date:20101207

Something like this should work, only Sept has 30 days, not 31:

 files - seq(as.Date('2008-09-30'), by = '1 day', length = 250)
 input - paste(.jsp?y=
+ , format(files, %Y)
+ , m=
+ , format(files, %m)
+ , d=
+ , format(files, %d)
+ , sep = 
+ )

 head(input)
[1] .jsp?y=2008m=09d=30 .jsp?y=2008m=10d=01
[3] .jsp?y=2008m=10d=02 .jsp?y=2008m=10d=03
[5] .jsp?y=2008m=10d=04 .jsp?y=2008m=10d=05


On Mon, Dec 6, 2010 at 8:18 PM, Arsenio Staer arsenio.st...@gmail.com wrote:

 So i realized putting this
 raw.untr - as.data.frame(read.csv(file='...jsp?y=2008m=09d=30',
 header = TRUE, as.is = TRUE))

 reads the file prefectly, so i was wondering how can i create a set of
 dataframes each with observations for one day (say for 3 years with 250 days
 of data each year = 750 datasets) ? And in each loop of a cycle the text
 string in the url would change to reflect the new date, say from
 y=2008m=09d=30 to y=2008m=09d=31?

 Any help would be great, i somewhat expected more of a reaction given how
 vibrant R community is,

 Arsenio

 --
 View this message in context: 
 http://r.789695.n4.nabble.com/Downloading-a-csv-through-a-jsp-url-with-variable-parameters-R-code-or-package-tp3070541p3075644.html
 Sent from the R help mailing list archive at Nabble.com.

 __
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 PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
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-- 
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Data Munger Guru

What is the problem that you are trying to solve?

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[R] understanding output of tapply/by cumsum

2010-12-07 Thread Gerrit Draisma


Dear R-users,

I have a dataset with categories and numbers.
I would like to compute and add cumulative numbers
to the dataset.
I do not understand the structure of by(...) or
tapply(...) output enough to handle it.

Here a small example
--
d-expand.grid(a=1:5,b=1:3,c=1:2)
d$n = 10 * d$a + d$b +0.1* d$c
Sn-by(d$n,list(d$a,d$c),cumsum)
str(Sn)
-
List of 10
 $ : num [1:3] 11.1 23.2 36.3
 $ : num [1:3] 21.1 43.2 66.3
 $ : num [1:3] 31.1 63.2 96.3
 $ : num [1:3]  41.1  83.2 126.3
 $ : num [1:3]  51.1 103.2 156.3
 $ : num [1:3] 11.2 23.4 36.6
 $ : num [1:3] 21.2 43.4 66.6
 $ : num [1:3] 31.2 63.4 96.6
 $ : num [1:3]  41.2  83.4 126.6
 $ : num [1:3]  51.2 103.4 156.6
 - attr(*, dim)= int [1:2] 5 2
 - attr(*, dimnames)=List of 2
  ..$ : chr [1:5] 1 2 3 4 ...
  ..$ : chr [1:2] 1 2
 - attr(*, call)= language by.default(data = d$n, INDICES = list(d$a, 
d$c), FUN = cumsum)

 - attr(*, class)= chr by
-
# these give (a) lists of one numerical vector(a)
Sn[5,2]
Sn[cbind(d$a,d$c)]
# how to access the individual cumsum values?
# and assign them to d$Sn?
--

Thanks,
Gerrit.

---
Gerrit Draisma
Department of Public Health
Erasmus MC, University Medical Center Rotterdam
Room AE-235
P.O. Box 2040 3000 CA  Rotterdam The Netherlands
Phone: +31 10 7043787 Fax: +31 10 7038474
http://mgzlx4.erasmusmc.nl/pwp/?gdraisma

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

Re: [R] Time out for a R Function

2010-12-07 Thread Mike Marchywka

 Date: Tue, 7 Dec 2010 16:11:42 +1100
 From: michael.bedw...@gmail.com
 To: santosh.srini...@gmail.com; r-help@r-project.org
 Subject: Re: [R] Time out for a R Function

 Below is a toy function with one way of doing it. There are bound to
 be better ways :)

This seems to just check total time on each iteration.
Presumably for this you would be better off setting an
ieration limit, not a timeout. A watchdog timer to insure
that pretend task doesn't hang would need another thread
and you would either need the target task to check
a shared variable that you set from the watchdog or
you just kill the target thread from watchdog but
this may be hard to do in a way that lets you continue
in any meaningful way. 

 function(niter = 10, time.out = 3) {
 pretend.task - function() {
 Sys.sleep(0.5)
 }

 start - proc.time()

 for (iter in 1:niter) {
 pretend.task()
 cur - proc.time() - start
 if (cur[3]  time.out) return(timed out)
 }

 return(completed)
 }

 On 7 December 2010 13:04, Santosh Srinivas  wrote:
  Hello Group,

  I have an R-function that works fine for most part but sometime runs into a
  long loop! (I'm lazy and short on time to debug right now so want to do
  something easy)
  For my purpose, it is ok to make few errors  is there a way I can put a
  timeout on a function and the r-process needes to move on to the next step?

  Thank you.
  S

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

Re: [R] what is this averaging function called ?, has R a built in function for it ?


my input is from csv file:

fname= 'test'
csvdata = read.table(file=paste(fname,'.csv',sep=''),head=FALSE)
x = csvdata$V1
y = csvdata$V2

I know that this group is not about php , but I managed to make function
from above a lot faster, and I'm still cannot operate R at a sufficient
level to recreate in in this language. Fit function in R that kinda
resembles it is smooth.spline, but that function seems to go in a strange
directions when data density becomes low

But I think I would get desired behaviour in smooth.spline if for x axis I
put matrix of simple 1,2,3 sequence

here is revised php code:

function smooth($in,$smooth=500) {
if (count(current($in))!=2) exit('wrong array');
timer();

foreach($in as $v) {
$v[0] = (string)$v[0];
if (!isset($y[$v[0]])) {
$y[$v[0]] = 0;
$z[$v[0]] = 0;
}
$y[$v[0]] += $v[1];
++$z[$v[0]];
}
unset($in);
ksort($y, SORT_NUMERIC);
ksort($z, SORT_NUMERIC);
$x = array_keys($z);
$y = array_values($y);
$z = array_values($z);
$count = count($z);
echo n.$count.' : ';
for ($k=0;$k$count;++$k) {
if (!($k % 1000)) echo $k.' ';
$u = $d = 0;
$usum = $dsum = 0;
$walk = 0;
while (true) {
++$walk;
if (isset($z[$k-$walk]) and 
$z[$k]+$z[$k-$walk]+$usum$smooth) {
$usum += $z[$k-$walk];
$u += $y[$k-$walk];
}
else break;
}
$walk = 0;
while (true) {
++$walk;
if (isset($z[$k+$walk]) and 
$z[$k]+$z[$k+$walk]+$dsum$smooth) {
$dsum += $z[$k+$walk];
$d += $y[$k+$walk];
}
else break;
}
$out[$k] = ($y[$k]+$u+$d)/($z[$k]+$usum+$dsum);
}
echo ' : '.timer().n;
return array($x,$out);
}
-- 
View this message in context: 
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
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Re: [R] How to find out if a data frame has automatic row names?

2010-12-07 Thread Sarah Goslee

If as.matrix distinguishes between the two options, then you could look
at the code for that function and see how it's done.

Sarah

On Tue, Dec 7, 2010 at 6:34 AM, Andreas Borg
andreas.b...@unimedizin-mainz.de wrote:
 Hi all,

 I just stumbled across the difference between explicit and 'automatic' row
 names for data frames. Is there a quick way to find out if a data frame has
 automatic row names or not? Of course I know this for data frames that I
 create myself, but what if a function needs to extract this information for
 an argument?

 The following example shows that this property is not easily visible:

 # data frame with automatic row names
 df - data.frame(a=1:2, b=3:4)
 # data frame with explicit row names
 df2 - data.frame(a=1:2, b=3:4, row.names=1:2)
 # printing does not reveal any difference
 df
  a b
 1 1 3
 2 2 4
 df2
  a b
 1 1 3
 2 2 4
 # both are considered equal
 all.equal(df, df2)
 [1] TRUE
 identical(df, df2)
 [1] TRUE
 # calling rownames gives the same result
 rownames(df)
 [1] 1 2
 rownames(df2)
 [1] 1 2
 # when converting to a matrix, it makes a difference
 as.matrix(df)
    a b
 [1,] 1 3
 [2,] 2 4
 as.matrix(df2)
  a b
 1 1 3
 2 2 4

 Thanks for any suggestion,

 Andreas

 --
 Andreas Borg
 Medizinische Informatik

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

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

[R] How can I change the longitude scale on the plotMap() PBSmapping package

2010-12-07 Thread Fabio Caltabellotta

Dear R-helpers,

I am using the package PBSmapping and I have a question on the fuction plotMap:


How can I change the longitude scale on data(worldLLhigh)? I changed
the coordinates of ylim () and xlim () on the function plotMap to plot
my study area. I would like to plot the Brazilian coastline that goes
from the latitude -26.0°S to -21.0°S and longitude from -48.0°W to
-45.0°W. However how can I change the scale value of xlim ranging from
312.0 to 315.0 for the scale of longitude 48.0W-and-45.0W to the same
area?. Can anyone help me?


Thanks,






-- 
Fabio Prior Caltabellotta, M.Sc.
Doutorando em Zoologia - Campus Botucatu
Universidade Estadual Paulista Júlio de Mesquita Filho - UNESP
Laboratório de Pesquisa em Elasmobrânquios
http://lattes.cnpq.br/2735351775616389
End: Praça Infante Dom Henrique s/n - Parque Bitarú
11330-150 - São Vicente-SP, Brasil
Tel: +5511 81649946

:(_)))( `·.¸¸.·´¯`·.¸.· _:(_

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

Re: [R] How to find out if a data frame has automatic row names?

2010-12-07 Thread Peter Ehlers


On 2010-12-07 03:34, Andreas Borg wrote:

Hi all,

I just stumbled across the difference between explicit and 'automatic'
row names for data frames. Is there a quick way to find out if a data
frame has automatic row names or not? Of course I know this for data
frames that I create myself, but what if a function needs to extract
this information for an argument?

The following example shows that this property is not easily visible:

# data frame with automatic row names
df- data.frame(a=1:2, b=3:4)
# data frame with explicit row names
df2- data.frame(a=1:2, b=3:4, row.names=1:2)
# printing does not reveal any difference
df
   a b
1 1 3
2 2 4
df2
   a b
1 1 3
2 2 4
# both are considered equal
all.equal(df, df2)
[1] TRUE
identical(df, df2)
[1] TRUE
# calling rownames gives the same result
rownames(df)
[1] 1 2
rownames(df2)
[1] 1 2
# when converting to a matrix, it makes a difference
as.matrix(df)
  a b
[1,] 1 3
[2,] 2 4
as.matrix(df2)
   a b
1 1 3
2 2 4

Thanks for any suggestion,



Check ?.row_names_info

Peter Ehlers


Andreas



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[R] Encoding problem - I fails to read Hebrew text from online

2010-12-07 Thread Tal Galili

Hello all,

# I am trying to read the text in this URL:
u -
http://google.com/complete/search?output=toolbarq=%d7%a9%d7%9c%d7%95%d7%9d
# By using this command:
readLines(u)

And no matter what variation I tried, I keep getting this output:
[1] ?xml version=\1.0\?toplevelCompleteSuggestionsuggestion
data=\#x5E9;#x5DC;#x5D5;#x5DD;\/   (etc...)
Instead of this output:
?xml version=1.0?toplevelCompleteSuggestionsuggestion
data=×©×××/num_queries
int=1680//CompleteSuggestionCompleteSuggestionsuggestion 
data=×©×××
×× ××/num_queries
int=232000//CompleteSuggestionCompleteSuggestionsuggestion
data=×©××× ×¢××××/
(etc)

I tried:
  readLines(u, encoding= latin1)
  readLines(u, encoding= UTF-8)
And also changing Sys.setlocale:
  Sys.setlocale(LC_ALL, Hebrew) # must be done for Hebrew to work.
  Sys.setlocale(LC_ALL, English) # must be done for Hebrew to work.

Are there any more options I could try to get this text properly encoded?

Thanks!
Tal



Contact
Details:---
Contact me: tal.gal...@gmail.com |  972-52-7275845
Read me: www.talgalili.com (Hebrew) | www.biostatistics.co.il (Hebrew) |
www.r-statistics.com (English)
--

[[alternative HTML version deleted]]

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

Re: [R] understanding output of tapply/by cumsum

Maybe 'ave' is what you were looking for:

 d$cum - ave(d$n, d$a, d$c, FUN = cumsum)
 d
   a b cn   cum
1  1 1 1 11.1  11.1
2  2 1 1 21.1  21.1
3  3 1 1 31.1  31.1
4  4 1 1 41.1  41.1
5  5 1 1 51.1  51.1
6  1 2 1 12.1  23.2
7  2 2 1 22.1  43.2
8  3 2 1 32.1  63.2
9  4 2 1 42.1  83.2
10 5 2 1 52.1 103.2
11 1 3 1 13.1  36.3
12 2 3 1 23.1  66.3
13 3 3 1 33.1  96.3
14 4 3 1 43.1 126.3
15 5 3 1 53.1 156.3
16 1 1 2 11.2  11.2
17 2 1 2 21.2  21.2
18 3 1 2 31.2  31.2
19 4 1 2 41.2  41.2
20 5 1 2 51.2  51.2
21 1 2 2 12.2  23.4
22 2 2 2 22.2  43.4
23 3 2 2 32.2  63.4
24 4 2 2 42.2  83.4
25 5 2 2 52.2 103.4
26 1 3 2 13.2  36.6
27 2 3 2 23.2  66.6
28 3 3 2 33.2  96.6
29 4 3 2 43.2 126.6
30 5 3 2 53.2 156.6



On Tue, Dec 7, 2010 at 6:39 AM, Gerrit Draisma gdrai...@xs4all.nl wrote:
 Dear R-users,

 I have a dataset with categories and numbers.
 I would like to compute and add cumulative numbers
 to the dataset.
 I do not understand the structure of by(...) or
 tapply(...) output enough to handle it.

