date:20091123




Erin Hodgess wrote:

Hello again.

I did the update and it worked well, except for one thing:  the Rcmdr
doesn't work anymore.

Has that come up before please?



Not that I know. But what does doesn't work anymore mean precisely?

Uwe



Thank you for all of your help.

Sincerely,
Erin



 Ligges lig...@statistik.tu-dortmund.de:


Erin Hodgess wrote:

Dear R People:

I just installed R-2.10.0 on Karmic Koala Ubuntu, via the sudo apt-get
install r-base, etc.

However, when I try to install an Rcmdr Plugin package, I get the
following:


install.packages(RcmdrPlugin.qual,depen=TRUE)

Warning in install.packages(RcmdrPlugin.qual, depen = TRUE) :
 argument 'lib' is missing: using '/usr/local/lib/R/site-library'
Warning in install.packages(RcmdrPlugin.qual, depen = TRUE) :
 'lib = /usr/local/lib/R/site-library' is not writable
Would you like to create a personal library
'~/R/i486-pc-linux-gnu-library/2.10'
to install packages into?  (y/n) y
--- Please select a CRAN mirror for use in this session ---
Loading Tcl/Tk interface ... done
also installing the dependency ‘qcc’

trying URL
'http://cran.opensourceresources.org/src/contrib/qcc_2.0.tar.gz'
Content type 'application/x-gzip' length 163556 bytes (159 Kb)
opened URL
==
downloaded 159 Kb

trying URL
'http://cran.opensourceresources.org/src/contrib/RcmdrPlugin.qual_0.4.0.tar.gz'
Content type 'application/x-gzip' length 3545 bytes
opened URL
==
downloaded 3545 bytes

* installing *source* package ‘qcc’ ...
** R
** data
** demo
** inst
** preparing package for lazy loading
** help
*** installing help indices
** building package indices ...
* DONE (qcc)
* installing *source* package ‘RcmdrPlugin.qual’ ...
** R
** inst
** preparing package for lazy loading
Warning: package 'Rcmdr' was built under R version 2.8.1 and help may
not work correctly
Loading required package: tcltk
Loading Tcl/Tk interface ... done
The Commander GUI is launched only in interactive sessions

Attaching package: 'Rcmdr'


   The following object(s) are masked from package:tcltk :

tclvalue

Warning: package 'rgl' was built under R version 2.8.1 and help may
not work correctly
Warning: package 'abind' was built under R version 2.8.1 and help may
not work correctly
Error : This is R 2.10.0, package 'MASS' needs = 2.9.2
ERROR: lazy loading failed for package ‘RcmdrPlugin.qual’
* removing ‘/home/erin/R/i486-pc-linux-gnu-library/2.10/RcmdrPlugin.qual’

The downloaded packages are in
   ‘/tmp/RtmpioC2DR/downloaded_packages’
Warning message:
In install.packages(RcmdrPlugin.qual, depen = TRUE) :
 installation of package 'RcmdrPlugin.qual' had non-zero exit status

The line that looks particularly strange is the 'MASS needs = 2.9.2.

Erin,

the MASS you have installed in one of your current libraries needs R =
2.9.2 (perhaps in /usr/local/lib/R/site-library ?)

Please run
 update.packages(checkBuilt=TRUE)
and packages in your library will be updated.

If the one in /usr/local/lib/R/site-library can't be updated, because you do
not have write permissions there, please remove it from the library path.


Best,
Uwe Ligges






Could that be the problem, please?

Thank you in advance for any help.

Sincerely,
Erin








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[R] Odp: FUN argument to return a vector in aggregate function

2009-11-23 Thread Petr PIKAL

Hi

r-help-boun...@r-project.org napsal dne 23.11.2009 09:15:02:

 Hi All,
 
 I am currently doing the following to compute summary statistics of 
 aggregated data:
 a = aggregate(warpbreaks$breaks, warpbreaks[,-1], mean)
 b = aggregate(warpbreaks$breaks, warpbreaks[,-1], sum)
 c = aggregate(warpbreaks$breaks, warpbreaks[,-1], length)
 ans = cbind(a, b[,3], c[,3])
 
 This seems unnecessarily complex to me so I tried
   aggregate(warpbreaks$breaks, warpbreaks[,-1], function(z) 
 c(mean(z),sum(z),length(z)))
 but aggregate doesn't allow FUN argument to return a vector.
 
 I tried by, tapply and several other functions as well but the 
 output needed further modifications to get the same format as ans 
above.


It is work for sapply/split.

sapply(split(warpbreaks$breaks, warpbreaks[,-1]), function(x) c(mean(x), 
sum(x), length(x)))
   A.L   B.L A.M   B.M   A.H   B.H
[1,]  44.6  28.2  24  28.8  24.6  18.8
[2,] 401.0 254.0 216 259.0 221.0 169.0
[3,]   9.0   9.0   9   9.0   9.0   9.0
 

Regards
Petr


 
 Is there any other function same as aggregate which allow FUN argument 
 to return vector.
 
 Regards
 Utkarsh
 
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Re: [R] how to tell if its better to standardize your data matrix first when you do principal

masterinex wrote:
Hi Hadley ,

I really apreciate the suggestions you gave, It was helpful , but I still
didnt quite get it all. and I really want to do a good job , so any
comments would sure come helpful, please understand me .

Well, we try to understand you, but we do not either. I think you really
nedc to consult some statistics textbook on PCA if my answer was not
sufficient. Given your questions, I doubt you understand what PCA does
and how it works. It does not predict anything.

Uwe Ligges

hadley wrote:

You've asked the same question on stackoverflow.com and received the
same answer. This is rude because it duplicates effort. If you
urgently need a response to a question, perhaps you should consider
paying for it.

Hadley

On Sun, Nov 22, 2009 at 12:04 PM, masterinex xevilgan...@hotmail.com
wrote:

so under which cases is it better to standardize the data matrix first
?
also is PCA generally used to predict the response variable , should I
keep that variable in my data matrix ?

Uwe Ligges-3 wrote:

masterinex wrote:

Hi guys ,

Im trying to do principal component analysis in R . There is 2 ways of
doing
it , I believe.
One is doing principal component analysis right away the other way is
standardizing the matrix first using s = scale(m)and then apply
principal
component analysis.
How do I tell what result is better ? What values in particular should
i
look at . I already managed to find the eigenvalues and eigenvectors ,
the
proportion of variance for each eigenvector using both methods.

Generally, it is better to standardize. But in some cases, e.g. for the
same units in your variables indicating also the importance, it might
make sense not to do so.
You should think about the analysis, you cannot know which result is
`better' unless you know an interpretation.

I noticed that the proportion of the variance for the first pca
without
standardizing had a larger value . Is there a meaning to it ? Isnt
this
always the case?
At last , if I am supposed to predict a variable ie weight should I
drop
the variable ie weight from my data matrix when I do principal
component
analysis ?

This sounds a bit like homework. If that is the case, please ask your
teacher rather than this list.
Anyway, it does not make sense to predict weight using a linear
combination (principle component) that contains weight, does it?

Uwe Ligges

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--
View this message in context:
http://old.nabble.com/how-to-tell-if-its-better-to-standardize-your-data-matrix-first-when-you-do-principal-tp26462070p26466400.html
Sent from the R help mailing list archive at Nabble.com.

--
http://had.co.nz/

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Re: [R] Computing multivariate normal probabilities. Was: Re: Problem with Numerical derivatives (numDeriv) and mvtnorm

2009-11-23 Thread Torsten Hothorn



On Sun, 22 Nov 2009, Ravi Varadhan wrote:



Hi Torsten,



Hi Ravi,


It would be useful to warn the users that the multivariate normal probability
calculated by pmvnorm using the GenzBretz algorithm is random, i.e. 
the result can vary between repeated executions of the function.


only if a different seed is used.

This 
would prevent inappropriate use of pmvnorm such as computing derivatives 
of it (see this email thread).




?pmvt has Randomized quasi-Monte Carlo methods are used for the
computations. and appropriate references. In addition, the new book by 
Alan Genz and Frank Bretz covers all technical details in depth, so

the procedures are well documented.

Anyway, I'll add a statement to ?pmvnorm.

Best wishes,

Torsten

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Re: [R] how to tell if its better to standardize your data matrix first when you do principal




masterinex wrote:

this is how my data matrix looks like . This is just for the first 10
observations , but the pattern is similar for the other observations.  



112.3 154.25  67.75 36.2  93.1  85.2  94.5  59.0 37.3  21.9   32.0   
27.4  17.1
2 6.1 173.25  72.25 38.5  93.6  83.0  98.7  58.7 37.3  23.4   30.5   
28.9  18.2
325.3 154.00  66.25 34.0  95.8  87.9  99.2  59.6 38.9  24.0   28.8   
25.2  16.6
410.4 184.75  72.25 37.4 101.8  86.4 101.2  60.1 37.3  22.8   32.4   
29.4  18.2
528.7 184.25  71.25 34.4  97.3 100.0 101.9  63.2 42.2  24.0   32.2   
27.7  17.7
620.9 210.25  74.75 39.0 104.5  94.4 107.8  66.0 42.0  25.6   35.7   
30.6  18.8
719.2 181.00  69.75 36.4 105.1  90.7 100.3  58.4 38.3  22.9   31.9   
27.8  17.7
812.4 176.00  72.50 37.8  99.6  88.5  97.1  60.0 39.4  23.2   30.5   
29.0  18.8
9 4.1 191.00  74.00 38.1 100.9  82.5  99.9  62.9 38.3  23.8   35.9   
31.1  18.2
10   11.7 198.25  73.50 42.1  99.6  88.6 104.1  63.1 41.7  25.0   35.6   
30.0  19.2



and after standardizing it  . 


1   -0.831228836 -0.898881671 -0.98330178 -0.77420686 -0.952294055
-0.712961621 -0.814552365 -0.0625400993 -0.53901713 -0.825399059 -0.08244945
2   -1.588060506 -0.185928394  0.75868364  0.23560461 -0.889886435
-0.931523054 -0.155497233 -0.1252522485 -0.53901713  0.295114747 -0.59529632
30.755676279 -0.908262635 -1.56396359 -1.74011349 -0.615292906
-0.444727135 -0.077038289  0.0628841989  0.15515266  0.743320270 -1.17652277
4   -1.063161122  0.245595958  0.75868364 -0.24734870  0.133598535
-0.593746294  0.236797489  0.1674044475 -0.53901713 -0.153090775  0.05430971
51.170713001  0.226834030  0.37157577 -1.56449410 -0.428070046 
0.757360745  0.346640011  0.8154299886  1.58687786  0.743320270 -0.01406987
60.218569932  1.202454304  1.72645331  0.45512884  0.470599683 
0.201022552  1.27244  1.4007433805  1.50010664  1.938534997  1.18257281

70.011051571  0.104881496 -0.20908604 -0.68639717  0.545488828
-0.166558039  0.095571389 -0.1879643976 -0.10516101 -0.078389855 -0.11663925
8   -0.819021874 -0.082737788  0.85546060 -0.07172932 -0.140994994
-0.385119472 -0.406565855  0.1465003978  0.37208072  0.145712907 -0.59529632
9   -1.832199755  0.480120063  1.43612241  0.05998522  0.021264819
-0.981196107  0.032804234  0.7527178395 -0.10516101  0.593918429  1.25095239
10  -0.904470611  0.752168024  1.24256848  1.81617909 -0.140994994
-0.375184861  0.691859366  0.7945259389  1.36994980  1.490329474  1.14838302



this is the result of applying PCA to the data matrix

Standard deviations:
 [1] 30.6645414  7.5513852  3.6927427  2.8703435  2.5363007  1.9136933 
1.5624131  1.3689630  1.2976189

[10]  1.1633458  1.1118231  0.7847148  0.4802303

Rotation:
PC1 PC2 PC3  PC4  PC5 
PC6  PC7 PC8

var1  0.18110712 -0.74864138 -0.46070566 -0.365658769  0.192810075
-0.132529979  0.023764851  0.03674873
var2  0.86458284  0.34243386 -0.05766909 -0.235504989 -0.046075934 
0.001493006 -0.024535011  0.13439659

var3  0.03765598  0.20097537 -0.15709612 -0.343218776 -0.295201121
-0.073295697 -0.086930370 -0.54389141
var40.05965733  0.01737951  0.09854179 -0.030801791  0.125735684 
0.341795876 -0.001735808  0.37152696

var5   0.23845698 -0.20616399  0.68948870  0.025904812  0.391188182
-0.428933369 -0.101780281 -0.16965893
var6   0.29928369 -0.47394636  0.24791449  0.341235161 -0.511378719 
0.447071255 -0.077534385 -0.13198544

var7 0.19503685  0.01385823 -0.24126047  0.531403827 -0.127426510
-0.410568454  0.608163973 -0.01265457
var8   0.13261863  0.06839078 -0.37740589  0.535332339  0.366103479 
0.032376851 -0.574484605 -0.05645694
var90.06246705  0.04407384 -0.09545362  0.037993146 -0.036651080 
0.012347288 -0.192976142 -0.13027876

var10   0.03027791  0.05533988 -0.03749859 -0.009257423  0.011026593
-0.010770032 -0.104041067  0.12125263
var11  0.07435322  0.04334969 -0.02666944  0.032036374  0.464035624 
0.454970952  0.347507539 -0.60527541
var12 0.04328710  0.04731771  0.00360668 -0.054200633  0.275901346 
0.297800123  0.324323749  0.30487145
var13   0.02095652  0.02146485  0.03598618 -0.022510780  0.005192075 
0.103988977  0.031541374  0.07877455


   PC9 PC10 PC11PC12 PC13
var1   -0.005328345  0.030549780 -0.049283616 -0.02211988  0.015660892
var2   0.170766596 -0.144031738  0.028862963  0.06984674  0.006293703
var3  -0.282549313  0.548650592  0.131284937 -0.14740722 -0.002384605
var4 0.024070488  0.614154008 -0.551480394 -0.03446124 -0.178123011
var5   -0.157551008  0.147685248  0.008044148 -0.04068258  0.007778992
var6   -0.058675551  0.006344813  0.130814072 -0.04088919 -0.028655330
var7 -0.099243751  0.171852216 -0.149231752 -0.06690208 -0.014693444
var80.006629025  0.199158097  0.187226774 -0.02511968  0.070896819
var9-0.658214712 -0.320120384 -0.53990  0.37630539 -0.023642902
var10   -0.259704149 -0.273030750 -0.074006053 -0.83676032

Re: [R] Downloading clim.pact package




ZABLONE OWITI wrote:

Hi all,

I am relatively a new user of R. I need assistance on how to download and
install the  clim.pact_2.2-39.zip. I am running R on Windows.



See the manual R Installation and Administration and afterwards read 
?install.packages


Uwe Ligges

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[R] Question about S4

2009-11-23 Thread Hun S. Tesatte

Dear R-ers,

I don't understand the following, maybe someone will help me 
explain:

 setClasss('A')
[1] A
 new('a')
Error in new(a) : 
  trying to generate an object from a virtual class (a)

 setClass('b', contains='a')
[1] b
 new('b')
An object of class “b”
S4 Type Object

In what way is B more concrete than A so that it's possible do 
instantiate B but not A? I don't quite get it. B adds nothing to 
nothing, and yet it's instantiable, while it's base is not. Makes 
no sense to me.

-- Hun

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Re: [R] HTML help missing?

2009-11-23 Thread Antje


Thanks a lot for the hint!

Antje



Duncan Murdoch wrote:

On 20/11/2009 7:11 AM, Antje wrote:

Hey there,

I'm running R 2.10 on Windows XP (Professional) and I was wondering 
where the HTML help window disappeared?
With earlier versions everything was fine. Now I get only this 
old-fashioned text windows without any links when I type

?some_function
Can anybody help me?

Edit the R_HOME/etc/Rprofile.site file to uncomment the line

options(help_type=html)

(or just install the patched version, this installer bug was 
discovered after release).


Duncan Murdoch



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Re: [R] Question about S4

2009-11-23 Thread Hun S. Tesatte

PS. All class names were upper-case, I messed up while copying the 
code, but it has no effect on the result. Thanks for help.

--Hun


On Mon, 23 Nov 2009 10:28:37 +0100 Hun S. Tesatte 
hunsynte...@hush.com wrote:
Dear R-ers,

I don't understand the following, maybe someone will help me 
explain:

 setClasss('A')
[1] A
 new('a')
Error in new(a) : 
  trying to generate an object from a virtual class (a)

 setClass('b', contains='a')
[1] b
 new('b')
An object of class “b”
S4 Type Object

In what way is B more concrete than A so that it's possible do 
instantiate B but not A? I don't quite get it. B adds nothing to 
nothing, and yet it's instantiable, while it's base is not. Makes 
no sense to me.

-- Hun

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Re: [R] MASS loading error

2009-11-23 Thread Erin Hodgess

The Rcmdr menu does not appear.

However, I tried an experiment: I downloaded the R-2.10.0.tar.gz file
and installed via ./configure, make, and all was well for Rcmdr.

Thanks,
Erin


2009/11/23 Uwe Ligges lig...@statistik.tu-dortmund.de:


 Erin Hodgess wrote:

 Hello again.

 I did the update and it worked well, except for one thing:  the Rcmdr
 doesn't work anymore.

 Has that come up before please?


 Not that I know. But what does doesn't work anymore mean precisely?

 Uwe


 Thank you for all of your help.

 Sincerely,
 Erin



  Ligges lig...@statistik.tu-dortmund.de:

 Erin Hodgess wrote:

 Dear R People:

 I just installed R-2.10.0 on Karmic Koala Ubuntu, via the sudo apt-get
 install r-base, etc.

 However, when I try to install an Rcmdr Plugin package, I get the
 following:

 install.packages(RcmdrPlugin.qual,depen=TRUE)

 Warning in install.packages(RcmdrPlugin.qual, depen = TRUE) :
  argument 'lib' is missing: using '/usr/local/lib/R/site-library'
 Warning in install.packages(RcmdrPlugin.qual, depen = TRUE) :
  'lib = /usr/local/lib/R/site-library' is not writable
 Would you like to create a personal library
 '~/R/i486-pc-linux-gnu-library/2.10'
 to install packages into?  (y/n) y
 --- Please select a CRAN mirror for use in this session ---
 Loading Tcl/Tk interface ... done
 also installing the dependency ‘qcc’

 trying URL
 'http://cran.opensourceresources.org/src/contrib/qcc_2.0.tar.gz'
 Content type 'application/x-gzip' length 163556 bytes (159 Kb)
 opened URL
 ==
 downloaded 159 Kb

 trying URL

 'http://cran.opensourceresources.org/src/contrib/RcmdrPlugin.qual_0.4.0.tar.gz'
 Content type 'application/x-gzip' length 3545 bytes
 opened URL
 ==
 downloaded 3545 bytes

 * installing *source* package ‘qcc’ ...
 ** R
 ** data
 ** demo
 ** inst
 ** preparing package for lazy loading
 ** help
 *** installing help indices
 ** building package indices ...
 * DONE (qcc)
 * installing *source* package ‘RcmdrPlugin.qual’ ...
 ** R
 ** inst
 ** preparing package for lazy loading
 Warning: package 'Rcmdr' was built under R version 2.8.1 and help may
 not work correctly
 Loading required package: tcltk
 Loading Tcl/Tk interface ... done
 The Commander GUI is launched only in interactive sessions

 Attaching package: 'Rcmdr'


       The following object(s) are masked from package:tcltk :

        tclvalue

 Warning: package 'rgl' was built under R version 2.8.1 and help may
 not work correctly
 Warning: package 'abind' was built under R version 2.8.1 and help may
 not work correctly
 Error : This is R 2.10.0, package 'MASS' needs = 2.9.2
 ERROR: lazy loading failed for package ‘RcmdrPlugin.qual’
 * removing
 ‘/home/erin/R/i486-pc-linux-gnu-library/2.10/RcmdrPlugin.qual’

 The downloaded packages are in
       ‘/tmp/RtmpioC2DR/downloaded_packages’
 Warning message:
 In install.packages(RcmdrPlugin.qual, depen = TRUE) :
  installation of package 'RcmdrPlugin.qual' had non-zero exit status

 The line that looks particularly strange is the 'MASS needs = 2.9.2.

