[R] retrieving object names passed indirectly to a function

2010-09-15 Thread Michael Bedward
Hi folks,

I'm stuck with a problem that I suspect has a trivial solution...

I have a function, call it foo, that takes a number of arguments which
it uses as parameters for a simulation. For later analysis, foo stores
the names of the objects passed to it along with the simulation
results in its output (written to a database). The objects names are
accessed with deparse(substitute(argname)).

A second function, call it foo.driver, takes as input a data.frame
with cols corresponding to foo arguments containing the character
names of objects. For each row of the data.frame in turn, it retrieves
the required objects from the global environment and passes them to
function foo.

Where I am stuck is in how foo.driver should pass an object to foo
such that calling deparse(substitute(argname)) in foo will retrieve
the correct name (ie. that specified in foo.driver's input
data.frame).

Hopefully, the following toy code example illustrates what I'm trying to do...

# some objects in the global environment to provide param values to foo
p1.1 - 1
p1.2 - 2
p2.1 - 3
p2.2 - 4

# controlling table
arg.table - data.frame(p1=c(p1.1, p2.2), p2=c(p2.1, p2.2),
replicates=c(100, 100))

foo.driver - function( arg.table ) {
  for (i in 1:nrow(arg.table)) {
for (j in 1:arg.table$replicates[i]) {
  # somehow retrieve objects named in cols p1 and p2 and
  # pass them to foo with their original names
  # (e.g. some clever use of get and assign ?)
}
  }
}

foo - function(p1, p2) {
  # retrieve and store arg names
  p1name - deparse(substitute(p1))
  p2name - deparse(substitute(p2))

  # run simulation with values of p1 and p2 then
  # store results together with names of objects
  # passed as args to output database
}

My actual simulation function involves a large number of arguments of
various classes and if at all possible I prefer to avoid hacking it
(e.g to accept object names rather than objects).

Any tips much appreciated.

Michael

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[R] HDP and 99% contour lines

2010-09-15 Thread Peter Spencer
Dear all,
 
I have a very simple question about how I can include HPD confidence
lines in a Marginal posterior distribution scatterplot
I have the following code to draw the scatterplot(s) from an output
table;
 
par(mfrow=c(2,2))
 
plot(temp1[,5],  temp1[,6],  xlab=log10(r),
ylab=log10(tf), pch=46)
plot(temp2[,5],  temp2[,6],  xlab=log10(r),
ylab=log10(tf), pch=46)
plot(temp3[,5],  temp3[,6],  xlab=log10(r),
ylab=log10(tf), pch=46)
plot(temp4[,5],  temp4[,6],  xlab=log10(r),
ylab=log10(tf), pch=46)
 
dev.print(device=pdf, height=12, width=10,
file=Marginal.pdf)
 
I would very much like to include say 0.1, 0.5 and 0.99 contours onto
the graphs, but I can't figure it out using any examples of how to do
it.  Can someone offer any suggestions?
 
Many thanks
Peter
 
 

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Re: [R] Using McNemar's test to detect shifts in response from pre- topost-treatment

2010-09-15 Thread Peter Dalgaard
On 09/14/2010 08:58 PM, Robert Baer wrote:
 McNemar is good for paired data. See
 ?mcnemar.test
 
 You need to get your data into the form of a matrix (e.g., help example), 
 and you will need data organized by concordant and discordant pairing.
 
 This means that you will need to organize your data differently than you 
 show us:
 # pairs YES YES pre-post
 # pairs YES NO pre-post
 # pairs NO-YES pre-post
 # pairs NO-NO pre-post0
 
 Finally, you will need to decide whether to apply the usual continuity 
 correction with your sample size (TRUE by default).

Also notice that this is a group A vs. group B situation, whereas
McNemar is for whether there is a pre-post difference within a single group.

Curiously, there appears to be two different schools on how to handle
this sort of data:

(A) In cancer epidemiology you would say that this is like a matched
case-control study and that in a model with logit-additive effects of
pair and that the pre-post odds ratio is estimated by the ratio of the
off-diagonal terms. To compare two such ORs between two tables, it is a
fairly obvious idea to form the 2x2 table of the two sets of
off-diagonal terms and test for independence. Rather nicely, you can use
Fisher's exact test in small-sample situations, and its OR estimate is
really the ORR - the ratio of the two ORs.

(B) However, in clinical trial literature, people focus on the
probability _difference_ and correct that for correlations within pairs.
One advantage of this is that estimates have the same interpretation
whether the experiment was paired or unpaired. Technically, what you do
is little different from scoring +/-1 for increase, and 0 for no change,
and then compare the average scores with a t-test or a permutation test.

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

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Re: [R] extracting objects from lists

2010-09-15 Thread Roman Luštrik

You can explore the structure of the list with str(). If you use Eclipse, a
list of available options will be created for you as you type (see
http://imagepaste.nullnetwork.net/viewimage.php?id=1293).

Cheers,
Roman

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Re: [R] Multiple CPU HowTo in Linux?

2010-09-15 Thread Edwin Groot
Hello all,
Thanks for your input, and helping to clear things up on where to go.
I will try out the multicore package and see if there are further
bottlenecks. It looks like some loops might need special treatment with
parallelization.
I have been pampered with the excellent walk-through vignettes of the
packages I have used so far. The HPC package guides lacked something in
the practical aspects of their usage.
lapply - mclapply at the beginning of my script? Well, I never would
have thought of such a thing. Thanks!

I might be back on the list when I run out of physical RAM ;-)

Edwin
-- 
On Tue, 14 Sep 2010 12:11:17 -0700
 Martin Morgan mtmor...@fhcrc.org wrote:
 On 09/14/2010 08:36 AM, Christian Raschke wrote:
  Edwin,
  
  I'm not sure what you mean by adapting; other than installing
  multicore, there is nothing else to set up. How and whether you
 could
  then parallelise your code strongly depends on the specific problem
 you
  are facing.
  
  What have done in the past was to look at the source of the
 functions
  from whatever package I was using that produced the bottleneck. If
 what
  is taking the longest time is actually embarrassingly parallel,
  mclapply() from package multicore can help. In the simplest case
 you
  could simply replace lapply() in the with an appropriate
 mclapply().
  Check out ?mclapply. But then again you might have to get a little
 more
  creative, depending on exactly what in the code is taking so long
 to
  run. If your problem is inherently sequential then even multicore
 won't
  help.
  
  Christian
  
  On 09/14/2010 09:35 AM, Edwin Groot wrote:
  Hello Cedrick,
  Ah, yes, that looks like it would apply to my situation. I was
  previously reading on snow, which is tailored for clusters, rather
 than
  a single desktop computer.
  Anyone with experience adapting multicore to an R-script?
  I have to admit I know little about parallel processing,
  multiprocessing and cluster processing.
 
  Edwin
 
  On Tue, 14 Sep 2010 10:15:42 -0400
Johnson, Cedrick W.cedr...@cedrickjohnson.com  wrote:

 ?multicore perhaps
 
  On 09/14/2010 10:01 AM, Edwin Groot wrote:
  
  Hello all,
  I upgraded my R workstation, and to my dismay, only one core
 
  appears to
  
  be used during intensive computation of a bioconductor function.
 
 Hi Edwin -- Since you have a Bioconductor package,  you might ask on
 the
 Bioconductor list, as the authors of some computationally intensive
 tasks have provided facilities for relatively transparent use of,
 e.g.,
 multicore or Rmpi. In ShortRead, for instance, loading multicore is
 enough to distribute some tasks across cores, and the srapply
 function
 can help (or not; things might be as easy as lapply - mclapply at
 the
 top of your script) with your own lapply-like code.
 
 http://bioconductor.org/help/mailing-list/
 
 Martin
  What I have now is two dual-core Xeon 5160 CPUs and 10 GB RAM.
 When
 
  I
  
  fully load it, top reports about 25% user, 75% idle and 0.98
 
  short-term
  
  load.
  The archives gave nothing helpful besides mention of snow. I
 
  thought of
  
  posting to HPC, but this system is fairly modest WRT processing
 
  power.
  
  Any pointers of where to start?
  ---
  #Not running anything at the moment

  sessionInfo()
   
  R version 2.11.1 (2010-05-31)
  x86_64-pc-linux-gnu
 
  locale:
 [1] LC_CTYPE=en_GB.UTF-8   LC_NUMERIC=C
 [3] LC_TIME=en_GB.UTF-8LC_COLLATE=en_GB.UTF-8
 [5] LC_MONETARY=C  LC_MESSAGES=en_GB.UTF-8
 [7] LC_PAPER=en_GB.UTF-8   LC_NAME=C
 [9] LC_ADDRESS=C   LC_TELEPHONE=C
  [11] LC_MEASUREMENT=en_GB.UTF-8 LC_IDENTIFICATION=C
 
  attached base packages:
  [1] stats graphics  grDevices utils datasets  methods
 
 base
  
 
  loaded via a namespace (and not attached):
  [1] tools_2.11.1
  ---
  $ uname -a
  Linux laux29 2.6.26-2-amd64 #1 SMP Sun Jun 20 20:16:30 UTC 2010
 
  x86_64
  
  GNU/Linux
  ---
  Thanks for your help,
  Edwin
 
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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.
   
  Dr. Edwin Groot, postdoctoral associate
  AG Laux
  Institut fuer Biologie III
  Schaenzlestr. 1
  79104 Freiburg, Deutschland
  +49 761-2032945
 
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 PLEASE do read the posting 

Re: [R] Problem with cat() == A related question

2010-09-15 Thread Martin Maechler
 William Revelle li...@revelle.net
 on Tue, 14 Sep 2010 14:52:32 -0500 writes:

 At 3:45 PM -0400 9/14/10, jim holtman wrote:
 The problem is the 'cat' enclosing the 'print'; just get rid of the
 'cat' -- that is what is causing the extra output
 
 On Tue, Sep 14, 2010 at 3:34 PM, Peng, C cpeng@gmail.com wrote:
 
 It is still visible even it is set invisible(NULL):
 
 fn1 - function(n = 5){
 +  mat - matrix(rnorm(5*5), 5, 5)
 +   cat(print(mat))
 +  invisible(NULL)}
 fn1()
 [,1][,2]   [,3][,4]   [,5]
 [1,] -1.22767085 -1.41468587 -2.0156231  0.29732942  0.5755600
 [2,] -0.16775996 -0.03780596 -0.9461079  0.91289175  0.1254273
 [3,]  0.09696032 -0.75522210 -0.7494442 -0.21341669  1.7088194
 [4,]  0.13535505 -1.09011005 -0.6074198  0.05342614 -1.1996344
 [5,]  0.66474083 -2.62206248  0.1329972  0.06132865  0.5124778
 -1.227671 -0.1677600 0.09696032 0.1353550 0.6647408 -1.414686 
-0.03780596
 -0.7552221 -1.09011 -2.622062 -2.015623 -0.9461079 -0.7494442 -0.6074198
 0.1329972 0.2973294 0.9128917 -0.2134167 0.05342614 0.06132865 0.57556
 0.1254273 1.708819 -1.199634 0.5124778
 
 --
 View this message in context: 
 http://r.789695.n4.nabble.com/Problem-with-cat-tp2538811p2539551.html
 Sent from the R help mailing list archive at Nabble.com.
 


 I think that what Peng was trying to do was print it without the 
 column and row names

 mat -  matrix(1:25,5,5)
 colnames(mat) - rownames(mat) - rep(,5)
 print(mat)
  
 1  6 11 16 21
 2  7 12 17 22
 3  8 13 18 23
 4  9 14 19 24
 5 10 15 20 25

Yes.