 Here a small example
 --
 d-expand.grid(a=1:5,b=1:3,c=1:2)
 d$n = 10 * d$a + d$b +0.1* d$c
 Sn-by(d$n,list(d$a,d$c),cumsum)
 str(Sn)
 -
 List of 10
  $ : num [1:3] 11.1 23.2 36.3
  $ : num [1:3] 21.1 43.2 66.3
  $ : num [1:3] 31.1 63.2 96.3
  $ : num [1:3]  41.1  83.2 126.3
  $ : num [1:3]  51.1 103.2 156.3
  $ : num [1:3] 11.2 23.4 36.6
  $ : num [1:3] 21.2 43.4 66.6
  $ : num [1:3] 31.2 63.4 96.6
  $ : num [1:3]  41.2  83.4 126.6
  $ : num [1:3]  51.2 103.4 156.6
  - attr(*, dim)= int [1:2] 5 2
  - attr(*, dimnames)=List of 2
  ..$ : chr [1:5] 1 2 3 4 ...
  ..$ : chr [1:2] 1 2
  - attr(*, call)= language by.default(data = d$n, INDICES = list(d$a,
 d$c), FUN = cumsum)
  - attr(*, class)= chr by
 -
 # these give (a) lists of one numerical vector(a)
 Sn[5,2]
 Sn[cbind(d$a,d$c)]
 # how to access the individual cumsum values?
 # and assign them to d$Sn?
 --

 Thanks,
 Gerrit.

 ---
 Gerrit Draisma
 Department of Public Health
 Erasmus MC, University Medical Center Rotterdam
 Room AE-235
 P.O. Box 2040 3000 CA  Rotterdam The Netherlands
 Phone: +31 10 7043787 Fax: +31 10 7038474
 http://mgzlx4.erasmusmc.nl/pwp/?gdraisma

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




-- 
Jim Holtman
Data Munger Guru

What is the problem that you are trying to solve?

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

Re: [R] understanding output of tapply/by cumsum

You can also use 'split' to separate each group:

 split(d, list(d$a, d$c))
$`1.1`
   a b cn  cum
1  1 1 1 11.1 11.1
6  1 2 1 12.1 23.2
11 1 3 1 13.1 36.3

$`2.1`
   a b cn  cum
2  2 1 1 21.1 21.1
7  2 2 1 22.1 43.2
12 2 3 1 23.1 66.3

$`3.1`
   a b cn  cum
3  3 1 1 31.1 31.1
8  3 2 1 32.1 63.2
13 3 3 1 33.1 96.3

$`4.1`
   a b cn   cum
4  4 1 1 41.1  41.1
9  4 2 1 42.1  83.2
14 4 3 1 43.1 126.3

$`5.1`
   a b cn   cum
5  5 1 1 51.1  51.1
10 5 2 1 52.1 103.2
15 5 3 1 53.1 156.3

$`1.2`
   a b cn  cum
16 1 1 2 11.2 11.2
21 1 2 2 12.2 23.4
26 1 3 2 13.2 36.6

$`2.2`
   a b cn  cum
17 2 1 2 21.2 21.2
22 2 2 2 22.2 43.4
27 2 3 2 23.2 66.6

$`3.2`
   a b cn  cum
18 3 1 2 31.2 31.2
23 3 2 2 32.2 63.4
28 3 3 2 33.2 96.6

$`4.2`
   a b cn   cum
19 4 1 2 41.2  41.2
24 4 2 2 42.2  83.4
29 4 3 2 43.2 126.6

$`5.2`
   a b cn   cum
20 5 1 2 51.2  51.2
25 5 2 2 52.2 103.4
30 5 3 2 53.2 156.6




On Tue, Dec 7, 2010 at 6:39 AM, Gerrit Draisma gdrai...@xs4all.nl wrote:
 Dear R-users,

 I have a dataset with categories and numbers.
 I would like to compute and add cumulative numbers
 to the dataset.
 I do not understand the structure of by(...) or
 tapply(...) output enough to handle it.

 Here a small example
 --
 d-expand.grid(a=1:5,b=1:3,c=1:2)
 d$n = 10 * d$a + d$b +0.1* d$c
 Sn-by(d$n,list(d$a,d$c),cumsum)
 str(Sn)
 -
 List of 10
  $ : num [1:3] 11.1 23.2 36.3
  $ : num [1:3] 21.1 43.2 66.3
  $ : num [1:3] 31.1 63.2 96.3
  $ : num [1:3]  41.1  83.2 126.3
  $ : num [1:3]  51.1 103.2 156.3
  $ : num [1:3] 11.2 23.4 36.6
  $ : num [1:3] 21.2 43.4 66.6
  $ : num [1:3] 31.2 63.4 96.6
  $ : num [1:3]  41.2  83.4 126.6
  $ : num [1:3]  51.2 103.4 156.6
  - attr(*, dim)= int [1:2] 5 2
  - attr(*, dimnames)=List of 2
  ..$ : chr [1:5] 1 2 3 4 ...
  ..$ : chr [1:2] 1 2
  - attr(*, call)= language by.default(data = d$n, INDICES = list(d$a,
 d$c), FUN = cumsum)
  - attr(*, class)= chr by
 -
 # these give (a) lists of one numerical vector(a)
 Sn[5,2]
 Sn[cbind(d$a,d$c)]
 # how to access the individual cumsum values?
 # and assign them to d$Sn?
 --

 Thanks,
 Gerrit.

 ---
 Gerrit Draisma
 Department of Public Health
 Erasmus MC, University Medical Center Rotterdam
 Room AE-235
 P.O. Box 2040 3000 CA  Rotterdam The Netherlands
 Phone: +31 10 7043787 Fax: +31 10 7038474
 http://mgzlx4.erasmusmc.nl/pwp/?gdraisma

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




-- 
Jim Holtman
Data Munger Guru

What is the problem that you are trying to solve?

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

Re: [R] Appearance of Forest Plot

2010-12-07 Thread s306


Hi Josh,

Thank you for your response. I have recreated my issue using the Olkin 95
data and subgrouping by year of study - I hope it shows where my problem
lies:


library(meta)

data(Olkin95)

metaeg-metabin(event.e, n.e, event.c, n.c, studlab = author,
data=Olkin95, method=MH,
sm=OR, MH.exact=FALSE, RR.cochrane=FALSE, comb.fixed=TRUE,
comb.random=TRUE,
title=, complab=, outclab=,
label.e=Active, label.c=Control,
print.CMH=FALSE, warn=FALSE)

Olkin95$year-Olkin95$year
metaeg$year-addvar(metaeg,Olkin95,year,by.y=author)
as.data.frame(metaeg)

print.meta(metaeg)
forest (metaeg,byvar=year)

Any ideas would be very much appreciated. Thanks again!
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[R] Plotting own function

2010-12-07 Thread Tobias Schultze

Hello,
I have a simple question but didn't find a solution.
How do I plot a custom function.

I have defined this function

func - function(x) {
y - exp(-0.5*((x-63.9)/23.2)^2)
if(x  63.9)
return(2 - y)
else {
return(y)
}
}

and want to plot it in a specified range like

x = seq(-10,150,length=500)
y - func(x)
plot(x,y,type=l,lwd=2,col=red)

But it doesn't work because it expects a vector but I need to compare the
scalar x.
How do I do it correctly?

Thanks a lot
Tobias

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Re: [R] less than full rank contrast methods

'On Tue, Dec 7, 2010 at 5:19 AM, mxkuhn mxk...@gmail.com wrote:
 Greg and Frank,

 Thanks for the replies. I didn't express myself very well; I'm not interest 
 in the model fitting aspect. I'd just like to get the full set of dummy 
 variables (optimally from model.matrix)


Try this:

levels(when$time) - c(morning, afternoon, night)
levels(when$day) - c(Mon, Tue, Wed, Thu, Fri, Sat, Sun)

contr.dummy - function(n, ...) diag(n)
mm - suppressWarnings(model.matrix(~ day + time, when))
mm[is.na(mm)] - 1
mm

You might also want to set the levels of your factors first so that it
includes levels that are not in the data and so that the levels are
sorted in an order other than alphabetical:


etc.


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

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Re: [R] Plotting own function

You need to use 'ifelse' in this case since it will handle a vector
and 'if' is only for single values:

func - function(x) {
   y - exp(-0.5*((x-63.9)/23.2)^2)
   ifelse(x  63.9, 2 - y, y)
}


x = seq(-10,150,length=500)
y - func(x)
plot(x,y,type=l,lwd=2,col=red)



On Tue, Dec 7, 2010 at 6:57 AM, Tobias Schultze webmas...@tubo-world.de wrote:
 Hello,
 I have a simple question but didn't find a solution.
 How do I plot a custom function.

 I have defined this function

 func - function(x) {
        y - exp(-0.5*((x-63.9)/23.2)^2)
        if(x  63.9)
                return(2 - y)
        else {
                return(y)
        }
 }

 and want to plot it in a specified range like

 x = seq(-10,150,length=500)
 y - func(x)
 plot(x,y,type=l,lwd=2,col=red)

 But it doesn't work because it expects a vector but I need to compare the
 scalar x.
 How do I do it correctly?

 Thanks a lot
 Tobias

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What is the problem that you are trying to solve?

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[R] increase or decrease variable by 1


many languages have shorthands for that operation like:

variable += 1
or
++variable

is there something like that in R ?
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[R] latex() hangs R console

2010-12-07 Thread Michael Friendly


[Env: R 2.11.1, Win Xp, Miktex 2.7]

I've just started using Hmisc::latex and friends, and find that running 
latex() produces the .tex and .dvi files, but hangs,

presumably trying to run yap.  An example is below.  What could be wrong?

My PATH seems OK:
 strsplit(Sys.getenv(path),';')
$path
 [1] c:\\program files\\imagemagick-6.4.4-q16   
c:\\Rtools\\bin
 [3] c:\\Rtools\\perl\\bin  
c:\\Rtools\\MinGW\\bin
 [5] C:\\Program Files\\MiKTeX 2.7\\miktex\\bin 
C:\\WINDOWS\\system32
 [7] C:\\WINDOWS
C:\\WINDOWS\\System32\\Wbem
 [9] C:\\Program Files\\Intel\\DMIX 
C:\\Program Files\\ATI Technologies\\ATI.ACE\\
[11] C:\\Program Files\\Common Files\\Roxio Shared\\DLLShared\\ 
C:\\WINDOWS\\system32\\nls
[13] C:\\WINDOWS\\system32\\nls\\ENGLISH
C:\\Program Files\\SecureCRT\\
[15] C:\\Program Files\\IDM Computer Solutions\\UltraEdit-32
C:\\Program Files\\SAS\\Shared Files\\Formats
[17] C:\\Program Files\\Graphviz2.20\\Bin   
C:\\Program Files\\MATLAB\\R2008a\\bin
[19] C:\\Program Files\\MATLAB\\R2008a\\bin\\win32  
C:\\Program Files\\TortoiseSVN\\bin
[21] C:\\Program Files\\HTML Help Workshop  
C:\\Program Files\\QuickTime\\QTSystem\\


Example:

 x - matrix(1:6, nrow=2, dimnames=list(c('a','b'),c('c','d','this 
that')))

 latex(x, file=)
% latex.default(x, file = )
%
\begin{table}[!tbp]
 \begin{center}
 \begin{tabular}{lrrr}\hline\hline
\multicolumn{1}{l}{x}\multicolumn{1}{c}{c}\multicolumn{1}{c}{d}\multicolumn{1}{c}{this 
that}\tabularnewline

\hline
a$1$$3$$5$\tabularnewline
b$2$$4$$6$\tabularnewline
\hline
\end{tabular}
\end{center}
\end{table}

 cd(c:/r/test)
 latex(x)
This is pdfTeX, Version 3.1415926-1.40.9 (MiKTeX 2.7)
entering extended mode
(C:/WINDOWS/TEMP/Rtmpz0QkT8/file311f289a.tex
LaTeX2e 2005/12/01
Babel v3.8l and hyphenation patterns for english, dumylang, 
nohyphenation, ge

rman, ngerman, french, loaded.
(C:\Program Files\MiKTeX 2.7\tex\latex\base\report.cls
Document Class: report 2005/09/16 v1.4f Standard LaTeX document class
(C:\Program Files\MiKTeX 2.7\tex\latex\base\size10.clo))
(C:\Program Files\MiKTeX 2.7\tex\latex\geometry\geometry.sty
(C:\Program Files\MiKTeX 2.7\tex\latex\graphics\keyval.sty)
(C:\Program Files\MiKTeX 2.7\tex\latex\geometry\geometry.cfg))
No file file311f289a.aux.
[1] (file311f289a.aux) )
Output written on file311f289a.dvi (1 page, 372 bytes).
Transcript written on file311f289a.log.

At this point, I have to press ESC after a few minutes to regain the 
console, and no .dvi (yap) window appears.
If I go to the temp directory, C:/WINDOWS/TEMP/Rtmpz0QkT8/, the .dvi 
file is there and double-click

launches yap.

  optionsCmds(xdvi)
[1] yap

Reading the code of print.latex - show.latex - show.dvi, I tried 
executing  system() directly, in the forms


 system(paste(yap, C:/WINDOWS/TEMP/Rtmpz0QkT8/file311f289a.dvi), 
intern=TRUE)


 system(paste(yap, C:/WINDOWS/TEMP/Rtmpz0QkT8/file311f289a.dvi))

However, both of these result in the same behavior -- R console hangs 
until I press ESC.




--
Michael Friendly Email: friendly AT yorku DOT ca
Professor, Psychology Dept.
York University  Voice: 416 736-5115 x66249 Fax: 416 736-5814
4700 Keele StreetWeb:   http://www.datavis.ca
Toronto, ONT  M3J 1P3 CANADA

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[R] Package install error: undefined symbol: _intel_fast_memcpy

2010-12-07 Thread rhelp . 20 . trevva

Hi,

I am trying to install the ncdf package on my Ubuntu 8.04 (Karmic)
box in the basement. I have installed the most recent distribution of
R, 2.12.0 without too many problems. However, when it comes to install
the ncdf netcdf package, I unfortunately get the following error:

Error in dyn.load(file, DLLpath = DLLpath, ...) :
 unable to load shared object
'/home/mpa/R/i486-pc-linux-gnu-library/2.12/ncdf/libs/ncdf.so':
 /home/mpa/R/i486-pc-linux-gnu-library/2.12/ncdf/libs/ncdf.so:
undefined symbol: _intel_fast_memcpy

I also get a similar error when trying to install the ncdf4 package.