 Erin,

 the MASS you have installed in one of your current libraries needs R =
 2.9.2 (perhaps in /usr/local/lib/R/site-library ?)

 Please run
  update.packages(checkBuilt=TRUE)
 and packages in your library will be updated.

 If the one in /usr/local/lib/R/site-library can't be updated, because you
 do
 not have write permissions there, please remove it from the library path.


 Best,
 Uwe Ligges





 Could that be the problem, please?

 Thank you in advance for any help.

 Sincerely,
 Erin









-- 
Erin Hodgess
Associate Professor
Department of Computer and Mathematical Sciences
University of Houston - Downtown
mailto: erinm.hodg...@gmail.com

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Re: [R] boxplot question

2009-11-23 Thread Antje


Peter Ehlers wrote:

If there's been an answer to this, I've missed it.
Here's my take.

Antje wrote:

Hi there,

I was wondering if anybody can explain to me why the boxplot ends up 
with different results in the following case:


I have some integer data as a vector and I compare the stats of 
boxplot with the same data divided by a factor.


I've attached a csv file with both data present (d1, d2). The factor 
is 34.16667.


If I run the boxplot function on d1 I get the following stats:

0.848...
0.907...
0.936...
0.965...
1.024...

For d2 I get these stats:

29
31
32
33
36


If I convert the stats of d1 with the factor, I get

29
31
32
33
35

Obviously different for the upper whisker. But why???

Antje


Antje:

Three comments:
1. I think your 'factor' is actually 205/6, not 34.16667.

2. This looks like another case of FAQ 7.31:

# Let's take your d2 and create d1; I'll call them x and y:
x - rep(c(29:38, 40), c(7, 24, 50, 71, 24, 12, 14, 7, 13, 5, 1))
y - x * 6 / 205

# x is your d2, sorted
# y is your d1, sorted

# The critical values are x[202:203] and y[202:203];
x[201:204]
#[1] 35 35 36 36

# The boxplot stats are:
sx - boxplot.stats(x)$stats
sy - boxplot.stats(y)$stats

# Calculate potential extent of upper whisker:
ux - sx[4] + (sx[4] - sx[2]) * 1.5  #36
uy - sy[4] + (sy[4] - sy[2]) * 1.5  #1.053658536585366

# Is y[203] = uy?
y[203] = uy
#[1] FALSE  #!!!

y[202] = uy
#[1] TRUE

# For x:
x[203] = ux
#[1] TRUE

And there's your answer: for y the whisker
goes to y[202], not y[203], due to the inevitable
imprecision in machine calculation.

3. last comment: I would not use boxplots for data like this.

 -Peter Ehlers



Hi Peter,

thanks a lot for your explanation! Now I understand the difference. I 
was using the boxplot statistic to filter outliers from my data. Do you 
have any suggestion for me what to use instead? (I tried to improve the 
estimation of mean and sd, when iteratively removing outliers by boxplot 
stats...)


Antje

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[R] unable to use coplot function with zoo package

2009-11-23 Thread e-letter

Readers,

I have tried to use the zoo package to merge datasets and then use the
coplot function, but the graph is not fully created. Only the panel
data is shown. Command terminal output below, with csv files. What is
the meaning of the warning message? Can anyone help please?

rhelpatconference.jabber.org
r 251
mandriva 2008

 library(chron)
 library(zoo)
 z1-read.zoo(test1.csv,header=FALSE,sep=,,FUN=times)
 z2-read.zoo(test2.csv,header=FALSE,sep=,,FUN=times)
 z3-read.zoo(test3.csv,header=FALSE,sep=,,FUN=times)
 z4-(na.approx(merge(z1,z2,z3),time(z1)))
 coplot(z4$z2~z4$z1|z4$z3)
Warning messages:
1: Index vectors are of different classes: integer times in:
merge.zoo(e1, e2, all = FALSE, retclass = NULL)
2: Index vectors are of different classes: integer times in:
merge.zoo(e1, e2, all = FALSE, retclass = NULL)
3: Index vectors are of different classes: integer times in:
merge.zoo(e1, e2, all = FALSE, retclass = NULL)
4: Index vectors are of different classes: integer times in:
merge.zoo(e1, e2, all = FALSE, retclass = NULL)
5: Index vectors are of different classes: integer times in:
merge.zoo(e1, e2, all = FALSE, retclass = NULL)
6: Index vectors are of different classes: integer times in:
merge.zoo(e1, e2, all = FALSE, retclass = NULL)
 z4$z1
09:50:01 09:50:30 09:50:44 09:51:00 09:51:27 09:51:30 09:52:00 09:52:10
  320.61   325.43   320.48   315.53   314.63   313.73   316.11   323.88
09:52:30 09:52:54
  331.65   328.48
 z4$z2
09:50:01 09:50:30 09:50:44 09:51:00 09:51:27 09:51:30 09:52:00 09:52:10
404.7500 402.7450 400.7400 300.2450 199.7500 266.7867 333.8233 400.8600
09:52:30 09:52:54
402.8350 404.8100
 z4$z3
09:50:01 09:50:30 09:50:44 09:51:00 09:51:27 09:51:30 09:52:00 09:52:10
 166.120  175.770  185.420  198.395  211.370  214.820  218.270  221.720
09:52:30 09:52:54
 228.370  235.020

test1.csv:
09:50:00,315.79
09:50:30,325.43
09:51:00,315.53
09:51:30,313.73
09:52:00,316.11
09:52:30,331.65
09:53:00,325.31
09:53:30,334.33
09:54:00,334.54
09:54:30,336.55
09:55:00,339.34
09:55:30,341.93
test2.csv:
09:50:01,404.75
09:50:44,400.74
09:51:27,199.75
09:52:10,400.86
09:52:54,404.81
09:53:41,404.36
09:54:20,406.11
09:55:03,408.85
test3.csv:
09:50:01,166.12
09:50:44,185.42
09:51:27,211.37
09:52:10,221.72
09:52:54,235.02
09:53:41,262.85
09:54:20,306.64
09:55:03,346.3

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Re: [R] Computing multivariate normal probabilities. Was: Re: Problem with Numerical derivatives (numDeriv) and mvtnorm

2009-11-23 Thread SL

Hi Torsten,

Thanks for you comment.

If you have some free time to spare, partial derivatives with respect
to bounds and correlation coefficients would be great for pmvnorm! In
complex problems, optim is not very good at estimating the hessian
numerically and first order derivatives help to build an OPG
estimator, which is not very good as compared to an analytical hessian
but still much better than the numerical hessian provided by optim i
have found the problems I study.

Best,
Stephane

2009/11/23 Torsten Hothorn torsten.hoth...@stat.uni-muenchen.de:

 On Sun, 22 Nov 2009, Ravi Varadhan wrote:


 Hi Torsten,


 Hi Ravi,

 It would be useful to warn the users that the multivariate normal
 probability
 calculated by pmvnorm using the GenzBretz algorithm is random, i.e.
 the result can vary between repeated executions of the function.

 only if a different seed is used.

 This would prevent inappropriate use of pmvnorm such as computing
 derivatives of it (see this email thread).


 ?pmvt has Randomized quasi-Monte Carlo methods are used for the
 computations. and appropriate references. In addition, the new book by Alan
 Genz and Frank Bretz covers all technical details in depth, so
 the procedures are well documented.

 Anyway, I'll add a statement to ?pmvnorm.

 Best wishes,

 Torsten

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Re: [R] interaction effects in a Linear model

2009-11-23 Thread Chuck Cleland

On 11/23/2009 1:06 AM, ashok varma wrote:
 hello,
 
 i am fitting a Linear model using R. I have to fit the model considering all
 the interaction effects of order 1 of the independent variables. But I have
 9 variables. So, it will be difficult for me to write all the 36
 combinations in the model. Can anyone please help how to get this done in
 much smarter way??

  This will include all main effects and two-way interactions:

lm(Y ~ (A + B + C + D + E + F + G + H + I)^2, data=mydata)

 thanks a lot,
 Ashok Varma.
 
   [[alternative HTML version deleted]]
 
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-- 
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New York, NY 10010
tel: (212) 845-4495 (Tu, Th)
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[R] Speed up R code

2009-11-23 Thread AnnaFowler


Hi, Im new to R and having some trouble with my code - it works, its just
very slow! Ive tried lots of things, but nothing quite seems to work, so any
help and suggestions would be really appreciated!

I want to calculate the marginal likelihood for every element of a row of a
matrix and the corresponding element in every other row. Then sum these for
each row, so I get an upper triangular matrix which consists of the sum of
the marginal likelihoods for row i and row j. 


X-read.table()
X1-as.matrix(X)

mlr-function(X1,p,q,Vinv=1){
X2-X1^2
  P-length(p)
  Q-length(q)

  Res-matrix(ncol=P,nrow=P)

  for(i in 1:(P-1)){

  lpi-p[i]
Y1-matrix(nrow=(P-i),ncol=Q)
Y2-matrix(nrow=(P-i),ncol=Q)
tB-matrix(nrow=(P-i),ncol=Q)

 for(j in (i+1):P){
 lpj-p[j]
 for(k in 1:Q){
Y1[j-i,k]-sum(X1[c(i,j),k])
Y2[j-i,k]-sum(X2[c(i,j),k])

}
   tB[j-i,]-(lpi+lpj)*q
  }

V1- (Vinv + tB)^(-1)
c-tB/2
a1-1 + c
b1-1 + (Y2 - V1*(Y1^2))/2
   z-log(V1)/2 + lgamma(a1) - c*log.pi - a1*log(b1)
   Res[i,]-c(rep(NA,i),apply(z,1,sum))
  }
  return(Res)
}


Its fine for a matrix of 100x100, but the data Im working with is 3538x116,
and so this can take hours!

Any help would be really appreciated!
Thanks,
Anna
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Re: [R] Speed up R code

2009-11-23 Thread SL

Loops tend to dramatically increase computation time. You may re-write
a vectorized version of your code if possible, i.e. use matrix
algebra. Calculus is a lot faster if one can avoid loops (at least
some of them) .

Best,
Stephane

2009/11/23 AnnaFowler a.fowle...@imperial.ac.uk:

 Hi, Im new to R and having some trouble with my code - it works, its just
 very slow! Ive tried lots of things, but nothing quite seems to work, so any
 help and suggestions would be really appreciated!

 I want to calculate the marginal likelihood for every element of a row of a
 matrix and the corresponding element in every other row. Then sum these for
 each row, so I get an upper triangular matrix which consists of the sum of
 the marginal likelihoods for row i and row j.


 X-read.table()
 X1-as.matrix(X)

 mlr-function(X1,p,q,Vinv=1){
 X2-X1^2
  P-length(p)
  Q-length(q)

  Res-matrix(ncol=P,nrow=P)

  for(i in 1:(P-1)){

      lpi-p[i]
        Y1-matrix(nrow=(P-i),ncol=Q)
        Y2-matrix(nrow=(P-i),ncol=Q)
        tB-matrix(nrow=(P-i),ncol=Q)

         for(j in (i+1):P){
         lpj-p[j]
         for(k in 1:Q){
            Y1[j-i,k]-sum(X1[c(i,j),k])
            Y2[j-i,k]-sum(X2[c(i,j),k])

            }
           tB[j-i,]-(lpi+lpj)*q
          }

        V1- (Vinv + tB)^(-1)
        c-tB/2
        a1-1 + c
        b1-1 + (Y2 - V1*(Y1^2))/2
       z-log(V1)/2 + lgamma(a1) - c*log.pi - a1*log(b1)
       Res[i,]-c(rep(NA,i),apply(z,1,sum))
  }
  return(Res)
 }


 Its fine for a matrix of 100x100, but the data Im working with is 3538x116,
 and so this can take hours!

 Any help would be really appreciated!
 Thanks,
 Anna
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[R] Calibration score for survival probability

2009-11-23 Thread Eleni Rapsomaniki


Good afternoon!

I need to evaluate the goodness-of-fit (aka calibration) for survival 
probability estimates from a Cox model. 
I tried to use 'calibrate' in the Design package but I'm not sure if it 
should/would produce what I need (ie a chi-sq type statistic with a table of 
expected vs observed probabilities). Any other functions I should be aware of?

Also, has anybody come across an implementation of the statistic described in:
A global goodness of fit statistic for Cox regression models by Parzen  
Lpisitz, Biometrics 55, 1999 

Many thanks in advance

Eleni Rapsomaniki

Research Associate
Strangeways Research Laboratory
Department of Public Health and Primary Care

University of Cambridge
 

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[R] randomForest bug in predict.randomForest()

2009-11-23 Thread J Sousa

The NEWS of the randomForest R library mention that version 4.5-13
fixed a bug in predict.randomForest() when newdata is a matrix with no
rownames.

Can anyone explain what were the consequences of this bug in the predictions?

I think the bug was fixed with the new line

if (is.null(rn)) rn - keep

in file
predict.randomForest.R

I got predictions with the new version using a RF trained before
version 4.5-13 and I can't see differences...

Thanks,
jaslqa

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Re: [R] non-intuitive behaviour after type conversion

2009-11-23 Thread Peter Ehlers



Alan Kelly wrote:

Deal list,
I have a data frame (birth) with mixed variables (numeric and 
alphanumeric).  One variable t1stvisit was originally coded as numeric 
with values 1,2, and 3.  After attaching the data frame, this  is what I 
see when I use str(t1stvisit)

actually, str(birth), I suspect, but not important.


$ t1stvisit: int  1 1 1 1 1 1 1 1 2 2 ...

This is as expected.
I then convert t1stvisit to a factor and to avoid creating a second copy 
of this variable independent of the data frame I use:

birth$t1stvisit = as.factor(birth$t1stvisit)
if I check that the conversion has worked:
is.factor(t1stvisit)
[1] FALSE
Now the only object present in the workspace in the data frame birth 
and, as noted,  I have not created any new variables. So why does R 
still treat t1stvisit as numeric?

is.factor(t1stvisit)
[1] FALSE

Yet when I try the following:
  is.factor(birth$t1stvisit)
[1] TRUE
So, there appears to be two versions of t1stvisit  - the original 
numeric version and the correct factor version although ls() only shows 
birth as present in the workspace.

If I type:
  summary(t1stvisit)
   Min. 1st Qu.  MedianMean 3rd Qu.Max.NA's
  1.000   1.000   2.000   1.574   2.000   3.000  29.000
I get the numeric version, but if I try
summary(birth$t1stvisit)
   123 NA's
 180  169   22   29
I get the factor version.

Frankly I feel that this behaviour is non-intuitive and potentially 
problematic. Nor have I seen warnings about this in the various text 
books on R.

Can anyone comment on why this should occur?


I haven't looked at discussions of 'attach()' for a while,
since I rarely use it nowadays (I find with() more convenient
most of the time), but Chapter 6 in 'An Introduction to R'
does discuss it.

There are indeed two versions of 'birth'.
Your basic problem is which version of 'birth' is being modified.
Hint: it's NOT the attached version.
Small example:

 dat - data.frame(x=1:3)
 attach(dat)
 dat$y - 4:6
 y
 #Error: object 'y' not found
 dat$y
 #[1] 4 5 6

BTW, you don't need as.factor(); use factor().

 -Peter Ehlers



Many thanks,
Alan Kelly

Dr. Alan Kelly
Department of Public Health  Primary Care
Trinity College Dublin

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Re: [R] how do i persuade IT to install R on PCs ?? ...and should I ??

2009-11-23 Thread stephen's mailinglist account

On Sun, Nov 22, 2009 at 11:15 PM, David Winsemius dwinsem...@comcast.netwrote:

On Nov 22, 2009, at 4:45 PM, stephen's mailinglist account wrote:

On Sun, Nov 22, 2009 at 11:14 AM, frenchcr frenc...@btinternet.com
wrote:

Please help me persuade IT to install R on my computer!

All suggestions welcome.

Our IT department run scared when you mention software that they have no
working experience of.

I need to know the pros and cons of having R on corporate desktops.

Please no funny stuff, this is quite a serious issue for us.

Pros and cons would be good.

Thanks.
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I requested to have R installed at work.
For me it helped that I have a lot of non-standard technical packages
anyway
that are off radar for support from the IT department anyway - they only
support for original install rights anyway.
They wanted to know what the licence was - GPL is recognised and they
don't
run a mile.
I did my homework and found some other people on a company research site
were already using R so I could use that as justification.
I had some code ready to run that could produce graphs easily that are
very
hard to do in Excel and require a lot of custom code (and even then aren't
good).
We do use some other stats packages anyway and are being encouraged to use
proper packages rather than kludging through in Excel
References like this (below) have been circulated at work which adds
weight
to arguments that we should not just accept the 'standard' Office install.
Although I did not use this in my justification.

@ARTICLE{,
author = {B.D. McCullough and David A. Heiser},

I'm not surprised to see McCollough and Heiser's names on such an article.
They have both a long track record of pointing out Excel's statistical
deficiencies. (I don't they did so together in the past.) MS has turned a
deaf ear to their efforts to point the way to correct methods. It is truly
amazing that MS continues to ignore constrictive criticism and that such
arrogance is compounded by corporate policies encouraging reliance on
demonstrably faulty tools. The full list of articles documenting MS's
resistance to statistical corrections would be much longer that just this
one article and extends back more than a decade.

title = {On the accuracy of statistical procedures in Microsoft Excel
2007},
journal = {Computational Statistics \ Data Analysis},
year = {2008},
volume = {52},
pages = {4570--4578},
number = {10}
}

( http://dx.doi.org/10.1016/j.csda.2008.03.004)
I use R via TINN-R (http://www.sciviews.org/Tinn-R/) on a Windows
desktop.

Stephen

David Winsemius, MD
Heritage Laboratories
West Hartford, CT

Further similar articles can be found at
http://www.practicalstats.com/xlsstats/excelstats.html and in the same issue
of Computional Statistics and data analysis

regards
Stephen

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Re: [R] boxplot question

2009-11-23 Thread Peter Ehlers



Antje wrote:

Peter Ehlers wrote:

If there's been an answer to this, I've missed it.
Here's my take.

Antje wrote:

Hi there,

I was wondering if anybody can explain to me why the boxplot ends up 
with different results in the following case:


I have some integer data as a vector and I compare the stats of 
boxplot with the same data divided by a factor.


I've attached a csv file with both data present (d1, d2). The factor 
is 34.16667.


If I run the boxplot function on d1 I get the following stats:

0.848...
0.907...
0.936...
0.965...
1.024...

For d2 I get these stats:

29
31
32
33
36


If I convert the stats of d1 with the factor, I get

29
31
32
33
35

Obviously different for the upper whisker. But why???

Antje


Antje:

Three comments:
1. I think your 'factor' is actually 205/6, not 34.16667.

