date:20110425

Hi all,

How can I tell RGL to set the center for the rotation to the origin of  the 
coordinate system (0,0,0).
It seems that the default is to use the center of the display not the origin of 
the coordinate system.

open3d()
lines3d(c(0, 1), c(0,0), c(0,0))
lines3d(c(0,0), c(0, 1), c(0,0))
lines3d(c(0,0), c(0,0), c(0, 1))

TIA
Mark

–––
Mark Heckmann
Blog: www.markheckmann.de
R-Blog: http://ryouready.wordpress.com

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Re: [R] Random Normal Variable Correlated to an Existing Binomial Variable

2011-04-25 Thread Petr Savicky

On Sun, Apr 24, 2011 at 07:00:26PM -0400, Shane Phillips wrote:
 Hi, R-Helpers!
 
 I have a dataframe that contains a binomial variable.  I need to add another 
 random variable drawn from a normal distribution with a specific mean and 
 standard deviation.  This variable also needs to be correlated with the 
 existing binomial variable with a specific correlation (say .75).  Any ideas?

Hi.

If X, Y are dependent random variables and we want to generate y, so
that (x, y) is a pair from their joint distribution with known x,
then y should be generated from the conditional distribution P(Y|X=x).
If the probability P(X=x) is not too small, then this may be done by
rejection sampling: Generate pairs (X, Y) until the condition X=x is
satisfied and use the corresponding Y.

It remains to generate pairs (X, Y), where Y is a normal variable
and X a binomial one. The parameters of Y are known, the parameters
of X should be chosen somehow and the correlation of X and Y is
known. I suggest the following. Compute the distribution of X as a
vector of probabilities p_0, ..., p_n (see ?dbinom). Find a nondecreasing
function f() from reals to {0, .., n} such that f(Y) has distribution
p_0, ..., p_n. The function may be determined by a sequence of
cutpoints a_1, ..., a_n defining f(y) as follows

  y  f(y)  
  (-infty, a_1)  0
  [a_1, a_2) 1
  ...
  [a_n, infty)   n

For each i, the cutpoint a_i is the (p_0 + ... + p_{i-1})-quantile of Y
(see ?qnorm). See ?cut for computing f().

The pair (f(Y), Y) has the required marginal distributions and, in my
opinion, the maximal possible correlation. If this correlation is lower
than the requested one, then i think there is no solution.

If the correlation of (f(Y), Y) is at least the required one, then use
a mixture of the distribution (f(Y), Y) and (X, Y), where X has the 
required marginal distribution of X, but is generated independently
from Y. The mixture parameter may be determined as a solution of an
equation with one variable.

Hope this helps.

Petr Savicky.

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[R] regular expression for nth character in a string

2011-04-25 Thread Gonçalo Ferraz

Hi, I have a string

InTrouble

and want to extract, say, the first two characters: In
or the last three: blee
or the 3rd, 4th, and 5th: Trou

Is there an easy way of doing this quickly with regular expressions in gsub, 
grep or similar?

Thank you for any help.

Gonçalo

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Re: [R] regular expression for nth character in a string

2011-04-25 Thread jim holtman

will this do it:


 x - InTrouble
 sub(^(..).*, \\1, x)  # first two
[1] In
 sub(.*(...)$, \\1, x)  # last three
[1] ble
 sub(^..(...).*, \\1, x)  # 3rd,4th,5th char
[1] Tro



2011/4/25 Gonçalo Ferraz gferra...@gmail.com:
 Hi, I have a string

 InTrouble

 and want to extract, say, the first two characters: In
 or the last three: blee
 or the 3rd, 4th, and 5th: Trou

 Is there an easy way of doing this quickly with regular expressions in gsub, 
 grep or similar?

 Thank you for any help.

 Gonçalo

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-- 
Jim Holtman
Data Munger Guru

What is the problem that you are trying to solve?

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Re: [R] regular expression for nth character in a string

2011-04-25 Thread Jim Lemon


On 04/25/2011 08:17 PM, Gonçalo Ferraz wrote:

Hi, I have a string

InTrouble

and want to extract, say, the first two characters: In
or the last three: blee
or the 3rd, 4th, and 5th: Trou

Is there an easy way of doing this quickly with regular expressions in gsub, 
grep or similar?


Hi Gonçalo,
You could always try:

Im-InTrouble
paste(unlist(strsplit(Im,))[c(7,4,1,2,9,8,5,6,3)],collapse=)

Jim

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Re: [R] regular expression for nth character in a string

2011-04-25 Thread David Winsemius



On Apr 25, 2011, at 6:17 AM, Gonçalo Ferraz wrote:


Hi, I have a string

InTrouble

and want to extract, say, the first two characters: In
or the last three: blee
or the 3rd, 4th, and 5th: Trou

Is there an easy way of doing this quickly with regular expressions  
in gsub, grep or similar?



Not greppish but seems to be the obvious approach:

 substr(Trouble, 1,2)
[1] Tr
 substr(Trouble, 3,5)
[1] oub

The greppish ways:

 sub((^..)(.*$), \\1, Troubles)
[1] Tr

 sub((^..)(...)(.*$), \\2, Troubles)
[1] oub
 sub((^..)(.{3})(.*$), \\2, Troubles)
[1] oub

--

David Winsemius, MD
West Hartford, CT

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Re: [R] filling array with functions

2011-04-25 Thread derek

Richard,

that way I will have to write functions manually and that is not possible
for large number of functions.

derek

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Re: [R] filling array with functions

2011-04-25 Thread peter dalgaard


On Apr 25, 2011, at 12:15 , derek wrote:

 Richard,
 
 that way I will have to write functions manually and that is not possible
 for large number of functions.
 

Well do what he means:

fv - vector(list,10) 
for (i...
{
  ...
  fv[[i]] - ...
  ...
}

(Your code still won't work as written; V[i] will not be evaluated until you 
call the function(s), likely to V[10]. Try something like fv[[i]] - local({zz 
- V[i]; function(x) zz - b*a*x}), or play around with eval(substitute()) )


 derek
 
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Re: [R] average among one factor in a nested dataframe

2011-04-25 Thread Kenneth Roy Cabrera Torres

Hi Junquian:

I try your code (there is a typo, I believe)

a-rnorm(6)
b-rnorm(9)
f1-c(x1,x2,x3)
f2-c(y1,y2)
id-c(1:6)
a_df-data.frame(cbind(id,f1,y1,a))
id-c(1:9)
b_df-data.frame(cbind(id,f1,y2,b))

But I don't understand the nested databases.
I see that both have f1 variable but I do not see f2 variable in any of
them. So, what do you mean with collapse f2?

Maybe you need to first merge() de databases and then aggregate()
them.

Have a nice day!

El dom, 24-04-2011 a las 23:42 -0500, Junqian Gordon Xu escribió:
 I have two nested data frames:
 
 a-rnorm(6)
 b-rnorm(9)
 f1-c(x1,x2,x3))
 f2-c(y1,y2)
 id-c(1:6)
 a_df-data.frame(cbind(id,f1,y1,a))
 id-c(1:9)
 b_df-data.frame(cbind(id,f1,y2,b))
 
 I want to preserve id and f1, but want to collapse f2 and take the
 corresponding mean values of a and b. Missing value in either dataframe
 should be handled properly (i.e., just take the non-missing number
 without dividing by 2).
 
 I had a look at rowSum/Means and s/l/tapply, but couldn't figure out how
 to handle this case cleanly. Any suggestions?
 
 Thanks
 Gordon
 
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Re: [R] average among one factor in a nested dataframe

2011-04-25 Thread Junqian Gordon Xu

Never mind, I find a generic solution:

require(reshape)
melted-melt(dataframe, id=c(id,f1,f2))
averaged=cast(melted,id+f1~variable,mean)

which collapses away f2, and it's easy to generalize this to collapse
any factors.

Thanks anyway
Gordon

On 4/25/11 6:14 AM, Kenneth Roy Cabrera Torres wrote:
 Hi Junquian:
 
 I try your code (there is a typo, I believe)
 
 a-rnorm(6)
 b-rnorm(9)
 f1-c(x1,x2,x3)
 f2-c(y1,y2)
 id-c(1:6)
 a_df-data.frame(cbind(id,f1,y1,a))
 id-c(1:9)
 b_df-data.frame(cbind(id,f1,y2,b))
 
 But I don't understand the nested databases.
 I see that both have f1 variable but I do not see f2 variable in any of
 them. So, what do you mean with collapse f2?
 
 Maybe you need to first merge() de databases and then aggregate()
 them.
 
 Have a nice day!
 
 El dom, 24-04-2011 a las 23:42 -0500, Junqian Gordon Xu escribió:
 I have two nested data frames:

 a-rnorm(6)
 b-rnorm(9)
 f1-c(x1,x2,x3))
 f2-c(y1,y2)
 id-c(1:6)
 a_df-data.frame(cbind(id,f1,y1,a))
 id-c(1:9)
 b_df-data.frame(cbind(id,f1,y2,b))

 I want to preserve id and f1, but want to collapse f2 and take the
 corresponding mean values of a and b. Missing value in either dataframe
 should be handled properly (i.e., just take the non-missing number
 without dividing by 2).

 I had a look at rowSum/Means and s/l/tapply, but couldn't figure out how
 to handle this case cleanly. Any suggestions?

 Thanks
 Gordon

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Re: [R] regular expression for nth character in a string

2011-04-25 Thread Gabor Grothendieck

2011/4/25 Gonçalo Ferraz gferra...@gmail.com:
 Hi, I have a string

 InTrouble

 and want to extract, say, the first two characters: In
 or the last three: blee
 or the 3rd, 4th, and 5th: Trou

 Is there an easy way of doing this quickly with regular expressions in gsub, 
 grep or similar?


strapply in gsubfn can readily do that.  It returns the matched part
or, if parentheses are used, only the part in parentheses:

 library(gsubfn)
 strapply(InTrouble, ^.., simplify = TRUE)
[1] In
 strapply(InTrouble, ...$, simplify = TRUE)
[1] ble
 strapply(InTrouble, ^..(...), simplify = TRUE)
[1] Tro
 strapply(InTrouble, ^.{2}(.{3}), simplify = TRUE)
[1] Tro


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

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Re: [R] Defining origin for rotation in RGL device


On 25/04/2011 5:46 AM, Mark Heckmann wrote:

Hi all,

How can I tell RGL to set the center for the rotation to the origin of  the 
coordinate system (0,0,0).
It seems that the default is to use the center of the display not the origin of 
the coordinate system.

open3d()
lines3d(c(0, 1), c(0,0), c(0,0))
lines3d(c(0,0), c(0, 1), c(0,0))
lines3d(c(0,0), c(0,0), c(0, 1))



You can attach any transformation you like to a mouse button.  See the 
mouseCallbacks demo for R implementations of the standard ones, and 
modify the mouseTrackball function there to choose the position of the 
origin of the coordinate system as the centre of rotation.


Duncan Murdoch

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Re: [R] help with \ in strings

2011-04-25 Thread Nutter, Benjamin

Depending on what else you're writing around the %, you might consider
using the latexTranslate() function in Hmisc.

-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org]
On Behalf Of viostorm
Sent: Sunday, April 24, 2011 8:48 AM
To: r-help@r-project.org
Subject: Re: [R] help with \ in strings

Josh,

Thank you so much!!! Works perfectly!

-Rob

Robert Schutt III, MD, MCS
Resident - Department of Internal Medicine University of Virginia,
Charlottesville, Virginia 

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Re: [R] Using Java methods in R

2011-04-25 Thread hill0093

So the first few posts show that I found out how to 
get Java functions to return type double numbers to R.
The arrays are still a problem.

Here is another of my attempts to understand how to get java arrays into R.
The Java code in class CalqsLin for an array 
of constants named conArr and for a function returning an array is:
public static double[][]conArr=new double[][] { { 10001,10002,10003,10004
},{ 20001,20002,20003,20004 },{ 30001,30002,30003,30004 } }; 
public final static double[][] arReturnTEST() { 
  double[][]retArr=new double[3][4]; for(int i=0;i3;i++)for(int
j=0;j4;j++)retArr[i][j]=i*1000+j; return(retArr); 
} 

I evidently do not know how to retrieve them in R, even with .jevalArray():
 arj34Ret - .jcall(qsLin,returnSig=[[D,arReturnTEST)
 ar34Ret - .jevalArray(arj34Ret)
Error in .jevalArray(arj34Ret) : 
  object is not a Java object reference (jobjRef/jarrayRef).
 connArr - .jevalArray(qsLin.conArr)
Error in .jevalArray(qsLin.conArr) : object 'qsLin.conArr' not found

I need help from someone who calls Java from R.
The whole session was:
 library(rJava)
 .jinit()
 .jaddClassPath(C:/ad/j)
 print(.jclassPath())
[1] C:\\Users\\ENVY17\\Documents\\R\\win-library\\2.12\\rJava\\java
[2] C:\\ad\\j  
 qsLin - .jnew(CalqsLin)
 calStg - 20110424235959
 print(calStg)
[1] 20110424235959
 dblTim -
 .jcall(qsLin,returnSig=D,linTimOfCalqsStgIsLev,calStg,as.integer(-4))
 print(dblTim,digits=20)
[1] 63470908799
 calStg -
 .jcall(qsLin,returnSig=S,calqsStgOfLinTimIsLev,dblTim,as.integer(-4))
 print(calStg)
[1] 20110424235959
 dblTim -
 .jcall(qsLin,returnSig=D,linTimOfCalqsStgIsLev,calStg,as.integer(-4))
 print(dblTim,digits=20)
[1] 63470908799
 arj34Ret - .jcall(qsLin,returnSig=[[D,arReturnTEST)
 ar34Ret - .jevalArray(arj34Ret)
Error in .jevalArray(arj34Ret) : 
  object is not a Java object reference (jobjRef/jarrayRef).
 connArr - .jevalArray(qsLin.conArr)
Error in .jevalArray(qsLin.conArr) : object 'qsLin.conArr' not found
 


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Re: [R] random typing over text

2011-04-25 Thread Kenn Konstabel

On Mon, Apr 25, 2011 at 4:22 AM, Jim Lemon j...@bitwrit.com.au wrote:
 On 04/24/2011 08:13 AM, derek wrote:

 Thank you very much. It was the Insert key. It was very annoying. Actually
 is
 this owerwrite function of any use?