A (slightly more general and efficient) version is

  sfsmisc::empty.dimnames( matrix(1:40, 5,8) )

  1  6 11 16 21 26 31 36
  2  7 12 17 22 27 32 37
  3  8 13 18 23 28 33 38
  4  9 14 19 24 29 34 39
  5 10 15 20 25 30 35 40
  

--- where the  empty.dimnames() function is just one small reason to
  get the sfsmisc package ;-)

Martin Maechler,
ETH Zurich

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Re: [R] Problems with pdf device using plot glht function on multcomp library.

2010-09-15 Thread Torsten Hothorn


Dear Ken,

it works for me on Linux. However, the print() commands around plot() are 
not necessary.


Torsten

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[R] difftime problems with daylight saving time?

2010-09-15 Thread ravi
Hi,
I seem to be having problems in calculating the period between two times, one 
with and the other without daylight saving time. Consider the following command 
:

 as.numeric(difftime(as.POSIXct(2010-04-01 06:00:00 
CEST),as.POSIXct(2010-03-03 08:27:00 CET),units=hours))
[1] 692.55

No problem here. Strangely enough, I seem to be having problems in using the 
same command when applied to a data frame. Let me explain with the following 
code :
df-structure(list(d1 = structure(c(1263279420, 1263885000, 1264492200,
1265094120, 1265784900, 1266391860, 1266908220, 1267515240, 1268117220,
1268722020, 1269327240, 1269841740, 1270532400, 1271138400, 1271833380,
1272349740, 1272953820, 1273470060, 1274075640, 1274768100, 1275285060,
1275976500, 1276494480, 1277100300, 1277703480, 1278309420, 1278913620,
NA, NA, NA, NA, 1282198020, 1282630920, 1283149380), class = c(POSIXt,
POSIXct), tzone = ), d2 = structure(c(1263367500, 1263972540,
1264578600, 1265182740, 1265872260, 1266479460, 1266996720, 1267601220,
1268204100, 1268811960, 1269415200, 1269929340, 1270620300, 1271224800,
1271918520, 1272436020, 1273045800, 1273558740, 1274163720, 1274854140,
1275373620, 1276062900, 1276583400, 1277187060, 1277792100, 1278397800,
1279002840, NA, NA, NA, NA, 1282285380, 1282718460, 1283237940
), class = c(POSIXt, POSIXct), tzone = )), .Names = c(d1,
d2), class = data.frame, row.names = c(NA, -34L))
# get the first of corresponding month for d1
g1-as.POSIXlt(df$d1,format=%Y-%m-%d %H:%M:%S)
g1$mday-1;g1$hour-6;g1$min-0;
df$m1-as.POSIXct(g1)
# period between d1 and m1
df$nh1-with(df,round(as.numeric(difftime(d1,m1,units=hours)),1))
# get the first of next month for d2
g1$mon-g1$mon+1
df$m2-g1
#period between m2 and d2
df$nh2-with(df,round(as.numeric(difftime(m2,d2,units=hours)),1))
str(df)

I get the following result :
 df
    d1  d2  m1   
nh1  m2   nh2
1  2010-01-12 07:57:00 2010-01-13 08:25:00 2010-01-01 06:00:00 265.9 2010-02-01 
06:00:00 453.6
2  2010-01-19 08:10:00 2010-01-20 08:29:00 2010-01-01 06:00:00 434.2 2010-02-01 
06:00:00 285.5
3  2010-01-26 08:50:00 2010-01-27 08:50:00 2010-01-01 06:00:00 602.8 2010-02-01 
06:00:00 117.2
4  2010-02-02 08:02:00 2010-02-03 08:39:00 2010-02-01 06:00:00  26.0 2010-03-01 
06:00:00 621.4
5  2010-02-10 07:55:00 2010-02-11 08:11:00 2010-02-01 06:00:00 217.9 2010-03-01 
06:00:00 429.8
6  2010-02-17 08:31:00 2010-02-18 08:51:00 2010-02-01 06:00:00 386.5 2010-03-01 
06:00:00 261.1
7  2010-02-23 07:57:00 2010-02-24 08:32:00 2010-02-01 06:00:00 530.0 2010-03-01 
06:00:00 117.5
8  2010-03-02 08:34:00 2010-03-03 08:27:00 2010-03-01 06:00:00  26.6 2010-04-01 
06:00:00    NA
9  2010-03-09 07:47:00 2010-03-10 07:55:00 2010-03-01 06:00:00 193.8 2010-04-01 
06:00:00    NA
10 2010-03-16 07:47:00 2010-03-17 08:46:00 2010-03-01 06:00:00 361.8 2010-04-01 
06:00:00    NA
11 2010-03-23 07:54:00 2010-03-24 08:20:00 2010-03-01 06:00:00 529.9 2010-04-01 
06:00:00    NA
12 2010-03-29 07:49:00 2010-03-30 08:09:00    NA    NA 2010-04-01 
06:00:00  45.9
13 2010-04-06 07:40:00 2010-04-07 08:05:00 2010-04-01 06:00:00 121.7 2010-05-01 
06:00:00 573.9
14 2010-04-13 08:00:00 2010-04-14 08:00:00 2010-04-01 06:00:00 290.0 2010-05-01 
06:00:00 406.0
15 2010-04-21 09:03:00 2010-04-22 08:42:00 2010-04-01 06:00:00 483.1 2010-05-01 
06:00:00 213.3
16 2010-04-27 08:29:00 2010-04-28 08:27:00 2010-04-01 06:00:00 626.5 2010-05-01 
06:00:00  69.5
17 2010-05-04 08:17:00 2010-05-05 09:50:00 2010-05-01 06:00:00  74.3 2010-06-01 
06:00:00 644.2
18 2010-05-10 07:41:00 2010-05-11 08:19:00 2010-05-01 06:00:00 217.7 2010-06-01 
06:00:00 501.7
19 2010-05-17 07:54:00 2010-05-18 08:22:00 2010-05-01 06:00:00 385.9 2010-06-01 
06:00:00 333.6
20 2010-05-25 08:15:00 2010-05-26 08:09:00 2010-05-01 06:00:00 578.2 2010-06-01 
06:00:00 141.8
21 2010-05-31 07:51:00 2010-06-01 08:27:00 2010-05-01 06:00:00 721.9 2010-06-01 
06:00:00  -2.5
22 2010-06-08 07:55:00 2010-06-09 07:55:00 2010-06-01 06:00:00 169.9 2010-07-01 
06:00:00 526.1
23 2010-06-14 07:48:00 2010-06-15 08:30:00 2010-06-01 06:00:00 313.8 2010-07-01 
06:00:00 381.5
24 2010-06-21 08:05:00 2010-06-22 08:11:00 2010-06-01 06:00:00 482.1 2010-07-01 
06:00:00 213.8
25 2010-06-28 07:38:00 2010-06-29 08:15:00 2010-06-01 06:00:00 649.6 2010-07-01 
06:00:00  45.8
26 2010-07-05 07:57:00 2010-07-06 08:30:00 2010-07-01 06:00:00  98.0 2010-08-01 
06:00:00 621.5
27 2010-07-12 07:47:00 2010-07-13 08:34:00 2010-07-01 06:00:00 265.8 2010-08-01 
06:00:00 453.4
28    NA    NA    NA    
NA    NA    NA
29    NA    NA    NA    
NA    NA    NA
30    NA    NA    NA    
NA    NA    NA
31    NA    NA    NA    
NA    NA    NA
32 2010-08-19 08:07:00 2010-08-20 08:23:00 2010-08-01 06:00:00 434.1 2010-09-01 
06:00:00 285.6
33 2010-08-24 08:22:00 

Re: [R] Package build and install under Windows

2010-09-15 Thread Uwe Ligges



On 14.09.2010 21:05, Tao wrote:


Any other method besides using RTOOLS? I am using R 2.10.1, which version of
Rtools should I use?


The most recent one that supports R-2.10.1? According to the webpage 
this is Rtools211.exe.


 It seems that there are two versions on the website.

The commands like R CMD build should be used in the CMD window, right?
Thanks.


Actually on your OS's command shell.

Uwe Ligges

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[R] Get File Names in Folder, Read Files, Update, and Write

2010-09-15 Thread Suphajak Ngamlak
Dear All,

 

Could you please recommend how I can do this?

 

I have several text files in one folder. Let's name them A0801.RSK,
A0802.RSK, 

I would like R to 

1)  Know all file names in this folder

2)  Update value in one column of these files

3)  Write results in another text file with _xval in the file names

 

Below is R code for read, update, and write one file

 

Import-C:/A0810.RSK

Table-read.table(file= Import, sep = ,, head=TRUE, na.strings = NA)

Table$VALUE -0

Export-C:/A_XVal0810.RSK

write.table(Table, file= Export, sep = ,, col.names = TRUE)

 

Thank you

Best Regards, 

Suphajak Ngamlak
Equity and Derivatives Trading
Phatra Securities Public Company Limited
Tel: +662-305-9179  
Email: supha...@phatrasecurities.com 


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Re: [R] multiple cores/building fails

2010-09-15 Thread Uwe Ligges



On 14.09.2010 21:50, Gregory Ryslik wrote:

Hi Everyone,

I have written debugged and tested my code and it works (yay!). I recently tried to 
parallelize it (1 core/per tree) and when I run the code I get an error saying the object 
control was not found. I do have an object control in my code but it always 
works fine as long as I am not running it on multiple cores (which I am doing via the 
help of sleigh and the 'nws' package).

Think that this could be an error in my source statements, I tried to build 
my package (with the hope of replace source with library()) and I get the following 
(where XXX is my package name that I've removed for privacy reasons):

* creating vignettes ... ERROR

Error: processing vignette '.Rnw' failed with diagnostics:
  chunk 4
Error : object control not found

Again it gives me that this control object is not found.

Does anyone have an idea how I can isolate what's going on. It's slightly tough 
to do because it all works fine in the single processor case!



Perhaps you forgot to populate the object to the client nodes?
Hard to say without any reproducible example code.

Uwe Ligges



Thanks!

Kind regards,
Greg
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Re: [R] Get File Names in Folder, Read Files, Update, and Write

2010-09-15 Thread Uwe Ligges



On 15.09.2010 10:55, Suphajak Ngamlak wrote:

Dear All,



Could you please recommend how I can do this?



I have several text files in one folder. Let's name them A0801.RSK,
A0802.RSK, 

I would like R to

1)  Know all file names in this folder



?list.files



2)  Update value in one column of these files
3)  Write results in another text file with _xval in the file names



?paste

Uwe Ligges






Below is R code for read, update, and write one file



Import-C:/A0810.RSK

Table-read.table(file= Import, sep = ,, head=TRUE, na.strings = NA)

Table$VALUE-0

Export-C:/A_XVal0810.RSK

write.table(Table, file= Export, sep = ,, col.names = TRUE)



Thank you

Best Regards,

Suphajak Ngamlak
Equity and Derivatives Trading
Phatra Securities Public Company Limited
Tel: +662-305-9179
Email: supha...@phatrasecurities.com


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Re: [R] converting string vector to integer/numeric vector

2010-09-15 Thread Uwe Ligges



On 05.09.2010 14:48, rajesh j wrote:

Hi,

Is it possible to convert a string vector to integer or numeric vector? In
my situation I receive data in a string vector and have to convert it based
on a given type.


?as.interger ?as.numeric

Uwe Ligges

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Re: [R] assignment by value or reference

2010-09-15 Thread Uwe Ligges
See the R Language Definition manual. Since R knows about lazy 
evaluation, it is sometimes neither by reference nor by value.
If you want to think binary, then by value fits better than by 
reference.


Uwe Ligges



On 05.09.2010 17:19, Xiaobo Gu wrote:

Hi Team,

  Can you please tell me the rules of assignment in R, by value or by 
reference.