2. This looks like another case of FAQ 7.31:

# Let's take your d2 and create d1; I'll call them x and y:
x - rep(c(29:38, 40), c(7, 24, 50, 71, 24, 12, 14, 7, 13, 5, 1))
y - x * 6 / 205

# x is your d2, sorted
# y is your d1, sorted

# The critical values are x[202:203] and y[202:203];
x[201:204]
#[1] 35 35 36 36

# The boxplot stats are:
sx - boxplot.stats(x)$stats
sy - boxplot.stats(y)$stats

# Calculate potential extent of upper whisker:
ux - sx[4] + (sx[4] - sx[2]) * 1.5  #36
uy - sy[4] + (sy[4] - sy[2]) * 1.5  #1.053658536585366

# Is y[203] = uy?
y[203] = uy
#[1] FALSE  #!!!

y[202] = uy
#[1] TRUE

# For x:
x[203] = ux
#[1] TRUE

And there's your answer: for y the whisker
goes to y[202], not y[203], due to the inevitable
imprecision in machine calculation.

3. last comment: I would not use boxplots for data like this.

 -Peter Ehlers



Hi Peter,

thanks a lot for your explanation! Now I understand the difference. I 
was using the boxplot statistic to filter outliers from my data. Do you 
have any suggestion for me what to use instead? (I tried to improve the 
estimation of mean and sd, when iteratively removing outliers by boxplot 
stats...)


Antje



Removing outliers to 'improve ...' is always problematic.
Perhaps you should not use mean or sd? Consider robust alternatives,
e.g. median/IQR. This very much depends on the purpose of the
analysis. See the taskview on Robust Statistical Methods.

For outliers, there's pkg:outliers. I haven't used it.
There seems to be quite a bit more: I got 277 hits from:

library(sos)
???outlier

 -Peter Ehlers

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Re: [R] non-intuitive behaviour after type conversion



On Nov 23, 2009, at 7:34 AM, Peter Ehlers wrote:



Alan Kelly wrote:

Deal list,
I have a data frame (birth) with mixed variables (numeric and  
alphanumeric).  One variable t1stvisit was originally coded as  
numeric with values 1,2, and 3.  After attaching the data frame,  
this  is what I see when I use str(t1stvisit)

actually, str(birth), I suspect, but not important.

$ t1stvisit: int  1 1 1 1 1 1 1 1 2 2 ...
This is as expected.
I then convert t1stvisit to a factor and to avoid creating a second  
copy of this variable independent of the data frame I use:

birth$t1stvisit = as.factor(birth$t1stvisit)
if I check that the conversion has worked:
is.factor(t1stvisit)
[1] FALSE
Now the only object present in the workspace in the data frame  
birth and, as noted,  I have not created any new variables. So  
why does R still treat t1stvisit as numeric?

is.factor(t1stvisit)
[1] FALSE
Yet when I try the following:
 is.factor(birth$t1stvisit)
[1] TRUE
So, there appears to be two versions of t1stvisit  - the original  
numeric version and the correct factor version although ls() only  
shows birth as present in the workspace.

If I type:
 summary(t1stvisit)
  Min. 1st Qu.  MedianMean 3rd Qu.Max.NA's
 1.000   1.000   2.000   1.574   2.000   3.000  29.000
I get the numeric version, but if I try
summary(birth$t1stvisit)
  123 NA's
180  169   22   29
I get the factor version.
Frankly I feel that this behaviour is non-intuitive and potentially  
problematic. Nor have I seen warnings about this in the various  
text books on R.

Can anyone comment on why this should occur?


I haven't looked at discussions of 'attach()' for a while,
since I rarely use it nowadays (I find with() more convenient
most of the time), but Chapter 6 in 'An Introduction to R'
does discuss it.

There are indeed two versions of 'birth'.
Your basic problem is which version of 'birth' is being modified.
Hint: it's NOT the attached version.
Small example:

dat - data.frame(x=1:3)
attach(dat)
dat$y - 4:6
y
#Error: object 'y' not found
dat$y
#[1] 4 5 6

BTW, you don't need as.factor(); use factor().

-Peter Ehlers


Alan;

Let me second Peter's advice. Attach creates more problems than it  
solves. When I ran his code above, I got output from y but it was not  
the 4:6 vector but something else that was in my workspace from a  
prior project. You should also be wary, however, of unexpected (to  
some of us newbies anyway) behavior with with:


 with(dat, z- x + y)
 dat
  x y
1 1 4
2 2 5
3 3 6
Since with is a function the assignment to z was local within that  
environment.


More effective this way.
 dat$z - with(dat, x+y)
 dat
  x y z
1 1 4 5
2 2 5 7
3 3 6 9

--
David





Many thanks,
Alan Kelly
Dr. Alan Kelly
Department of Public Health  Primary Care
Trinity College Dublin
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David Winsemius, MD
Heritage Laboratories
West Hartford, CT

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Re: [R] dynlm predict with newdata?

2009-11-23 Thread Gabor Grothendieck

The dyn package has a predict method that can typically be used to
predict one step ahead (and you can use a loop to get multiple steps).
 To use just preface lm (or glm or any model fitting function in R
that uses model.frame in the same way as lm).  It works with zoo,
zooreg, ts, its and irts class series.

library(dyn) # this also pulls in zoo

# generate test data
set.seed(123)
x - zooreg(rep(1, 10))
for(i in 2:10) x[i] - x[i-1] + rnorm(1)

# fit model
Lag - function(x, k = 1) lag(x, -k)
mod - dyn$lm(x ~ Lag(x))

# perform prediction and plot
x. - predict(mod, list(x = x))
plot(cbind(x, x.), col = 1:2, screen = 1)

# get more info
package?dyn
?dyn

On Sun, Nov 22, 2009 at 9:43 PM, zubin binab...@bellsouth.net wrote:
 Hello, can one use predict, as you can with other model objects like lm,
 with dynlm to predict a new data set that is identical in field names,
 just a different time period.

 Be nice if you could, I don't really want to create a new data set with
 all the lags, hoping it would generate dynamically.  Does not seem to
 work, get a # of column error.  Any suggestions?


 R str(dfz)
 An 'xts' object from 2009-09-25 09:45:06 to 2009-10-19 15:00:57 containing:
  Data: num [1:28232, 1:8] 0.54771 -0.00825 1.27406 0.69705 1.08107 ...
  - attr(*, dimnames)=List of 2
  ..$ : NULL
  ..$ : chr [1:8] PC1 PC2 PC3 PC4 ...
  Indexed by objects of class: [POSIXt,POSIXct] TZ: GMT
  xts Attributes:
  NULL

 R str(z)
 An 'xts' object from 2009-10-21 09:45:04 to 2009-10-21 15:00:56 containing:
  Data: num [1:2304, 1:8] -0.5044 1.237 -0.7764 0.3931 0.0629 ...
  - attr(*, dimnames)=List of 2
  ..$ : NULL
  ..$ : chr [1:8] PC1 PC2 PC3 PC4 ...
  Indexed by objects of class: [POSIXt,POSIXct] TZ: GMT
  xts Attributes:
  NULL


 dols = dynlm(FAS0 ~ L(FAS0,1:10) + L(PC1,0:10) + L(PC2,0:10) +
 L(PC3,0:10) + L(PC4,0:10) + L(PC5,0:10) + L(PC6,0:10) + L(PC7,0:10),
 data=dfz)

 R predict(dols,newdata=z)

 /*Error in fix.by(by.x, x) : 'by' must match numbers of columns*/

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Re: [R] Calibration score for survival probability

2009-11-23 Thread Frank E Harrell Jr


Eleni Rapsomaniki wrote:

Good afternoon!

I need to evaluate the goodness-of-fit (aka calibration) for survival probability estimates from a Cox model. 
I tried to use 'calibrate' in the Design package but I'm not sure if it should/would produce what I need (ie a chi-sq type statistic with a table of expected vs observed probabilities). Any other functions I should be aware of?


Also, has anybody come across an implementation of the statistic described in:
A global goodness of fit statistic for Cox regression models by Parzen  Lpisitz, Biometrics 55, 1999 


Many thanks in advance

Eleni Rapsomaniki


Eleni,

The Design package, and its replacement, the rms package, produces 
calibration curves but no chi-square test because we do not have a 
corresponding method for that.  Formal tests are overused in this 
context anyway.  An index such as the maximum or 90th percentile of 
absolute calibration error are often more useful.  I have learned 
however that any statistical method that categorizes continuous 
variables (in this case, the predictions or the covariate space) is 
arbitrary and has many other problems.  The calibrate functions in the 
rms package have a new option to obtain smooth calibration curves 
without grouping, by fitting spline hazard models during validation.


Note that if you have done any model/variable selection you have to 
re-run such model building from scratch for each resample of the data, 
or the calibration plot will be over optimistic.  calibrate() makes this 
automatic if doing backward stepdown variable selection.  Many 
statisticians make the mistake of only validating the final selected 
model, which can only be done by one-time data splitting (which requires 
 tens of thousands of observations to perform adequately).


Frank



Research Associate
Strangeways Research Laboratory
Department of Public Health and Primary Care

University of Cambridge
 


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--
Frank E Harrell Jr   Professor and Chair   School of Medicine
 Department of Biostatistics   Vanderbilt University

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Re: [R] how do i persuade IT to install R on PCs ?? ...and should I ??


On Nov 23, 2009, at 7:47 AM, stephen's mailinglist account wrote:


 On Sun, Nov 22, 2009 at 11:15 PM, David Winsemius dwinsem...@comcast.net 
  wrote:

 On Nov 22, 2009, at 4:45 PM, stephen's mailinglist account wrote:

 On Sun, Nov 22, 2009 at 11:14 AM, frenchcr frenc...@btinternet.com  
 wrote:



 Please help me persuade IT to install R on my computer!

 All suggestions welcome.

 Our IT department run scared when you mention software that they  
 have no
 working experience of.

 I need to know the pros and cons of having R on corporate desktops.

 Please no funny stuff, this is quite a serious issue for us.

 Pros and cons would be good.

 Thanks.
 --
 I requested to have R installed at work.
 For me it helped that I have a lot of non-standard technical  
 packages anyway
 that are off radar for support from the IT department anyway - they  
 only
 support for original install rights anyway.
 They wanted to know what the licence was - GPL is recognised and  
 they don't
 run a mile.
 I did my homework and found some other people on a company research  
 site
 were already using R so I could use that as justification.
 I had some code ready to run that could produce graphs easily that  
 are very
 hard to do in Excel and require a lot of custom code (and even then  
 aren't
 good).
 We do use some other stats packages anyway and are being encouraged  
 to use
 proper packages rather than kludging through in Excel
 References like this (below) have been circulated at work which adds  
 weight
 to arguments that we should not just accept the 'standard' Office  
 install.
 Although I did not use this in my justification.

 @ARTICLE{,
  author = {B.D. McCullough and David A. Heiser},

 I'm not surprised to see McCollough and Heiser's names on such an  
 article. They have both a long track record of pointing out Excel's  
 statistical deficiencies. (I don't they did so together in the  
 past.)  MS has turned a deaf ear to their efforts to point the way  
 to correct methods. It is truly amazing that MS continues to ignore  
 constrictive criticism and that such arrogance is compounded by  
 corporate policies encouraging reliance on demonstrably faulty  
 tools. The full list of articles documenting MS's resistance to  
 statistical corrections would be much longer that just this one  
 article and extends back more than a decade.


  title = {On the accuracy of statistical procedures in Microsoft Excel
 2007},
  journal = {Computational Statistics \ Data Analysis},
  year = {2008},
  volume = {52},
  pages = {4570--4578},
  number = {10}
 }

 -- 

 David Winsemius, MD



 Further similar articles can be found at 
 http://www.practicalstats.com/xlsstats/excelstats.html 
  and in the same issue of Computional Statistics and data analysis

It was a good read. We had a recent example submitted to r-help where  
I had occasion to test their solution 2 (use OO.org' Calc) and found  
it to be just as bad at curve fitting for a polynomial as had been  
Excel. Take a look at the pdf attached to this r-help item:

https://stat.ethz.ch/pipermail/r-help/2009-November/218005.html



 regards
 Stephen

David Winsemius, MD
Heritage Laboratories
West Hartford, CT


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Re: [R] R: Re: chol( neg.def.matrix ) WAS: Re: Choleski and Choleski with pivoting of matrix fails

2009-11-23 Thread Peter Dalgaard

simona.racio...@libero.it wrote:
 It works! But Once I have the square root of this matrix, how do I convert it 
 to a real (not imaginary) matrix which has the same property? Is that 
 possible?

No. That is theoretically impossible.

If A = B'B, then x'Ax = ||Bx||^2 = 0

for any x, which implies in particular that all eigenvalues of A should
be nonnegative.

 
 Best,
 Simon
 
 Messaggio originale
 Da: p.dalga...@biostat.ku.dk
 Data: 21-nov-2009 18.56
 A: Charles C. Berrycbe...@tajo.ucsd.edu
 Cc: simona.racio...@libero.itsimona.racio...@libero.it, r-h...@r-
 project.org
 Ogg: Re: [R] chol( neg.def.matrix ) WAS: Re: Choleski and Choleski with 
 pivoting of matrix fails
 Charles C. Berry wrote:
 On Sat, 21 Nov 2009, simona.racio...@libero.it wrote:

 Hi Everyone,

 I need to take the square root of the following matrix:

[,1]   [,2][,3]
 [1,]  0.5401984 -0.3998675 -1.3785897
 [2,] -0.3998675  1.0561872  0.8158639
 [3,] -1.3785897  0.8158639  1.6073119

 I tried Choleski which fails. I then tried Choleski with pivoting, but
 unfortunately the square root I get is not valid. I also tried eigen
 decomposition but i did no get far.

 Any clue on how to do it?!

 If you want to take the square root of a negative definite matrix, you 
 could use

 sqrtm( neg.def.mat )

 from the expm package on rforge:

 http://r-forge.r-project.org/projects/expm/
 But that matrix is not negative definite! It has 2 positive and one 
 negative eigenvalue. It is non-positive definite.

 It is fairly easy in any case to get a matrix square root from the eigen 
 decomposition:

 v%*%diag(sqrt(d+0i))%*%t(v)
   [,1]  [,2]  [,3]
 [1,]  0.5164499+0.4152591i -0.1247682-0.0562317i -0.7257079+0.3051868i
 [2,] -0.1247682-0.0562317i  0.9618445+0.0076145i  0.3469916-0.0413264i
 [3,] -0.7257079+0.3051868i  0.3469916-0.0413264i  1.0513849+0.2242912i
 ch - v%*%diag(sqrt(d+0i))%*%t(v)
 t(ch)%*% ch
   [,1]  [,2]  [,3]
 [1,]  0.5401984+0i -0.3998675-0i -1.3785897-0i
 [2,] -0.3998675-0i  1.0561872+0i  0.8158639-0i
 [3,] -1.3785897-0i  0.8158639-0i  1.6073119-0i

 A triangular square root is, er, more difficult, but hardly impossible.

 -- 
O__   Peter Dalgaard Øster Farimagsgade 5, Entr.B
   c/ /'_ --- Dept. of Biostatistics PO Box 2099, 1014 Cph. K
  (*) \(*) -- University of Copenhagen   Denmark  Ph:  (+45) 35327918
 ~~ - (p.dalga...@biostat.ku.dk)  FAX: (+45) 35327907

 
 


-- 
   O__   Peter Dalgaard Øster Farimagsgade 5, Entr.B
  c/ /'_ --- Dept. of Biostatistics PO Box 2099, 1014 Cph. K
 (*) \(*) -- University of Copenhagen   Denmark  Ph:  (+45) 35327918
~~ - (p.dalga...@biostat.ku.dk)  FAX: (+45) 35327907

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Re: [R] unable to use coplot function with zoo package

2009-11-23 Thread Gabor Grothendieck

You have no complete rows so how could it plot anything?
If you had some complete rows then this should work:
coplot(z1 ~ z2 | z3, as.data.frame(z4))

Lines1 - 09:50:00,315.79
09:50:30,325.43
09:51:00,315.53
09:51:30,313.73
09:52:00,316.11
09:52:30,331.65
09:53:00,325.31
09:53:30,334.33
09:54:00,334.54
09:54:30,336.55
09:55:00,339.34
09:55:30,341.93
Lines2 -
09:50:01,404.75
09:50:44,400.74
09:51:27,199.75
09:52:10,400.86
09:52:54,404.81
09:53:41,404.36
09:54:20,406.11
09:55:03,408.85
Lines3 -
09:50:01,166.12
09:50:44,185.42
09:51:27,211.37
09:52:10,221.72
09:52:54,235.02
09:53:41,262.85
09:54:20,306.64
09:55:03,346.3
library(zoo)
library(chron)
z1 - read.zoo(textConnection(Lines1), sep = ,, FUN = times)
z2 - read.zoo(textConnection(Lines2), sep = ,, FUN = times)
z3 - read.zoo(textConnection(Lines3), sep = ,, FUN = times)
z4 - merge(z1, z2, z3)

 z4
 z1 z2 z3
09:50:00 315.79 NA NA
09:50:01 NA 404.75 166.12
09:50:30 325.43 NA NA
09:50:44 NA 400.74 185.42
09:51:00 315.53 NA NA
09:51:27 NA 199.75 211.37
09:51:30 313.73 NA NA
09:52:00 316.11 NA NA
09:52:10 NA 400.86 221.72
09:52:30 331.65 NA NA
09:52:54 NA 404.81 235.02
09:53:00 325.31 NA NA
09:53:30 334.33 NA NA
09:53:41 NA 404.36 262.85
09:54:00 334.54 NA NA
09:54:20 NA 406.11 306.64
09:54:30 336.55 NA NA
09:55:00 339.34 NA NA
09:55:03 NA 408.85 346.30
09:55:30 341.93 NA NA
 packageDescription(zoo)$Version
[1] 1.6-1
 R.version.string # Windows Vista
[1] R version 2.10.0 Patched (2009-11-21 r50532)




On Mon, Nov 23, 2009 at 5:25 AM, e-letter inp...@gmail.com wrote:
 Readers,

 I have tried to use the zoo package to merge datasets and then use the
 coplot function, but the graph is not fully created. Only the panel
 data is shown. Command terminal output below, with csv files. What is
 the meaning of the warning message? Can anyone help please?

 rhelpatconference.jabber.org
 r 251
 mandriva 2008

 library(chron)
 library(zoo)
 z1-read.zoo(test1.csv,header=FALSE,sep=,,FUN=times)
 z2-read.zoo(test2.csv,header=FALSE,sep=,,FUN=times)
 z3-read.zoo(test3.csv,header=FALSE,sep=,,FUN=times)
 z4-(na.approx(merge(z1,z2,z3),time(z1)))
 coplot(z4$z2~z4$z1|z4$z3)
 Warning messages:
 1: Index vectors are of different classes: integer times in:
 merge.zoo(e1, e2, all = FALSE, retclass = NULL)
 2: Index vectors are of different classes: integer times in:
 merge.zoo(e1, e2, all = FALSE, retclass = NULL)
 3: Index vectors are of different classes: integer times in:
 merge.zoo(e1, e2, all = FALSE, retclass = NULL)
 4: Index vectors are of different classes: integer times in:
 merge.zoo(e1, e2, all = FALSE, retclass = NULL)
 5: Index vectors are of different classes: integer times in:
 merge.zoo(e1, e2, all = FALSE, retclass = NULL)
 6: Index vectors are of different classes: integer times in:
 merge.zoo(e1, e2, all = FALSE, retclass = NULL)
 z4$z1
 09:50:01 09:50:30 09:50:44 09:51:00 09:51:27 09:51:30 09:52:00 09:52:10
  320.61   325.43   320.48   315.53   314.63   313.73   316.11   323.88
 09:52:30 09:52:54
  331.65   328.48
 z4$z2
 09:50:01 09:50:30 09:50:44 09:51:00 09:51:27 09:51:30 09:52:00 09:52:10
 404.7500 402.7450 400.7400 300.2450 199.7500 266.7867 333.8233 400.8600
 09:52:30 09:52:54
 402.8350 404.8100
 z4$z3
 09:50:01 09:50:30 09:50:44 09:51:00 09:51:27 09:51:30 09:52:00 09:52:10
  166.120  175.770  185.420  198.395  211.370  214.820  218.270  221.720
 09:52:30 09:52:54
  228.370  235.020

 test1.csv:
 09:50:00,315.79
 09:50:30,325.43
 09:51:00,315.53
 09:51:30,313.73
 09:52:00,316.11
 09:52:30,331.65
 09:53:00,325.31
 09:53:30,334.33
 09:54:00,334.54
 09:54:30,336.55
 09:55:00,339.34
 09:55:30,341.93
 test2.csv:
 09:50:01,404.75
 09:50:44,400.74
 09:51:27,199.75
 09:52:10,400.86
 09:52:54,404.81
 09:53:41,404.36
 09:54:20,406.11
 09:55:03,408.85
 test3.csv:
 09:50:01,166.12
 09:50:44,185.42
 09:51:27,211.37
 09:52:10,221.72
 09:52:54,235.02
 09:53:41,262.85
 09:54:20,306.64
 09:55:03,346.3

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[R] categorisation of continuous variables in R

2009-11-23 Thread joris meys

Dear all,

I'm looking for a function comparable to switch, to categorize a
continuous variable in a few levels. Off course that can be done with
a series of ifelse statements, but that looks rather clumsy. I looked
at switch, but couldn't figure out how to use it for this. I guess
that's not possible, as it only works with characters or integers, not
with intervals.