I have tried to find the meaning of the error, but I can't find
anything - I'm assuming that its a dependency problem. Can anyone help
me out here please?

Best wishes,

Mark

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Re: [R] How to find out if a data frame has automatic row names?

2010-12-07 Thread Andreas Borg


That does it. Thanks!


Check ?.row_names_info

Peter Ehlers


Andreas







--
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Medizinische Informatik

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

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

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Re: [R] Sparse matrix performance question

2010-12-07 Thread Douglas Bates

On Mon, Dec 6, 2010 at 1:11 PM, scott white distributedin...@gmail.com wrote:
 Btw, forgot to mention I am using the standard Matrix package and I am
 running version 2.10.1 of R.

 On Mon, Dec 6, 2010 at 11:04 AM, scott white 
 distributedin...@gmail.comwrote:

 I have a very sparse square matrix which is  20K rows  columns and I am
 trying to row standardize the matrix for the rows that have non-missing
 value as follows:

 row_sums - rowSums(M,na.rm=TRUE)
 nonzero_idxs - which(row_sums0)
 nonzero_M - M[nonzero_idxs,]/row_sums[nonzero_idxs]
 M[nonzero_idxs,] - nonzero_M

Assignment of submatrices in a sparse matrix can be slow because there
is so much checking that needs to be done.  It is probably easier to
do the calculation directly on the data component of the matrix and
generate a new one.  The tricky bit to remember is that the indices in
the sparse matrix representation are 0-based so you need to add 1 when
using them in R.

I enclose a transcript.


 Each line completes well under a second except the last line which takes
 well over 10 seconds which is simply assigning the sub-matrix of rows that
 have non-missing values to the complete matrix. I am curious to know why it
 is so slow and how to speed it up. Should I be doing this differently or try
 a different sparse matrix library?

 Any feedback is appreciated.

 thanks,
 Scott



        [[alternative HTML version deleted]]

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R version 2.12.0 (2010-10-15)
Copyright (C) 2010 The R Foundation for Statistical Computing
ISBN 3-900051-07-0
Platform: x86_64-pc-linux-gnu (64-bit)

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

  Natural language support but running in an English locale

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

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

 library(Matrix)
Loading required package: lattice

Attaching package: 'Matrix'

The following object(s) are masked from 'package:base':

det

 set.seed(1234)
 M - sparseMatrix(i=sample(5000, 1000, replace=TRUE),
+   j=sample(5000, 1000, replace=TRUE),
+   x=rnorm(1000), dims=c(5000, 5000))
 str(M)
Formal class 'dgCMatrix' [package Matrix] with 6 slots
  ..@ i   : int [1:1000] 2014 549 1098 3137 130 1523 2198 3921 4323 931 ...
  ..@ p   : int [1:5001] 0 0 0 0 0 0 0 0 0 0 ...
  ..@ Dim : int [1:2] 5000 5000
  ..@ Dimnames:List of 2
  .. ..$ : NULL
  .. ..$ : NULL
  ..@ x   : num [1:1000] -0.4236 -0.5322 0.0675 -0.4105 -2.3708 ...
  ..@ factors : list()
 range(m...@i)
[1]1 4996
 str(rs - rowSums(M, na.rm=TRUE))
 num [1:5000] 0 0.501 0 0.598 -0.957 ...
 res - sparseMatrix(i...@i, p...@p, dim...@dim,
+ x...@x/r...@i + 1L], index1=FALSE)
 str(res)
Formal class 'dgCMatrix' [package Matrix] with 6 slots
  ..@ i   : int [1:1000] 2014 549 1098 3137 130 1523 2198 3921 4323 931 ...
  ..@ p   : int [1:5001] 0 0 0 0 0 0 0 0 0 0 ...
  ..@ Dim : int [1:2] 5000 5000
  ..@ Dimnames:List of 2
  .. ..$ : NULL
  .. ..$ : NULL
  ..@ x   : num [1:1000] 1 1 1 -0.655 1 ...
  ..@ factors : list()
 table(rowSums(res))

   01 
4082  918 
 
 proc.time()
   user  system elapsed 
  3.010   0.120   3.612 
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Re: [R] less than full rank contrast methods

On Tue, Dec 7, 2010 at 7:54 AM, Gabor Grothendieck
ggrothendi...@gmail.com wrote:
 'On Tue, Dec 7, 2010 at 5:19 AM, mxkuhn mxk...@gmail.com wrote:
 Greg and Frank,

 Thanks for the replies. I didn't express myself very well; I'm not interest 
 in the model fitting aspect. I'd just like to get the full set of dummy 
 variables (optimally from model.matrix)


 Try this:

 levels(when$time) - c(morning, afternoon, night)
 levels(when$day) - c(Mon, Tue, Wed, Thu, Fri, Sat, Sun)

 contr.dummy - function(n, ...) diag(n)
 mm - suppressWarnings(model.matrix(~ day + time, when))
 mm[is.na(mm)] - 1
 mm

 You might also want to set the levels of your factors first so that it
 includes levels that are not in the data and so that the levels are
 sorted in an order other than alphabetical:


The levels(when$day) lines should be:

levels(when$day) - c(Monday, Tuesday, Wednesday, Thursday,
Friday, Saturday, Sunday)



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

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[R] fast subsetting of lists in lists

Hello,


my data is contained in nested lists (which seems not necessarily to be
the best approach). What I need is a fast way to get subsets from the data.

An example:

test - list(list(a = 1, b = 2, c = 3), list(a = 4, b = 5, c = 6),
list(a = 7, b = 8, c = 9))

Now I would like to have all values in the named variables a, that is
the vector c(1, 4, 7). The best I could come up with is:

val - sapply(1:3, function (i) {test[[i]]$a})

which is unfortunately not very fast. According to R-inferno this is due
to the fact that apply and its derivates do looping in R rather than
rely on C-subroutines as the common [-operator.

Does someone now a trick to do the same as above with the faster
built-in subsetting? Something like:

test[somesubsettingmagic]


Thank you for your advice


Alex

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Re: [R] fast subsetting of lists in lists

On Tue, Dec 7, 2010 at 9:47 AM, Alexander Senger
sen...@physik.hu-berlin.de wrote:
 Hello,


 my data is contained in nested lists (which seems not necessarily to be
 the best approach). What I need is a fast way to get subsets from the data.

 An example:

 test - list(list(a = 1, b = 2, c = 3), list(a = 4, b = 5, c = 6),
 list(a = 7, b = 8, c = 9))

 Now I would like to have all values in the named variables a, that is
 the vector c(1, 4, 7). The best I could come up with is:

 val - sapply(1:3, function (i) {test[[i]]$a})

 which is unfortunately not very fast. According to R-inferno this is due
 to the fact that apply and its derivates do looping in R rather than
 rely on C-subroutines as the common [-operator.

 Does someone now a trick to do the same as above with the faster
 built-in subsetting? Something like:

 test[somesubsettingmagic]



This does not involve apply.  You could time it to see if its any faster:

 test.un - unlist(test)
 unname(test.un[names(test.un) == a])
[1] 1 4 7

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

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Re: [R] fast subsetting of lists in lists

2010-12-07 Thread Gerrit Eichner


Hello, Alexander,

does

utest - unlist(test)
utest[ names( utest) == a]

come close to what you need?

Hth,

Gerrit


On Tue, 7 Dec 2010, Alexander Senger wrote:


Hello,


my data is contained in nested lists (which seems not necessarily to be
the best approach). What I need is a fast way to get subsets from the data.

An example:

test - list(list(a = 1, b = 2, c = 3), list(a = 4, b = 5, c = 6),
list(a = 7, b = 8, c = 9))

Now I would like to have all values in the named variables a, that is
the vector c(1, 4, 7). The best I could come up with is:

val - sapply(1:3, function (i) {test[[i]]$a})

which is unfortunately not very fast. According to R-inferno this is due
to the fact that apply and its derivates do looping in R rather than
rely on C-subroutines as the common [-operator.

Does someone now a trick to do the same as above with the faster
built-in subsetting? Something like:

test[somesubsettingmagic]


Thank you for your advice


Alex

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[R] string

2010-12-07 Thread Benoit Wastine


Hi,

I'm running R 2.11
Does anyone know if it possible to transform one character vector to one 
character string ?


Many thanks
Benoit

--
Benoit Wastine
Laboratoire des Sciences du Climat et de l’Environnement (LSCE/IPSL)
CEA-CNRS-UVSQ
CE Saclay
Orme des merisiers
Bât 703 - Pte 13A
91191 Gif sur Yvette Cedex
France

Tel : 33 (0)1 69 08 21 97
Fax : 33 (0)1 69 08 77 16

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[R] Package install error: _intel_fast_memcpy

2010-12-07 Thread r-help . 20 . trevva

Hi,

I am trying to install the ncdf package on my Ubuntu 8.04 (Karmic)
box in the basement. I have installed the most recent distribution of
R, 2.12.0 without too many problems. However, when it comes to install
the ncdf netcdf package, I unfortunately get the following error:

Error in dyn.load(file, DLLpath = DLLpath, ...) :
 unable to load shared object
'/home/mpa/R/i486-pc-linux-gnu-library/2.12/ncdf/libs/ncdf.so':
 /home/mpa/R/i486-pc-linux-gnu-library/2.12/ncdf/libs/ncdf.so:
undefined symbol: _intel_fast_memcpy

I also get a similar error when trying to install the ncdf4 package.

I have tried to find the meaning of the error, but I can't find
anything - I'm assuming that its a dependency problem. Can anyone help
me out here please?

Best wishes,

Mark

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Re: [R] string

2010-12-07 Thread Jonathan P Daily

?paste

collapse = T may be necessary.
--
Jonathan P. Daily
Technician - USGS Leetown Science Center
11649 Leetown Road
Kearneysville WV, 25430
(304) 724-4480
Is the room still a room when its empty? Does the room,
 the thing itself have purpose? Or do we, what's the word... imbue it.
 - Jubal Early, Firefly

r-help-boun...@r-project.org wrote on 12/07/2010 10:11:41 AM:

 [image removed] 
 
 [R] string
 
 Benoit Wastine 
 
 to:
 
 r-help
 
 12/07/2010 10:14 AM
 
 Sent by:
 
 r-help-boun...@r-project.org
 
 Hi,
 
 I'm running R 2.11
 Does anyone know if it possible to transform one character vector to one 

 character string ?
 
 Many thanks
 Benoit
 
 -- 
 Benoit Wastine
 Laboratoire des Sciences du Climat et de l’Environnement (LSCE/IPSL)
 CEA-CNRS-UVSQ
 CE Saclay
 Orme des merisiers
 Bât 703 - Pte 13A
 91191 Gif sur Yvette Cedex
 France
 
 Tel : 33 (0)1 69 08 21 97
 Fax : 33 (0)1 69 08 77 16
 
 __
 R-help@r-project.org mailing list
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 PLEASE do read the posting guide 
http://www.R-project.org/posting-guide.html
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Re: [R] string

2010-12-07 Thread Ivan Calandra


Hi,
If I understand what you mean (no example...), see ?paste and the 
collpase argument

Ivan

Le 12/7/2010 16:11, Benoit Wastine a écrit :

Hi,

I'm running R 2.11
Does anyone know if it possible to transform one character vector to 
one character string ?


Many thanks
Benoit



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

**
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Re: [R] string



On Dec 7, 2010, at 10:11 AM, Benoit Wastine wrote:


Hi,

I'm running R 2.11
Does anyone know if it possible to transform one character vector to  
one character string ?


?gsub

Also look at the even more powerful gsubfn package. There is also the  
stringr package.


--
David Winsemius, MD
West Hartford, CT

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Re: [R] string

2010-12-07 Thread Ted Harding

Ivan's advice is good, but understanding clearly what
character string to separate the results might mean is
a bit tricky!

Example:

  cvec - c(J,e, ,m,',a,p,p,e,l,l,e,
 ,B,e,n,o,i,t)
  cstring - paste(cvec,collapse=)
  cstring
  # [1] Je m'appelle Benoit

Now try it without the 'collapse='.

Hoping this helps,
Ted.

On 07-Dec-10 15:19:39, Ivan Calandra wrote:
 Hi,
 If I understand what you mean (no example...), see ?paste and the 
 collpase argument
 Ivan
 
 Le 12/7/2010 16:11, Benoit Wastine a écrit :
 Hi,

 I'm running R 2.11
 Does anyone know if it possible to transform one character vector to 
 one character string ?

 Many thanks
 Benoit
 -- 
 Ivan CALANDRA


E-Mail: (Ted Harding) ted.hard...@wlandres.net
Fax-to-email: +44 (0)870 094 0861
Date: 07-Dec-10   Time: 15:38:44
-- XFMail --

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Re: [R] increase or decrease variable by 1

2010-12-07 Thread Uwe Ligges




On 07.12.2010 14:43, madr wrote:


many languages have shorthands for that operation like:

variable += 1
or
++variable

is there something like that in R ?


No.

Uwe Ligges

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Re: [R] string

2010-12-07 Thread Ivan Calandra

It took me quite some time to understand the difference between sep and 
collapse.


The examples in Phil Spector's book (2008) helped me to get it:

paste(c('X','Y'), 1:5, sep='_')
X_1 Y_2 X_3 Y_4 X_5
paste(c('X','Y'), 1:5, collapse='|')  ## sep=  by default
[1] X 1|Y 2|X 3|Y 4|X 5
paste(c('X','Y'), 1:5, sep='_', collapse='|')
[1] X_1|Y_2|X_3|Y_4|X_5

Ivan

Le 12/7/2010 16:38, (Ted Harding) a écrit :

Ivan's advice is good, but understanding clearly what
character string to separate the results might mean is
a bit tricky!

Example:

   cvec- c(J,e, ,m,',a,p,p,e,l,l,e,
  ,B,e,n,o,i,t)
   cstring- paste(cvec,collapse=)
   cstring
   # [1] Je m'appelle Benoit

Now try it without the 'collapse='.

Hoping this helps,
Ted.

On 07-Dec-10 15:19:39, Ivan Calandra wrote:

Hi,
If I understand what you mean (no example...), see ?paste and the
collpase argument
Ivan

Le 12/7/2010 16:11, Benoit Wastine a écrit :

Hi,

I'm running R 2.11
Does anyone know if it possible to transform one character vector to
one character string ?

