[R] recommended way to group function calls in Sweave

2012-04-25 Thread Liviu Andronic
Dear all
When using Sweave, I'm always hitting the same bump: I want to group
repetitive calls in a function, but I want both the results and the
function calls in the printed output. Let me explain myself.

Consider the following computation in an Sweave document:
summary(iris[,1:2])
cor(iris[,1:2])

When using these two calls directly, I obtain the following output:
 summary(iris[,1:2])
  Sepal.LengthSepal.Width
 Min.   :4.300   Min.   :2.000
 1st Qu.:5.100   1st Qu.:2.800
 Median :5.800   Median :3.000
 Mean   :5.843   Mean   :3.057
 3rd Qu.:6.400   3rd Qu.:3.300
 Max.   :7.900   Max.   :4.400
 cor(iris[,1:2])
 Sepal.Length Sepal.Width
Sepal.Length1.000  -0.1175698
Sepal.Width-0.1175698   1.000


However, if I try to group the calls in a function:
f - function(d, ind){
  print(summary(d[ , ind]))
  print(cor(d[ , ind]))
  return(invisible(NULL))
}


Then I get a different output in the Sweave PDF:
 f(iris, 1:2)
  Sepal.LengthSepal.Width
 Min.   :4.300   Min.   :2.000
 1st Qu.:5.100   1st Qu.:2.800
 Median :5.800   Median :3.000
 Mean   :5.843   Mean   :3.057
 3rd Qu.:6.400   3rd Qu.:3.300
 Max.   :7.900   Max.   :4.400
 Sepal.Length Sepal.Width
Sepal.Length1.000  -0.1175698
Sepal.Width-0.1175698   1.000


Of course I can use 'echo=F' to remove the ' f(iris, 1:2)' in the
above, but how can I do to keep the original calls 'summary(d[ ,
ind])' and 'cor(d[ , ind])'? Or even better, could the actual calls be
used, after the replacement of arguments by their values:
'summary(iris[,1:2])' and 'cor(iris[,1:2])'.

Or is the recommended way to use cat()-ed statements:
f - function(d, ind){
  cat('Variable summary:\n')
  print(summary(d[ , ind]))
  cat('\nCorrelation table:\n')
  print(cor(d[ , ind]))
  return(invisible(NULL))
}


To obtain such output:
 f(iris, 1:2)
Variable summary:
  Sepal.LengthSepal.Width
 Min.   :4.300   Min.   :2.000
 1st Qu.:5.100   1st Qu.:2.800
 Median :5.800   Median :3.000
 Mean   :5.843   Mean   :3.057
 3rd Qu.:6.400   3rd Qu.:3.300
 Max.   :7.900   Max.   :4.400

Correlation table:
 Sepal.Length Sepal.Width
Sepal.Length1.000  -0.1175698
Sepal.Width-0.1175698   1.000


What is the recommended way of grouping repetitive function calls?
What do you usually use in your own documents? Regards
Liviu


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Re: [R] recommended way to group function calls in Sweave

2012-04-25 Thread Liviu Andronic
On Wed, Apr 25, 2012 at 3:41 PM, Duncan Murdoch
murdoch.dun...@gmail.com wrote:
 I would use the last method, or if the calls were truly repetitive (i.e.
 always identical, not just the same pattern), use a named chunk.

Labeled chunks are indeed what I was looking for [1]. As far as I
understand, this is what Sweave functions (or are these macros?)
look like:


=
d - iris
ind - 1:2
@

sw=
  summary(d[ , ind])
  cor(d[ , ind])
@

=
d - iris
ind - 2:4
sw
@


Regards
Liviu

[1] vignette('Sweave', 'utils')

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Re: [R] Organizations where IT has approved the use of R software

2012-04-13 Thread Liviu Andronic
On Fri, Apr 13, 2012 at 8:56 AM, Michael Bibo
michael_b...@health.qld.gov.au wrote:
 Have you seen this from the archives?

 http://www.mail-archive.com/r-help@r-project.org/msg45768.html

In addition see [1], which highlights the likes of Google, Merck or
Stanford University.

Liviu

[1] http://www.r-project.org/foundation/memberlist.html

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[R] plyr: set '.progress' argument to default to text

2012-04-10 Thread Liviu Andronic
Dear all
Is it possible to set globally the option .progress = text to all
the apply functions in 'plyr'. For example, current default is
daply(..., .progress = none). I would like to set it to daply(...,
.progress = text), so as to avoid writing the argument every time I
call such a function. I looked into ?daply and ?create_progress_bar
without much luck.

Regards
Liviu


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Re: [R] plyr: set '.progress' argument to default to text

2012-04-10 Thread Liviu Andronic
On Tue, Apr 10, 2012 at 5:32 PM, R. Michael Weylandt
michael.weyla...@gmail.com wrote:
 You might try the Defaults package.


Thanks for the hint. Unfortunately
 library(Defaults)
 setDefaults('create_progress_bar', name = text)

doesn't do the trick. But one can set defaults for each *ply()
function individually:
 setDefaults('ddply', .progress = text)
 setDefaults('daply', .progress = text)

Regards
Liviu

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Re: [R] R generated means are different from the boxplot!

2012-04-06 Thread Liviu Andronic
Hello

On Fri, Apr 6, 2012 at 10:35 AM, Jhope jeanwaij...@gmail.com wrote:
 Hi R-listers,

 1) I am having trouble understanding why the means I have calculated from
 Aeventexhumed (A, B, and C) are different from the means showing on the
 boxplot I generated (see attached).  I have added the script as to how my
 data is organized.

For starters, boxplots do not display means but medians. [1]
[1] http://en.wikipedia.org/wiki/Boxplot


 2) Also when I went through the data manually the means I calculated  for
 each nesting event are slightly different than what I generated through R
 (see below).
 A,  B,  C
 0.2155051, 0.1288241, 0.1124618

Just guessing, perhaps it is due to some missing values.

Regards
Liviu


 Thanks in advance,

 Jean

 ---
 require(plyr)
 Loading required package: plyr
 resp - read.csv(file.choose())
 envir - read.csv(file.choose())
 resp - resp[!is.na(resp$Aeventexhumed), ]
 resp$QuadratEvent - paste(resp$QuadratID, resp$Aeventexhumed, sep=)
 resp$QuadratEvent - as.character(resp$QuadratEvent)
 envir - envir[!is.na(envir$Aeventexhumed), ]
 envir$QuadratEvent - paste(envir$QuadratID, envir$Aeventexhumed, sep=)
 envir$QuadratEvent - as.character(envir$QuadratEvent)
 ExDate - Sector - Quadrat - Aeventexhumed - NULL
 ST1 - ST2 - ST3 - ST4 - ST0 - NULL
 Shells - Hatchlings - MaxHatch - DeadHatch - NULL
 Oldeggs - TotalEggs - QuadratEvent - NULL
 for (q in unique(as.character(resp$QuadratEvent))) {
 +     s - resp[as.character(resp$QuadratEvent) == q, ]
 +     ExDate - c(ExDate, as.character(s$ExDate[1]))
 +     Sector - c(Sector, as.character(s$Sector[1]))
 +     Quadrat - c(Quadrat, as.character(s$Quadrat[1]))
 +     Aeventexhumed - as.character(c(Aeventexhumed,
 as.character(s$Aeventexhumed[1])))
 +     QuadratEvent- c(QuadratEvent, q)
 +     ST1 - c(ST1, sum(s$ST1, na.rm=TRUE))
 +     ST2 - c(ST2, sum(s$ST2, na.rm=TRUE))
 +     ST3 - c(ST3, sum(s$ST3, na.rm=TRUE))
 +     ST4 - c(ST4, sum(s$ST4, na.rm=TRUE))
 +     ST0 - c(ST0, sum(s$ST0, na.rm=TRUE))
 +     Shells - c(Shells, sum(s$Shells, na.rm=TRUE))
 +     Hatchlings - c(Hatchlings, sum(s$Hatchlings, na.rm=TRUE))
 +     MaxHatch - c(MaxHatch, sum(s$MaxHatch, na.rm=TRUE))
 +     DeadHatch - c(DeadHatch, sum(s$DeadHatch, na.rm=TRUE))
 +     Oldeggs - c(Oldeggs, sum(s$Oldeggs, na.rm=TRUE))
 +     TotalEggs - c(TotalEggs, sum(s$TotalEggs, na.rm=TRUE))
 + }
 responses - data.frame(QuadratEvent, ExDate, Sector, Quadrat,
 +                         Aeventexhumed, ST0, ST1, ST2, ST3, ST4, Shells,
 +                         Hatchlings, MaxHatch, DeadHatch, Oldeggs,
 +                         TotalEggs, stringsAsFactors=FALSE)
 responses$QuadratEvent - as.character(responses$QuadratEvent)
 data.to.analyze - join(responses, envir, by=QuadratEvent)
 data.to.analyze$NotHatched - data.to.analyze$TotalEggs -
 data.to.analyze$Shells
 data.to.analyze$Rayos - paste(Rayos, data.to.analyze$Rayos, sep=.)

 Hsuccess - Shells/TotalEggs
 tapply(Hsuccess, Aeventexhumed, mean, na.rm=TRUE)
        A         B         C
 0.2156265 0.1288559 0.1124327
 boxplot(HSuccess ~ Aeventexhumed, data = data.to.analyze, col = blue,
 +         main = Hatching Success of Arribadas in 2010,
 +         xlab = Arribada Event,
 +         ylab = Hatching Success % (Shells / Total Eggs))

 http://r.789695.n4.nabble.com/file/n4536926/hatch_Aeventexhumed.png

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Re: [R] reclaiming lost memory in R

2012-04-06 Thread Liviu Andronic
On Fri, Apr 6, 2012 at 2:21 PM, Ramiro Barrantes
ram...@precisionbioassay.com wrote:
 Please let me know if you have any other suggestions or clues.

See this older post by Brian [1] and check ?Memory-limits.
Otherwise, I remember someone suggesting that even if R releases the
memory internally, the OS may still keep it reserved for the R
process. (Unfortunately I cannot find the reference.)

Regards
Liviu

[1] http://tolstoy.newcastle.edu.au/R/e10/help/10/05/3851.html

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Re: [R] spaghetti plots in R

2012-04-04 Thread Liviu Andronic
On Wed, Apr 4, 2012 at 4:22 PM, uday uday_143...@hotmail.com wrote:
 I would like to plat some spaghetti plots from my data , ma data is as

See:
require(sos)
findFn('spaghetti')


Liviu


 follows
 ak[1:3,]
          [,1]      [,2]      [,3]      [,4]      [,5]      [,6]      [,7]
 [,8]      [,9]
 [1,] 0.3211745 0.4132568 0.5649930 0.6920562 0.7760113 0.8118568 0.8609301
 0.9088819 0.9326736
 [2,] 0.3159234 0.4071270 0.5579212 0.6844584 0.7684690 0.8243702 0.8677043
 0.8931288 0.9261926
 [3,] 0.3075260 0.3993699 0.5493242 0.6765600 0.7614591 0.8127050 0.8537816
 0.8884786 0.9343690
         [,10] [,11]    [,12]
 [1,] 0.9605178     1 1.003940
 [2,] 0.9647617     1 1.012930
 [3,] 0.9618874     1 1.007103
 dim(ak[1:3,])
 [1]  3 12

 pre[1:3,]
         [,1]     [,2]     [,3]     [,4]     [,5]     [,6]     [,7]     [,8]
 [,9]    [,10]
 [1,] 10.34615 52.02116 146.1736 243.2864 347.4150 431.6711 521.4271 629.0045
 729.9594 827.8628
 [2,] 10.34615 52.02539 146.3670 244.3871 350.1785 454.6706 546.5499 638.3344
 741.9849 842.5700
 [3,] 10.34615 52.02754 146.4656 244.9480 351.5865 457.1768 550.1341 643.0880
 748.1114 850.0670
        [,11]    [,12]
 [1,] 921.5508 956.4445
 [2,] 953.9648 995.8201
 [3,] 951.6384 987.9105

 dim(pre) 3 12

 I have tried
 plot(ak[1,],pre[1,],type=l)
 lines(ak[2,],pre[2,],type=l,col=red)
 but it only works for few data, but I have very big list and data is in
 matrix format. I would be very glad if someone can help me to fix this
 issue.


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Re: [R] spaghetti plots in R

2012-04-04 Thread Liviu Andronic
On Wed, Apr 4, 2012 at 5:04 PM, uday uday_143...@hotmail.com wrote:
 Hi Liviu ,
 thanks for post , but I could not find findFn('spaghetti') , I can see the
 following functions in sos package
 Extract.findFn
 findFn

After installing 'sos', use the 'findFn()' function. For example, run
findFn('spaghetti')

See the documentation of ?findFn. Alternatively, look into RcmdrPlugin.sos.

Regards
Liviu

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Re: [R] How to use Latex code in R loop?

2012-04-01 Thread Liviu Andronic
On Sun, Apr 1, 2012 at 4:48 PM, Manish Gupta mandecent.gu...@gmail.com wrote:
 Hi,

 I am newbie in Latex and R. I am working on one report in which i need to
 read file and display content of file by formatting (adding color boxes and
 colorful text for each record). For this i need to use latex code in R loop.
 How can i use Latex code in R loop. Any example will help me a lot.

See Sweave FAQ [1].
Liviu

[1] http://www.statistik.lmu.de/~leisch/Sweave/FAQ.html#x1-11000A.9

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[R] proper order of calls when estimating nested models

2012-03-24 Thread Liviu Andronic
Dear all
How should one proceed when estimating nested models containing
missing data. What I would like to do is to first estimate the model
with the control variables only, and then estimate the model
containing also the variables of interest. For example,
 summary(reg.a - lm(IMC ~ STYLE + SEXE + AGE, imc))

Call:
lm(formula = IMC ~ STYLE + SEXE + AGE, data = imc)

Residuals:
Min  1Q  Median  3Q Max
-10.720  -2.428  -0.550   1.712  32.206

Coefficients:
Estimate Std. Error t value Pr(|t|)
(Intercept) 24.858780.23407 106.202   2e-16 ***
STYLE0.095440.12057   0.7920.429
SEXE-1.829980.11490 -15.926   2e-16 ***
AGE  0.456650.02839  16.087   2e-16 ***
---
Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

Residual standard error: 3.746 on 5969 degrees of freedom
  (576 observations deleted due to missingness)
Multiple R-squared: 0.07776,Adjusted R-squared: 0.07729
F-statistic: 167.8 on 3 and 5969 DF,  p-value:  2.2e-16

 summary(reg.b - update(reg.a, . ~ . + DEMANDEC + LATITUDEC + SUPSUPC))

Call:
lm(formula = IMC ~ STYLE + SEXE + AGE + DEMANDEC + LATITUDEC +
SUPSUPC, data = imc)

Residuals:
Min  1Q  Median  3Q Max
-10.687  -2.424  -0.552   1.677  32.258

Coefficients:
Estimate Std. Error t value Pr(|t|)
(Intercept) 24.916000.24010 103.773   2e-16 ***
STYLE0.101650.12455   0.816   0.4144
SEXE-1.871950.11654 -16.063   2e-16 ***
AGE  0.455430.02932  15.530   2e-16 ***
DEMANDEC 0.220410.10960   2.011   0.0444 *
LATITUDEC   -0.273780.11106  -2.465   0.0137 *
SUPSUPC -0.142350.08169  -1.743   0.0815 .
---
Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

Residual standard error: 3.739 on 5785 degrees of freedom
  (757 observations deleted due to missingness)
Multiple R-squared: 0.08293,Adjusted R-squared: 0.08197
F-statistic: 87.18 on 6 and 5785 DF,  p-value:  2.2e-16


Note that both have been estimated on different samples. If I try the
following I get an error:
 anova(reg.a, reg.b)
Error in anova.lmlist(object, ...) :
  models were not all fitted to the same size of dataset


Should I reverse the order of calls? Should I always first estimate
the complete model, and then the model containing only the control
vars?

Regards
Liviu

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Re: [R] Recommendations regarding textbooks

2012-03-22 Thread Liviu Andronic
On Thu, Mar 22, 2012 at 1:17 PM, richard willey
richard.wil...@gmail.com wrote:
 I was hoping to get some advice regarding teaching R in an academic 
 environment.


 thinks regarding “An R Companion to Applied Regression”.

CAR features an excellent introduction to R programming.

Liviu

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Re: [R] vif function using lm object

2011-12-23 Thread Liviu Andronic
On Wed, Dec 21, 2011 at 9:28 AM, arunkumar akpbond...@gmail.com wrote:
 Hi,

   can anyone please explain why the vif should have more than 2 terms.

 *vif.lm(lmobj) : model contains fewer than 2 terms*

 why it is throwng error if it is one variable.

