to an extra column for this row.  This would
look something like this:
 Chromosome3 position Value binary3
 [1,] 1 12 a   1
 [2,] 1 13 b   1
 [3,] 1 14 c   1
 [4,] 1 15 d   1
 [5,] 1 16 e   1
 [6,] 1 18 f   1
 [7,] 1 20 g   0
 [8,] 1 21 h   0
 [9,] 1 22 i   0
[10,] 1 23 j   0
[11,] 1 25 k   0
[12,] 1 35 l   0
[13,] 2 12 m   1
[14,] 2 13 n   1
[15,] 2 14 o   1
[16,] 2 15 p   1
[17,] 2 16 q   1
[18,] 2 17 s   0
[19,] 2 18 d   0
[20,] 2 19 f   0


Many thanks in advance,

Niels

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[R] make a shortlist of data based on blocks of binary values in one column

2011-04-20 Thread baboon2010

My question is twofold.

Part 1:
My data looks like this:

(example set, real data has 2*10^6 rows)
binary-c(1,1,1,0,0,0,1,1,1,0,0)
Chromosome-c(1,1,1,1,1,1,2,2,2,2,2)
start-c(12,17,18,20,25,36,12,15,16,17,19)
Table-cbind(Chromosome,start,binary)
  Chromosome start binary
 [1,]  112  1
 [2,]  117  1
 [3,]  118  1
 [4,]  120  0
 [5,]  125  0
 [6,]  136  0
 [7,]  212  1
 [8,]  215  1
 [9,]  216  1
[10,]  217  0
[11,]  219  0

As output I need a shortlist for each binary block: giving me the starting
and ending position of each block.
Which for these example would look like this:
 Chromosome2 position_start position_end binary2
[1,]   1 12   18   1
[2,]   1 20   36   0
[3,]   2 12   16   1
[4,]   2 17   19   0

Part 2:
Based on the output of part 1, I need to assign the binary to rows of
another data set. If the position value in this second data set falls in one
of the blocks defined in the shortlist made in part1,the binary value of the
shortlist should be assigned to an extra column for this row.  This would
look something like this:
 Chromosome3 position Value binary3
 [1,] 1 12 a   1
 [2,] 1 13 b   1
 [3,] 1 14 c   1
 [4,] 1 15 d   1
 [5,] 1 16 e   1
 [6,] 1 18 f   1
 [7,] 1 20 g   0
 [8,] 1 21 h   0
 [9,] 1 22 i   0
[10,] 1 23 j   0
[11,] 1 25 k   0
[12,] 1 35 l   0
[13,] 2 12 m   1
[14,] 2 13 n   1
[15,] 2 14 o   1
[16,] 2 15 p   1
[17,] 2 16 q   1
[18,] 2 17 s   0
[19,] 2 18 d   0
[20,] 2 19 f   0

Many thanks in advance,

Niels

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Re: [R] Yearly aggregates and matrices

2011-04-20 Thread mathijsdevaan

As a follow up on this post, I am trying to slightly adjust the solution
kindly provided by Gabor. However, I am getting some results that I do not
understand. Example:

# devel version of zoo 
install.packages(zoo, repos = http://r-forge.r-project.org;) 
library(zoo)

DF1 = data.frame(read.table(textConnection(B  C  D  E  F  G
8025  1995  0  4  1  2
8025  1997  1  1  3  4
8026  1995  0  7  0  0
8026  1996  1  2  3  0
8026  1997  1  2  3  1
8026  1998  6  0  0  4
8026  1999  3  7  0  3
8027  1997  1  2  3  9
8027  1998  1  2  3  1
8027  1999  6  0  0  2
8028  1999  3  7  0  0
8029  1995  0  2  3  3
8029  1998  1  2  3  2
8029  1999  6  0  0  1),head=TRUE,stringsAsFactors=FALSE))

a - read.zoo(DF1, split = 1, index = 2, FUN = identity)
sum.na - function(x) if (any(!is.na(x))) sum(x, na.rm = TRUE) else NA
b - rollapply(a, 3,  sum.na, align = right, partial = TRUE)
newDF - lapply(1:nrow(b), function(i)
   prop.table(na.omit(matrix(b[i,], nc = 4, byrow = TRUE,
   dimnames = list(unique(DF1$B), names(DF1)[-1:-2]))), 1))
names(newDF) - time(a)
c-lapply(newDF, function(mat) tcrossprod(mat / sqrt(rowSums(mat^2

Now I would like the elements e in c to be equal to 1-e. However,

c-lapply(newDF, function(mat) 1 - tcrossprod(mat / sqrt(rowSums(mat^2

gives a value  of 2.220446e-16 for as.data.frame(c['1999'])[2,2] instead of
0

What am I doing wrong here? Thanks a lot!


 First we use read.zoo to reform DF into a multivariate time series and 
 use rollapply (where we have used the devel version of zoo since it 
 supports the partial= argument on rollapply).  We then reform each 
 resulting row into a matrix converting each row of each matrix to 
 proportions.  Finally we form the desired scaled cross product. 
 
 # devel version of zoo 
 install.packages(zoo, repos = http://r-forge.r-project.org;) 
 library(zoo) 
 
 z - read.zoo(DF, split = 2, index = 3, FUN = identity) 
 
 sum.na - function(x) if (any(!is.na(x))) sum(x, na.rm = TRUE) else NA 
 r - rollapply(z, 3,  sum.na, align = right, partial = TRUE) 
 
 newDF - lapply(1:nrow(r), function(i) 
prop.table(na.omit(matrix(r[i,], nc = 4, byrow = TRUE, 
dimnames = list(unique(DF$B), names(DF)[-2:-3]))[, -1]),
 1)) 
 names(newDF) - time(z) 
 
 lapply(newDF, function(mat) tcrossprod(mat / sqrt(rowSums(mat^2



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[R] How to replace 'star (*)' with blank space?

2011-04-20 Thread Ruby Chu

Hi There

I'm not very sure how to replace the stars in a character vector

For example:

a character vector (n rows by 1 col)

[1] -27  -21  -25-28  *  -29
[2] -27   -28  *  -29
.
.
.
.
.
.
[n] -1***

I wish to replace all the *s with a blank, result as per below

[1] -27  -21  -25-28   -29
[2] -27 -28   -29
.
.
.
.
.
.
[n] -1

I have tried to use gsub( )
like: gsub(*, , x)
but doesn't seem to be working...please advise

Thank you

Ruby

[[alternative HTML version deleted]]

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Re: [R] Integrate na.rm in own defined functions

2011-04-20 Thread vioravis

This should work!!

rmse-function (x){ 
dquared-x^2 
sum1-sum(x^2,na.rm=TRUE) 
rmse-sqrt((1/length(x))*sum1) 
rmse}



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Re: [R] Fibonacci

2011-04-20 Thread Erich Neuwirth

The easy solution to compute the Fibonacci numbers is
 
fibo - function(n,a=1,b=1){
 if (n == 1) return(a)
 if (n == 2) return(b)
 return(fibo(n-1,b,a+b)) 
}

It avoids double recursion.
It is, however, not as resource efficient as a loop since R does not do
tail recursion elimination.




On Apr 20, 2011, at 11:42 AM, Georgina Imberger wrote:

 Hi!
 
 I am trying to work out the code to get a Fibonacci sequence, using the
 while() loop and only one variable. And I can't figure it out.
 
 Fibonacci-c(1,1)
 while (max(Fibonacci)500){
 Fibonacci-c(Fibonacci, (max(Fibonacci) + ?(Fibanacci)))
 }
 
 
 How can I tell R to take the value one before the max value? (Without
 defining another variable)
 
 (Probably super easy... I am a beginner...)
 
 Thanks,
 Georgie
 
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Re: [R] Fibonacci

2011-04-20 Thread Petr Savicky

On Wed, Apr 20, 2011 at 11:42:38AM +0200, Georgina Imberger wrote:
 Hi!
 
 I am trying to work out the code to get a Fibonacci sequence, using the
 while() loop and only one variable. And I can't figure it out.
 
 Fibonacci-c(1,1)
 while (max(Fibonacci)500){
 Fibonacci-c(Fibonacci, (max(Fibonacci) + ?(Fibanacci)))
 }
 
 
 How can I tell R to take the value one before the max value? (Without
 defining another variable)

Is it allowed to use length() function? If so, then try
the following

  Fibonacci-c(1,1)
  while (max(Fibonacci)500){
  Fibonacci-c(Fibonacci, Fibonacci[length(Fibonacci) - 1] + 
Fibonacci[length(Fibonacci)])
  }

Petr Savicky.

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Re: [R] cut histogram

2011-04-20 Thread Jim Lemon


On 04/20/2011 12:32 AM, Santosh wrote:

Dear Rxperts,

Below is a small sample of values (cut short due to space considerations
while posting).. I was wondering if it is possible to construct boundaries
(or intervals) based on the distribution of points.  Is it anything similar
to boundary detection of distributions?
x1-
c(0.00,0.25,0.50,1.00,2.00,3.00,4.00,5.00,7.03,10.00,14.01,21.00,28.02,14.00,28.00,7.00,29.00,7.01,10.02,6.97,10.06,21.03,28.01,14.10,28.14,
10.03,7.05,35.00,21.01,10.05,14.02,21.07,7.02,14.12,10.04,0.02,0.06,0.10,0.27,0.52,1.02,2.02,3.02,4.02,5.02,21.02,21.10,0.11,28.04,28.12,
27.98,14.04,21.06,14.05,28.03,8.00,13.99,27.99,21.04,5.28,28.11,28.05,13.98,2.01,5.99,15.02,28.06,3.07,1.01,6.04,6.01,7.99,13.96,0.99,3.01,
20.98,34.99,13.97,6.14,7.07,28.10,2.99,6.00,6.99,14.03,34.98,4.01,4.99,20.99,27.97,0.98,6.07,28.09,20.97,0.93,1.93,5.01,6.96,20.94,0.92,
20.95,8.07,14.07,27.95,34.95,2.97,4.96,20.96,27.96,34.96,2.96,33.99,34.00,6.72,27.71,6.91,25.91,26.02,7.14,26.16,26.19,6.94,28.95,28.98,6.82,
27.85,6.85,27.88,6.86,28.88,6.93,27.91,11.92,20.92,7.94,28.92,9.76,32.81,10.94,31.92,6.95,29.94,29.96,8.94,28.94,6.89,28.96,9.69,30.79,30.80,
6.90,29.91,9.93,30.95,6.88,7.89,31.97,31.98,3.92,7.91,31.72,6.83,29.88,29.80,29.81,27.90,32.94,28.93,28.97,7.83,28.76,28.77,6.98,34.01,10.97)

hist(x1,brea=200), # arbitrarily chosen high number of breakpoints

Would highly appreciate your thoughts/suggestions...


Hi Santosh,
Have a look at the classInt package.

Jim

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Re: [R] Integrate na.rm in own defined functions

2011-04-20 Thread Alexander Engelhardt


Am 20.04.2011 10:59, schrieb Mauro:

It`s probably an easy question, but couldn`t figure it out.

I`ve defined a function like:

rmse-function (x){
dquared-x^2
sum1-sum(x^2)
rmse-sqrt((1/length(x))*sum1)
rmse}

My problem is, that I have NA Values in x and the above function returns NA.