From my about 3 months of experience of part time job of R, it seems most times 
it is by value, especially in function parameter and return values assignment; 
and it is by reference when referencing container sub-objects of container 
objects, such as elements of List objects and row/column objects of DataFrame 
objectes; but it is by value when referencing the smallest unit of element of a 
container object, such as cell of data frame objects.






Xiaobo.Gu




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Re: [R] CRAN Mirror definition error.

2010-09-15 Thread Uwe Ligges
You may have a proxy in your network that has to be used. Since this 
seems to be Windows: Please see the R for Windows FAQs for more information.


Best wishes,
Uwe



On 01.09.2010 19:43, Lenart, Joseph wrote:

Hello,

I was wondering if anyone has run across this problem in the past and might be 
able to suggest a solution.

I did a review of the FAQ's link trying to find a topic that closely related to 
the experience I was having, however I didn't see
anything that might help in diagnosing the problem.

While invoking the options under the package menu I get the following errors:


chooseCRANmirror()

Warning message:
In open.connection(con, r) : cannot open: HTTP status was '0 (nil)'

setRepositories()
update.packages(ask='graphics')

Warning: unable to access index for repository 
http://www.ibiblio.org/pub/languages/R/CRAN/bin/windows/contrib/2.11
Warning: unable to access index for repository 
http://www.stats.ox.ac.uk/pub/RWin/bin/windows/contrib/2.11

utils:::menuInstallPkgs()

Warning: unable to access index for repository 
http://www.ibiblio.org/pub/languages/R/CRAN/bin/windows/contrib/2.11
Warning: unable to access index for repository 
http://www.stats.ox.ac.uk/pub/RWin/bin/windows/contrib/2.11
Error in install.packages(NULL, .libPaths()[1L], dependencies = NA, type = 
type) :
   no packages were specified

setRepositories()


Does anyone have any insights into what might cause this issue, and how I might 
fix the problem?
Any thoughts you may have would be greatly appreciated.

Thanks in advance - Joe Lenart

p.s. -  I have confirmed that I can access the mirror sites using a regular 
browser, so I do not believe this to be a security
or firewall-related issue.

Ad Astra...
Joseph P. Lenart

Capacity Management Unit - Non Mainframe
THE AETNA
151 Farmington Ave.  Mailstop: ASA1
Hartford, CT 06156
Phone - 860-273-7085  Fax - 860-273-5053

If you hear a voice within you say, 'You cannot paint,' then by all means paint and 
that voice will be silenced.
- Vincent van Gogh




This e-mail may contain confidential or privileged information. If
you think you have received this e-mail in error, please advise the
sender by reply e-mail and then delete this e-mail immediately.
Thank you. Aetna
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Re: [R] Using feather.plot to try and generate a stick plot of current velocity data (and having issues)

2010-09-15 Thread Jim Lemon

On 09/15/2010 01:29 PM, Hughes, Ed wrote:

Hello All,



I am attempting to use the feather.plot function from the plotrix
package to graph current velocity data as I have speed and direction.  I
let r be the first 10 rows of current speed data and theta be the
first 10 rows of directional data in radians.  I had tried this with 10
measurements, but keep getting the following error message:






feather.plot(r,theta,1:10,yref=0,use.arrows=FALSE, fp.type=m)




Error in segments(xpos, yref, xpos + x, y, ...) : invalid third argument



My goal was to trouble shoot this smaller data set and see if I could
ramp up to few thousand entries to basically generate a stick plot of
current flow data.  Curious if I was doing anything obviously wrong with
my arguments or if I should be using an entirely different function.


Hi Eddie,
I think the problem is with the format of your data. I tried a whole 
bunch of variants of vectors, matrices and data frames and couldn't 
reproduce your error. Could you post a sample data set that causes this 
problem? Thanks.


Jim

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Re: [R] ggplot: stat_smooth() and nls method

2010-09-15 Thread Benoit Boulinguiez

great help, thanks a lot

Le 14/09/2010 21:44, Brian Diggs a écrit :

On 9/11/2010 7:52 AM, Benoit Boulinguiez wrote:

Hi all,

Does one of you know if there is any way to combine a nls method in
the stat_smooth of ggplot?

Regards


According to the documentation for predict.nls, it is unable to create 
standard errors for the predictions, so that has to be turned off in 
the stat_smooth call.  Also, the default formula isn't valid for nls 
and has to be overridden, and the start values can be passed.  I used 
geom_smooth rather than stat_smooth, but either work.



library(ggplot2)

DF - data.frame(x=1:20, y=rnorm(20))

ggplot(DF, aes(x=x, y=y)) +
  geom_smooth(method=nls, formula=y~b*x+c, se=FALSE, 
start=list(b=0,c=1)) +

  geom_point()

ggplot(DF, aes(x=x, y=y)) +
  geom_smooth(method=nls, formula=y~sin(b*x), se=FALSE, 
start=list(b=1)) +

  geom_point()



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[R] Programming: loop versus vector oriented

2010-09-15 Thread Jan private
Dear all,

I am new to R and to it's programming philosophy. The following function
is supposed to work on a vector, but I can't figure out how to do that
without looping through every element of it. Is there a more elegant
way?

Note: I have shortened it, so it is NOT correct from the pipe hydraulics
point of view

# Calculate wall friction factor
# reynolds: Reynolds number
# dk: relative roughness of pipe
lambda_wall - function (reynolds, dk) {
  z - 1
  result - 0

  for (Re in reynolds) {
if (Re = 2320) {
  # Laminar flow
  lambda - 64/Re
} else if (Re  65 * dk[z]) {
  # Turbulent flow
  if (Re  1e+5) {
lambda - 0.3164 / sqrt(sqrt(Re))
  } else {
lambda - 0.309/(log10(Re/7))^2
  }
} else {
  # Intermediate area
  lambdanew - 1 / (2 * log10(3.71 * dk[z]))^2 # Start value
  iter - 0

  repeat {
lambda - lambdanew
lambdanew - 1 / (2 * log10(2.51/(Re * sqrt(lambda)) + 0.27/dk[z]))^2
iter - iter + 1
if ((abs(lambdanew - lambda)  0.001) || (iter  100)) break
  }

  lambda = lambdanew
}

result[z] - lambda
z - z + 1
  }

  result
} # lambda_wall()

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[R] Interpolate? a line

2010-09-15 Thread Alaios
Hello everyone. 
I have created a 100*100 matrix in R.
Let's now say that I have a line that starts from (2,3) point and ends to the 
(62,34) point. In other words this line starts at cell (2,3) and ends at cell 
(62,34).

Is it possible to get by some R function all the matrix's cells that this line 
transverses?

I would like to thank you for your feedback.

Best Regards
Alex


  
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[R] R install in Fedora

2010-09-15 Thread wesley mathew
 Dear All

As you suggested I tried to install  was trying to install
R-2.11.1-3.fc13.1686 in  Fedora-13-i386 but it also shows error.
I wrote  yum install R-2.11.1-3fc13.i686
Errors
Cannot retrieve repository metadata (repond.xml) for repository: fedora.
Please verify its path and try again.
Could you please help me to solve this problem.
Thanks in advance
Kind Regards

Wesley C Mathew

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[R] cochran-grubbs tests results

2010-09-15 Thread miguelgonzalez

Hello, 

I'm new in this R world and I don't know much about statistics, but now I
have to analize some data and I've got some first queries yet:

I have 5 sets of area mesures and each set has 5 repetitions.
My first step is to check data looking for outliers. I've used the outliers
package. I have to use the cochran test and the grubbs test in case I find
any outlier. The problem is that I don't know how to interpret the results.
Cochran test finds an outlier among variances. Grubbs test finds an outlier
within the set of measures wich corresponds to the variance that is an
outlier. I repeat the Cochran test discarding that measure. And it finds a
new outlier and so on. This is what I've done first:

-
 varianza
[1] 9.979937e-07 1.131360e-07 4.140418e-06 2.184604e-07 3.662077e-07
 data1-c(rep(4,5))
 cochran.test(varianza, data1, inlying = FALSE)

Cochran test for outlying variance

data:  varianza 
C = 0.7094, df = 4, k = 5, p-value = 0.007645
alternative hypothesis: Group 3 has outlying variance 
sample estimates:
   12345 
9.979937e-07 1.131360e-07 4.140418e-06 2.184604e-07 3.662077e-07   


It would be really helpful if someone could tell me what C, df, K and
p-value mean. 
Besides, I don't know if alternative hypothesis (in the grubbs test) means
that the highest value is an outlier to be discarded or just the worst
value, but not necessarily to be discarded.


 grubbs.test(serie3, type = 10, opposite = FALSE, two.sided = FALSE) 

Grubbs test for one outlier

data:  serie3 
G = 1.3384, U = 0.2039, p-value = 0.2155
alternative hypothesis: highest value 0.238877332 is an outlier 
-

As you see I'm quite lost. 
Thank you in advance and I'll keep trying.

Gracias
 
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Re: [R] Interpolate? a line

2010-09-15 Thread Michael Bedward
Hello Alex,

Here is one way to do it. It works but it's not pretty :)

interp - approx(c(2, 62), c(3, 34), method=linear, xout=2:62)
m - matrix(c(interp$x, round(interp$y)), ncol=2)
tie - m[,2] == c(-Inf, m[-nrow(m),2])
m - m[ !tie, ]

You might want to examine the result like this...

plot(m)  # plots points
lines(c(2,26), c(3, 34))  # overlay line for comparison

Michael

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Re: [R] Package build and install under Windows

2010-09-15 Thread Duncan Murdoch

Tao wrote:

Any other method besides using RTOOLS? I am using R 2.10.1, which version of
Rtools should I use? It seems that there are two versions on the website.
  


What website are you talking about?  The Rtools website 
http://www.murdoch-sutherland.com/Rtools/ has 7 different versions of 
the tools, and clearly labels which versions of R are supported by each.


Duncan Murdoch


The commands like R CMD build should be used in the CMD window, right?
Thanks.



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Re: [R] Interpolate? a line

2010-09-15 Thread Alaios
Thank you very much. I am trying to execute it line by line to get some 
understanding how this works.
Could you please explain me what these two lines do?

tie - m[,2] == c(-Inf, m[-nrow(m),2])
m - m[ !tie, ]

I would like to thank you in advance for your help

Best Regards
Alex





From: Michael Bedward michael.bedw...@gmail.com

Cc: Rhelp r-help@r-project.org
Sent: Wed, September 15, 2010 12:52:45 PM
Subject: Re: [R] Interpolate? a line

Hello Alex,

Here is one way to do it. It works but it's not pretty :)

interp - approx(c(2, 62), c(3, 34), method=linear, xout=2:62)
m - matrix(c(interp$x, round(interp$y)), ncol=2)
tie - m[,2] == c(-Inf, m[-nrow(m),2])
m - m[ !tie, ]

You might want to examine the result like this...

plot(m)  # plots points
lines(c(2,26), c(3, 34))  # overlay line for comparison

Michael



  
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Re: [R] Model averaging with (and without) interaction terms

2010-09-15 Thread Ben Bolker
Leslie Young leslie.young101 at gmail.com writes:

 
 I’ve used logistic regression to create models to assess the effect of
 3 variables on the presence or absence of a species, including the
 interaction terms between variables and model averaging using MuMI:
 model.avg
 
 The top models (delta4) include several models with interaction terms
 and some models without; model weights are quite low for all models
 (0.25). My problem is that the models with interactions have negative
 coefficients on the variables with a positive interaction term whereas
 the same model without an interaction has positive coefficients.
 MuMIn: model.avg averages all these models together, so the
 relationship is washed out (CI overlaps 0).
 