The _VIF_ is a measure of _multicollinearity_, which occurs between
two or more predictors in a regression model. In other words, you
cannot have collinearity in a single predictor.

Also, to compute the VIF you need to regress one regressor on all the
remaining regressors. If you have a single regressor, then you cannot
estimate a VIF.

See the related articles on Wikipedia.

Regards
Liviu






 -
 Thanks in Advance
        Arun
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Re: [R] Multicollinearty in logistic regression models

2011-12-16 Thread Liviu Andronic
On Fri, Dec 16, 2011 at 1:47 PM, David Winsemius dwinsem...@comcast.net wrote:
 Harrell's rms package has a vif function that is intended for use with fits
 from his logistic regression model function, lrm. This uses the variance

Also see vif() in 'car'.

Liviu

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Re: [R] nice report generator?

2011-12-07 Thread Liviu Andronic
On Wed, Dec 7, 2011 at 7:43 PM, Janko Thyson
janko.thyson.rst...@googlemail.com wrote:
 I like knitr. IMHO Yihui really came up with a killer package there:
 http://yihui.github.com/knitr/

If we're talking about nice, I'll chip in for LyX. It has support for
Sweave, and will soon support knitr. This eliminates (most of) the
burden of learning LaTeX.

Liviu

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Re: [R] how to view/edit large matrix/array in R?

2011-12-05 Thread Liviu Andronic
On Tue, Dec 6, 2011 at 4:42 AM, Michael comtech@gmail.com wrote:
 I didn't mean the speed of saving the file is slow... I meant the manual
 procedures of exporting and then opening Excel, etc. is slow and
 inconvenient and unproductive...

You may want to use RExcel then.

Liviu


 On Mon, Dec 5, 2011 at 9:38 PM, R. Michael Weylandt 
 michael.weyla...@gmail.com michael.weyla...@gmail.com wrote:

 I think what most everyone is getting at is that the visual identification
 of numeric outliers is an exceedingly difficult task and one we humans are
 not well evolved for. Rather they are all suggesting you use visual
 techniques to spot and fix outliers individually. This practice has a long
 and reputable history in statistics and has been shown to be far more
 efficient than simply scanning pages of numbers for a single misplaced
 decimal. In conjunction, I'd also recommend use of the identify() function,
 which serves just this purpose.

 If you have so much data that the csv export is unbearably slow it seems
 unlikely you can check it all by hand.

 Another, more general, methodology if you are worried about data
 corruption is to use robust statistics when applicable.

 Michael

 On Dec 5, 2011, at 10:27 PM, Michael comtech@gmail.com wrote:

  In R-Studio, edit gives non-grid based format which is similar to
 fix...
 
  But exporting to Excel is time-consuming...
 
  On Mon, Dec 5, 2011 at 9:17 PM, jim holtman jholt...@gmail.com wrote:
 
  'edit' does allow you to change it.  If all else fails, export to
  Excel, split the screen and then synchronize the two displays.
 
  On Mon, Dec 5, 2011 at 9:52 PM, Michael comtech@gmail.com wrote:
  View doesn't allow editing? And not multi-window so I cannot put
  variables
  side-by-side for comparsion? Thanks!
 
  On Mon, Dec 5, 2011 at 7:37 PM, jim holtman jholt...@gmail.com
 wrote:
 
  Have you tried
 
  ?View
  ?edit
 
  On Mon, Dec 5, 2011 at 8:12 PM, Bert Gunter gunter.ber...@gene.com
  wrote:
  ... and do you really think perusing thousands of numbers by eye is
  any way to edit/check data?!
 
  Personal viewpoint: I would say that this is a large area of
  statistics and data analysis that the discipline fails to address in
  any systematic way ... perhaps because there is no way to address it
  systematically? Contrary views and corrections -- especially
  references! -- would be very welcome.
 
  My advice would be: graphics! -- but I can't provide anything more
  useful without the specifics of the problem.
 
  Some packages may provide the interactivity you seek -- check the
 CRAN
  GUI task view, R Commander, etc.
 
  -- Bert
 
  On Mon, Dec 5, 2011 at 5:01 PM, Michael comtech@gmail.com
  wrote:
  head, tail and fix commands don't really work well if I have large
  matrix/array for which I would like to be able to scroll up and dow,
  left
  and right ...
 
  Could anybody please help me?
 
  Thanks
 
        [[alternative HTML version deleted]]
 
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  --
 
  Bert Gunter
  Genentech Nonclinical Biostatistics
 
  Internal Contact Info:
  Phone: 467-7374
  Website:
 
 
 
 http://pharmadevelopment.roche.com/index/pdb/pdb-functional-groups/pdb-biostatistics/pdb-ncb-home.htm
 
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  --
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  Data Munger Guru
 
  What is the problem that you are trying to solve?
  Tell me what you want to do, not how you want to do it.
 
 
 
 
 
  --
  Jim Holtman
  Data Munger Guru
 
  What is the problem that you are trying to solve?
  Tell me what you want to do, not how you want to do it.
 
 
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Re: [R] Help! Big problem when using browser() to do R debugging?

2011-12-03 Thread Liviu Andronic
On Sat, Dec 3, 2011 at 2:38 AM, Michael comtech@gmail.com wrote:
 Hi all,

 Could you please help me?

 I am having the following weird problem when debugging R programs
 using browser():

 In my function, I've inserted a browser() in front of Step 1. My
 function has 3 steps and at the end of each step, it will print out
 the message Step i is done...

 However, after I hit ENTER when the program stopped before Step 1
 and entered into the debugging mode, it not only executed the next
 line(i.e. the Step 1), but also all the (many) remaining lines in that
 function, as shown below:


 Browse[1]
 [1] Step 1 is done..
 [1] Step 2 is done..
 [1] Step 3 is done..

 Then it automatically quited the debugging mode and when I tried to
 check the value of myobj, I've got the following error message:

 names(myobj)
 Error: object 'myobj' not found
 No suitable frames for recover()

 

 So my question is: why did one key stroke ENTER lead it to execute
 all the remaining lines in that function and then returned from the
 function and quited the debugging mode?

As soon as you get into 'debug' mode, type 'n', hit enter, then you
can simply hit enter to evaluate the fun step by step. To exit the
step-by-step, type 'c' and hit enter: this will evaluate the rest of
the function.

Regards
Liviu


 Thanks a lot!

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Re: [R] detecting autocorrelation structure in panel data

2011-11-28 Thread Liviu Andronic
On Mon, Nov 28, 2011 at 3:54 AM, Matteo Richiardi
matteo.richia...@gmail.com wrote:
 Hello,
 I'm a newby in R. I have created a data.frame holding panel data, with

Take a look at the 'plm' package and its vignette.

Regards
Liviu


 the following columns: id,time,y, say:

 periods = 100
 numcases = 100
 df = data.frame(
  id = rep(1:numcases,periods),
  time = rep(1:periods, each = numcases)
 )
 df = transform(df,y=c(rnorm(numcases*periods)+id)

 I want to check whether y is autocorrelated. I came across the acf()
 function, but I cannot understand how to specify that my time variable
 is time and that individuals are identified by their id. I have
 also tried to convert my data.frame in a ts or a xts object, but got
 really lost there. I'd really apreciate help...

 Thank you,
 Matteo

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Re: [R] First read (was: Re: Looping and paste)

2011-11-24 Thread Liviu Andronic
On Thu, Nov 24, 2011 at 9:27 AM, Patrick Burns pbu...@pburns.seanet.com wrote:
 It's very seldom that I disagree with
 Bert, but here is one time.

 I don't think An Introduction to R is
 a suitable first read for people with
 little computational experience.

I must agree with Patrick here. The 'Intro to R' may be appropriate
for someone already versed in statistics and/or programming, but it is
hardly useful as a first read for the neophytes.


 Better (I modestly suggest) would be:

 http://www.burns-stat.com/pages/Tutor/hints_R_begin.html

To chip in, the first two chapters of Fox and Weisberg (2011) make for
an excellent introduction to R programming. It's gentle, but also
covers many of the difficulties and misunderstandings that one would
encounter in R.

Regards
Liviu




 which includes some other references.
 'Hints' is imperfect and incomplete but
 it suffers slightly less from the curse of
 knowledge than a lot of other R documentation.

 Pat

 On 24/11/2011 00:15, Bert Gunter wrote:

 ... and you can of course do the assignment:

 Bndy-  paste (BndY,to,50+seq_len(BndY), mN, sep =  )

 An Introduction to R tells you about such fundamentals and should be
 a first read for anyone learning R.

 --- Bert

 On Wed, Nov 23, 2011 at 4:10 PM, Bert Gunterbgun...@gene.com  wrote:

 Don't do this!  paste() is vectorized.

 paste (BndY,to,50+seq_len(BndY), mN, sep =  )

 Cheers,
 Bert

 On Wed, Nov 23, 2011 at 3:31 PM, B77Sbps0...@auburn.edu  wrote:

 out- vector(list)
 Ylab- for(i in 1:length(BndY))
 {
 out[i]- paste(BndY[i], to ,BndY[i],mN)
 }

 Ylab- do.call(c, out)






 markm0705 wrote

 Dear R helpers

 I'm trying to make up some labels for plot from this vector

 BndY-seq(from = 18900,to= 19700, by = 50)

 using

 Ylab-for(i in BndY) {c((paste(i, to ,i+50,mN)))}

 but the vector created is NULL

 However if i use

 for(i in BndY) {print(c(paste(i, to ,i+50,mN)))}

 I can see the for loop is making the labels I'm looking for but not
 sure
 on my error in assigning them to a vector

 Thanks in advance



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 --

 Bert Gunter
 Genentech Nonclinical Biostatistics

 Internal Contact Info:
 Phone: 467-7374
 Website:

 http://pharmadevelopment.roche.com/index/pdb/pdb-functional-groups/pdb-biostatistics/pdb-ncb-home.htm





 --
 Patrick Burns
 pbu...@pburns.seanet.com
 twitter: @portfolioprobe
 http://www.portfolioprobe.com/blog
 http://www.burns-stat.com
 (home of 'Some hints for the R beginner'
 and 'The R Inferno')

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[R] understanding all.equal() output: Mean relative difference

2011-11-24 Thread Liviu Andronic
Dear all
How should one parse all.equal() output? I'm specifically referring to
the 'mean relative difference' messages. For example,
 all.equal(pi, 355/113)
[1] Mean relative difference: 8.491368e-08

But I'm not sure how to understand these messages. When they're close
to 0 (or 1xe-16), then it's intuitive. But when they're big,
 all.equal(1, 4)
[1] Mean relative difference: 3
 all.equal(2, 4)
[1] Mean relative difference: 1
 all.equal(3, 4)
[1] Mean relative difference: 0.333

the messages start making much less sense. I tried Wikipedia [1], but
the description is cryptic, as is the help page. Also, Fox and
Weisberg (2011) don't explain this particular message.

Regards
Liviu

[1] http://en.wikipedia.org/wiki/Mean_difference#Relative_mean_difference


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[R] proper work-flow with 'formula' objects and lm()

2011-11-24 Thread Liviu Andronic
Dear all
I have a work-flow issue with lm(). When I use
 lm(y1~x1, anscombe)

Call:
lm(formula = y1 ~ x1, data = anscombe)

Coefficients:
(Intercept)   x1
 3.0001   0.5001


I get as expected the formula, y1 ~ x1, in the print()ed results or
summary(). However, if I pass through a formula object
 (form - formula(y1~x1))
y1 ~ x1
 lm(form, anscombe)

Call:
lm(formula = form, data = anscombe)

Coefficients:
(Intercept)   x1
 3.0001   0.5001


then I only get the object name in the call, 'form', instead of the
formula. When passing a 'formula' object to lm, is it possible to
retain in the resulting 'lm' object the actual formula used for the
regression: y1 ~ x1?

Regards
Liviu


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Re: [R] proper work-flow with 'formula' objects and lm()

2011-11-24 Thread Liviu Andronic
On Thu, Nov 24, 2011 at 9:14 PM, Duncan Murdoch
murdoch.dun...@gmail.com wrote:
 It is retained.  terms(fit) will give it to you, if fit is an lm object.


Thank you. The following works nicely.
 (form - formula(y1~x1))
y1 ~ x1
 x - lm(form, anscombe)
 formula(terms(x))
y1 ~ x1


However, I was hoping that there was a way to input the 'form' object
so that summary(x) would print the underlying formula used, not the
formula object name. I was thinking of something in the style of
x - lm(deparse(form), anscombe)
summary(x)

Can this be done? Regards
Liviu

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Re: [R] proper work-flow with 'formula' objects and lm()

2011-11-24 Thread Liviu Andronic
On Thu, Nov 24, 2011 at 10:25 PM, Prof Brian Ripley
rip...@stats.ox.ac.uk wrote:
 Yes.  That's a job for substitute (the second time today).

 form - formula(y1~x1)
 x - eval(substitute(lm(f, anscombe), list(f = form)))
 summary(x)

 Call:
 lm(formula = y1 ~ x1, data = anscombe)

That's what I wanted. Thanks!

However, I do want to simplify the syntax and define a new function:
x.lm -
  function(formula, data, ...)
{
  eval(substitute(lm(f, data, ...), list(f = formula)))
}

For the simple case it works just fine
 (form - formula(y1~x1))
y1 ~ x1
 x - x.lm(form, anscombe)

But it fails when I try to input more lm() arguments:
 (x - x.lm(form, anscombe, subset=-5))
Error in eval(expr, envir, enclos) :
  ..1 used in an incorrect context, no ... to look in

Am I doing something obviously wrong? Regards
Liviu

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Re: [R] proper work-flow with 'formula' objects and lm()

2011-11-24 Thread Liviu Andronic
On Thu, Nov 24, 2011 at 11:55 PM, Prof Brian Ripley
rip...@stats.ox.ac.uk wrote:
 You would get exactly the same problem with ...,, anway.

 Here's a commonly used approach in R sources:

 x.lm - function(formula, data, ...)
 {
    Call - match.call(expand.dots = TRUE)
    Call[[1]] - as.name(lm)
    Call$formula - as.formula(terms(formula))
    eval(Call)
 }

From limited testing this seems to do exactly what I wanted!

I thank all those who came up with suggestions. Regards
Liviu




 On Thu, 24 Nov 2011, Liviu Andronic wrote:

 On Thu, Nov 24, 2011 at 10:25 PM, Prof Brian Ripley
 rip...@stats.ox.ac.uk wrote:

 Yes.  That's a job for substitute (the second time today).

 form - formula(y1~x1)
 x - eval(substitute(lm(f, anscombe), list(f = form)))
 summary(x)

 Call:
 lm(formula = y1 ~ x1, data = anscombe)

 That's what I wanted. Thanks!

 However, I do want to simplify the syntax and define a new function:
 x.lm -
  function(formula, data, ...)
 {
  eval(substitute(lm(f, data, ...), list(f = formula)))
 }

 For the simple case it works just fine

 (form - formula(y1~x1))

 y1 ~ x1

 x - x.lm(form, anscombe)

 But it fails when I try to input more lm() arguments:

 (x - x.lm(form, anscombe, subset=-5))

 Error in eval(expr, envir, enclos) :
  ..1 used in an incorrect context, no ... to look in

 Am I doing something obviously wrong? Regards
 Liviu


 --
 Brian D. Ripley,                  rip...@stats.ox.ac.uk
 Professor of Applied Statistics,  http://www.stats.ox.ac.uk/~ripley/
 University of Oxford,             Tel:  +44 1865 272861 (self)
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Re: [R] [OT] 1 vs 2-way anova technical question

2011-11-22 Thread Liviu Andronic
On Tue, Nov 22, 2011 at 2:09 PM, Giovanni Azua brave...@gmail.com wrote:
 Mr. Gunter did not read/understand my problem, and there were no useful tips 
 but only ad hominem attacks. By your side-taking I suspect you are in the 
 same party club if you want to defend him maybe you should start by tying 
 better your dog so to speak.

I believe that most of the readers of this thread got put off by your
offending and misplaced remarks. To echo other posters, it would be
nice to get your e-mail address banned from the list.

Regards
Liviu

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Re: [R] changelog for MASS?

2011-11-17 Thread Liviu Andronic
On Thu, Nov 17, 2011 at 7:33 AM, R. Michael Weylandt
michael.weyla...@gmail.com wrote:
 Hmmm...sorry -- the only thing I can suggest is maybe striking some
 sort of deal that you change when it gets however many months out of
 date:  if you look here
 (http://cran.r-project.org/src/contrib/Archive/MASS/), you can see the
 last time each version of MASS was updated and by seeing what you

You can also get this info on crantastic (scroll to 'prev versions'):
http://crantastic.org/packages/MASS

Regards
Liviu


 have, you can see about how out of date you are. In the context of
 MASS, I wouldn't worry so much: it's tied to a book, not active
 research, so I don't think it gets updated too often in big ways.