I`m looking for a way to use na.rm=TRUE like in for instance
mean(x,na.rm=TRUE).


If you had no sum function to give the na.rm=TRUE parameter, you can 
add the tiny line to your code:

x - x[!is.na(x)]

Or give your function a na.rm parameter yourself:

rmse-function (x, na.rm=FALSE){

if(na.rm==TRUE){
 x - x[!is.na(x)]   
}

 dquared-x^2
 sum1-sum(x^2)
 rmse-sqrt((1/length(x))*sum1)
 rmse
}

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Re: [R] How to replace 'star (*)' with blank space?


gsub(\\*,  ,x)

should do. Please read about regular expression as used by gsub().

Uwe Ligges


On 20.04.2011 12:00, Ruby Chu wrote:

Hi There

I'm not very sure how to replace the stars in a character vector

For example:

a character vector (n rows by 1 col)

[1] -27  -21  -25-28  *  -29
[2] -27   -28  *  -29
.
.
.
.
.
.
[n] -1***

I wish to replace all the *s with a blank, result as per below

[1] -27  -21  -25-28   -29
[2] -27 -28   -29
.
.
.
.
.
.
[n] -1

I have tried to use gsub( )
like: gsub(*, , x)
but doesn't seem to be working...please advise

Thank you

Ruby

[[alternative HTML version deleted]]

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and provide commented, minimal, self-contained, reproducible code.


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Re: [R] How to replace 'star (*)' with blank space?



On Apr 20, 2011, at 6:00 AM, Ruby Chu wrote:


Hi There

I'm not very sure how to replace the stars in a character vector

For example:

a character vector (n rows by 1 col)

[1] -27  -21  -25-28  *  -29
[2] -27   -28  *  -29
.
.
.
.
.
.
[n] -1***

I wish to replace all the *s with a blank, result as per below

[1] -27  -21  -25-28   -29
[2] -27 -28   -29
.
.
.
.
.
.
[n] -1

I have tried to use gsub( )
like: gsub(*, , x)


* is an operator in regex, so you need to double escape it in the  
pattern argument (but may not need to escape it in the replace argument.


 gsub(\\*, , test)
[1] test

 gsub(\\., *, test......)
[1] test******


but doesn't seem to be working...please advise

Thank you

Ruby

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David Winsemius, MD
West Hartford, CT

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[R] S: expert mailing list for general statistical questions

2011-04-20 Thread Sascha Vieweg


Hello R users and experts

Once in a while I have got questions that are not so much 
R-related but related to (social scientific) statistics in 
general. R-help would be the wrong list for such posts, and I am 
looking for a similar mailing list or newsgroup (nntp). (It is 
just a personal taste that I don't like web forums.) I have 
googled and found a variety here: 
http://mathstore.ac.uk/?q=node/1447 -- e.g. the list stats-discuss 
(which failed to load), and the newsgroups sci.stat.consult and 
sci.stat.edu.


However, before stumbling around, could you give a good 
recommendation on a mailing list or newsgroup appropriate for such 
issues?


Thanks for your tips.

Regards, *S*

--
Sascha Vieweg, saschav...@gmail.com

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Re: [R] S: expert mailing list for general statistical questions



On Apr 20, 2011, at 7:44 AM, Sascha Vieweg wrote:


Hello R users and experts

Once in a while I have got questions that are not so much R-related  
but related to (social scientific) statistics in general.

snipped


However, before stumbling around, could you give a good  
recommendation on a mailing list or newsgroup appropriate for such  
issues?


http://stats.stackexchange.com/

David Winsemius, MD
West Hartford, CT

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Re: [R] user input

2011-04-20 Thread Peter Ehlers


On 2011-04-20 02:51, Andreas Borg wrote:

Hi Ivan,

there are also data.entry() and edit(), which are more tailored to data
frames and might look awkward for entering a single value. You could
also take a look at CRAN, maybe there is a GUI package that does what
you want.



Ivan,
have a look at the rpanel package. Might give you some ideas.

Peter Ehlers


Andreas

Ivan Calandra schrieb:

Dear users,

I have looked on different sources and found different functions to
prompt the user to provide input. However, I couldn't find one that
does exactly what I'm looking for.

select.list() and menu() are nice because a graphic window appears to
prompt the user. However, the user can only choose from a predefined
list of choices. readline() and scan() are more free in the input but
prompt the user in the console.

Examples:
a- select.list(choices=c(0.0,0.1,0.2), title=select number)
a## I cannot choose 0.15
[1] 0.1

b- readline(prompt=select number )
select number 0.2## in the console, not really visible
b
[1] 0.2


I would like both free input and a pop-up window. Is there a
function for this?

Thanks in advance,
Ivan






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Re: [R] Fibonacci

2011-04-20 Thread Nutter, Benjamin

Fibonacci - c(1, 1)
while (max (Fibonacci)  500){
  Fibonacci - c(Fibonacci, sum(tail(Fibonacci, 2))) } 

-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org]
On Behalf Of Georgina Imberger
Sent: Wednesday, April 20, 2011 5:43 AM
To: r-help@r-project.org
Subject: [R] Fibonacci

Hi!

I am trying to work out the code to get a Fibonacci sequence, using the
while() loop and only one variable. And I can't figure it out.

Fibonacci-c(1,1)
while (max(Fibonacci)500){
Fibonacci-c(Fibonacci, (max(Fibonacci) + ?(Fibanacci))) }

How can I tell R to take the value one before the max value? (Without
defining another variable)

(Probably super easy... I am a beginner...)

Thanks,
Georgie

[[alternative HTML version deleted]]

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[R] Extrapolating data points for individuals who lack them

2011-04-20 Thread Ellis, David

Hi,

We have an experiment where individuals responses were measured over 5 days. 
Some responses were not obtained because we only allowed individuals to respond 
within a limited time-frame. These individuals are given the maximum response 
time as they did not respond, yet we feel they may have done if given time (and 
by looking at the rest of their responses over time, the non-response days 
stand out).

We therefore want to extrapolate data points for individuals, on days when they 
didn't respond, using a regression of days when they did.

Does anyone know how we could do this quickly and easily in R?

Thanks very much
Dave

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Re: [R] SPDEP Package, neighbours list for Moran's I on large grid dataset

2011-04-20 Thread Roger Bivand

There is a recent thread on the R-sig-geo list that addresses your need,
using a focal function in the raster package. The thread is at:

https://stat.ethz.ch/pipermail/r-sig-geo/2011-April/011444.html

That list may be a better place to post on this topic.

Hope this helps,

Roger

Laurent Jégou wrote:

Hello list members, i'd like to calculate the Moran I on a large dataset,
a
8640x3432 grid of values.

When i try to create the neighbours list with the cell2nb function, on
such
a scale, R works for several hours and seems to crash. I'm using the last
version (2.13), 64 bits, on a mac pro with 4go of memory.

I think that i'm doing it wrong :-)

I'd like to use a moving window weight matrix instead of a full scale
one,
but i was not lucky enough to find a solution in the docs.

Thanks for any help,

Laurent

--
Laurent Jégou
Cartographe et enseignant
UTM - Dépt. Géographie
31058 TOULOUSE Cedex 9 - 05.61.50.43.89
http://www.univ-tlse2.fr/geoprdc

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-
Roger Bivand
Economic Geography Section
Department of Economics
Norwegian School of Economics and Business Administration
Helleveien 30
N-5045 Bergen, Norway

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[R] lapply sequence

2011-04-20 Thread Dean Marks

Good day,

My question is: Does the lapply function guarantee a particular sequence in
which elements are mapped? And, are we guaranteed that lapply will always be
sequential (i.e. never map elements in parallel) ?

The reason I ask is if I use lapply with the mapping function set to
something that has side-effects that need to be executed in a particular
sequence.

If this is not possible, is there an alternate method other than using a for
loop?

-- 
Dean Marks

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Re: [R] Yearly aggregates and matrices

2011-04-20 Thread Gabor Grothendieck

On Wed, Apr 20, 2011 at 5:49 AM, mathijsdevaan mathijsdev...@gmail.com wrote:
 As a follow up on this post, I am trying to slightly adjust the solution
 kindly provided by Gabor. However, I am getting some results that I do not
 understand. Example:

 # devel version of zoo
 install.packages(zoo, repos = http://r-forge.r-project.org;)
 library(zoo)

 DF1 = data.frame(read.table(textConnection(    B  C  D  E  F  G
 8025  1995  0  4  1  2
 8025  1997  1  1  3  4
 8026  1995  0  7  0  0
 8026  1996  1  2  3  0
 8026  1997  1  2  3  1
 8026  1998  6  0  0  4
 8026  1999  3  7  0  3
 8027  1997  1  2  3  9
 8027  1998  1  2  3  1
 8027  1999  6  0  0  2
 8028  1999  3  7  0  0
 8029  1995  0  2  3  3
 8029  1998  1  2  3  2
 8029  1999  6  0  0  1),head=TRUE,stringsAsFactors=FALSE))

 a - read.zoo(DF1, split = 1, index = 2, FUN = identity)
 sum.na - function(x) if (any(!is.na(x))) sum(x, na.rm = TRUE) else NA
 b - rollapply(a, 3,  sum.na, align = right, partial = TRUE)
 newDF - lapply(1:nrow(b), function(i)
       prop.table(na.omit(matrix(b[i,], nc = 4, byrow = TRUE,
               dimnames = list(unique(DF1$B), names(DF1)[-1:-2]))), 1))
 names(newDF) - time(a)
 c-lapply(newDF, function(mat) tcrossprod(mat / sqrt(rowSums(mat^2

 Now I would like the elements e in c to be equal to 1-e. However,

 c-lapply(newDF, function(mat) 1 - tcrossprod(mat / sqrt(rowSums(mat^2

 gives a value  of 2.220446e-16 for as.data.frame(c['1999'])[2,2] instead of
 0

 What am I doing wrong here? Thanks a lot!


See FAQ 7.31 at:

http://cran.r-project.org/doc/FAQ/R-FAQ.html#Why-doesn_0027t-R-think-these-numbers-are-equal_003f


-- 
Statistics  Software Consulting
GKX Group, GKX Associates Inc.
tel: 1-877-GKX-GROUP
email: ggrothendieck at gmail.com

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Re: [R] problem reading csv file

2011-04-20 Thread Petr PIKAL

Thank you. 

r-help-boun...@r-project.org napsal dne 19.04.2011 22:00:56:

 Not attached. You might succeed if you rename the file with a .txt 
 extension and re-post.
 
 Almost surely an encoding issue. We may need your session Info to get 
 locale.

Maybe you are right. After some further thorough documentation search I 
tried encoding, which was not success. After that I tried fileEncoding = 
UCS-2LE and bingo, that was it.

From file help page

The encodings ‘UCS-2LE’ and ‘UTF-16LE’ are treated specially,
 as they are appropriate values for Windows ‘Unicode’ text files.
 If the first two bytes are the Byte Order Mark ‘0xFFFE’ then these
 are removed as some implementations of ‘iconv’ do not accept BOMs.

˙ţ1 looks in hex editor like that 0xFFFE so the problem is solved for now.