 Hi derek,
 As Duncan mentioned, it is very useful when one wishes to type over existing
 text. However, this is a fairly uncommon wish in the typical GUI user
 environment, where the typist can highlight a group of characters and then
 begin typing. The highlighted characters disappear and the replacement is
 accomplished without changing the behavior of the keyboard. Many
 applications now attempt to guess what you want to highlight by performing
 the operation on words. To me this is not an advantage, for it often means
 that I delete one or more characters beyond what I wish. As Rolf noted, the
 adjacency of the Insert and Delete keys makes it far too easy to switch
 unwittingly to Insert mode. I sometimes wonder whether keyboard designers
 ever have to type, or whether they simply dictate their design inspirations
 into a microphone as some non-typists of my acquaintance do. When I received
 a new PC at work, some bright spark had placed an extra Backslash key next
 to the left Shift key _and_ reduced the size of the Shift key. After a few
 days of typing backslashes every time I wanted a capital letter, I popped
 both keys off and extended the Shift key to mostly cover the useless
 Backslash. I once programmed a special keyboard for a one-handed typist that
 allowed one to reprogram the meaning of keys, thereby accommodating
 individual preference rather than the lumbering Frankenstein of the average
 user. Now that would be a worthwhile innovation.

A reprogrammable Lego keyboard would be good, thanks in advance. If
you do it, would you please consider adding an extra safety feature
for windows users: every time the victim either holds Shift for 8
seconds or presses Shift for 5 times in a row, there would be a nice
voice saying: You just pressed Shift for N times. Ordinarily that
means that Windows will activate StickyKeys(TM) which will make your
keyboard useless until you restart the computer. If you really wish to
activate StickyKeys(TM), press Ctrl-Backspace-Esc for 12 times, then
RightAlt-Tab-F8 for 3 times, and then smile in your webcam for at
least 45 seconds without making any facial movements not directly
required to perform the aforementioned task..

KK

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Re: [R] problem with library tseriesChaos

2011-04-25 Thread Fukushima Shintaro

Hello.

Is it possible to determine time delay with other function's output or I
can choose any random value?

There are several ways to estimate time delay in chaotic time series
analysis.
Please refer to the book Nonlinear time series analysis by Holger Kantz
and Thomas Shreiber.
http://www.amazon.com/Nonlinear-Time-Analysis-Holger-Kantz/dp/0521529026
http://www.amazon.com/Nonlinear-Time-Analysis-Holger-Kantz/dp/0521529026

For example, time delay is defined as the time when average mutual
information reaches the first local minimum.

Using the package tseriesChaos, it is estimated as follows:

library(tseriesChaos)
x.mu - mutual(x, plot=FALSE)
x.mu.d - diff(x.mu)
d - min(which(x.mu.d 0)) - 1

The last two can also be simplied.
d - which.min(x.mu) - 1

Recently, in Japan, a book has been published on useful packages of R.
http://www.amazon.co.jp/R%E3%83%91%E3%83%83%E3%82%B1%E3%83%BC%E3%82%B8%E3%82%AC%E3%82%A4%E3%83%89%E3%83%96%E3%83%83%E3%82%AF-%E5%B2%A1%E7%94%B0-%E6%98%8C%E5%8F%B2/dp/448902097X
http://www.amazon.co.jp/R%E3%83%91%E3%83%83%E3%82%B1%E3%83%BC%E3%82%B8%E3%82%AC%E3%82%A4%E3%83%89%E3%83%96%E3%83%83%E3%82%AF-%E5%B2%A1%E7%94%B0-%E6%98%8C%E5%8F%B2/dp/448902097X

I wrote an article on the packages RTisean and tseriesChaos.
There, I showed an example of estimating the maximum Lyapunov exponent,
using the NH3 laser data.
The source code is available from the following website.
https://gist.github.com/4efa66686c24f158f7e9
https://gist.github.com/4efa66686c24f158f7e9

I would be happy to be of any service to you.

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[R] probit regression marginal effects

2011-04-25 Thread Quark

Dear R-community,

I am currently replicating a study and obtain mostly the same results as the 
author. At one point, however, I calculate marginal effects that seem to be 
unrealistically small. I would greatly appreciate if you could have a look at 
my reasoning and the code below and see if I am mistaken at one point or 
another.

My sample contains 24535 observations, the dependent variable x028bin is a 
binary variable taking on the values 0 and 1, and there are furthermore 10 
explaining variables. Nine of those independent variables have numeric levels, 
the independent variable f025grouped is a factor consisting of country names.

I would like to run a probit regression including country dummies and then 
compute marginal effects. In order to do so, I first eliminate missing values 
and use cross-tabs between the dependent and independent variables to verify 
that there are no small or 0 cells. Then I run the probit model which works 
fine and I obtain reasonable results:

 probit4AKIE - glm(x028bin ~ x003 + x003squ + x025secv2 + x025terv2 + x007bin 
 + x04chief + x011rec + a009bin + x045mod + c001bin + f025grouped, 
 family=binomial(link=probit), data=wvshm5red2delna, na.action=na.pass) 

 summary(probit4AKIE)


However, when calculating marginal effects with all variables at their means 
from the probit coefficients and a scale factor, the marginal effects I obtain 
are much too small (e.g. 2.6042e-78). My code looks like this:

 ttt - cbind(wvshm5red2delna$x003,
wvshm5red2delna$x003squ,
wvshm5red2delna$x025secv2,
wvshm5red2delna$x025terv2,
wvshm5red2delna$x007bin,
wvshm5red2delna$x04chief,
wvshm5red2delna$x011rec,
wvshm5red2delna$a009bin,
wvshm5red2delna$x045mod,
wvshm5red2delna$c001bin,
wvshm5red2delna$f025grouped,
wvshm5red2delna$f025grouped,
wvshm5red2delna$f025grouped,
wvshm5red2delna$f025grouped,
wvshm5red2delna$f025grouped,
wvshm5red2delna$f025grouped,
wvshm5red2delna$f025grouped,
wvshm5red2delna$f025grouped,
wvshm5red2delna$f025grouped) #I put variable f025grouped 9 times because this 
variable consists of 9 levels

 ttt - as.data.frame(ttt)

 xbar - as.matrix(mean(cbind(1,ttt[1:19]))) #1:19 position of variables in 
 dataframe ttt

 betaprobit4AKIE - probit4AKIE$coefficients

 zxbar - t(xbar) %*% betaprobit4AKIE

 scalefactor - dnorm(zxbar)

 marginprobit4AKIE - scalefactor * betaprobit4AKIE[2:20] #2:20 are the 
 positions of variables in the output of the probit model 'probit4AKIE' 
 (variables need to be in the same ordering as in data.frame ttt), the 
 constant in the model occupies the first position

 marginprobit4AKIE #in this step I obtain values that are much too small


I apologize that I can not provide you with a working example as my dataset is 
much too large. Any comment would be greatly appreciated. Thanks a lot.

Best,

Tobias
-- 
GMX DSL Doppel-Flat ab 19,99 Euro/mtl.! Jetzt mit 
gratis Handy-Flat! http://portal.gmx.net/de/go/dsl

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Re: [R] Two Copies of Each Message

2011-04-25 Thread david.jessop

I have the same problem - it's happened before and then just fixed itself.  But 
rather annoying. 

-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On 
Behalf Of Thomas Lumley
Sent: 23 April 2011 00:34
To: Stephen P Molnar
Cc: R-help
Subject: Re: [R] Two Copies of Each Message

On Sat, Apr 23, 2011 at 12:25 AM, Stephen P Molnar s.mol...@sbcglobal.net 
wrote:
 I receive two copies of every message posted to this list.

 How can I stop this?  I have read the Primary Help web page and 
 searched the achieves without finding an answer.

Unsubscribe from the list.  You will then either receive one copy (because you 
have another address subscribed), in which case you are done, or no copies, in 
which case you should be able to resubscribe and get just one copy.

-thomas

--
Thomas Lumley
Professor of Biostatistics
University of Auckland

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Issued by UBS AG or affiliates to professional investors...{{dropped:30}}

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[R] blank space escape sequence in R?

Is there a blank space escape sequence in R, i.e. something like \sp etc. to 
produce a blank space?

TIA
Mark
–––
Mark Heckmann
Blog: www.markheckmann.de
R-Blog: http://ryouready.wordpress.com

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Re: [R] blank space escape sequence in R?


On 25/04/2011 9:01 AM, Mark Heckmann wrote:

Is there a blank space escape sequence in R, i.e. something like \sp etc. to 
produce a blank space?


You need to give some context.  A blank in a character vector will be 
printed as a blank, so you are probably talking about something else, 
but what?


Duncan Murdoch

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Re: [R] blank space escape sequence in R?

I use a function that inserts line breaks (\n as escape sequence) according 
to some criterion when there are blanks in the string.
e.g. some text \nand some more text.

What I want now is another form of a blank, so my function will not insert a 
\n at that point.
e.g. some text\spaceand some more text

Here \space stands for some escape sequence for a  blank, which is what I am 
looking for. 
So what I need is something that will appear as a blank when printed but not in 
the string itself.

TIA

Am 25.04.2011 um 15:05 schrieb Duncan Murdoch:

 On 25/04/2011 9:01 AM, Mark Heckmann wrote:
 Is there a blank space escape sequence in R, i.e. something like \sp etc. to 
 produce a blank space?
 
 You need to give some context.  A blank in a character vector will be printed 
 as a blank, so you are probably talking about something else, but what?
 
 Duncan Murdoch


Mark Heckmann
Blog: www.markheckmann.de
R-Blog: http://ryouready.wordpress.com





[[alternative HTML version deleted]]

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Re: [R] Survival analysis: same subject with multiple treatments

2011-04-25 Thread Terry Therneau

---Begin inclusion --
I need some help to figure out what is the proper model in survival
analysis
for my data.
Subjects were randomized to 3 treatments in trial 1, some of them
experience
the event during the trial;
After period of time those subjects were randomized to 3 treatments
again in
trial 2, but different from what they got in 1st trial, some of them
experience the event during the 2nd trial (I think the carryover effect
can
be ignored since the time between two trials is long enough.)
---end inclusion --

If it were me I would add cluster(subject) to the model. Multiple
events/subject is not simple, however, and you should find a book
chapter to read.

Terry Therneau

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Re: [R] blank space escape sequence in R?


On 25/04/2011 9:13 AM, Mark Heckmann wrote:

I use a function that inserts line breaks (\n as escape sequence) according 
to some criterion when there are blanks in the string.
e.g. some text \nand some more text.

What I want now is another form of a blank, so my function will not insert a 
”\n at that point.
e.g. some text\spaceand some more text

Here \space stands for some escape sequence for a  blank, which is what I am 
looking for.
So what I need is something that will appear as a blank when printed but not in 
the string itself.


I don't think R has anything like that built in.   You'll need to attach 
a class to your vector of strings, and write a print method for it that 
does the substitution before printing.


Duncan Murdoch


TIA

Am 25.04.2011 um 15:05 schrieb Duncan Murdoch:

  On 25/04/2011 9:01 AM, Mark Heckmann wrote:
  Is there a blank space escape sequence in R, i.e. something like \sp etc. 
to produce a blank space?

  You need to give some context.  A blank in a character vector will be 
printed as a blank, so you are probably talking about something else, but what?

  Duncan Murdoch

–––
Mark Heckmann
Blog: www.markheckmann.de
R-Blog: http://ryouready.wordpress.com







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[R] R workshop in Hamilton, Ontario, May 24 and 25

2011-04-25 Thread John Fox

Dear r-help list members,

I'll be teaching a two-day introductory R workshop at McMaster University in
Hamilton, Ontario, on May 24 and 25. The workshop will largely be based on
materials from Fox and Weisberg, An R Companion to Applied Regression,
Second Edition (Sage, 2011). Further information about the workshop is
available at
https://www.socialsciences.mcmaster.ca/registrations/john-fox-introduction-
to-r/fg_base_view_p3. I've asked that a few spaces in the workshop be
reserved for non-McMaster attendees and made available at a modest cost.

Regards,
 John


John Fox
Senator William McMaster
  Professor of Social Statistics
Department of Sociology
McMaster University
Hamilton, Ontario, Canada
http://socserv.mcmaster.ca/jfox

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Re: [R] random typing over text

2011-04-25 Thread peter dalgaard


On Apr 25, 2011, at 14:52 , Kenn Konstabel wrote:
...
 means that Windows will activate StickyKeys(TM) which will make your
 keyboard useless until you restart the computer. 

...or, apparently, press both shift keys simultaneously, whichever comes first.

Now, whoever decided that CapsLock should be 3rd or 4th largest key on my 
keyboard

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

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Re: [R] blank space escape sequence in R?

2011-04-25 Thread Jan van der Laan


There exists a non-breaking space:

http://en.wikipedia.org/wiki/Non-breaking_space

Perhaps you could use this. In R on Linux under gnome-terminal I can 
enter it with CTRL+SHIFT+U00A0. This seems to work: it prints as a 
space, but is not equal to ' '. I don't know if there are any 
difficulties using, for example, utf8 encoding in source files (which 
you'll probably need).


Jan



On 04/25/2011 03:28 PM, Duncan Murdoch wrote:

On 25/04/2011 9:13 AM, Mark Heckmann wrote:
I use a function that inserts line breaks (\n as escape sequence) 
according to some criterion when there are blanks in the string.

e.g. some text \nand some more text.