 The other thing is to actually compare differences in the source code,
 though that might be more trouble than it's worth.

 Michael

 On Mon, Nov 14, 2011 at 4:30 PM, Xu Wang xuwang...@gmail.com wrote:
 Thanks Michael,

 But I can't see the dates on the NEWS so I have no idea what changed from
 last version or from whichever version we actually have installed. Do you
 see what I mean?

 Thanks,

 Xu

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Re: [R] Running R with browser without installing anything

2011-10-20 Thread Liviu Andronic
On Thu, Oct 20, 2011 at 7:29 PM, Bogaso Christofer
bogaso.christo...@gmail.com wrote:
 Dear all, the company I work for has Matlab installed for
 statistical/mathematical calculations and really not ready to go with R
 (even installing exe file for R). Therefore I was wondering is it possible
 to do analysis R using browser like IE, without installing anything?

Look into RStudio Server.
Liviu




 Thanks for your suggestion.






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Re: [R] R and Forex

2011-10-11 Thread Liviu Andronic
On Wed, Oct 12, 2011 at 3:29 AM, Yves S. Garret
yoursurrogate...@gmail.com wrote:
 Hi all,

   I recently started learning about Forex and found this O'Reilly book in
 Barnes  Nobles about R.  I bought it out of pure curiosity.  I like what I
 see.  However, I have a question.  Has anyone tried to bring these two ideas
 together in a financial and trading sense?  Are there any libraries or
 modules in R that can aid in this venture?


 fortune('equity')

I have never heard anyone (knowledgable or otherwise) claim that, in the
absence of transition costs, SAS is better than R for equity modeling. If you
come across any such claim, I would be happy to refute it.
   -- David Kane
  R-SIG-Finance (December 2004)


You may want to address this question to r-sig-finance, and check out
the Finance Task View [1]. Regards
Liviu

[1] http://cran.at.r-project.org/web/views/Finance.html


 --Yves

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Re: [R] Understanding the workflow between sweave, R and Latex

2011-10-03 Thread Liviu Andronic
On Fri, Sep 30, 2011 at 2:41 PM, Duncan Murdoch
murdoch.dun...@gmail.com wrote:
 As an aside, I don't recommend the workflow you describe:  it's very slow
 and cumbersome.  It's much better to tell your text editor how to run both
 Sweave and Latex in one command.  In the upcoming release of R 2.14.0, this

Another approach is to use LyX. The latest stable release comes with
an Sweave module that provides out-of-the box support for Sweave
documents. Once everything is configured, and on a Mac it should be
fairly straightforward in this case, then compiling documents is
usually a matter of pressing a button or activating a key combination.
LyX takes care of a lot of automation for you, including BibTeX et al.

Regards
Liviu

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[R] extracting p-values in scientific notation

2011-10-03 Thread Liviu Andronic
Dear all
How does print.htest display the p-value in scientific notation?
 (x - cor.test(iris[[1]], iris[[3]]))

Pearson's product-moment correlation

data:  iris[[1]] and iris[[3]]
t = 21.65, df = 148, p-value  2.2e-16
alternative hypothesis: true correlation is not equal to 0
95 percent confidence interval:
 0.8270 0.9055
sample estimates:
   cor
0.8718

Above the p-value comes as ' 2.2e-16', while inspecting the object I
get a good old '0'.
 x$p.value
[1] 0

I tried to inspect print.htest but couldn't find it. I also played
with format, round and the like to no avail. Any pointers?

Regards
Liviu


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Re: [R] extracting p-values in scientific notation

2011-10-03 Thread Liviu Andronic
Thanks all for your pointers. The following does trick:
 base::format.pval(x$p.value)  ##Hmisc also has such a function
[1] 2e-16


On Mon, Oct 3, 2011 at 10:25 AM, Rolf Turner rolf.tur...@xtra.co.nz wrote:
 Isn't it true that 0  2.2e-16?


Yes, but it doesn't mean that the p-value actually hits absolute zero.
And cor.test, as Ted noticed, returns
 identical(x$p.value, 0)
[1] TRUE

Not that this makes a great practical difference in my case, but I
would still prefer to print 2e-16 in my Sweave document.

Regards
Liviu

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Re: [R] Supporting R/Membership

2011-09-25 Thread Liviu Andronic
On Sat, Sep 24, 2011 at 5:43 PM, Joshua Wiley jwiley.ps...@gmail.com wrote:
 I printed the form to mail off, but anecdotally I had planned on this
 for well over a year and kept putting it off.  If this is something
 that useRs and the R Foundation would be interested in, I would be
 happy to help work on setting something up and maintaining it.

If the R Core accepts this, I would very much prefer to have an
on-line system to make donations. Personally I would like to make a
donation, but I'm uncomfortable with sending my credit card details by
mail.

Regards
Liviu

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Re: [R] Reading R Code aloud

2011-09-14 Thread Liviu Andronic
On Wed, Sep 14, 2011 at 4:50 AM, cwdillon cwdil...@gmail.com wrote:
 Well, I've bee reading the R Manual, Nathan Yau's book on Flowing Data, and a
 few others like this to my kids to get them to go to sleep. I started off
 doing it the *wrong* way, just to keep it boring but slipped into doing it
 the *correct* way for my own sanity. It puts the kids to sleep in no time
 and I actually get some meaningful reading in. The familiar narrative style
 and topics such as hotdog (hdb) eating contests are not as effective as R
 manuals and getting the kids to sleep; even with the code.

This is a pearl! May I nominate this as a fortune?
Liviu

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Re: [R] Solve your R problems

2011-09-12 Thread Liviu Andronic
On Mon, Sep 12, 2011 at 9:41 AM, Patrick Burns pbu...@pburns.seanet.com wrote:
 R-help is all about solving R problems.
 So here ya go:
 http://www.portfolioprobe.com/2011/09/12/solve-your-r-problems/

Sweet. :)

May I suggest a font change: anything but the default CM should do the
trick. For one I prefer Palatino  Optima (you'll most likely have
access to Palladio or Pagella, and Classico), but it tends to be heavy
on longer documents. You could use Aldum, if you have access to it,
instead of Palatino: it's the book-design version of Palatino. I hear
that Minion Pro fonts are very good (again, if you have access), and
there are some very free alternatives such as Libertine  Biolinum.

Although the default CM tend to be very hard to read, I guess they
could fit in handily with the theme of your document. Regards
Liviu

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Re: [R] Latex + R + sweave

2011-09-12 Thread Liviu Andronic
On Mon, Sep 12, 2011 at 9:21 AM, Twaha Mlwilo uddessy2...@hotmail.com wrote:

 Hello all,
 Good day,
 I have problem on how to remove the source code from the pdf output.Here I 
 mean this.
  code in sweave Rnw files

  =
  x-c(1,2,3,4,5,6)
 x
  mean(x)
 sd(x)
 @
 then would like it appear as
 mean = 3.5
 sd=1.3
 x=1,2,3,4,5,6

Try this:
echo=F=
  x-c(1,2,3,4,5,6)
 x
  mean(x)
 sd(x)
 @

See ?RweaveLatex. Also, try
mean = \Sexpr{mean(x)}

in your document. You could also play with ?paste in the code chunk,
or with generating a matrix with appropriate row names, and printing
that.

Regards
Liviu

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Re: [R] problem with plm package

2011-09-03 Thread Liviu Andronic
(please reply to the list, too, as it increases your chances of
getting a helpful answer)


On Sat, Sep 3, 2011 at 2:51 AM, loth lorien lothlorie...@hotmail.com wrote:
 Hi Liviu,

 Thank you for your suggestions, but they didn't seem to work.

 I got R to estimate a fixed effects model for me:
 formula-plm(RoE~RoA+NIM+Cash.TL+CAR+NPL.Loans, data=ComBank,
 model=within)

 But when I try to estimate a random effects model for a panel data set I
 still get the error message:  missing value where TRUE/FALSE needed

I am suspecting that something may relate to the data. Do you have
persistent variables that change seldom? Maybe that's causing a
problem for the 'random' estimator. Is your panel balanced? If not,
does it contain very few observations for some ids?

To make sure that the functions work as expected on your system, try
to replicate some of the examples in the 'plm' vignette.

Regards
Liviu


 There must be something simple that I am missing, but at the moment, I can't
 figure out what it is.

 Thanks again,

 Ashraf

 From: landronim...@gmail.com
 Date: Thu, 1 Sep 2011 10:37:57 +0200
 Subject: Re: [R] problem with plm package
 To: lothlorie...@hotmail.com
 CC: r-help@r-project.org

 On Thu, Sep 1, 2011 at 4:25 AM, Ash lothlorie...@hotmail.com wrote:
  Hi,
 
  I am trying to complete a very simple panel analysis on some bank data.
 
  Call:
  Formula-plm(RoE~RoA+CAR+Inc.Dep+Cash.TL+NPL.Loans, data=Banks1,
  model=random, index=c(Bank.I.D.,Year))
  summary(Formula)
 
  I get the following error code:
 
  ERROR:
   missing value where TRUE/FALSE needed
 
  Does anyone know what true/false field I am missing?
 
 I am not sure what goes wrong, but try first to
 Banks1.p - pdata.frame(Banks1, index=c(Bank.I.D.,Year)

 and see if that went fine, and then
 Banks1.fit - plm(RoE~RoA+CAR+Inc.Dep+Cash.TL+NPL.Loans,
 data=Banks1.p, model=random)

 It may help if you posted
 str(Banks1)

 Liviu




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Re: [R] problem with plm package

2011-09-01 Thread Liviu Andronic
On Thu, Sep 1, 2011 at 4:25 AM, Ash lothlorie...@hotmail.com wrote:
 Hi,

 I am trying to complete a very simple panel analysis on some bank data.

 Call:
 Formula-plm(RoE~RoA+CAR+Inc.Dep+Cash.TL+NPL.Loans, data=Banks1,
 model=random, index=c(Bank.I.D.,Year))
 summary(Formula)

 I get the following error code:

 ERROR:
  missing value where TRUE/FALSE needed

 Does anyone know what true/false field I am missing?

I am not sure what goes wrong, but try first to
Banks1.p - pdata.frame(Banks1, index=c(Bank.I.D.,Year)

and see if that went fine, and then
Banks1.fit - plm(RoE~RoA+CAR+Inc.Dep+Cash.TL+NPL.Loans,
data=Banks1.p, model=random)

It may help if you posted
str(Banks1)

Liviu

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Re: [R] !!!function to do the knn!!!

2011-09-01 Thread Liviu Andronic
I would nominate the following fortune:


On Thu, Sep 1, 2011 at 12:16 PM, Petr PIKAL petr.pi...@precheza.cz wrote:
 Try to call 112 if you are in Europe.


David Winsemius: Thank your for your entry in the Poorly Capitalized
and Inadequately Searched Posting Contest.
mark: help, help ,help!!!
Petr PIKAL: Try to call 112 if you are in Europe.

Liviu

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Re: [R] Treat an Unquoted Character String as a Data Frame

2011-09-01 Thread Liviu Andronic
On Thu, Sep 1, 2011 at 5:48 PM, Jean V Adams jvad...@usgs.gov wrote:
 Try this

 data - eval(parse(text=paste(study, level, ., population, sep=)))

..or this:
data - get(paste(study, level, ., population, sep=))

Liviu


 Jean

 -

 dbateman wrote on 08/31/2011 17:44:44:

 I have several datasets that come from different studies (fv02 and fv03),
 they represent different levels (patients and lesions), and they have
 different patient populations (itt, mitt, mitt3).  I wanted to write some
 code that would pass my three requirements into a function I wrote,
 produce the output, but not have to require me to also pass a unique plot
 title or output filename for each function call.  The title and file name
 would be created in the function according to the three input parameters.

 My datasets are named like this (all six are repeated for fv03 in place
 of fv02):
   fv02patients.itt
   fv02patients.mitt
   fv02patients.mitt3
   fv02lesions.itt
   fv02lesions.mitt
   fv02lesions.mitt3

 Taking the first dataset as an example, I currently have this code:
   study=fv02
   level=patients
   population=itt
   noquote(paste(study,level,.,population,sep=))

 This produces the desired fv02patients.itt, but it is just an unquoted
 character string rather than a data frame.

 Here is a condensed look at my function:

 waterfall=function(data,title,file)
 {       barplot(height=data$brpercent,main=paste(Best Change in Tumor
 Volume\n,title)),
        savePlot(filename=file,type=pdf)
 }
 waterfall(data=fv02patients.itt,title=EC-FV-02 ITT
 Patients,file=fv02_itt_patients_waterfall)


 It may be easier to leave it the way I have it, but if this is possible, I
 would still be interested in knowing.

        [[alternative HTML version deleted]]

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Re: [R] how to split a data frame by two variables

2011-09-01 Thread Liviu Andronic
On Thu, Sep 1, 2011 at 7:53 PM, Changbin Du changb...@gmail.com wrote:
 HI, Dear R community,

 I want to split a data frame by using two variables: let and g

It's not clear what you want to do, but investigate the following:

 require(plyr)
Loading required package: plyr
 ddply(x, .(let, g), function(y) mean(y$g))
   let g V1
1a 1  1
2a 2  2
3b 1  1
4b 2  2
5c 1  1
6c 2  2
7d 1  1
8d 2  2
9e 1  1
10   e 2  2

This splits the df in groups of unique 'let' and 'g', and computes the
mean for each such group.

Regards
Liviu

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Re: [R] Grouping variables in a data frame

2011-08-27 Thread Liviu Andronic
On Sat, Aug 27, 2011 at 7:26 AM, Andra Isan andra_i...@yahoo.com wrote:
 Hi All,

 I have a data frame as follow:

 user_id time age location gender
 .

 and I learn a logistic regression to learn the weights (glm with family= 
 (link = logit))), my response value is either zero or one. I would like to 
 group the users based on user_id and time and see the y values and predicted 
 y values at the same time. Or plot them some how. Is there any way to somehow 
 group them together so that I can learn more about my data by grouping them?

It's very difficult to help you because you haven't followed the
posting guide. But I suspect you're looking for the following:

 require(plyr)
Loading required package: plyr
 data(mtcars)
 ##considering 'gear' as 'id' and 'carb' as time
 ddply(mtcars, .(gear, carb), function(x) mean(x$hp))
   gear carbV1
1 31 104.0
2 32 162.5
3 33 180.0
4 34 228.0
5 41  72.5
6 42  79.5
7 44 116.5
8 52 102.0
9 54 264.0
1056 175.0
1158 335.0

This will compute the mean of 'hp' for each group of id  time.
Liviu


 I would like to get these at the end
 user_id time y predicted_y

 Thanks a lot,
 Andra

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Re: [R] obtaining p-values for lm.ridge() coefficients (package 'MASS')

2011-08-24 Thread Liviu Andronic
Dear Michael
Thank you for your pointers.


On Tue, Aug 23, 2011 at 4:05 PM, Michael Friendly frien...@yorku.ca wrote:
 First, you should be using rms::ols, as Design is old.

Good to know. I've always wondered why Design and rms, in many cases,
were providing similar functions and (upon cursory inspection)
identical documentation.


 Second, penalty in ols() is not the same as the ridge constant in lm.ridge,
 but rather a penalty on the log likelihood.  The documentation
 for ols refers to ?lrm for the description.

I see. At the same time ?rms::ols contains: The penalized maximum
likelihood estimate (penalized least squares or ridge estimate) of
beta is (X'X + P)^{-1} X'Y., while ?rms::lrm defines P as penalty
\times diag(pf) \times penalty.matrix \times diag(pf), where pf is the
vector of square roots of penalty factors computed from penalty. This
is suspiciously similar to the definition of the classical Ridge
Regression beta estimates (X'X + kI)^{-1} X'Y, where k is the lambda
constant and I the identity matrix.

Hence I was wondering if rms::ols could be forced to use 'P = kI', and
thus compute the classical Ridge estimates? I tried to hack rms::ols
to ensure that it passed 'kI' to lm.pfit(..., penalty.matrix = ...),
but I get strange results so I must be missing something. (See
attached.)


 Third, other ridge estimators (e.g., ElemStatLearn::simple.ridge also
 give slightly different results for the same data due to differences in the
 way scaling is handled.

And ?ElemStatLearn::prostate points to several other 'ridge'-related functions.

However, similar to MASS::lm.ridge, simple.ridge() roughly stops at
estimating the betas, hence my question: is the purpose of Ridge
Regression to stabilize the coefficients (when the design matrix is
collinear) and use the results to infer the relative importance of the
coefficients on the response, while keeping the unpenalized model for
prediction purposes? Or am I missing something?