Basically, I'm looking for a clean way to do :
test - runif(10,1,100)

testFunc -function(x){
x -  ifelse(test10,lowest,
ifelse(10=test  test 50,low,
  ifelse(50=test  test 90,high,
ifelse(90=test  test100,highest,NA)
  )
)
  )
return(as.factor(x))
}

testFunc(test)

Thank you in advance

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Re: [R] FUN argument to return a vector in aggregate function

2009-11-23 Thread Gabor Grothendieck

Try this:

 library(doBy)
 summaryBy(breaks ~ ., warpbreaks, FUN = c(mean, sum, length))

  wool tension breaks.mean breaks.sum breaks.length
1A   L44.6401 9
2A   M24.0216 9
3A   H24.6221 9
4B   L28.2254 9
5B   M28.8259 9
6B   H18.8169 9

On Mon, Nov 23, 2009 at 3:15 AM, utkarshsinghal
utkarsh.sing...@global-analytics.com wrote:
 Hi All,

 I am currently doing the following to compute summary statistics of
 aggregated data:
 a = aggregate(warpbreaks$breaks, warpbreaks[,-1], mean)
 b = aggregate(warpbreaks$breaks, warpbreaks[,-1], sum)
 c = aggregate(warpbreaks$breaks, warpbreaks[,-1], length)
 ans = cbind(a, b[,3], c[,3])

 This seems unnecessarily complex to me so I tried
 aggregate(warpbreaks$breaks, warpbreaks[,-1], function(z)
 c(mean(z),sum(z),length(z)))
 but aggregate doesn't allow FUN argument to return a vector.

 I tried by, tapply and several other functions as well but the output
 needed further modifications to get the same format as ans above.

 Is there any other function same as aggregate which allow FUN argument to
 return vector.

 Regards
 Utkarsh

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Re: [R] categorisation of continuous variables in R

2009-11-23 Thread joris meys

Never mind, found the function :

cut(test,breaks=c(0,10,50,90,100),labels=c(lowest,low,high,highest),include.lowest=T,right=F)

Cheers
Joris

On Mon, Nov 23, 2009 at 2:14 PM, joris meys jorism...@gmail.com wrote:
 Dear all,

 I'm looking for a function comparable to switch, to categorize a
 continuous variable in a few levels. Off course that can be done with
 a series of ifelse statements, but that looks rather clumsy. I looked
 at switch, but couldn't figure out how to use it for this. I guess
 that's not possible, as it only works with characters or integers, not
 with intervals.

 Basically, I'm looking for a clean way to do :
 test - runif(10,1,100)

 testFunc -function(x){
 x -  ifelse(test10,lowest,
        ifelse(10=test  test 50,low,
          ifelse(50=test  test 90,high,
            ifelse(90=test  test100,highest,NA)
          )
        )
      )
 return(as.factor(x))
 }

 testFunc(test)

 Thank you in advance


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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
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Re: [R] categorisation of continuous variables in R



On Nov 23, 2009, at 8:14 AM, joris meys wrote:


Dear all,

I'm looking for a function comparable to switch, to categorize a
continuous variable in a few levels.


?cut



Off course that can be done with
a series of ifelse statements, but that looks rather clumsy. I looked
at switch, but couldn't figure out how to use it for this. I guess
that's not possible, as it only works with characters or integers, not
with intervals.

Basically, I'm looking for a clean way to do :
test - runif(10,1,100)


 test - runif(10,1,100)
 testcut - cut(test, breaks=c(-Inf, 10, 50, 90, 100))
 table(testcut)
testcut
(-Inf,10]   (10,50]   (50,90]  (90,100]
1 4 5 0




testFunc -function(x){
x -  ifelse(test10,lowest,
   ifelse(10=test  test 50,low,
 ifelse(50=test  test 90,high,
   ifelse(90=test  test100,highest,NA)
 )
   )
 )
return(as.factor(x))
}

testFunc(test)

Thank you in advance



--
David Winsemius, MD
Heritage Laboratories
West Hartford, CT

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Re: [R] how do i persuade IT to install R on PCs ?? ...and should I ??

2009-11-23 Thread Barry Rowlingson

On Mon, Nov 23, 2009 at 1:01 PM, David Winsemius dwinsem...@comcast.net wrote:

 It was a good read. We had a recent example submitted to r-help where
 I had occasion to test their solution 2 (use OO.org' Calc) and found
 it to be just as bad at curve fitting for a polynomial as had been
 Excel. Take a look at the pdf attached to this r-help item:

 https://stat.ethz.ch/pipermail/r-help/2009-November/218005.html

 Well, the OO.org guys are trying to make something compatible with a
piece of MS software, but maybe this is taking it too far.

 I love the story that sections of Microsoft XML spec for Office says
things like Do whatever Excel does, thus setting into stone the bugs
inherent in that package as an ISO standard...

Barry

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

[R] Check if string has all alphabets or numbers

2009-11-23 Thread Harsh

Hi R users,
I'd like to know if anyone has come across problems wherein it was necessary
to check if strings contained all alphabets, some numbers or all numbers?

In my attempt to test if a string is numeric, alpha-numeric (also includes
if string is only alphabets) :

# Reproducible R code below
mywords- c(harry,met,sally,subway10,1800Movies,12345)

mywords.alphanum
-lapply(sapply(mywords,function(x)strsplit(x,NULL)),function(y)
ifelse(sum(is.na(sapply(y,as.numeric))) == 0  length(y) 
0,numeric,alpha-numeric))

names(mywords.alphanum)[(which(mywords.alphanum == numeric))]


I understand that such one-liners  (the second line of code above) that
make multiple calls are discouraged, but I seem to find then fascinating.

Looking forward to alternate solutions/packages  for the above problem.

Thanks
Harsh Singhal

[[alternative HTML version deleted]]

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

Re: [R] Calibration score for survival probability

2009-11-23 Thread Eleni Rapsomaniki

Dear Prof. Harrell,
Thank you very much for your prompt and very helpful response. I guess
that since a global statistic such as a chi-sq test is not applicable in
this case, the calibration curve itself from the calibration() is the
most informative alternative (most graphical methods reveal more
information than a single statistic anyway!). I will try the updated
version in the rms package to compare.

Best Wishes 

Eleni Rapsomaniki
 
Research Associate
Strangeways Research Laboratory
Department of Public Health and Primary Care
University of Cambridge
 

-Original Message-
From: Frank E Harrell Jr [mailto:f.harr...@vanderbilt.edu] 
Sent: 23 November 2009 13:01
To: Eleni Rapsomaniki
Cc: r-help@r-project.org
Subject: Re: [R] Calibration score for survival probability

Eleni Rapsomaniki wrote:
 Good afternoon!
 
 I need to evaluate the goodness-of-fit (aka calibration) for survival
probability estimates from a Cox model. 
 I tried to use 'calibrate' in the Design package but I'm not sure if
it should/would produce what I need (ie a chi-sq type statistic with a
table of expected vs observed probabilities). Any other functions I
should be aware of?
 
 Also, has anybody come across an implementation of the statistic
described in:
 A global goodness of fit statistic for Cox regression models by
Parzen  Lpisitz, Biometrics 55, 1999 
 
 Many thanks in advance
 
 Eleni Rapsomaniki

Eleni,

The Design package, and its replacement, the rms package, produces 
calibration curves but no chi-square test because we do not have a 
corresponding method for that.  Formal tests are overused in this 
context anyway.  An index such as the maximum or 90th percentile of 
absolute calibration error are often more useful.  I have learned 
however that any statistical method that categorizes continuous 
variables (in this case, the predictions or the covariate space) is 
arbitrary and has many other problems.  The calibrate functions in the 
rms package have a new option to obtain smooth calibration curves 
without grouping, by fitting spline hazard models during validation.

Note that if you have done any model/variable selection you have to 
re-run such model building from scratch for each resample of the data, 
or the calibration plot will be over optimistic.  calibrate() makes this

automatic if doing backward stepdown variable selection.  Many 
statisticians make the mistake of only validating the final selected 
model, which can only be done by one-time data splitting (which requires

  tens of thousands of observations to perform adequately).

Frank

 
 Research Associate
 Strangeways Research Laboratory
 Department of Public Health and Primary Care
 
 University of Cambridge
  
 
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 R-help@r-project.org mailing list
 https://stat.ethz.ch/mailman/listinfo/r-help
 PLEASE do read the posting guide
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 and provide commented, minimal, self-contained, reproducible code.
 


-- 
Frank E Harrell Jr   Professor and Chair   School of Medicine
  Department of Biostatistics   Vanderbilt
University

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Re: [R] Check if string has all alphabets or numbers

2009-11-23 Thread jim holtman

try this:

 mywords- c(harry,met,sally,subway10,1800Movies,12345)
 grep(^[[:alpha:]]*$, mywords)  # letters
[1] 1 2 3
 grep(^[[:digit:]]*$, mywords)  # numbers
[1] 6


On Mon, Nov 23, 2009 at 8:28 AM, Harsh singhal...@gmail.com wrote:
 Hi R users,
 I'd like to know if anyone has come across problems wherein it was necessary
 to check if strings contained all alphabets, some numbers or all numbers?

 In my attempt to test if a string is numeric, alpha-numeric (also includes
 if string is only alphabets) :

 # Reproducible R code below
 mywords- c(harry,met,sally,subway10,1800Movies,12345)

 mywords.alphanum
 -lapply(sapply(mywords,function(x)strsplit(x,NULL)),function(y)
 ifelse(sum(is.na(sapply(y,as.numeric))) == 0  length(y) 
 0,numeric,alpha-numeric))

 names(mywords.alphanum)[(which(mywords.alphanum == numeric))]


 I understand that such one-liners  (the second line of code above) that
 make multiple calls are discouraged, but I seem to find then fascinating.

 Looking forward to alternate solutions/packages  for the above problem.

 Thanks
 Harsh Singhal

        [[alternative HTML version deleted]]

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




-- 
Jim Holtman
Cincinnati, OH
+1 513 646 9390

What is the problem that you are trying to solve?

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

Re: [R] Check if string has all alphabets or numbers

2009-11-23 Thread jim holtman

Added a little more:

 mywords- c(harry,met,sally,subway10,1800Movies,12345, not 
 correct 123)
 all.letters - grep(^[[:alpha:]]*$, mywords)
 all.numbers - grep(^[[:digit:]]*$, mywords)  # numbers
 mixed - grep(^[[:digit:][:alpha:]]*$, mywords)
 all.letters
[1] 1 2 3
 all.numbers
[1] 6
 # mixed
 setdiff(mixed, c(all.numbers, all.letters))
[1] 4 5
 # not any of the above
 setdiff(seq(length(mywords)), c(mixed, all.numbers, all.letters))
[1] 7



On Mon, Nov 23, 2009 at 8:28 AM, Harsh singhal...@gmail.com wrote:
 Hi R users,
 I'd like to know if anyone has come across problems wherein it was necessary
 to check if strings contained all alphabets, some numbers or all numbers?

 In my attempt to test if a string is numeric, alpha-numeric (also includes
 if string is only alphabets) :

 # Reproducible R code below
 mywords- c(harry,met,sally,subway10,1800Movies,12345)

 mywords.alphanum
 -lapply(sapply(mywords,function(x)strsplit(x,NULL)),function(y)
 ifelse(sum(is.na(sapply(y,as.numeric))) == 0  length(y) 
 0,numeric,alpha-numeric))

 names(mywords.alphanum)[(which(mywords.alphanum == numeric))]


 I understand that such one-liners  (the second line of code above) that
 make multiple calls are discouraged, but I seem to find then fascinating.

 Looking forward to alternate solutions/packages  for the above problem.

 Thanks
 Harsh Singhal

        [[alternative HTML version deleted]]

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




-- 
Jim Holtman
Cincinnati, OH
+1 513 646 9390

What is the problem that you are trying to solve?

__
R-help@r-project.org mailing list
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and provide commented, minimal, self-contained, reproducible code.

Re: [R] how do i persuade IT to install R on PCs ?? ...and should I ??

2009-11-23 Thread Marc Schwartz



On Nov 23, 2009, at 7:24 AM, Barry Rowlingson wrote:

On Mon, Nov 23, 2009 at 1:01 PM, David Winsemius dwinsem...@comcast.net 
 wrote:



It was a good read. We had a recent example submitted to r-help where
I had occasion to test their solution 2 (use OO.org' Calc) and  
found

it to be just as bad at curve fitting for a polynomial as had been
Excel. Take a look at the pdf attached to this r-help item:

https://stat.ethz.ch/pipermail/r-help/2009-November/218005.html


Well, the OO.org guys are trying to make something compatible with a
piece of MS software, but maybe this is taking it too far.

I love the story that sections of Microsoft XML spec for Office says
things like Do whatever Excel does, thus setting into stone the bugs
inherent in that package as an ISO standard...

Barry




Just to reinforce Baz' statement about OO.org, for those who have been  
around here for a while, you may recall this discussion back in 2003:


https://stat.ethz.ch/pipermail/r-help/2003-June/034565.html

https://stat.ethz.ch/pipermail/r-help/2003-June/034860.html


Nothing has changed since then in OO.org's Calc version 3.1.1, which  
is the current release.



On OSX, I can add Numbers from Apple's iWork '09 the mix:

Formula: =4.145 * 100 + 0.5
Result:  415.00

Formula: =0.5 - 0.4 - 0.1
Result:  -0.27755575615629

Formula: =(0.5 - 0.4 - 0.1)
Result: -0.27755575615629


So FWIW, Apple has not made the same errors as MS and OO.org, at least  
in this narrow example.


Also, here are two additional Excel resources:

David Heiser's page: http://www.daheiser.info/excel/frontpage.html

Patrick Burns' Spreadsheet Addition: 
http://www.burns-stat.com/pages/Tutor/spreadsheet_addiction.html


HTH,

Marc Schwartz

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Re: [R] Over-coloring facets on persp() plot

2009-11-23 Thread Marc Chiarini (Tufts)

Thanks Duncan (and David).  I couldn't get back to my computer until 
today.  I understand it pretty well now and I'm able to get what I need.


On a side note, I have had a hard time getting rgl to work.  For the 
moment, I have to compile without libpng and ftfonts, and when I load 
the package I get an error related to OpenGL (from memory, something 
like no GLX extension found).  I'll write more as soon as I'm able.


Regards,
Marc Chiarini

Duncan Murdoch wrote:

On 22/11/2009 1:07 AM, Marc Chiarini (Tufts) wrote:

Dear R Community:

Recently, I have managed to plot some really useful graphs of my 
research data using persp().  I have even figured out how to overplot 
rectangular regions (corresponding to submatrices) with a different 
color.  This is accomplished by using par(new=T).  I am now searching 
for a way to highlight a set of (possibly non-contiguous) facets 
with a specific color, e.g., the facet between each set of four 
points whose values are all above a certain threshold.  An example 
would be coloring the raised corners of the classic sombrero (found 
in example(persp)) differently from the rest of the sombrero.  I feel 
like the last example in persp() is pointing me in the right 
direction, but I'm not quite getting it.  Any help is much appreciated.


Think of the facets as an nx-1 by ny-1 matrix.  Pass the col arg by 
creating a matrix of this shape.  (A vector version of the data in the 
matrix would also be good enough.)


If you pass something shorter, it will be recycled to that length.

You could also use persp3d from the rgl package, but an important 
difference is that it colours all nx by ny vertices, and interpolates 
colours on the facets.  So you can't use the same colour matrix as in 
persp.


Duncan Murdoch
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Re: [R] How to add a top level title to multiple plots

2009-11-23 Thread Jason Rupert

In the example below, is there any way to get the top title, i.e. Distribution 
Comparisons, in a bit from the top margin?  

Thanks agian


 


- Original Message 
From: Jason Rupert jasonkrup...@yahoo.com
To: Peter Ehlers ehl...@ucalgary.ca
Cc: R Project Help R-help@r-project.org; Me jasonkrup...@yahoo.com
Sent: Sat, November 21, 2009 12:31:04 AM
Subject: Re: [R] How to add a top level title to multiple plots

Cool. 

I ended up with the following:



par(mfrow = c(2, 2))
# Plot 1
plot(rnorm(10),type=l,col=red)
title(main = list(paste(Normal),
      col=black, cex = 1.0))


# Plot 2
plot(rpois(10, 4),type=l,col=blue)
title(main = list(paste(Poison),
      col=black, cex = 1.0))


# Plot 3
plot(rnbinom(10, mu = 4, size = 100),type=l,col=green)
title(main = list(paste(Binomial),
      col=black, cex = 1.0))

# Plot 4
plot(runif(10),type=l,col=black)
title(main = list(paste(Uniform),
      col=black, cex = 1.0))


title(Distribution Comparisons, outer = TRUE)

# c(bottom, left, top, right)
par(oma=c(2,2,3,2))


Thanks again. 





- Original Message 
From: Peter Ehlers ehl...@ucalgary.ca
To: Jason Rupert jasonkrup...@yahoo.com
Cc: R-help@r-project.org
Sent: Fri, November 20, 2009 5:23:33 PM
Subject: Re: [R] How to add a top level title to multiple plots


Jason Rupert wrote:
 How can I add an overall plot title to these four plots? 
 I would like to have something that says, Distribution Comparisons:
 
title(Distribution Comparisons, outer = TRUE)

But you may have to make room for it with par(oma=...).

-Peter Ehlers

 
 par(mfrow = c(2, 2))
 # Plot 1
 plot(rnorm(10),type=l,col=red)
 title(main = list(paste(Normal),
                          col=black, cex = 1.0))
 
 
 # Plot 2
 plot(rpois(10, 4),type=l,col=blue)
 title(main = list(paste(Poison),
                          col=black, cex = 1.0))
 
 
 # Plot 3
 plot(rnbinom(10, mu = 4, size = 100),type=l,col=green)
 title(main = list(paste(Binomial),
                          col=black, cex = 1.0))
 
 # Plot 4
 plot(runif(10),type=l,col=black)
 title(main = list(paste(Uniform),
                          col=black, cex = 1.0))
 
 
 Thanks again for any feedback and insights. 
 