 Eg.
 
 mod1-glm(presence ~ x1*x2, family=”binomial”)
 coefficients: -0.661 x1, -0.043 x2, 0.02 x1:x2
 
 mod2 - glm(presence ~ x1 + x2, family=”binomial”)
 coefficients: 0.245 x1, 0.021 x2
 
 I’ve read that it is difficult to compare models with and without
 interaction terms, but nothing regarding how one might go about doing
 so. Should interaction models be averaged differently or separately
 than models without interaction terms?  Is there another way to
 approach this?

  The tricky aspect of comparing models with and without interaction
terms is that the main effect parameters have different meanings
when the interactions are included.  It looks like your parameters
are both continuous, so I'll discuss things in this context.
In the interaction model, the x1 parameter gives the expected change
in log-odds (logit probability) for a 1-unit increase in the
corresponding predictor variable **when x2 is equal to 0**. In the
non-interaction model, the x1 parameter gives the *average* expected
change in log-odds across the distribution of x2 values observed in
the data sets.
   It will help a bit if you follow Schielzeth [Schielzeth, Holger. 2010. Simple
means to improve the interpretability of regression coefficients. Methods in
Ecology and Evolution , no. . doi:10./j.2041-210X.2010.00012.x.] and
mean-correct your predictors
before fitting the model.
   I have a bigger problem (which perhaps luckily for you is not shared
with much of the ecological community), which is that usually people
who are model-averaging *parameters* (rather than *predictions*) are
essentially trying to use information-theoretic approaches to test
hypotheses ...

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Re: [R] HDP and 99% contour lines

2010-09-15 Thread Ben Bolker
Peter Spencer P.Spencer at murdoch.edu.au writes:

 
 Dear all,
 
 I have a very simple question about how I can include HPD confidence
 lines in a Marginal posterior distribution scatterplot
 I have the following code to draw the scatterplot(s) from an output
 table;

  I think the code in HPDregionplot in the emdbook package
might help.

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Re: [R] Interpolate? a line

2010-09-15 Thread Michael Bedward
Ah, now that you ask that I realize I was making an assumption
assumption about what you want to achieve. Those two lines remove
coords where the y value is equal to that of the preceding coord.

That may or may not be what you want to do. To visualize what is
happening, type plot(m) before doing those two lines (you'll see most
points arranged in horizontal pairs), the execute the two lines and do
plot(m) again.

Michael

On 15 September 2010 21:05, Alaios ala...@yahoo.com wrote:
 Thank you very much. I am trying to execute it line by line to get some
 understanding how this works.
 Could you please explain me what these two lines do?
 tie - m[,2] == c(-Inf, m[-nrow(m),2])
 m - m[ !tie, ]


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Re: [R] Interpolate? a line

2010-09-15 Thread David Winsemius

Replacing context:


Hello everyone.
I have created a 100*100 matrix in R.
Let's now say that I have a line that starts from (2,3) point and  
ends to the
(62,34) point. In other words this line starts at cell (2,3) and  
ends at cell

(62,34).

Is it possible to get by some R function all the matrix's cells  
that this line

transverses?

I would like to thank you for your feedback.

Best Regards
Alex


On Sep 15, 2010, at 6:52 AM, Michael Bedward wrote:


Hello Alex,

Here is one way to do it. It works but it's not pretty :)


If you want an alternative, consider that produces the Y cell indices  
(since the x cell indices are already 2:62):


 linefn - function(x) 3+((34-3)/(62-2)) *(x-2)
 findInterval(linefn(2:62), 3:34)
 [1]  1  1  2  2  3  3  4  4  5  5  6  6  7  7  8  8  9  9 10 10 11  
11 12 12 13 13 14
[28] 14 15 15 16 17 17 18 18 19 19 20 20 21 21 22 22 23 23 24 24 25 25  
26 26 27 27 28

[55] 28 29 29 30 30 31 32
# that seems off by two
 linefn(62)
[1] 34
 linefn(2)
[1] 3 # but that checks out and I realized those were just indices for  
the 3:34 findInterval vector


 (3:34)[findInterval(linefn(2:62), 3:34)]
 [1]  3  3  4  4  5  5  6  6  7  7  8  8  9  9 10 10 11 11 12 12 13  
13 14 14 15 15 16
[28] 16 17 17 18 19 19 20 20 21 21 22 22 23 23 24 24 25 25 26 26 27 27  
28 28 29 29 30

[55] 30 31 31 32 32 33 34

( no rounding and I think the logic is clearer.)

--
David.



interp - approx(c(2, 62), c(3, 34), method=linear, xout=2:62)
m - matrix(c(interp$x, round(interp$y)), ncol=2)
tie - m[,2] == c(-Inf, m[-nrow(m),2])
m - m[ !tie, ]

You might want to examine the result like this...

plot(m)  # plots points
lines(c(2,26), c(3, 34))  # overlay line for comparison

you can add a grid with
abline(v=2:62, h=3:34)


Michael

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David Winsemius, MD
West Hartford, CT

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[R] A question on modelling binary response data using factors

2010-09-15 Thread Cormac Long
 Dear all,

 A question on modelling proportional data in R.  I have a test experiment
that was designed in a particular way, and which I can analyse by hand to
 an extent.  I am really struggling to get R to give me sensible results in
modelling it properly, so must be doing something wrong here.

 As background, I conduct a series of experiments and count the hits and
misses - I believe this could be modelled using a glm.  The experiments
 are done under a range of conditions, altering three different factors,
lets call them A, B, and C.  Factors A and B each have 2 levels
(A1,A2,B1,B2)
 while factor C has 5 levels (C1,C2,C3,C4,C5).  The experiment has only
partial coverage, that is not every A is tested with every B and every C.
However,
 I was careful in the experimental design to ensure that every A and every B
was tested against at least one C.

 Here is my experimental data:

  FactorA   FactorB  FactorCHit   Miss
  A1  B1 C1 17  83
  A1  B1 C2 17  83
  A1  B1 C3 18  82
  A1  B1 C4 NA NA
  A1  B1 C5 NA NA
  A1  B2 C1 11  89
  A1  B2 C2 17  83
  A1  B2 C3 17  83
  A1  B2 C4 NA NA
  A1  B2 C5 NA NA
  A2  B1 C1 NA NA
  A2  B1 C2 NA NA
  A2  B1 C3 23  77
  A2  B1 C4 19  81
  A2  B1 C5 29  71
  A2  B2 C1 NA NA
  A2  B2 C2 NA NA
  A2  B2 C3 13  87
  A2  B2 C4 20  80
  A2  B2 C5 24  76


 I to compare individual rates against the overall rate, and look at
difference across pooled rate for each factor

 Doing hand calculations, it is possible to calculate the individual hit
rates, the overall hit rate of the pool, and the standard errors on these
 using hitrate = hits/(hits+misses) and hitrateerror =
sqrt((hits+misses)*(hits/(hits+misses))*(misses/(hits+misses)))/(hits+misses)

 Again doing hand calculations, it is possible to extract the influence of
each variant of each factor, by producing a pool and comparing the
influence
 of each variant to the pool.  For the overall pool, there were 225 hits
from 1200 trials, for a pooled hitrate of 18.8%.

 For factor C the analysis is relatively straightforward, in that we
calculate the hitrate for each C1 (28 hits from 200 trials = 14.0%), C2, ...
etc.  We then
 calculate the influence of a given factor compared to the pool - for
example C1 has a hitrate of 14.0% compared to a pooled hitrate of 18.8%, so
has an
 influence of (14.0%/18.8%)-100% = -25%.

 For factor A, the analysis is slightly complicated by the fact that factor
C3 is the only variant that fairly tests factor A
 (none of factors C1,C2,C4,C5 do not have experiments for both factors A1
and A2).  Therefore, just taking the factor C3 data, we can show that the
pool
 of 400 trials had 71 hits, for a hitrate of 17.8%.  We can then calculate
the hitrate for A1 and A2 as 17.5% and 18.0% respectively.  From this we can
 say that the effect of A1 and A2 was -1.4% and +1.4%, that is to say
(17.5%/17.8%)-100% and (18.0/17.8%)-100%.

 The trick here is that the pooled hitrate for factor A is 17.8%, i.e. the
hitrate for factor C3 alone, not 18.8% which is the overall pooled hitrate.
 However we have taken out the influence of C3 by normalising to the C3
hitrate, i.e. we could calculate the expected hitrate of the combination
 A2 B2 C1, which was never tested, by combining the influence of A2, B2 and
C1 on the overall pool hitrate of 18.8%.

 It should be possible to model this using e.g. a glm with appropriate
contrasts and model, however for the life of me I cannot work out how to get
 this to work. Getting R to provide a sensible intercept to the model,
i.e. the overall hitrate of 18.8%, and also get sensible indications of the
 importance or otherwise of each factor.

 In particular, I am really struggling to get R to test the relevance of
factor A - this is because when tested using all factor C, not just C3,
there
 is a huge bias in that C1 and C2 have comparitively low hit rates, and were
only tested against A1, whereas C4 and C5 have comparitively high hitrates,
 but were only tested against A2.  That means that it appears, unless you
look at the C3 data only, that factor A is highly important - whereas it
clearly
 is not.

 The reason for using R to model this correctly, rather than just doing
the hand 

[R] barplot: space between axis and bars

2010-09-15 Thread Daniel Stepputtis

Hi all,
I have a problem with a rather simple plot (which I have used several 
times) - barplot.


I want to create a barplot, where no space is between the axis and the bars.

Example:
barplot(rnorm(10), space=0)

creates a vertical axis at the left and bars. I want to skip the space 
between both.


Thank you for your help.
Daniel

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


Re: [R] Interpolate? a line

2010-09-15 Thread David Winsemius


On Sep 15, 2010, at 7:24 AM, David Winsemius wrote:


Replacing context:


Hello everyone.
I have created a 100*100 matrix in R.
Let's now say that I have a line that starts from (2,3) point and  
ends to the
(62,34) point. In other words this line starts at cell (2,3) and  
ends at cell

(62,34).

Is it possible to get by some R function all the matrix's cells  
that this line

transverses?

I would like to thank you for your feedback.

Best Regards
Alex


On Sep 15, 2010, at 6:52 AM, Michael Bedward wrote:


Hello Alex,

Here is one way to do it. It works but it's not pretty :)


If you want an alternative, consider that produces the Y cell  
indices (since the x cell indices are already 2:62):


 linefn - function(x) 3+((34-3)/(62-2)) *(x-2)
 findInterval(linefn(2:62), 3:34)
[1]  1  1  2  2  3  3  4  4  5  5  6  6  7  7  8  8  9  9 10 10 11  
11 12 12 13 13 14
[28] 14 15 15 16 17 17 18 18 19 19 20 20 21 21 22 22 23 23 24 24 25  
25 26 26 27 27 28

[55] 28 29 29 30 30 31 32
# that seems off by two
 linefn(62)
[1] 34
 linefn(2)
[1] 3 # but that checks out and I realized those were just indices  
for the 3:34 findInterval vector


 (3:34)[findInterval(linefn(2:62), 3:34)]
[1]  3  3  4  4  5  5  6  6  7  7  8  8  9  9 10 10 11 11 12 12 13  
13 14 14 15 15 16
[28] 16 17 17 18 19 19 20 20 21 21 22 22 23 23 24 24 25 25 26 26 27  
27 28 28 29 29 30

[55] 30 31 31 32 32 33 34

( no rounding and I think the logic is clearer.)