Thank you
Liviu
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Re: [R] obtaining p-values for lm.ridge() coefficients (package 'MASS')

2011-08-24 Thread Liviu Andronic
The attachment seems to have been dropped, so I'm pasting the code
below. Sorry for that
Liviu


On Wed, Aug 24, 2011 at 1:44 PM, Liviu Andronic landronim...@gmail.com wrote:
 Second, penalty in ols() is not the same as the ridge constant in lm.ridge,
 but rather a penalty on the log likelihood.  The documentation
 for ols refers to ?lrm for the description.

 I see. At the same time ?rms::ols contains: The penalized maximum
 likelihood estimate (penalized least squares or ridge estimate) of
 beta is (X'X + P)^{-1} X'Y., while ?rms::lrm defines P as penalty
 \times diag(pf) \times penalty.matrix \times diag(pf), where pf is the
 vector of square roots of penalty factors computed from penalty. This
 is suspiciously similar to the definition of the classical Ridge
 Regression beta estimates (X'X + kI)^{-1} X'Y, where k is the lambda
 constant and I the identity matrix.

 Hence I was wondering if rms::ols could be forced to use 'P = kI', and
 thus compute the classical Ridge estimates? I tried to hack rms::ols
 to ensure that it passed 'kI' to lm.pfit(..., penalty.matrix = ...),
 but I get strange results so I must be missing something. (See
 attached.)


ols.ridge -
function (formula, data, weights, subset, na.action = na.delete,
method = qr, model = FALSE, x = FALSE, y = FALSE, se.fit = FALSE,
linear.predictors = TRUE, penalty = 0, penalty.matrix, tol = 1e-07,
sigma = NULL, var.penalty = c(simple, sandwich), ridge=TRUE, ...)
{
call - match.call()
var.penalty - match.arg(var.penalty)
m - match.call(expand = FALSE)
mc - match(c(formula, data, subset, weights, na.action),
names(m), 0)
m - m[c(1, mc)]
m$na.action - na.action
m$drop.unused.levels - TRUE
m[[1]] - as.name(model.frame)
if(ridge) lambda - penalty
if (length(attr(terms(formula), term.labels))) {
dul - .Options$drop.unused.levels
if (!length(dul) || dul) {
on.exit(options(drop.unused.levels = dul))
options(drop.unused.levels = FALSE)
}
X - Design(eval.parent(m))
options(drop.unused.levels = dul)
atrx - attributes(X)
atr - atrx$Design
nact - atrx$na.action
Terms - atrx$terms
assig - DesignAssign(atr, 1, Terms)
penpres - FALSE
if (!missing(penalty)  any(unlist(penalty) != 0))
penpres - TRUE
if (!missing(penalty.matrix)  any(penalty.matrix !=
0))
penpres - TRUE
if (penpres  missing(var.penalty))
warning(default for var.penalty has changed to \simple\)
if (method == model.frame)
return(X)
scale - as.character(formula[2])
attr(Terms, formula) - formula
weights - model.extract(X, weights)
if (length(weights)  penpres)
stop(may not specify penalty with weights)
Y - model.extract(X, response)
n - length(Y)
if (model)
m - X
X - model.matrix(Terms, X)
if (length(atr$colnames))
dimnames(X)[[2]] - c(Intercept, atr$colnames)
else dimnames(X)[[2]] - c(Intercept, dimnames(X)[[2]][-1])
if (method == model.matrix)
return(X)
}
else {
if (length(weights))
stop(weights not implemented when no covariables are present)
assig - NULL
yy - attr(terms(formula), variables)[1]
Y - eval(yy, sys.parent(2))
nmiss - sum(is.na(Y))
if (nmiss == 0)
nmiss - NULL
else names(nmiss) - as.character(yy)
Y - Y[!is.na(Y)]
yest - mean(Y)
coef - yest
n - length(Y)
if (!length(sigma))
sigma - sqrt(sum((Y - yest)^2)/(n - 1))
cov - matrix(sigma * sigma/n, nrow = 1, ncol = 1, dimnames =
list(Intercept,
Intercept))
fit - list(coefficients = coef, var = cov, non.slopes = 1,
fail = FALSE, residuals = Y - yest, df.residual = n -
1, intercept = TRUE)
if (linear.predictors) {
fit$linear.predictors - rep(yest, n)
names(fit$linear.predictors) - names(Y)
}
if (model)
fit$model - m
if (x)
fit$x - matrix(1, ncol = 1, nrow = n, dimnames = list(NULL,
Intercept))
if (y)
fit$y - Y
fit$fitFunction - c(ols, lm)
oldClass(fit) - c(ols, rms, lm)
return(fit)
}
if (!penpres) {
fit - if (length(weights))
lm.wfit(X, Y, weights, method = method, ...)
else lm.fit(X, Y, method = method, ...)
cov.unscaled - chol2inv(fit$qr$qr)
r - fit$residuals
yhat - Y - r
if (length(weights)) {
sse - sum(weights * r^2)
m - sum(weights * yhat/sum(weights))
ssr - sum(weights * (yhat - m)^2)
r2 - ssr/(ssr + sse)
if (!length(sigma))
sigma - sqrt(sse/fit$df.residual

Re: [R] debugging functions in R

2011-08-24 Thread Liviu Andronic
On Wed, Aug 24, 2011 at 4:20 PM, Eran Eidinger e...@taykey.com wrote:
 Hi,

 I am not sure if this is the right list to ask this question (though I did
 not find a more appropriate one).
 I've started using R a month ago, and small scripts work fine. However, when
 I start writing more complex code, it gets messy.

 1. Is there any way to debug normally, with breakpoints?


 fortune('browser')

My solution when I run into mysteries like this is to put 'browser()' in the
function just before or after the line of interest. The magnitude and direction
of my stupidity usually become clear quickly.
   -- Patrick Burns
  R-help (February 2006)


Use browser() to inspect the environment and execute the code one step
at a time.
Liviu


2. I am using the Eclipse plugin (StatET), and tried JGR(). Is there an IDE
 that enables breakpoints?
 3. Is there an equivalent to include in other programming languages? So
 many functions in one file are very messy. I would like to break it to
 several files.
 4. Any way to create a local context of variables inside a function?
 Otherwise I have to be careful to give different names inside functions, to
 those in the workspace.

 I should point that I am a long time Matlab user and am probably expecting
 some things that don't necessarily exist in R...

 I know it's a lot, if there is a more appropriate forum to ask these, please
 point me in that direction.

 Thanks,
 Eran.

 *

 *

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[R] obtaining p-values for lm.ridge() coefficients (package 'MASS')

2011-08-23 Thread Liviu Andronic
Dear all
I'm familiarising myself with Ridge Regressions in R and the following
is bugging me: How does one get p-values for the coefficients obtained
from MASS::lm.ridge() output (for a given lambda)? Consider the
example below (adapted from PRA [1]):

 require(MASS)
 data(longley)
 gr - lm.ridge(Employed ~ .,longley,lambda = seq(0,0.1,0.001))
 plot(gr)
 select(gr)
modified HKB estimator is 0.004275
modified L-W estimator is 0.0323
smallest value of GCV  at 0.003
 ##let's choose 0.03 for lambda
 coef(gr)[gr$lam == 0.03,]
 GNP.deflator  GNP   Unemployed Armed.Forces
Population Year
-1620.429355 0.021060 0.007994-0.013146-0.007752
-0.101880 0.869116


But how does one obtain the customary 'lm' summary information for the
model above? I tried supplying the chosen lambda to Design::ols()
using its 'penalty' argument, but strangely the results differ. See
below.

 require(Design)
 Design::ols(Employed ~ GNP.deflator + GNP + Unemployed + Armed.Forces + 
 Population + Year,data=longley,penalty = 0.03,tol=1e-12)

Linear Regression Model

Design::ols(formula = Employed ~ GNP.deflator + GNP + Unemployed +
Armed.Forces + Population + Year, data = longley, penalty = 0.03,
tol = 1e-12)

 n Model L.R.   d.f. R2  Sigma
16  78.22   7.05 0.9929 0.3438

Residuals:
 Min   1Q   Median   3Q  Max
-0.42480 -0.17774 -0.02169  0.16834  0.77203

Coefficients:
Value Std. Error   t Pr(|t|)
Intercept-1580.022688 518.501881 -3.0473 0.016006
GNP.deflator 0.023161   0.068862  0.3363 0.745322
GNP  0.008351   0.015160  0.5509 0.596839
Unemployed  -0.013057   0.002651 -4.9255 0.001178
Armed.Forces-0.007691   0.002097 -3.6671 0.006406
Population  -0.096757   0.144240 -0.6708 0.521355
Year 0.847932   0.269614  3.1450 0.013812

Adjusted R-Squared: 0.9866

 ##the above seems more similar to a lambda of 0.032 than the one required 
 (0.03)
 coef(gr)[gr$lam %in% c(0.03, 0.032),]  ##lm.ridge output
GNP.deflator  GNP Unemployed Armed.Forces Population   Year
0.030 -1620  0.02106 0.007994   -0.01315-0.007752   -0.10188 0.8691
0.032 -1580  0.02316 0.008351   -0.01306-0.007691   -0.09676 0.8479


The difference between the coefficients of MASS::lm.ridge and
Design::ols is small, but they're different nonetheless and I'm
wondering if I'm doing something wrong. Perhaps Design::ols uses a
penalty different from the one specified? Alternatively, is there some
other way to perform diagnostics on Ridge Regression models?

Thank you
Liviu

[1] http://cran.r-project.org/doc/contrib/Faraway-PRA.pdf

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Re: [R] How to navigate (zoom, pan) in a plot/graph

2011-08-21 Thread Liviu Andronic
On Sun, Aug 21, 2011 at 3:04 PM, Eran Eidinger e...@taykey.com wrote:
 Thank you Uwe,

 That solves the second question.
 Still looking for some solution to zooming in and out dynamically.

Try package 'playwith'.
Liviu

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[R] Hmisc::rcorr on a 'data.frame'?

2011-08-19 Thread Liviu Andronic
Dear all
?Hmisc::rcorr states that it takes as main argument a numeric
matrix. But is it normal that it fails in such an ugly way on a data
frame? (See below.) If the function didn't attempt any conversion to a
matrix, I would have expected it to state that in the error message
that it didn't accept 'data.frame' objects in its input. Also, I
vaguely remember having used in the past rcorr() on data frames.

Regards
Liviu

 require(Hmisc)
 rcorr(mtcars[ , 1:4])
Error in storage.mode(x) - if (.R.) double else single :
  (list) object cannot be coerced to type 'double'
 rcorr(as.matrix(mtcars[ , 1:4]))
   mpg   cyl  disphp
mpg   1.00 -0.85 -0.85 -0.78
cyl  -0.85  1.00  0.90  0.83
disp -0.85  0.90  1.00  0.79
hp   -0.78  0.83  0.79  1.00

n= 32


P
 mpg cyl disp hp
mpg   0   00
cyl   0   00
disp  0   00
hp0   0   0




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[R] display only the top-right half of a correlation matrix?

2011-08-19 Thread Liviu Andronic
Dear all
Is there an easy way to display only one half (top-right or
bottom-left) of a correlation matrix?

 require(Hmisc)
 rcorr(as.matrix(mtcars[ , 1:4]))
   mpg   cyl  disphp
mpg   1.00 -0.85 -0.85 -0.78
cyl  -0.85  1.00  0.90  0.83
disp -0.85  0.90  1.00  0.79
hp   -0.78  0.83  0.79  1.00

n= 32


P
 mpg cyl disp hp
mpg   0   00
cyl   0   00
disp  0   00
hp0   0   0


Since the two sides are identical, there is little value in having
both displayed at the same time. Moreover, it considerably slows down
the inspection of the results.

Thank you
Liviu



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Re: [R] display only the top-right half of a correlation matrix?

2011-08-19 Thread Liviu Andronic
On Fri, Aug 19, 2011 at 9:02 PM, Peter Langfelder
peter.langfel...@gmail.com wrote:
 Use as.dist: here's an example.

Seems promising, but for one issue: I would like to keep the diagonal
and thus specify 'diag=T', but then as.dist() replaces the diagonal
values with zero. (See below.) Is there a way to prevent it from doing
that? Either keep the original values, or not display anything in the
diagonal (as for the upper part)?

Regards
Liviu


 (xb - rcorr(as.matrix(mtcars[ , 1:4])))
   mpg   cyl  disphp
mpg   1.00 -0.85 -0.85 -0.78
cyl  -0.85  1.00  0.90  0.83
disp -0.85  0.90  1.00  0.79
hp   -0.78  0.83  0.79  1.00

n= 32


P
 mpg cyl disp hp
mpg   0   00
cyl   0   00
disp  0   00
hp0   0   0
 round(as.dist(xb$r, T), 2)
   mpg   cyl  disphp
mpg   0.00
cyl  -0.85  0.00
disp -0.85  0.90  0.00
hp   -0.78  0.83  0.79  0.00


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Re: [R] display only the top-right half of a correlation matrix?

2011-08-19 Thread Liviu Andronic
On Fri, Aug 19, 2011 at 9:38 PM, Peter Langfelder
peter.langfel...@gmail.com wrote:
 if as.dist doesn't work, use brute force:

 x = matrix(rnorm(5*100), 100, 5)
 mat = signif(cor(x), 2);
 mat[lower.tri(mat)] = 

 data.frame(mat)

Yes, brute force works. This isn't quite how I wanted to do this, but
the following seems to work for me.

Thanks all
Liviu

require(Hmisc)
print.rcorr -
function (x, upper=FALSE, ...)
{
r - format(round(x$r, 2))
if(!is.null(upper)) r[if(!upper) upper.tri(r) else lower.tri(r)] - ''
print(data.frame(r))
n - x$n
if (all(n == n[1, 1]))
cat(\nn=, n[1, 1], \n\n)
else {
cat(\nn\n)
print(n)
}
cat(\nP\n)
P - x$P
P - ifelse(P  0.0001, 0, P)
p - format(round(P, 4))
p[is.na(P)] - 
if(!is.null(upper)) p[if(!upper) upper.tri(p) else lower.tri(p)] - ''
print(p, quote = FALSE)
invisible()
}

 (xb - rcorr(as.matrix(mtcars[ , 1:4])))
   mpg   cyl  disphp
mpg   1.00
cyl  -0.85  1.00
disp -0.85  0.90  1.00
hp   -0.78  0.83  0.79  1.00

n= 32


P
 mpg cyl disp hp
mpg
cyl   0
disp  0   0
hp0   0   0

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Re: [R] round() a data frame containing 'character' variables?

2011-08-11 Thread Liviu Andronic
On Wed, Aug 10, 2011 at 5:52 PM, Jean V Adams jvad...@usgs.gov wrote:

 As it says in

 ?format

 the digits argument specifies ... how many significant digits are to be
 used ... enough decimal places will be used so that the smallest (in
 magnitude) number has this many significant digits ...

 In your example, the last value in column a is 0.06348058 which is written
 as 0.063 to two significant digits.

I did inspect the help page (several times), but the significant
digits formulation was a little too cryptic to me. Now I understand
better, thanks.


 There is no way (that I know of) to make the format() function do the same
 sort of thing as round().

 If digits won't meet your needs, you could try something like this

 data.frame(lapply(x, function(y) if(is.numeric(y)) round(y, 2) else y))

Quite awkward, but this is more or less what I was looking for. Again, thanks.

Regards
Liviu

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Re: [R] round() a data frame containing 'character' variables?

2011-08-11 Thread Liviu Andronic
Hello

On Thu, Aug 11, 2011 at 4:34 PM, Martin Maechler
maech...@stat.math.ethz.ch wrote:
 Well, as you say it's  View() only, so why don't you just
 view the result of

 something like
          print(data, digits = 2)

Unfortunately print(..., digits = 2) suffers of the same deficiencies
as format(..., digits = 2): they both display a minimal number of
_significant_ digits. What I would need, however, is to specify the
number of decimal places as in round(..., digits = 2).


 (it does correspond to  signif(), rather than round(),
  but I think that that's more sensible anyway).

Although I do appreciate the rationale behind ?signif, for several
subjective reasons---easiness of inspecting round()-ed data or
aesthetics in printing round()-ed data in Sweave (as in I cannot send
format()-ed data to my advisor)---I would prefer to ?round my data.

Is there any chance that 'base' would feature a round() function that
doesn't stumble when processing non-numeric data (simply by ignoring
it), or a format() or print() function with a 'round=NULL' argument?

Regards
Liviu

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Re: [R] round() a data frame containing 'character' variables?