 
 
 __
 R-help@r-project.org mailing list
 https://stat.ethz.ch/mailman/listinfo/r-help
 PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
 and provide commented, minimal, self-contained, reproducible code.
 
 




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

Re: [R] Over-coloring facets on persp() plot

2009-11-23 Thread Duncan Murdoch


On 23/11/2009 8:55 AM, Marc Chiarini (Tufts) wrote:
Thanks Duncan (and David).  I couldn't get back to my computer until 
today.  I understand it pretty well now and I'm able to get what I need.


On a side note, I have had a hard time getting rgl to work.  For the 
moment, I have to compile without libpng and ftfonts, and when I load 
the package I get an error related to OpenGL (from memory, something 
like no GLX extension found).  I'll write more as soon as I'm able.



You likely need to install OpenGL support on your system.  (It might be 
in MesaGL.)


Duncan Murdoch


Regards,
Marc Chiarini

Duncan Murdoch wrote:

On 22/11/2009 1:07 AM, Marc Chiarini (Tufts) wrote:

Dear R Community:

Recently, I have managed to plot some really useful graphs of my 
research data using persp().  I have even figured out how to overplot 
rectangular regions (corresponding to submatrices) with a different 
color.  This is accomplished by using par(new=T).  I am now searching 
for a way to highlight a set of (possibly non-contiguous) facets 
with a specific color, e.g., the facet between each set of four 
points whose values are all above a certain threshold.  An example 
would be coloring the raised corners of the classic sombrero (found 
in example(persp)) differently from the rest of the sombrero.  I feel 
like the last example in persp() is pointing me in the right 
direction, but I'm not quite getting it.  Any help is much appreciated.
Think of the facets as an nx-1 by ny-1 matrix.  Pass the col arg by 
creating a matrix of this shape.  (A vector version of the data in the 
matrix would also be good enough.)


If you pass something shorter, it will be recycled to that length.

You could also use persp3d from the rgl package, but an important 
difference is that it colours all nx by ny vertices, and interpolates 
colours on the facets.  So you can't use the same colour matrix as in 
persp.


Duncan Murdoch



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


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

Re: [R] Check if string has all alphabets or numbers

2009-11-23 Thread hadley wickham

 mywords- c(harry,met,sally,subway10,1800Movies,12345, not 
 correct 123)
 all.letters - grep(^[[:alpha:]]*$, mywords)
 all.numbers - grep(^[[:digit:]]*$, mywords)  # numbers
 mixed - grep(^[[:digit:][:alpha:]]*$, mywords)

mywords- c(harry,met,sally,subway10,1800Movies,12345,
not correct 123, )
mywords[grepl(^[[:digit:][:alpha:]]*$, mywords)]

So maybe you should use

mywords[grepl(^[[:digit:][:alpha:]]+$, mywords)]


And grepl is highly recommended over grep.

Hadley

-- 
http://had.co.nz/

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Re: [R] other decriptive stats packages

2009-11-23 Thread Matthieu Lesnoff

  are there any more packages that help decribe and explore data

 sets 

package tdisplay at http://forums.cirad.fr/logiciel-R/viewtopic.php?t=2409

Details:

 display-packageTool Box for Data Importation and Display

 import Import Data From External Data Bases
(Access, Excel or Dbf formats) using package RODBC

 displayContents of Variables of a Data frame

 aggstatComplement to Aggregate
 univar Descriptive Statistics for a Continuous variable
 quant  Quantiles for a Continuous variable

 ctab   Cross Tabulation (table outputs)
 ctab2  Cross Tabulation (data frame outputs)

 splitbin   Splits Binomial Data into Bernoulli Data (used in
functions ctab and ctab2)

[[alternative HTML version deleted]]

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Re: [R] How to add a top level title to multiple plots



On Nov 23, 2009, at 8:58 AM, Jason Rupert wrote:

In the example below, is there any way to get the top title, i.e.  
Distribution Comparisons, in a bit from the top margin?




You could use instead:

title(\nDistribution Comparisons, outer = TRUE)

And please correct the spelling of Poisson.

--
David


Thanks agian





- Original Message 
From: Jason Rupert jasonkrup...@yahoo.com
To: Peter Ehlers ehl...@ucalgary.ca
Cc: R Project Help R-help@r-project.org; Me jasonkrup...@yahoo.com
Sent: Sat, November 21, 2009 12:31:04 AM
Subject: Re: [R] How to add a top level title to multiple plots

Cool.

I ended up with the following:



par(mfrow = c(2, 2))
# Plot 1
plot(rnorm(10),type=l,col=red)
title(main = list(paste(Normal),
  col=black, cex = 1.0))


# Plot 2
plot(rpois(10, 4),type=l,col=blue)
title(main = list(paste(Poison),
  col=black, cex = 1.0))


# Plot 3
plot(rnbinom(10, mu = 4, size = 100),type=l,col=green)
title(main = list(paste(Binomial),
  col=black, cex = 1.0))

# Plot 4
plot(runif(10),type=l,col=black)
title(main = list(paste(Uniform),
  col=black, cex = 1.0))


title(Distribution Comparisons, outer = TRUE)

# c(bottom, left, top, right)
par(oma=c(2,2,3,2))


Thanks again.





- Original Message 
From: Peter Ehlers ehl...@ucalgary.ca
To: Jason Rupert jasonkrup...@yahoo.com
Cc: R-help@r-project.org
Sent: Fri, November 20, 2009 5:23:33 PM
Subject: Re: [R] How to add a top level title to multiple plots


Jason Rupert wrote:

How can I add an overall plot title to these four plots?
I would like to have something that says, Distribution Comparisons:


title(Distribution Comparisons, outer = TRUE)

But you may have to make room for it with par(oma=...).

-Peter Ehlers



par(mfrow = c(2, 2))
# Plot 1
plot(rnorm(10),type=l,col=red)
title(main = list(paste(Normal),
  col=black, cex = 1.0))


# Plot 2
plot(rpois(10, 4),type=l,col=blue)
title(main = list(paste(Poison),
  col=black, cex = 1.0))


# Plot 3
plot(rnbinom(10, mu = 4, size = 100),type=l,col=green)
title(main = list(paste(Binomial),
  col=black, cex = 1.0))

# Plot 4
plot(runif(10),type=l,col=black)
title(main = list(paste(Uniform),
  col=black, cex = 1.0))


Thanks again for any feedback and insights.



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David Winsemius, MD
Heritage Laboratories
West Hartford, CT

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Re: [R] Speed up R code

2009-11-23 Thread AnnaFowler


Thanks Stephane, Thats a great help!


SL-16 wrote:
 
 Loops tend to dramatically increase computation time. You may re-write
 a vectorized version of your code if possible, i.e. use matrix
 algebra. Calculus is a lot faster if one can avoid loops (at least
 some of them) .
 
 Best,
 Stephane
 
 2009/11/23 AnnaFowler a.fowle...@imperial.ac.uk:

 Hi, Im new to R and having some trouble with my code - it works, its just
 very slow! Ive tried lots of things, but nothing quite seems to work, so
 any
 help and suggestions would be really appreciated!

 I want to calculate the marginal likelihood for every element of a row of
 a
 matrix and the corresponding element in every other row. Then sum these
 for
 each row, so I get an upper triangular matrix which consists of the sum
 of
 the marginal likelihoods for row i and row j.


 X-read.table()
 X1-as.matrix(X)

 mlr-function(X1,p,q,Vinv=1){
 X2-X1^2
  P-length(p)
  Q-length(q)

  Res-matrix(ncol=P,nrow=P)

  for(i in 1:(P-1)){

      lpi-p[i]
        Y1-matrix(nrow=(P-i),ncol=Q)
        Y2-matrix(nrow=(P-i),ncol=Q)
        tB-matrix(nrow=(P-i),ncol=Q)

         for(j in (i+1):P){
         lpj-p[j]
         for(k in 1:Q){
            Y1[j-i,k]-sum(X1[c(i,j),k])
            Y2[j-i,k]-sum(X2[c(i,j),k])

            }
           tB[j-i,]-(lpi+lpj)*q
          }

        V1- (Vinv + tB)^(-1)
        c-tB/2
        a1-1 + c
        b1-1 + (Y2 - V1*(Y1^2))/2
       z-log(V1)/2 + lgamma(a1) - c*log.pi - a1*log(b1)
       Res[i,]-c(rep(NA,i),apply(z,1,sum))
  }
  return(Res)
 }


 Its fine for a matrix of 100x100, but the data Im working with is
 3538x116,
 and so this can take hours!

 Any help would be really appreciated!
 Thanks,
 Anna
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[R] Re adline()

2009-11-23 Thread yonosoyelmejor


Hello, I would like to ask you a question.I have a program in R and I use the
readline method to ask the user some things,but i don´t use the R console
but I use Win console then not appear what I put.I put the code as you look
for:

cat(1- 24horas\n)
cat(2- 12horas\n)
cat(3- 8horas\n)
selection-readline(prompt=\nSelecciona numero de horas:)

if(selection==1){
prediccion=exp(x.reconstruida[1441:1450])
}
if(selection==2){
prediccion=exp(x.reconstruida[720:729])
}
if(selection==3){
prediccion=exp(x.reconstruida[481:491])
}

write.table(prediccion,C:\\Temp\\prePrueba.txt,quote=F,row.names=F,append=T,col.names=F)


//end of code.

Excuse me but my English is bad, i hope explained me well,

A greeting,
Ignacio.


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[R] R: chol( neg.def.matrix ) WAS: Re: Choleski and Choleski with pivoting of matrix fails

2009-11-23 Thread simona.racio...@libero.it

Thanks Chuck!
It works! But Once I have the square root of this matrix, how do I convert it 
to a real (not imaginary) matrix which has the same property? Is that 
possible?

Best,
Simon

Messaggio originale
Da: cbe...@tajo.ucsd.edu
Data: 21-nov-2009 18.11
A: simona.racio...@libero.itsimona.racio...@libero.it
Cc: r-help@r-project.org
Ogg: chol( neg.def.matrix ) WAS: Re: [R] Choleski and Choleski with pivoting 
of matrix fails

On Sat, 21 Nov 2009, simona.racio...@libero.it wrote:

 Hi Everyone,

 I need to take the square root of the following matrix:

[,1]   [,2][,3]
 [1,]  0.5401984 -0.3998675 -1.3785897
 [2,] -0.3998675  1.0561872  0.8158639
 [3,] -1.3785897  0.8158639  1.6073119

 I tried Choleski which fails. I then tried Choleski with pivoting, but
 unfortunately the square root I get is not valid. I also tried eigen
 decomposition but i did no get far.

 Any clue on how to do it?!


If you want to take the square root of a negative definite matrix, you 
could use

   sqrtm( neg.def.mat )

from the expm package on rforge:

   http://r-forge.r-project.org/projects/expm/

HTH,

Chuck



 Thanks,
 Simon

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


Charles C. Berry(858) 534-2098
 Dept of Family/Preventive 
Medicine
E mailto:cbe...@tajo.ucsd.edu  UC San Diego
http://famprevmed.ucsd.edu/faculty/cberry/  La Jolla, San Diego 92093-0901




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[R] R: Re: chol( neg.def.matrix ) WAS: Re: Choleski and Choleski with pivoting of matrix fails

2009-11-23 Thread simona.racio...@libero.it

It works! But Once I have the square root of this matrix, how do I convert it 
to a real (not imaginary) matrix which has the same property? Is that 
possible?

Best,
Simon

Messaggio originale
Da: p.dalga...@biostat.ku.dk
Data: 21-nov-2009 18.56
A: Charles C. Berrycbe...@tajo.ucsd.edu
Cc: simona.racio...@libero.itsimona.racio...@libero.it, r-h...@r-
project.org
Ogg: Re: [R] chol( neg.def.matrix ) WAS: Re: Choleski and Choleski with 
pivoting of matrix fails

Charles C. Berry wrote:
 On Sat, 21 Nov 2009, simona.racio...@libero.it wrote:
 
 Hi Everyone,

 I need to take the square root of the following matrix:

[,1]   [,2][,3]
 [1,]  0.5401984 -0.3998675 -1.3785897
 [2,] -0.3998675  1.0561872  0.8158639
 [3,] -1.3785897  0.8158639  1.6073119

 I tried Choleski which fails. I then tried Choleski with pivoting, but
 unfortunately the square root I get is not valid. I also tried eigen
 decomposition but i did no get far.

 Any clue on how to do it?!
 
 
 If you want to take the square root of a negative definite matrix, you 
 could use
 
 sqrtm( neg.def.mat )
 
 from the expm package on rforge:
 
 http://r-forge.r-project.org/projects/expm/

But that matrix is not negative definite! It has 2 positive and one 
negative eigenvalue. It is non-positive definite.

It is fairly easy in any case to get a matrix square root from the eigen 
decomposition:

  v%*%diag(sqrt(d+0i))%*%t(v)
   [,1]  [,2]  [,3]
[1,]  0.5164499+0.4152591i -0.1247682-0.0562317i -0.7257079+0.3051868i
[2,] -0.1247682-0.0562317i  0.9618445+0.0076145i  0.3469916-0.0413264i
[3,] -0.7257079+0.3051868i  0.3469916-0.0413264i  1.0513849+0.2242912i
  ch - v%*%diag(sqrt(d+0i))%*%t(v)
  t(ch)%*% ch
   [,1]  [,2]  [,3]
[1,]  0.5401984+0i -0.3998675-0i -1.3785897-0i
[2,] -0.3998675-0i  1.0561872+0i  0.8158639-0i
[3,] -1.3785897-0i  0.8158639-0i  1.6073119-0i

A triangular square root is, er, more difficult, but hardly impossible.

-- 
O__   Peter Dalgaard Øster Farimagsgade 5, Entr.B
   c/ /'_ --- Dept. of Biostatistics PO Box 2099, 1014 Cph. K
  (*) \(*) -- University of Copenhagen   Denmark  Ph:  (+45) 35327918
~~ - (p.dalga...@biostat.ku.dk)  FAX: (+45) 35327907


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[R] Excel-Export

2009-11-23 Thread koj


Dear all,

i use the following package/syntax to export data to excel:

library(xlsReadWrite)
write.xls( exportdata,pfad,colNames = TRUE,sheet = 1,from = 1,rowNames =
FALSE )

Everything is fine, but the format of the export is not the best. For
example, I every time have to adjust the column width. Furthermore there is
no possibility to highlight some cell or make them colourful. 

I create a lot of such outputs and I want to have the sheets in a nicer
format to make them better readable, especially for the people, who works
with my files.

Have you any suggestions to reach this target? Ideally I can do this within
R and not in a second step via a VBA-Makro or something like this. Is there
a possibility in R, e. g. with a package?

I have the same problem with the package xtable for export to Latex. 

I am very happy about the packages, but it would be perfect, if I can adjust
some things in this direction. Probably there are somes ways for a solution.
I am very glad about references. 

Thank you very much in advance,

Jens.




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Re: [R] R-help Digest, Vol 81, Issue 23

2009-11-23 Thread Thushyanthan Baskaran


Hi,

keine ahnung. Das liegt jetzt bei Hr. Feld. Frag mal bei ihm nach.

Grüße

Thushyanthan



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Today's Topics:

   1. Define return values of a function (soeren.vo...@eawag.ch)
   2. Re: Define return values of a function (baptiste auguie)
   3. Re: python (Stefan Evert)
   4. Re: other decriptive stats packages (Jakson A. Aquino)
   5. Re: python (Jean Legeande)
   6. Re: Over-coloring facets on persp() plot (Duncan Murdoch)
   7. Re: other decriptive stats packages (Tal Galili)
   8. Re: other decriptive stats packages (Tal Galili)
   9. Re: Define return values of a function (David Winsemius)
  10. Re: other decriptive stats packages (Liviu Andronic)
  11. Re: Over-coloring facets on persp() plot (David Winsemius)
  12. Re: Over-coloring facets on persp() plot (David Winsemius)
  13. How to make a matrix of a number of factors? (Peng Yu)
  14. Repeated measures unbalanced in a split-split design
  (Marcelo Laia)
  15. Re: MASS loading error (Uwe Ligges)
  16. How to make the assignment in a for-loop not affect variables
  outside the loop? (Peng Yu)
  17. Re: How to make a matrix of a number of factors? (baptiste auguie)
  18. Re: Metaplot Axis Annotation (Uwe Ligges)
  19. Re: How to make the assignment in a for-loop not affect
  variables outside the loop? (Uwe Ligges)
  20. Why F value and Pr are not show in summary() of an aov()
  result? (Peng Yu)
  21. Re: how to tell if its better to standardize your data matrix
  first when you do principal (Uwe Ligges)
  22. how do i persuade IT to install R on PCs ?? ...and should I
  ?? (frenchcr)
  23. Computing multivariate normal probabilities. Was: Re: Problem
  with Numerical derivatives (numDeriv) and mvtnorm (Ravi Varadhan)
  24. Re: consecutive numbering of elements in a matrix (Jim Bouldin)
  25. Re: Why F value and Pr are not show in summary() of an aov()
  result? (Sundar Dorai-Raj)
  26. scatter plot  equation (Rofizah Mohammad)
  27. Re: python (Peter Ehlers)
  28. Re: scatter plot  equation (Duncan Murdoch)
  29. Re: consecutive numbering of elements in a matrix
  (David Winsemius)
  30. Re: scatter plot  equation (David Winsemius)
  31. Do you keep an archive of useful R code? and if so - how?
  (Tal Galili)
  32. Re: python (Jean Legeande)
  33. Re: how do i persuade IT to install R on PCs ?? ...and should
  I ?? (Marc Schwartz)
  34. contour(): lines  labels in different colours? ( (Ted Harding))
  35. Re: Do you keep an archive of useful R code? and if so -
  how? (Marc Schwartz)
  36. Re: how do i persuade IT to install R on PCs ?? ...and should
  I ?? (Ben Bolker)
  37. Re: scatter plot  equation (Rofizah Mohammad)
  38. Re: AKIMA: z values at a set coordinate (Rhelp wanted)
  39. Re: Do you keep an archive of useful R code? and if so -
  how? (Tal Galili)
  40. Re: Do you keep an archive of useful R code? and if so -
  how? (Barry Rowlingson)
  41. Re: Do you keep an archive of useful R code? and if so -
  how? (Marc Schwartz)
  42. Re: Do you keep an archive of useful R code? and if so -
  how? (Tal Galili)
  43. Re: consecutive numbering of elements in a matrix (Jim Bouldin)
  44. serialized plot object (2 years later) (Jack Tanner)
  45. Re: consecutive numbering of elements in a matrix
  (Dimitris Rizopoulos)
  46. Re: Do you keep an archive of useful R code? and if so -
  how? (Hans-Peter Suter)
  47. Re: Computing multivariate normal probabilities. Was: Re:
  Problem with Numerical derivatives (numDeriv) and mvtnorm
  (stephane Luchini)
  48. mac os X: mprobit fails to install (stephane Luchini)
  49. Re: Do you keep an archive of useful R code? and if so -
  how? (Joe King)
  50. Re: Do you keep an archive of useful R code? and if so -
  how? (Carlos J. Gil Bellosta)
  51. Re: mac os X: mprobit fails to install (Phil Spector)
  52. Re: how to tell if its better to standardize your data matrix
  first when you do principal (masterinex)
  53. Re: consecutive numbering of elements in a matrix (Jim Bouldin)
  54. Re: mac os X: mprobit fails to install (SL)
  55. How to get the factor level means with interaction term? (Peng Yu)
  56. Re: Do you keep an archive of useful R code? and if so -
  how? (Nikhil Kaza)
  57. where is lme() that is referred by aov() help page? (Peng Yu)
  58. Re: where is lme() that is referred by aov() help page?
  (Duncan Murdoch)
  59. Re: mac os X: mprobit fails to install (David Winsemius)
  60. Re: where

[R] testing of randomness of residuals in the lm function in R

2009-11-23 Thread ashok varma

Hello,

I am fitting a Linear model using R. Now, i want to know how to test whether
my model is following the homoscadasticity and residuals being
random assumptions or not while fitting the model.

thanks a lot in advance,

Ashok Varma

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[R] heatmap.2 adapting the colour scale and text overlay

2009-11-23 Thread Daniel Brewer


Hello,

I am using heatmap.2 from the gplots library to plot a small symmetrical 
matrix.