Many thanks
Benoit

--
Ivan CALANDRA


E-Mail: (Ted Harding)ted.hard...@wlandres.net
Fax-to-email: +44 (0)870 094 0861
Date: 07-Dec-10   Time: 15:38:44
-- XFMail --

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

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

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[R] tableGrob and properties of a cell

2010-12-07 Thread Alex Bird

Hello there,

  could you please help to modify gpar() properties of a cell inside
tableGrob() output.
  In the following example I want to have different color for one out of 4 cells

require(gridExtra)
z-matrix(1:4,2,2)
grid.draw(tableGrob(z))

  The only way I found for now how to do it is to do something like this
downViewport(current.vpTree()$children[[1]]$children[[6]]$name)
grid.text('2',gp=gpar(col='red',fontface='bold'))

But for some reasons when I try to print the table into jpeg the last
grid.text() prints desired label text somewhere but not in the place I
want it to be.
Is there another more smart way to do it?

Thanks in advance!

Kind regards,
Alex

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Re: [R] increase or decrease variable by 1

On Tue, Dec 7, 2010 at 8:43 AM, madr madra...@interia.pl wrote:

 many languages have shorthands for that operation like:

 variable += 1
 or
 ++variable

 is there something like that in R ?


You can do this:

 x - 3
 `+`(x) - 1
 x
[1] 4



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

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[R] How to switch label font from Chinese to English

2010-12-07 Thread Jianhong Wang

Hi

I tried to plot something simple

x=c('2010-08-20', '2010-08-30')
y = c(1,2)
t = strptime(x, %Y-%m-%d)
plot(t,y)

At the xlab, the fonts are Chinese.  How can I switch it to English?  I am
working under Windows 7 64 bits Home Edition and R is Win32 version 2.11.1

Thanks!

[[alternative HTML version deleted]]

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Re: [R] Creating binary variable depending on strings of two dataframes



On Dec 7, 2010, at 11:30 AM, Pete Pete wrote:



Hi,
consider the following two dataframes:
x1=c(232,3454,3455,342,13)
x2=c(1,1,1,0,0)
data1=data.frame(x1,x2)

y1=c(232,232,3454,3454,3455,342,13,13,13,13)
y2=c(E1,F3,F5,E1,E2,H4,F8,G3,E1,H2)
data2=data.frame(y1,y2)

I need a new column in dataframe data1 (x3), which is either 0 or 1
depending if the value E1 in y2 of data2 is true while x1=y1. The  
result

of data1 should look like this:
  x1 x2 x3
1 232   1   1
2 3454 1   1
3 3455 1   0
4 342   0   0
5 13 0   1

I think a SQL command could help me but I am too inexperienced with  
it to

get there.


 dat3 - merge(data1, data2[data2$y2==E1, ], by.x=x1, by.y=y1,  
all.x=TRUE)

 dat3$y2 - 0 + (dat3$y2 %in% E1)
 dat3
x1 x2 y2
1   13  0  1
2  232  1  1
3  342  0  0
4 3454  1  1
5 3455  1  0

(Admittedly not in the original order, but in my hands the R merge  
operation doesn't lend itself well to maintaining the original order.  
I see that Grothendieck's solution is better in this respect, a  
typical occurrence in comparison of our respective efforts with R.)


--

David Winsemius, MD
West Hartford, CT

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Re: [R] please show me simple example how to plot


Well, maybe I didn't write it clear.
I know how to create scatterplot, and how to import data from csv file.
But I do not know how to add this fitting that mentioned in sumbject to a
plot.
I do not know for what function name to look for in R. I played some time
with ls, and it didn't create a line. I know the name for the fitting I ned
from Statistica, and this is from help from this program:

A curve is fitted to the XY coordinate data according to the
distance-weighted least squares smoothing procedure (the influence of
individual points decreases with the horizontal distance from the respective
points on the curve). For more information, see Distance-Weighted Least
Squares.

Now I need to recreate it in R.

Sorry again for not being clear.
-- 
View this message in context: 
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Sent from the R help mailing list archive at Nabble.com.

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Re: [R] increase or decrease variable by 1

On Tue, Dec 7, 2010 at 12:25 PM, Bert Gunter gunter.ber...@gene.com wrote:
 Ted:

 Inline below...

 On Tue, Dec 7, 2010 at 8:42 AM, Ted Harding ted.hard...@wlandres.net wrote:
 Indeed!

  x - x + 1

 (and being generous with unnecessary spaces) uses 10 characters.

  `+`(x)-1

 (being mean with them) uses 9. The mean version of the first
 uses only 6: x-x+1

 However, I suppose there is merit in the spiritual exercise
 of contemplating how `+`(x)-1 gets worked out!

 AFAICS it doesn't.
 `+`(x)-1
 Error in +x - 1 : could not find function +-

Sorry, my code was missing the first line:

 `+-` - `+`

 x - 3
 `+`(x) - 1
 x
[1] 4



-- 
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tel: 1-877-GKX-GROUP
email: ggrothendieck at gmail.com

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[R] please show me simple example how to plot Distance-Weighted Least Squares fitting


I got simple x,y pairs of data and simple scatterplot and just cannot figure
how to do it , there are many examples but always there is error popping out

please show me an example stripped with additional data just core of what I
need to do to get this damn line
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Re: [R] fast subsetting of lists in lists

I tried to hide the gory details as the structure of my datasets is
rather complicated. Basically its a long list of lists which in turn
contain character vectors, dates, numerics and dataframes, all named.
While the hierarchy is fixed neither the number of elements nor their
ordering is. But if I try to access a certain element, then I know it is
there and contains sensible data.
For a typical day of measurements the whole package weights around 1
GiB. How often and what I need to extract varies as the analyses is
rather dynamic.

As far as I can see a thorough refactoring of the datasets so that
everything is contained in one large dataframe might be a solution. But
I wouldn't be too unhappy if I could avoid this rather tedious work.

Alex


Am 07.12. 18:26, schrieb William Dunlap:
 To find the fastest method you need to tell more
 about the constraints on your problem.
Do you always have a list of lists of scalars
   or are the lists buried at various depths
   or do the numeric vectors at the leaves have
   various lengths?
If you always have a list of lists of scalars,
   do the names always come in the same order?
   (It may be faster to select by numeric position
   than by name).
Do all the lists of numeric vectors contain an
   element by the given name?
What is a typical size for the problem?  How
   many times do you typically need to repeat
   the solution?
 
 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 Alexander Senger
 Sent: Tuesday, December 07, 2010 9:12 AM
 To: r-help@r-project.org
 Subject: Re: [R] fast subsetting of lists in lists

 Hello Gerrit, Gabor,


 thank you for your suggestion.

 Unfortunately unlist seems to be rather expensive. A short 
 test with one
 of my datasets gives 0.01s for an extraction based on my approach and
 5.6s for unlist alone. The reason seems to be that unlist relies on
 lapply internally and does so recursively?

 Maybe there is still another way to go?

 Alex

 Am 07.12.2010 15:59, schrieb Gerrit Eichner:
 Hello, Alexander,

 does

 utest - unlist(test)
 utest[ names( utest) == a]

 come close to what you need?

 Hth,

 Gerrit


 On Tue, 7 Dec 2010, Alexander Senger wrote:

 Hello,


 my data is contained in nested lists (which seems not 
 necessarily to be
 the best approach). What I need is a fast way to get 
 subsets from the
 data.

 An example:

 test - list(list(a = 1, b = 2, c = 3), list(a = 4, b = 5, c = 6),
 list(a = 7, b = 8, c = 9))

 Now I would like to have all values in the named variables 
 a, that is
 the vector c(1, 4, 7). The best I could come up with is:

 val - sapply(1:3, function (i) {test[[i]]$a})

 which is unfortunately not very fast. According to 
 R-inferno this is due
 to the fact that apply and its derivates do looping in R 
 rather than
 rely on C-subroutines as the common [-operator.

 Does someone now a trick to do the same as above with the faster
 built-in subsetting? Something like:

 test[somesubsettingmagic]


 Thank you for your advice


 Alex

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Re: [R] coxph failure

2010-12-07 Thread Therneau, Terry M., Ph.D.


Larry,
   You found a data set that kills coxph.  I'll have to think about what
to do since on the one hand it's your own fault for trying to fit a very
bad model, and on the other I'd like the routine to give a nice error
message before it dies.

   In the data set you sent me the predictor variable is very skewed:

 quantile(anomaly1$CREAT, c(0, .5, .9, .999, 1))
0%50%90%  99.9%   100% 
0.5000 1.4000 2.2000 4.6027 7.3000 

There are 1000 observations, the largest is 7.3 and the second largest
is 4.6.  The subject with a 7.3 was a failure.  When you fit a 4 degree
of freedom polynomial to this there is enough space between the two
largest obvservations for the fit to turn nearly straight up and give a
probability of failure of approx 1.0 to this point.  This is the actual,
true maximum likelihood solution, so the routine is doing the right
thing to try and attain it.   At iteration 1 the linear predictor from
the Cox model has the following distribution.   

quantile(lp, c(0, .5, .9, .999, 1))
  0%  50%  90%99.9% 100% 
 -1.65321482  -1.19164206  -0.08964025  31.05389866 900.67627778 

Unfortunately exp(900) = Inf, and that pretty much does in my C routine
right there; within another iteration there are NA values littered about
and the return value for the coefficients is all NA.  

If instead you fit a smoothing spline to the data
   coxph(Surv(TTFAIL, FFAIL) ~ pspline(CREAT6MO, df=4), data=anomaly1)
the fitted response is essentially linear from 0 to 5 before turning
upward, but it never gets outrageous due to the shrinkage towards
linearity.

Catching this type of issue with an extremely skewed variable
(creatinine^4) would be very hard to do reliably. 

The primary message is that polynomials and biology are a bad mix.

Terry Therneau

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Re: [R] latex() hangs R console

2010-12-07 Thread RICHARD M. HEIBERGER

Michael,

The easiest workaround is to assign the result of the latex() command.

myfilename - latex(x)
print.default(myfilename)


It looks to me like the insides of the dvi.latex function aren't quite right
for Windows.

Rich

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[R] Creating binary variable depending on strings of two dataframes

2010-12-07 Thread Pete Pete


Hi,
consider the following two dataframes:
x1=c(232,3454,3455,342,13)
x2=c(1,1,1,0,0)
data1=data.frame(x1,x2)

y1=c(232,232,3454,3454,3455,342,13,13,13,13)
y2=c(E1,F3,F5,E1,E2,H4,F8,G3,E1,H2)
data2=data.frame(y1,y2)

I need a new column in dataframe data1 (x3), which is either 0 or 1
depending if the value E1 in y2 of data2 is true while x1=y1. The result
of data1 should look like this:
   x1 x2 x3
1 232   1   1
2 3454 1   1
3 3455 1   0
4 342   0   0
5 13 0   1

I think a SQL command could help me but I am too inexperienced with it to
get there.

Thanks for your help!

-- 
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http://r.789695.n4.nabble.com/Creating-binary-variable-depending-on-strings-of-two-dataframes-tp3076724p3076724.html
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Re: [R] fast subsetting of lists in lists

On Tue, Dec 7, 2010 at 12:12 PM, Alexander Senger
sen...@physik.hu-berlin.de wrote:
 Hello Gerrit, Gabor,


 thank you for your suggestion.

 Unfortunately unlist seems to be rather expensive. A short test with one
 of my datasets gives 0.01s for an extraction based on my approach and
 5.6s for unlist alone. The reason seems to be that unlist relies on
 lapply internally and does so recursively?

 Maybe there is still another way to go?


Too bad it wasn't faster but note that unlist does not rely on lapply.
  Here is the complete source of unlist and from it we see the key
portion is done internally, i.e. in C:

 unlist
function (x, recursive = TRUE, use.names = TRUE)
{
if (.Internal(islistfactor(x, recursive))) {
lv - unique(.Internal(unlist(lapply(x, levels), recursive,
FALSE)))
nm - if (use.names)
names(.Internal(unlist(x, recursive, use.names)))
res - .Internal(unlist(lapply(x, as.character), recursive,
FALSE))
res - match(res, lv)
structure(res, levels = lv, names = nm, class = factor)
}
else .Internal(unlist(x, recursive, use.names))
}
environment: namespace:base


-- 
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Re: [R] please show me simple example how to plot

2010-12-07 Thread Peter Ehlers


On 2010-12-07 09:37, madr wrote:


Well, maybe I didn't write it clear.
I know how to create scatterplot, and how to import data from csv file.
But I do not know how to add this fitting that mentioned in sumbject to a
plot.
I do not know for what function name to look for in R. I played some time
with ls, and it didn't create a line. I know the name for the fitting I ned
from Statistica, and this is from help from this program:

A curve is fitted to the XY coordinate data according to the
distance-weighted least squares smoothing procedure (the influence of
individual points decreases with the horizontal distance from the respective
points on the curve). For more information, see Distance-Weighted Least
Squares.


Sounds like loess() to me.

Peter Ehlers



Now I need to recreate it in R.

Sorry again for not being clear.


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Re: [R] How to formulate constraint like abs(x) = y in constrOptim (or other)

2010-12-07 Thread Benjamin B.

Thanks a lot for your responses.
Alabama and Rsolnp packages look promising.
My first tests worked well.

For others searching for alike problems I now use something like:

#equality constraint function
eqFkt-function(x){
  #for sum(x)=sum_x
  sm-sum(x)-sum_x
  return(c(sm))
}  
  
ineqFkt-function(x){
  #for -max_x  x  max_x
  ubound- -x+max_x
  lbound- x+max_x
  #for sum of all negative value to be  min_x_neg
  negbound  - sum(x[x0])-min_x_neg
  return(c(ubound, lbound, negbound))
}  
  
#define start  
start_weights-seq(0, 0, length.out=len_data)
#where mini_func is the minimizing function
res -auglag(start_weights, mini_func, heq=eqFkt, hin=ineqFkt)

Benjamin

-Ursprüngliche Nachricht-
Von: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] Im
Auftrag von Ravi Varadhan
Gesendet: Montag, 6. Dezember 2010 23:53
An: 'Hans W Borchers'; r-h...@stat.math.ethz.ch
Betreff: Re: [R] How to formulate constraint like abs(x) = y in constrOptim
(or other)

Hi Benjamin,

If you just had abs(x_i)  c_i, it will reduce to linear inequalities, but
your constraint cannot be reduced to that.