2011-08-11 Thread Liviu Andronic
On Thu, Aug 11, 2011 at 4:58 PM, David Winsemius dwinsem...@comcast.net wrote:
 numVars - sapply(iris, is.numeric)
 iris[numVars] - lapply(iris[numVars], round, digits = 2)
 head(iris)

 Then drop the assignment operation and just print:

 ( lapply(iris[numVars], round, digits = 2) )

That's a nice catch, thanks. The only problem with this approach is
that non-numeric variables are simply discarded, while I would prefer
for them to be simply ignored.


 Or set your digits argument to print:

 print( head(iris), digits=1)

 Or did I miss something?

As I explained in today's reply to Martin, print(..., digits = 2) and
format(..., digits = 2) both display a minimal number of
_significant_ digits. However, for various reasons, I need to specify
the number of decimal places as in round(..., digits = 2). Both
print() and format() don't provide an obvious way to do that.

Regards
Liviu

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Re: [R] round() a data frame containing 'character' variables?

2011-08-11 Thread Liviu Andronic
On Wed, Aug 10, 2011 at 5:52 PM, Jean V Adams jvad...@usgs.gov wrote:
 data.frame(lapply(x, function(y) if(is.numeric(y)) round(y, 2) else y))

Building on this, I came up with a 'data.frame' method for round().
I'm not sure how orthodox it is, but it seems to do the job as
intended:
round.data.frame -
function(x, digits = 0)
{
data.frame(lapply(x, function(y) if(is.numeric(y)) round(y, digits) else y))
}

Here's an example:
 set.seed(2)
 x - data.frame(a=rnorm(10), b=rnorm(10), c=rnorm(10), d=letters[1:10])
 print(x, digits=2)  ##it displays 2, 3 or 4 digits, instead of the desired 2
   a  b   c d
1  -0.90  0.418  2.0908 a
2   0.18  0.982 -1.1999 b
3   1.59 -0.393  1.5896 c
4  -1.13 -1.040  1.9547 d
5  -0.08  1.782  0.0049 e
6   0.13 -2.311 -2.4517 f
7   0.71  0.879  0.4772 g
8  -0.24  0.036 -0.5966 h
9   1.98  1.013  0.7922 i
10 -0.14  0.432  0.2896 j
 round(x, 2)
   a b c d
1  -0.90  0.42  2.09 a
2   0.18  0.98 -1.20 b
3   1.59 -0.39  1.59 c
4  -1.13 -1.04  1.95 d
5  -0.08  1.78  0.00 e
6   0.13 -2.31 -2.45 f
7   0.71  0.88  0.48 g
8  -0.24  0.04 -0.60 h
9   1.98  1.01  0.79 i
10 -0.14  0.43  0.29 j

Would such a method get any chance of inclusion in 'base'? Regards
Liviu

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Re: [R] convert 'list' to 'vector'?

2011-08-11 Thread Liviu Andronic
On Wed, Aug 10, 2011 at 11:39 PM, David Winsemius
dwinsem...@comcast.net wrote:
 is.vector(x)
 [1] TRUE

 You probably didn't realize that lists _are_ vectors before this.

Nice to know.


 Hopefully this expected invitation to grilling will not result in severe
 burns. I note that one of the *apply family would have been successful in
 returning the desired object in conjunction with c() :

 rapply(x, c)
  [1] a b c d e f g h i j k l

 `rapply` is designed to do a complete traversal of the list structure and
 its default how argument is unlist. It was linked from the page where
 you were looking at `simplify2array`.

This also does the job nicely. As does

 c(x, recursive=T)
 [1] a b c d e f g h i j k l

per Peter's suggestion. Thanks all for all the solutions.

Regards
Liviu

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[R] round() a data frame containing 'character' variables?

2011-08-10 Thread Liviu Andronic
Dear all
It is difficult to use round(..., digits=2) on a data frame since one
has to first take care to remove non-numeric variables such as
'character' or 'factor':
 head(round(iris, 2))
Error in Math.data.frame(list(Sepal.Length = c(5.1, 4.9, 4.7, 4.6, 5,  :
  non-numeric variable in data frame: Species
 head(round(iris[1:4], 2))
  Sepal.Length Sepal.Width Petal.Length Petal.Width
1  5.1 3.5  1.4 0.2
2  4.9 3.0  1.4 0.2
3  4.7 3.2  1.3 0.2
4  4.6 3.1  1.5 0.2
5  5.0 3.6  1.4 0.2
6  5.4 3.9  1.7 0.4

Is there an elegant way to use round() on a data frame containing
'character' variables without removing them? Thank you
Liviu



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Re: [R] round() a data frame containing 'character' variables?

2011-08-10 Thread Liviu Andronic
Hello

On Wed, Aug 10, 2011 at 11:41 AM, Dimitris Rizopoulos
d.rizopou...@erasmusmc.nl wrote:
 One approach is the following:

 numVars - sapply(iris, is.numeric)
 iris[numVars] - lapply(iris[numVars], round, digits = 2)
 head(iris)

That's interesting, but still doesn't do what I need. Since it's a
read-only View() operation I would like to avoid at all cost modifying
the original data frame. And since my data frames are relatively big,
I would like to avoid generating unnecessary copies. Basically I would
need round() to ignore objects that it knows it cannot handle.

Any other ideas? Thanks
Liviu

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Re: [R] round() a data frame containing 'character' variables?

2011-08-10 Thread Liviu Andronic
Hello

On Wed, Aug 10, 2011 at 2:31 PM, Jean V Adams jvad...@usgs.gov wrote:
 The function format() might serve your needs.

This looks very promising, but yields some strange results. See below:

 x - data.frame(a=rnorm(10), b=rnorm(10), c=rnorm(10), d=letters[1:10])
 x
 a   bc d
1   0.54114449 -0.11195580  1.526279364 a
2   3.27109063  0.50848249 -0.215760332 b
3  -0.27064475 -1.04749725  0.082319811 c
4  -0.06638611 -0.58600572  0.004148253 d
5  -0.06170739 -0.37885203  0.689125494 e
6   0.53211363 -0.09150913 -0.463972307 f
7  -0.43314431 -0.28981614 -0.973410994 g
8   0.52137857 -1.15077343  0.163120205 h
9  -1.39581552  1.27378389  0.136708313 i
10  0.06348058 -0.00369746 -0.570214119 j
 format(x, digits=2)  ##it displays 3 or 4 digits, instead of the required 2
a   b   c d
1   0.541 -0.1120  1.5263 a
2   3.271  0.5085 -0.2158 b
3  -0.271 -1.0475  0.0823 c
4  -0.066 -0.5860  0.0041 d
5  -0.062 -0.3789  0.6891 e
6   0.532 -0.0915 -0.4640 f
7  -0.433 -0.2898 -0.9734 g
8   0.521 -1.1508  0.1631 h
9  -1.396  1.2738  0.1367 i
10  0.063 -0.0037 -0.5702 j
 format(x, digits=2, nsmall=1, scientific=7)  ##no change when setting related 
 arguments
a   b   c d
1   0.541 -0.1120  1.5263 a
2   3.271  0.5085 -0.2158 b
3  -0.271 -1.0475  0.0823 c
4  -0.066 -0.5860  0.0041 d
5  -0.062 -0.3789  0.6891 e
6   0.532 -0.0915 -0.4640 f
7  -0.433 -0.2898 -0.9734 g
8   0.521 -1.1508  0.1631 h
9  -1.396  1.2738  0.1367 i
10  0.063 -0.0037 -0.5702 j
 round(x[1:3], digits=2)  ##works as expected
   a b c
1   0.54 -0.11  1.53
2   3.27  0.51 -0.22
3  -0.27 -1.05  0.08
4  -0.07 -0.59  0.00
5  -0.06 -0.38  0.69
6   0.53 -0.09 -0.46
7  -0.43 -0.29 -0.97
8   0.52 -1.15  0.16
9  -1.40  1.27  0.14
10  0.06  0.00 -0.57

Any ideas why format() and round() give so different results? Can
format() be set to behave similarly to round? Regards
Liviu

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[R] convert 'list' to 'vector'?

2011-08-10 Thread Liviu Andronic
Dear all
How does one convert a non-symmetric list to a vector? See below:

 x - list()
 x[[1]] - letters[1:5]
 x[[2]] - letters[6:10]
 x[[3]] - letters[11:12]
 x
[[1]]
[1] a b c d e

[[2]]
[1] f g h i j

[[3]]
[1] k l

 paste(x)
[1] c(\a\, \b\, \c\, \d\, \e\) c(\f\, \g\, \h\,
\i\, \j\)
[3] c(\k\, \l\)
 as.vector(x)
[[1]]
[1] a b c d e

[[2]]
[1] f g h i j

[[3]]
[1] k l

 simplify2array(x)
[[1]]
[1] a b c d e

[[2]]
[1] f g h i j

[[3]]
[1] k l


What I would need to get instead is:
 letters[1:12]
 [1] a b c d e f g h i j k l

Any ideas? Regards
Liviu


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Re: [R] convert 'list' to 'vector'?

2011-08-10 Thread Liviu Andronic
On Wed, Aug 10, 2011 at 9:02 PM, R. Michael Weylandt
michael.weyla...@gmail.com wrote:
 unlist()

Thanks all! This is perfect, and very R-ish: never where a novice
would expect it to be.

Cheers
Liviu

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Re: [R] convert 'list' to 'vector'?

2011-08-10 Thread Liviu Andronic
On Wed, Aug 10, 2011 at 9:32 PM, David Winsemius dwinsem...@comcast.net wrote:
 Thanks all! This is perfect, and very R-ish: never where a novice
 would expect it to be.

 Well, since `unlist` is linked in the See Also on the help page for `list`,
 I can only hope you meant that in complete jest.

More or less. I would have expected that to transform a 'list' into a
'vector' I should look into 'as.vector' (or its See Also), and I would
have never guessed to look for 'unlist'.

R documentation is sometimes (often?) hard to parse, and when learning
R more often than not you're looking in the wrong place. But yes, it
was intended as humour (although I did expect to get grilled).


 As the Posting Guide says: ... sometimes `read the manual' is the
 appropriate response.

I did, but I was on the wrong track. It actually hasn't occurred to me
to check ?list, but See Also in both ?as.vector and ?simplify2array
does not link to 'unlist'. Since these are the two places where I
turned to in the first place, and I have also played extensively with
sapply(..., simplify=...) arguments, and there was nothing obvious in
their respective See Also, I figured that I did my homework reasonably
well.

Regards
Liviu

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Re: [R] Very silent R

2011-08-07 Thread Liviu Andronic
On Fri, Aug 5, 2011 at 3:54 PM, khadeeja ismail haj...@yahoo.com wrote:
 other attached packages:
 [1] Rcmdr_1.6-2

I bet that this is what generates the lack of error messages. Check
Rcmdr documentation on how to turn it off.
Liviu

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Re: [R] Life Cycle Assessment with R.

2011-07-26 Thread Liviu Andronic
On Mon, Jul 25, 2011 at 8:03 PM, Jose Bustos Melo jbustosm...@yahoo.es wrote:
 Who knows if there's people working in Life Cycle Assesment (carbon emition) 
 with R? or If there's someone interested in doing a package about it, please 
 let me know!

Some of the places to check would be:
rseek.org
http://cran.at.r-project.org/web/packages/
http://cran.at.r-project.org/web/views/ (which includes an
'Environmetrics' View)

I would also check Bioconductor.
http://en.wikipedia.org/wiki/Bioconductor

Regards
Liviu

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[R] add label attribute to objects?

2011-07-21 Thread Liviu Andronic
Dear all
I know that the R way of documenting things is to work on your project
in package development mode, and document each object (such as data
frames) in a *.Rd files. This should work for gurus. What about a
simpler way to document things, geared for mere mortals?

I was thinking of a label() or tag() function that could store and
retrieve an alphanumeric comment for a given object (for example,
either on 'iris' or on 'iris$Species'). Using 'sos' I found some
potential candidates.
 require(sos)
 xb - findFn('label attribute')
found 307 matches;  retrieving 16 pages
2 3 4 5 6 7 8 9 10 11 12 13 14 15 16
 grepFn( 'label', xb)


There is Hmisc::label, which seems to address my concern, but I cannot
get it to work on data frames. Am I doing something wrong here?
 require(Hmisc)
 #works fine on vectors, but loses the label upon conversion
 age - c(21,65,43)
 y   - 1:3
 label(age) - Age in Years
 str(age)
Class 'labelled'  atomic [1:3] 21 65 43
  ..- attr(*, label)= chr Age in Years
 #fails on df
 x - iris
 label(x)
Sepal.Length  Sepal.Width Petal.Length  Petal.Width  Species
  
Warning message:
In mapply(FUN = label, x = x, default = default, MoreArgs = list(self
= TRUE),  :
  longer argument not a multiple of length of shorter
 label(x) - 'some random comment'
 label(x)
Sepal.Length  Sepal.Width Petal.Length  Petal.Width  Species
  
Warning message:
In mapply(FUN = label, x = x, default = default, MoreArgs = list(self
= TRUE),  :
  longer argument not a multiple of length of shorter
 #however the label attribute exists!!
 str(x)
'data.frame':   150 obs. of  5 variables:
 $ Sepal.Length: num  5.1 4.9 4.7 4.6 5 5.4 4.6 5 4.4 4.9 ...
 $ Sepal.Width : num  3.5 3 3.2 3.1 3.6 3.9 3.4 3.4 2.9 3.1 ...
 $ Petal.Length: num  1.4 1.4 1.3 1.5 1.4 1.7 1.4 1.5 1.4 1.5 ...
 $ Petal.Width : num  0.2 0.2 0.2 0.2 0.2 0.4 0.3 0.2 0.2 0.1 ...
 $ Species : Factor w/ 3 levels setosa,versicolor,..: 1 1 1 1
1 1 1 1 1 1 ...
 - attr(*, label)= chr some random comment


I've also found mefa::label(), which actually does what it advertises,
but as 'label' in Hmisc it has one significant drawback: Inheritance
is not implemented, so by any transformation of the object into
another one, the label is lost.
 #works fine on vectors, but loses the label upon conversion
 (x - 1:10)
 [1]  1  2  3  4  5  6  7  8  9 10
 (label(x) - x is a vector)
[1] x is a vector
 str(x)
 atomic [1:10] 1 2 3 4 5 6 7 8 9 10
 - attr(*, label)= chr x is a vector
 str(as.numeric(x))
 num [1:10] 1 2 3 4 5 6 7 8 9 10
 #works fine on df, but loses the label upon conversion
 x - iris
 (label(x) - x is a vector)
[1] x is a vector
 str(x)
'data.frame':   150 obs. of  5 variables:
 $ Sepal.Length: num  5.1 4.9 4.7 4.6 5 5.4 4.6 5 4.4 4.9 ...
 $ Sepal.Width : num  3.5 3 3.2 3.1 3.6 3.9 3.4 3.4 2.9 3.1 ...
 $ Petal.Length: num  1.4 1.4 1.3 1.5 1.4 1.7 1.4 1.5 1.4 1.5 ...
 $ Petal.Width : num  0.2 0.2 0.2 0.2 0.2 0.4 0.3 0.2 0.2 0.1 ...
 $ Species : Factor w/ 3 levels setosa,versicolor,..: 1 1 1 1
1 1 1 1 1 1 ...
 - attr(*, label)= chr x is a vector
 str(as.matrix(x))
 chr [1:150, 1:5] 5.1 4.9 4.7 4.6 5.0 5.4 ...
 - attr(*, dimnames)=List of 2
  ..$ : NULL
  ..$ : chr [1:5] Sepal.Length Sepal.Width Petal.Length Petal.Width ...


Are any of you aware of a label() function that implements
inheritance? Also, is there a reason for not having a similar label()
implementation in base R? (Folks coming from the SPSS world would
expect it.)

Regards
Liviu


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Re: [R] Strange graphs

2011-07-18 Thread Liviu Andronic
On Mon, Jul 18, 2011 at 5:04 PM, UnitRoot akhussa...@gmail.com wrote:
 Hello,
 I am facing problems with plotting graphs in R. It was working well since
 today, but I cannot get some graphs for unknown reasons. Initially it gave
 me diagonal straight lines. Now it is giving me very strange curves.  I have
 tried to re-install the softwear. It still did not help.

 Please, help me out. Oh, I am using fGarch package outputs and based on them
 trying to plot the graphs. I do not know if it is because of the package
 that I am using.

You have provided too little information. To get any significant help:
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.

Regards
Liviu

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Re: [R] Low Pain Unicode Characters in pdf graph?

2011-07-04 Thread Liviu Andronic
On Sun, May 15, 2011 at 3:06 PM, ivo welch ivo.we...@gmail.com wrote:
 Dear R-experts---is there a relatively low-pain way to get unicode
 characters into a plot to a pdf device?