This is the command:

heatmap.2(tempHeat,symm=T,trace=none,cexRow=0.7,cexCol=0.7,col=redgreen,density.info=none)

I have a couple of questions:
1) The range is from -0.2 to 0.4 and the colour scheme I am using is 
redgreen.  What I would like is that a zero value to be black, 0.4 
strong green and -0.2 red.  Is this possible?  At the moment around 0.1 
is black.
2) Is it possible to overlay text on the squares of colour?  How ould 
one go about doing that.


I am happy to use another heatmap function if ti would be better.  I am 
not using the original heatmap as it does not provide a colour key.


Thanks

Dan
--
**
Daniel Brewer, Ph.D.

Institute of Cancer Research
Molecular Carcinogenesis
Email: daniel.bre...@icr.ac.uk
**

The Institute of Cancer Research: Royal Cancer Hospital, a charitable Company 
Limited by Guarantee, Registered in England under Company No. 534147 with its 
Registered Office at 123 Old Brompton Road, London SW7 3RP.

This e-mail message is confidential and for use by the a...{{dropped:2}}

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Re: [R] How to add a top level title to multiple plots

2009-11-23 Thread Peter Ehlers



Jason Rupert wrote:
In the example below, is there any way to get the top title, i.e. Distribution Comparisons, in a bit from the top margin?  


Thanks agian


Your par(oma=...) needs to come _before_ plotting; make it the
first statement.
In addition, title() takes a line= argument; try line=0.5 or
even line=-1.0.

 -Peter Ehlers



 



- Original Message 
From: Jason Rupert jasonkrup...@yahoo.com
To: Peter Ehlers ehl...@ucalgary.ca
Cc: R Project Help R-help@r-project.org; Me jasonkrup...@yahoo.com
Sent: Sat, November 21, 2009 12:31:04 AM
Subject: Re: [R] How to add a top level title to multiple plots

Cool. 


I ended up with the following:



par(mfrow = c(2, 2))
# Plot 1
plot(rnorm(10),type=l,col=red)
title(main = list(paste(Normal),
  col=black, cex = 1.0))


# Plot 2
plot(rpois(10, 4),type=l,col=blue)
title(main = list(paste(Poison),
  col=black, cex = 1.0))


# Plot 3
plot(rnbinom(10, mu = 4, size = 100),type=l,col=green)
title(main = list(paste(Binomial),
  col=black, cex = 1.0))

# Plot 4
plot(runif(10),type=l,col=black)
title(main = list(paste(Uniform),
  col=black, cex = 1.0))


title(Distribution Comparisons, outer = TRUE)

# c(bottom, left, top, right)
par(oma=c(2,2,3,2))


Thanks again. 






- Original Message 
From: Peter Ehlers ehl...@ucalgary.ca
To: Jason Rupert jasonkrup...@yahoo.com
Cc: R-help@r-project.org
Sent: Fri, November 20, 2009 5:23:33 PM
Subject: Re: [R] How to add a top level title to multiple plots


Jason Rupert wrote:
How can I add an overall plot title to these four plots? 
I would like to have something that says, Distribution Comparisons:



title(Distribution Comparisons, outer = TRUE)

But you may have to make room for it with par(oma=...).

-Peter Ehlers


par(mfrow = c(2, 2))
# Plot 1
plot(rnorm(10),type=l,col=red)
title(main = list(paste(Normal),
  col=black, cex = 1.0))


# Plot 2
plot(rpois(10, 4),type=l,col=blue)
title(main = list(paste(Poison),
  col=black, cex = 1.0))


# Plot 3
plot(rnbinom(10, mu = 4, size = 100),type=l,col=green)
title(main = list(paste(Binomial),
  col=black, cex = 1.0))

# Plot 4
plot(runif(10),type=l,col=black)
title(main = list(paste(Uniform),
  col=black, cex = 1.0))


Thanks again for any feedback and insights. 




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[R] anova() method for coxph objects with robust standard errors

2009-11-23 Thread Mura Tamakou

Hi all,

what is the reason why the anova() method for coxph objects does not work when 
robust standard errors have been requested, e.g.,

fit - coxph(Surv(futime, fustat) ~ resid.ds *rx + ecog.ps, data = ovarian, 
robust = T)
anova(fit)


any pointers will be much appreciated.

Sincerely,

Mura

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[R] calling R from java eclipse

2009-11-23 Thread thierry CHEKOUO T.

Dear R  Java users

I have the problems with calling R from Java (JRE6) on Windows. After
running, we have:

#

# A fatal error has been detected by the Java Runtime Environment:

#

# EXCEPTION_ACCESS_VIOLATION (0xc005) at pc=0x6c7330c0, pid=1436,
tid=6200

#

# JRE version: 6.0_17-b04

# Java VM: Java HotSpot(TM) Client VM (14.3-b01 mixed mode, sharing
windows-x86 )

# Problematic frame:

# C [R.dll+0x330c0]

#

# An error report file with more information is saved as:

# C:\Users\chekouo\workspace\SIM\hs_err_pid1436.log

#

# If you would like to submit a bug report, please visit:

# http://java.sun.com/webapps/bugreport/crash.jsp

# The crash happened outside the Java Virtual Machine in native code.

# See problematic frame for where to report the bug.

help me!Thanks!

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[R] find distance between two points with different scale

2009-11-23 Thread marten


i have a dataset with x and y values. i want to randomly select a point on
the x-axis between the smallest and largest x-value and use a y value of
zero for them.
Next, for each randomly selected point {x[i],0}i want to find the point in
the {x,y} space which is closest to them. 
However, as seen below, the scale of y-values is not the same as the
x-values, using normal distance would always end up in a value that is very
close to zero, therefore generating always the same. how can i overcome this
problem?

My code is;

#randomly select 100 values between 0 and 1
y = runif(100,0,1)
x = 1:length(x)

plot(x,y)

#randomly sample 20 values of 'x'
x.rand = sample(x, 20)
y.rand = rep(0, length(x.rand))

points(x.rand, y.rand, col=blue)

#find a point which has minimal distance between the {x,y} and the
{x.rand,0}
dist.indices = c()

for(i in 1:length(x.rand)){
  dist.values = ((x.rand[i]-y.rand[i])^2+(x-y)^2)^0.5
  dist.indices = c(indices,which(dist.values == min(dist.values)))
}

This will always end up with one dist.indices that is very close to zero.

Thanks in advance,
Marten


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Re: [R] Re adline()

2009-11-23 Thread Peter Dalgaard

yonosoyelmejor wrote:
 Hello, I would like to ask you a question.I have a program in R and I use the
 readline method to ask the user some things,but i don´t use the R console
 but I use Win console then not appear what I put.I put the code as you look
 for:
 
 cat(1- 24horas\n)
 cat(2- 12horas\n)
 cat(3- 8horas\n)
 selection-readline(prompt=\nSelecciona numero de horas:)
 
 if(selection==1){
 prediccion=exp(x.reconstruida[1441:1450])
 }
 if(selection==2){
 prediccion=exp(x.reconstruida[720:729])
 }
 if(selection==3){
 prediccion=exp(x.reconstruida[481:491])
 }
 
 write.table(prediccion,C:\\Temp\\prePrueba.txt,quote=F,row.names=F,append=T,col.names=F)
 
 
 //end of code.
 
 Excuse me but my English is bad, i hope explained me well,
 
 A greeting,
 Ignacio.
 
 

I think this is a matter of flushing the console before the readline
call. See ?flush.console

-pd

-- 
   O__   Peter Dalgaard Øster Farimagsgade 5, Entr.B
  c/ /'_ --- Dept. of Biostatistics PO Box 2099, 1014 Cph. K
 (*) \(*) -- University of Copenhagen   Denmark  Ph:  (+45) 35327918
~~ - (p.dalga...@biostat.ku.dk)  FAX: (+45) 35327907

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Re: [R] heatmap.2 adapting the colour scale and text overlay



On Nov 23, 2009, at 10:02 AM, Daniel Brewer wrote:


Hello,

I am using heatmap.2 from the gplots library to plot a small  
symmetrical matrix.


This is the command:

heatmap. 
2 
(tempHeat 
,symm 
= 
T 
,trace 
=none,cexRow=0.7,cexCol=0.7,col=redgreen,density.info=none)


I have a couple of questions:
1) The range is from -0.2 to 0.4 and the colour scheme I am using is  
redgreen.  What I would like is that a zero value to be black, 0.4  
strong green and -0.2 red.  Is this possible?  At the moment around  
0.1 is black.


Perhaps:
color.palette = colorRampPalette(c(green, black, pink, red))
 heatmap.2(tempHeat,symm=T,trace=none,cexRow=0.7,cexCol=0.7,  
col=color.palette, density.info=none)


2) Is it possible to overlay text on the squares of colour?  How  
ould one go about doing that.


I am happy to use another heatmap function if ti would be better.  I  
am not using the original heatmap as it does not provide a colour key.


Thanks

Dan
--
**
Daniel Brewer, Ph.D.

Institute of Cancer Research
Molecular Carcinogenesis
Email: daniel.bre...@icr.ac.uk
**

The Institute of Cancer Research: Royal Cancer Hospital, a  
charitable Company Limited by Guarantee, Registered in England under  
Company No. 534147 with its Registered Office at 123 Old Brompton  
Road, London SW7 3RP.


This e-mail message is confidential and for use by the a...{{dropped: 
2}}


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Heritage Laboratories
West Hartford, CT

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Re: [R] Excel-Export

2009-11-23 Thread Henrique Dallazuanna

Use the RDCOMClient package from omegahat.org.

On Mon, Nov 23, 2009 at 11:02 AM, koj jens.k...@gmx.li wrote:

 Dear all,

 i use the following package/syntax to export data to excel:

 library(xlsReadWrite)
 write.xls( exportdata,pfad,colNames = TRUE,sheet = 1,from = 1,rowNames =
 FALSE )

 Everything is fine, but the format of the export is not the best. For
 example, I every time have to adjust the column width. Furthermore there is
 no possibility to highlight some cell or make them colourful.

 I create a lot of such outputs and I want to have the sheets in a nicer
 format to make them better readable, especially for the people, who works
 with my files.

 Have you any suggestions to reach this target? Ideally I can do this within
 R and not in a second step via a VBA-Makro or something like this. Is there
 a possibility in R, e. g. with a package?

 I have the same problem with the package xtable for export to Latex.

 I am very happy about the packages, but it would be perfect, if I can adjust
 some things in this direction. Probably there are somes ways for a solution.
 I am very glad about references.

 Thank you very much in advance,

 Jens.




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-- 
Henrique Dallazuanna
Curitiba-Paraná-Brasil
25° 25' 40 S 49° 16' 22 O

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Re: [R] Excel-Export

2009-11-23 Thread Marc Schwartz


On Nov 23, 2009, at 7:02 AM, koj wrote:



Dear all,

i use the following package/syntax to export data to excel:

library(xlsReadWrite)
write.xls( exportdata,pfad,colNames = TRUE,sheet = 1,from =  
1,rowNames =

FALSE )

Everything is fine, but the format of the export is not the best. For
example, I every time have to adjust the column width. Furthermore  
there is

no possibility to highlight some cell or make them colourful.

I create a lot of such outputs and I want to have the sheets in a  
nicer
format to make them better readable, especially for the people, who  
works

with my files.

Have you any suggestions to reach this target? Ideally I can do this  
within
R and not in a second step via a VBA-Makro or something like this.  
Is there

a possibility in R, e. g. with a package?

I have the same problem with the package xtable for export to Latex.

I am very happy about the packages, but it would be perfect, if I  
can adjust
some things in this direction. Probably there are somes ways for a  
solution.

I am very glad about references.

Thank you very much in advance,

Jens.


The ability to adjust columns widths in a predictable fashion is not  
easy to implement outside of Excel (or an equivalent application). The  
Auto-fit function in Excel, for example, is a run-time only feature,  
which means that Excel must be available and running.


The problem is that assumptions have to be made as to the cell content  
on a per column basis, including the font and font formatting being  
used and is not guaranteed to be accurate outside of Excel. I had  
looked at this for the WriteXLS package, but it is problematic.


With respect to the formatting of individual cells (colors, borders,  
etc.), you might want to look at the RDCOMClient package:


  http://www.omegahat.org/RDCOMClient/

See examples in the Introduction for that package.

HTH,

Marc Schwartz

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[R] get.gpar, grid.Call

2009-11-23 Thread Christian Hoffmann


Hi

I have been unable to find information on the following problem:

Error in get.gpar(): unable to find function grid.Call

Reinstalling R-2.10.0, platform: i386-pc-mingw32 did not help.

Before this message there was a get.gpar missing. We then transported 
the source code from another machine and source()-ed it.


Any ideas. It seems to belong in the base package, but this is 
unavailable for install.packages or update.packages.


Thanks for help.

C.Hoffmann

--
Dr. Christian W. Hoffmann,
Swiss Federal Research Institute WSL
Zuercherstrasse 111, CH-8903 Birmensdorf, Switzerland 
Tel +41-44-7392-277 (office), -111(exchange), -215  (fax)

christian.hoffm...@wsl.ch,  www.wsl.ch/personal_homepages/hoffmann/
Jene, die grundlegende Freiheit aufgeben wuerden, um eine geringe voruebergehende 
Sicherheit zu erwerben, verdienen weder Freiheit noch Sicherheit. - Benjamin 
Franklin

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[R] Friedman test for replicated blocked data

2009-11-23 Thread Pepe Trujillo

Dear Marco:

Look in agricolae package.

There is a Friedman test with and without replicates.

In this package is used the Conover's book as a reference.

Greetings,

Dr. D. José Trujillo Carmona
Dpt. of Mathematics
Fac. of Science of Badajoz
Univ. of Extremadura
Spain



From: Marco Chiarandini 
machud_at_intellektik.informatik.tu-darmstadt.de 
mailto:machud_at_intellektik.informatik.tu-darmstadt.de?Subject=Re:%20%5BR%5D%20Friedman%20test%20for%20replicated%20blocked%20data
 

Date: Tue 14 Dec 2004 - 04:20:51 EST


Hi,

I would need to extend the Friedman test to a replicated design. 
Currently the function: friedman.test(y, ...) only works for 
unreplicated designs.

I found in Conover 1999 Practical Nonparamteric statistics an 
extension of the formula to my case.

Nevertheless, other sources, like Sheskin 2000 Parametric and 
Nonparametric statistical Procedures and Daniel 1990 Applied 
nonparametric statistics give a different formula from that of Conover 
in the unreplicated case. Since they do not provide the extension of 
this formula to the replicated case I would like to know if someone 
could give me a reference where I could find such extension. The formula 
is indeed very simple:

F= z \sqrt{bk(k+1)/6}

where z is a quantile from the normal distribution, b the number of 
blocks and k the number of treatments.

Unfortunately, the sources cited do not provide indications from where 
the formula comes from.

Thank you for consideration,

Greetings,

Marco



Marco Chiarandini,
Technische Universitaet Darmstadt

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[R] List Name help

2009-11-23 Thread Rajasekaramya


Hi There,

I have a named List object.I want to access all the list elements that has
the same name
for example
The List 
test - list()
$d2
v1 v2 v3 v4
1 2 3 4 5

$d2
v1 v2 v3 v4
1 2 3 4 5

$d3
v1 v2 v3 v4
8 9 19 10

$d1
v1 v2 v3 v4
12 14 15 16

so if i say test[[d2]] or test[d2] i should get the first two that
matches the name rite but i am not getting them jus getting the first one. 

Any suggestions on it

Ramya






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[R] Natural colours for topographic data

2009-11-23 Thread Karl Ove Hufthammer

Dear list members

I'm currently working on some topographic (elevation) data, and was 
somewhat surprised that the 'topo.colors' and 'terrain.colors' are of 
little to no use here.

The problem is that these functions only return a palette of colours; 
they don't map depth values to colours. So if I plot (using 'image', 
'persp' or similar functions) and specify these palettes, ocean areas 
may coloured green (indicating) land, which may be quite confusing.

I have looked through various packages, and have found several colour 
palette functions, but none that do what I need.

Basically, I just need a function that takes a vector of elevation 
values as input, and outputs a vector of 'natural' colours. For negative 
values (i.e., ocean), the 'blue' colours of 'topo.colors' would be OK, 
and for positive values either the colours of 'terrain.colors' or the 
non-blue colours of 'topo.colors' would look nice.

It is of course not very difficult to create such a function myself, 
e.g. using the 'cut' function and a standard palette. But perhaps 
somebody has already has made one?

My ideal topographic colour mapping function would support separate 
colour levels for water and land (so that you can specifiy for example 5 
colours of water, from a depth of 5000 meters to 0 meters, and 20 
colours of land, from a depth of 0 meters to 2000 meters), support 
several nice palettes, and also support logarithmic colour mapping. But 
I would be happy with a simple one, that just mapped negative values to 
water colours and positive values to land colours.

-- 
Karl Ove Hufthammer

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Re: [R] List Name help

2009-11-23 Thread Henrique Dallazuanna

Try this:

test[grep(d2, names(test))]

On Mon, Nov 23, 2009 at 2:53 PM, Rajasekaramya ramya.vict...@gmail.com wrote:

 Hi There,

 I have a named List object.I want to access all the list elements that has
 the same name
 for example
 The List
 test - list()
 $d2
 v1 v2 v3 v4
 1 2 3 4 5

 $d2
 v1 v2 v3 v4
 1 2 3 4 5

 $d3
 v1 v2 v3 v4
 8 9 19 10

 $d1
 v1 v2 v3 v4
 12 14 15 16

 so if i say test[[d2]] or test[d2] i should get the first two that
 matches the name rite but i am not getting them jus getting the first one.

 Any suggestions on it

 Ramya






 --
 View this message in context: 
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 __
 R-help@r-project.org mailing list
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 PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
 and provide commented, minimal, self-contained, reproducible code.




-- 
Henrique Dallazuanna
Curitiba-Paraná-Brasil
25° 25' 40 S 49° 16' 22 O

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Re: [R] get.gpar, grid.Call




Christian Hoffmann wrote:

Hi

I have been unable to find information on the following problem:

Error in get.gpar(): unable to find function grid.Call

Reinstalling R-2.10.0, platform: i386-pc-mingw32 did not help.

Before this message there was a get.gpar missing. We then transported 
the source code from another machine and source()-ed it.



You need to load package grid (which contains both).

Uwe Ligges



Any ideas. It seems to belong in the base package, but this is 
unavailable for install.packages or update.packages.