Thanks again

Petr

 
 -- 
 David
 On Apr 19, 2011, at 2:47 PM, Petr Pikal wrote:
 
  Dear all
 
  I have several files which claim to be *.csv (one attached, maybe it
  will come through) . They can be read to Open Office without much
  problem, however I can not read them into R. I tried
  read.table(H2O.CSV, sep=,, dec=.)
  V1
  1 ˙ţ1
  2
  3
  
  read.table(H2O.CSV, sep=,, dec=., skip=1)
  Error in read.table(H2O.CSV, sep = ,, dec = ., skip = 1) :
  empty beginning of file
 
  readLines(H2O.CSV, 1)
  [1] ˙ţ1
 
  readLines(H2O.CSV, 5)
  [1] ˙ţ1
 
 
  readChar(H2O.CSV, 10)
  [1] ˙ţ1
  readBin(H2O.CSV, 10)
  [1] 9.456937e-308
 
 
  This is how first two lines appear in Notepad
 
  1,1.77436423301697,BV
  , 
  
91.0779418945313,7.35872077941895,0.178956836462021,1.70007145404816,1.90102112293243
  2,1.94783389568329,VV
  , 
  
1341.51489257812,9.04244899749756,1.76539707183838,1.90102112293243,3.52783703804016
  
 
  The problem seems to be in first item ˙ţ1 which somehow blocks
  further values to be read.
 
  Does anybody have idea what to do or where to look for some help? I do
  not want to transfer files through spreadsheet manually.
 
  Best regards
  Petr
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 David Winsemius, MD
 West Hartford, CT
 
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Re: [R] BMA, logistic regression, odds ratio, model reduction etc

2011-04-20 Thread Frank Harrell

Deleting variables is a bad idea unless you make that a formal part of the
BMA so that the attempt to delete variables is penalized for.  Instead of
BMA I recommend simple penalized maximum likelihood estimation (see the lrm
function in the rms package) or pre-modeling data reduction that is blinded
to the outcome variable.
Frank


細田弘吉 wrote:
 
 Hi everybody,
 I apologize for long mail in advance.
 
 I have data of 104 patients, which consists of 15 explanatory variables
 and one binary outcome (poor/good). The outcome consists of 25 poor
 results and 79 good results. I tried to analyze the data with logistic
 regression. However, the 15 variables and 25 events means events per
 variable (EPV) is much less than 10 (rule of thumb). Therefore, I used R
 package, BMA to perform logistic regression with BMA to avoid this
 problem.
 
 model 1 (full model):
 x1, x2, x3, x4 are continuous variables and others are binary data.
 
 x16.bic.glm - bic.glm(outcome ~ ., data=x16.df,
 glm.family=binomial, OR20, strict=FALSE)
 summary(x16.bic.glm)
 (The output below has been cut off at the right edge to save space)
 
   62  models were selected
  Best  5  models (cumulative posterior probability =  0.3606 ):
 
  p!=0EV SDmodel 1model2
 Intercept100-5.1348545  1.652424-4.4688  -5.15
 -5.1536
 age3.3   0.0001634  0.007258  .
 sex4.0
.M   -0.0243145  0.220314  .
 side  10.8
 .R   0.0811227  0.301233  .
 procedure 46.9  -0.5356894  0.685148  .  -1.163
 symptom3.8  -0.0099438  0.129690  .  .
 stenosis   3.4  -0.0003343  0.005254  .
 x13.7  -0.0061451  0.144084  .
 x2   100.0   3.1707661  0.892034 3.2221 3.11
 x351.3  -0.4577885  0.551466-0.9154 .
 HT 4.6
   .positive  0.0199299  0.161769  .  .
 DM 3.3
   .positive -0.0019986  0.105910  .  .
 IHD3.5
.positive 0.0077626  0.122593  .  .
 smoking9.1
.positive 0.0611779  0.258402  .  .
 hyperlipidemia16.0
   .positive  0.1784293  0.512058  .  .
 x4 8.2   0.0607398  0.267501  .  .
 
 
 nVar   2  2
  1  3  3
 BIC   -376.9082
 -376.5588  -376.3094  -375.8468  -374.5582
 post prob0.104
 0.087  0.077  0.061  0.032
 
 [Question 1]
 Is it O.K to calculate odds ratio and its 95% confidence interval from
 EV (posterior distribution mean) and“SD”(posterior distribution
 standard deviation)?
 For example, 95%CI of EV of x2 can be calculated as;
 exp(3.1707661)
 [1] 23.82573 - odds ratio
 exp(3.1707661+1.96*0.892034)
 [1] 136.8866
 exp(3.1707661-1.96*0.892034)
 [1] 4.146976
 -- 95%CI (4.1 to 136.9)
 Is this O.K.?
 
 [Question 2]
 Is it permissible to delete variables with small value of p!=0 and
 EV, such as age (3.3% and 0.0001634) to reduce the number of
 explanatory variables and reconstruct new model without those variables
 for new session of BMA?
 
 model 2 (reduced model):
 I used R package, pvclust, to reduce the model. The result suggested
 x1, x2 and x4 belonged to the same cluster, so I picked up only x2.
 Based on the subject knowledge, I made a simple unweighted sum, by
 counting the number of clinical features. For 9 features (sex, side,
 HT2, hyperlipidemia, DM, IHD, smoking, symptom, age), the sum ranges
 from 0 to 9. This score was defined as ClinicalScore. Consequently, I
 made up new data set (x6.df), which consists of 5 variables (stenosis,
 x2, x3, procedure, and ClinicalScore) and one binary outcome
 (poor/good). Then, for alternative BMA session...
 
 BMAx6.glm - bic.glm(postopDWI_HI ~ ., data=x6.df,
 glm.family=binomial, OR=20, strict=FALSE)
 summary(BMAx6.glm)
 (The output below has been cut off at the right edge to save space)
 Call:
 bic.glm.formula(f = postopDWI_HI ~ ., data = x6.df, glm.family =
 binomial, strict = FALSE, OR = 20)
 
 
   13  models were selected
  Best  5  models (cumulative posterior probability =  0.7626 ):
 
 p!=0EV SD   model 1model 2
 Intercept   100-5.6918362  1.81220-4.4688-6.3166
 stenosis  8.1  -0.0008417  0.00815  .  .
 x2  100.0   3.0606165  0.87765 3.2221 3.1154
 x3   46.5  -0.3998864  0.52688-0.9154  .
 procedure   49.3   0.5747013

Re: [R] lapply sequence

2011-04-20 Thread Duncan Murdoch


On 20/04/2011 7:26 AM, Dean Marks wrote:

Good day,

My question is: Does the lapply function guarantee a particular sequence in
which elements are mapped? And, are we guaranteed that lapply will always be
sequential (i.e. never map elements in parallel) ?


No.


The reason I ask is if I use lapply with the mapping function set to
something that has side-effects that need to be executed in a particular
sequence.


Use a for loop.

If this is not possible, is there an alternate method other than using a for
loop?


while or repeat.

Duncan Murdoch

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[R] Two Questions

2011-04-20 Thread Stephen P Molnar

Sorry for the somewhat nondescript subject line, but I have two questions:

 

1.What is a really good book on R for a nonprogrammer?

2.   How do I open more than one R Graphics: Device 2(ACTIVE).  That
what is the R command that I can use to keep more than one plot open.  I am
running a script from a book on Chemometrics that results in more than one
graph during the execution, but it seems that R deletes each graph when the
script calls for the next plot.

 

Thanks in advance

 

Stephen P. Molnar, Ph.D.  Life is a
fuzzy set

Foundation for Chemistry Stochastic
and multivariate

http://www.FoundationForChemistry.com

 


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Re: [R] Simple question about symbols()

2011-04-20 Thread Ben Bolker

David Winsemius dwinsemius at comcast.net writes:

 
 
 On Apr 19, 2011, at 10:51 PM, murilofm wrote:
 
  Thanks for the answer; I see that col=c(blue,red)[inv$c+1]  
  creates a
  vector of red and blue associated with the binnary c.
  But still I got everything red.
 
 If you want tested solution, submit test data.
 

  David also suggests (off-lists) that you use bg instead of col as 
the argument name.  (Another strategy is to go to ?symbols
and search for colour -- I wouldn't expect a new user
to guess the c(blue,red)[inv$c+1] stuff, but you probably
could go to the help page and figure out that the appropriate
argument name is bg ...)

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Re: [R] Simple question about symbols()

2011-04-20 Thread murilofm

The link to the csv file is

http://www.filedropper.com/data_5

I use the d variable to create the radius:

radius - sqrt( inv$d/ pi )

and i tried

symbols(inv$a, inv$b, circles=radius, inches=0.35, fg=white,
   bg=red, xlab=aa, ylab=bb,
   col=c(blue,red)[inv$c+1])

Thanks for the help.

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Re: [R] SPDEP Package, neighbours list for Moran's I on large grid dataset

2011-04-20 Thread Ben Haller

Laurent Jégou wrote:

 Hello list members, i'd like to calculate the Moran I on a large dataset,
 a 8640x3432 grid of values.
 
 When i try to create the neighbours list with the cell2nb function, on
 such a scale, R works for several hours and seems to crash. I'm using the last
 version (2.13), 64 bits, on a mac pro with 4go of memory.
 
 I think that i'm doing it wrong :-)
 
 I'd like to use a moving window weight matrix instead of a full scale
 one, but i was not lucky enough to find a solution in the docs.

  I confronted this problem recently, and decided to roll my own.  For a large 
dataset, an exhaustive computation of Moran's I is very time-consuming, as you 
say; but an estimate of it, based on a subset of pairs chosen randomly, seemed 
(for my data, at least) to converge quite nicely.  Here's my code.  It assumes 
a square grid, so you'll need to adapt it for your non-square grid, but that 
should be trivial.  To determine the right number of pairs for a good estimate 
in my application, I called this function with various numbers of pairs, and 
plotted the estimate produced as a function of the number of pairs used, and 
observed good convergence by 200,000 pairs.  You will probably want to do this 
yourself, for your dataset.  200,000 pairs took just a second or two to run, on 
my machine, so this is much, much faster than an exhaustive calculation.  As a 
bonus, this code also gives you Geary's C.  Oh, and the code below assumes a 
wraparound lattice (i.e. a torus, i.e. periodic boundaries), which may not be 
what you want; just get rid of the d_wraparound stuff and the following pmax() 
call if you want non-wraparound boundaries, and that should work fine, I think. 
 Not sure what you mean by the moving window thing, so I leave that as an 
exercise for the reader.  :-

  I've never made an R package, and I'm not sure there's much demand for this 
code (is there?), so I'm presently unlikely to package it up.  However, if 
someone who owns a package related to this problem would like to adopt this 
code, generalize it to non-square lattices, add a flag to choose periodic or 
non-periodic boundaries, etc., feel free to do so; I hereby place it in the 
public domain.  I'd just like to hear about it if someone does this, and 
receive credit somewhere, that's all.  I'm not super-good in R, either, so if 
there's a better way to do some things, like the somewhat hacked random index 
generation (trying to avoid floating point issues when generating a random 
integer), please let me know.  I'm always interested in learning how to do 
things better.

  And of course this code is provided without warranty, may have bugs, etc.

  Enjoy!