But I also realized it didn't enumerate all the the cells were crossed  
either, only indicating which cell was associated with an integer  
value of x. Also would have even more serious problems if the slope  
were greater than unity. To enumerate the cell indices that were  
crossed, try:


unique( floor( cbind( seq(2,62, by=0.1), linefn(seq(2,62,  
by=0.1)) ) )  )

  [,1] [,2]
 [1,]23
 [2,]33
 [3,]44
 [4,]54
 [5,]55
 [6,]65
 [7,]75
 [8,]76
 snipping interior results
[83,]   58   32
[84,]   59   32
[85,]   60   32
[86,]   60   33
[87,]   61   33
[88,]   62   34

That could probably be passed to rect() to illustrate (and check logic):

rect(cellidxs[,1], cellidxs[,2], cellidxs[,1]+1, cellidxs[,2]+1,  
col=red)


#redraw line :
 lines(2:62, 3+(34-3)/(62-2)*(0:60))




--
David.



interp - approx(c(2, 62), c(3, 34), method=linear, xout=2:62)
m - matrix(c(interp$x, round(interp$y)), ncol=2)
tie - m[,2] == c(-Inf, m[-nrow(m),2])
m - m[ !tie, ]

You might want to examine the result like this...

plot(m)  # plots points
lines(c(2,26), c(3, 34))  # overlay line for comparison

you can add a grid with
abline(v=2:62, h=3:34)


Michael

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David Winsemius, MD
West Hartford, CT

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David Winsemius, MD
West Hartford, CT

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Re: [R] content analysis

2010-09-15 Thread Bob Green

Christian,

Have you looked at TM?

There is also Quantitative Corpus Linguistics with R by Stefan Gries, 
which you may find of interest.


Tau, Readme and rattle might be worth looking into.

See also: http://ses.telecom-paristech.fr/lebart/

regards

Bob

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Re: [R] barplot: space between axis and bars

2010-09-15 Thread Thomas Stewart
Try

par(xaxs=i)
barplot(rnorm(10), space=0)

-tgs

On Wed, Sep 15, 2010 at 7:54 AM, Daniel Stepputtis 
daniel.stepput...@vti.bund.de wrote:

 Hi all,
 I have a problem with a rather simple plot (which I have used several
 times) - barplot.

 I want to create a barplot, where no space is between the axis and the
 bars.

 Example:
 barplot(rnorm(10), space=0)

 creates a vertical axis at the left and bars. I want to skip the space
 between both.

 Thank you for your help.
 Daniel

 __
 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.


[[alternative HTML version deleted]]

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Re: [R] retrieving object names passed indirectly to a function

2010-09-15 Thread Michael Bedward
Answering my own question (in case it's of use to anyone else) here is
the example code again but now using call and eval to accomplish the
desired output:

foo - function(p1, p2) {
  p1name - deparse(substitute(p1))
  p2name - deparse(substitute(p2))

  # The real function runs a simulation, but
  # here we just confirm the object names
  # and values passed as args
  cat(passed, p1name, with value, p1, \n)
  cat(passed, p2name, with value, p2, \n)
}

foo.driver - function(param.table) {
  for (i in 1:nrow(param.table)) {
foo.call - call(foo,
as.name(param.table$p1[i]),
as.name(param.table$p2[i]) )

for (rep in 1:param.table$repl[i]) {
  eval(foo.call)
}
  }
}

# some objects in the global env which will be used as args to foo
p1a - 1.1
p1b - 1.2
p2a - 2.1
p2b - 2.2

# the controlling table for foo.driver
param.table - data.frame(p1=c(p1a, p1b), p2=c(p2a,p2b),
   replicates=c(1,1), stringsAsFactors=FALSE)

# Now run foo.driver with param.table...
foo.driver(param.table)

The code above produces the following output:
passed p1a with value 1.1
passed p2a with value 2.1
passed p1b with value 1.2
passed p2b with value 2.2

Happy :)

Michael

On 15 September 2010 16:07, Michael Bedward michael.bedw...@gmail.com wrote:
 Hi folks,

 I'm stuck with a problem that I suspect has a trivial solution...

 I have a function, call it foo, that takes a number of arguments which
 it uses as parameters for a simulation. For later analysis, foo stores
 the names of the objects passed to it along with the simulation
 results in its output (written to a database). The objects names are
 accessed with deparse(substitute(argname)).

 A second function, call it foo.driver, takes as input a data.frame
 with cols corresponding to foo arguments containing the character
 names of objects. For each row of the data.frame in turn, it retrieves
 the required objects from the global environment and passes them to
 function foo.

 Where I am stuck is in how foo.driver should pass an object to foo
 such that calling deparse(substitute(argname)) in foo will retrieve
 the correct name (ie. that specified in foo.driver's input
 data.frame).

 Hopefully, the following toy code example illustrates what I'm trying to do...

 # some objects in the global environment to provide param values to foo
 p1.1 - 1
 p1.2 - 2
 p2.1 - 3
 p2.2 - 4

 # controlling table
 arg.table - data.frame(p1=c(p1.1, p2.2), p2=c(p2.1, p2.2),
 replicates=c(100, 100))

 foo.driver - function( arg.table ) {
  for (i in 1:nrow(arg.table)) {
    for (j in 1:arg.table$replicates[i]) {
      # somehow retrieve objects named in cols p1 and p2 and
      # pass them to foo with their original names
      # (e.g. some clever use of get and assign ?)
    }
  }
 }

 foo - function(p1, p2) {
  # retrieve and store arg names
  p1name - deparse(substitute(p1))
  p2name - deparse(substitute(p2))

  # run simulation with values of p1 and p2 then
  # store results together with names of objects
  # passed as args to output database
 }

 My actual simulation function involves a large number of arguments of
 various classes and if at all possible I prefer to avoid hacking it
 (e.g to accept object names rather than objects).

 Any tips much appreciated.

 Michael


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Re: [R] barplot: space between axis and bars

2010-09-15 Thread Jim Lemon

On 09/15/2010 09:54 PM, Daniel Stepputtis wrote:

Hi all,
I have a problem with a rather simple plot (which I have used several
times) - barplot.

I want to create a barplot, where no space is between the axis and the
bars.

Example:
barplot(rnorm(10), space=0)

creates a vertical axis at the left and bars. I want to skip the space
between both.


Hi Daniel,
Try the barp function in the plotrix package.
Jim

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


[R] Enumerating unit cell traversals Re: Interpolate? a line

2010-09-15 Thread David Winsemius


On Sep 15, 2010, at 7:55 AM, David Winsemius wrote:



On Sep 15, 2010, at 7:24 AM, David Winsemius wrote:


Replacing context:


Hello everyone.
I have created a 100*100 matrix in R.
Let's now say that I have a line that starts from (2,3) point and  
ends to the
(62,34) point. In other words this line starts at cell (2,3) and  
ends at cell

(62,34).

Is it possible to get by some R function all the matrix's cells  
that this line

transverses?

I would like to thank you for your feedback.

Best Regards
Alex


On Sep 15, 2010, at 6:52 AM, Michael Bedward wrote:


Hello Alex,

Here is one way to do it. It works but it's not pretty :)


If you want an alternative, consider that produces the Y cell  
indices (since the x cell indices are already 2:62):


 linefn - function(x) 3+((34-3)/(62-2)) *(x-2)
 findInterval(linefn(2:62), 3:34)
[1]  1  1  2  2  3  3  4  4  5  5  6  6  7  7  8  8  9  9 10 10 11  
11 12 12 13 13 14
[28] 14 15 15 16 17 17 18 18 19 19 20 20 21 21 22 22 23 23 24 24 25  
25 26 26 27 27 28

[55] 28 29 29 30 30 31 32
# that seems off by two
 linefn(62)
[1] 34
 linefn(2)
[1] 3 # but that checks out and I realized those were just indices  
for the 3:34 findInterval vector


 (3:34)[findInterval(linefn(2:62), 3:34)]
[1]  3  3  4  4  5  5  6  6  7  7  8  8  9  9 10 10 11 11 12 12 13  
13 14 14 15 15 16
[28] 16 17 17 18 19 19 20 20 21 21 22 22 23 23 24 24 25 25 26 26 27  
27 28 28 29 29 30

[55] 30 31 31 32 32 33 34

( no rounding and I think the logic is clearer.)


But I also realized it didn't enumerate all the the cells were  
crossed either, only indicating which cell was associated with an  
integer value of x. Also would have even more serious problems if  
the slope were greater than unity. To enumerate the cell indices  
that were crossed, try:


unique( floor( cbind( seq(2,62, by=0.1), linefn(seq(2,62,  
by=0.1)) ) )  )

 [,1] [,2]
[1,]23
[2,]33
[3,]44
[4,]54
[5,]55
[6,]65
[7,]75
[8,]76
snipping interior results
[83,]   58   32
[84,]   59   32
[85,]   60   32
[86,]   60   33
[87,]   61   33
[88,]   62   34

That could probably be passed to rect() to illustrate (and check  
logic):


rect(cellidxs[,1], cellidxs[,2], cellidxs[,1]+1, cellidxs[,2]+1,  
col=red)


#redraw line :
lines(2:62, 3+(34-3)/(62-2)*(0:60))


And then I got to wondering if every 0.1 was sufficient to catch all  
the corners and discovered I could identify 3 more cell traversals  
with by=0.01


 cellid2 -unique( floor(cbind(seq(2,62, by=0.01), linefn(seq(2,62,  
by=0.01) )) ) )

 NROW(cellid2) # 91 cells
 rect(cellid2[,1], cellid2[,2], cellid2[,1]+1, cellid2[,2]+1,  
col=blue)
 rect(cellidxs[,1], cellidxs[,2], cellidxs[,1]+1, cellidxs[,2]+1,  
col=red)

 lines(2:62, 3+(34-3)/(62-2)*(0:60))

(Trying by=0.001 did not change the count, but did take longer)

--
David.


-
David.



interp - approx(c(2, 62), c(3, 34), method=linear, xout=2:62)
m - matrix(c(interp$x, round(interp$y)), ncol=2)
tie - m[,2] == c(-Inf, m[-nrow(m),2])
m - m[ !tie, ]

You might want to examine the result like this...

plot(m)  # plots points
lines(c(2,26), c(3, 34))  # overlay line for comparison

you can add a grid with
abline(v=2:62, h=3:34)


Michael





David Winsemius, MD
West Hartford, CT

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R-help@r-project.org mailing list
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Re: [R] Enumerating unit cell traversals Re: Interpolate? a line

2010-09-15 Thread Michael Bedward
Nice posts David. I like the way that 'simple' problems such as this
one give rise to an interesting assortment of approaches and gotchas
:)

Michael

On 15 September 2010 22:41, David Winsemius dwinsem...@comcast.net wrote:

 On Sep 15, 2010, at 7:55 AM, David Winsemius wrote:


 On Sep 15, 2010, at 7:24 AM, David Winsemius wrote:

 Replacing context:

 Hello everyone.
 I have created a 100*100 matrix in R.
 Let's now say that I have a line that starts from (2,3) point and ends
 to the
 (62,34) point. In other words this line starts at cell (2,3) and ends
 at cell
 (62,34).

 Is it possible to get by some R function all the matrix's cells that
 this line
 transverses?

 I would like to thank you for your feedback.

 Best Regards
 Alex

 On Sep 15, 2010, at 6:52 AM, Michael Bedward wrote:

 Hello Alex,

 Here is one way to do it. It works but it's not pretty :)

 If you want an alternative, consider that produces the Y cell indices
 (since the x cell indices are already 2:62):

  linefn - function(x) 3+((34-3)/(62-2)) *(x-2)
  findInterval(linefn(2:62), 3:34)
 [1]  1  1  2  2  3  3  4  4  5  5  6  6  7  7  8  8  9  9 10 10 11 11 12
 12 13 13 14
 [28] 14 15 15 16 17 17 18 18 19 19 20 20 21 21 22 22 23 23 24 24 25 25 26
 26 27 27 28
 [55] 28 29 29 30 30 31 32
 # that seems off by two
  linefn(62)
 [1] 34
  linefn(2)
 [1] 3 # but that checks out and I realized those were just indices for
 the 3:34 findInterval vector

  (3:34)[findInterval(linefn(2:62), 3:34)]
 [1]  3  3  4  4  5  5  6  6  7  7  8  8  9  9 10 10 11 11 12 12 13 13 14
 14 15 15 16
 [28] 16 17 17 18 19 19 20 20 21 21 22 22 23 23 24 24 25 25 26 26 27 27 28
 28 29 29 30
 [55] 30 31 31 32 32 33 34

 ( no rounding and I think the logic is clearer.)