What I want now is another form of a blank, so my function will not 
insert a ”\n at that point.

e.g. some text\spaceand some more text

Here \space stands for some escape sequence for a  blank, which is 
what I am looking for.
So what I need is something that will appear as a blank when printed 
but not in the string itself.


I don't think R has anything like that built in.   You'll need to 
attach a class to your vector of strings, and write a print method for 
it that does the substitution before printing.


Duncan Murdoch


TIA

Am 25.04.2011 um 15:05 schrieb Duncan Murdoch:

  On 25/04/2011 9:01 AM, Mark Heckmann wrote:
  Is there a blank space escape sequence in R, i.e. something like 
\sp etc. to produce a blank space?


  You need to give some context.  A blank in a character vector will 
be printed as a blank, so you are probably talking about something 
else, but what?


  Duncan Murdoch

–––
Mark Heckmann
Blog: www.markheckmann.de
R-Blog: http://ryouready.wordpress.com







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Re: [R] blank space escape sequence in R?

2011-04-25 Thread Jan van der Laan


There exists a non-breaking space:

http://en.wikipedia.org/wiki/Non-breaking_space

Perhaps you could use this. In R on Linux under gnome-terminal I can 
enter it with CTRL+SHIFT+U00A0. This seems to work: it prints as a 
space, but is not equal to ' '. I don't know if there are any 
difficulties using, for example, utf8 encoding in source files (which 
you'll probably need).


Jan




On 04/25/2011 03:28 PM, Duncan Murdoch wrote:

On 25/04/2011 9:13 AM, Mark Heckmann wrote:
I use a function that inserts line breaks (\n as escape sequence) 
according to some criterion when there are blanks in the string.

e.g. some text \nand some more text.

What I want now is another form of a blank, so my function will not 
insert a ”\n at that point.

e.g. some text\spaceand some more text

Here \space stands for some escape sequence for a  blank, which is 
what I am looking for.
So what I need is something that will appear as a blank when printed 
but not in the string itself.


I don't think R has anything like that built in.   You'll need to 
attach a class to your vector of strings, and write a print method for 
it that does the substitution before printing.


Duncan Murdoch


TIA

Am 25.04.2011 um 15:05 schrieb Duncan Murdoch:

  On 25/04/2011 9:01 AM, Mark Heckmann wrote:
  Is there a blank space escape sequence in R, i.e. something like 
\sp etc. to produce a blank space?


  You need to give some context.  A blank in a character vector will 
be printed as a blank, so you are probably talking about something 
else, but what?


  Duncan Murdoch

–––
Mark Heckmann
Blog: www.markheckmann.de
R-Blog: http://ryouready.wordpress.com







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Re: [R] problem with library tseriesChaos

2011-04-25 Thread Uwe Ligges

Dear Fukushima Shintaro,

please also reply to the person who asked the question (she or he might
not be subscribed to the mailing list). Additionally, please always
quote the original question. Other mailing list readers won't lerned
from an answer without corresponding question.

Thank you for providing help, best wishes,
Uwe Ligges

On 25.04.2011 14:32, Fukushima Shintaro wrote:

Hello.

Is it possible to determine time delay with other function's output or I
can choose any random value?

For example, time delay is defined as the time when average mutual
information reaches the first local minimum.

Using the package tseriesChaos, it is estimated as follows:

library(tseriesChaos)
x.mu- mutual(x, plot=FALSE)
x.mu.d- diff(x.mu)
d- min(which(x.mu.d 0)) - 1

The last two can also be simplied.

d- which.min(x.mu) - 1

I would be happy to be of any service to you.

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Re: [R] Random Relabelling

2011-04-25 Thread kmatthews

Thanks to everyone... this helps a lot.  Just a quick question about
etiquette in this forum (as it my first time questioning)... are notes of
gratitude usually given in these forums?

On Wed, Apr 20, 2011 at 1:26 PM, jthetzel [via R] 
ml-node+3463799-950416470-231...@n4.nabble.com wrote:

 Kevin,

 The following follows John's suggestion, but without the loop.  It's quick
 for me.

 Jeremy


 Jeremy T. Hetzel
 Boston University



 ## Generate sample data
 n - 4000
 rep - 1000
 rate - rnorm(n, mean = 15, sd = 2) / 10 # Mortality rates around
 15/100k

 ## Create an empty matrix with appropriate dimensions
 permutations - matrix(ncol = n, nrow = rep)

 ## Use apply() to resample
 permutations - apply(permutations, 1, function(x)
 {
 sample(rate, size = n, replace = F)
 })

 ## Look at the matrix
 dim(permutations)
 head(permutations)

 ## Find the column means
 means - apply(permutations, 1, mean)
 means





 On Wednesday, April 20, 2011 1:56:35 PM UTC-4, John Kane wrote:

 
  There is probably a better way to do this but a for loop like this should

  work. You would just need to change the numbers to yours and then add on
 the
  locations
  =
 
  scores  - 1:5
  mydata - matrix(data=NA, nrow=5, ncol=10)
 
  for(i in 1:10) {
  mydata[,i] - sample(scores, 5, replace=FALSE)
  }
 
  =
  --- On Wed, 4/20/11, Kevin Matthews [hidden 
  email]http://user/SendEmail.jtp?type=nodenode=3463799i=0by-user=t
 wrote:
 
  From: Kevin Matthews [hidden 
  email]http://user/SendEmail.jtp?type=nodenode=3463799i=1by-user=t

  Subject: Re: [R] Random Relabelling
  To: John Kane [hidden 
  email]http://user/SendEmail.jtp?type=nodenode=3463799i=2by-user=t

  Cc: [hidden 
  email]http://user/SendEmail.jtp?type=nodenode=3463799i=3by-user=t
  Received: Wednesday, April 20, 2011, 1:22 PM
 
  I have a map of Iowa of with 4000 locations.  At each location, I have a
  cancer mortality rate.  I need to test my null hypothesis; that the
 spatial
  distribution of the mortality rates is  random.  For this test, I need to

  establish a spatial reference distribution.
 
 
  My reference distribution will be created by some random relabelling
  algorithm.  The 4000 locations would remain fixed, but the observed
  mortality rates would be randomly redistributed.  Then, I want 1000
  permutations of the same algorithm.  For each of those 1000 times, I
 would
  record the redistributed mortality rate at each location.  Then,  I would

  calculate the mean of the 1000 points.  The result would be a spatial
  reference distribution with a mean value of the random permutations at
 each
  of the 4000 locations.
 
  Thanks for the response,Kevin
 
  On Wed, Apr 20, 2011 at 11:08 AM, John Kane [hidden 
  email]http://user/SendEmail.jtp?type=nodenode=3463799i=4by-user=t
 wrote:
 
 
  Can you explain this a bit more. At the moment I don't see what you are
  trying to achieve.   calculate the mean of the 1000 values at each of
 the
  4000 points does not seem to make sense.
 
  --- On Wed, 4/20/11, kmatthews [hidden 
  email]http://user/SendEmail.jtp?type=nodenode=3463799i=5by-user=t
 wrote:
 
   From: kmatthews [hidden 
   email]http://user/SendEmail.jtp?type=nodenode=3463799i=6by-user=t

 
   Subject: [R] Random Relabelling
 
   To: [hidden 
   email]http://user/SendEmail.jtp?type=nodenode=3463799i=7by-user=t
 
   Received: Wednesday, April 20, 2011, 10:04 AM
 
   I have 4000 observations that I need
 
   to randomly relabel 1000 times and then
 
   calculate the mean of the 1000 values at each of the 4000
 
   points.  Any ideas
 
   for where to begin?
 
  
 
   Thanks
 
   Kevin
 
  
 
 
  [[alternative HTML version deleted]]
 
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  email]http://user/SendEmail.jtp?type=nodenode=3463799i=8by-user=tmailing
   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.
 
 
 __
 [hidden 
 email]http://user/SendEmail.jtp?type=nodenode=3463799i=9by-user=tmailing 
 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.
 Jeremy T. Hetzel
 Boston University


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Re: [R] blank space escape sequence in R?

2011-04-25 Thread Mike Miller


On Mon, 25 Apr 2011, Mark Heckmann wrote:

I use a function that inserts line breaks (\n as escape sequence) 
according to some criterion when there are blanks in the string. e.g. 
some text \nand some more text.


What I want now is another form of a blank, so my function will not 
insert a ?\n at that point. e.g. some text\spaceand some more text


Here \space stands for some escape sequence for a blank, which is what 
I am looking for. So what I need is something that will appear as a 
blank when printed but not in the string itself.



Is it possible to use \x20 or some similar way to evoke the hexadecimal 
ascii form of blank?  That works in perl as does \040 for the octal form.


Mike

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Re: [R] filling array with functions

2011-04-25 Thread derek

Still I haven't had any luck yet.
How about defining new function and its domain, is it somehow possible?
Like this:
a-function(x) x beolongs to natural numbers 0,100

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Re: [R] Questions about lrm, validate, pentrace (Re: BMA, logistic regression, odds ratio, model reduction etc)

2011-04-25 Thread Frank Harrell

You've done a lot of good work on this.  Yes I would say you have moderate
overfitting with the first model.  The only thing that saved you from having
severe overfitting is that there seems to be a signal present [I am assume
this model is truly pre-specified and was not developed at all by looking at
patterns of responses Y.]

The use of backwards stepdown demonstrated much worse overfitting.  This is
in line with what we know about the damage of stepwise selection methods
that do not incorporate shrinkage.  I would throw away the stepwise
regression model.  You'll find that the model selected is entirely
arbitrary.  And you can't use the selected variables in any re-fit of the
model, i.e., you can't use lrm pretending that the two remaining variables
were pre-specified.  Stepwise regression methods only seem to help.  When
assessed properly we see that is an illusion.

You are using penalizing properly but you did not print the full table of
penalties vs. effective AIC.  We don't have faith that you penalized enough. 
I tend to run pentrace using a very wide range of possible penalties to make
sure I've found the global optimum.

Penalization somewhat solves the EPV problem but there is no substitute for
getting more data.

You can run validate specifying your final penalty as an argument.

Frank



細田弘吉 wrote:
 
 According to the advice, I tried rms package.
 Just to make sure, I have data of 104 patients (x6.df), which consists 
 of 5 explanatory variables and one binary outcome (poor/good) (previous 
 model 2 strategy). The outcome consists of 25 poor results and 79 good 
 results. Therefore, My events per variable (EPV) is only 5 (much less 
 than the rule of thumb of 10).
 
 My questions are about validate and pentrace in rms package.
 I present some codes and results.
 I appreciate anybody's help in advance.
 
   x6.lrm - lrm(outcome ~ stenosis+x1+x2+procedure+ClinicalScore, 
 data=x6.df, x=T, y=T)
 
   x6.lrm
 ...
 Obs  104LR chi2  29.24R2   0.367C   0.816
   negative 79d.f. 5g1.633Dxy 0.632
   positive 25Pr( chi2) 0.0001   gr5.118gamma   0.632
 max |deriv| 1e-08gp0.237tau-a   0.233
   Brier   0.127
 
 CoefS.E.   Wald Z Pr(|Z|)
 Intercept  -5.5328 2.6287 -2.10  0.0353
 stenosis   -0.0150 0.0284 -0.53  0.5979
 x1  3.0425 0.9100  3.34  0.0008
 x2 -0.7534 0.4519 -1.67  0.0955
 procedure   1.2085 0.5717  2.11  0.0345
 ClinicalScore   0.3762 0.2287  1.65  0.0999
 
 It seems not too bad. Next, validation by bootstrap ...
 
   validate(x6.lrm, B=200, bw=F)
index.orig trainingtest optimism index.corrected   n
 Dxy   0.6324   0.6960  0.5870   0.1091  0.5233 200
 R20.3668   0.4370  0.3154   0.1216  0.2453 200
 Intercept 0.   0. -0.2007   0.2007 -0.2007 200
 Slope 1.   1.  0.7565   0.2435  0.7565 200
 Emax  0.   0.  0.0999   0.0999  0.0999 200
 D 0.2716   0.3368  0.2275   0.1093  0.1623 200
 U-0.0192  -0.0192  0.0369  -0.0561  0.0369 200
 Q 0.2908   0.3560  0.1906   0.1654  0.1254 200
 B 0.1272   0.1155  0.1384  -0.0229  0.1501 200
 g 1.6328   2.0740  1.4647   0.6093  1.0235 200
 gp0.2367   0.2529  0.2189   0.0341  0.2026 200
 
 The apparent Dxy is 0.63, and bias-corrected Dxy is 0.52. The maximum 
 absolute error is estimated to be 0.099. The changes in slope and 
 intercept are also more substantial. In all, there is evidence that I am 
 somewhat overfitting the data, right?.
 
 Furthermore, using step-down variable selection ...
 
   validate(x6.lrm, B=200, bw=T)
 
   Backwards Step-down - Original Model
 
   DeletedChi-Sq d.f. P  Residual d.f. P  AIC
   stenosis   0.28   10.5979 0.28 10.5979 -1.72
   ClinicalScore  2.60   10.1068 2.88 20.2370 -1.12
   x2 2.86   10.0910 5.74 30.1252 -0.26
 
 Approximate Estimates after Deleting Factors
 
   Coef   S.E. Wald Z P
 Intercept  -5.865 1.4136 -4.149 3.336e-05
 x1  2.915 0.8685  3.357 7.889e-04
 procedure   1.072 0.5590  1.918 5.508e-02
 
 Factors in Final Model
 
 [1] x1 procedure
index.orig trainingtest optimism index.corrected   n
 Dxy   0.5661   0.6755  0.5559   0.1196  0.4464 200
 R20.2876   0.4085  0.2784   0.1301  0.1575 200
 Intercept 0.   0. -0.2459   0.2459 -0.2459 200
 Slope 1.   1.  0.7300   0.2700  0.7300 200
 Emax  0.   0.  0.1173   0.1173  0.1173 200
 D 0.2038   0.3130  0.1970   0.1160  0.0877 200
 U-0.0192  -0.0192  0.0382  -0.0574  0.0382 200
 Q

Re: [R] Random Relabelling

2011-04-25 Thread David Winsemius



On Apr 25, 2011, at 10:53 AM, kmatthews wrote:


Thanks to everyone... this helps a lot.  Just a quick question about
etiquette in this forum (as it my first time questioning)... are  
notes of

gratitude usually given in these forums?