You might try alabama or Rsolnp packages.

Ravi.

---
Ravi Varadhan, Ph.D.
Assistant Professor,
Division of Geriatric Medicine and Gerontology School of Medicine Johns
Hopkins University

Ph. (410) 502-2619
email: rvarad...@jhmi.edu


-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
Behalf Of Hans W Borchers
Sent: Monday, December 06, 2010 1:08 PM
To: r-h...@stat.math.ethz.ch
Subject: Re: [R] How to formulate constraint like abs(x) = y in constrOptim
(or other)

Benjamin B. benj.bad.ac at googlemail.com writes:

 Hello list reader,
 
 I am trying to form some constraints for an optimization I am working on.
 I think I have understand the use of the constraints in matrix form. I use
 them like:
 
 [...]
 
 Now I would like to formulate a constraint like
 
 Sum(Abs(x_i))=limit
 
 But I have no idea how I could get such a condition by using the matrix
 constraint formulation.
 
 I have already searched the help, web and all searchable R resources I
could
 find, but got no helping result.
 
 Is constrOptim the right function to use?
 
 Maybe you can give me a hint or can point me to a good description of
 constraint handling.
 
 Greetings and thanks in advance,
 
 Benjamin
 
 Benjamin B.
 Hamburg, Germany

With 'constrOptim' you can formulate linear constraints only, and in most
cases abs() is not linear. You might try one of the nonlinear optimization
packages.

Another possibility is to reformulate the absolute value with two binary
values, if your optimization function itself is linear; you didn't tell us
the whole story.

Hans Werner

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Re: [R] increase or decrease variable by 1



On Dec 7, 2010, at 12:42 PM, Gabor Grothendieck wrote:

On Tue, Dec 7, 2010 at 12:25 PM, Bert Gunter  
gunter.ber...@gene.com wrote:

Ted:

Inline below...

On Tue, Dec 7, 2010 at 8:42 AM, Ted Harding  
ted.hard...@wlandres.net wrote:

Indeed!

 x - x + 1

(and being generous with unnecessary spaces) uses 10 characters.

 `+`(x)-1

(being mean with them) uses 9. The mean version of the first
uses only 6: x-x+1

However, I suppose there is merit in the spiritual exercise
of contemplating how `+`(x)-1 gets worked out!


AFAICS it doesn't.

`+`(x)-1

Error in +x - 1 : could not find function +-


Sorry, my code was missing the first line:


`+-` - `+`

x - 3
`+`(x) - 1
x

[1] 4


One can also attempt further creative violence to the language that  
shows that is possible to construct unary operators like the C  
prefix-+ that do not require shift-9 operand shift-0.


 `+`
function (e1, e2)  .Primitive(+)
 `+` - `-`# in a different universe perhaps
  4 + 2
[1] 2
 `+` - function (e1, e2)  .Primitive(+) # attempt to restore
  4 + 2# test
function (e1, e2)  .Primitive(+)  ##  oops
 `+` -  .Primitive(+)
  4 + 2
[1] 6  # works
 `!` - function(x) x + 1   # hijacking a logical NOT definition
 !3
[1] 4   # so this is a redefinition of one of the two unary operators
# ( of which I am aware, I'm not an R-guru.)

 `!` - .Primitive(!) ## hoping to restore expected behavior
 !3
[1] FALSE   # whew. That seemed dangerous but I appear to have survived.

--

David Winsemius, MD
West Hartford, CT

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[R] randomForest: How to append ID column along with predictions

2010-12-07 Thread Dennis

Hi all,

When running a prediction using RF on another data, I get two columns
returned: row number(?) and predicted class. Is there a way of
appending the unique row value from an ID column in the dataframe to
the predictions instead of the row number? I'm assuming that the
returned results follow the data frame in that the first result
returned equals the first entry in the dataframe.

i.e., instead of a prediction output like this:

1, ants
2, ants
3, bees
4, ants

I'd like the first column to pull IDs from the dataframe associated
with each row (row number in parenthesis for illustration):

(1) 1130, ants
(2) 1130, ants
(3) 2139, bees
(4) 1130, ants

This is likely a simple procedure, but I haven't been able to get
anything to work. Any help would be appreciated!

Cheers,

Dennis

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Re: [R] fast subsetting of lists in lists

2010-12-07 Thread Henrik Bengtsson

First, subset 'test' once, e.g.

testT - test[1:3];

and then use sapply() on that, e.g.

val - sapply(testT, FUN=function (x) { x$a })

Then you can avoid one level of function calls, by

val - sapply(testT, FUN=[[, a)

Second, there is some overhead in [[, $ etc.  You can use
.subset2() to avoid this, e.g.

val - sapply(testT, FUN=.subset2, a)

Third, it may be that using sapply() to structure you results is a bit
overkill.  If you know that the 'a' element is always of the same
dimension, you can do it yourself, e.g.

val - lapply(testT, FUN=.subset2, a)
val - unlist(val, use.names=FALSE)   # use.names=FALSE is much faster than TRUE

See what that does

/Henrik

On Tue, Dec 7, 2010 at 6:47 AM, Alexander Senger
sen...@physik.hu-berlin.de wrote:
 Hello,


 my data is contained in nested lists (which seems not necessarily to be
 the best approach). What I need is a fast way to get subsets from the data.

 An example:

 test - list(list(a = 1, b = 2, c = 3), list(a = 4, b = 5, c = 6),
 list(a = 7, b = 8, c = 9))

 Now I would like to have all values in the named variables a, that is
 the vector c(1, 4, 7). The best I could come up with is:

 val - sapply(1:3, function (i) {test[[i]]$a})

 which is unfortunately not very fast. According to R-inferno this is due
 to the fact that apply and its derivates do looping in R rather than
 rely on C-subroutines as the common [-operator.

 Does someone now a trick to do the same as above with the faster
 built-in subsetting? Something like:

 test[somesubsettingmagic]


 Thank you for your advice


 Alex

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Re: [R] fast subsetting of lists in lists

2010-12-07 Thread Matthew Dowle

Hello Alex,

Assuming it was just an inadequate example (since a data.frame would suffice 
in that case), did you know that a data.frames' columns do not have to be 
vectors but can be lists?  I don't know if that helps.

 DF = data.frame(a=1:3)
 DF$b = list(pi, 2:3, letters[1:5])
 DF
  a b
1 1  3.141593
2 2  2, 3
3 3 a, b, c, d, e
 DF$b
[[1]]
[1] 3.141593

[[2]]
[1] 2 3

[[3]]
[1] a b c d e
 sapply(DF,class)
a b
integerlist


That is still regular though in the sense that each row has a value for all 
the columns, even if that value is NA, or NULL in lists.

If your data is not regular then one option is to flatten it into 
(row,column,value) tuple similar to how sparse matrices are stored.  Your 
value column may be list rather than vector.

Then (and yes you guessed this was coming) ... you can use data.table to 
query the flat structure quickly by setting a key on the first two columns, 
or maybe just the 2nd column when you need to pick out the values for one 
'column' quickly for all 'rows'.

There was a thread about using list() columns in data.table here :

http://r.789695.n4.nabble.com/Suggest-a-cool-feature-Use-data-table-like-a-sorted-indexed-data-list-tp2544213p2544213.html

 Does someone now a trick to do the same as above with the faster built-in 
 subsetting? Something like:
 test[somesubsettingmagic]

So in data.table if you wanted all the 'b' values,  you might do something 
like this :

setkey(DT,column)
DT[J(b), value]

which should return the list() quickly from the irregular data.

Matthew


Alexander Senger sen...@physik.hu-berlin.de wrote in message 
news:4cfe6aee.6030...@physik.hu-berlin.de...
 Hello Gerrit, Gabor,


 thank you for your suggestion.

 Unfortunately unlist seems to be rather expensive. A short test with one
 of my datasets gives 0.01s for an extraction based on my approach and
 5.6s for unlist alone. The reason seems to be that unlist relies on
 lapply internally and does so recursively?

 Maybe there is still another way to go?

 Alex

 Am 07.12.2010 15:59, schrieb Gerrit Eichner:
 Hello, Alexander,

 does

 utest - unlist(test)
 utest[ names( utest) == a]

 come close to what you need?

 Hth,

 Gerrit


 On Tue, 7 Dec 2010, Alexander Senger wrote:

 Hello,


 my data is contained in nested lists (which seems not necessarily to be
 the best approach). What I need is a fast way to get subsets from the
 data.

 An example:

 test - list(list(a = 1, b = 2, c = 3), list(a = 4, b = 5, c = 6),
 list(a = 7, b = 8, c = 9))

 Now I would like to have all values in the named variables a, that is
 the vector c(1, 4, 7). The best I could come up with is:

 val - sapply(1:3, function (i) {test[[i]]$a})

 which is unfortunately not very fast. According to R-inferno this is due
 to the fact that apply and its derivates do looping in R rather than
 rely on C-subroutines as the common [-operator.

 Does someone now a trick to do the same as above with the faster
 built-in subsetting? Something like:

 test[somesubsettingmagic]


 Thank you for your advice


 Alex

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Re: [R] increase or decrease variable by 1

On Tue, Dec 7, 2010 at 1:12 PM, David Winsemius dwinsem...@comcast.net wrote:

 On Dec 7, 2010, at 12:42 PM, Gabor Grothendieck wrote:

 On Tue, Dec 7, 2010 at 12:25 PM, Bert Gunter gunter.ber...@gene.com
 wrote:

 Ted:

 Inline below...

 On Tue, Dec 7, 2010 at 8:42 AM, Ted Harding ted.hard...@wlandres.net
 wrote:

 Indeed!

  x - x + 1

 (and being generous with unnecessary spaces) uses 10 characters.

  `+`(x)-1

 (being mean with them) uses 9. The mean version of the first
 uses only 6: x-x+1

 However, I suppose there is merit in the spiritual exercise
 of contemplating how `+`(x)-1 gets worked out!

 AFAICS it doesn't.

 `+`(x)-1

 Error in +x - 1 : could not find function +-

 Sorry, my code was missing the first line:

 `+-` - `+`

 x - 3
 `+`(x) - 1
 x

 [1] 4

 One can also attempt further creative violence to the language that shows
 that is possible to construct unary operators like the C prefix-+ that do
 not require shift-9 operand shift-0.

 `+`
 function (e1, e2)  .Primitive(+)
 `+` - `-`    # in a different universe perhaps
  4 + 2
 [1] 2
 `+` - function (e1, e2)  .Primitive(+) # attempt to restore
  4 + 2    # test
 function (e1, e2)  .Primitive(+)  ##  oops
 `+` -  .Primitive(+)
  4 + 2
 [1] 6          # works
 `!` - function(x) x + 1   # hijacking a logical NOT definition
 !3
 [1] 4   # so this is a redefinition of one of the two unary operators
        # ( of which I am aware, I'm not an R-guru.)

 `!` - .Primitive(!) ## hoping to restore expected behavior
 !3
 [1] FALSE   # whew. That seemed dangerous but I appear to have survived.

Unlike the above, +- is not predefined so its definition does not
overwrite existing definitions.

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

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Re: [R] increase or decrease variable by 1

2010-12-07 Thread Ted Harding

Indeed!

  x - x + 1

(and being generous with unnecessary spaces) uses 10 characters.

  `+`(x)-1

(being mean with them) uses 9. The mean version of the first
uses only 6: x-x+1

However, I suppose there is merit in the spiritual exercise
of contemplating how `+`(x)-1 gets worked out!

Ted.

On 07-Dec-10 16:23:17, Ivan Calandra wrote:
 x+1 is not that complicated... Am I missing something here?
 
 Le 12/7/2010 16:55, Gabor Grothendieck a écrit :
 On Tue, Dec 7, 2010 at 8:43 AM, madrmadra...@interia.pl  wrote:
 many languages have shorthands for that operation like:

 variable += 1
 or
 ++variable

 is there something like that in R ?

 You can do this:

 x- 3
 `+`(x)- 1
 x
 [1] 4
 -- 
 Ivan CALANDRA


E-Mail: (Ted Harding) ted.hard...@wlandres.net
Fax-to-email: +44 (0)870 094 0861
Date: 07-Dec-10   Time: 16:42:44
-- XFMail --

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[R] how to get vector of data from line ?


I have created a density line
d- density(X)
now I need to read values from that line

for example what is the value of this line at x = 1, 2, 3 etc...
-- 
View this message in context: 
http://r.789695.n4.nabble.com/how-to-get-vector-of-data-from-line-tp3076943p3076943.html
Sent from the R help mailing list archive at Nabble.com.

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Re: [R] tableGrob and properties of a cell

2010-12-07 Thread baptiste auguie

Hi,

Embarrassingly enough, it was quite straight-forward in the first
versions of grid.table(). You might want to try with version r11 for
example,

source(http://gridextra.googlecode.com/svn-history/r11/trunk/R/tableGrob.r;)
library(grid)

 tc  = textConnection(
  carat   VeryLongWordIndeed color clarity depth
 14513  1.35 Ideal J VS2  61.4
 28685  0.30  Good GVVS1  64.0
 50368  0.75 Ideal F SI2  59.2)
 d = read.table(tc,head=T)
 close(tc)
 grid.newpage()
 g = grid.table(d)
 grid.ls(g)
 grid.edit(top-head-fill-5, gp=gpar(fill=red))
 grid.edit(cells-label-33, label=expression(alpha),gp=gpar(col=orange))

As I recall, the reason why this is more difficult now (if at all
possible, I have no idea), is that I placed the actual drawing code in
a drawDetails method, so that it gets updated on-the-fly, etc.

I hope this simpler version will suit you, I might re-include it in
the package as an alternative.

HTH,

baptiste






On 7 December 2010 16:53, Alex Bird sund...@gmail.com wrote:
 Hello there,

  could you please help to modify gpar() properties of a cell inside
 tableGrob() output.
  In the following example I want to have different color for one out of 4 
 cells

 require(gridExtra)
 z-matrix(1:4,2,2)
 grid.draw(tableGrob(z))

  The only way I found for now how to do it is to do something like this
 downViewport(current.vpTree()$children[[1]]$children[[6]]$name)
 grid.text('2',gp=gpar(col='red',fontface='bold'))

 But for some reasons when I try to print the table into jpeg the last
 grid.text() prints desired label text somewhere but not in the place I
 want it to be.
 Is there another more smart way to do it?