Thanks for help.

C.Hoffmann



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Re: [R] Natural colours for topographic data



On Nov 23, 2009, at 11:52 AM, Karl Ove Hufthammer wrote:


Dear list members

I'm currently working on some topographic (elevation) data, and was
somewhat surprised that the 'topo.colors' and 'terrain.colors' are of
little to no use here.

The problem is that these functions only return a palette of colours;
they don't map depth values to colours. So if I plot (using 'image',
'persp' or similar functions) and specify these palettes, ocean areas
may coloured green (indicating) land, which may be quite confusing.

I have looked through various packages, and have found several colour
palette functions, but none that do what I need.

Basically, I just need a function that takes a vector of elevation
values as input, and outputs a vector of 'natural' colours. For  
negative

values (i.e., ocean), the 'blue' colours of 'topo.colors' would be OK,
and for positive values either the colours of 'terrain.colors' or the
non-blue colours of 'topo.colors' would look nice.

It is of course not very difficult to create such a function myself,
e.g. using the 'cut' function and a standard palette. But perhaps
somebody has already has made one?

My ideal topographic colour mapping function would support separate
colour levels for water and land (so that you can specifiy for  
example 5

colours of water, from a depth of 5000 meters to 0 meters, and 20
colours of land, from a depth of 0 meters to 2000 meters), support
several nice palettes, and also support logarithmic colour mapping.  
But
I would be happy with a simple one, that just mapped negative values  
to

water colours and positive values to land colours.


Searching with the strategy color positive negative zero in r-search  
and limiting it to r-help replies,  I get this Jim Lemon reply using  
(naturally) plotrix's color.scale:


http://finzi.psych.upenn.edu/R/Rhelp02/archive/90837.html

The application to your needs looks pretty immediate.




--
Karl Ove Hufthammer

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David Winsemius, MD
Heritage Laboratories
West Hartford, CT

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Re: [R] Define return values of a function

If you want to control what gets printed in the output of your function, then 
you need to get into creating classes and methods.  Choose a class to return 
from your function, and create a print method for that class, then when you run 
your function the output will cause the print method to be called if you don't 
store the object, but the entire object will be available if stored.  E.g.:

testfunc - function(x) {
out - list( a=summary(x), b=quantile(x), c=range(x) )
class(out) - 'myclass'
return(out)
}

print.myclass - function(x) {
cat('my output:\n')
print(x$a)
cat('\n\n')
cat('Q1: ',x$b[2], '\n')
return(invisible(x))
}

testfunc(rnorm(100))
tmp - testfunc(rnorm(100))
tmp
str(tmp)

Hope this helps,

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


 -Original Message-
 From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-
 project.org] On Behalf Of soeren.vo...@eawag.ch
 Sent: Sunday, November 22, 2009 4:27 AM
 To: r-help@r-project.org
 Subject: [R] Define return values of a function
 
 I have created a function to do something:
 
 i - factor(sample(c(A, B, C, NA), 793, rep=T, prob=c(8, 7, 5,
 1)))
 k - factor(sample(c(X, Y, Z, NA), 793, rep=T, prob=c(12, 7, 9,
 1)))
 mytable - function(x){
xtb - x
btx - x
# do more with x, not relevant here
cat(The table has been created, see here:\n)
print(xtb)
list(table=xtb, elbat=btx)
 }
 tbl - table(i, k)
 mytable(tbl) # (1)
 z - mytable(tbl) # (2)
 str(z) # (3)
 
 (1) Wanted: outputs the string and the table properly. *Unwanted*:
 outputs the list elements.
 
 (2) and (3) Wanted: outputs the string properly. Wanted: assigns the
 list properly.
 
 How can I get rid of the *unwanted* part? That is, how do I define
 what the functions prints and -- on the other hand -- what it returns
 without printing?
 
 Thanks
 
 Sören
 
 --
 Sören Vogel, Dipl.-Psych. (Univ.), PhD-Student, Eawag, Dept. SIAM
 http://www.eawag.ch, http://sozmod.eawag.ch
 
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 guide.html
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Re: [R] Fitting a mixture of poisson

2009-11-23 Thread Jim Silverton

Hello,
I would like to find out how to get the fit from a mixture of poisson. For
example, suppose I know the
 Y = 0.25*Poisson(2) + 0.75*Poisson(10)
then if I have say y = 4, how do I get the fitted value for y?

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Re: [R] testing of randomness of residuals in the lm function in R




ashok varma wrote:

Hello,

I am fitting a Linear model using R. Now, i want to know how to test whether
my model is following the homoscadasticity and residuals being
random assumptions or not while fitting the model.



See ?lm. For accessing residuals, see ?residuals. Or use plot() on a lm 
object to see diagnostic plots.


Uwe Ligges




thanks a lot in advance,

Ashok Varma

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Re: [R] non-intuitive behaviour after type conversion

2009-11-23 Thread Don MacQueen

When you attach() something, it loads it into memory and there it 
stays. It is not a link, reference, or pointer to the original. 
Changing the original (the version in the dataframe), which is what 
you did, does not change the attached copy in memory. In essence, you 
did a type conversion on one copy, but afterwards started looking at 
the other copy.


See also an interjected comments below.

-Don

At 8:54 AM + 11/23/09, Alan Kelly wrote:

Deal list,
I have a data frame (birth) with mixed variables (numeric and 
alphanumeric).  One variable t1stvisit was originally coded as 
numeric with values 1,2, and 3.  After attaching the data frame, 
this  
is what I see when I use str(t1stvisit)


$ t1stvisit: int  1 1 1 1 1 1 1 1 2 2 ...

This is as expected.
I then convert t1stvisit to a factor and to avoid creating a second 
copy of this variable independent of the data frame I use:

birth$t1stvisit = as.factor(birth$t1stvisit)
if I check that the conversion has worked:
is.factor(t1stvisit)
[1] FALSE
Now the only object present in the workspace in the data frame 
birth and, as noted,  I have not created any new variables. So why 
does R still treat t1stvisit as numeric?

is.factor(t1stvisit)
[1] FALSE

Yet when I try the following:

 is.factor(birth$t1stvisit)

[1] TRUE
So, there appears to be two versions of t1stvisit  - the original 
numeric version and the correct factor version although ls() only 
shows birth as present in the workspace.


Right.
  find('t1stvisit')
will show you there are two of them, and where in memory they are located.
If you type
   t1stvisit
at the prompt, you always get the first one. The one in the attached 
dataframe is the second one. Use the

  search()
function to show you the different locations in memory where objects 
can be found.


When you did the attach(), did you get a message like:


 attach(tmp)


The following object(s) are masked _by_ .GlobalEnv :

 x

(yours would have referred to your variables, not the x in my example).
That message tells you you have two variables of the same name, 
stored in two different locations in the search path.


As a general rule, it's just plain confusing to have more than one 
object of the same name in more than one location. In your situation, 
I would get rid of the one that's not in the dataframe. But even 
then, if you change it in the dataframe you'll still need to detach 
and re-attach the dataframe, so using attach() is probably not the 
best choice in the long run. Maybe the with() function would meet 
your needs.



If I type:

 summary(t1stvisit)

   Min. 1st Qu.  MedianMean 3rd Qu.Max.NA's
  1.000   1.000   2.000   1.574   2.000   3.000  29.000
I get the numeric version, but if I try
summary(birth$t1stvisit)
   123 NA's
 180  169   22   29
I get the factor version.

Frankly I feel that this behaviour is non-intuitive and potentially 
problematic. Nor have I seen warnings about this in the various text 
books on R.

Can anyone comment on why this should occur?
Many thanks,
Alan Kelly

Dr. Alan Kelly
Department of Public Health  Primary Care
Trinity College Dublin

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[R] NAs introduced by coercion warning?

2009-11-23 Thread Omar Gonzalez Post






I'm running cluster analysis on a data frame but when I calculate the distance 
I get this warning NAs introduced by coercion.  What does this mean?

  
[[alternative HTML version deleted]]

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Re: [R] Error using 32-bit R and RODBC package on 64-bit Windows Server OS with R version 2.10

2009-11-23 Thread helpme

Thank you for the link to the 32 bit Windows Oracle downloads for the
Instant Client.

That solved the problem. I'm now able to connect to Oracle database from R
using the RODBC library.

Thank you Mark!

-Melanie

On Wed, Nov 18, 2009 at 6:23 PM, Marc Schwartz marc_schwa...@me.com wrote:

 Just to clarify on your first point, it is not that RODBC cannot work with
 64 bit ODBC drivers. Given your particular configuration of 64 bit Windows,
 32 bit R and 64 bit ODBC drivers, you are likely running into compatibility
 issues.

 From the error message below, it would seem that you are also either
 missing the requisite Oracle client software, or your system configuration
 variables are not set or are not set to the proper paths.

 The 32 bit Windows Oracle downloads are available from:


 http://www.oracle.com/technology/software/tech/oci/instantclient/htdocs/winsoft.html

 Now, I don't run Windows (have not in a long time), so I am not clear as to
 the subtleties that may be in play here given that you may be installing 32
 bit drivers over an existing 64 bit installation. You may need to remove the
 64 bit install, in order to have a clean install of the 32 bit Oracle client
 apps. If they install into different locations, that might help to solve the
 problem, in which case, you need to be careful in configuring any system
 environment variables so that they point to the proper location.

 If you have access to in-house tech support or an Oracle SysAdmin, I would
 highly recommend that you seek them out to aid in ensuring that you end up
 with a clean 32 bit Oracle client installation. As I noted previously, I
 would be sure that you can connect to the Oracle server using the 32 bit
 Oracle Instant Client application as a test to ensure that OS and Oracle
 related configuration issues have been resolved. Then test the RODBC
 connection within R. That two step process has helped me to debug local
 configuration issues on both Linux and OSX.

 HTH,

 Marc Schwartz


 On Nov 18, 2009, at 3:12 PM, helpme wrote:

  Now that I know RODBC only works with 32-bit ODBC drivers this explains
 the
 problem I was having.

 The system has a 64 bit ODBC driver is definitely installed. I can tell
 because when you go to system32 folder and click on odbcad32.exe it goes
 to
 the Microsoft ODBC manager where I can select the driver installed for the
 64-bit Oracle system.

 The system32 folder contains the 64 bit driver for ODBC. When I go to the
 syswow64 directory and click on the odbcad32.exe it does not take me to
 the
 Microsoft ODBC manager. Instead I get this error:

 Navigate to C:\Windows\syswow64\odbcad32.exe
 2.) Select System DSN
 3.) Add Microsoft ODBC for Oracle
 I receive this error: The Oracle(tm) client and networking components were
 not found. These components are supplied by Oracle Corporation and are
 part
 of the Oracle Version 7.3 (or greater) client software installation. You
 will be unable to use this driver until these components have been
 installed.



 I don't believe the 32-bit ODBC driver is present. What is the best way to
 tell if the 32-bit Oracle client software isn't installed and I'm
 wondering
 if anyone has experience to install it on a 64-bit system and call it from
 RODC?


 On Mon, Nov 16, 2009 at 4:54 PM, Marc Schwartz marc_schwa...@me.com
 wrote:

  On Nov 16, 2009, at 2:39 PM, helpme wrote:

 I am receiving an error when trying to connect to the Oracle Database

 using
 RODBC on a 64-bit Windows Server OS. The version of R is
 2.10.0-win32.exe

 Is this the wrong version. Does RODBC only work with 32-bit ODBC
 drivers?

 've read over all the posts and documentation manuals.
 The system is Windows Server 2003 with R 2.81. and the latest
 downloadable
 RODBC package. The Oracle SID/DSN is mfopdw. I made sure to add it to
 Control Panel-Administrative Priviledges-Microsoft ODBC system/user
 DNS.

 I've also tried the following in no particular order:

 1.) Turn on all oracle services in control panel-administrative
 priviledges.
 2.) Checked tsnnames.ora for SID.
 3.) Add microsoft ODBC service to Control Panel services for SID
 4.) Use Sqldeveler to test connection another way besides R (It was
 successful)
 5.) channel-odbcDriverConnect(
 connection=Driver={Microsoft ODBC for Oracle};
 DSN=abc,UID=abc;PWD=abc;case=oracle)

 received error drivers SQLAllocHandle on SQL_HANDLE_ENV failed one time;
 another time I got the error that Oracle client and networking
 components
 7.3 or greater is not found.

 6.) tnsping mfopdw

 lsnrctl start mfopdw

 tried to add oracle/bin to path

 Nothing is working.


 Three quick comments:

 1. A better place to post these types of queries would be on the R-SIG-DB
 e-mail list, which is focused in this domain. More info here:

 https://stat.ethz.ch/mailman/listinfo/r-sig-db

 2. Prof. Ripley will be a more definitive resource, so I would wait until
 he might respond.

 3. If you have not yet, be sure to read the RODBC vignette, which is

Re: [R] How do I specify a partially completed survival analysis model?



On Nov 20, 2009, at 1:27 PM, David Winsemius wrote:



On Nov 20, 2009, at 11:07 AM, RWilliam wrote:



In reply to suggestion by David W., setting an offset parameter  
doesn't seem

to work as R is not recognizing the X2 part of  coxph(
Surv(Time,Censor)~X1, offset=log(4.3*X2), data= a ). Also, here's  
some

sample data:



The problem, arising as a result of not having a dataset against  
which to test my memories of syntactic niceties, is that glm and  
coxph use different methods of supplying offsets.


It's been pointed out to me that coxph()'s required syntactic  
incorporation of offsets is the same as glm()'s preferred inclusion in  
the formula, and that my erroneous impression that a separate offset  
argument is necessary might have be the result of SAS poisoning.


I suspect that infection is the more correct biomedical analogy,  
since I copied my use from another who was probably the index case.  
That usage was also similar to the separate specification of offsets  
(e.g.  $CAL LPY=%LOG(PY) $OFFSET LPY) in GLIM which was my statistical  
upbringing.


--
David.


Thereau and Gramsch's book has examples, but if you did not have the  
book you still had alternatives. A bit of searching with the terms:  
coxph Therneau offset;  produced lots of hits for the occurrence of  
offset in warning messages so adding -warning to that search then  
produced a hit to the Google books look at TG's text with a worked  
example:


 a$logX2 - log(a$X2)
 coxph(Surv(Time,Censor)~X1 + offset(logX2), data= a )
Call:
coxph(formula = Surv(Time, Censor) ~ X1 + offset(logX2), data = a)


coef exp(coef) se(coef) zp
X1 -0.885 0.413 1.43 -0.62 0.54

#Or just:

 coxph(Surv(Time,Censor)~X1 + offset(log(4.3*X2)), data= a )




 X1 X2 TimeCensor
1   1 0.40619454  77.00666  0
2   1 0.20717868 100.0  0
3   1 0.77360963  79.03463  1
4   1 0.62221954 100.0  0
5   1 0.32191280 100.0  0
6   1 0.73790704  72.84842  0
7   1 0.65012237 100.0  0
8   1 0.71596105 100.0  0
9   1 0.74787202  84.00172  0
10  1 0.66803790  41.65760  0
11  1 0.79922364  92.41999  0
12  1 0.76433736  90.99983  0
13  1 0.57014524 100.0  0
14  1 0.39642235 100.0  0
15  1 0.55756045 100.0  0
16  0 0.60079340 100.0  0
17  0 0.43630695 100.0  0
18  0 0.09388013 100.0  0
19  0 0.55956791 100.0  0
20  0 0.52491597  97.71884  1

where we set the coefficient of X2 to be 8.



RWilliam wrote:


Sorry for being impatient but is there really no way of doing this  
at all?

It's quite urgent so any help is very much appreciated. Thank you.



RWilliam wrote:


Hello,

I just started using R to do epidemiologic simulation research  
using the
Cox proportional hazard model. I have 2 covariates X1 and X2  
which I want
to model as h(t,X)=h0(t)*exp(b1*X1+b2*X2). I assume independence  
of X

from t.

After I simulate Time and Censor data vectors denoting the  
censoring time
and status respectively, I can call the following function to fit  
the
data into the Cox model (a is a data.frame containing 4 columns  
X1, X2,

Time and Censor):
b = coxph (Surv (Time, Censor) ~ X1 + X2, data = a, method =  
breslow);


Now the purpose of me doing simulation is that I have another  
mechanism
to generate the number b2. From the given b2 (say it's 4.3), Cox  
model
can be fit to generate b1 and check how feasible the new model  
is. Thus,
my question is, how do I specify such a model that is partially  
completed
(as in b2 is known). I tried things like  
Surv(Time,Censor)~X1+4.3*X2, but

it's not working. Thanks very much.






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David Winsemius, MD
Heritage Laboratories
West Hartford, CT

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West Hartford, CT

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[R] Analyzing XML data

2009-11-23 Thread skyjo


I would like to analyze XML data from MLB's website. I do the following
steps: 

library(XML) 
dat-http://gd2.mlb.com/components/game/mlb/year_2007/month_05/day_02/gid_2007_05_02_arimlb_lanmlb_1/inning/inning_1.xml;
 
example- xmlTreeParse(dat, useInternalNodes=T) 
ex-xpathApply(example, //inning, xmlToList)  #could also use
//atbat 

This brings information from the first inning of the Arizona/LA game played
on 5/2/2007 into R. 
Now that I have it in R, I don't understand how to 'get at' the data. For
example, how could I calculate the average start speed (
[[1]]$top$atbat$pitch[start_speed] ) of all pitches thrown in the inning? 

Thanks, 
Skyler 
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[R] Loess Fit

2009-11-23 Thread Christian Miner

I'm working on Loess fit models using R, once I have the fit accomplished,
I'm looking to back-out the equation of the fitted non-linear curve,
wondering if there is a way to determine this equation in R? I've been
looking but can't find any literature. For me, the graph of the function is
great, but without the equation of the graph, I'm kinda dead in the water.

Christian Miner
503-866-6977

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[R] Different fonts on the same axis

2009-11-23 Thread MichelleJ


Hi,

I need to have both italics and standard font on the y axis. 
This is my script:
plot(x,y,pch=16,xlab=metric,ylab=species CPUE)

I want species in italics and CPUE in standard text.
Any tips would be appreciated.
Thanks. 
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Re: [R] interaction effects in a Linear model

2009-11-23 Thread ashok varma

Hi Cleland and Dennis,

thanks a lot for the answer.


Ashok Varma

On Mon, Nov 23, 2009 at 5:05 PM, Chuck Cleland cclel...@optonline.netwrote:

  On 11/23/2009 1:06 AM, ashok varma wrote:
  hello,
 
  i am fitting a Linear model using R. I have to fit the model considering
 all
  the interaction effects of order 1 of the independent variables. But I
 have
  9 variables. So, it will be difficult for me to write all the 36
  combinations in the model. Can anyone please help how to get this done in
  much smarter way??

  This will include all main effects and two-way interactions:

 lm(Y ~ (A + B + C + D + E + F + G + H + I)^2, data=mydata)

  thanks a lot,
  Ashok Varma.
 
 [[alternative HTML version deleted]]
 
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  and provide commented, minimal, self-contained, reproducible code.

 --
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 NDRI, Inc. (www.ndri.org)
 71 West 23rd Street, 8th floor
 New York, NY 10010
 tel: (212) 845-4495 (Tu, Th)
 tel: (732) 512-0171 (M, W, F)
 fax: (917) 438-0894


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[R] NAs introduced by coercion

2009-11-23 Thread post009


I'm running cluster analysis on a data frame but when I calculate the
distance I get this warning NAs introduced by coercion.  What does this
mean?
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[R] vegan package: goodness.cca function

2009-11-23 Thread L Quinn


Hello,

I am posting this to R-help and Ecolog-L, in hopes that enough Ecologgers are R 
users and enough R-help users are ecologists. Please forgive the cross-posting.