Ben Haller
McGill University


correlation_stats - function(p, n_pairs=20)
{
# Compute Moran's I and Geary's C for the landscape p.  This is tricky 
because the exact calculation
# would be far too time-consuming, as it involves comparison of every 
point to every other point.
# So I am trying to estimate it here with a small subset of all 
possible point comparisons.
p_size - NROW(p)   # 512
p_length - length(p)   # 262144
mean_p - mean(p)
pv - as.vector(p)

# select points and look up their values; for speed, points are 
selected by their vector index
p1ix - floor(runif(n_pairs, min=0, max=p_length - 0.001))
p1x - (p1ix %/% p_size) + 1
p1y - (p1ix %% p_size) + 1
p1ix - p1ix + 1

p2ix - floor(runif(n_pairs, min=0, max=p_length - 0.001))
p2x - (p2ix %/% p_size) + 1
p2y - (p2ix %% p_size) + 1
p2ix - p2ix + 1

v1 - pv[p1ix] - mean_p
v2 - pv[p2ix] - mean_p

# Calculate distances between point pairs, both directly and wrapping 
around
# The average direct distance is much smaller than the average 
wraparound distance,
# because points near the center vertically have a maximum direct 
distance near 256,
# but a minimum wraparound distance near 256.  The Moran's I estimate 
from wraparound
# distances is different, as a result.  Rupert recommends taking the 
shorter of the
# two distances, whichever it is, because that keeps us within the 
meaningful scale
# of our space, before it just starts repeating due to periodicity.
d_direct - 1 / sqrt(((p1x - p2x) ^ 2) + ((p1y - p2y) ^ 2))
d_direct[is.infinite(d_direct)] - 0

d_wraparound - 1 / sqrt(((p1x - p2x) ^ 2) + ((p_size - abs(p1y - p2y)) 
^ 2))
d_wraparound[is.infinite(d_wraparound)] - 0

d - pmax(d_direct, d_wraparound)   # max because we want the min 
distance, and these are 1/distance

# precalculate some shared terms
sum_d - sum(d)
sum_v1_sq - sum(v1 ^ 2)

# Moran's I: -1 is perfect dispersion, 1 is perfect correlation, 0 is a 
random spatial pattern
M_I - (n_pairs

Re: [R] Two Questions

2011-04-20 Thread Duncan Murdoch


On 20/04/2011 9:23 AM, Stephen P Molnar wrote:

Sorry for the somewhat nondescript subject line, but I have two questions:



1.What is a really good book on R for a nonprogrammer?


Any book that teaches you the basics of programming would be good, it 
doesn't need to be about R.  If you want to use R and remain as a 
nonprogrammer, you will not have any easy time.



2.   How do I open more than one R Graphics: Device 2(ACTIVE).  That
what is the R command that I can use to keep more than one plot open.  I am
running a script from a book on Chemometrics that results in more than one
graph during the execution, but it seems that R deletes each graph when the
script calls for the next plot.



dev.new()  will open a new plot window, and subsequent plotting commands 
will be drawn there.  dev.set() lets you switch back to drawing on the 
original one.


Duncan Murdoch




Thanks in advance



Stephen P. Molnar, Ph.D.  Life is a
fuzzy set

Foundation for Chemistry Stochastic
and multivariate

http://www.FoundationForChemistry.com




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Re: [R] Two Questions



On Apr 20, 2011, at 9:23 AM, Stephen P Molnar wrote:

Sorry for the somewhat nondescript subject line, but I have two  
questions:




1.What is a really good book on R for a nonprogrammer?

2.   How do I open more than one R Graphics: Device 2(ACTIVE).   
That

what is the R command that I can use to keep more than one plot open.


You can have more than one device available, but you need to address  
them serially. Only one device can receive input at a time.


?Devices
?dev.set


 I am
running a script from a book on Chemometrics that results in more  
than one
graph during the execution, but it seems that R deletes each graph  
when the

script calls for the next plot.


More likely you are seeing one graph displayed at a time on the screen  
device. On my screen device the cmd-left-arrow will bring up prior  
plots to a depth of 15 earlier results.


--

David Winsemius, MD
West Hartford, CT

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[R] Test email: Please disregard

2011-04-20 Thread Paul Miller

Having trouble posting. Thought it might have something to do with the 
particular message I'm sending. Sending this message to test that possibility.

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Re: [R] Yearly aggregates and matrices

2011-04-20 Thread mathijsdevaan

Thanks for clarifying that.

Best


Gabor Grothendieck wrote:
 
 On Wed, Apr 20, 2011 at 5:49 AM, mathijsdevaan
 lt;mathijsdev...@gmail.comgt; wrote:
 As a follow up on this post, I am trying to slightly adjust the solution
 kindly provided by Gabor. However, I am getting some results that I do
 not
 understand. Example:

 # devel version of zoo
 install.packages(zoo, repos = http://r-forge.r-project.org;)
 library(zoo)

 DF1 = data.frame(read.table(textConnection(    B  C  D  E  F  G
 8025  1995  0  4  1  2
 8025  1997  1  1  3  4
 8026  1995  0  7  0  0
 8026  1996  1  2  3  0
 8026  1997  1  2  3  1
 8026  1998  6  0  0  4
 8026  1999  3  7  0  3
 8027  1997  1  2  3  9
 8027  1998  1  2  3  1
 8027  1999  6  0  0  2
 8028  1999  3  7  0  0
 8029  1995  0  2  3  3
 8029  1998  1  2  3  2
 8029  1999  6  0  0  1),head=TRUE,stringsAsFactors=FALSE))

 a - read.zoo(DF1, split = 1, index = 2, FUN = identity)
 sum.na - function(x) if (any(!is.na(x))) sum(x, na.rm = TRUE) else NA
 b - rollapply(a, 3,  sum.na, align = right, partial = TRUE)
 newDF - lapply(1:nrow(b), function(i)
       prop.table(na.omit(matrix(b[i,], nc = 4, byrow = TRUE,
               dimnames = list(unique(DF1$B), names(DF1)[-1:-2]))), 1))
 names(newDF) - time(a)
 c-lapply(newDF, function(mat) tcrossprod(mat / sqrt(rowSums(mat^2

 Now I would like the elements e in c to be equal to 1-e. However,

 c-lapply(newDF, function(mat) 1 - tcrossprod(mat /
 sqrt(rowSums(mat^2

 gives a value  of 2.220446e-16 for as.data.frame(c['1999'])[2,2] instead
 of
 0

 What am I doing wrong here? Thanks a lot!

 
 See FAQ 7.31 at:
 
 http://cran.r-project.org/doc/FAQ/R-FAQ.html#Why-doesn_0027t-R-think-these-numbers-are-equal_003f
 
 
 -- 
 Statistics  Software Consulting
 GKX Group, GKX Associates Inc.
 tel: 1-877-GKX-GROUP
 email: ggrothendieck at gmail.com
 
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Re: [R] Simple question about symbols()



On Apr 20, 2011, at 8:45 AM, murilofm wrote:


The link to the csv file is

http://www.filedropper.com/data_5

I use the d variable to create the radius:

radius - sqrt( inv$d/ pi )

and i tried

symbols(inv$a, inv$b, circles=radius, inches=0.35, fg=white,
  bg=red, xlab=aa, ylab=bb,
  col=c(blue,red)[inv$c+1])


You should follow the Posting Guide's advice and offer the full code  
you used to read in that data. It's _not_ comma separated, and for  
some reason even using read.csv2 to properly read a semi-colon  
separated file, I also needed to to convert the first two variables  
from factor to numeric  ... please also read the FAQ item on this topic.


Once that was done this works, since col is _not_ the proper  
argument for coloring with that function:


?symbols

plot.new()
 symbols(inv$a, inv$b, circles=radius, inches=0.35,
 bg=c(blue,red)[inv$c+1])

--

David Winsemius, MD
West Hartford, CT

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Re: [R] Extrapolating data points for individuals who lack them

2011-04-20 Thread Mike Marchywka

 From: de...@exeter.ac.uk
 To: r-help@r-project.org
 Date: Wed, 20 Apr 2011 12:41:29 +0100
 Subject: [R] Extrapolating data points for individuals who lack them

 Hi,

 We have an experiment where individuals responses were measured over 5 days. 
 Some responses were not obtained because we only allowed individuals to 
 respond within a limited time-frame. These individuals are given the maximum 
 response time as they did not respond, yet we feel they may have done if 
 given time (and by looking at the rest of their responses over time, the 
 non-response days stand out).

 We therefore want to extrapolate data points for individuals, on days when 
 they didn't respond, using a regression of days when they did.

 Does anyone know how we could do this quickly and easily in R?

You are probably talking about right censoring. See things like this, 
( you may have good luck just with R rather than CRAN ) 

http://www.google.com/#sclient=psyhl=ensource=hpq=CRAN+informative+%22right+censoring%22

If you post data maybe someone can try a few things. It isn't hard to take data
subsets, fit models, and replace data with model predictions but easier and more
interesting to illustrate with your data.

Personally I would avoid making up data and of course extrapolation tends
to be the most error prone way of doing that. 

 Thanks very much
 Dave

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Re: [R] BMA, logistic regression, odds ratio, model reduction etc

2011-04-20 Thread khosoda


Dear Prof. Harrel,

Thank you very much for your quick advice.
I will try rms package.

Regarding model reduction, is my model 2 method (clustering and recoding 
that are blinded to the outcome) permissible?


Sincerely,

--
KH

(11/04/20 22:01), Frank Harrell wrote:

Deleting variables is a bad idea unless you make that a formal part of the
BMA so that the attempt to delete variables is penalized for.  Instead of
BMA I recommend simple penalized maximum likelihood estimation (see the lrm
function in the rms package) or pre-modeling data reduction that is blinded
to the outcome variable.
Frank


細田弘吉 wrote:


Hi everybody,
I apologize for long mail in advance.

I have data of 104 patients, which consists of 15 explanatory variables
and one binary outcome (poor/good). The outcome consists of 25 poor
results and 79 good results. I tried to analyze the data with logistic
regression. However, the 15 variables and 25 events means events per
variable (EPV) is much less than 10 (rule of thumb). Therefore, I used R
package, BMA to perform logistic regression with BMA to avoid this
problem.

model 1 (full model):
x1, x2, x3, x4 are continuous variables and others are binary data.


x16.bic.glm- bic.glm(outcome ~ ., data=x16.df,

glm.family=binomial, OR20, strict=FALSE)

summary(x16.bic.glm)

(The output below has been cut off at the right edge to save space)

   62  models were selected
  Best  5  models (cumulative posterior probability =  0.3606 ):

  p!=0EV SDmodel 1model2
Intercept100-5.1348545  1.652424-4.4688  -5.15
-5.1536
age3.3   0.0001634  0.007258  .
sex4.0
.M   -0.0243145  0.220314  .
side  10.8
 .R   0.0811227  0.301233  .
procedure 46.9  -0.5356894  0.685148  .  -1.163
symptom3.8  -0.0099438  0.129690  .  .
stenosis   3.4  -0.0003343  0.005254  .
x13.7  -0.0061451  0.144084  .
x2   100.0   3.1707661  0.892034 3.2221 3.11
x351.3  -0.4577885  0.551466-0.9154 .
HT 4.6
   .positive  0.0199299  0.161769  .  .
DM 3.3
   .positive -0.0019986  0.105910  .  .
IHD3.5
.positive 0.0077626  0.122593  .  .
smoking9.1
.positive 0.0611779  0.258402  .  .
hyperlipidemia16.0
   .positive  0.1784293  0.512058  .  .
x4 8.2   0.0607398  0.267501  .  .


nVar   2  2
  1  3  3
BIC   -376.9082
-376.5588  -376.3094  -375.8468  -374.5582
post prob0.104
0.087  0.077  0.061  0.032

[Question 1]
Is it O.K to calculate odds ratio and its 95% confidence interval from
EV (posterior distribution mean) and“SD”(posterior distribution
standard deviation)?
For example, 95%CI of EV of x2 can be calculated as;

exp(3.1707661)

[1] 23.82573 -  odds ratio

exp(3.1707661+1.96*0.892034)

[1] 136.8866

exp(3.1707661-1.96*0.892034)

[1] 4.146976
--  95%CI (4.1 to 136.9)
Is this O.K.?