The practice varies, some people do appreciate it. Doing so when one  
is not subscribed, however, adds many additional mouse-maneuvers to  
the moderator workload, ...  wasted time in my opinion. I would  
suggest a private thank you message in that instance (or even more to  
be preferred ... subscribing.)


--
David.

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[R] R equivalent to (D)QDAWO in Fortran?

2011-04-25 Thread Bao, Wenkai

Hi useRs,

I have a set of fortran code that was passed down from previous students, and I 
am converting its algorithm into R codes.

I encounter this function in Fortran (D)QDAWO, which numerically integrates a 
function f with a user-specified cosine or sine weight. It is used because the 
original function that I want to integrate is f(x)*cos(x).

I tried in R to directly integrate by integrate(g(x)), where g(x)=f(x)*cos(x), 
but a lot of error messages and warnings were given. So is there some function 
in R that is equivalent to this QDAWO in Fortran, to specifically integrate a 
function with a sine or cosine component?

Thank you very much!

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[R] Problem with grid's text-based units and TrueType fonts

2011-04-25 Thread Al Roark


Hi all:
I'm using grid to create a layout in R that will include text mixed with 
graphics.  In the layout, the positions of certain graphical elements depend on 
the number of lines in adjacent text blocks (which will vary from case to 
case).  I was hoping to use grid's built in functions to automate the placement 
of the graphical elements.  I can get this to work if I use the standard pdf 
device, but when using the Cairo_pdf device (from the cairoDevice package), I 
run into problems.  I need the TrueType functionality of Cairo.
The example below illustrates my problem. The rectangle in SizeMatch.pdf 
matches the size of the text block, while the rectangle in SizeMismatch.pdf 
is too small.  Why is this the case?  I'm using R-2.13.0 (32bit) on Windows 7.  
Thanks a bunch.  Cheers, Al
---
library(cairoDevice)library(grid)
Cairo_pdf(SizeMismatch.pdf,11,8.5,pointsize=10)pushViewport(viewport(gp=gpar(fontfamily=Arial,font=1,fontsize=8,lineheight=0.9)))grid.text(The
 quick\nbrown fox\njumps\nover\nthe lazy 
dog)grid.rect(width=0.5,height=unit(5,lines))dev.off()
pdf(SizeMatch.pdf,11,8.5,pointsize=10)pushViewport(viewport(gp=gpar(fontfamily=Times,font=1,fontsize=8,lineheight=0.9)))grid.text(The
 quick\nbrown fox\njumps\nover\nthe lazy 
dog)grid.rect(width=0.5,height=unit(5,lines))dev.off()   
   
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Re: [R] Defining origin for rotation in RGL device


On 25/04/2011 7:54 AM, Duncan Murdoch wrote:

On 25/04/2011 5:46 AM, Mark Heckmann wrote:
  Hi all,

  How can I tell RGL to set the center for the rotation to the origin of  the 
coordinate system (0,0,0).
  It seems that the default is to use the center of the display not the origin 
of the coordinate system.

  open3d()
  lines3d(c(0, 1), c(0,0), c(0,0))
  lines3d(c(0,0), c(0, 1), c(0,0))
  lines3d(c(0,0), c(0,0), c(0, 1))


You can attach any transformation you like to a mouse button.  See the
mouseCallbacks demo for R implementations of the standard ones, and
modify the mouseTrackball function there to choose the position of the
origin of the coordinate system as the centre of rotation.


This was a little trickier than I was thinking because of the weird 
coordinate system.  You have to remember to transpose translationMatrix 
when you're planning to work in the coordinates of userMatrix.   Here's 
a function (modified from mouseTrackball in the demo) that I think does 
what you want.


Just call

mouseTrackballOrigin()

to set it up on button 1 on the current device with center of rotation 
at (0,0,0).


Duncan Murdoch

mouseTrackballOrigin - function(button = 1, dev = rgl.cur(), 
origin=c(0,0,0) ) {

width - height - rotBase - NULL
userMatrix - list()
cur - rgl.cur()
offset - NULL
scale - NULL

screenToVector - function(x, y) {
  radius - max(width, height)/2
  centre - c(width, height)/2
  pt - (c(x, y) - centre)/radius
  len - vlen(pt)

  if (len  1.e-6) pt - pt/len

  maxlen - sqrt(2)
  angle - (maxlen - len)/maxlen*pi/2
  z - sin(angle)
  len - sqrt(1 - z^2)
  pt - pt * len
  return (c(pt, z))
}

trackballBegin - function(x, y) {
vp - par3d(viewport)
width - vp[3]
height - vp[4]
cur - rgl.cur()
bbox - par3d(bbox)
center - c(sum(bbox[1:2])/2, sum(bbox[3:4])/2, sum(bbox[5:6])/2)
scale - par3d(scale)
offset - (center - origin)*scale
for (i in dev) {
if (inherits(try(rgl.set(i, TRUE)), try-error)) dev - 
dev[dev != i]

else userMatrix[[i]] - par3d(userMatrix)
}
rgl.set(cur, TRUE)
rotBase - screenToVector(x, height - y)
}

trackballUpdate - function(x,y) {
rotCurrent - screenToVector(x, height - y)
angle - angle(rotBase, rotCurrent)
axis - xprod(rotBase, rotCurrent)
mouseMatrix - rotationMatrix(angle, axis[1], axis[2], axis[3])
for (i in dev) {
if (inherits(try(rgl.set(i, TRUE)), try-error)) dev - 
dev[dev != i]
else par3d(userMatrix = t(translationMatrix(-offset[1], 
-offset[2], -offset[3])) %*% mouseMatrix  %*% 
t(translationMatrix(offset[1], offset[2], offset[3])) %*%userMatrix[[i]])

}
rgl.set(cur, TRUE)
}

for (i in dev) {
rgl.set(i, TRUE)
rgl.setMouseCallbacks(button, begin = trackballBegin, update = 
trackballUpdate, end = NULL)

}
rgl.set(cur, TRUE)
}

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Re: [R] Pass String from R to C

2011-04-25 Thread Jaimin Dave

I tried using char ** but it is printing some random string.

On Sat, Apr 23, 2011 at 6:08 PM, Duncan Murdoch murdoch.dun...@gmail.comwrote:

 On 11-04-23 7:04 PM, Jaimin Dave wrote:

 Hi,
 I am using a function which accepts the string from R and prints it.
 But when I am calling .C(main,hello);
 it is printing any random thing.
 My C function is
 void main(char *str)


 See Writing R Extensions.  The declaration should be char **str.

 Duncan Murdoch

  {
 Rprintf(%s,str);
 }

 Can you help how to achieve this using .C interface?

 Thanks
 Jaimin

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Re: [R] Pass String from R to C


On 25/04/2011 12:51 PM, Jaimin Dave wrote:

I tried using char ** but it is printing some random string.


str is a pointer to an array of pointers to strings.  That's what char** 
means.  So you need to declare it that way, and use it that way.


This works for me:

File test.c:

void test(char **str)
{
  Rprintf(%s,*str);
}

Duncan Murdoch

On Sat, Apr 23, 2011 at 6:08 PM, Duncan Murdochmurdoch.dun...@gmail.comwrote:

  On 11-04-23 7:04 PM, Jaimin Dave wrote:

  Hi,
  I am using a function which accepts the string from R and prints it.
  But when I am calling .C(main,hello);
  it is printing any random thing.
  My C function is
  void main(char *str)


  See Writing R Extensions.  The declaration should be char **str.

  Duncan Murdoch

   {
  Rprintf(%s,str);
  }

  Can you help how to achieve this using .C interface?

  Thanks
  Jaimin

 [[alternative HTML version deleted]]

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Re: [R] Pass String from R to C


On 25/04/2011 1:03 PM, Duncan Murdoch wrote:

On 25/04/2011 12:51 PM, Jaimin Dave wrote:
  I tried using char ** but it is printing some random string.

str is a pointer to an array of pointers to strings.  That's what char**
means.  So you need to declare it that way, and use it that way.

This works for me:

File test.c:

void test(char **str)
{
Rprintf(%s,*str);
}


Oops, I just noticed that the include was missing.  The full file should 
have


#include R.h

at the beginning.

Duncan Murdoch



Duncan Murdoch
  On Sat, Apr 23, 2011 at 6:08 PM, Duncan 
Murdochmurdoch.dun...@gmail.comwrote:

 On 11-04-23 7:04 PM, Jaimin Dave wrote:
  
 Hi,
 I am using a function which accepts the string from R and prints it.
 But when I am calling .C(main,hello);
 it is printing any random thing.
 My C function is
 void main(char *str)
  
  
 See Writing R Extensions.  The declaration should be char **str.
  
 Duncan Murdoch
  
  {
 Rprintf(%s,str);
 }
  
 Can you help how to achieve this using .C interface?
  
 Thanks
 Jaimin
  
[[alternative HTML version deleted]]
  
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Re: [R] merge with origin information in new variable names

2011-04-25 Thread Eric Fail

Is there anyone out there who can suggest a way to solve this problem?

Thanks,
Esben

On Sun, Apr 24, 2011 at 8:53 PM, Jeff Newmiller
jdnew...@dcn.davis.ca.us wrote:
 Merge only lets you combine two tables at a time, but it does have a
 suffix argument that is intended to address your concern, but only for
 variable names that would conflict.

 In your example, the id variables are all sequenced exactly the same, so you
 could actually use cbind rather than merge.

 However, whether you use merge or cbind, I think the most direct route to
 your desired result is to rename the data columns before you combine them,
 using the names function on the left hand side of an assignment with a
 vector of new names on the right.
 ---
 Jeff Newmiller The . . Go Live...
 DCN:jdnew...@dcn.davis.ca.us Basics: ##.#. ##.#. Live Go...
 Live: OO#.. Dead: OO#.. Playing
 Research Engineer (Solar/Batteries O.O#. #.O#. with
 /Software/Embedded Controllers) .OO#. .OO#. rocks...1k
 ---
 Sent from my phone. Please excuse my brevity.

 Eric Fail eric.f...@gmx.com wrote:

 Dear R-list,

 Here is my simple question,

 I have n data frames that I would like to merge, but I can't figure out
 how to add information about the origin of the variable(s).

 Here is my problem,

 DF.wave.1 - data.frame(id=1:10,var.A=sample(letters[1:4],10,TRUE))
 DF.wave.2 - data.frame(id=1:10,var.M=sample(letters[5:8],10,TRUE))
 DF.wave.3 - data.frame(id=1:10,var.A=sample(letters[5:8],10,TRUE))

 Now; I would like to merge the three dataframes into one, but append a
 suffix to the individual variables names about thir origin.

 DF.wave.all - merge(DF.wave.1,DF.wave.2,DF.wave.3,by=id, [what to do
 here])

 In other words, I would like it to loook like this.

 DF.wave.all
id var.A.wave.1 var.M.wave.2 var.A.wave.3
 1   1chj
 2   2cej
 3   3cgk
 4   4cej
 5   5cgi
 6   6dek
 7   7chk
 8   8bgj
 9   9bfi
 10 10dhi


 Is there a command I can use directly in merge? 'suffixes' isn't really
 handy here.

 Thanks,
 Eric
 
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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.


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[R] Problem with ddply in the plyr-package: surprising output of a date-column

2011-04-25 Thread Christoph Jäckel

Hi Together,

I have a problem with the plyr package - more precisely with the ddply
function - and would be very grateful for any help. I hope the example
here is precise enough for someone to identify the problem. Basically,
in this step I want to identify observations that are identical in
terms of certain identifiers (ID1, ID2, ID3) and just want to save
those observations (in this step, without deleting any rows or
manipulating any data) in a separate data.frame. However, I get the
warning message below and the column with dates is messed up.
Interestingly, the value column (the type is factor here, but if you
change that with as.integer it doesn't make any difference) is handled
correctly. Any idea what I do wrong?

df - 
data.frame(cbind(ID1=c(1,2,2,3,3,4,4),ID2=c('a','b','b','c','d','e','e'),ID3=c(v1,v1,v1,v1,v2,v1,v1),

Date=c(1985-05-1,1985-05-2,1985-05-3,1985-05-4,1985-05-5,1985-05-6,1985-05-7),
 Value=c(1,2,3,4,5,6,7)))
df[,1] - as.character(df[,1])
df[,2] - as.character(df[,2])
df$Date   - strptime(df$Date,%Y-%m-%d)

#Apparently there are two observation that have the same IDs: ID1=2 and ID1=4
ddply(df,.(ID1,ID2,ID3),nrow)
#I want to save those IDs in a separate data.frame, so the desired output is:
df[c(2:3,6:7),]

#My idea: Write a custom function that only returns observations with
multiple rows.
#Seems to work except that the Date column doesn't make any sense anymore
#Warning message: In output[[var]][rng] - df[[var]]: number of items
to replace is not a multiple of replacement length
ddply(df,.(ID1,ID2,ID3),function(df) if(nrow(df)=1){NULL}else{df})

#Notice that it works perfectly if I only have one observation with
multiple rows
ddply(df[1:6,],.(ID1,ID2,ID3),function(df) if(nrow(df)=1){NULL}else{df})

Thanks in advance,

Christoph



Christoph Jäckel (Dipl.-Kfm.)