I deleted a number of species from a canonical correspondence analysis (CCA) 
model after using the function goodness.cca in vegan.  In his explanation of 
the diagnostic tools for CCA 
(http://cc.oulu.fi/~jarioksa/softhelp/vegan/html/goodness.cca.html), Jari 
Oksanen states that It is a common practise to use goodness statistics to 
remove
species from ordination plots. However, I have not been able to find 
references on this common practice in the literature. I may be using 
inappropriate search terms?? However, if anybody can point me towards some 
cite-able references/rules of thumb on this, I would very much appreciate it.

And yes, I am also aware that Jari goes on to say that deleting species isn't 
always a good idea. 

Thank you!

Lauren Quinn
  
_
Hotmail: Trusted email with powerful SPAM protection.

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Re: [R] Different fonts on the same axis

2009-11-23 Thread Peter Alspach

Tena koe Michelle

?plotmath

For your particular example, try:

plot(1:10, 1:10, pch=16, xlab=metric, ylab=quote(~italic(species)~ '
CPUE'))

Depending on your default font, you might need to set the family
argument of par():

par(family='serif')

HTH 

Peter Alspach

 -Original Message-
 From: r-help-boun...@r-project.org 
 [mailto:r-help-boun...@r-project.org] On Behalf Of MichelleJ
 Sent: Tuesday, 24 November 2009 6:28 a.m.
 To: r-help@r-project.org
 Subject: [R] Different fonts on the same axis
 
 
 Hi,
 
 I need to have both italics and standard font on the y axis. 
 This is my script:
 plot(x,y,pch=16,xlab=metric,ylab=species CPUE)
 
 I want species in italics and CPUE in standard text.
 Any tips would be appreciated.
 Thanks. 
 --
 View this message in context: 
 http://old.nabble.com/Different-fonts-on-the-same-axis-tp26482
 460p26482460.html
 Sent from the R help mailing list archive at Nabble.com.
 
 __
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 https://stat.ethz.ch/mailman/listinfo/r-help
 PLEASE do read the posting guide 
 http://www.R-project.org/posting-guide.html
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Re: [R] Check if string has all alphabets or numbers

2009-11-23 Thread jim holtman

Here is the way you can use grepl to get the various combinations:

 mywords- c(harry,met,sally,subway10,1800Movies,12345,
+ not correct 123, )

 numbers - grepl(^[[:digit:]]+$, mywords)
 letters - grepl(^[[:alpha:]]+$, mywords)
 both - grepl(^[[:digit:][:alpha:]]+$, mywords)

 mywords[letters]
[1] harry met   sally
 mywords[numbers]
[1] 12345
 mywords[xor((letters | numbers), both)] # letters  numbers mixed
[1] subway10   1800Movies




On Mon, Nov 23, 2009 at 9:17 AM, hadley wickham h.wick...@gmail.com wrote:
 mywords- c(harry,met,sally,subway10,1800Movies,12345, not 
 correct 123)
 all.letters - grep(^[[:alpha:]]*$, mywords)
 all.numbers - grep(^[[:digit:]]*$, mywords)  # numbers
 mixed - grep(^[[:digit:][:alpha:]]*$, mywords)

 mywords- c(harry,met,sally,subway10,1800Movies,12345,
 not correct 123, )
 mywords[grepl(^[[:digit:][:alpha:]]*$, mywords)]

 So maybe you should use

 mywords[grepl(^[[:digit:][:alpha:]]+$, mywords)]


 And grepl is highly recommended over grep.

 Hadley

 --
 http://had.co.nz/




-- 
Jim Holtman
Cincinnati, OH
+1 513 646 9390

What is the problem that you are trying to solve?

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Re: [R] Loess Fit



On Nov 23, 2009, at 2:03 PM, Christian Miner wrote:

I'm working on Loess fit models using R, once I have the fit  
accomplished,

I'm looking to back-out the equation of the fitted non-linear curve,
wondering if there is a way to determine this equation in R? I've been
looking but can't find any literature. For me, the graph of the  
function is
great, but without the equation of the graph, I'm kinda dead in the  
water.


So you want a global equation from a procedure that does local fitting  
for each point? Maybe you should re-think your strategy. The points  
that form the graph _are_ the function. Maybe you should re-think your  
strategy. Perhaps you want spline fiting.


--

David Winsemius, MD
Heritage Laboratories
West Hartford, CT

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Re: [R] Loess Fit

2009-11-23 Thread Thomas S. Dye

Hi Christian,

Do you have a reference to a publication where this has been done?

All the best,
Tom

On Nov 23, 2009, at 10:15 AM, Christian Miner wrote:

 it's a tricky maneuver. When I finish the fit, the predict function  
 will give me the values, and I can smooth this out so it looks like  
 a nice non-linear equation. The last thing I need to do is find the  
 equation, knowing I know have the x and y values, as they will  
 relate to each other in a non-linear fashion.

 Matlab can do this, but I'd rather not go into that realm.

 On Mon, Nov 23, 2009 at 12:11 PM, Thomas S. Dye t...@tsdye.com wrote:
 Hi Christian,

 I haven't seen an equation associated with a loess fit before.  Do  
 you have a reference to one of these?

 All the best,
 Tom


 On Nov 23, 2009, at 9:03 AM, Christian Miner wrote:

 I'm working on Loess fit models using R, once I have the fit  
 accomplished,
 I'm looking to back-out the equation of the fitted non-linear curve,
 wondering if there is a way to determine this equation in R? I've been
 looking but can't find any literature. For me, the graph of the  
 function is
 great, but without the equation of the graph, I'm kinda dead in the  
 water.

 Christian Miner
 503-866-6977

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 -- 
 Christian Miner
 503-866-6977



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Re: [R] Loess Fit

Check out the loess.demo function in the TeachingDemos package.  Using this 
function with your data will show the scatterplot with the loess curve.  Then 
if you click in the plot it will show the line/curve used to do the prediction 
for that x-value, click on another x-value and you will see that the line/curve 
used there is different from before.  There is not a single function created by 
the loess algorithm, rather a different approximation for every possible 
x-value (theoretically infinite, practically still a big number).

You could fit a function to the results of loess (or approximate the points 
with a function), but the function will probably be complicated enough that it 
would not tell you much that the points themselves don't.

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


 -Original Message-
 From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-
 project.org] On Behalf Of Christian Miner
 Sent: Monday, November 23, 2009 12:04 PM
 To: r-help@r-project.org
 Subject: [R] Loess Fit
 
 I'm working on Loess fit models using R, once I have the fit
 accomplished,
 I'm looking to back-out the equation of the fitted non-linear curve,
 wondering if there is a way to determine this equation in R? I've been
 looking but can't find any literature. For me, the graph of the
 function is
 great, but without the equation of the graph, I'm kinda dead in the
 water.
 
 Christian Miner
 503-866-6977
 
   [[alternative HTML version deleted]]
 
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[R] using contents of one column to direct addition of new rows in dataframe

2009-11-23 Thread Jessi Brown

Hello, R gurus. I've been trying to get R to do some data manipulation
for me, and so far have been stumped in figuring out any elegant way
to do so. Searches through the R-help archive haven't helped, so now
I'm trying plan B.

Suppose I have a dataframe that summarizes events that took place
between some interval of time in days, Time. I want to expand the
dataframe to repeat each row the appropriate number of days so that
Time is now = 1 for each original record. Actually, I really don't
care if Time=1 is displayed in the new dataframe, as it will no longer
be relevant.

Also, I want to have R keep track of a time-variant variable, Date, so
that it advances along one day for each new row. Here's an example of
what the starting record would be like, and what I'd like to end up
with:

 foo.df - data.frame(ID=N1, Date=2, Time=3)
 foo.df
  ID Date Time
1 N123

# insert R wizardry here to get new.foo.df

new.foo.df
ID  DateTime
N1  2   1
N1  3   1
N1  4   1


I know, or at least suspect, that there is a fairly straightforward
way of doing so with building some ugly loop. But even after
considering versions of apply and related commands, I just can't
figure it out. Any help would be greatly appreciated.

cheers, Jessi Brown

University of Nevada, Reno

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Re: [R] Loess Fit

2009-11-23 Thread Thomas S. Dye

Thanks Christian.  The client is always right ...

Tom

On Nov 23, 2009, at 10:32 AM, Christian Miner wrote:

 sorry, I don't. I have data that I can estimate, but my client wants  
 to What if... 20 separate scenarios, and that requires a function.

 On Mon, Nov 23, 2009 at 12:22 PM, Thomas S. Dye t...@tsdye.com wrote:
 Hi Christian,

 Do you have a reference to a publication where this has been done?

 All the best,
 Tom

 On Nov 23, 2009, at 10:15 AM, Christian Miner wrote:

 it's a tricky maneuver. When I finish the fit, the predict function  
 will give me the values, and I can smooth this out so it looks like  
 a nice non-linear equation. The last thing I need to do is find the  
 equation, knowing I know have the x and y values, as they will  
 relate to each other in a non-linear fashion.

 Matlab can do this, but I'd rather not go into that realm.

 On Mon, Nov 23, 2009 at 12:11 PM, Thomas S. Dye t...@tsdye.com  
 wrote:
 Hi Christian,

 I haven't seen an equation associated with a loess fit before.  Do  
 you have a reference to one of these?

 All the best,
 Tom


 On Nov 23, 2009, at 9:03 AM, Christian Miner wrote:

 I'm working on Loess fit models using R, once I have the fit  
 accomplished,
 I'm looking to back-out the equation of the fitted non-linear curve,
 wondering if there is a way to determine this equation in R? I've  
 been
 looking but can't find any literature. For me, the graph of the  
 function is
 great, but without the equation of the graph, I'm kinda dead in the  
 water.

 Christian Miner
 503-866-6977

[[alternative HTML version deleted]]

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




 -- 
 Christian Miner
 503-866-6977





 -- 
 Christian Miner
 503-866-6977



[[alternative HTML version deleted]]

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Re: [R] testing of randomness of residuals in the lm function in R

2009-11-23 Thread Tal Galili

Hi Ashok,
You can try :
*shapiro.test(**my.lm$resid**)*
But I remember reading that good tests for the normality assumptions
(specifically that of the residuals), do not exist as we would have hoped
them to.


Good luck,
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)
--




2009/11/23 Uwe Ligges lig...@statistik.tu-dortmund.de



 ashok varma wrote:

 Hello,

 I am fitting a Linear model using R. Now, i want to know how to test
 whether
 my model is following the homoscadasticity and residuals being
 random assumptions or not while fitting the model.



 See ?lm. For accessing residuals, see ?residuals. Or use plot() on a lm
 object to see diagnostic plots.

 Uwe Ligges




  thanks a lot in advance,

 Ashok Varma

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Re: [R] How to make a cartesian pairlist from a vector?

2009-11-23 Thread newbyr


This is very helpful. Thank you very much.


Charles C. Berry wrote:
 
 On Fri, 20 Nov 2009, newbyr wrote:
 

 Hi,

 I'm looking for a function that will take a list of columns or data.frame
 and corvert it to cartesian pairlist. For example for this data.frame
 (see
 below), I'd like to get a list of all possible pairs:

 sound
  cs rs ns
  7  5  2
  4  5  4
  6  3  1
  8  4  1
  6  4  2
  6  7  1
  2  2  5
  9  2  5

 
 how can I get this?
 mylist
  ((cs,rs), (cs,ns), (rs,ns))
 
   str(combn(colnames(df),2,function(x) df[,x],simplify=F))
 
 see
 
   ?combn
 
 HTH,
 
 Chuck
 
 p.s. Following the posting guide's hints you might try
 
 ?combinations  # no go, but it says try ??combinations
 ??combinations # shows utils:combn
 ?combn #  bingo!
 
 
 
 

 This list contains a set of all possible combinations of column pairs.

 Thank you for your help.

 -- 
 View this message in context:
 http://old.nabble.com/How-to-make-a-cartesian-pairlist-from-a-vector--tp26452861p26452861.html
 Sent from the R help mailing list archive at Nabble.com.

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

 
 Charles C. Berry(858) 534-2098
  Dept of Family/Preventive
 Medicine
 E mailto:cbe...@tajo.ucsd.edu UC San Diego
 http://famprevmed.ucsd.edu/faculty/cberry/  La Jolla, San Diego 92093-0901
 
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 R-help@r-project.org mailing list
 https://stat.ethz.ch/mailman/listinfo/r-help
 PLEASE do read the posting guide
 http://www.R-project.org/posting-guide.html
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-- 
View this message in context: 
http://old.nabble.com/How-to-make-a-cartesian-pairlist-from-a-vector--tp26452861p26485572.html
Sent from the R help mailing list archive at Nabble.com.

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Re: [R] Loess Fit

If the client is always right, then what do they need us for?

If you need a function to reproducibly generate predictions, then use loess to 
generate a set of predictions for a reasonably dense set of x-values, then use 
approxfun or splinefun to create a function to interpolate for you.  Then this 
function can give the predictions for various x values.

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


 -Original Message-
 From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-
 project.org] On Behalf Of Thomas S. Dye
 Sent: Monday, November 23, 2009 1:39 PM
 To: Christian Miner
 Cc: r-help@r-project.org
 Subject: Re: [R] Loess Fit
 
 Thanks Christian.  The client is always right ...
 
 Tom
 
 On Nov 23, 2009, at 10:32 AM, Christian Miner wrote:
 
  sorry, I don't. I have data that I can estimate, but my client wants
  to What if... 20 separate scenarios, and that requires a function.
 
  On Mon, Nov 23, 2009 at 12:22 PM, Thomas S. Dye t...@tsdye.com
 wrote:
  Hi Christian,
 
  Do you have a reference to a publication where this has been done?
 
  All the best,
  Tom
 
  On Nov 23, 2009, at 10:15 AM, Christian Miner wrote:
 
  it's a tricky maneuver. When I finish the fit, the predict function
  will give me the values, and I can smooth this out so it looks like
  a nice non-linear equation. The last thing I need to do is find the
  equation, knowing I know have the x and y values, as they will
  relate to each other in a non-linear fashion.
 
  Matlab can do this, but I'd rather not go into that realm.
 
  On Mon, Nov 23, 2009 at 12:11 PM, Thomas S. Dye t...@tsdye.com
  wrote:
  Hi Christian,
 
  I haven't seen an equation associated with a loess fit before.  Do
  you have a reference to one of these?
 
  All the best,
  Tom
 
 
  On Nov 23, 2009, at 9:03 AM, Christian Miner wrote:
 
  I'm working on Loess fit models using R, once I have the fit
  accomplished,
  I'm looking to back-out the equation of the fitted non-linear curve,
  wondering if there is a way to determine this equation in R? I've
  been
  looking but can't find any literature. For me, the graph of the
  function is
  great, but without the equation of the graph, I'm kinda dead in the
  water.
 
  Christian Miner
  503-866-6977
 
 [[alternative HTML version deleted]]
 
  __
  R-help@r-project.org mailing list
  https://stat.ethz.ch/mailman/listinfo/r-help
  PLEASE do read the posting guide http://www.R-project.org/posting-
 guide.html
  and provide commented, minimal, self-contained, reproducible code.
 
 
 
 
  --
  Christian Miner
  503-866-6977
 
 
 
 
 
  --
  Christian Miner
  503-866-6977
 
 
 
   [[alternative HTML version deleted]]
 
 __
 R-help@r-project.org mailing list
 https://stat.ethz.ch/mailman/listinfo/r-help
 PLEASE do read the posting guide http://www.R-project.org/posting-
 guide.html
 and provide commented, minimal, self-contained, reproducible code.

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Re: [R] Loess Fit

2009-11-23 Thread Christian Miner

Thanks, I'll try that. What is the output of approxfun?

On Mon, Nov 23, 2009 at 1:03 PM, Greg Snow greg.s...@imail.org wrote:

 If the client is always right, then what do they need us for?

 If you need a function to reproducibly generate predictions, then use loess
 to generate a set of predictions for a reasonably dense set of x-values,
 then use approxfun or splinefun to create a function to interpolate for you.
  Then this function can give the predictions for various x values.

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


  -Original Message-
  From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-
  project.org] On Behalf Of Thomas S. Dye
  Sent: Monday, November 23, 2009 1:39 PM
  To: Christian Miner
  Cc: r-help@r-project.org
  Subject: Re: [R] Loess Fit
 
  Thanks Christian.  The client is always right ...
 
  Tom
 
  On Nov 23, 2009, at 10:32 AM, Christian Miner wrote:
 
   sorry, I don't. I have data that I can estimate, but my client wants
   to What if... 20 separate scenarios, and that requires a function.
  
   On Mon, Nov 23, 2009 at 12:22 PM, Thomas S. Dye t...@tsdye.com
  wrote:
   Hi Christian,
  
   Do you have a reference to a publication where this has been done?
  
   All the best,
   Tom
  
   On Nov 23, 2009, at 10:15 AM, Christian Miner wrote:
  
   it's a tricky maneuver. When I finish the fit, the predict function
   will give me the values, and I can smooth this out so it looks like
   a nice non-linear equation. The last thing I need to do is find the
   equation, knowing I know have the x and y values, as they will
   relate to each other in a non-linear fashion.
  
   Matlab can do this, but I'd rather not go into that realm.
  
   On Mon, Nov 23, 2009 at 12:11 PM, Thomas S. Dye t...@tsdye.com
   wrote:
   Hi Christian,
  
   I haven't seen an equation associated with a loess fit before.  Do
   you have a reference to one of these?
  
   All the best,
   Tom
  
  
   On Nov 23, 2009, at 9:03 AM, Christian Miner wrote:
  
   I'm working on Loess fit models using R, once I have the fit
   accomplished,
   I'm looking to back-out the equation of the fitted non-linear curve,
   wondering if there is a way to determine this equation in R? I've
   been
   looking but can't find any literature. For me, the graph of the
   function is
   great, but without the equation of the graph, I'm kinda dead in the
   water.
  
   Christian Miner
   503-866-6977
  
  [[alternative HTML version deleted]]
  
   __
   R-help@r-project.org mailing list
   https://stat.ethz.ch/mailman/listinfo/r-help
   PLEASE do read the posting guide http://www.R-project.org/posting-
  guide.html
   and provide commented, minimal, self-contained, reproducible code.
  
  
  
  
   --
   Christian Miner
   503-866-6977
  
  
  
  
  
   --
   Christian Miner
   503-866-6977
  
 
 
[[alternative HTML version deleted]]
 
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  guide.html
  and provide commented, minimal, self-contained, reproducible code.




-- 
Christian Miner
503-866-6977

[[alternative HTML version deleted]]

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Re: [R] serialized plot object (2 years later)

2009-11-23 Thread Tobias Verbeke


Hi Jack,

Jack Tanner wrote:

About 2 years ago, Tobias Verbeke asked:

I am looking for a way to capture the binary string that in normal use of 
graphics devices will bewritten to (most commonly) a file connection... Is there a 
way of capturing the binary `jpeg string'
[generated by jpeg()]?

http://tolstoy.newcastle.edu.au/R/e2/devel/07/09/4276.html

Brian Ripley's answer was Nope, unfortunately, they write to files not connections and no R object 
is involved.


Is this still the case?


Yes (assuming your question wasn't a rhetorical question).

I still think it would be a useful feature if graphics devices
could write to connections.

Best,
Tobias

P.S. Your question seems more appropriate for R-devel.

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Re: [R] Loess Fit