 Thanks in advance!

 Kind regards,
 Alex

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-- 


Dr. Baptiste Auguié

Departamento de Química Física,
Universidade de Vigo,
Campus Universitario, 36310, Vigo, Spain

tel: +34 9868 18617
http://webs.uvigo.es/coloides

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[R] Efficient way to use data frame of indices to initialize matrix

2010-12-07 Thread Cutler, Gene

I have a data frame with three columns, x, y, and a.  I want to create a matrix 
from these values such that for matrix m:
m[x,y] == a

Obviously, I can go row by row through the data frame and insert the value a at 
the correct x,y location in the matrix.  I can make that slightly more 
efficient (perhaps), by doing something like this:
 for (each.x in unique(df$x)) m[each.x, df$y[df$x == each.x]] - df$a[df$x == 
 each.x]

But I feel that there must be a more efficient, or at least more elegant way to 
do this.

--
Gene

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Re: [R] please show me simple example how to plot

2010-12-07 Thread Greg Snow

Like Peter says, this sounds like loess, there are examples on the help page 
for scatter.smooth, you could also do this with lattice graphics using 
type=c('p','smooth'), or ggplot2 graphics (probably something like geom_smooth 
or geom_loess, I don't know ggplot2 that well yet).

If you want to learn more about how loess works then use loess.demo in the 
TeachingDemos package.  

There are also other smoothers available, but we would need more information 
before knowing where to point you.

For adding a general curve or line to a scatterplot also look at functions like 
lines, abline, and curve, as well as the entire ggplot2 package.

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


 -Original Message-
 From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-
 project.org] On Behalf Of madr
 Sent: Tuesday, December 07, 2010 10:37 AM
 To: r-help@r-project.org
 Subject: Re: [R] please show me simple example how to plot
 
 
 Well, maybe I didn't write it clear.
 I know how to create scatterplot, and how to import data from csv file.
 But I do not know how to add this fitting that mentioned in sumbject to
 a
 plot.
 I do not know for what function name to look for in R. I played some
 time
 with ls, and it didn't create a line. I know the name for the fitting I
 ned
 from Statistica, and this is from help from this program:
 
 A curve is fitted to the XY coordinate data according to the
 distance-weighted least squares smoothing procedure (the influence of
 individual points decreases with the horizontal distance from the
 respective
 points on the curve). For more information, see Distance-Weighted Least
 Squares.
 
 Now I need to recreate it in R.
 
 Sorry again for not being clear.
 --
 View this message in context: http://r.789695.n4.nabble.com/please-
 show-me-simple-example-how-to-plot-Distance-Weighted-Least-Squares-
 fitting-tp3076824p3076863.html
 Sent from the R help mailing list archive at Nabble.com.
 
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[R] tweedie in gam

2010-12-07 Thread Payal Bal

Hello,

I have count data with a large number of zeros. To model this data using
gams I want to try the Tweedie distribution instead. (The zero-inflated
poisson willbe my next step)

How do i decide on the value of the power 'p' parameter in the formula:
family=Tweedie(p, link) ? For a glm, this can be found using the
tweedie.profile function. But it doesnot work with the gam model formula
where explantory variables are specified as s(x1) instead of x1.

Will using a tweedie on discrete data affect my results?

If my question has already been addressed before, a reference to the
relevant information will be very much appreciated.

Thanks,
payal.

[[alternative HTML version deleted]]

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Re: [R] How to switch label font from Chinese to English

2010-12-07 Thread Prof Brian Ripley


On Tue, 7 Dec 2010, Jianhong Wang wrote:


Hi

I tried to plot something simple

x=c('2010-08-20', '2010-08-30')
y = c(1,2)
t = strptime(x, %Y-%m-%d)
plot(t,y)

At the xlab, the fonts are Chinese.  How can I switch it to English?  I am
working under Windows 7 64 bits Home Edition and R is Win32 version 2.11.1


Is this covered by the Windows FAQ at
http://cran.r-project.org/bin/windows/base/rw-FAQ.html

especially Qs 3.2 and 3.3?  (We don't know what you see in Chinese.) 
I suspect the issue is the locale category LC_TIME which is used to 
'print' date-times (and in my UK locale I get an x axis annotated ny 
'Aug').  If I do



Sys.setlocale('LC_TIME', 'zh_CN.utf8')

[1] zh_CN.utf8

plot(t,y)


I get Chinese labels 

If so, you can also use a different format for the annotations: see 
?axis.POSIXct, e.g.


plot(t, y, format='%m/%d')




Thanks!

[[alternative HTML version deleted]]

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--
Brian D. Ripley,  rip...@stats.ox.ac.uk
Professor of Applied Statistics,  http://www.stats.ox.ac.uk/~ripley/
University of Oxford, Tel:  +44 1865 272861 (self)
1 South Parks Road, +44 1865 272866 (PA)
Oxford OX1 3TG, UKFax:  +44 1865 272595

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Re: [R] Efficient way to use data frame of indices to initialize matrix

2010-12-07 Thread Whit Armstrong

index m as a vector and do the assignment in one step

i - df$row + (df$col-1)*nrow(m)
m[i] - df$a

or something along those lines.

-Whit


On Tue, Dec 7, 2010 at 1:31 PM, Cutler, Gene gcut...@amgen.com wrote:
 I have a data frame with three columns, x, y, and a.  I want to create a 
 matrix from these values such that for matrix m:
 m[x,y] == a

 Obviously, I can go row by row through the data frame and insert the value a 
 at the correct x,y location in the matrix.  I can make that slightly more 
 efficient (perhaps), by doing something like this:
 for (each.x in unique(df$x)) m[each.x, df$y[df$x == each.x]] - df$a[df$x == 
 each.x]

 But I feel that there must be a more efficient, or at least more elegant way 
 to do this.

 --
 Gene

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Re: [R] How to switch label font from Chinese to English

2010-12-07 Thread Jianhong Wang

Thanks! Brian,

As you said, if locale changes, it is fine now.  What I add is
Sys.setlocale(LC_TIME, American)

On Tue, Dec 7, 2010 at 12:14 PM, Prof Brian Ripley rip...@stats.ox.ac.ukwrote:

 On Tue, 7 Dec 2010, Jianhong Wang wrote:

  Hi

 I tried to plot something simple

 x=c('2010-08-20', '2010-08-30')
 y = c(1,2)
 t = strptime(x, %Y-%m-%d)
 plot(t,y)

 At the xlab, the fonts are Chinese.  How can I switch it to English?  I am
 working under Windows 7 64 bits Home Edition and R is Win32 version 2.11.1


 Is this covered by the Windows FAQ at
 http://cran.r-project.org/bin/windows/base/rw-FAQ.html

 especially Qs 3.2 and 3.3?  (We don't know what you see in Chinese.) I
 suspect the issue is the locale category LC_TIME which is used to 'print'
 date-times (and in my UK locale I get an x axis annotated ny 'Aug').  If I
 do

  Sys.setlocale('LC_TIME', 'zh_CN.utf8')

 [1] zh_CN.utf8

 plot(t,y)


 I get Chinese labels 

 If so, you can also use a different format for the annotations: see
 ?axis.POSIXct, e.g.

 plot(t, y, format='%m/%d')



 Thanks!

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 Brian D. Ripley,  rip...@stats.ox.ac.uk
 Professor of Applied Statistics,  
 http://www.stats.ox.ac.uk/~ripley/http://www.stats.ox.ac.uk/%7Eripley/
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Re: [R] how to get vector of data from line ?

2010-12-07 Thread Greg Snow

There are a few different options depending on what you are trying to do.

If you just need some data pairs (for plotting for example), then the return 
from density has a vector of x's and a vector of y's, just use those.

If you have specific x values that you need the height at and they follow a 
regular pattern, then you can modify the from, to, and n parameters to generate 
your x values, then they are in the return value.

You can approximate the values by passing the x and y from the return of 
density to the functions approx or approxfun.

You can estimate the density in a different way using the logspline package, it 
provides functions to give you the height of the curve at specified values.

You can just add the kernels together at your specified x values instead of 
using density (search the archives, I think Prof. Ripley recently posted a 
function to do this or similar).

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


 -Original Message-
 From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-
 project.org] On Behalf Of madr
 Sent: Tuesday, December 07, 2010 11:28 AM
 To: r-help@r-project.org
 Subject: [R] how to get vector of data from line ?
 
 
 I have created a density line
 d- density(X)
 now I need to read values from that line
 
 for example what is the value of this line at x = 1, 2, 3 etc...
 --
 View this message in context: http://r.789695.n4.nabble.com/how-to-get-
 vector-of-data-from-line-tp3076943p3076943.html
 Sent from the R help mailing list archive at Nabble.com.
 
 __
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 https://stat.ethz.ch/mailman/listinfo/r-help
 PLEASE do read the posting guide http://www.R-project.org/posting-
 guide.html
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Re: [R] Efficient way to use data frame of indices to initialize matrix

2010-12-07 Thread Greg Snow

tmpdf - data.frame( x = c(1,2,3), y=c(2,3,1), a=c(10,20,30) )
mymat - matrix(0, ncol=3, nrow=3)
mymat[ as.matrix(tmpdf[,c('x','y')]) ] - tmpdf$a


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


 -Original Message-
 From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-
 project.org] On Behalf Of Cutler, Gene
 Sent: Tuesday, December 07, 2010 11:31 AM
 To: r-help@r-project.org
 Subject: [R] Efficient way to use data frame of indices to initialize
 matrix
 
 I have a data frame with three columns, x, y, and a.  I want to create
 a matrix from these values such that for matrix m:
 m[x,y] == a
 
 Obviously, I can go row by row through the data frame and insert the
 value a at the correct x,y location in the matrix.  I can make that
 slightly more efficient (perhaps), by doing something like this:
  for (each.x in unique(df$x)) m[each.x, df$y[df$x == each.x]] -
 df$a[df$x == each.x]
 
 But I feel that there must be a more efficient, or at least more
 elegant way to do this.
 
 --
 Gene
 
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[R] randomForest: How to append ID column along with predictions

2010-12-07 Thread Dennis Duro

Hi all,

When running a prediction using RF on another data, I get two columns
returned: row number(?) and predicted class. Is there a way of
appending the unique row value from an ID column in the dataframe to
the predictions instead of the row number? I'm assuming that the
returned results follow the data frame in that the first result
returned equals the first entry in the dataframe.

i.e., instead of a prediction output like this:

1, ants
2, ants
3, bees
4, ants

I'd like the first column to pull IDs from the dataframe associated
with each row (row number in parenthesis for illustration):

(1) 1130, ants
(2) 1130, ants
(3) 2139, bees
(4) 1130, ants

This is likely a simple procedure, but I haven't been able to get
anything to work. Any help would be appreciated!

Cheers,

Dennis

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Re: [R] how to get vector of data from line ?



On Dec 7, 2010, at 1:27 PM, madr wrote:



I have created a density line
d- density(X)
now I need to read values from that line

for example what is the value of this line at x = 1, 2, 3 etc...


Learn to search:

http://finzi.psych.upenn.edu/Rhelp10/2010-September/253910.html



--

David Winsemius, MD
West Hartford, CT

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[R] Statistical Analysis with R Beginner's Guide Book

2010-12-07 Thread johnmquick

Hi Everyone,

I'm writing to announce my new R beginner's guide book and answer questions
related to it.

The primary focus of Statistical Analysis with R is helping new users become
accustomed to R and empowering them to apply R to suit their own needs. It
is a beginner's guide written for a broad audience and should be well
received by businesspeople, IT professionals, researchers, and students
alike. Statistical Analysis with R takes readers on a journey from their
first installation and launch of R, to analyzing and assessing data, to
communicating and visualizing results. You can
http://rtutorialseries.blogspot.com/2010/11/r-beginners-guide-book-update.html
learn more about the book on my R Tutorial Series blog. The book itself can
be found on the http://link.packtpub.com/or7f1u Packt Publishing website .

If you have questions about the book, such as its content coverage,
approach, audience, etc., please respond and I will do my best to clarify.

Sincerely,
John M. Quick

-
John M. Quick

* http://rTutorialSeries.blogspot.com R Tutorial Series Blog
* http://link.packtpub.com/or7f1u R Beginner's Guide
* http://www.johnmquick.com www.johnmquick.com

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Re: [R] tableGrob and properties of a cell

2010-12-07 Thread Alex Bird

Hi Baptiste!

 Grand-grand merci!!! It works!

Thanks!
Alex

2010/12/7 baptiste auguie baptiste.aug...@googlemail.com:
 baptiste

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Re: [R] fast subsetting of lists in lists

Hello,

Matthew's hint is interesting:

Am 07.12.2010 19:16, schrieb Matthew Dowle:
 Hello Alex,
 
 Assuming it was just an inadequate example (since a data.frame would suffice 
 in that case), did you know that a data.frames' columns do not have to be 
 vectors but can be lists?  I don't know if that helps.
 
 DF = data.frame(a=1:3)
 DF$b = list(pi, 2:3, letters[1:5])
 DF
   a b
 1 1  3.141593
 2 2  2, 3
 3 3 a, b, c, d, e
 DF$b
 [[1]]
 [1] 3.141593
 
 [[2]]
 [1] 2 3
 
 [[3]]
 [1] a b c d e
 sapply(DF,class)
 a b
 integerlist

 
 That is still regular though in the sense that each row has a value for all 
 the columns, even if that value is NA, or NULL in lists.

My data is mostly regular, that is every sublist contains a data.frame
which is the major contribution to overall size. The reason I use lists
is mainly that I need also some bits of information about the
environment. I thought about putting these into additional columns of
the data.frame (and add redundancy and maybe 30% of overhead this way),
one column per variable. But as memory usage is already close to the
limit of my machine this might break things (the situation is a bit
tricky, isn't it?).
I didn't know that a column of a data.frame can be a list. So if I need
only let's say 10 entries in that list, but my data.frame has several
hundred rows, would the empty parts of the column-list be filled
with cycled values or would they be really empty and thus not use
additional memory?
Secondly as I mentioned in another email to this topic: a whole day of
data contains about 100 chunks of data that is 100 of the sublists
described above. I could put them all into one large data.frame, but
then I would have to extract the environmental data from the long
list, now containing repeated occurrences of variables with the same
name. I guess subsetting could become tricky here (dependend on name and
position, I assume), but I'm eager to learn an easy way of doing so.