Research Assistant

Chair for Financial Management and Capital Markets | Lehrstuhls für
Finanzmanagement und Kapitalmärkte

TUM School of Management | Technische Universität München

Arcisstr. 21 | D-80333 München | Germany

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Re: [R] blank space escape sequence in R?

2011-04-25 Thread Matt Shotwell

You can embed hex escapes in strings (except \x00). The value(s) that
you embed will depend on the character encoding used on you platform. If
this is UTF-8, or some other ASCII compatible encoding, \x20 will work:

 foo\x20bar
[1] foo bar


For other locales, you might try charToRaw( ) to see the binary (hex)
representation for the space character on your platform, and substitute
this sequence instead.

On Mon, 2011-04-25 at 15:01 +0200, Mark Heckmann wrote:
 Is there a blank space escape sequence in R, i.e. something like \sp etc. to 
 produce a blank space?
 
 TIA
 Mark
 –––
 Mark Heckmann
 Blog: www.markheckmann.de
 R-Blog: http://ryouready.wordpress.com
 
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Re: [R] merge with origin information in new variable names

2011-04-25 Thread Phil Spector


Eric -
As others have said, you should change the names of the variables
in the data frames before you merge them.  Here's one implementation
of that idea:

   DF.wave.1 - data.frame(id=1:10,var.A=sample(letters[1:4],10,TRUE))
   DF.wave.2 - data.frame(id=1:10,var.M=sample(letters[5:8],10,TRUE))
   DF.wave.3 - data.frame(id=1:10,var.A=sample(letters[5:8],10,TRUE))

   nms = paste('wave',1:3,sep='.')
   dfs = list(DF.wave.1,DF.wave.2,DF.wave.3)
   names(dfs) = nms

   changenm = function(nm){
   df = dfs[[nm]]
   wh = names(df) != 'id'
   names(df)[wh] = paste(names(df)[wh],nm,sep='.')
   df
   }

   Reduce(function(x,y)merge(x,y,by='id'),lapply(names(dfs),changenm))

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




On Mon, 25 Apr 2011, Eric Fail wrote:


Is there anyone out there who can suggest a way to solve this problem?

Thanks,
Esben

On Sun, Apr 24, 2011 at 8:53 PM, Jeff Newmiller
jdnew...@dcn.davis.ca.us wrote:

Merge only lets you combine two tables at a time, but it does have a
suffix argument that is intended to address your concern, but only for
variable names that would conflict.

In your example, the id variables are all sequenced exactly the same, so you
could actually use cbind rather than merge.

However, whether you use merge or cbind, I think the most direct route to
your desired result is to rename the data columns before you combine them,
using the names function on the left hand side of an assignment with a
vector of new names on the right.
---
Jeff Newmiller The . . Go Live...
DCN:jdnew...@dcn.davis.ca.us Basics: ##.#. ##.#. Live Go...
Live: OO#.. Dead: OO#.. Playing
Research Engineer (Solar/Batteries O.O#. #.O#. with
/Software/Embedded Controllers) .OO#. .OO#. rocks...1k
---
Sent from my phone. Please excuse my brevity.

Eric Fail eric.f...@gmx.com wrote:


Dear R-list,

Here is my simple question,

I have n data frames that I would like to merge, but I can't figure out
how to add information about the origin of the variable(s).

Here is my problem,

DF.wave.1 - data.frame(id=1:10,var.A=sample(letters[1:4],10,TRUE))
DF.wave.2 - data.frame(id=1:10,var.M=sample(letters[5:8],10,TRUE))
DF.wave.3 - data.frame(id=1:10,var.A=sample(letters[5:8],10,TRUE))

Now; I would like to merge the three dataframes into one, but append a
suffix to the individual variables names about thir origin.

DF.wave.all - merge(DF.wave.1,DF.wave.2,DF.wave.3,by=id, [what to do
here])

In other words, I would like it to loook like this.

DF.wave.all
   id var.A.wave.1 var.M.wave.2 var.A.wave.3
1   1chj
2   2cej
3   3cgk
4   4cej
5   5cgi
6   6dek
7   7chk
8   8bgj
9   9bfi
10 10dhi


Is there a command I can use directly in merge? 'suffixes' isn't really
handy here.

Thanks,
Eric

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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[R] Help with objects

2011-04-25 Thread Russ Abbott

Hi all,

I would appreciate some help in understanding how to find out about objects.
For example, to the extent that I understand R it seems to treat everything
as an object even without declaring them as objects. Everything has
attributes, for example, which are like instance variables in objects. In
addition, there are S3 and S3 category objects.  Is there a good
introductory description of how these are different from standard R objects
and how they are different from each other?

Also, how does one find out more about how objects are declared. For
example, Data Mining with
Rhttp://www.liaad.up.pt/~ltorgo/DataMiningWithR/code3.htmluses the
quantmod package. I am used to Java's JavaDoc where one can see how
classes are declared, what the instance variables and methods are, etc. I
don't see anything similar for this package.  How, for example, would one
find out what the instance variables are in a quantmod object and what the
methods are that are defined on quantmod objects? I know that there is the
standard quantmod documentation, but that doesn't seem
to distinguish between standard functions and class-based methods. Nor, as
far as I can see, does it describe the instance variables in a quantmod
object.

Thanks.

*-- Russ Abbott*
*_*
***  Professor, Computer Science*
*  California State University, Los Angeles*

*  Google voice: 747-*999-5105
*  blog: *http://russabbott.blogspot.com/
  vita:  http://sites.google.com/site/russabbott/
*_*

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Re: [R] blank space escape sequence in R?

2011-04-25 Thread Matt Shotwell

I may have misread your original email. Whether you use a hex escape or
a space character, the resulting string in memory is identical:

 identical(a\x20b, a b)
[1] TRUE

But, if you were to read a file containing the six characters a
\x20b (say with readLines), then the six characters would be read into
memory, and printed like this:

a\\x20b

That is, not with a space character substituted for \x20. So, now I'm
not sure this is a solution.

On Mon, 2011-04-25 at 12:24 -0500, Matt Shotwell wrote:
 You can embed hex escapes in strings (except \x00). The value(s) that
 you embed will depend on the character encoding used on you platform. If
 this is UTF-8, or some other ASCII compatible encoding, \x20 will work:
 
  foo\x20bar
 [1] foo bar
 
 
 For other locales, you might try charToRaw( ) to see the binary (hex)
 representation for the space character on your platform, and substitute
 this sequence instead.
 
 On Mon, 2011-04-25 at 15:01 +0200, Mark Heckmann wrote:
  Is there a blank space escape sequence in R, i.e. something like \sp etc. 
  to produce a blank space?
  
  TIA
  Mark
  –––
  Mark Heckmann
  Blog: www.markheckmann.de
  R-Blog: http://ryouready.wordpress.com
  
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Re: [R] blank space escape sequence in R?

2011-04-25 Thread Petr Savicky

On Mon, Apr 25, 2011 at 04:37:15PM +0200, Jan van der Laan wrote:
 There exists a non-breaking space:
 
 http://en.wikipedia.org/wiki/Non-breaking_space
 
 Perhaps you could use this. In R on Linux under gnome-terminal I can 
 enter it with CTRL+SHIFT+U00A0. This seems to work: it prints as a 
 space, but is not equal to ' '.

This character may be specified as \u00A0.

  a - abc\u00A0def
  a
  [1] abc def
  
The utf-8 representation of the obtained string is

  charToRaw(a)
  [1] 61 62 63 c2 a0 64 65 66

Using Unicode package, the string may be analyzed as follows 

  library(Unicode)
  u_char_inspect(as.u_char_seq(a, ))

  Code Name Char
  1 U+0061 LATIN SMALL LETTER Aa
  2 U+0062 LATIN SMALL LETTER Bb
  3 U+0063 LATIN SMALL LETTER Cc
  4 U+00A0   NO-BREAK SPACE 
  5 U+0064 LATIN SMALL LETTER Dd
  6 U+0065 LATIN SMALL LETTER Ee
  7 U+0066 LATIN SMALL LETTER Ff

Hope this helps.

Petr Savicky.

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[R] extracting names from matrix according to a condition

Dear Community,

I have a matrix with assigned colnames and rolnames as follows:

AB
NR0.15  0,05
AL 0,05  0,05
.   ..
.   ..
.   ..

I want to extract the names of the rows for which A0,1 and B0,1. In
the above example this would be observation NR only. Hence the output
should write for instance:

names:
NR

Is this possible? Thank you very much for your help.

Best Regards

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Re: [R] Problem with ddply in the plyr-package: surprising output of a date-column

2011-04-25 Thread Brian Diggs


On 4/25/2011 10:19 AM, Christoph Jäckel wrote:

Hi Together,

I have a problem with the plyr package - more precisely with the ddply
function - and would be very grateful for any help. I hope the example
here is precise enough for someone to identify the problem. Basically,
in this step I want to identify observations that are identical in
terms of certain identifiers (ID1, ID2, ID3) and just want to save
those observations (in this step, without deleting any rows or
manipulating any data) in a separate data.frame. However, I get the
warning message below and the column with dates is messed up.
Interestingly, the value column (the type is factor here, but if you
change that with as.integer it doesn't make any difference) is handled
correctly. Any idea what I do wrong?

df- 
data.frame(cbind(ID1=c(1,2,2,3,3,4,4),ID2=c('a','b','b','c','d','e','e'),ID3=c(v1,v1,v1,v1,v2,v1,v1),

Date=c(1985-05-1,1985-05-2,1985-05-3,1985-05-4,1985-05-5,1985-05-6,1985-05-7),
  Value=c(1,2,3,4,5,6,7)))
df[,1]- as.character(df[,1])
df[,2]- as.character(df[,2])
df$Date- strptime(df$Date,%Y-%m-%d)

#Apparently there are two observation that have the same IDs: ID1=2 and ID1=4
ddply(df,.(ID1,ID2,ID3),nrow)
#I want to save those IDs in a separate data.frame, so the desired output is:
df[c(2:3,6:7),]

#My idea: Write a custom function that only returns observations with
multiple rows.
#Seems to work except that the Date column doesn't make any sense anymore
#Warning message: In output[[var]][rng]- df[[var]]: number of items
to replace is not a multiple of replacement length
ddply(df,.(ID1,ID2,ID3),function(df) if(nrow(df)=1){NULL}else{df})

#Notice that it works perfectly if I only have one observation with
multiple rows
ddply(df[1:6,],.(ID1,ID2,ID3),function(df) if(nrow(df)=1){NULL}else{df})


Works for me:

 df[c(2:3,6:7),]
  ID1 ID2 ID3  Date Value
2   2   b  v1 1985-05-2 2
3   2   b  v1 1985-05-3 3
6   4   e  v1 1985-05-6 6
7   4   e  v1 1985-05-7 7
 ddply(df,.(ID1,ID2,ID3),function(df) if(nrow(df)=1){NULL}else{df})
  ID1 ID2 ID3  Date Value
1   2   b  v1 1985-05-2 2
2   2   b  v1 1985-05-3 3
3   4   e  v1 1985-05-6 6
4   4   e  v1 1985-05-7 7
 sessionInfo()
R version 2.13.0 (2011-04-13)
Platform: x86_64-pc-mingw32/x64 (64-bit)

locale:
[1] LC_COLLATE=English_United States.1252
[2] LC_CTYPE=English_United States.1252
[3] LC_MONETARY=English_United States.1252
[4] LC_NUMERIC=C
[5] LC_TIME=English_United States.1252

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

other attached packages:
[1] plyr_1.5.2

loaded via a namespace (and not attached):
[1] tools_2.13.0

A couple of things: there was just an update of plyr to 1.5.2; maybe 
that fixes what you are seeing?  Also, your df consists of only factors. 
 cbind-ing the data before turning it into a data.frame makes it a 
character matrix which gets converted to factors.


 str(df)
'data.frame':   7 obs. of  5 variables:
 $ ID1  : Factor w/ 4 levels 1,2,3,4: 1 2 2 3 3 4 4
 $ ID2  : Factor w/ 5 levels a,b,c,d,..: 1 2 2 3 4 5 5
 $ ID3  : Factor w/ 2 levels v1,v2: 1 1 1 1 2 1 1
 $ Date : Factor w/ 7 levels 1985-05-1,1985-05-2,..: 1 2 3 4 5 6 7
 $ Value: Factor w/ 7 levels 1,2,3,4,..: 1 2 3 4 5 6 7

Maybe that has something to do with the odd dates since they are not 
really dates at all, just string representations of factor levels. 
Compare with:


DF - data.frame(ID1=c(1,2,2,3,3,4,4),
ID2=c('a','b','b','c','d','e','e'),
ID3=c(v1,v1,v1,v1,v2,v1,v1),
Date=as.Date(c(1985-05-1,1985-05-2,1985-05-3,
1985-05-4,1985-05-5,1985-05-6,1985-05-7)),
Value=c(1,2,3,4,5,6,7))
str(DF)
#'data.frame':   7 obs. of  5 variables:
# $ ID1  : num  1 2 2 3 3 4 4
# $ ID2  : Factor w/ 5 levels a,b,c,d,..: 1 2 2 3 4 5 5
# $ ID3  : Factor w/ 2 levels v1,v2: 1 1 1 1 2 1 1
# $ Date : Date, format: 1985-05-01 1985-05-02 ...
# $ Value: num  1 2 3 4 5 6 7

This version also works for me.

ddply(DF,.(ID1,ID2,ID3),function(df) if(nrow(df)=1){NULL}else{df})
#  ID1 ID2 ID3   Date Value
#1   2   b  v1 1985-05-02 2
#2   2   b  v1 1985-05-03 3
#3   4   e  v1 1985-05-06 6
#4   4   e  v1 1985-05-07 7


Thanks in advance,

Christoph



Christoph Jäckel (Dipl.-Kfm.)