[Question 2]
Is it permissible to delete variables with small value of p!=0 and
EV, such as age (3.3% and 0.0001634) to reduce the number of
explanatory variables and reconstruct new model without those variables
for new session of BMA?

model 2 (reduced model):
I used R package, pvclust, to reduce the model. The result suggested
x1, x2 and x4 belonged to the same cluster, so I picked up only x2.
Based on the subject knowledge, I made a simple unweighted sum, by
counting the number of clinical features. For 9 features (sex, side,
HT2, hyperlipidemia, DM, IHD, smoking, symptom, age), the sum ranges
from 0 to 9. This score was defined as ClinicalScore. Consequently, I
made up new data set (x6.df), which consists of 5 variables (stenosis,
x2, x3, procedure, and ClinicalScore) and one binary outcome
(poor/good). Then, for alternative BMA session...


BMAx6.glm- bic.glm(postopDWI_HI ~ ., data=x6.df,

glm.family=binomial, OR=20, strict=FALSE)

summary(BMAx6.glm)

(The output below has been cut off at the right edge to save space)
Call:
bic.glm.formula(f = postopDWI_HI ~ ., data = x6.df, glm.family =
binomial, strict = FALSE, OR = 20)


   13  models were selected
  Best  5  models (cumulative posterior probability =  0.7626 ):

 p!=0EV SD   model 1model 2
Intercept   100-5.6918362  1.81220-4.4688-6.3166
stenosis  8.1

[R] Random Relabelling

2011-04-20 Thread kmatthews

I have 4000 observations that I need to randomly relabel 1000 times and then
calculate the mean of the 1000 values at each of the 4000 points.  Any ideas
for where to begin? 

Thanks
Kevin 

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Re: [R] Fibonacci

2011-04-20 Thread Bart Joosen

Another solution:

while (Fibonacci[1]  500)  Fibonacci - c(sum(Fibonacci[c(1,2)]),
Fibonacci)

While this adds the sum before the existing values, the length or tail
function or avoided, but even with reordering, its faster
(Fibonacci[length(Fibonacci):1])

Best regards

Bart


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[R] Matrix package transpose

2011-04-20 Thread Tobias Abenius


Hi,

Since I installed R 2.13 I cannot use the transpose method t on sparse 
matrices inside my package. Outside the package works. Is there 
something new that I have to import methods? Can I then import 
everything from the Matrix package? The problem is that R tries to use 
t.default which of course doesn't work.


Happy easter, Tobias
--
Tobias Abenius
Ph.D. Student, M.Sc. in Computer Science

Mathematical Statistics
Mathematical Sciences
University of Gothenburg
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[R] Unlist command drops all my column names in the first row and adopts NAs

2011-04-20 Thread Haillie

Hi Everyone, 

I am having trouble turning my data.frame into a matrix format. Because I
wanted to change my data.frame with mostly factor variables into a numeric
matrix, I used the following code --
UN2010frame-data.matrix(lapply(UN2010,as.numeric))

However when i checked the mode of the UN2010frame, it still showed up as a
list. Because the code I want to run (Ordrating) does not accept data in a
list format, I used UN2010matrix - unlist(UN2010frame) to unlist my matrix.
When I did this, my first row ( which was formerly a row with column names)
turned into NAs. This was a problem for me because when I tried to run an
ordinal IRT model using this data set, I got the following error message. 

Error in 1:nrow(Y) : argument of length 0

I think it is because all the values in my first row are now gone 

If you could help me on any front, It would be deeply appreciated. Thank you
very much!

Haillie 


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Re: [R] grid.table + splom: how to nicely align panel entries

Dear Baptiste,

very nice, indeed! 

Two minor issues that remain, are:
(1) I tried to omit the decimal dot for those numbers that do not have digits 
after the decimal dot. But somehow it does not work...
(2) Do you know how one can decrease the text size for the text appearing in 
the 
lower panel? I tried to work with cex=0.5... but it was ignored all the 
time.

Cheers,

Marius


library(lattice) 
library(grid)
library(gridExtra)

## function for correct digit alignment
align.digits - function(l){
sp - strsplit(as.character(l), \\.)
chars - sapply(sp, function(x) nchar(x)[1])
n - max(chars)-chars
l0 - sapply(n, function(x) paste(rep(0, x), collapse=))
sapply(seq_along(sp), function(i){
if(length(sp[[1]])==1){
as.expression(bquote(phantom(.(l0[i])) * .(sp[[i]][1])))
}else{
as.expression(bquote(phantom(.(l0[i])) * 
.(sp[[i]][1])*.*.(sp[[i]][2])))
}
})
}

## splom with customized lower.panel
## x: data
## arr: array of containing expressions which are plotted in a grid table in 
the 
##  lower panel (i,j)]
splom2 - function(x, arr, nr){
## function for creating table 
table.fun - function(vec){ # vector containing lines for table for *one* 
panel
grid.table(matrix(vec, nrow=nr, byrow=TRUE),
   parse=TRUE, # parse labels as expressions
   theme=theme.list(
   gpar.corefill=gpar(fill=NA, col=NA), # make bg transparent
   core.just=left, padding.h=unit(0,mm)) # justification of 
labels
   ) 
}
## splom
splom(x, varname.cex=1.2,
  superpanel=function(z, ...){
  panel.pairs(z, upper.panel=panel.splom, lower.panel=function(i,j){
  table.fun(arr[i,j,])
  }, ...)
  })
}

## create data and array of expressions
d - 4
x - matrix(runif(d*1000), ncol=d) # data to be plotted with splom
nr - 3 # number of rows for the panel entries
nc - 3 # number of cols for the panel entries
arr - array(list(rep(NA,nr*nc)), dim=c(d,d,nr*nc), dimnames=c(i,j,val)) 
# array containing the table entries per panel
f - function(i,j) (i+j)*10 # dummy function
eq - phantom()==phantom()
for(i in 1:d){
for(j in 1:d){
numbers - align.digits(c(round(pi,4), round(pi, 6), f(i,j)))
arr[i,j,] - c(alpha, eq, numbers[1],
   italic(bbb), eq, numbers[2],
   gamma, eq, numbers[3])
}
}

## plot
splom2(x, arr, nr=3)


On 2011-04-20, at 11:56 , baptiste auguie wrote:

 On 20 April 2011 21:16, Marius Hofert m_hof...@web.de wrote:
 Dear expeRts,
 
 is there a way to get the entries in each panel correctly aligned according 
 to the
 equality signs?
 
 Here is the wish-list:
 (1) the equality signs in each panel should be vertically aligned
 
 You can put the equal signs in their own column,
 
 library(gridExtra)
 d = matrix(c(italic(a), phantom()==phantom(), round(pi,4),
 italic(b), phantom()==phantom(), round(pi,6)), ncol=3, byrow=T)
 grid.table(d, parse=T,theme=theme.list(core.just=left))
 
 (2) the numbers should be aligned on the decimal point
 
 You could place some phantom()s to do this,
 
 align.digits = function(l)
 {
 
 sp - strsplit(as.character(l), \\.)
 chars - sapply(sp, function(x) nchar(x)[1])
 n = max(chars) - chars
 l0 = sapply(n, function(x) paste(rep(0, x), collapse=))
 labels = sapply(seq_along(sp), function(i) {
  as.expression(bquote(phantom(.(l0[i])) * .(sp[[i]][1])*.*.(sp[[i]][2])))})
 
 return(labels)
 }
 
 library(gridExtra)
 
 d - align.digits(l = c(125.3, 1.2344))
 grid.table(d, parse=T,core.just=left)
 
 HTH,
 
 baptiste
 
 One could adjust the phantom()-arguments by hand to achieve (1), but is 
 there a
 simpler solution? For (2) I have no idea.
 
 Cheers,
 
 Marius
 
 
 library(lattice)
 library(grid)
 library(gridExtra)
 
 ## splom with customized lower.panel
 ## x: data
 ## arr: array of containing expressions which are plotted in a grid table in 
 the
 ##  lower panel (i,j)]
 splom2 - function(x, arr){
## function for creating table
table.fun - function(vec){ # vector containing lines for table for *one* 
 panel
grid.table(matrix(vec, ncol=2, byrow=TRUE),
   parse=TRUE, # parse labels as expressions
   theme=theme.list(
   gpar.corefill=gpar(fill=NA, col=NA), # make bg transparent
   core.just=left, padding.h=unit(0,mm)) # justification 
 of labels
   )
}
## splom
splom(x, varname.cex=1.4,
  superpanel=function(z, ...){
  panel.pairs(z, upper.panel=panel.splom, 
 lower.panel=function(i,j){
  table.fun(arr[i,j,])
  }, ...)
  })
 }
 
 ## create data and array of expressions
 d - 4
 x - matrix(runif(d*1000), ncol=d) # data to be plotted with splom
 arr - array(list(rep(NA, 3*2)), dim=c(d,d,3*2), dimnames=c(i,j,val)) 
 # array containing the table

Re: [R] How to get R plots with FastRweb

2011-04-20 Thread Simon Urbanek

Devi,

FastRWeb doesn't use Java - it is entirely R based so all you need is a web 
server with either CGI or PHP. The client it uses is either a C++ client (part 
of FastRWeb in the cgi-bin directory of the installed package - just copy to 
you server's cgi-bin) or a PHP client (in Rserve/src/client/php/simple.php - 
simply uncomment process_FastRWeb()). The location of the R scripts it serves 
is defined by PROJECT_ROOT at the time you compile FastRWeb which is by default 
/var/FastRWeb so by default foo.png.R would be /var/FastRWeb/web.R/foo.png.R 

Clearly, you could use Java to connect to the Rserve serving FastRWeb but it 
sort of defeats the purpose, because FastRWeb does all the work for you 
consisting of parsing the request and running the scripts.

If you still have issues, I can point you to a sample configuration including 
the setup etc. (for a unix server) if that helps.

Cheers,
Simon

PS: please use stats-rosuda-devel mailing list - I'm not monitoring R-help.