Have you tried Cairo package or cairo_pdf()? Both are making use of
Cairo, which uses UTF-8 and automatically embeds fonts.

Regards
Liviu

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Re: [R] GUI Experience - Box Plots

2011-06-03 Thread Liviu Andronic
On Fri, Jun 3, 2011 at 6:46 AM, Vikas Garud information4vi...@gmail.com wrote:
 Hi,

 This is to share my experience of using GUI's for using various
 Quality Management/Assurance tools.
 A few days back, I had I had written a mail about Box plots and
 received some very good suggestions, including that for simple
 scripts.  Thanks everybody for the same.

 Based on the suggestions, I am trying three GUI's:  R Commander,
 JGR-Deducer combination and RKward.

 For Box Plots, I found the JGR-Deducer combination the best choice.

 I could not get box plots - Factors and Dodging - in R Commander and
 RKward, though the reply on RKward mailing list suggests that the
 newer version of RKward should solve the problem.

 Thank you everybody for all the help.  I'll be asking questions about
 various tools that I would like to deploy through GUI's for R.  I'll
 also be sharing my experiences.

A good place would be r-sig-gui. Regards
Liviu


 Vikas Garud

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Re: [R] Box Plot under GUI (R Commander/RKward)

2011-05-17 Thread Liviu Andronic
On Tue, May 17, 2011 at 2:15 PM, Jannis bt_jan...@yahoo.de wrote:
  Otherwise I would stick to SPSS/Excel/Sigmaplot or browse for some R GUIs 
 with some basics beeing implemented on a click basis like RStudio, Rkward 
 or TinnR. All R GUI implementations are however limited to basic 
 plotting/analysis routines and I fear you will get stuck pretty soon when 
 relying on the GUI.

Also Deducer.
Liviu

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Re: [R] Powerful PC to run R

2011-05-16 Thread Liviu Andronic
On Fri, May 13, 2011 at 5:25 PM, John C Frain fra...@gmail.com wrote:
 Given the price of the Thinkpad X201 I would be very upset if it
 overheated and would contact my supplier for a replacement.

Consider getting an aluminium cooling pad, such as those provided by
Zalman. It helps keeping my HP from overheating and shutting down.
Liviu


 I also
 understand that this CPU may slow down if it is overheated.  You would
 probably get a better performance from a desktop workstation or a
 larger laptop with better ventilation.  As already stated a large
 mumer of cpu's and multithreading are of little advantage when you are
 running standard R.

 John

 On 13 May 2011 11:38, Michael Haenlein haenl...@escpeurope.eu wrote:
 Dear all,

 I'm currently running R on my laptop -- a Lenovo Thinkpad X201 (Intel Core
 i7 CPU, M620, 2.67 Ghz, 8 GB RAM). The problem is that some of my
 calculations run for several days sometimes even weeks (mainly simulations
 over a large parameter space). Depending on the external conditions, my
 laptop sometimes shuts down due to overheating.

 I'm now thinking about buying a more powerful desktop PC or laptop. Can
 anybody advise me on the best configuration to run R as fast as possible? I
 will use this PC exclusively for R so any other factors are of limited
 importance.

 Thanks,

 Michael


 Michael Haenlein
 Assocaite Professor of Marketing
 ESCP Europe
 Paris, France

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Re: [R] What does the - operator mean?

2011-04-21 Thread Liviu Andronic
On Thu, Apr 21, 2011 at 9:59 PM, Thomas Lumley tlum...@uw.edu wrote:
 The Evil and Wrong use is to modify variables in the global environment.

I'm a bit at a loss. Why is it so wrong and evil? In other words, how
should one modify variables in the global environment? Through the use
of return()?

Regards
Liviu

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Re: [R] CairoPDF, Fonts, and Windows 7

2011-04-16 Thread Liviu Andronic
On Sat, Apr 16, 2011 at 6:39 PM, Al Roark hrbuil...@hotmail.com wrote:

 Hi All:
 I have some basic questions about Cairo graphics engine. I'm trying to use 
 the Cairo package to produce PDF output, mainly because I perceive it to be 
 easy to use with a wide variety of fonts.

You may also want to try cairo_pdf() that, I think, supports Windows
since 2.13.0.


 But right now, I'm stuck trying to figure out what fonts are available to be 
 used with Cairo, specifically the CairoPDF function.  I've been able to 
 successfully produce some test PDFs with grid graphics and CairoPDF (and the 
 font Georgia), so my basic setup is working.
 I've tried to experiment with the CairoFonts() function but get the message:
 the R Cairo package was not installed with fontconfig. Please consider 
 installing the cairo graphics engine (www.cairographics.org) with freetype 
 and fontconfig support
 I'm not quite sure how to interpret this.  My version of Cairo comes from the 
 GTK+ 2.22 bundle from the GTK+ Project website.

It is possible that you need to install fontconfig separately. On
Linux it is a separate package, but I know that it is widely used in
LaTeX, so it might be possible that you need to have a LaTeX
distribution installed, such as miktex. Or install LyX (complete);
this should also take care of fontconfig. But these are mere guesses.

Regards
Liviu


 I'm running R 2.12.2 on Windows 7.
 Any feedback or direction toward informative resources is much appreciated.
 Cheers,Al
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[R] trouble with \SweaveOpts{grdevice=...}

2011-04-14 Thread Liviu Andronic
Dear all
I've just tried the brand new 'grdevice' option in Sweave but couldn't
make it work. When I declare
\SweaveOpts{grdevice=pdf}

or
\SweaveOpts{grdevice=cairo_pdf}

trying to plot something simple
fig=T, echo=T=
plot(1:10,1:10,main='Some title')
@

would result in an Sweave error:
18:16:47.299:  1 : term verbatim
18:16:47.308:  2 : echo term verbatim pdf pdf
18:16:47.391: Error in dev(name = chunkprefix, width = options$width,
height = options$height,  :
18:16:47.392:   unused argument(s) (name = chunkprefix)
18:16:47.393: Calls: Sweave - Anonymous - dev
18:16:47.396: Execution halted
Systemcall.cpp(238): Systemcall: 'Rscript --verbose --no-save
--no-restore /usr/local/share/lyx-svn/scripts/lyxsweave.R
/tmp/lyx_tmpdir.J15281/lyx_tmpbuf5/newfile2.Rnw
/tmp/lyx_tmpdir.J15281/lyx_tmpbuf5/newfile2.tex ISO-8859-15
/tmp/' finished with exit code 1
Error: Cannot convert file

An error occurred while running:
Rscript --verbose --no-save --no-restore
/usr/local/share/lyx-svn/scripts/lyxsweave.R
/tmp/lyx_tmpdir.J15281/lyx_tmpbuf5/newfile2.Rnw
/tmp/lyx_tmpdir.J15281/lyx_tmpbuf5/newfile2.tex ISO-8859-15 /tmp/
18:16:47.978: Error while previewing format: pdf2


If I comment out the statement
%\SweaveOpts{grdevice=pdf}

the compilation proceeds as expected. Hoping that it gets through, I'm
attaching the Sweave file. Am I doing something wrong?

Regards
Liviu


 sessionInfo()
R version 2.13.0 (2011-04-13)
Platform: x86_64-pc-linux-gnu (64-bit)

locale:
 [1] LC_CTYPE=en_US.UTF-8   LC_NUMERIC=C
 [3] LC_TIME=en_US.UTF-8LC_COLLATE=en_US.UTF-8
 [5] LC_MONETARY=C  LC_MESSAGES=en_US.UTF-8
 [7] LC_PAPER=en_US.UTF-8   LC_NAME=C
 [9] LC_ADDRESS=C   LC_TELEPHONE=C
[11] LC_MEASUREMENT=en_US.UTF-8 LC_IDENTIFICATION=C

attached base packages:
[1] stats graphics  grDevices utils datasets  methods   base



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Re: [R] Sweave and Slides (Beamer)

2011-04-14 Thread Liviu Andronic
Hello
I sympathise with you, since Beamer can quickly drive one mad.


On Fri, Apr 15, 2011 at 3:11 AM, Brett Presnell presn...@stat.ufl.edu wrote:
 So, is this useful to anyone besides me?  What trick(s) am I missing
 that would make it easier/better, or that would obviate altogether the
 need for such manipulations?

Not a solution that would work in all cases, have you tried reducing
the font size? Default Beamer font size is quite generous and leaves
some space for manoeuvring. For example, in my Beamer Sweave
presentations I put this in the preamble:
\AtBeginDocument{
\DefineVerbatimEnvironment{Sinput}{Verbatim} {xleftmargin=2em,fontsize=
\footnotesize}
\DefineVerbatimEnvironment{Soutput}{Verbatim}{xleftmargin=2em,fontsize=
\footnotesize}
\DefineVerbatimEnvironment{Scode}{Verbatim}{xleftmargin=2em,fontsize=
\footnotesize}
}

Regards
Liviu

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Re: [R] By function

2011-04-09 Thread Liviu Andronic
Hello

On Fri, Apr 8, 2011 at 12:38 AM, Raoni Rodrigues
caciquesamu...@gmail.com wrote:
 Hello all!

 I have a data frame with nine variables and 293 cases. (attached goes
 the csv file).

The CSV didn't get through so it's difficult to replicate your
example. Please post the output of:
str(cpue)


 I need to calculate a index using the sum of one variable (N) divided
 by the length of other variable (Fisherman), but for each day (Date).

 I tried to use that codes:

 by (cpue, cpue$Date, function (x) sum (cpue$N)/length(cpue$Fisherman))

[..]

 In both codes, as result, I received the same value for all Date
 value. And, oddly, this result is the value of calculation for all
 data frame, as I used just:

 sum (cpue$N)/length(cpue$Fishermans)

From the code above, this is expected since in the function()
statement you use the entire df (cpue) and not the subset (x). Try
this:
by (cpue, cpue$Date, function (x) sum (x$N)/length(x$Fisherman))


 Sorry for the basic question, but I worked on this simple code all day
 and can't do it work out! And have no idea why...

 I'm using R 2.11 in Windows XP.

It is often a good a idea to post the output of
sessionInfo()

Regards
Liviu



 Since now, thanks for the attention and all help,

 Best,

 Raoni

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Re: [R] R licence

2011-04-07 Thread Liviu Andronic
On Thu, Apr 7, 2011 at 9:45 AM, Stanislav Bek
stanislav.pavel@gmail.com wrote:
 Hi,

 is it possible to use some statistic computing by R in proprietary software?

I don't know if this covers your case, but SAS and SPSS provide
interfaces to R. Regards
Liviu


 Our software is written in c#, and we intend to use
 http://rdotnet.codeplex.com/
 to get R work there. Especially we want to use loess function.

 Thanks,

 Best regards,
 Stanislav

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[R] plyr workaround to converting by() to a data frame

2011-04-07 Thread Liviu Andronic
Dear all
Is there a clean plyr version of the following by() and do.call(rbind,
...) construct:
 df-data.frame(a=1:10,b=11:20,c=21:30,grp1=c(x,y),grp2=c(x,y),grp3=c(x,y))
 dfsum-by(df[c(a,b,c)], df[c(grp1,grp2,grp3)], range)
 as.data.frame(dfsum)
Error in as.data.frame.default(dfsum) :
  cannot coerce class 'by' into a data.frame
 do.call(rbind, dfsum)
 [,1] [,2]
[1,]1   29
[2,]2   30


Thank you
Liviu


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[R] Sweave Cairo driver?

2011-04-06 Thread Liviu Andronic
Dear all
I would like to use Sweave with the Cairo() graphics device instead of
pdf(), since the former supports Unicode, allows for easier font
selection out of a greater range of available fonts, and automatically
embeds fonts into the resulting PDF.

Following this older discussion [1], has anyone come up with an Sweave
driver that uses the Cairo graphics device? Thank you
Liviu

[1] http://tolstoy.newcastle.edu.au/R/e12/help/10/10/0986.html



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Re: [R] Sweave Cairo driver?

2011-04-06 Thread Liviu Andronic
On Thu, Apr 7, 2011 at 6:39 AM, Prof Brian Ripley rip...@stats.ox.ac.uk wrote:
 This is simple in 2.13.0 RC, which allows you to add graphics devices for
 Sweave.

Interesting. Which help files should I look into?


 On Wed, 6 Apr 2011, Liviu Andronic wrote:
 since the former supports Unicode, allows for easier font
 selection out of a greater range of available fonts, and automatically
 embeds fonts into the resulting PDF.

 As does cairo_pdf(), builtin to R and more reliable in my experience.

Thank you, I didn't know that, and it looks nice, indeed. Regards
Liviu

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Re: [R] Sweave Cairo driver?

2011-04-06 Thread Liviu Andronic
Dear Brian

On Thu, Apr 7, 2011 at 6:39 AM, Prof Brian Ripley rip...@stats.ox.ac.uk wrote:
 allows for easier font
 selection out of a greater range of available fonts,
 As does cairo_pdf(), builtin to R and more reliable in my experience.

I've looked into the help pages and searched the lists, but could only
find out that The default font family is Helvetica.

How could I use a different  font with cairo_pdf? In Cairo (the R
package), there is a ?CairoFontMatch and ?CairoFonts. Are there
equivalents for cairo_pdf?

Thank you
Liviu

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Re: [R] Syntax coloring in R console

2011-04-01 Thread Liviu Andronic
Hello


On Fri, Apr 1, 2011 at 12:32 PM, January Weiner
january.wei...@mpiib-berlin.mpg.de wrote:
 I have found the packages xterm256 and highlight, but I was not
 able to figure out how to use it to highlight the syntax in console
 output.

I've been in this position before, without finding a solution for
syntax highlighting in the default R console.


 Also, I tried several GUI interfaces, but I was not able to find
 something that suits me better than the default R console. R cmdr is
 definitely not for me, as I don't want to fundamentally change the way
 I am managing my data in R. Rkwrd seems to be nice (from screenshots),
 but its installation requires all the base KDE libraries, which I
 don't want to install.

Recently RStudio was introduced [1] and, although beta, it received
quasi-unanimous acclaim from the community, so you risk finding it
useful too.
[1] http://alternativeto.net/software/rstudio/about

Regards
Liviu


 I tried JGR, the GUI for R, but I have found the following problems
 with this package:

 - I was not able to change the background color from a repulsive grey,
 - apparently, GNU readline is not implemented in that package, that
 is, there is no functionality similar to ctrl-r (which searches
 through the history for matching commands), something I use
 frequently, and
 - tab expansion is of limited use (e.g. doesn't browse files in the
 current directory when expanding quoted arguments e.g. in
 read.table).

 All in all, I'd be happy to continue using the plain R console, but
 syntax highlighting would be nice. Any advice would be extremely
 welcome.

 Kind regards,

 January



 sessionInfo()
 R version 2.12.2 (2011-02-25)
 Platform: i486-pc-linux-gnu (32-bit)

 locale:
  [1] LC_CTYPE=en_US.utf8       LC_NUMERIC=C
 LC_TIME=en_US.utf8        LC_COLLATE=en_US.utf8     LC_MONETARY=C
  [6] LC_MESSAGES=en_US.utf8    LC_PAPER=en_US.utf8       LC_NAME=C
            LC_ADDRESS=C              LC_TELEPHONE=C
 [11] LC_MEASUREMENT=en_US.utf8 LC_IDENTIFICATION=C

 attached base packages:
 [1] stats     graphics  grDevices utils     datasets  methods   base


 --

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Re: [R] Syntax coloring in R console

2011-04-01 Thread Liviu Andronic
On Fri, Apr 1, 2011 at 3:48 PM, January Weiner
january.wei...@mpiib-berlin.mpg.de wrote:
 RStudio might be a fine program, but it does not feature syntax
 highlighting, which is the only thing I am missing from R Console (it
 only colors the commands typed).

The idea is that you shouldn't use the R console for you main
programming needs, but only for quick and dirty checks. For the bulk
of programming tasks you are invited to use the integrated editor
(File  New  Script). The editor window does feature syntax
highlighting, and a very helpful completion mechanism (via tab).
Sending lines for execution to the terminal is as easy as clicking
'run lines' or ctrl+enter. If you're not a fan of keeping scripts for
your projects you can easily use temporary files that you don't save.


 Moreover, the very idea of squeezing
 all R windows into one window-desktop would be counterproductive in
 my particular case.

Notice that all panes are freely resizable and can be resized to the
point of becoming hidden. Future releases will give more control over
the panes layout (I think).

Regards
Liviu


 Thank you anyways!

 j.