Research Assistant

Chair for Financial Management and Capital Markets | Lehrstuhls für
Finanzmanagement und Kapitalmärkte

TUM School of Management | Technische Universität München

Arcisstr. 21 | D-80333 München | Germany




--
Brian S. Diggs, PhD
Senior Research Associate, Department of Surgery
Oregon Health  Science University

Re: [R] blank space escape sequence in R?

Thanks, Matt!
\x20 works great for me!

Am 25.04.2011 um 19:42 schrieb Matt Shotwell:

 I may have misread your original email. Whether you use a hex escape or
 a space character, the resulting string in memory is identical:
 
 identical(a\x20b, a b)
 [1] TRUE
 
 But, if you were to read a file containing the six characters a
 \x20b (say with readLines), then the six characters would be read into
 memory, and printed like this:
 
 a\\x20b
 
 That is, not with a space character substituted for \x20. So, now I'm
 not sure this is a solution.
 
 On Mon, 2011-04-25 at 12:24 -0500, Matt Shotwell wrote:
 You can embed hex escapes in strings (except \x00). The value(s) that
 you embed will depend on the character encoding used on you platform. If
 this is UTF-8, or some other ASCII compatible encoding, \x20 will work:
 
 foo\x20bar
 [1] foo bar
 
 
 For other locales, you might try charToRaw( ) to see the binary (hex)
 representation for the space character on your platform, and substitute
 this sequence instead.
 
 On Mon, 2011-04-25 at 15:01 +0200, Mark Heckmann wrote:
 Is there a blank space escape sequence in R, i.e. something like \sp etc. 
 to produce a blank space?
 
 TIA
 Mark
 
 Mark Heckmann
 Blog: www.markheckmann.de
 R-Blog: http://ryouready.wordpress.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
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 __
 R-help@r-project.org mailing list
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Mark Heckmann
Blog: www.markheckmann.de
R-Blog: http://ryouready.wordpress.com





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Re: [R] Problem with ddply in the plyr-package: surprising output of a date-column

2011-04-25 Thread Peter Ehlers


On 2011-04-25 10:19, Christoph Jäckel wrote:

Hi Together,

I have a problem with the plyr package - more precisely with the ddply
function - and would be very grateful for any help. I hope the example
here is precise enough for someone to identify the problem. Basically,
in this step I want to identify observations that are identical in
terms of certain identifiers (ID1, ID2, ID3) and just want to save
those observations (in this step, without deleting any rows or
manipulating any data) in a separate data.frame. However, I get the
warning message below and the column with dates is messed up.
Interestingly, the value column (the type is factor here, but if you
change that with as.integer it doesn't make any difference) is handled
correctly. Any idea what I do wrong?

df- 
data.frame(cbind(ID1=c(1,2,2,3,3,4,4),ID2=c('a','b','b','c','d','e','e'),ID3=c(v1,v1,v1,v1,v2,v1,v1),

Date=c(1985-05-1,1985-05-2,1985-05-3,1985-05-4,1985-05-5,1985-05-6,1985-05-7),
  Value=c(1,2,3,4,5,6,7)))
df[,1]- as.character(df[,1])
df[,2]- as.character(df[,2])
df$Date- strptime(df$Date,%Y-%m-%d)

#Apparently there are two observation that have the same IDs: ID1=2 and ID1=4
ddply(df,.(ID1,ID2,ID3),nrow)
#I want to save those IDs in a separate data.frame, so the desired output is:
df[c(2:3,6:7),]

#My idea: Write a custom function that only returns observations with
multiple rows.
#Seems to work except that the Date column doesn't make any sense anymore
#Warning message: In output[[var]][rng]- df[[var]]: number of items
to replace is not a multiple of replacement length
ddply(df,.(ID1,ID2,ID3),function(df) if(nrow(df)=1){NULL}else{df})

#Notice that it works perfectly if I only have one observation with
multiple rows
ddply(df[1:6,],.(ID1,ID2,ID3),function(df) if(nrow(df)=1){NULL}else{df})


I would characterize your problem as:
a) using strptime - this is what gives ddply() fits;

b) not using str() to check whether R agrees with
   you with respect to your data;

c) using cbind() inside data.frame(). This isn't
   wrong, but is rarely (in my experience) useful.

If you use as.Date (or even nothing) on your Date
variable, you'll find that ddply does what you want.
To see why it doesn't work with strptime, check
str(df) and then ?Posixlt. You've converted Date
values to lists.

My comment about cbind() is to warn you that your
Values variable, as you have constructed it, is
a factor.

Peter Ehlers



Thanks in advance,

Christoph



Christoph Jäckel (Dipl.-Kfm.)



Research Assistant

Chair for Financial Management and Capital Markets | Lehrstuhls für
Finanzmanagement und Kapitalmärkte

TUM School of Management | Technische Universität München

Arcisstr. 21 | D-80333 München | Germany

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Re: [R] Help with objects


Russ Abbott wrote:

Hi all,

I would appreciate some help in understanding how to find out about objects.
For example, to the extent that I understand R it seems to treat everything
as an object even without declaring them as objects.


That's correct.

 Everything has

attributes, for example, which are like instance variables in objects.


That's not quite true.  There are a few special kinds of objects which 
can't have attributes.  The NULL object is the only one you're likely to 
meet in regular code.


 In

addition, there are S3 and S3 category objects.  Is there a good
introductory description of how these are different from standard R objects
and how they are different from each other?


S3 and S4 are layers put on top of the regular system for handling 
objects.  S3 objects just have a class attribute, which causes some 
functions to handle them specially; this is introduced in An 
Introduction to R, and more fully described in the R Language 
Definition.  S4 is a more complex system, partly described in the 
?Methods help topic, more completely in the references listed on that page.


Duncan Murdoch


Also, how does one find out more about how objects are declared. For
example, Data Mining with
Rhttp://www.liaad.up.pt/~ltorgo/DataMiningWithR/code3.htmluses the
quantmod package. I am used to Java's JavaDoc where one can see how
classes are declared, what the instance variables and methods are, etc. I
don't see anything similar for this package.  How, for example, would one
find out what the instance variables are in a quantmod object and what the
methods are that are defined on quantmod objects? I know that there is the
standard quantmod documentation, but that doesn't seem
to distinguish between standard functions and class-based methods. Nor, as
far as I can see, does it describe the instance variables in a quantmod
object.

Thanks.

*-- Russ Abbott*
*_*
***  Professor, Computer Science*
*  California State University, Los Angeles*

*  Google voice: 747-*999-5105
*  blog: *http://russabbott.blogspot.com/
  vita:  http://sites.google.com/site/russabbott/
*_*

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Re: [R] extracting names from matrix according to a condition

2011-04-25 Thread Peter Ehlers


On 2011-04-25 10:58, ivan wrote:

Dear Community,

I have a matrix with assigned colnames and rolnames as follows:

 AB
NR0.15  0,05
AL 0,05  0,05
.   ..
.   ..
.   ..

I want to extract the names of the rows for which A0,1 and B0,1. In
the above example this would be observation NR only. Hence the output
should write for instance:

names:
NR

Is this possible? Thank you very much for your help.


Call the matrix m. Then

 rownames(m[ m[, A]  0.1  m[, B]  0.1, , drop=FALSE ])

should do what you want.

Peter Ehlers

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Re: [R] [SPAM] - R equivalent to (D)QDAWO in Fortran? - Found word(s) list error in the Text body

2011-04-25 Thread David Reiner

dqdawo is an IMSL routine which is similar to QUADPACK's dqawo. (The initial d 
is for double precision),
and maybe similar to NAG's d01anf.
If you search for 'imsl qdawo' you should find some more description.

On the other hand, the error messages may give someone enough information to 
help you directly.
Maybe your function f() is not vectorized?
See the posting guide linked to at the bottom of every posting here.
Follow it well to get incredible help here; don't follow it to get guesses or 
no response.

HTH,
-- David

-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On 
Behalf Of Bao, Wenkai
Sent: Monday, April 25, 2011 11:04 AM
To: r-help@r-project.org
Subject: [SPAM] - [R] R equivalent to (D)QDAWO in Fortran? - Found word(s) list 
error in the Text body

Hi useRs,

I have a set of fortran code that was passed down from previous students, and I 
am converting its algorithm into R codes.

I encounter this function in Fortran (D)QDAWO, which numerically integrates a 
function f with a user-specified cosine or sine weight. It is used because the 
original function that I want to integrate is f(x)*cos(x).

I tried in R to directly integrate by integrate(g(x)), where g(x)=f(x)*cos(x), 
but a lot of error messages and warnings were given. So is there some function 
in R that is equivalent to this QDAWO in Fortran, to specifically integrate a 
function with a sine or cosine component?

Thank you very much!

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Re: [R] Problem with ddply in the plyr-package: surprising output of a date-column

2011-04-25 Thread William Dunlap



Bill Dunlap
Spotfire, TIBCO Software
wdunlap tibco.com  

 -Original Message-
 From: r-help-boun...@r-project.org 
 [mailto:r-help-boun...@r-project.org] On Behalf Of Brian Diggs
 Sent: Monday, April 25, 2011 11:05 AM
 To: christoph.jaec...@wi.tum.de
 Cc: r-help@r-project.org
 Subject: Re: [R] Problem with ddply in the plyr-package: 
 surprising output of a date-column
 
 On 4/25/2011 10:19 AM, Christoph Jäckel wrote:
  Hi Together,
 
  I have a problem with the plyr package - more precisely 
 with the ddply
  function - and would be very grateful for any help. I hope 
 the example
  here is precise enough for someone to identify the problem. 
 Basically,
  in this step I want to identify observations that are identical in
  terms of certain identifiers (ID1, ID2, ID3) and just want to save
  those observations (in this step, without deleting any rows or
  manipulating any data) in a separate data.frame. However, I get the
  warning message below and the column with dates is messed up.
  Interestingly, the value column (the type is factor here, but if you
  change that with as.integer it doesn't make any difference) 
 is handled
  correctly. Any idea what I do wrong?
 
  df- 
 data.frame(cbind(ID1=c(1,2,2,3,3,4,4),ID2=c('a','b','b','c','d
','e','e'),ID3=c(v1,v1,v1,v1,v2,v1,v1),
 
  
 Date=c(1985-05-1,1985-05-2,1985-05-3,1985-05-4,1985-0
 5-5,1985-05-6,1985-05-7),
Value=c(1,2,3,4,5,6,7)))
  df[,1]- as.character(df[,1])
  df[,2]- as.character(df[,2])
  df$Date- strptime(df$Date,%Y-%m-%d)
 
  #Apparently there are two observation that have the same 
 IDs: ID1=2 and ID1=4
  ddply(df,.(ID1,ID2,ID3),nrow)
  #I want to save those IDs in a separate data.frame, so the 
 desired output is:
  df[c(2:3,6:7),]
 
  #My idea: Write a custom function that only returns 
 observations with
  multiple rows.
  #Seems to work except that the Date column doesn't make any 
 sense anymore
  #Warning message: In output[[var]][rng]- df[[var]]: number of items
  to replace is not a multiple of replacement length
  ddply(df,.(ID1,ID2,ID3),function(df) if(nrow(df)=1){NULL}else{df})
 
  #Notice that it works perfectly if I only have one observation with
  multiple rows
  ddply(df[1:6,],.(ID1,ID2,ID3),function(df) 
 if(nrow(df)=1){NULL}else{df})
 
 Works for me:
 
   df[c(2:3,6:7),]
ID1 ID2 ID3  Date Value
 2   2   b  v1 1985-05-2 2
 3   2   b  v1 1985-05-3 3
 6   4   e  v1 1985-05-6 6
 7   4   e  v1 1985-05-7 7
   ddply(df,.(ID1,ID2,ID3),function(df) if(nrow(df)=1){NULL}else{df})
ID1 ID2 ID3  Date Value
 1   2   b  v1 1985-05-2 2
 2   2   b  v1 1985-05-3 3
 3   4   e  v1 1985-05-6 6
 4   4   e  v1 1985-05-7 7
 [ ... version info elided ... ] 
 A couple of things: there was just an update of plyr to 1.5.2; maybe 
 that fixes what you are seeing?  Also, your df consists of 
 only factors. 
   cbind-ing the data before turning it into a data.frame makes it a 
 character matrix which gets converted to factors.
 
   str(df)
 'data.frame':   7 obs. of  5 variables:
   $ ID1  : Factor w/ 4 levels 1,2,3,4: 1 2 2 3 3 4 4
   $ ID2  : Factor w/ 5 levels a,b,c,d,..: 1 2 2 3 4 5 5
   $ ID3  : Factor w/ 2 levels v1,v2: 1 1 1 1 2 1 1
   $ Date : Factor w/ 7 levels 1985-05-1,1985-05-2,..: 1 2 
 3 4 5 6 7
   $ Value: Factor w/ 7 levels 1,2,3,4,..: 1 2 3 4 5 6 7

The OP's data.frame contained a POSIXlt (not factor) object
in the Date column
   str(df)
  'data.frame':   7 obs. of  5 variables:
   $ ID1  : chr  1 2 2 3 ...
   $ ID2  : chr  a b b c ...
   $ ID3  : Factor w/ 2 levels v1,v2: 1 1 1 1 2 1 1
   $ Date : POSIXlt, format: 1985-05-01 1985-05-02 ...
   $ Value: Factor w/ 7 levels 1,2,3,4,..: 1 2 3 4 5 6 7
and apparently plyr's equivalent of rbind doesn't support that class.