On Apr 20, 2011, at 3:14 AM, MLSC MANIPAL wrote:

 Dear friends,
 
 I am working in a web service project which uses integration of Java with R. 
 I have used RJava to connect with Java and that is working fine. As R 
 produces more interactive plots, I would also like to pipe plots generate 
 from R on web page. I came to know that FastRWeb, R2HTML, brew and 
 WebGraphics, Cairo together can be used to do that. I have installed all 
 those successfully. I am developing my web service project with NetBeans. For 
 testing purpose I have used the excersie (kmeans) given in the FastRWeb 
 document.I have put the kmeans.png.R program in my netbeansProject/R 
 directory where other java code runs properly.
 
 student@mlscubl30:~/NetBeansProjects/R$ ls
 build  build.xml  dist  kmeans.png.R  nbproject  src  test  web
 student@mlscubl30:~/NetBeansProjects/R$ 
 
 When i try to execute it with http://localhost:8080/R/kmeans.png in browser 
 it does not execute it. Hence can you please tell me what exactly i have to 
 do in order to make it up and run?
 
 thanks in advance.
 
 Regards,
 Devi
 Manipal Life Sciences Center

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Re: [R] Unlist command drops all my column names in the first row and adopts NAs

Why not just data.matrix(UN2010) ?

--- On Wed, 4/20/11, Haillie n...@princeton.edu wrote:

 From: Haillie n...@princeton.edu
 Subject: [R] Unlist command drops all my column names in the first row and 
 adopts NAs
 To: r-help@r-project.org
 Received: Wednesday, April 20, 2011, 11:05 AM
 Hi Everyone, 

 I am having trouble turning my data.frame into a matrix
 format. Because I
 wanted to change my data.frame with mostly factor variables
 into a numeric
 matrix, I used the following code --
 UN2010frame-data.matrix(lapply(UN2010,as.numeric))

 However when i checked the mode of the UN2010frame, it
 still showed up as a
 list. Because the code I want to run (Ordrating) does not
 accept data in a
 list format, I used UN2010matrix - unlist(UN2010frame)
 to unlist my matrix.
 When I did this, my first row ( which was formerly a row
 with column names)
 turned into NAs. This was a problem for me because when I
 tried to run an
 ordinal IRT model using this data set, I got the following
 error message. 

 Error in 1:nrow(Y) : argument of length 0

 I think it is because all the values in my first row are
 now gone 

 If you could help me on any front, It would be deeply
 appreciated. Thank you
 very much!

 Haillie 

 --
 View this message in context: 
 http://r.789695.n4.nabble.com/Unlist-command-drops-all-my-column-names-in-the-first-row-and-adopts-NAs-tp3463294p3463294.html
 Sent from the R help mailing list archive at Nabble.com.

 __
 R-help@r-project.org
 mailing list
 https://stat.ethz.ch/mailman/listinfo/r-help
 PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
 and provide commented, minimal, self-contained,
 reproducible code.

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Re: [R] Simple question about symbols()

2011-04-20 Thread murilofm

Thank you for the answer and sorry about the bad post i'll remember that
in the future.
By the way, the line code i used to read the data was

inv - read.csv(data.csv, header=TRUE, sep=;)

I tried before to use the bg, but for some reason it wasn't working out for
me.
But now i got it.

Thanks

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[R] How can I 'predict' from an nls model with a fit specified for separate groups?

2011-04-20 Thread Stuart Rosen

Following an example on p 111 in 'Nonlinear Regression with R' by Ritz  
Streibig, I have been fitting nls models using square brackets with the 
grouping variable inside. In their book is this example, in which 
'state' is a factor indicating whether a treatment has been used or not:


 Puromycin.m1 - nls(rate ~ Vm[state] *
+ conc/(K[state] + conc), data = Puromycin,
+ start = list(K = c(0.1, 0.1),
+ Vm = c(200, 200)))

What I cannot figure out is how to specify the value of the grouping 
variable in a 'predict' statement. In my own example, I can only seem to 
get the predictions for the 1st specified level of the grouping 
variable. I promise that I have read the documentation, and have tried a 
number of things, but cannot get the correct predictions.


Thank you for any help.

Yours - Stuart

--
/**/
Stuart Rosen, PhD
Professor of Speech and Hearing Science
UCL Speech, Hearing and Phonetic Sciences

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[R] Test email #2: Please ignore

2011-04-20 Thread Paul Miller

Having trouble posting. This is a second test email to help determine if the 
problem has been resolved.

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Re: [R] Random Relabelling

Can you explain this a bit more. At the moment I don't see what you are trying 
to achieve.   calculate the mean of the 1000 values at each of the 4000 
points does not seem to make sense.

--- On Wed, 4/20/11, kmatthews kevin-matth...@uiowa.edu wrote:

 From: kmatthews kevin-matth...@uiowa.edu
 Subject: [R] Random Relabelling
 To: r-help@r-project.org
 Received: Wednesday, April 20, 2011, 10:04 AM
 I have 4000 observations that I need
 to randomly relabel 1000 times and then
 calculate the mean of the 1000 values at each of the 4000
 points.  Any ideas
 for where to begin? 

 Thanks
 Kevin 

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Re: [R] Sweave

2011-04-20 Thread R Heberto Ghezzo, Dr

Hello,
I never used Sweave before but now I try and got:
 
  rnwfile - system.file(Sweave, example-1.Rnw, package = utils)
  Sweave(rnwfile)
Writing to file example-1.tex
Processing code chunks with options ...
 1 : echo term verbatim
 2 : term verbatim pdf
You can now run (pdf)latex on 'example-1.tex'
  tools::texi2dvi(example-1.tex, pdf = TRUE)
Error in tools::texi2dvi(example-1.tex, pdf = TRUE) : 
  pdflatex is not available

The code is from Sweave.pdf file in utils/doc
I am using R-2.13.0 in Win7  Toshiba laptop
Please any help?

R.Heberto Ghezzo Ph.D.
Montreal - Canada
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Re: [R] Sweave

2011-04-20 Thread Sarah Goslee

Hi,

On Wed, Apr 20, 2011 at 12:11 PM, R Heberto Ghezzo, Dr
heberto.ghe...@mcgill.ca wrote:
 Hello,
 I never used Sweave before but now I try and got:

Have you installed LaTeX on your computer? It doesn't come installed on
Windows machines by default. I would suggest MiKTeX http://www.miktex.org/
but people who use Windows more often than I do may have other ideas.

Sarah


  rnwfile - system.file(Sweave, example-1.Rnw, package = utils)
  Sweave(rnwfile)
 Writing to file example-1.tex
 Processing code chunks with options ...
  1 : echo term verbatim
  2 : term verbatim pdf
 You can now run (pdf)latex on 'example-1.tex'
  tools::texi2dvi(example-1.tex, pdf = TRUE)
 Error in tools::texi2dvi(example-1.tex, pdf = TRUE) :
  pdflatex is not available

 The code is from Sweave.pdf file in utils/doc
 I am using R-2.13.0 in Win7  Toshiba laptop
 Please any help?

 R.Heberto Ghezzo Ph.D.
 Montreal - Canada
 __



-- 
Sarah Goslee
http://www.functionaldiversity.org

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[R] Class htest with non-numeric p-values

2011-04-20 Thread JiHO

Hi everyone,

For some tests, tables of critical values are readily available while
the algorithm to compute those critical values is not. In such cases,
only a range for the p-value can be evaluated (e.g. 0.05  p  0.1)
from the table of critical values and the statistic.

Is there anyway to include that in an object of class htest? I see
in print.htest() that the p.value element goes through format.pval()
which expect a numeric argument. Is there any workaround?

Thank you in advance. Sincerely,

JiHO
---
http://maururu.net

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Re: [R] Two Questions

 From: Stephen P Molnar s.mol...@sbcglobal.net
 Subject: [R] Two Questions
 To: R-help r-help@r-project.org
 Received: Wednesday, April 20, 2011, 9:23 AM

 1.        What is a really good book on
 R for a nonprogrammer?
Have a look at the books listed on the R website.

Books by Peter Dalgaard, Phil Spector, Michael Crawley  John Verzani are all 
possibilities.  Also haved a look at the Contributed Documentation page on the 
site.  It has some very useful material.

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[R] survexp with weights

2011-04-20 Thread Mike Harwood

Hello,

I probably have a syntax error in trying to generate an expected
survival curve from a weighted cox model, but I can't see it.  I used
the help sample code to generate a weighted model, with the addition
of a weights=albumin argument (I only chose albumin because it had
no missing values, not because of any real relevance).  Below are my
code with the resulting error messages.  Thanks in advance!

 pfit - coxph(Surv(time,status0) ~ trt + log(bili) + log(protime) + age +
+ + platelet,  data=pbc
+ )

 pfit
Call:
coxph(formula = Surv(time, status  0) ~ trt + log(bili) +
log(protime) +
age + +platelet, data = pbc)


  coef exp(coef) se(coef)z  p
trt  -0.000624 0.999  0.17304 -0.00360 1.
log(bili) 0.985497 2.679  0.08949 11.01262 0.
log(protime)  2.79400116.346  0.95289  2.93215 0.0034
age   0.020590 1.021  0.00805  2.55666 0.0110
platelet -0.001699 0.998  0.00085 -2.00130 0.0450

Likelihood ratio test=164  on 5 df, p=0  n= 308, number of events=
143
   (110 observations deleted due to missingness)

 plot(survfit(Surv(time, status0) ~ trt, data=pbc))
 lines(survexp( ~ trt, ratetable=pfit, data=pbc), col='purple')

 pfit.wtd - coxph(Surv(time,status0) ~ trt + log(bili) + log(protime) + age +
+ + platelet,  weights=albumin, data=pbc
+ )

 pfit.wtd
Call:
coxph(formula = Surv(time, status  0) ~ trt + log(bili) +
log(protime) +
age + +platelet, data = pbc, weights = albumin)


 coef exp(coef) se(coef)  z   p
trt  -0.01354 0.987 0.094204 -0.144 8.9e-01
log(bili) 0.99282 2.699 0.048690 20.391 0.0e+00
log(protime)  2.5413612.697 0.525797  4.833 1.3e-06
age   0.01913 1.019 0.004398  4.350 1.4e-05
platelet -0.00165 0.998 0.000462 -3.578 3.5e-04

Likelihood ratio test=535  on 5 df, p=0  n= 308, number of events=
143
   (110 observations deleted due to missingness)
 plot(survfit(Surv(time, status0) ~ trt, data=pbc))
 lines(survexp( ~ trt, ratetable=pfit.wtd, data=pbc), col='purple')
Error in eval(expr, envir, enclos) : object 'albumin' not found
 lines(survexp( ~ trt, ratetable=pfit.wtd, weights=albumin, data=pbc), 
 col='purple')
Error in eval(expr, envir, enclos) : object 'albumin' not found
In addition: Warning message:
In survexp(~trt, ratetable = pfit.wtd, weights = albumin, data =
pbc) :
  Weights ignored
 lines(survexp( ~ trt, ratetable=pfit.wtd, weights=pbc$albumin, data=pbc), 
 col='purple')
Error in eval(expr, envir, enclos) : object 'albumin' not found
In addition: Warning message:
In survexp(~trt, ratetable = pfit.wtd, weights = pbc$albumin, data =
pbc) :
  Weights ignored


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[R] Partial Least Squares Regression independent variables

2011-04-20 Thread tommym

Hi,

I'm running PLSR on the PLS package.  I have 507 independent and one
dependent variable.

model-plsr(y~x1+x2+x3.., data = mydata, validation = CV)

the problem with this is writing in 507 variable names is not realistic as I
run out of line space in the command window.