 But I also realized it didn't enumerate all the the cells were crossed
 either, only indicating which cell was associated with an integer value of
 x. Also would have even more serious problems if the slope were greater than
 unity. To enumerate the cell indices that were crossed, try:

 unique( floor( cbind( seq(2,62, by=0.1), linefn(seq(2,62, by=0.1)) ) )  )
     [,1] [,2]
 [1,]    2    3
 [2,]    3    3
 [3,]    4    4
 [4,]    5    4
 [5,]    5    5
 [6,]    6    5
 [7,]    7    5
 [8,]    7    6
 snipping interior results
 [83,]   58   32
 [84,]   59   32
 [85,]   60   32
 [86,]   60   33
 [87,]   61   33
 [88,]   62   34

 That could probably be passed to rect() to illustrate (and check logic):

 rect(cellidxs[,1], cellidxs[,2], cellidxs[,1]+1, cellidxs[,2]+1,
 col=red)

 #redraw line :
 lines(2:62, 3+(34-3)/(62-2)*(0:60))

 And then I got to wondering if every 0.1 was sufficient to catch all the
 corners and discovered I could identify 3 more cell traversals with by=0.01

 cellid2 -unique( floor(cbind(seq(2,62, by=0.01), linefn(seq(2,62,
 by=0.01) )) ) )
 NROW(cellid2) # 91 cells
 rect(cellid2[,1], cellid2[,2], cellid2[,1]+1, cellid2[,2]+1, col=blue)
 rect(cellidxs[,1], cellidxs[,2], cellidxs[,1]+1, cellidxs[,2]+1,
 col=red)
 lines(2:62, 3+(34-3)/(62-2)*(0:60))

 (Trying by=0.001 did not change the count, but did take longer)

 --
 David.

 -
 David.


 interp - approx(c(2, 62), c(3, 34), method=linear, xout=2:62)
 m - matrix(c(interp$x, round(interp$y)), ncol=2)
 tie - m[,2] == c(-Inf, m[-nrow(m),2])
 m - m[ !tie, ]

 You might want to examine the result like this...

 plot(m)  # plots points
 lines(c(2,26), c(3, 34))  # overlay line for comparison

 you can add a grid with
 abline(v=2:62, h=3:34)

 Michael



 David Winsemius, MD
 West Hartford, CT

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


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Re: [R] Enumerating unit cell traversals Re: Interpolate? a line

2010-09-15 Thread Rainer M Krug
-BEGIN PGP SIGNED MESSAGE-
Hash: SHA1

On 15/09/10 14:46, Michael Bedward wrote:
 Nice posts David. I like the way that 'simple' problems such as this
 one give rise to an interesting assortment of approaches and gotchas
 :)
 
 Michael
 
 On 15 September 2010 22:41, David Winsemius dwinsem...@comcast.net wrote:

 On Sep 15, 2010, at 7:55 AM, David Winsemius wrote:


 On Sep 15, 2010, at 7:24 AM, David Winsemius wrote:

 Replacing context:

 Hello everyone.
 I have created a 100*100 matrix in R.
 Let's now say that I have a line that starts from (2,3) point and ends
 to the
 (62,34) point. In other words this line starts at cell (2,3) and ends
 at cell
 (62,34).

 Is it possible to get by some R function all the matrix's cells that
 this line
 transverses?

This is a known problem in computer graphics and the algorythm used to
identify these cells is the Bresenham's Line Drawing Algorithm

You can check out one version at

http://www.falloutsoftware.com/tutorials/dd/dd4.htm

or

http://en.wikipedia.org/wiki/Bresenham's_line_algorithm

Cheers,

Rainer


 I would like to thank you for your feedback.

 Best Regards
 Alex

 On Sep 15, 2010, at 6:52 AM, Michael Bedward wrote:

 Hello Alex,

 Here is one way to do it. It works but it's not pretty :)

 If you want an alternative, consider that produces the Y cell indices
 (since the x cell indices are already 2:62):

 linefn - function(x) 3+((34-3)/(62-2)) *(x-2)
 findInterval(linefn(2:62), 3:34)
 [1]  1  1  2  2  3  3  4  4  5  5  6  6  7  7  8  8  9  9 10 10 11 11 12
 12 13 13 14
 [28] 14 15 15 16 17 17 18 18 19 19 20 20 21 21 22 22 23 23 24 24 25 25 26
 26 27 27 28
 [55] 28 29 29 30 30 31 32
 # that seems off by two
 linefn(62)
 [1] 34
 linefn(2)
 [1] 3 # but that checks out and I realized those were just indices for
 the 3:34 findInterval vector

 (3:34)[findInterval(linefn(2:62), 3:34)]
 [1]  3  3  4  4  5  5  6  6  7  7  8  8  9  9 10 10 11 11 12 12 13 13 14
 14 15 15 16
 [28] 16 17 17 18 19 19 20 20 21 21 22 22 23 23 24 24 25 25 26 26 27 27 28
 28 29 29 30
 [55] 30 31 31 32 32 33 34

 ( no rounding and I think the logic is clearer.)

 But I also realized it didn't enumerate all the the cells were crossed
 either, only indicating which cell was associated with an integer value of
 x. Also would have even more serious problems if the slope were greater than
 unity. To enumerate the cell indices that were crossed, try:

 unique( floor( cbind( seq(2,62, by=0.1), linefn(seq(2,62, by=0.1)) ) )  )
 [,1] [,2]
 [1,]23
 [2,]33
 [3,]44
 [4,]54
 [5,]55
 [6,]65
 [7,]75
 [8,]76
 snipping interior results
 [83,]   58   32
 [84,]   59   32
 [85,]   60   32
 [86,]   60   33
 [87,]   61   33
 [88,]   62   34

 That could probably be passed to rect() to illustrate (and check logic):

 rect(cellidxs[,1], cellidxs[,2], cellidxs[,1]+1, cellidxs[,2]+1,
 col=red)

 #redraw line :
 lines(2:62, 3+(34-3)/(62-2)*(0:60))

 And then I got to wondering if every 0.1 was sufficient to catch all the
 corners and discovered I could identify 3 more cell traversals with by=0.01

 cellid2 -unique( floor(cbind(seq(2,62, by=0.01), linefn(seq(2,62,
 by=0.01) )) ) )
 NROW(cellid2) # 91 cells
 rect(cellid2[,1], cellid2[,2], cellid2[,1]+1, cellid2[,2]+1, col=blue)
 rect(cellidxs[,1], cellidxs[,2], cellidxs[,1]+1, cellidxs[,2]+1,
 col=red)
 lines(2:62, 3+(34-3)/(62-2)*(0:60))

 (Trying by=0.001 did not change the count, but did take longer)

 --
 David.

 -
 David.


 interp - approx(c(2, 62), c(3, 34), method=linear, xout=2:62)
 m - matrix(c(interp$x, round(interp$y)), ncol=2)
 tie - m[,2] == c(-Inf, m[-nrow(m),2])
 m - m[ !tie, ]

 You might want to examine the result like this...

 plot(m)  # plots points
 lines(c(2,26), c(3, 34))  # overlay line for comparison

 you can add a grid with
 abline(v=2:62, h=3:34)

 Michael



 David Winsemius, MD
 West Hartford, CT

 __
 R-help@r-project.org mailing list
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 R-help@r-project.org mailing list
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Biology, UCT), Dipl. Phys. (Germany)

Centre of Excellence for Invasion Biology
Natural Sciences Building
Office Suite 2039
Stellenbosch University
Main Campus, Merriman Avenue
Stellenbosch
South Africa

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Cell:   +27 - (0)8 39 47 90 42
Fax (SA):   +27 - (0)8 65 16 27 82
Fax (D) :   +49 - (0)3 21 21 25 22 44
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email:  rai...@krugs.de

Skype:  RMkrug
-BEGIN PGP 

Re: [R] R install in Fedora

2010-09-15 Thread Marc Schwartz
On Sep 15, 2010, at 5:14 AM, wesley mathew wrote:

 Dear All
 
 As you suggested I tried to install  was trying to install
 R-2.11.1-3.fc13.1686 in  Fedora-13-i386 but it also shows error.
 I wrote  yum install R-2.11.1-3fc13.i686
 Errors
 Cannot retrieve repository metadata (repond.xml) for repository: fedora.
 Please verify its path and try again.
 Could you please help me to solve this problem.
 Thanks in advance
 Kind Regards
 
 Wesley C Mathew


The command that I told you to use was:

  yum install R

to be used as root. Note that there is nothing after the R. 

You have to have root privileges to use yum and to install the R RPMs. So you 
either need to use:

  su - 

then enter the root password and then use the command above, or if you have set 
up sudo, you can use:

  sudo yum install R

after which you will need to enter the root password.

If you still get the error, you have a configuration problem with your yum repo 
setup, which means that you really need to be looking to the Fedora (not R) 
lists for assistance, since it is not an R issue.

That being said, as I also noted, if you post to the R-SIG-Fedora list, there 
are folks there who can offer Fedora specific assistance for R, as this thread 
is getting to be OT for R-Help.

Marc Schwartz

__
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] decision tree display

2010-09-15 Thread Uwe Ligges

If you are talking about an rpart object you are going to plot, see

?plot.rpart

and

?text.rpart


Uwe Ligges


On 19.08.2010 23:10, Olga Shaganova wrote:

I am using  plot and text commands to display the decision tree I built,
here is the code:

plot(fit, compress=TRUE)
  text(fit, use.n=TRUE)
but the the result of this code is not readable. Text doesn't get fully
displayed (missing on the margines and overlapping in the middle).  Where
can I get info on how to adjust the tree display and text size display?
Please help.

Thank you!
Olga

[[alternative HTML version deleted]]

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Re: [R] About choosing file

2010-09-15 Thread Uwe Ligges



On 26.08.2010 09:50, Stephen Liu wrote:

Hi folks,


Following command only works on Windows

Test01=read.table(file.choose(), header=TRUE)


It popup open a dialog box for choosing file.


But it doesn't work on Linux (Ubuntu);

Test01=read.table(file.choose(), header=TRUE)

Enter file name:


Entering file name doesn't work.  Please advise.  TIA



(assuming R-2.11.1)

Why not? What happens? What is the error message? Which file did you 
specify and does

list.files()
lists the file? Please show the output.

Best,
Uwe Ligges



B.R.
Stephen L




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Re: [R] binary package build error:target 'xf-' is not a directory

2010-09-15 Thread Uwe Ligges


You have to build a *source* package under Linux in order to install it 
under Windows afterwards. A Linux binary package cannot be used under 
Windows.


Best,
Uwe Ligges


On 31.08.2010 09:59, raje...@cse.iitm.ac.in wrote:



Hi,



I built a package in linux and generated its binary using R CMD build --binary 
testpack. This generated testpack.tar.gz.

I tried to install this package in windows using R CMD INSTALL package.tar.gz 
and I got the error

*installing to library 'H:/R-2.11.1/library'

*installing *binary* package 'testpack' ...
/cygdrive/RTools/bin/cp: target 'xf-' is not a directory

ERROR: installing binary package failed



can someone help me out. what does this mean?