If you want to continue using POSIXlt objects you can get your
immediate result without ddply; subscripting will do the job:
   nDups - with(df, ave(rep(0,nrow(df)), ID1, ID2, ID3, FUN=length))
   print(nDups)
  [1] 1 2 2 1 1 2 2
   df[nDups1, ]
ID1 ID2 ID3   Date Value
  2   2   b  v1 1985-05-02 2
  3   2   b  v1 1985-05-03 3
  6   4   e  v1 1985-05-06 6
  7   4   e  v1 1985-05-07 7
   str(.Last.value)
  'data.frame':   4 obs. of  5 variables:
   $ ID1  : chr  2 2 4 4
   $ ID2  : chr  b b e e
   $ ID3  : Factor w/ 2 levels v1,v2: 1 1 1 1
   $ Date : POSIXlt, format: 1985-05-02 1985-05-03 ...
   $ Value: Factor w/ 7 levels 1,2,3,4,..: 2 3 6 7

If you need plyr for other tasks you ought to use a different
class for your date data (or wait until plyr can deal with
POSIXlt objects).

Bill Dunlap
Spotfire, TIBCO Software
wdunlap tibco.com 

 
 Maybe that has something to do with the odd dates since 
 they are not 
 really dates at all, just string representations of factor levels. 
 Compare with:
 
 DF - data.frame(ID1=c(1,2,2,3,3,4,4),
   ID2=c('a','b','b','c','d','e','e'),
   ID3=c(v1,v1,v1,v1,v2,v1,v1),
   Date=as.Date(c(1985-05-1,1985-05-2,1985-05-3,

Re: [R] extracting names from matrix according to a condition

thank you very much. worked great for me.

On Mon, Apr 25, 2011 at 8:22 PM, Peter Ehlers ehl...@ucalgary.ca wrote:
 On 2011-04-25 10:58, ivan wrote:

 Dear Community,

 I have a matrix with assigned colnames and rolnames as follows:

             A            B
 NR    0.15      0,05
 AL     0,05      0,05
 .           .            .
 .           .            .
 .           .            .

 I want to extract the names of the rows for which A0,1 and B0,1. In
 the above example this would be observation NR only. Hence the output
 should write for instance:

 names:
 NR

 Is this possible? Thank you very much for your help.

 Call the matrix m. Then

  rownames(m[ m[, A]  0.1  m[, B]  0.1, , drop=FALSE ])

 should do what you want.

 Peter Ehlers


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Re: [R] Problem with ddply in the plyr-package: surprising output of a date-column

2011-04-25 Thread Brian Diggs


On 4/25/2011 11:55 AM, William Dunlap wrote:



Bill Dunlap
Spotfire, TIBCO Software
wdunlap tibco.com


-Original Message-
From: r-help-boun...@r-project.org
[mailto:r-help-boun...@r-project.org] On Behalf Of Brian Diggs
Sent: Monday, April 25, 2011 11:05 AM
To: christoph.jaec...@wi.tum.de
Cc: r-help@r-project.org
Subject: Re: [R] Problem with ddply in the plyr-package:
surprising output of a date-column

On 4/25/2011 10:19 AM, Christoph Jäckel wrote:

Hi Together,

I have a problem with the plyr package - more precisely

with the ddply

function - and would be very grateful for any help. I hope

the example

here is precise enough for someone to identify the problem.

Basically,

in this step I want to identify observations that are identical in
terms of certain identifiers (ID1, ID2, ID3) and just want to save
those observations (in this step, without deleting any rows or
manipulating any data) in a separate data.frame. However, I get the
warning message below and the column with dates is messed up.
Interestingly, the value column (the type is factor here, but if you
change that with as.integer it doesn't make any difference)

is handled

correctly. Any idea what I do wrong?

df-

data.frame(cbind(ID1=c(1,2,2,3,3,4,4),ID2=c('a','b','b','c','d

','e','e'),ID3=c(v1,v1,v1,v1,v2,v1,v1),




Date=c(1985-05-1,1985-05-2,1985-05-3,1985-05-4,1985-0
5-5,1985-05-6,1985-05-7),

   Value=c(1,2,3,4,5,6,7)))
df[,1]- as.character(df[,1])
df[,2]- as.character(df[,2])
df$Date- strptime(df$Date,%Y-%m-%d)

#Apparently there are two observation that have the same

IDs: ID1=2 and ID1=4

ddply(df,.(ID1,ID2,ID3),nrow)
#I want to save those IDs in a separate data.frame, so the

desired output is:

df[c(2:3,6:7),]

#My idea: Write a custom function that only returns

observations with

multiple rows.
#Seems to work except that the Date column doesn't make any

sense anymore

#Warning message: In output[[var]][rng]- df[[var]]: number of items
to replace is not a multiple of replacement length
ddply(df,.(ID1,ID2,ID3),function(df) if(nrow(df)=1){NULL}else{df})

#Notice that it works perfectly if I only have one observation with
multiple rows
ddply(df[1:6,],.(ID1,ID2,ID3),function(df)

if(nrow(df)=1){NULL}else{df})

Works for me:

df[c(2:3,6:7),]
ID1 ID2 ID3  Date Value
2   2   b  v1 1985-05-2 2
3   2   b  v1 1985-05-3 3
6   4   e  v1 1985-05-6 6
7   4   e  v1 1985-05-7 7
ddply(df,.(ID1,ID2,ID3),function(df) if(nrow(df)=1){NULL}else{df})
ID1 ID2 ID3  Date Value
1   2   b  v1 1985-05-2 2
2   2   b  v1 1985-05-3 3
3   4   e  v1 1985-05-6 6
4   4   e  v1 1985-05-7 7
[ ... version info elided ... ]
A couple of things: there was just an update of plyr to 1.5.2; maybe
that fixes what you are seeing?  Also, your df consists of
only factors.
   cbind-ing the data before turning it into a data.frame makes it a
character matrix which gets converted to factors.

str(df)
'data.frame':   7 obs. of  5 variables:
   $ ID1  : Factor w/ 4 levels 1,2,3,4: 1 2 2 3 3 4 4
   $ ID2  : Factor w/ 5 levels a,b,c,d,..: 1 2 2 3 4 5 5
   $ ID3  : Factor w/ 2 levels v1,v2: 1 1 1 1 2 1 1
   $ Date : Factor w/ 7 levels 1985-05-1,1985-05-2,..: 1 2
3 4 5 6 7
   $ Value: Factor w/ 7 levels 1,2,3,4,..: 1 2 3 4 5 6 7


The OP's data.frame contained a POSIXlt (not factor) object
in the Date column
 str(df)
   'data.frame':   7 obs. of  5 variables:
$ ID1  : chr  1 2 2 3 ...
$ ID2  : chr  a b b c ...
$ ID3  : Factor w/ 2 levels v1,v2: 1 1 1 1 2 1 1
$ Date : POSIXlt, format: 1985-05-01 1985-05-02 ...
$ Value: Factor w/ 7 levels 1,2,3,4,..: 1 2 3 4 5 6 7


Thanks, Bill. Somehow I missed that, despite the OP having it in his 
code; I even copied it into my testing window.  It was my error for not 
running it and noting it.



and apparently plyr's equivalent of rbind doesn't support that class.


plyr uses rbind.fill primarily.  And it doesn't handle columns of 
POSIXlt based on testing that directly. (Although with only one 
argument, it just passes the data.frame back, which is why when there 
was just a single duplicate, it worked; that bypassed the code that 
couldn't handle POSIXlt's.)



If you want to continue using POSIXlt objects you can get your
immediate result without ddply; subscripting will do the job:
 nDups- with(df, ave(rep(0,nrow(df)), ID1, ID2, ID3, FUN=length))
 print(nDups)
   [1] 1 2 2 1 1 2 2
 df[nDups1, ]
 ID1 ID2 ID3   Date Value
   2   2   b  v1 1985-05-02 2
   3   2   b  v1 1985-05-03 3
   6   4   e  v1 1985-05-06 6
   7   4   e  v1 1985-05-07 7
 str(.Last.value)
   'data.frame':   4 obs. of  5 variables:
$ ID1  : chr  2 2 4 4
$ ID2  : chr  b b e e
$ ID3  : Factor w/ 2 levels v1,v2: 1 1 1 1
$ Date : POSIXlt, format: 1985-05-02 1985-05-03 ...
$ Value: Factor w/ 7 levels 1,2,3,4,..: 2 3 6 7

If you need plyr for other tasks you ought to use a different
class for your date data (or

Re: [R] Problem with ddply in the plyr-package: surprising output of a date-column

2011-04-25 Thread Hadley Wickham

 If you need plyr for other tasks you ought to use a different
 class for your date data (or wait until plyr can deal with
 POSIXlt objects).

How do you get POSIXlt objects into a data frame?

 df - data.frame(x = as.POSIXlt(as.Date(c(2008-01-01
 str(df)
'data.frame':   1 obs. of  1 variable:
 $ x: POSIXct, format: 2008-01-01

 df - data.frame(x = I(as.POSIXlt(as.Date(c(2008-01-01)
 str(df)
'data.frame':   1 obs. of  1 variable:
 $ x: AsIs, format: 0

Hadley

-- 
Assistant Professor / Dobelman Family Junior Chair
Department of Statistics / Rice University
http://had.co.nz/

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[R] Generalized Linear Model

2011-04-25 Thread Megan

Hello,

I am trying to run a generalized linear model but do not know where to
begin. I have attached my data to R but do not know where to go from there.
I have two independent variables (each has two factors associated with them)
and two dependent variables, each with either a yes/no response which I've
valued either 0 or 1 in the data set. Any input would be greatly
appreciated. 

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Re: [R] Help needed!

2011-04-25 Thread shuangyan

Hello, i got this package from the paper: Nonparametric Covariance Function
estimation for Functional and Longitudinal data.
http://stat.wharton.upenn.edu/~tcai/paper/html/Covariance-Function.html.
Thanks lot!

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[R] About How to check if a html directory exists

2011-04-25 Thread Wendy Han

Dear all,



I want to let R automatically download available files from a website in
certain folders. Since the files I need, if exist, are in a certain
directory, I used download.file() function with a predesigned directory.
However, some of the folders do not have the file I want. When this occurs,
my program returns error and terminates. So I wonder if there is any way
that I can check if a directory exists before downloading that file. I
appreciate your help!



Thanks!

Wendy

[[alternative HTML version deleted]]

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[R] Factor function

Dear All,

I just want to remove “NA” from the levels of a factor.  For example:

d-data.frame(matrix(c(ww,ww,xx,yy,ww,yy,xx,yy,NA),
ncol=3, byrow=TRUE))

 factor(d[, 3], exclude=NA)
[1] xx yy NA
Levels: NA xx yy

But “NA” is still listed in the levels. How can I solve this problem? Thanks
in advance.

Lisa


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Re: [R] Problem with ddply in the plyr-package: surprising output of a date-column

2011-04-25 Thread Christoph Jäckel

Hi together,

thank you so much for your help! The problem was indeed the
strptime-function. Replacing that with as.Date solves the problem,
both in the example I provided and in my actual data set.

I think this is a lesson for me to not use types I'm not really
familiar with (POSIXlt in this case).

Thanks again!

Christoph

On Mon, Apr 25, 2011 at 10:07 PM, Hadley Wickham had...@rice.edu wrote:

  If you need plyr for other tasks you ought to use a different
  class for your date data (or wait until plyr can deal with
  POSIXlt objects).

 How do you get POSIXlt objects into a data frame?

  df - data.frame(x = as.POSIXlt(as.Date(c(2008-01-01
  str(df)
 'data.frame':   1 obs. of  1 variable:
  $ x: POSIXct, format: 2008-01-01

  df - data.frame(x = I(as.POSIXlt(as.Date(c(2008-01-01)
  str(df)
 'data.frame':   1 obs. of  1 variable:
  $ x: AsIs, format: 0

 Hadley

 --
 Assistant Professor / Dobelman Family Junior Chair
 Department of Statistics / Rice University
 http://had.co.nz/



--


Christoph Jäckel (Dipl.-Kfm.)



Research Assistant

Chair for Financial Management and Capital Markets | Lehrstuhl für
Finanzmanagement und Kapitalmärkte

TUM School of Management | Technische Universität München

Arcisstr. 21 | D-80333 München | Germany

Mailto: christoph.jaec...@wi.tum.de | Web: www.fm.wi.tum.de

Phone: +49 89 289 25482 | Fax: +49 89 289 25488



Head of Chair:

Univ.-Prof. Dr. Christoph Kaserer

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Re: [R] Factor function

2011-04-25 Thread Ista Zahn

Hi Lisa,

NA != NA

The first represents a missing observation, the second represents a
character string.

HTH,
Ista


On Mon, Apr 25, 2011 at 3:53 PM, Lisa lisa...@gmail.com wrote:
 Dear All,

 I just want to remove “NA” from the levels of a factor.  For example:

 d-data.frame(matrix(c(ww,ww,xx,yy,ww,yy,xx,yy,NA),
 ncol=3, byrow=TRUE))

 factor(d[, 3], exclude=NA)
 [1] xx yy NA
 Levels: NA xx yy

 But “NA” is still listed in the levels. How can I solve this problem? Thanks
 in advance.

 Lisa


 --
 View this message in context: 
 http://r.789695.n4.nabble.com/Factor-function-tp3473984p3473984.html
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-- 
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University of Rochester
Department of Clinical and Social Psychology
http://yourpsyche.org

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Re: [R] Generalized Linear Model

2011-04-25 Thread Alexander Engelhardt


Am 25.04.2011 21:28, schrieb Megan:

Hello,

I am trying to run a generalized linear model but do not know where to
begin. I have attached my data to R but do not know where to go from there.
I have two independent variables (each has two factors associated with them)


What do you mean by this? You have two input variables, who are binary, 
meaning yes/no (or male/female, high/low, ...) variables?



and two dependent variables, each with either a yes/no response which I've
valued either 0 or 1 in the data set. Any input would be greatly
appreciated.


If your dependent variable is binary, you might want to google for 
logistic regression (this belongs to generalized linear models). The 
R-function who handles this is glm(), with the parameter family=binomial().


Have fun,
 Alex

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Re: [R] Generalized Linear Model

2011-04-25 Thread Kehl Dániel


Hi!