I cannot run the model thus:

model-plsr(dataframeY~dataframeX...)

as is it is specifically warned against in the documentation.

In the example the independent variables have been collated and are
represented by a single name (NIR, represents numerous variables labelled
NIR.1, NIR,2 etc).  Can anyone advise on how this is done?

Thanks

Tommy McDermott

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[R] Fibonacci

2011-04-20 Thread Georgina Imberger

Thank-you all!!
Very helpful.

-- Forwarded message --
From: Bart Joosen bartjoo...@hotmail.com
Date: 20 April 2011 15:46
Subject: Re: [R] Fibonacci
To: r-help@r-project.org

Another solution:

while (Fibonacci[1]  500)  Fibonacci - c(sum(Fibonacci[c(1,2)]),
Fibonacci)

While this adds the sum before the existing values, the length or tail
function or avoided, but even with reordering, its faster
(Fibonacci[length(Fibonacci):1])

Best regards

Bart

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[[alternative HTML version deleted]]

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Re: [R] 'Record' row values every time the binary value in a collumn changes

2011-04-20 Thread jim holtman

Here is an answer to part 1:

 binary-c(1,1,1,0,0,0,1,1,1,0,0)
 Chromosome-c(1,1,1,1,1,1,2,2,2,2,2)
 start-c(12,17,18,20,25,36,12,15,16,17,19)
 Table-cbind(Chromosome,start,binary)
 # determine where the start/end of each group is
 # use indices since the size is large
 startEnd - lapply(split(seq(nrow(Table))
+  , list(Table[, Chromosome], Table[, 'binary'])
+  , drop = TRUE
+  )
+   , function(.indx){
+ se - range(.indx)
+ c(Chromosome2 = unname(Table[se[1L], Chromosome])
+   , position_start = unname(Table[se[1L], 'start'])
+   , position_end = unname(Table[se[2L], 'start'])
+   , binary2 = unname(Table[se[1L], 'binary'])
+   )
+ })
 do.call(rbind, startEnd)
Chromosome2 position_start position_end binary2
1.0   1 20   36   0
2.0   2 17   19   0
1.1   1 12   18   1
2.1   2 12   16   1




On Wed, Apr 20, 2011 at 5:01 AM, baboon2010 nielsvande...@live.be wrote:
 My question is twofold.

 Part 1:
 My data looks like this:

 (example set, real data has 2*10^6 rows)
 binary-c(1,1,1,0,0,0,1,1,1,0,0)
 Chromosome-c(1,1,1,1,1,1,2,2,2,2,2)
 start-c(12,17,18,20,25,36,12,15,16,17,19)
 Table-cbind(Chromosome,start,binary)
      Chromosome start binary
  [1,]          1    12      1
  [2,]          1    17      1
  [3,]          1    18      1
  [4,]          1    20      0
  [5,]          1    25      0
  [6,]          1    36      0
  [7,]          2    12      1
  [8,]          2    15      1
  [9,]          2    16      1
 [10,]          2    17      0
 [11,]          2    19      0

 As output I need a shortlist for each binary block: giving me the starting
 and ending position of each block.
 Which for these example would look like this:
     Chromosome2 position_start position_end binary2
 [1,]           1             12           18       1
 [2,]           1             20           36       0
 [3,]           2             12           16       1
 [4,]           2             17           19       0

 Part 2:
 Based on the output of part 1, I need to assign the binary to rows of
 another data set. If the position value in this second data set falls in one
 of the blocks defined in the shortlist made in part1,the binary value of the
 shortlist should be assigned to an extra column for this row.  This would
 look something like this:
     Chromosome3 position Value binary3
  [1,] 1         12     a   1
  [2,] 1         13     b   1
  [3,] 1         14     c   1
  [4,] 1         15     d   1
  [5,] 1         16     e   1
  [6,] 1         18     f   1
  [7,] 1         20     g   0
  [8,] 1         21     h   0
  [9,] 1         22     i   0
 [10,] 1         23     j   0
 [11,] 1         25     k   0
 [12,] 1         35     l   0
 [13,] 2         12     m   1
 [14,] 2         13     n   1
 [15,] 2         14     o   1
 [16,] 2         15     p   1
 [17,] 2         16     q   1
 [18,] 2         17     s   0
 [19,] 2         18     d   0
 [20,] 2         19     f   0


 Many thanks in advance,

 Niels

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 View this message in context: 
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-- 
Jim Holtman
Data Munger Guru

What is the problem that you are trying to solve?

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[R] Expanding a VCV Matrix

2011-04-20 Thread Maithula Chandrashekhar

Dear all, I have special task to expand a given VCV matrix, however
could not accomplice yet. Let say I have following VCV matrix

 mat  - matrix(c(1,2,0,2,5,0.5,0,0.5,3), 3, 3)
 colnames(mat) - rownames(mat) - paste(variable, 1:3)
 mat
   variable 1 variable 2 variable 3
variable 1  12.00.0
variable 2  25.00.5
variable 3  00.53.0

Now, say I have a general string vector like this:

 Str - c(paste(variable, 1:4), variable2)
 Str
[1] variable 1 variable 2 variable 3 variable 4 variable 2

Now according to this string, I want my previous VCV matrix also
expands. Therefore, as variable 4 is not there in VCV matrix, so it
will be ignored. Therefore final VCV matrix will be of order 4, and
will look like:

variable 1  variable 2  variable 3  variable 2
variable 1  1   2   0   2
variable 2  2   5   0.5 5
variable 3  0   0.5 3   0.5
variable 2  2   5   0.5 5

However, I do not think it is just some straightforward expansion,
which could be done just by the subsetting mechanism of R. Is there
any idea on how can I do it for general case?

Thanks for your time,

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Re: [R] 'Record' row values every time the binary value in acollumn changes

2011-04-20 Thread William Dunlap

 -Original Message-
 From: r-help-boun...@r-project.org 
 [mailto:r-help-boun...@r-project.org] On Behalf Of jim holtman
 Sent: Wednesday, April 20, 2011 9:59 AM
 To: baboon2010
 Cc: r-help@r-project.org
 Subject: Re: [R] 'Record' row values every time the binary 
 value in acollumn changes
 
 Here is an answer to part 1:
 
  binary-c(1,1,1,0,0,0,1,1,1,0,0)
  Chromosome-c(1,1,1,1,1,1,2,2,2,2,2)
  start-c(12,17,18,20,25,36,12,15,16,17,19)
  Table-cbind(Chromosome,start,binary)
  # determine where the start/end of each group is
  # use indices since the size is large
  startEnd - lapply(split(seq(nrow(Table))
 +  , list(Table[, Chromosome], Table[, 
 'binary'])
 +  , drop = TRUE
 +  )
 +   , function(.indx){
 + se - range(.indx)
 + c(Chromosome2 = unname(Table[se[1L], Chromosome])
 +   , position_start = unname(Table[se[1L], 'start'])
 +   , position_end = unname(Table[se[2L], 'start'])
 +   , binary2 = unname(Table[se[1L], 'binary'])
 +   )
 + })
  do.call(rbind, startEnd)
 Chromosome2 position_start position_end binary2
 1.0   1 20   36   0
 2.0   2 17   19   0
 1.1   1 12   18   1
 2.1   2 12   16   1

The following will likely be quicker way to find where
a column changes values than that lapply() when there
are lots of rows:

  f1 - function (Table) {
  isFirstInRun - function(x) c(TRUE, x[-1] != x[-length(x)])
  isLastInRun - function(x) c(x[-1] != x[-length(x)], TRUE)
  with(data.frame(Table), {
  first - isFirstInRun(binary)
  last - isLastInRun(binary)
  cbind(Chromosome2 = Chromosome[first], position_start = start[first], 
  position_end = start[last], binary2 = binary[first])
})
  }

E.g.,

   f1(Table)
   Chromosome2 position_start position_end binary2
  [1,]   1 12   18   1
  [2,]   1 20   36   0
  [3,]   2 12   16   1
  [4,]   2 17   19   0

Bill Dunlap
Spotfire, TIBCO Software
wdunlap tibco.com  

 
 
 
 
 On Wed, Apr 20, 2011 at 5:01 AM, baboon2010 
 nielsvande...@live.be wrote:
  My question is twofold.
 
  Part 1:
  My data looks like this:
 
  (example set, real data has 2*10^6 rows)
  binary-c(1,1,1,0,0,0,1,1,1,0,0)
  Chromosome-c(1,1,1,1,1,1,2,2,2,2,2)
  start-c(12,17,18,20,25,36,12,15,16,17,19)
  Table-cbind(Chromosome,start,binary)
       Chromosome start binary
   [1,]          1    12      1
   [2,]          1    17      1
   [3,]          1    18      1
   [4,]          1    20      0
   [5,]          1    25      0
   [6,]          1    36      0
   [7,]          2    12      1
   [8,]          2    15      1
   [9,]          2    16      1
  [10,]          2    17      0
  [11,]          2    19      0
 
  As output I need a shortlist for each binary block: giving 
 me the starting
  and ending position of each block.
  Which for these example would look like this:
      Chromosome2 position_start position_end binary2
  [1,]           1             12           18       1
  [2,]           1             20           36       0
  [3,]           2             12           16       1
  [4,]           2             17           19       0
 
  Part 2:
  Based on the output of part 1, I need to assign the binary 
 to rows of
  another data set. If the position value in this second data 
 set falls in one
  of the blocks defined in the shortlist made in part1,the 
 binary value of the
  shortlist should be assigned to an extra column for this 
 row.  This would
  look something like this:
      Chromosome3 position Value binary3
   [1,] 1         12     a   1
   [2,] 1         13     b   1
   [3,] 1         14     c   1
   [4,] 1         15     d   1
   [5,] 1         16     e   1
   [6,] 1         18     f   1
   [7,] 1         20     g   0
   [8,] 1         21     h   0
   [9,] 1         22     i   0
  [10,] 1         23     j   0
  [11,] 1         25     k   0
  [12,] 1         35     l   0
  [13,] 2         12     m   1
  [14,] 2         13     n   1
  [15,] 2         14     o   1
  [16,] 2         15     p   1
  [17,] 2         16     q   1
  [18,] 2         17     s   0
  [19,] 2         18     d   0
  [20,] 2         19     f   0
 
 
  Many thanks in advance,
 
  Niels
 
  --
  View this message in context: 
 http://r.789695.n4.nabble.com/Record-row-values-every-time-the
-binary-value-in-a-collumn-changes-tp3462496p3462496.html
  Sent from the R help mailing list archive at Nabble.com.
 
  __
  R-help@r-project.org mailing list
  https://stat.ethz.ch/mailman/listinfo/r-help
  PLEASE do read the posting guide 
 http://www.R-project.org/posting-guide.html
  and provide commented, minimal, self-contained, reproducible code.

Re: [R] How can I 'predict' from an nls model with a fit specified for separate groups?