 [1] http://alternativeto.net/software/rstudio/about

 Regards
 Liviu


 I tried JGR, the GUI for R, but I have found the following problems
 with this package:

 - I was not able to change the background color from a repulsive grey,
 - apparently, GNU readline is not implemented in that package, that
 is, there is no functionality similar to ctrl-r (which searches
 through the history for matching commands), something I use
 frequently, and
 - tab expansion is of limited use (e.g. doesn't browse files in the
 current directory when expanding quoted arguments e.g. in
 read.table).

 All in all, I'd be happy to continue using the plain R console, but
 syntax highlighting would be nice. Any advice would be extremely
 welcome.

 Kind regards,

 January



 sessionInfo()
 R version 2.12.2 (2011-02-25)
 Platform: i486-pc-linux-gnu (32-bit)

 locale:
  [1] LC_CTYPE=en_US.utf8       LC_NUMERIC=C
 LC_TIME=en_US.utf8        LC_COLLATE=en_US.utf8     LC_MONETARY=C
  [6] LC_MESSAGES=en_US.utf8    LC_PAPER=en_US.utf8       LC_NAME=C
            LC_ADDRESS=C              LC_TELEPHONE=C
 [11] LC_MEASUREMENT=en_US.utf8 LC_IDENTIFICATION=C

 attached base packages:
 [1] stats     graphics  grDevices utils     datasets  methods   base


 --

 __
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 --
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 Max Planck Institute for Infection Biology
 Charitéplatz 1
 D-10117 Berlin, Germany
 Web   : www.mpiib-berlin.mpg.de
 Tel     : +49-30-28460514

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Re: [R] Syntax coloring in R console

2011-04-01 Thread Liviu Andronic
On Fri, Apr 1, 2011 at 5:05 PM, January Weiner
january.wei...@mpiib-berlin.mpg.de wrote:
 Dear Liviu,

 thanks for the programming tip! However, I do all my editing in vim,
 which has had syntax highlighting for quite a while, as well as
 auto-completion and a number of other goodies. But while I do most of
 my programing in vim,

For info, vim has plug-in for R.


 I do most of my scientific studies simply using
 the R interface and keeping a manually edited lab book apart from
 the scripts. Clearly, one can do as much with Rstudio -- I just don't
 see any advantages, but I do see a disadvantage in my specific case:
 all sub-windows are confined to the desktop. In other words, having
 multiple plots on one monitor and a terminal with R command line is
 not possible.

I see. This would be more of a hack, but you could use playwith as an
external graphics device. Otherwise, there seems to be an easy hack:

 x11()
 plot(1:10)


 A question, though. Given that I have projects assorted in various
 directories, how can I start RStudio opening a project stored in
 .RData and .Rhistory of a given directory? I.e., how can I make
 RStudio open the current directory (like R does), and not $HOME?

This is a known issue and the RStudio devels plan to address this.
Currently I simply open RStudio in $HOME (and make sure that there is
no existent .RData there), then open the .RData from my project's
directory (by clicking on it in the right-hand Files pane) and then in
the same pane I hit More  Set as working dir. A bit cumbersome, but
seems to work, and a small price to pay for all the available
functionality.

Regards
Liviu


 Thanks nonetheless,

 j.


 On Fri, Apr 1, 2011 at 4:20 PM, Liviu Andronic landronim...@gmail.com wrote:
 On Fri, Apr 1, 2011 at 3:48 PM, January Weiner
 january.wei...@mpiib-berlin.mpg.de wrote:
 RStudio might be a fine program, but it does not feature syntax
 highlighting, which is the only thing I am missing from R Console (it
 only colors the commands typed).

 The idea is that you shouldn't use the R console for you main
 programming needs, but only for quick and dirty checks. For the bulk
 of programming tasks you are invited to use the integrated editor
 (File  New  Script). The editor window does feature syntax
 highlighting, and a very helpful completion mechanism (via tab).
 Sending lines for execution to the terminal is as easy as clicking
 'run lines' or ctrl+enter. If you're not a fan of keeping scripts for
 your projects you can easily use temporary files that you don't save.


 Moreover, the very idea of squeezing
 all R windows into one window-desktop would be counterproductive in
 my particular case.

 Notice that all panes are freely resizable and can be resized to the
 point of becoming hidden. Future releases will give more control over
 the panes layout (I think).

 Regards
 Liviu


 Thank you anyways!

 j.



 [1] http://alternativeto.net/software/rstudio/about

 Regards
 Liviu


 I tried JGR, the GUI for R, but I have found the following problems
 with this package:

 - I was not able to change the background color from a repulsive grey,
 - apparently, GNU readline is not implemented in that package, that
 is, there is no functionality similar to ctrl-r (which searches
 through the history for matching commands), something I use
 frequently, and
 - tab expansion is of limited use (e.g. doesn't browse files in the
 current directory when expanding quoted arguments e.g. in
 read.table).

 All in all, I'd be happy to continue using the plain R console, but
 syntax highlighting would be nice. Any advice would be extremely
 welcome.

 Kind regards,

 January



 sessionInfo()
 R version 2.12.2 (2011-02-25)
 Platform: i486-pc-linux-gnu (32-bit)

 locale:
  [1] LC_CTYPE=en_US.utf8       LC_NUMERIC=C
 LC_TIME=en_US.utf8        LC_COLLATE=en_US.utf8     LC_MONETARY=C
  [6] LC_MESSAGES=en_US.utf8    LC_PAPER=en_US.utf8       LC_NAME=C
            LC_ADDRESS=C              LC_TELEPHONE=C
 [11] LC_MEASUREMENT=en_US.utf8 LC_IDENTIFICATION=C

 attached base packages:
 [1] stats     graphics  grDevices utils     datasets  methods   base


 --

 __
 R-help@r-project.org mailing list
 https://stat.ethz.ch/mailman/listinfo/r-help
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 http://www.R-project.org/posting-guide.html
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 --
  Dr. January Weiner 3 --
 Max Planck Institute for Infection Biology
 Charitéplatz 1
 D-10117 Berlin, Germany
 Web   : www.mpiib-berlin.mpg.de
 Tel     : +49-30-28460514

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Re: [R] Performance Difference? Windows vs. Linux

2011-03-18 Thread Liviu Andronic
On Fri, Mar 18, 2011 at 11:00 PM, Brigid Mooney bkmoo...@gmail.com wrote:
 I'm not trying to start a Windows vs. Linux debate, but I've been
 using R on a Windows machine for a while, and was recently wondering
 if R's performance would be faster on a Linux machine.  And similarly,
 if any incremental increase in processing speed would be worth the
 time it would take me to migrate my entire system to Linux (including
 a database that I access via an R package.)

It would help to know what is your system. A sessionInfo() would be a start.
Liviu


 I don't know how much it matters what R is doing - but I've got R
 pulling a large amount data from a database, performing many complex
 computations on that data, and then writing output data to a database.

 Thanks so much for the input,
 Brigid

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Re: [R] Teaching R: To quote, or not to quote?

2011-03-09 Thread Liviu Andronic
On Wed, Mar 9, 2011 at 9:57 AM, Ingmar Visser i.vis...@uva.nl wrote:
 FWIW, the style for JSS now enforces (via Achimitization) the use of quotes
 as you show above.


 An excellent candidate for the fortunes package!

We should probably address Achim for that. :)

Regards
Liviu


 best, Ingmar

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Re: [R] sorting subsetting a data.frame

2011-03-07 Thread Liviu Andronic
On Mon, Mar 7, 2011 at 1:38 AM, David Winsemius dwinsem...@comcast.net wrote:
 subset(x[order(x$Species1), ],  Sepal.Length==6.7 )

Thank you all for the suggestions. Now I can do exactly what I wanted. Regards
Liviu

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[R] sorting subsetting a data.frame

2011-03-06 Thread Liviu Andronic
Dear all
This may be obvious, but I cannot get it working. I'm trying to subset
 sort a data frame in one go.
x - iris
x$Species1 - as.character(x$Species)
##subsetting alone works fine
with(x, x[Sepal.Length==6.7,])
##sorting alone works fine
with(x, x[order(Sepal.Length, rev(sort(Species1))),])
##gets subsetted, but not sorted as expected
with(x, x[(Sepal.Length==6.7)  order(Sepal.Length, rev(sort(Species1))),])
##gets subsetted, but sorts very strangely
xa - with(x, x[Sepal.Length==6.7,]); with(xa, xa[order(Sepal.Length,
rev(sort(Species1))),])
xa - with(x, x[Sepal.Length==6.7,]); with(xa, xa[order(rev(sort(Species1))),])

I've checked The R Inferno, Quick-R and several other places with no
obvious solution.

Any ideas? Regards
Liviu


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Re: [R] sorting subsetting a data.frame

2011-03-06 Thread Liviu Andronic
On Sun, Mar 6, 2011 at 11:49 PM, Liviu Andronic landronim...@gmail.com wrote:
 Dear all
 This may be obvious, but I cannot get it working. I'm trying to subset
  sort a data frame in one go.
 x - iris
 x$Species1 - as.character(x$Species)
 ##subsetting alone works fine
 with(x, x[Sepal.Length==6.7,])
 ##sorting alone works fine
 with(x, x[order(Sepal.Length, rev(sort(Species1))),])
 ##gets subsetted, but not sorted as expected
 with(x, x[(Sepal.Length==6.7)  order(Sepal.Length, rev(sort(Species1))),])
 ##gets subsetted, but sorts very strangely
 xa - with(x, x[Sepal.Length==6.7,]); with(xa, xa[order(Sepal.Length,
 rev(sort(Species1))),])
 xa - with(x, x[Sepal.Length==6.7,]); with(xa, 
 xa[order(rev(sort(Species1))),])

And of course I found the culprit after sending the e-mail: wrong call
order. The following does what I want, although it's a bit messy:
xa - with(x, x[Sepal.Length==6.7,]); with(xa, xa[rev(order(sort(Species1))),])
xa - subset(x, Sepal.Length==6.7); with(xa, xa[rev(order((sort(Species1,])

Regards
Liviu


 I've checked The R Inferno, Quick-R and several other places with no
 obvious solution.

 Any ideas? Regards
 Liviu


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Re: [R] sorting subsetting a data.frame

2011-03-06 Thread Liviu Andronic
On Sun, Mar 6, 2011 at 11:53 PM, Liviu Andronic landronim...@gmail.com wrote:
 On Sun, Mar 6, 2011 at 11:49 PM, Liviu Andronic landronim...@gmail.com 
 wrote:
 Dear all
 This may be obvious, but I cannot get it working. I'm trying to subset
  sort a data frame in one go.
 x - iris
 x$Species1 - as.character(x$Species)
 ##subsetting alone works fine
 with(x, x[Sepal.Length==6.7,])
 ##sorting alone works fine
 with(x, x[order(Sepal.Length, rev(sort(Species1))),])
 ##gets subsetted, but not sorted as expected
 with(x, x[(Sepal.Length==6.7)  order(Sepal.Length, rev(sort(Species1))),])
 ##gets subsetted, but sorts very strangely
 xa - with(x, x[Sepal.Length==6.7,]); with(xa, xa[order(Sepal.Length,
 rev(sort(Species1))),])
 xa - with(x, x[Sepal.Length==6.7,]); with(xa, 
 xa[order(rev(sort(Species1))),])

 And of course I found the culprit after sending the e-mail: wrong call
 order. The following does what I want, although it's a bit messy:
 xa - with(x, x[Sepal.Length==6.7,]); with(xa, 
 xa[rev(order(sort(Species1))),])
 xa - subset(x, Sepal.Length==6.7); with(xa, 
 xa[rev(order((sort(Species1,])

But it still doesn't work on my data! Any ideas for a different approach?
Liviu


 Regards
 Liviu


 I've checked The R Inferno, Quick-R and several other places with no
 obvious solution.

 Any ideas? Regards
 Liviu


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Re: [R] problems with playwith

2011-03-01 Thread Liviu Andronic
On Tue, Mar 1, 2011 at 5:58 PM, R Heberto Ghezzo, Dr
heberto.ghe...@mcgill.ca wrote:
 hello, i tried to run playwith but :

 library(playwith)
 Loading required package: lattice
 Loading required package: cairoDevice
 Loading required package: gWidgetsRGtk2
 Loading required package: gWidgets
 Error in inDL(x, as.logical(local), as.logical(now), ...) :
  unable to load shared object 'H:/R/cran/RGtk2/libs/i386/RGtk2.dll':
  LoadLibrary failure:  The specified procedure could not be found.
 Failed to load RGtk2 dynamic library, attempting to install it.

Did you install RGtk2? [1]
Liviu

[1] https://code.google.com/p/playwith/


 Learn more about GTK+ at http://www.gtk.org
 If the package still does not load, please ensure that GTK+ is installed and 
 that it is on your PATH environment variable
 IN ANY CASE, RESTART R BEFORE TRYING TO LOAD THE PACKAGE AGAIN
 Error : .onAttach failed in attachNamespace() for 'gWidgetsRGtk2', details:
  call: .Call(name, ..., PACKAGE = PACKAGE)
  error: C symbol name S_gtk_icon_factory_new not in DLL for package RGtk2
 Error: package 'gWidgetsRGtk2' could not be loaded

 Sys.getenv(PATH)
                                                                               
                                                                               
                                                                               
                                                                               
            PATH
 H:\\R/GTK/bin;H:\\R/GTK/lib;H:\\R/ImageMagick;C:\\windows\\system32;C:\\windows;C:\\windows\\System32\\Wbem;C:\\windows\\System32\\WindowsPowerShell\\v1.0\\;C:\\Program
  Files\\Common Files\\Ulead Systems\\MPEG;C:\\Program 
 Files\\QuickTime\\QTSystem\\;H:\\R\\GTK\\GTK2-Runtime\\bin;H:\\PortableUSB/PortableApps/MikeTex/miktex/bin

 packages(lattice, cairoDevice, gWidgetsRGtk2, gWidgets, RGtk2, playwith) were 
 reinstalled
 program GTK was reinstalled.
 using R-2-12-2 on Windows 7
 Can anybody suggest a solution?
 thanks

 R.Heberto Ghezzo Ph.D.
 Montreal - Canada
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Re: [R] Is there any Command showing correlation of all variables in a dataset?

2011-03-01 Thread Liviu Andronic
On Tue, Mar 1, 2011 at 11:41 AM, JoonGi joo...@hanmail.net wrote:

 Thanks in advance.

 I want to derive correlations of variables in a dataset

 Specifically

 library(Ecdat)
 data(Housing)
 attach(Housing)
 cor(lotsize, bathrooms)

 this code results only the correlationship between two variables.
 But I want to examine all the combinations of variables in this dataset.
 And I will finally make a table in Latex.

 How can I test correlations for all combinations of variables?
 with one simple command?

See Rcmdr for an example.
Liviu



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Re: [R] When is *interactive* data visualization useful to use?

2011-02-19 Thread Liviu Andronic
On Fri, Feb 18, 2011 at 8:00 PM, Tom Hopper tomhop...@gmail.com wrote:
 Tal,

 One interactive capability that I have repeatedly wished for (but
 never taken the time to develop with the existing R tools) is the
 ability to interactively zoom in on and out of a data set,

I believe that you can do this with playwith. See this [1]. Regards
Liviu

[1] 
http://code.google.com/p/playwith/wiki/Screenshots#Time_series_plot_(Lattice)


and to
 interactively create call-outs of sections of the data. Much of the
 data that I deal with takes the form of time series where both the
 full data and small section carry meaningful information.

 Some of the capabilities of Deducer approach interactive graphing,
 such as adjusting alpha values or smoothers, though the updates don't
 happen in quite real-time.

 - Tom

 On Friday, February 11, 2011, Tal Galili tal.gal...@gmail.com wrote:
 Hello all,

 Before getting to my question, I would like to apologize for asking this
 question here.  My question is not directly an R question, however, I still
 find the topic relevant to R community of users  - especially due to only *
 partial* (current) support for interactive data visualization (see here:
 http://cran.r-project.org/web/views/Graphics.html  were with iplots we are
 waiting for iplots extreme, and with rggobi, it currently can not run with R
 2.12 and windows 7 OS).

 And now for my question:

 While preparing for a talk I will give soon, I recently started digging into
 two major (Free) tools for interactive data visualization:
 GGobihttp://www.ggobi.org/
  and mondrian http://rosuda.org/mondrian/ - both offer a great range of
 capabilities (even if they're a bit buggy).

 I wish to ask for your help in articulating (both to myself, and for my
 future audience) *When is it helpful to use interactive plots? Either for
 data exploration (for ourselves) and data presentation (for a client)?*

 For when explaining the data to a client, I can see the value of animation
 for:

    - Using identify/linking/brushing for seeing which data point in the
    graph is what.
    - Presenting a sensitivity analysis of the data (e.g: if we remove this
    point, here is what we will get)
    - Showing the effect of different groups in the data (e.g: let's look at
    our graphs for males and now for the females)
    - Showing the effect of time (or age, or in general, offering another
    dimension to the presentation)

 For when exploring the data ourselves, I can see the value of
 identify/linking/brushing when exploring an outlier in a dataset we are
 working on.