~Aks
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Re: [R] Enumerating unit cell traversals Re: Interpolate? a line

2010-09-15 Thread David Winsemius


On Sep 15, 2010, at 8:58 AM, Rainer M Krug wrote:


-BEGIN PGP SIGNED MESSAGE-
Hash: SHA1

On 15/09/10 14:46, Michael Bedward wrote:

Nice posts David. I like the way that 'simple' problems such as this
one give rise to an interesting assortment of approaches and gotchas
:)

Michael

On 15 September 2010 22:41, David Winsemius  
dwinsem...@comcast.net wrote:


On Sep 15, 2010, at 7:55 AM, David Winsemius wrote:



On Sep 15, 2010, at 7:24 AM, David Winsemius wrote:


Replacing context:


Hello everyone.
I have created a 100*100 matrix in R.
Let's now say that I have a line that starts from (2,3) point  
and ends

to the
(62,34) point. In other words this line starts at cell (2,3)  
and ends

at cell
(62,34).

Is it possible to get by some R function all the matrix's  
cells that

this line
transverses?


This is a known problem in computer graphics and the algorythm used to
identify these cells is the Bresenham's Line Drawing Algorithm

You can check out one version at

http://www.falloutsoftware.com/tutorials/dd/dd4.htm

or

http://en.wikipedia.org/wiki/Bresenham's_line_algorithm


After viewing the Wikipedia description and the illustration I think  
that may not be the same problem. Note that the cells traversed only  
at their very outer corners do not get chosen. The goal is to  
approximate a line rather than to enumerate traversals.


--
David.




Cheers,

Rainer



I would like to thank you for your feedback.

Best Regards
Alex


On Sep 15, 2010, at 6:52 AM, Michael Bedward wrote:


Hello Alex,

Here is one way to do it. It works but it's not pretty :)


If you want an alternative, consider that produces the Y cell  
indices

(since the x cell indices are already 2:62):


linefn - function(x) 3+((34-3)/(62-2)) *(x-2)
findInterval(linefn(2:62), 3:34)
[1]  1  1  2  2  3  3  4  4  5  5  6  6  7  7  8  8  9  9 10 10  
11 11 12

12 13 13 14
[28] 14 15 15 16 17 17 18 18 19 19 20 20 21 21 22 22 23 23 24 24  
25 25 26

26 27 27 28
[55] 28 29 29 30 30 31 32
# that seems off by two

linefn(62)

[1] 34

linefn(2)
[1] 3 # but that checks out and I realized those were just  
indices for

the 3:34 findInterval vector


(3:34)[findInterval(linefn(2:62), 3:34)]
[1]  3  3  4  4  5  5  6  6  7  7  8  8  9  9 10 10 11 11 12 12  
13 13 14

14 15 15 16
[28] 16 17 17 18 19 19 20 20 21 21 22 22 23 23 24 24 25 25 26 26  
27 27 28

28 29 29 30
[55] 30 31 31 32 32 33 34

( no rounding and I think the logic is clearer.)


But I also realized it didn't enumerate all the the cells were  
crossed
either, only indicating which cell was associated with an integer  
value of
x. Also would have even more serious problems if the slope were  
greater than

unity. To enumerate the cell indices that were crossed, try:

unique( floor( cbind( seq(2,62, by=0.1), linefn(seq(2,62,  
by=0.1)) ) )  )

   [,1] [,2]
[1,]23
[2,]33
[3,]44
[4,]54
[5,]55
[6,]65
[7,]75
[8,]76
snipping interior results
[83,]   58   32
[84,]   59   32
[85,]   60   32
[86,]   60   33
[87,]   61   33
[88,]   62   34

That could probably be passed to rect() to illustrate (and check  
logic):


rect(cellidxs[,1], cellidxs[,2], cellidxs[,1]+1, cellidxs[,2]+1,
col=red)

#redraw line :
lines(2:62, 3+(34-3)/(62-2)*(0:60))


And then I got to wondering if every 0.1 was sufficient to catch  
all the
corners and discovered I could identify 3 more cell traversals  
with by=0.01



cellid2 -unique( floor(cbind(seq(2,62, by=0.01), linefn(seq(2,62,
by=0.01) )) ) )
NROW(cellid2) # 91 cells
rect(cellid2[,1], cellid2[,2], cellid2[,1]+1, cellid2[,2]+1,  
col=blue)

rect(cellidxs[,1], cellidxs[,2], cellidxs[,1]+1, cellidxs[,2]+1,
col=red)
lines(2:62, 3+(34-3)/(62-2)*(0:60))


(Trying by=0.001 did not change the count, but did take longer)

--
David.


-
David.



interp - approx(c(2, 62), c(3, 34), method=linear, xout=2:62)
m - matrix(c(interp$x, round(interp$y)), ncol=2)
tie - m[,2] == c(-Inf, m[-nrow(m),2])
m - m[ !tie, ]

You might want to examine the result like this...

plot(m)  # plots points
lines(c(2,26), c(3, 34))  # overlay line for comparison


you can add a grid with
abline(v=2:62, h=3:34)


Michael





David Winsemius, MD
West Hartford, CT

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- --
Rainer M. Krug, PhD (Conservation Ecology, SUN), MSc (Conservation
Biology, UCT), Dipl. Phys. (Germany)

Centre of Excellence for Invasion Biology
Natural Sciences Building
Office Suite 2039
Stellenbosch University
Main 

Re: [R] Error: cannot allocate vector of size X.0 Mb

2010-09-15 Thread John1983


Thank you all for the suggestions. We do intend to get more RAM space.
Meanwhile I shall have a look at the ShortRead package features.
-- 
View this message in context: 
http://r.789695.n4.nabble.com/Error-cannot-allocate-vector-of-size-X-0-Mb-tp2539031p2540518.html
Sent from the R help mailing list archive at Nabble.com.

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Re: [R] how to convert a unix R package to windows installable zip file?

2010-09-15 Thread Uwe Ligges



On 24.08.2010 17:35, Jack Luo wrote:

Hi,

I downloaded a package with .gz extension, which presumably works for Unix,
when I installed it directly on windows, it has the error msg:

Error in gzfile(file, r) : cannot open the connection
In addition: Warning messages:
1: In unzip(zipname, exdir = dest) : error 1 in extracting from zip file
2: In gzfile(file, r) :
   cannot open compressed file 'CustomCDF_1.1.0.tar.gz/DESCRIPTION', probable
reason 'No such file or directory'

Does anyone know how to convert the package into a windows installable zip
file?


Please see the R Installation and Administration manual in order to 
learn how to install source packages (such as the tar.gz file you got) 
under Windows.


Best,
Uwe Ligges



Thanks a bunch,

-Jack

[[alternative HTML version deleted]]

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[R] Keyword to clear the screen

2010-09-15 Thread Ron Michael
Hi all, can anyone please tell me what is the key-word to clear the screen? I 
am aware of the cont+L, however I need some R-syntax, that can be typed into 
the console to clear the screen.
 
Thanks


[[alternative HTML version deleted]]

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[R] Fwd: R install in Fedora

2010-09-15 Thread wesley mathew
Dear Sir

Actually I tried in both ways : yum install R and  sudo yum install R ,
but the error is same.
I have root privilege, means I used the command su - and enter the password.
 May be its problem from Fedora.
Thanks for your valuable help. I will post this issue in R-SIG-Fedora list.

Kind Regards

Wesley
-- Forwarded message --
From: Marc Schwartz marc_schwa...@me.com
Date: Wed, Sep 15, 2010 at 2:02 PM
Subject: Re: [R] R install in Fedora
To: wesley mathew wesleycmat...@gmail.com
Cc: r-help@r-project.org


On Sep 15, 2010, at 5:14 AM, wesley mathew wrote:

 Dear All

 As you suggested I tried to install  was trying to install
 R-2.11.1-3.fc13.1686 in  Fedora-13-i386 but it also shows error.
 I wrote  yum install R-2.11.1-3fc13.i686
 Errors
 Cannot retrieve repository metadata (repond.xml) for repository: fedora.
 Please verify its path and try again.
 Could you please help me to solve this problem.
 Thanks in advance
 Kind Regards

 Wesley C Mathew


The command that I told you to use was:

 yum install R

to be used as root. Note that there is nothing after the R.

You have to have root privileges to use yum and to install the R RPMs. So
you either need to use:

 su -

then enter the root password and then use the command above, or if you have
set up sudo, you can use:

 sudo yum install R

after which you will need to enter the root password.

If you still get the error, you have a configuration problem with your yum
repo setup, which means that you really need to be looking to the Fedora
(not R) lists for assistance, since it is not an R issue.

That being said, as I also noted, if you post to the R-SIG-Fedora list,
there are folks there who can offer Fedora specific assistance for R, as
this thread is getting to be OT for R-Help.

Marc Schwartz




-- 
Wesley C Mathew

[[alternative HTML version deleted]]

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


Re: [R] Enumerating unit cell traversals Re: Interpolate? a line

2010-09-15 Thread Rainer M Krug
-BEGIN PGP SIGNED MESSAGE-
Hash: SHA1

On 15/09/10 15:23, David Winsemius wrote:
 
 On Sep 15, 2010, at 8:58 AM, Rainer M Krug wrote:
 
 -BEGIN PGP SIGNED MESSAGE-
 Hash: SHA1

 On 15/09/10 14:46, Michael Bedward wrote:
 Nice posts David. I like the way that 'simple' problems such as this
 one give rise to an interesting assortment of approaches and gotchas
 :)

 Michael

 On 15 September 2010 22:41, David Winsemius dwinsem...@comcast.net
 wrote:

 On Sep 15, 2010, at 7:55 AM, David Winsemius wrote:


 On Sep 15, 2010, at 7:24 AM, David Winsemius wrote:

 Replacing context:

 Hello everyone.
 I have created a 100*100 matrix in R.
 Let's now say that I have a line that starts from (2,3) point
 and ends
 to the
 (62,34) point. In other words this line starts at cell (2,3) and
 ends
 at cell
 (62,34).

 Is it possible to get by some R function all the matrix's cells
 that
 this line
 transverses?

 This is a known problem in computer graphics and the algorythm used to
 identify these cells is the Bresenham's Line Drawing Algorithm

 You can check out one version at

 http://www.falloutsoftware.com/tutorials/dd/dd4.htm

 or

 http://en.wikipedia.org/wiki/Bresenham's_line_algorithm
 
 After viewing the Wikipedia description and the illustration I think
 that may not be the same problem. Note that the cells traversed only at
 their very outer corners do not get chosen. The goal is to approximate
 a line rather than to enumerate traversals.

You might actually be right.

Cheers,

Rainer
 


- -- 
Rainer M. Krug, PhD (Conservation Ecology, SUN), MSc (Conservation
Biology, UCT), Dipl. Phys. (Germany)

Centre of Excellence for Invasion Biology
Natural Sciences Building
Office Suite 2039
Stellenbosch University
Main Campus, Merriman Avenue
Stellenbosch
South Africa

Tel:+33 - (0)9 53 10 27 44
Cell:   +27 - (0)8 39 47 90 42
Fax (SA):   +27 - (0)8 65 16 27 82
Fax (D) :   +49 - (0)3 21 21 25 22 44
Fax (FR):   +33 - (0)9 58 10 27 44
email:  rai...@krugs.de

Skype:  RMkrug
-BEGIN PGP SIGNATURE-
Version: GnuPG v1.4.10 (GNU/Linux)
Comment: Using GnuPG with Mozilla - http://enigmail.mozdev.org/

iEYEARECAAYFAkyQyl8ACgkQoYgNqgF2egpdnACfS3zQX3MUAPraxIm5ExbFcnQu
bdYAn0vpYs3Gy2XmutNoJQKaVVZ+hnYH
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Re: [R] Keyword to clear the screen

2010-09-15 Thread Ivan Calandra
  The first result of RSiteSearch(clear console) gives 
PBSmodelling::clearRcon()

Ivan

Le 9/15/2010 15:27, Ron Michael a écrit :
 Hi all, can anyone please tell me what is the key-word to clear the screen? I 
 am aware of the cont+L, however I need some R-syntax, that can be typed 
 into the console to clear the screen.
   