2011-04-20 Thread peter dalgaard


On Apr 20, 2011, at 17:04 , Stuart Rosen wrote:

 Following an example on p 111 in 'Nonlinear Regression with R' by Ritz  
 Streibig, I have been fitting nls models using square brackets with the 
 grouping variable inside. In their book is this example, in which 'state' is 
 a factor indicating whether a treatment has been used or not:
 
  Puromycin.m1 - nls(rate ~ Vm[state] *
 + conc/(K[state] + conc), data = Puromycin,
 + start = list(K = c(0.1, 0.1),
 + Vm = c(200, 200)))
 
 What I cannot figure out is how to specify the value of the grouping variable 
 in a 'predict' statement. In my own example, I can only seem to get the 
 predictions for the 1st specified level of the grouping variable. I promise 
 that I have read the documentation, and have tried a number of things, but 
 cannot get the correct predictions.

What's the problem? If I continue your example with

plot(Puromycin$conc, predict(Puromycin.m1), col=(1:2)[Puromycin$state])

I get a black and a red set of points, distinctly different.

It's difficult to say what you did wrong in the number of things that you 
tried, but I suspect you didn't set up your newdata properly:

 myconc - seq(0,1.2,,11) lv - levels(Puromycin$state)
 predict(Puromycin.m1, 
 newdata=data.frame(conc=myconc,state=factor(untreated,levels=lv)))
 [1]   0. 114.6850 133.7022 141.5248 145.7897 148.4743 150.3196 151.6661
 [9] 152.6918 153.4993 154.1514
 predict(Puromycin.m1, 
 newdata=data.frame(conc=myconc,state=factor(treated,levels=lv)))
 [1]   0. 138.6154 167.8413 180.5289 187.6203 192.1490 195.2916 197.6000
 [9] 199.3674 200.7641 201.8956


-- 
Peter Dalgaard
Center for Statistics, Copenhagen Business School
Solbjerg Plads 3, 2000 Frederiksberg, Denmark
Phone: (+45)38153501
Email: pd@cbs.dk  Priv: pda...@gmail.com

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Re: [R] 'Record' row values every time the binary value in a collumn changes

2011-04-20 Thread Dennis Murphy

Hi:

Here are a couple more options using packages plyr and data.table. The
labels in the second part are changed because they didn't make sense
in a 2M line file (well, mine may not either, but it's a start). You
can always change them to something more pertinent.

# Question 1:
Table - data.frame(binary, chromosome = Chromosome, start)

library(plyr)
(df - ddply(Table, .(chromosome, binary), summarise, position_start =
min(start),
 position_end = max(start)))
  chromosome binary position_start position_end
1  1  0 20   36
2  1  1 12   18
3  2  0 17   19
4  2  1 12   16

library(data.table)
dTable - data.table(Table, key = 'chromosome, binary')
(dt - dTable[, list(position_start = min(start),
   position_end = max(start)), by = 'chromosome, binary'])
 chromosome binary position_start position_end
[1,]  1  0 20   36
[2,]  1  1 12   18
[3,]  2  0 17   19
[4,]  2  1 12   16

## Question 2:

For plyr, it's easy to write a function that takes a generic input data frame
(in this case, a single line) and then outputs a data frame with
positions and labels.

tfun - function(df) {
 diff - with(df, position_end - position_start + 1)
 position - with(df, seq(position_start, position_end))
 value - paste(df$chromosome, df$binary, letters[1:diff], sep = '.')
 data.frame(chromosome = df$chromosome, position, value, binary = df$binary)
}

# Then:

 ddply(df, .(chromosome, binary), tfun)
   chromosome position value binary
1   1   20 1.0.a  0
2   1   21 1.0.b  0
3   1   22 1.0.c  0
4   1   23 1.0.d  0
5   1   24 1.0.e  0
6   1   25 1.0.f  0
7   1   26 1.0.g  0
8   1   27 1.0.h  0
9   1   28 1.0.i  0
10  1   29 1.0.j  0
11  1   30 1.0.k  0
12  1   31 1.0.l  0
13  1   32 1.0.m  0
14  1   33 1.0.n  0
15  1   34 1.0.o  0
16  1   35 1.0.p  0
17  1   36 1.0.q  0
18  1   12 1.1.a  1
19  1   13 1.1.b  1
20  1   14 1.1.c  1
21  1   15 1.1.d  1
22  1   16 1.1.e  1
23  1   17 1.1.f  1
24  1   18 1.1.g  1
25  2   17 2.0.a  0
26  2   18 2.0.b  0
27  2   19 2.0.c  0
28  2   12 2.1.a  1
29  2   13 2.1.b  1
30  2   14 2.1.c  1
31  2   15 2.1.d  1
32  2   16 2.1.e  1

# For data.table, one can apply the internals of tfun directly:

dt[, list(chromosome = chromosome, position = seq(position_start, position_end),
value = paste(chromosome, binary,
  letters[1:(position_end - position_start + 1)],
sep = '.'),
binary = binary), by = 'chromosome, binary']
   chromosome binary chromosome.1 position value binary.1
1  01   20 1.0.a0
1  01   21 1.0.b0
1  01   22 1.0.c0
1  01   23 1.0.d0
1  01   24 1.0.e0
1  01   25 1.0.f0
1  01   26 1.0.g0
1  01   27 1.0.h0
1  01   28 1.0.i0
1  01   29 1.0.j0
1  01   30 1.0.k0
1  01   31 1.0.l0
1  01   32 1.0.m0
1  01   33 1.0.n0
1  01   34 1.0.o0
1  01   35 1.0.p0
1  01   36 1.0.q0
1  11   12 1.1.a1
1  11   13 1.1.b1
1  11   14 1.1.c1
1  11   15 1.1.d1
1  11   16 1.1.e1
1  11   17 1.1.f1
1  11   18 1.1.g1
2  02   17 2.0.a0
2  02   18 2.0.b0
2  02   19 2.0.c0
2  12   12 2.1.a1
2  12   13 2.1.b1
2  12   14

Re: [R] Two Questions

2011-04-20 Thread Greg Snow

When running a large number of commands from a script that produces multiple 
plots it is often best to send the plots to the pdf device (or other system) 
that you can then page through after it is finished.  You could also specify 
par(ask=TRUE) then you would be prompted before changing the plot (but other 
code would not execute either).

-- 
Gregory (Greg) L. Snow Ph.D.
Statistical Data Center
Intermountain Healthcare
greg.s...@imail.org
801.408.8111

 -Original Message-
 From: r-help-boun...@r-project.org [mailto:r-help-bounces@r-
 project.org] On Behalf Of Stephen P Molnar
 Sent: Wednesday, April 20, 2011 7:23 AM
 To: R-help
 Subject: [R] Two Questions
 
 Sorry for the somewhat nondescript subject line, but I have two
 questions:
 
 
 
 1.What is a really good book on R for a nonprogrammer?
 
 2.   How do I open more than one R Graphics: Device 2(ACTIVE).
 That
 what is the R command that I can use to keep more than one plot open.
 I am
 running a script from a book on Chemometrics that results in more than
 one
 graph during the execution, but it seems that R deletes each graph when
 the
 script calls for the next plot.
 
 
 
 Thanks in advance
 
 
 
 Stephen P. Molnar, Ph.D.  Life
 is a
 fuzzy set
 
 Foundation for Chemistry
 Stochastic
 and multivariate
 
 http://www.FoundationForChemistry.com
 
 
 
 
   [[alternative HTML version deleted]]
 
 __
 R-help@r-project.org mailing list
 https://stat.ethz.ch/mailman/listinfo/r-help
 PLEASE do read the posting guide http://www.R-project.org/posting-
 guide.html
 and provide commented, minimal, self-contained, reproducible code.

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Re: [R] 'Record' row values every time the binary value in a collumn changes

2011-04-20 Thread Phil Spector


Here's one way to do part 1:


rr = rle(Table[,'binary'])
cc = cumsum(rr$lengths)+1
thestarts =  c(1,cc[cc=nrow(Table)])
theends = cc-1
answer = 
cbind(Table[thestarts,'Chromosome'],Table[thestarts,'start'],Table[theends,'start'],rr$values)
answer

 [,1] [,2] [,3] [,4]
[1,]1   12   181
[2,]1   20   360
[3,]2   12   161
[4,]2   17   190

If I understand you correctly, here's a way to do part 2:


Next = 
matrix(c(rep(1,12),rep(2,8),c(12,13,14,15,16,18,20,21,22,23,25,35,12,13,14,15,16,17,18,19)),ncol=2)
apply(Next,1,function(x)answer[answer[,1]==x[1]  x[2] = answer[,2]  x[2] = 
answer[,3],4])

 [1] 1 1 1 1 1 1 0 0 0 0 0 0 1 1 1 1 1 0 0 0

- Phil Spector
 Statistical Computing Facility
 Department of Statistics
 UC Berkeley
 spec...@stat.berkeley.edu












On Wed, Apr 20, 2011 at 5:01 AM, baboon2010 nielsvande...@live.be wrote:

My question is twofold.

Part 1:
My data looks like this:

(example set, real data has 2*10^6 rows)
binary-c(1,1,1,0,0,0,1,1,1,0,0)
Chromosome-c(1,1,1,1,1,1,2,2,2,2,2)
start-c(12,17,18,20,25,36,12,15,16,17,19)
Table-cbind(Chromosome,start,binary)
     Chromosome start binary
 [1,]          1    12      1
 [2,]          1    17      1
 [3,]          1    18      1
 [4,]          1    20      0
 [5,]          1    25      0
 [6,]          1    36      0
 [7,]          2    12      1
 [8,]          2    15      1
 [9,]          2    16      1
[10,]          2    17      0
[11,]          2    19      0

As output I need a shortlist for each binary block: giving me the starting
and ending position of each block.
Which for these example would look like this:
    Chromosome2 position_start position_end binary2
[1,]           1             12           18       1
[2,]           1             20           36       0
[3,]           2             12           16       1
[4,]           2             17           19       0

Part 2:
Based on the output of part 1, I need to assign the binary to rows of
another data set. If the position value in this second data set falls in one
of the blocks defined in the shortlist made in part1,the binary value of the
shortlist should be assigned to an extra column for this row.  This would
look something like this:
    Chromosome3 position Value binary3
 [1,] 1         12     a   1
 [2,] 1         13     b   1
 [3,] 1         14     c   1
 [4,] 1         15     d   1
 [5,] 1         16     e   1
 [6,] 1         18     f   1
 [7,] 1         20     g   0
 [8,] 1         21     h   0
 [9,] 1         22     i   0
[10,] 1         23     j   0
[11,] 1         25     k   0
[12,] 1         35     l   0
[13,] 2         12     m   1
[14,] 2         13     n   1
[15,] 2         14     o   1
[16,] 2         15     p   1
[17,] 2         16     q   1
[18,] 2         17     s   0
[19,] 2         18     d   0
[20,] 2         19     f   0


Many thanks in advance,

Niels

--
View this message in context: 
http://r.789695.n4.nabble.com/Record-row-values-every-time-the-binary-value-in-a-collumn-changes-tp3462496p3462496.html
Sent from the R help mailing list archive at Nabble.com.

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--
Jim Holtman
Data Munger Guru

What is the problem that you are trying to solve?

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Re: [R] Random Relabelling