 But other then these two examples, I am not sure what other practical use
 these techniques offer. Especially for our own data exploration!

 It could be argued that the interactive part is good for exploring (For
 example) a different behavior of different groups/clusters in the data. But
 when (in practice) I approached such situation, what I tended to do was to
 run the relevant statistical procedures (and post-hoc tests) - and what I
 found to be significant I would then plot with colors clearly dividing the
 data to the relevant groups. From what I've seen, this is a safer approach
 then wondering around the data (which could easily lead to data dredging
 (were the scope of the multiple comparison needed for correction is not even
 clear).

 I'd be very happy to read your experience/thoughts on this matter.


 Thanks in advance,
 Tal


 Contact
 Details:---
 Contact me: tal.gal...@gmail.com |  972-52-7275845
 Read me: www.talgalili.com (Hebrew) | www.biostatistics.co.il (Hebrew) |
 www.r-statistics.com (English)
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Re: [R] R-/Text-editor for Windows?

2011-01-29 Thread Liviu Andronic
Hello

On Fri, Jan 28, 2011 at 8:03 PM, Jonathan P Daily jda...@usgs.gov wrote:
 Geany also handles a ton of filetypes, but lacks a direct interface to R
 in the windows version - I wrote an AutoHotKey script that did this for me
 in about 5 minutes.

Could you please expand on that? I prefer to use Geany when working on
Linux, but I always found it annoying the Windows limitation.

Thank you
Liviu

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Re: [R] how to get coefficient and scores of Principal component analysis in R?

2011-01-28 Thread Liviu Andronic
Hello

On Fri, Jan 28, 2011 at 8:16 PM, Zunqiu Chen che...@ohsu.edu wrote:
 It might be a simple question. But I could not find the answer from function 
 “prcomp” or “princomp”. Does anyone know what are the codes to get 
 coefficient and scores of Principal component analysis in R?

Try these:
https://stat.ethz.ch/pipermail/r-help/2010-November/261719.html
http://www.statmethods.net/advstats/factor.html
http://www.unt.edu/rss/class/Jon/R_SC/Module7/M7_PCAandFA.R
http://stats.stackexchange.com/questions/612/is-psychprincipal-function-still-pca-when-using-rotation
http://zoonek2.free.fr/UNIX/48_R/05.html#1

Liviu

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Re: [R] speed up in R apply

2011-01-06 Thread Liviu Andronic
On Wed, Jan 5, 2011 at 10:49 PM, Young Cho young.s...@gmail.com wrote:
 When introduced to R, I learned how to use *apply whenever I could to avoid
 for-loops and all. And, getting the habit, I think I somehow got the
 mis-conception that it is a magic source, always an optimal way of coding in
 R.

See [1] for an article on vectorisation and loops in R.
Liviu

[1] http://www.r-project.org/doc/Rnews/Rnews_2008-1.pdf


 Thanks a lot for all of your helpful advice and comment!

 Young

 On Wed, Jan 5, 2011 at 3:09 PM, David Winsemius dwinsem...@comcast.netwrote:


 On Jan 5, 2011, at 2:40 PM, Douglas Bates wrote:

  On Wed, Jan 5, 2011 at 1:22 PM, David Winsemius dwinsem...@comcast.net
 wrote:


 On Jan 5, 2011, at 10:03 AM, Young Cho wrote:

  Hi,

 I am doing some simulations and found a bottle neck in my R script. I
 made
 an example:

  a = matrix(rnorm(500),100,5)
 tt  = Sys.time(); sum(a[,1]*a[,2]*a[,3]*a[,4]*a[,5]); Sys.time() - tt


 [1] -1291.026
 Time difference of 0.2354031 secs


 tt  = Sys.time(); sum(apply(a,1,prod)); Sys.time() - tt


 [1] -1291.026
 Time difference of 20.23150 secs

 Is there a faster way of calculating sum of products (of columns, or of
 rows)?


 You should look at crossprod and tcrossprod.


 Hmm.  Not sure that would help, David.  You could use a matrix
 multiplication of a %*% rep(1, ncol(a)) if you wanted the row sums but
 of course you could also use rowSums to get those.


 Thanks for pointing  that out. I misread the OP's code.


  And is this an expected behavior?


 Yes. For loops and *apply strategies are slower than the proper use of
 vectorized functions.


 To expand a bit on David's point, the apply function isn't magic.  It
 essentially loops over the rows, in this case.  By multiplying columns
 together you are performing the looping over the rows in compiled
 code, which is much, much faster.  If you want to do this kind of
 operation effectively in R for a general matrix (i.e. not knowing in
 advance that it has exactly 5 columns) you could use Reduce

  a - matrix(rnorm(500),100,5)
 system.time(pr1 - a[,1]*a[,2]*a[,3]*a[,4]*a[,5])

  user  system elapsed
  0.15    0.09    0.37

 system.time(pr2 - apply(a, 1, prod))

  user  system elapsed
 22.090   0.140  22.902

 all.equal(pr1, pr2)

 [1] TRUE

 system.time(pr3 - Reduce(get(*), as.data.frame(a), rep(1, nrow(a


 Slightly faster would be:

 system.time(pr3 - Reduce(*, as.data.frame(a)))

 And thanks for the nice example. Using a data.frame to feed Reduce
 materially enhances its value to me.


   user  system elapsed
  0.410   0.010   0.575

 all.equal(pr3, pr2)

 [1] TRUE


 --
 David Winsemius, MD
 West Hartford, CT



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Re: [R] Cost-benefit/value for money analysis

2011-01-05 Thread Liviu Andronic
On Wed, Jan 5, 2011 at 12:29 PM, Graham Smith myotis...@gmail.com wrote:
 maximal choices would break the budget. This sounds like a homework problem
 and I don't see any student effort yet. Search terms include: decision
 analysis , cost-benefit analysis, or utility theory.


 Hopefully,  my response to Ben will clarify my question, and why I am asking
 it.  At the moment (and that may change) I'm not specifically interested in
 how you do it R, just as to whether there is a package aimed at this kind of
 Cost Benefit analysis.

Try this:
 require(sos)
 findFn('cost benefit')
found 12 matches

Regards
Liviu

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Re: [R] Sweave for big data analysis

2011-01-01 Thread Liviu Andronic
(slightly off-topic)
Hello


On Sat, Jan 1, 2011 at 12:20 AM, Yihui Xie x...@yihui.name wrote:
 I still recommend the pgfSweave package (as usual) -- you can cache
 both data objects (using cacheSweave) and graphics (using pgf).

Do these packages re-implement Sweave, or just use it as a dependency?
For example, if Sweave gets updated, will the changes be automatically
available for pgfSweave users or not?

Regards
Liviu


 Regards,
 Yihui
 --
 Yihui Xie xieyi...@gmail.com
 Phone: 515-294-2465 Web: http://yihui.name
 Department of Statistics, Iowa State University
 2215 Snedecor Hall, Ames, IA



 On Fri, Dec 31, 2010 at 2:35 PM, Lars Bishop lars...@gmail.com wrote:
 Hi,

 Maybe I'm missing the point here...but let's suppose you are working with
 large data sets and using functions that take a significant amount of time
 to run in R. I woulnd't like to run these functions every time I call
 Sweave(myfile.Rnw) within R. What is the common practice to use Sweave
 in these situations. I would just run the function once, save the results
 and only load them each time I run Sweave on the .Rnw file. Makes sense?

 Sorry, the question seems silly, but I'd appreciate your thoughts.

 Thanks,
 Lars.

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Re: [R] latex() etc.: How to nicely format a matrix for a LaTeX document?

2010-12-30 Thread Liviu Andronic
On Thu, Dec 30, 2010 at 9:48 PM, Marius Hofert m_hof...@web.de wrote:
 Thanks, John, I did.

 Here is an update [not all problems are solved; can they be?]:

 Consider:

 library(xtable)
 mat - matrix(c(1,NA,3,100,10012.23423,4), ncol = 3, byrow = TRUE)
 print(xtable(mat, digits = 1), floating = FALSE, only.contents = TRUE,
      include.rownames = FALSE, include.colnames = FALSE, hline.after = NULL)

 With this one obtains a clean output of the rows (maybe someone knows how to 
 suppress the % latex table generated... that is displayed).
 However, as you can see from the output, the  symbols are not vertically 
 aligned (that would be quite helpful in reading the table in a source file). 
 This is even worse if you have row names of different lengths...

If you used Sweave, would you still need to read the  table code in a
source file? Regards
Liviu


 So my questions (2) and (3) are solved, but (1) remains. Is there any way to 
 (maybe) first format the matrix entries to get the right alignment in 
 xtable()?

 Cheers,

 Marius

 On 2010-12-30, at 21:29 , John Kane wrote:

 Have a look at xtable.

 --- On Thu, 12/30/10, Marius Hofert m_hof...@web.de wrote:

 From: Marius Hofert m_hof...@web.de
 Subject: [R] latex() etc.: How to nicely format a matrix for a LaTeX 
 document?
 To: Help R r-h...@stat.math.ethz.ch
 Received: Thursday, December 30, 2010, 11:50 AM
 Dear (T)eXpeRts,

 I try to create a LaTeX table from an R matrix for the
 first time. I am not sure what the best approach is, I
 just read about latex() from Hmisc (toLatex() didn't work).

 Consider the following minimal example:

 library(Hmisc)
 mat - matrix(c(1,NA,3,100,1,4), ncol = 3, byrow =
 TRUE)
 latex(mat, file = , booktabs = TRUE, numeric.dollar =
 FALSE, table.env = FALSE)

 I am only interested in the part between \midrule and
 \bottomrule [I couldn't figure out how to remove the tabular
 environment]. It looks like this:

   13\tabularnewline
 10014\tabularnewline

 My questions/problems are:

 (1) if there are NA's, you can see that the output is not
 aligned according to the -symbols. That makes it hard
 to read in a LaTeX *source* file. How can I get something
 like:
   1
    3\tabularnewline
 10014\tabularnewline

 (2) it would even be nicer to read if the output was like
 this:
   1         3
 \tabularnewline
 100  1  4 \tabularnewline
 How can I achieve this?

 (3) is there another package/function to get output like
 this more easily? I wrote a one-liner which formats the
 lines separately, but it would be nice to have the columns
 aligned as given in (2). I believe it is most helpful to
 have the output in a form which is readable in a *source*
 file (i.e., .tex), since the headers/footers from tabular
 [or tabularx etc.] are often easy to put in the document.

 Cheers,

 Marius

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Re: [R] What is the best way to lag a time series?

2010-12-26 Thread Liviu Andronic
On Sun, Dec 26, 2010 at 8:49 AM, Christian Schoder
schoc...@newschool.edu wrote:
 Dear R-users,

 I've been using R for a while and I am very satisfied! Unfortunately, I
 still have not figured out an efficient and general way to construct and
 use lags of time series, especially when I need to work with different
 packages.

 Let me give an example. I have two time series x and y and I want to
 estimate a variaty of distributed lags models and run different tests
 (autocorrelation, etc). It is obvious that I need to be able to lag x
 and y in a flexible way. So far, my temporary solution was to construct
 the lags manually (x1,..,xn and y1,..,yn) in a spreadsheet and import it
 to R, which is not very satisfactory because it does not allow for much
 flexibility.

 Is there a straighforward command which allows me to easily construct a
 lag

Perhaps ?diff.

Liviu


 when required and which allows me to, for example, use the lm()
 command to fit a dynamic model and the bgtest() command to perform the
 breusch-godfrey test on the same model?

 Is it adviseable to use time series objects which consist of many time
 series (like a dataframe) or is it better to have it contain only one
 time series?

 I would be grateful for any hints and links.

 Thx!
 Christian

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Re: [R] matrix indexing in 'for' loop?

2010-12-22 Thread Liviu Andronic
On Wed, Dec 22, 2010 at 2:57 AM, Phil Spector spec...@stat.berkeley.edu wrote:
 To make your loop work, you need to learn about the get function.
 I'm not going to give you the details because there are better
 approaches available.
 First, let's make some data that will give values which can be verified.
  (All the correlations of the data you created
 are exactly equal to 1.)  And to make the code readable, I'll
 omit the ts.m prefix.

 set.seed(14)
 dmi = matrix(rnorm(20),4,5)
 soi = matrix(rnorm(20),4,5)
 pe = matrix(rnorm(20),4,5)
 allmats = list(dmi,soi,pe)

 Since cor.test won't automatically do the tests for all columns
 of a matrix, I'll write a little helper function:

 gettests = function(x)apply(x,2,function(col)cor.test(pe[,2],col)
 tests = lapply(allmats,gettests)

 Now tests is a list of length 2, with a list of the output from
 cor.test for the five columns of the each matrix with pe[,2]
 (Notice that in your program you made no provision to store the results
 anywhere.)

 Suppose you want the correlations:

 sapply(tests,function(x)sapply(x,function(test)test$estimate))

           [,1]       [,2]
 cor  0.12723615  0.1342751
 cor  0.07067819  0.6228158
 cor -0.28761533  0.6218661
 cor  0.83731828 -0.9602551
 cor -0.36050836  0.1170035

 The probabilities for the tests can be found similarly:

 sapply(tests,function(x)sapply(x,function(test)test$p.value))

          [,1]       [,2]
 [1,] 0.8727638 0.86572490
 [2,] 0.9293218 0.37718416
 [3,] 0.7123847 0.37813388
 [4,] 0.1626817 0.03974489
 [5,] 0.6394916 0.88299648

Hmisc already provides a function for doing this, rcorr(). Try to
compute correlations using Rcmdr for an example.

Regards
Liviu


 (Take a look at the Values section in the help file for cor.test
 to get the names of other quantities of interest.)

 The main advantage to this approach is that if you add more matrices
 to the allmats list, the other steps automaticall take it into account.

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





 On Tue, 21 Dec 2010, govin...@msu.edu wrote:



 Hi,

 I am having trouble with matrices. I?have 2 matrices as given below, and I
 am interested in using these matrices inside for loops used to calculate
 correlations. I am creating a list with the names of the matrices assuming
 this list could be indexed inside the 'for' loop to retrieve the matrix
 values. But, as expected the code throws out an error. Can someone suggest a
 better way to call these matrices inside the loops?

 ts.m.dmi - matrix(c(1:20), 4, 5)
 ts.m.soi - matrix(c(21:40), 4, 5)
 ts.m.pe - matrix(c(21:40), 4, 5)

 factors - c(ts.m.dmi, ts.m.soi)
 for (j in 0:1){
 y - factors[j+1]

 for (i in 1:5){

 cor.pe.y - cor(ts.m.pe[,2], y[,i])
 ct.tst - cor.test(ts.m.pe[,2], y[,i])
 }
 }

 Thanks for your time.

 --
 Regards,
 Maha
 Graduate Student
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Re: [R] matrix indexing in 'for' loop?

2010-12-22 Thread Liviu Andronic
On Wed, Dec 22, 2010 at 6:39 PM,  govin...@msu.edu wrote:
 Thank you both for your suggestions. I have another question - is there a
 specific way to access the individual elements of a 'list' variable? i.e.

 dmi = matrix(rnorm(20),4,5)
 soi = matrix(rnorm(20),4,5)
 pe = matrix(rnorm(20),4,5)
 y - list(dmi, soi, pe)

 y[[1]]   gives
 [,1]   [,2]   [,3]   [,4]   [,5]
 [1,] -0.54463900  1.6732445 -0.3807847 -1.0460530 -0.8142748
 [2,] -0.49654004 -0.9634258  0.9074139 -0.1576030 -1.2268558
 [3,] -1.61835766 -0.4240122  0.3626670  0.7182964  0.1576446
 [4,] -0.06313983  0.6743465 -1.9897107  0.8027337 -1.4372131

 But, I am interested in accessing the 1st column of 'dmi' matrix here - not
 as dmi[,1] but in terms of 'y'. Is it possible with 'list' function or
 something else?


Is this what you need?
 y[[1]]
   [,1][,2][,3]   [,4]  [,5]
[1,]  0.2523449 -1.26606999  0.27594223 -0.5158524 -0.599899
[2,] -1.9155409  0.01081565  0.44497873  0.1819517  1.187899
[3,]  0.7624774  0.31296468 -0.02187076  0.9389091  0.865527
[4,] -0.8082626 -1.44330148  1.32975075  0.1788399  1.204917

 y[[1]][,1]
[1]  0.2523449 -1.9155409  0.7624774 -0.8082626

Liviu

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