Sorry for not submitting an illustrative example, but I'm afraid that
would be quite lengthy and not so illustrative any more.

The data.table mentioned below seems to be an interesting alternative;
I'll definitely look into this. But it would also mean quite a bit of
homework, as far as I can see...

Thanks

Alex


 If your data is not regular then one option is to flatten it into 
 (row,column,value) tuple similar to how sparse matrices are stored.  Your 
 value column may be list rather than vector.
 
 Then (and yes you guessed this was coming) ... you can use data.table to 
 query the flat structure quickly by setting a key on the first two columns, 
 or maybe just the 2nd column when you need to pick out the values for one 
 'column' quickly for all 'rows'.
 
 There was a thread about using list() columns in data.table here :
 
 http://r.789695.n4.nabble.com/Suggest-a-cool-feature-Use-data-table-like-a-sorted-indexed-data-list-tp2544213p2544213.html
 
 Does someone now a trick to do the same as above with the faster built-in 
 subsetting? Something like:
 test[somesubsettingmagic]
 
 So in data.table if you wanted all the 'b' values,  you might do something 
 like this :
 
 setkey(DT,column)
 DT[J(b), value]
 
 which should return the list() quickly from the irregular data.
 
 Matthew
 
 
 Alexander Senger sen...@physik.hu-berlin.de wrote in message 
 news:4cfe6aee.6030...@physik.hu-berlin.de...
 Hello Gerrit, Gabor,


 thank you for your suggestion.

 Unfortunately unlist seems to be rather expensive. A short test with one
 of my datasets gives 0.01s for an extraction based on my approach and
 5.6s for unlist alone. The reason seems to be that unlist relies on
 lapply internally and does so recursively?

 Maybe there is still another way to go?

 Alex

 Am 07.12.2010 15:59, schrieb Gerrit Eichner:
 Hello, Alexander,

 does

 utest - unlist(test)
 utest[ names( utest) == a]

 come close to what you need?

 Hth,

 Gerrit


 On Tue, 7 Dec 2010, Alexander Senger wrote:

 Hello,


 my data is contained in nested lists (which seems not necessarily to be
 the best approach). What I need is a fast way to get subsets from the
 data.

 An example:

 test - list(list(a = 1, b = 2, c = 3), list(a = 4, b = 5, c = 6),
 list(a = 7, b = 8, c = 9))

 Now I would like to have all values in the named variables a, that is
 the vector c(1, 4, 7). The best I could come up with is:

 val - sapply(1:3, function (i) {test[[i]]$a})

 which is unfortunately not very fast. According to R-inferno this is due
 to the fact that apply and its derivates do looping in R rather than
 rely on C-subroutines as the common [-operator.

 Does someone now a trick to do the same as above with the faster
 built-in subsetting? Something like:

 test[somesubsettingmagic]


 Thank you for your advice


 Alex

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 PLEASE do read the posting guide

Re: [R] fast subsetting of lists in lists

Hello Gerrit, Gabor,


thank you for your suggestion.

Unfortunately unlist seems to be rather expensive. A short test with one
of my datasets gives 0.01s for an extraction based on my approach and
5.6s for unlist alone. The reason seems to be that unlist relies on
lapply internally and does so recursively?

Maybe there is still another way to go?

Alex

Am 07.12.2010 15:59, schrieb Gerrit Eichner:
 Hello, Alexander,
 
 does
 
 utest - unlist(test)
 utest[ names( utest) == a]
 
 come close to what you need?
 
 Hth,
 
 Gerrit
 
 
 On Tue, 7 Dec 2010, Alexander Senger wrote:
 
 Hello,


 my data is contained in nested lists (which seems not necessarily to be
 the best approach). What I need is a fast way to get subsets from the
 data.

 An example:

 test - list(list(a = 1, b = 2, c = 3), list(a = 4, b = 5, c = 6),
 list(a = 7, b = 8, c = 9))

 Now I would like to have all values in the named variables a, that is
 the vector c(1, 4, 7). The best I could come up with is:

 val - sapply(1:3, function (i) {test[[i]]$a})

 which is unfortunately not very fast. According to R-inferno this is due
 to the fact that apply and its derivates do looping in R rather than
 rely on C-subroutines as the common [-operator.

 Does someone now a trick to do the same as above with the faster
 built-in subsetting? Something like:

 test[somesubsettingmagic]


 Thank you for your advice


 Alex

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 PLEASE do read the posting guide
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Re: [R] Efficient way to use data frame of indices to initialize matrix



On Dec 7, 2010, at 1:49 PM, Greg Snow wrote:


tmpdf - data.frame( x = c(1,2,3), y=c(2,3,1), a=c(10,20,30) )
mymat - matrix(0, ncol=3, nrow=3)
mymat[ as.matrix(tmpdf[,c('x','y')]) ] - tmpdf$a


cbind is also useful for assembly of arguments to the  matrix-`[-`  
function:


tmpdf - data.frame( x = c(1,2,3), y=c(2,3,1), a=c(10,20,30) )
 mymat - matrix(NA, ncol=max(tmpdf$y), nrow=max(tmpdf$x))
 mymat[ cbind(tmpdf$x,tmpdf$y) ] - tmpdf$a

 mymat
 [,1] [,2] [,3]
[1,]   NA   10   NA
[2,]   NA   NA   20
[3,]   30   NA   NA



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



-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-
project.org] On Behalf Of Cutler, Gene
Sent: Tuesday, December 07, 2010 11:31 AM
To: r-help@r-project.org
Subject: [R] Efficient way to use data frame of indices to initialize
matrix

I have a data frame with three columns, x, y, and a.  I want to  
create

a matrix from these values such that for matrix m:
m[x,y] == a

Obviously, I can go row by row through the data frame and insert the
value a at the correct x,y location in the matrix.  I can make that
slightly more efficient (perhaps), by doing something like this:

for (each.x in unique(df$x)) m[each.x, df$y[df$x == each.x]] -

df$a[df$x == each.x]

But I feel that there must be a more efficient, or at least more
elegant way to do this.

--
Gene

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David Winsemius, MD
West Hartford, CT

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Re: [R] please show me simple example how to plot Distance-Weighted Least Squares fitting

2010-12-07 Thread Sarah Goslee

It's impossible to give you really good advice without an example of
your data, the R code you used and the error message.

But here's a simple scatterplot example:

fakedata - data.frame(x=runif(15), y=runif(15))
plot(fakedata$x, fakedata$y)

or since you mentioned a line:

fakedata - fakedata[order(fakedata$x),]
plot(fakedata$x, fakedata$y, type=l)

For anything more, we would need the information requested in the
R-help posting guide linked at the bottom of each and every R-help
message.

Sarah

On Tue, Dec 7, 2010 at 12:17 PM, madr madra...@interia.pl wrote:

 I got simple x,y pairs of data and simple scatterplot and just cannot figure
 how to do it , there are many examples but always there is error popping out

 please show me an example stripped with additional data just core of what I
 need to do to get this damn line

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-- 
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http://www.functionaldiversity.org

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[R] R problem

2010-12-07 Thread Bruce John

I'm using R version 2.12.0

I'm trying to open a stata file (.dta); but I keep getting the message:

Error in read.dta(c:/p4subset_biology.dta) :
Calloc could not allocate (-603966867 of 4) memory

What does this mean?

The file should contain lots of physiological data. 

Thank you for your help.

-Bruce

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Re: [R] Creating binary variable depending on strings of two dataframes

On Tue, Dec 7, 2010 at 11:30 AM, Pete Pete noxyp...@gmail.com wrote:

 Hi,
 consider the following two dataframes:
 x1=c(232,3454,3455,342,13)
 x2=c(1,1,1,0,0)
 data1=data.frame(x1,x2)

 y1=c(232,232,3454,3454,3455,342,13,13,13,13)
 y2=c(E1,F3,F5,E1,E2,H4,F8,G3,E1,H2)
 data2=data.frame(y1,y2)

 I need a new column in dataframe data1 (x3), which is either 0 or 1
 depending if the value E1 in y2 of data2 is true while x1=y1. The result
 of data1 should look like this:
   x1     x2 x3
 1 232   1   1
 2 3454 1   1
 3 3455 1   0
 4 342   0   0
 5 13     0   1

 I think a SQL command could help me but I am too inexperienced with it to
 get there.


Try this:

 library(sqldf)
 sqldf(select x1, x2, max(y2 = 'E1') x3 from data1 d1 left join data2 d2 on 
 (x1 = y1) group by x1, x2 order by d1.rowid)
x1 x2 x3
1  232  1  1
2 3454  1  1
3 3455  1  0
4  342  0  0
5   13  0  1


-- 
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GKX Group, GKX Associates Inc.
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email: ggrothendieck at gmail.com

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[R] help on timeseries

2010-12-07 Thread cameron


i have time series of momentum signal.  I want to get the date of each of the
-1  signal period.  for example , the first period of -1 signal begins on
2005-9-21 and ends on 2005-9-28.  2nd period of -1 signal begins on
2005-09-30 and ends on 2005-10-28.


Thx
Cameron

 datePx   200MA Signals
2005-09-15   26.27  25.838651
2005-09-16   26.07  25.832751
2005-09-19   26.00  25.827301
2005-09-20   25.84  25.820351
2005-09-21   25.49  25.81115   -1
2005-09-22   25.34  25.80250   -1
2005-09-23   25.27  25.79205   -1
2005-09-26   25.27  25.78225   -1
2005-09-27   25.34  25.77355   -1
2005-09-28   25.67  25.76565   -1
2005-09-29   25.94  25.759201
2005-09-30   25.73  25.75230   -1
2005-10-03   25.50  25.74400   -1
2005-10-04   24.98  25.73410   -1
2005-10-05   24.67  25.72270   -1
2005-10-06   24.73  25.71100   -1
2005-10-07   24.59  25.69910   -1
2005-10-10   24.46  25.68635   -1
2005-10-11   24.41  25.67415   -1
2005-10-12   24.30  25.66090   -1
2005-10-13   24.59  25.64935   -1
2005-10-14   24.67  25.63890   -1
2005-10-17   24.53  25.62795   -1
2005-10-18   24.57  25.61710   -1
2005-10-19   25.09  25.60835   -1
2005-10-20   24.79  25.59840   -1
2005-10-21   24.78  25.58855   -1
2005-10-24   25.10  25.58070   -1
2005-10-25   25.03  25.57185   -1
2005-10-26   25.11  25.56375   -1
2005-10-27   24.85  25.55410   -1
2005-10-28   25.53  25.55040   -1
2005-10-31   25.70  25.548301
2005-11-01   25.96  25.546501
2005-11-02   26.46  25.548901
2005-11-03   26.44  25.551801
2005-11-04   26.66  25.556851

-code

library(tseries) 
library(timeSeries)

msft - as.timeSeries(get.hist.quote(instrument=MSFT, 
start=1986-03-31,
end=2008-09-10, quote=c(C), provider=yahoo, retclass=zoo)) 
msft - cbind(msft,as.timeSeries(rollapply(as.zoo(msft
),width=200,mean,align=right)))
msft - cbind(msft,NA)
msft[,3] - (ifelse(msft[,1]msft[,2],1,-1))

print(msft)

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[R] Changing names of a string variable

2010-12-07 Thread Jahan

Hello,
Here is the data set I am working with:
dat=read.csv('http://dl.dropbox.com/u/1852742/relexpressions.csv')
names(dat)

#Under the 'target' column, I want to change all of the values into
all capital letters (e.g. fgf2 becomes FGF2).  I have taken a loop
approach but I think my Python background is confusing me.  Here is
what I attempted:

for(line in dat[,1]){if(line=='fgf2'){line-'FGF2'}}
#I made sure that it was looping through each line by entering:
for(line in dat[,1]){if(line=='fgf2'){print(line)}} #Which did exactly
what I thought it would.

#This doesn't change anything, though.  Also, after reading the
documentation I am still confused how to add multiple if
statements.  I would need that because I need to change the names of
fgf2, fgf7, and fgf10.

#I am sure there is a built-in function to do what I need, but I
haven't found it yet so I tried this loop method.  Any help is greatly
appreciated!

Jahan

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Re: [R] Dataframe from list of similar lists: not _a_ way, but _the best_ way

2010-12-07 Thread Brian Diggs


On 12/7/2010 1:03 AM, Nick Sabbe wrote:

Hi All.

I often find myself in this situation:

. Based on some vector (or list) of values, I need to calculate a
few new values for each of them, where some of the new values are numbers,
but some are more of descriptive nature (so: character strings)

. So I use e.g. sapply, passing a custom function that returns a
list with all the calculated values

. The result of this is: a list (=the return value of sapply) of
lists, that all have the same kind of named values

A silly example:

list.of.lists-sapply(1:10, function(nr){list(org=nr,
chr=as.character(nr))})


Actually, this is not a list of lists, but rather a list of vectors with 
dimensions.  I didn't know such a thing existed, but obviously it does.



It seems rather obvious that the result would be better structured as a
dataframe.

Now I know a few ways to do this (using do.call), but I fear most of these
are rather bad in performance: I suspect all the data is being repetitively
copied which may be slow.

So, my question to the specialists:

. Is the above way of working reasonable for this kind of problem?
Or would you suggest otherwise?

. What would be the best (as in: quickest) way of transforming this
list of lists to a dataframe? The answer to this is probably based upon
knowledge of the inner workings of R? Or is there any way in which this
depends on the specifics of my function (for nontrivial functions and list
sizes)?


I don't know that this is best (in terms of fastest and/or least memory 
usage), but to me the following is best in that it hands off the 
problem to a package that is designed to handle such problems, so 
presumably does a better job than any one-off approach.


library(plyr)

DF - ldply(1:10, function(nr){data.frame(org=nr, chr=as.character(nr))})

Note that the internal function returns a data.frame rather than a list, 
and the *dply functions automatically stitch the individual data.frames 
together.  Check out the documentation to the plyr package.



Thanks!