Try to read about the glm function, type:

?glm

in your R editor. It looks like you have contingency tables, maybe a 
loglin model would be good to start with.


D

2011-04-25 12:28 keltezéssel, Megan írta:

Hello,

I am trying to run a generalized linear model but do not know where to
begin. I have attached my data to R but do not know where to go from there.
I have two independent variables (each has two factors associated with them)
and two dependent variables, each with either a yes/no response which I've
valued either 0 or 1 in the data set. Any input would be greatly
appreciated.

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[R] My code is too loopy

2011-04-25 Thread Dimitri Liakhovitski

Hello!
I wrote a piece of code below that does the job but seems too loopy to me.
I was wondering if there is any way to make it more efficient/less loopy?
Thanks a lot for your hints!
Dimitri

### Creating example data set:

mygroups-c(rep(group1, 8),rep(group2, 8))
myweeks-seq(as.Date(2010-01-04), length = 8, by = week)
values.w-c(0,10,15,20,0,0,0,10,100,200,0,0,300,200,0,0)
mydata-data.frame(group=mygroups,mydates=myweeks,myvalue=values.w)
mydata$group-as.factor(mydata$group)
str(mydata)
(mydata)

### Doing the following within each level of the factor mydata$group:
### Create a new variable (new.value) that equals:
### myvalue in the same week * 0.5 +
### myvalue 1 week ago  * 0.35
### myvalue 2 weeks ago * 0.15

groups-levels(mydata$group)
(groups)

mydata[[new.value]]-mydata[[myvalue]]*0.5

for(i in groups){   # looping through groups
  temp.data-mydata[mydata$group %in% i,] # selecting values for one group
  
temp.data[2,new.value]-temp.data[[new.value]][2]+temp.data[1,myvalue]*0.35
# 2nd new value
  for(myrow in 3:nrow(temp.data)){  # Starting in row 3 and looping through rows

temp.data[myrow,new.value]-temp.data[[new.value]][myrow]+temp.data[(myrow-1),myvalue]*.35+temp.data[(myrow-2),myvalue]*.15
  }
  mydata[mydata$group %in% i,]-temp.data
}


-- 
Dimitri Liakhovitski
Ninah Consulting
www.ninah.com

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Re: [R] Factor function

Did you see the data frame d? Thanks.

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Re: [R] Factor function

2011-04-25 Thread Ista Zahn

Yes... did you understand that NA is not equal to NA?

Best,
Ista

On Mon, Apr 25, 2011 at 4:31 PM, Lisa lisa...@gmail.com wrote:
 Did you see the data frame d? Thanks.

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Re: [R] Factor function

Thank you for your reply again. I really know that NA is not NA. I just
want to figure out how to remove NA from the levels. Thanks again.

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Re: [R] Problem with ddply in the plyr-package: surprising output of a date-column

2011-04-25 Thread Brian Diggs


On 4/25/2011 1:07 PM, Hadley Wickham wrote:

If you need plyr for other tasks you ought to use a different
class for your date data (or wait until plyr can deal with
POSIXlt objects).


How do you get POSIXlt objects into a data frame?


df- data.frame(x = as.POSIXlt(as.Date(c(2008-01-01
str(df)

'data.frame':   1 obs. of  1 variable:
  $ x: POSIXct, format: 2008-01-01


df- data.frame(x = I(as.POSIXlt(as.Date(c(2008-01-01)
str(df)

'data.frame':   1 obs. of  1 variable:
  $ x: AsIs, format: 0

Hadley


Assigning to a column after the data.frame creation step

 df - data.frame(x = as.POSIXlt(as.Date(c(2008-01-01
 str(df)
'data.frame':   1 obs. of  1 variable:
 $ x: POSIXct, format: 2008-01-01
 dput(df)
structure(list(x = structure(1199145600, class = c(POSIXct,
POSIXt), tzone = UTC)), .Names = x, row.names = c(NA, -1L
), class = data.frame)
 df$x - as.POSIXlt(as.Date(c(2008-01-01)))
 str(df)
'data.frame':   1 obs. of  1 variable:
 $ x: POSIXlt, format: 2008-01-01
 dput(df)
structure(list(x = structure(list(sec = 0, min = 0L, hour = 0L,
mday = 1L, mon = 0L, year = 108L, wday = 2L, yday = 0L, isdst = 
0L), .Names = c(sec,

min, hour, mday, mon, year, wday, yday, isdst
), class = c(POSIXlt, POSIXt), tzone = UTC)), .Names = x, 
row.names = c(NA,

-1L), class = data.frame)

This is reminiscent of the 1d array problem; there are types that are 
coerced into other types when passed as part of a data.frame constructor 
(data.frame call), but are not coerced when assigned to a column.


Looking at help pages, calls to data.frame call as.data.frame on each 
argument; `[-.data.frame` has a section on coercion which starts The 
story over when replacement values are coerced is a complicated one, and 
one that has changed during R's development. This section is a guide 
only. which makes me think it is not all that well defined.


Digging more, there is a as.data.frame.POSIXlt, although the help page 
for it (DateTimeClasses in base) does not mention it or document it.  It 
is documented, though, in as.data.frame (which also has comments about 
coercing 1 dimensional arrays).


So, potentially, there could be differences with any class that has an 
as.data.frame method because it will be treated differently if passed to 
data.frame versus a column assignment with `[-.data.frame`


 methods(as.data.frame)
 [1] as.data.frame.aovproj*as.data.frame.array
 [3] as.data.frame.AsIsas.data.frame.character
 [5] as.data.frame.complex as.data.frame.data.frame
 [7] as.data.frame.Dateas.data.frame.default
 [9] as.data.frame.difftimeas.data.frame.factor
[11] as.data.frame.ftable* as.data.frame.function
[13] as.data.frame.idf*as.data.frame.integer
[15] as.data.frame.listas.data.frame.logical
[17] as.data.frame.logLik* as.data.frame.matrix
[19] as.data.frame.model.matrixas.data.frame.numeric
[21] as.data.frame.numeric_version as.data.frame.ordered
[23] as.data.frame.POSIXct as.data.frame.POSIXlt
[25] as.data.frame.raw as.data.frame.table
[27] as.data.frame.ts  as.data.frame.vector

So, I suppose it is working as documented.  Though I wonder how long ago 
it was that someone (who has been using R regularly for at least a year) 
actually read the entire help page for data.frame and/or as.data.frame. 
 It's one of those things you think you know and understand until you 
find out you don't.


--
Brian S. Diggs, PhD
Senior Research Associate, Department of Surgery
Oregon Health  Science University

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[R] [R-pkgs] Rook: software and specification for R web applications and servers

2011-04-25 Thread Jeffrey Horner

Dear  useRs,

Rook is a new package that does three things:

 - It provides a way to run R web applications on your desktop with the
 new internal R web server named Rhttpd. Please see the Rhttpd help page.

 - It provides a set of reference classes you can use to write you R
 web applications. The following help pages provide more information:
 Brewery, Builder, File, Middleware, Redirect, Request, Response, Static,
 URLMap, and Utils. Also see the example web applications located in
 'system(exampleApps,package=Rook)'.

 - It provides a specification for writing your R web applications to
 work on any web server that supports the Rook specification. Currently,
 only Rhttpd implements it, but rApache is close behind. See the Rook
 help page for more information.

You may not see the need for web applications written in R, but consider
using Rook to build a statistical engine that complements a front-end
web system, or consider creating elegant ggplot2 graphics on-demand from
a fresh data stream. Also, consider creating dynamic instructional content
for the classroom.

If you have other examples or ideas, please join in the discussion on
R-help or here:

http://groups.google.com/group/rrook

--
Jeffrey Horner  (author of rApache and brew)

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Re: [R] Factor function

2011-04-25 Thread Petr Savicky

On Mon, Apr 25, 2011 at 12:53:40PM -0700, Lisa wrote:
 Dear All,
 
 I just want to remove “NA” from the levels of a factor.  For example:
 
 d-data.frame(matrix(c(ww,ww,xx,yy,ww,yy,xx,yy,NA),
 ncol=3, byrow=TRUE))
 
  factor(d[, 3], exclude=NA)
 [1] xx yy NA
 Levels: NA xx yy
 
 But “NA” is still listed in the levels. How can I solve this problem?

The column d[, 3] is already a factor. It is possible to avoid
this using

  d-data.frame(matrix(c(ww,ww,xx,yy,ww,yy,xx,yy,NA),
  ncol=3, byrow=TRUE), stringsAsFactors=FALSE)

Then, we get

  factor(d[, 3], exclude=NA)

  [1] xx   yy   NA
  Levels: xx yy

Hope this helps.

Petr Savicky.

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[R] split data frame by factor

Dear Community,

I have a dataframe like this one:

A   B
5   1
6   1
7   1
8   1
9   2
10  2
11  2
12  2

I have a problem splitting up the above data frame in respect to the
factor represented by B, whereas the resulting vector should contain
the numeric values only. I tried split() but it produces this:
A   B
[1]
5   1
6   1
7   1
8   1
[2]
9   2
10  2
11  2
12  2

I want to achieve this,though:

A   
[1]
5   
6   
7   
8
[2] 
9   
10  
11  
12  

How is this done? Thank you very much in advance.

Best Regards.

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Re: [R] Problem with ddply in the plyr-package: surprising output of a date-column

2011-04-25 Thread Peter Ehlers


On 2011-04-25 13:07, Hadley Wickham wrote:

If you need plyr for other tasks you ought to use a different
class for your date data (or wait until plyr can deal with
POSIXlt objects).


How do you get POSIXlt objects into a data frame?


df- data.frame(x = as.POSIXlt(as.Date(c(2008-01-01
str(df)

'data.frame':   1 obs. of  1 variable:
  $ x: POSIXct, format: 2008-01-01


df- data.frame(x = I(as.POSIXlt(as.Date(c(2008-01-01)
str(df)

'data.frame':   1 obs. of  1 variable:
  $ x: AsIs, format: 0

Hadley



To mimic the OP's code

  df - data.frame(x = 2008-01-01)
  df$x - as.POSIXlt(df$x, %Y-%m-%d)
  str(df)
  #'data.frame':   1 obs. of  1 variable:
  # $ x: POSIXlt, format: 2008-01-01

Peter Ehlers

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Re: [R] split data frame by factor

sorry, the problem has been solved.

On Mon, Apr 25, 2011 at 11:28 PM, ivan i.pet...@gmail.com wrote:
 Dear Community,

 I have a dataframe like this one:

 A       B
 5       1
 6       1
 7       1
 8       1
 9       2
 10      2
 11      2
 12      2

 I have a problem splitting up the above data frame in respect to the
 factor represented by B, whereas the resulting vector should contain
 the numeric values only. I tried split() but it produces this:
 A       B
 [1]
 5       1
 6       1
 7       1
 8       1
 [2]
 9       2
 10      2
 11      2
 12      2

 I want to achieve this,though:

 A
 [1]
 5
 6
 7
 8
 [2]
 9
 10
 11
 12

 How is this done? Thank you very much in advance.

 Best Regards.


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Re: [R] My code is too loopy

2011-04-25 Thread Dennis Murphy

Hi:

I think the embed() function is your friend here. From its help page example,

 x - 1:10
 embed (x, 3)
 [,1] [,2] [,3]
[1,]321
[2,]432
[3,]543
[4,]654
[5,]765
[6,]876
[7,]987
[8,]   1098


Applying it to your test data,

# h() creates a weighted average of the observations in each row
h - function(x) embed(x, 3) %*% c(0.5, 0.35, 0.15)
library(plyr)
ddply(mydata, group, summarise, ma = h(myvalue))
group ma
1  group1  11.00
2  group1  16.75
3  group1   9.25
4  group1   3.00
5  group1   0.00
6  group1   5.00
7  group2  85.00
8  group2  30.00
9  group2 150.00
10 group2 205.00
11 group2 115.00
12 group2  30.00

Does that work for you? The rollapply() function in the zoo package
may also be applicable with a similar input function that computes a
weighted average.

HTH,
Dennis


On Mon, Apr 25, 2011 at 1:50 PM, Dimitri Liakhovitski
dimitri.liakhovit...@gmail.com wrote:
 Hello!
 I wrote a piece of code below that does the job but seems too loopy to me.
 I was wondering if there is any way to make it more efficient/less loopy?
 Thanks a lot for your hints!
 Dimitri

 ### Creating example data set:

 mygroups-c(rep(group1, 8),rep(group2, 8))
 myweeks-seq(as.Date(2010-01-04), length = 8, by = week)
 values.w-c(0,10,15,20,0,0,0,10,100,200,0,0,300,200,0,0)
 mydata-data.frame(group=mygroups,mydates=myweeks,myvalue=values.w)
 mydata$group-as.factor(mydata$group)
 str(mydata)
 (mydata)

 ### Doing the following within each level of the factor mydata$group:
 ### Create a new variable (new.value) that equals:
 ### myvalue in the same week * 0.5 +
 ### myvalue 1 week ago  * 0.35
 ### myvalue 2 weeks ago * 0.15

 groups-levels(mydata$group)
 (groups)

 mydata[[new.value]]-mydata[[myvalue]]*0.5

 for(i in groups){   # looping through groups
  temp.data-mydata[mydata$group %in% i,] # selecting values for one group
  temp.data[2,new.value]-temp.data[[new.value]][2]+temp.data[1,myvalue]*0.35
 # 2nd new value
  for(myrow in 3:nrow(temp.data)){  # Starting in row 3 and looping through 
 rows
    
 temp.data[myrow,new.value]-temp.data[[new.value]][myrow]+temp.data[(myrow-1),myvalue]*.35+temp.data[(myrow-2),myvalue]*.15
  }
  mydata[mydata$group %in% i,]-temp.data
 }


 --
 Dimitri Liakhovitski
 Ninah Consulting
 www.ninah.com

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Re: [R] Factor function