Nick Sabbe

--
ping: nick.sa...@ugent.be
link:http://biomath.ugent.be/  http://biomath.ugent.be
wink: A1.056, Coupure Links 653, 9000 Gent
ring: 09/264.59.36
-- Do Not Disapprove


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

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Re: [R] how to get vector of data from line ?


well, I found id the simplest answer is:

d- approx(density(x),xout=...)
where xout is a vector of the points where the data should be sampled
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Re: [R] Changing names of a string variable

2010-12-07 Thread Erik Iverson


Hello,

To do what you want, see ?toupper :

levels(dat$target) - toupper(levels(dat$target))

However, for clarity,

dat$target is not a string variable, it is a factor,
which you can verify with

 str(dat)

Factors are enumerated types, and have a discrete
set of 'levels' associated with them,
which can be queried with the ?levels function.

'Behind the scenes', R is storing this variable
as numeric codes, not the character string associated
with each numeric code.

This contrasts with character vectors.


--Erik

Jahan wrote:

Hello,
Here is the data set I am working with:
dat=read.csv('http://dl.dropbox.com/u/1852742/relexpressions.csv')
names(dat)

#Under the 'target' column, I want to change all of the values into
all capital letters (e.g. fgf2 becomes FGF2).  I have taken a loop
approach but I think my Python background is confusing me.  Here is
what I attempted:

for(line in dat[,1]){if(line=='fgf2'){line-'FGF2'}}
#I made sure that it was looping through each line by entering:
for(line in dat[,1]){if(line=='fgf2'){print(line)}} #Which did exactly
what I thought it would.

#This doesn't change anything, though.  Also, after reading the
documentation I am still confused how to add multiple if
statements.  I would need that because I need to change the names of
fgf2, fgf7, and fgf10.

#I am sure there is a built-in function to do what I need, but I
haven't found it yet so I tried this loop method.  Any help is greatly
appreciated!

Jahan

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Re: [R] Changing names of a string variable

2010-12-07 Thread Phil Spector


Jahan -
   Try

dat$target = toupper(dat$target)

I would not recommend a loop for something 
like this.


(You might also want to brush up on your python,
because what you're trying doesn't work in 
python either.)


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




On Tue, 7 Dec 2010, Jahan wrote:


Hello,
Here is the data set I am working with:
dat=read.csv('http://dl.dropbox.com/u/1852742/relexpressions.csv')
names(dat)

#Under the 'target' column, I want to change all of the values into
all capital letters (e.g. fgf2 becomes FGF2).  I have taken a loop
approach but I think my Python background is confusing me.  Here is
what I attempted:

for(line in dat[,1]){if(line=='fgf2'){line-'FGF2'}}
#I made sure that it was looping through each line by entering:
for(line in dat[,1]){if(line=='fgf2'){print(line)}} #Which did exactly
what I thought it would.

#This doesn't change anything, though.  Also, after reading the
documentation I am still confused how to add multiple if
statements.  I would need that because I need to change the names of
fgf2, fgf7, and fgf10.

#I am sure there is a built-in function to do what I need, but I
haven't found it yet so I tried this loop method.  Any help is greatly
appreciated!

Jahan

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Re: [R] increase or decrease variable by 1

2010-12-07 Thread Bert Gunter

Ted:

Inline below...

On Tue, Dec 7, 2010 at 8:42 AM, Ted Harding ted.hard...@wlandres.net wrote:
 Indeed!

  x - x + 1

 (and being generous with unnecessary spaces) uses 10 characters.

  `+`(x)-1

 (being mean with them) uses 9. The mean version of the first
 uses only 6: x-x+1

 However, I suppose there is merit in the spiritual exercise
 of contemplating how `+`(x)-1 gets worked out!

AFAICS it doesn't.
 `+`(x)-1
Error in +x - 1 : could not find function +-

`+`(x,1) does:

 `+`(x,1)
[1] 4

-- Bert




 Ted.

 On 07-Dec-10 16:23:17, Ivan Calandra wrote:
 x+1 is not that complicated... Am I missing something here?

 Le 12/7/2010 16:55, Gabor Grothendieck a écrit :
 On Tue, Dec 7, 2010 at 8:43 AM, madrmadra...@interia.pl  wrote:
 many languages have shorthands for that operation like:

 variable += 1
 or
 ++variable

 is there something like that in R ?

 You can do this:

 x- 3
 `+`(x)- 1
 x
 [1] 4
 --
 Ivan CALANDRA

 
 E-Mail: (Ted Harding) ted.hard...@wlandres.net
 Fax-to-email: +44 (0)870 094 0861
 Date: 07-Dec-10                                       Time: 16:42:44
 -- XFMail --

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-- 
Bert Gunter
Genentech Nonclinical Biostatistics

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Re: [R] Changing names of a string variable

2010-12-07 Thread Erik Iverson




Phil Spector wrote:

Jahan -
   Try

dat$target = toupper(dat$target)



Note that in this case, the above *will* coerce dat$target
to a character vector, which may or may not be what
is intended.

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Re: [R] Changing names of a string variable



On Dec 7, 2010, at 2:31 PM, Erik Iverson wrote:


Hello,

To do what you want, see ?toupper :

levels(dat$target) - toupper(levels(dat$target))

However, for clarity,

dat$target is not a string variable, it is a factor,
which you can verify with

 str(dat)


And to further clarify, it is a factor because of the default behavior  
of read.table and data.frame in creating factors when offered streams  
of character data. The stringsAsFactors=FALSE argument can be used  
within these functions or can be set globally with options().


--
David.



Factors are enumerated types, and have a discrete
set of 'levels' associated with them,
which can be queried with the ?levels function.

'Behind the scenes', R is storing this variable
as numeric codes, not the character string associated
with each numeric code.

This contrasts with character vectors.


--Erik

Jahan wrote:

Hello,
Here is the data set I am working with:
dat=read.csv('http://dl.dropbox.com/u/1852742/relexpressions.csv')
names(dat)
#Under the 'target' column, I want to change all of the values into
all capital letters (e.g. fgf2 becomes FGF2).  I have taken a loop
approach but I think my Python background is confusing me.  Here is
what I attempted:
for(line in dat[,1]){if(line=='fgf2'){line-'FGF2'}}
#I made sure that it was looping through each line by entering:
for(line in dat[,1]){if(line=='fgf2'){print(line)}} #Which did  
exactly

what I thought it would.
#This doesn't change anything, though.  Also, after reading the
documentation I am still confused how to add multiple if
statements.  I would need that because I need to change the names of
fgf2, fgf7, and fgf10.
#I am sure there is a built-in function to do what I need, but I
haven't found it yet so I tried this loop method.  Any help is  
greatly

appreciated!
Jahan



David Winsemius, MD
West Hartford, CT

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[R] More elegant magnitude method

2010-12-07 Thread Jonathan P Daily

I have a need to find the order of number to get a scaling parameter as a 
power of 10. I have a function that works *so far*, but it is ugly and 
probably buggy. In the interest of avoiding code-based outliers in my 
data, I thought I would ask if anyone here has a better way.

 scl - function(x){
+ length(charToRaw(format(trunc(x), scientific = F)))-1}
 a - 123456789
 b - 1E15
 c -  12.345
 scl(a)
[1] 8
 scl(b)
[1] 15
 scl(c)
[1] 1

Thanks
--
Jonathan P. Daily
Technician - USGS Leetown Science Center
11649 Leetown Road
Kearneysville WV, 25430
(304) 724-4480
Is the room still a room when its empty? Does the room,
 the thing itself have purpose? Or do we, what's the word... imbue it.
 - Jubal Early, Firefly

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Re: [R] fast subsetting of lists in lists

2010-12-07 Thread Bert Gunter

Alexander:

I'm not sure exactly what you want, so the following may be irrelevant...

BUT, noting that data frames ARE lists and IF what you have can then
be abstracted as lists of lists of lists of ... to various depths
AND IF what you want is just to pick out and combined all named
vectors (which could be  columns of data frames) with a given name at
whatever depth they they appear in the lists
THEN it is natural to do this recursively as follows:

vaccuum - function(x,nm)
## x is a named list (of lists of ...)
## nm in the name searched for
{
  y - NULL
  for(nmx in names(x)) y -  c(y,
  {
z - x[[nmx]]
if(nmx==nm)z
else if(is.list(z))Recall(z,nm)
})
 y
}

##Example

 test - list(a=1:3, be = list(a=4:6,c=data.frame(a=10:15,b=I(letters[1:6]

 vaccuum(test,b)
[1] a b c d e f

 vaccuum(test,a)
 [1]  1  2  3  4  5  6 10 11 12 13 14 15

caveat: If perchance this is (at least close to) what you want, it is
likely to be rather inefficient. It also needs some robustifying.

Cheers,
Bert









On Tue, Dec 7, 2010 at 10:54 AM, Alexander Senger
sen...@physik.hu-berlin.de wrote:
 Hello,

 Matthew's hint is interesting:

 Am 07.12.2010 19:16, schrieb Matthew Dowle:
 Hello Alex,

 Assuming it was just an inadequate example (since a data.frame would suffice
 in that case), did you know that a data.frames' columns do not have to be
 vectors but can be lists?  I don't know if that helps.

 DF = data.frame(a=1:3)
 DF$b = list(pi, 2:3, letters[1:5])
 DF
   a             b
 1 1      3.141593
 2 2          2, 3
 3 3 a, b, c, d, e
 DF$b
 [[1]]
 [1] 3.141593

 [[2]]
 [1] 2 3

 [[3]]
 [1] a b c d e
 sapply(DF,class)
         a         b
 integer    list


 That is still regular though in the sense that each row has a value for all
 the columns, even if that value is NA, or NULL in lists.

 My data is mostly regular, that is every sublist contains a data.frame
 which is the major contribution to overall size. The reason I use lists
 is mainly that I need also some bits of information about the
 environment. I thought about putting these into additional columns of
 the data.frame (and add redundancy and maybe 30% of overhead this way),
 one column per variable. But as memory usage is already close to the
 limit of my machine this might break things (the situation is a bit
 tricky, isn't it?).
 I didn't know that a column of a data.frame can be a list. So if I need
 only let's say 10 entries in that list, but my data.frame has several
 hundred rows, would the empty parts of the column-list be filled
 with cycled values or would they be really empty and thus not use
 additional memory?
 Secondly as I mentioned in another email to this topic: a whole day of
 data contains about 100 chunks of data that is 100 of the sublists
 described above. I could put them all into one large data.frame, but
 then I would have to extract the environmental data from the long
 list, now containing repeated occurrences of variables with the same
 name. I guess subsetting could become tricky here (dependend on name and
 position, I assume), but I'm eager to learn an easy way of doing so.

 Sorry for not submitting an illustrative example, but I'm afraid that
 would be quite lengthy and not so illustrative any more.

 The data.table mentioned below seems to be an interesting alternative;
 I'll definitely look into this. But it would also mean quite a bit of
 homework, as far as I can see...

 Thanks

 Alex


 If your data is not regular then one option is to flatten it into
 (row,column,value) tuple similar to how sparse matrices are stored.  Your
 value column may be list rather than vector.

 Then (and yes you guessed this was coming) ... you can use data.table to
 query the flat structure quickly by setting a key on the first two columns,
 or maybe just the 2nd column when you need to pick out the values for one
 'column' quickly for all 'rows'.

 There was a thread about using list() columns in data.table here :

 http://r.789695.n4.nabble.com/Suggest-a-cool-feature-Use-data-table-like-a-sorted-indexed-data-list-tp2544213p2544213.html

 Does someone now a trick to do the same as above with the faster built-in
 subsetting? Something like:
 test[somesubsettingmagic]

 So in data.table if you wanted all the 'b' values,  you might do something
 like this :

 setkey(DT,column)
 DT[J(b), value]

 which should return the list() quickly from the irregular data.

 Matthew


 Alexander Senger sen...@physik.hu-berlin.de wrote in message
 news:4cfe6aee.6030...@physik.hu-berlin.de...
 Hello Gerrit, Gabor,


 thank you for your suggestion.

 Unfortunately unlist seems to be rather expensive. A short test with one
 of my datasets gives 0.01s for an extraction based on my approach and
 5.6s for unlist alone. The reason seems to be that unlist relies on
 lapply internally and does so recursively?

 Maybe there is still another way to go?

 Alex

 Am 07.12.2010 15:59, schrieb Gerrit Eichner:
 Hello, Alexander,

 does

 utest -

Re: [R] increase or decrease variable by 1

2010-12-07 Thread Duncan Murdoch


On 07/12/2010 12:42 PM, Gabor Grothendieck wrote:

On Tue, Dec 7, 2010 at 12:25 PM, Bert Guntergunter.ber...@gene.com  wrote:

Ted:

Inline below...

On Tue, Dec 7, 2010 at 8:42 AM, Ted Hardingted.hard...@wlandres.net  wrote:

Indeed!

  x- x + 1

(and being generous with unnecessary spaces) uses 10 characters.

  `+`(x)-1

(being mean with them) uses 9. The mean version of the first
uses only 6: x-x+1

However, I suppose there is merit in the spiritual exercise
of contemplating how `+`(x)-1 gets worked out!


AFAICS it doesn't.

`+`(x)-1

Error in +x- 1 : could not find function +-


Sorry, my code was missing the first line:


`+-`- `+`

x- 3
`+`(x)- 1
x

[1] 4


Note that there are at least two subtle differences between your code 
and x += 1 or ++x:


First, the value of `+`(x)- 1 is 1, i.e.

 print(`+`(x)- 1)

will give 1 regardless of x, unlike ++x.

Another difference is that  `+`(x)- 1 is equivalent to x - x + 1, and 
in R, that doesn't necessarily increment x:  the x on the right hand 
side might be a global variable, and the result of x + 1 will be 
assigned to a new local variable.  For example, the sequence


x - 3
f - function() `+`(x)- 1
f()
x

leaves x set to 3.  (But it temporarily created a new variable called x 
in the evaluation frame of f().)


I don't think the first difference is documented; I haven't checked the 
source to know if it's intentional.  (It generally makes sense that foo 
- bar has value bar; the problem is that your assignment doesn't 
follow the rule that it assigns bar to foo.)


The second one is a basic fact of life in R, and the source of a few 
bugs:  the only way around it that I can see would be to allow users to 
declare things about variables (e.g. x is a global variable, don't 
create a local when I assign to it!).


Duncan Murdoch

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Re: [R] More elegant magnitude method