 Thanks


   [[alternative HTML version deleted]]



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

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

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


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Re: [R] relative risk regression with survey data

2010-09-15 Thread Ravi Varadhan
Dear Thomas,

You said, the log-binomial model is very non-robust when the fitted values
get close to 1, and there is some controversy over the best approach.
Could you please point me to a paper that discusses the issues?

I have written some code to do maximum likelihood estimation for relative,
additive, and mixed risk regression models with binomial model.  I have been
able to obtain good convergence.  I have used bootstrap to get standard
errors.  However, I am not sure if these standard errors are valid when
fitted values were close to 0 or 1. It seems to me that when the fitted
probabilities are close to 0 or 1, there is not a good way to estimate
standard errors.
   

Thanks,
Ravi.

-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
Behalf Of Thomas Lumley
Sent: Monday, September 13, 2010 10:41 PM
To: Daniel Nordlund
Cc: r-help@r-project.org
Subject: Re: [R] relative risk regression with survey data

On Mon, 13 Sep 2010, Daniel Nordlund wrote:

 I have been asked to look at options for doing relative risk regression on

 some survey data.  I have a binary DV and several predictor / adjustment 
 variables.  In R, would this be as simple as using the survey package to

 set up an appropriate design object and then running svyglm with 
 family=binomial(log) ?  Any other suggestions for covariate adjustment of 
 relative risk estimates?  Any and all suggestions welcomed.

If the fitted values don't get too close to 1 then svyglm(
,family=quasibinomial(log)) will do it.

The log-binomial model is very non-robust when the fitted values get close
to 1, and there is some controversy over the best approach.  You can still
use svyglm(  ,family=quasibinomial(log)) but you will probably need to set
the number of iterations much higher (perhaps 200).

Alternatively, you can use nonlinear least squares  [svyglm(,
family=gaussian(log))] or other quasilikelihood approaches, such as
family=quasipoisson(log).  These are all consistent for the same parameter
if the model is correctly specified and are much more robust to x-outliers.
I rather like nonlinear least squares, because it's easy to explain.

  -thomas


Thomas Lumley
Professor of Biostatistics
University of Washington, Seattle

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Re: [R] Keyword to clear the screen

2010-09-15 Thread Henrique Dallazuanna
Try this (for windows):


clear - function() {
library(RDCOMClient)
ws - COMCreate(WScript.Shell)
invisible(ws$SendKeys('^L'))
}

On Wed, Sep 15, 2010 at 10:27 AM, Ron Michael ron_michae...@yahoo.comwrote:

 Hi all, can anyone please tell me what is the key-word to clear the screen?
 I am aware of the cont+L, however I need some R-syntax, that can be typed
 into the console to clear the screen.

 Thanks


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 http://www.R-project.org/posting-guide.html
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-- 
Henrique Dallazuanna
Curitiba-Paraná-Brasil
25° 25' 40 S 49° 16' 22 O

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[R] help for shift.down

2010-09-15 Thread Malangi Chikkalingaiah, Ramya
Hi all,

I created a matrix by parsing a csv file when I shift rows I am getting the 
wrong output  as below my matrix has single column with multiple rows but it is 
splitting into multiple columns and not shifting row wise also can you please 
resolve this issue
CMA1-function()
{
PMA3- read.csv(c:/myfile2.csv, strip.white = TRUE,head=TRUE,sep=,)
PMA1-as.matrix(PMA3)
shift.down(PMA1,rows=47,fill=0)
}
CMA1()
[,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10] [,11] [,12] [,13] [,14] 
[,15] [,16] [,17] [,18] [,19] [,20] [,21] [,22] [,23] [,24] [,25] [,26]
[1,]000000000 0 0 0 0 0 
0 0 0 0 0 0 0 0 0 0 0 0
[2,]000000000 0 0 0 0 0 
0 0 0 0 0 0 0 0 0 0 0 0
 [,27] [,28] [,29] [,30] [,31] [,32] [,33] [,34] [,35] [,36] [,37] [,38] 
[,39] [,40] [,41] [,42] [,43] [,44] [,45] [,46] [,47] [,48] [,49] [,50] [,51]
[1,] 0 0 0 0 0 0 0 0 0 0 0 0
 0 0 0 0 0 0 0 0 0 0 0 0 0
[2,] 0 0 0 0 0 0 0 0 0 0 0 0
 0 0 0 0 0 0 0 0 0 0 0 0 0
 [,52] [,53] [,54] [,55] [,56] [,57] [,58] [,59] [,60] [,61] [,62] [,63] 
[,64] [,65] [,66] [,67] [,68] [,69] [,70] [,71] [,72] [,73] [,74] [,75] [,76]
[1,] 0 0 0 0 0 0 0 0 0 0 0 0
 0 0 0 0 0 0 0 0 0 0 0 0 0
[2,] 0 0 0 0 0 0 0 0 0 0 0 0
 0 0 0 0 0 0 0 0 0 0 0 0 0
 [,77] [,78] [,79] [,80] [,81] [,82] [,83] [,84] [,85] [,86] [,87] [,88] 
[,89] [,90] [,91] [,92] [,93] [,94] [,95] [,96] [,97] [,98] [,99] [,100]
[1,] 0 0 0 0 0 0 0 0 0 0 0 0
 0 0 0 0 0 0 0 0 0 0 0  0
[2,] 0 0 0 0 0 0 0 0 0 0 0 0
 0 0 0 0 0 0 0 0 0 0 0  0
 [,101] [,102] [,103] [,104] [,105] [,106] [,107] [,108] [,109] [,110] 
[,111] [,112] [,113] [,114] [,115] [,116] [,117] [,118] [,119] [,120] [,121]
[1,]  0  0  0  0  0  0  0  0  0  0  
0  0  0  0  0  0  0  0  0  0  0
[2,]  0  0  0  0  0  0  0  0  0  0  
0  0  0  0  0  0  0  0  0  0  0
 [,122] [,123] [,124] [,125] [,126] [,127] [,128] [,129] [,130] [,131] 
[,132] [,133] [,134] [,135] [,136] [,137] [,138] [,139] [,140] [,141] [,142]
[1,]  0  0  0  0  0  0  0  0  0  0  
0  0  0  0  0  0  0  0  0  0  0
[2,]  0  0  0  0  0  0  0  0  0  0  
0  0  0  0  0  0  0  0  0  0  0
 [,143] [,144] [,145] [,146] [,147] [,148] [,149] [,150] [,151] [,152] 
[,153] [,154] [,155] [,156] [,157] [,158] [,159] [,160] [,161] [,162] [,163]
[1,]  0  0  0  0  0  0  0  0  0  0  
0  0  0  0  0  0  0  0  0  0  0
[2,]  0  0  0  0  0  0  0  0  0  0  
0  0  0  0  0  0  0  0  0  0  0
 [,164] [,165] [,166] [,167] [,168] [,169] [,170] [,171] [,172] [,173] 
[,174] [,175] [,176] [,177] [,178] [,179] [,180] [,181] [,182] [,183] [,184]
[1,]  0  0  0  0  0  0  0  0  0  0  
0  0  0  0  0  0  0  0  0  0  0
[2,]  0  0  0  0  0  0  0  0  0  0  
0  0  0  0  0  0  0  0  0  0  0
 [,185] [,186] [,187] [,188] [,189] [,190] [,191] [,192] [,193] [,194] 
[,195] [,196] [,197] [,198] [,199] [,200] [,201] [,202] [,203] [,204] [,205]
[1,]  0  0  0  0  0  0  0  0  0  0  
0  0  0  0  0  0  0  0  0  0  0
[2,]  0  0  0  0  0  0  0  0  0  0  
0  0  0  0  0  0  0  0  0  0  0
 [,206] [,207] [,208] [,209] [,210] [,211] [,212] [,213] [,214] [,215] 
[,216] [,217] [,218] [,219] [,220] [,221] [,222] [,223] [,224] [,225] [,226]
[1,]  0  0  0  0  0  0  0  0  0  0  
0  0  0  0  0  0  0  0  0  0  0
[2,]  0  0  0  0  0  0  0  0  0  0  
0  0  0  0  0  0  0  0  0 

Re: [R] Keyword to clear the screen

2010-09-15 Thread Ron Michael
Thanks everyone who have looked into my problem. In the mean time I already 
found one tip here 
http://onertipaday.blogspot.com/2007/05/how-to-clear-screen-in-r.html to do 
that. I am yet to try it...perhaps it would solve me problem :)
 
Thanks,

--- On Wed, 15/9/10, Ron Michael ron_michae...@yahoo.com wrote:


From: Ron Michael ron_michae...@yahoo.com
Subject: [R] Keyword to clear the screen
To: r-h...@stat.math.ethz.ch
Date: Wednesday, 15 September, 2010, 8:27 PM


Hi all, can anyone please tell me what is the key-word to clear the screen? I 
am aware of the cont+L, however I need some R-syntax, that can be typed into 
the console to clear the screen.
 
Thanks


    [[alternative HTML version deleted]]


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[R] Re : content analysis

2010-09-15 Thread Sametrie Neurones
Bob,

Thank you very much for your help. I found TM and Readme.
I am going to look into Tau and rattle right now.

Thanks,

Christian





De : Bob Green bgr...@dyson.brisnet.org.au
À : r-help@r-project.org
Cc : Sametrie Neurones neuro...@ymail.com
Envoyé le : Mer 15 septembre 2010, 21h 00min 29s
Objet : Re: content analysis

Christian,

Have you looked at TM?

There is also Quantitative Corpus Linguistics with R by Stefan Gries, which you 
may find of interest.

Tau, Readme and rattle might be worth looking into.

See also: http://ses.telecom-paristech.fr/lebart/

regards

Bob


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


Re: [R] About choosing file

2010-09-15 Thread Stephen Liu
Hi Uwe,

 (assuming R-2.11.1)
Yes

 Test01=read.table(file.choose(), header=TRUE)
Enter file name: 

My problem here is ;
the file is NOT on /home/user directory.  I need to browse for it.

 list.files()
[1] boxplot.pdf Desktop FourPlots01.png hist.png   
[5] jranke_cran.asc M-x R   M-x R~ 


B.R.
Stephen L






- Original Message 
From: Uwe Ligges lig...@statistik.tu-dortmund.de
To: Stephen Liu sati...@yahoo.com
Cc: R-help@r-project.org
Sent: Wed, September 15, 2010 9:20:48 PM
Subject: Re: [R] About choosing file



On 26.08.2010 09:50, Stephen Liu wrote:
 Hi folks,


 Following command only works on Windows
 Test01=read.table(file.choose(), header=TRUE)

 It popup open a dialog box for choosing file.


 But it doesn't work on Linux (Ubuntu);
 Test01=read.table(file.choose(), header=TRUE)
 Enter file name:


 Entering file name doesn't work.  Please advise.  TIA


(assuming R-2.11.1)

Why not? What happens? What is the error message? Which file did you 
specify and does
list.files()
lists the file? Please show the output.

Best,
Uwe Ligges


 B.R.
 Stephen L




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




__
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
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Re: [R] assignment by value or reference

2010-09-15 Thread Xiaobo Gu
Thanks.



Xiaobo.Gu

-Original Message-
From: Uwe Ligges [mailto:lig...@statistik.tu-dortmund.de]
Sent: Wednesday, September 15, 2010 5:06 PM
To: Xiaobo Gu
Cc: r-help@r-project.org; '顾小波'
Subject: Re: [R] assignment by value or reference

See the R Language Definition manual. Since R knows about lazy
evaluation, it is sometimes neither by reference nor by value.
If you want to think binary, then by value fits better than by
reference.

Uwe Ligges



On 05.09.2010 17:19, Xiaobo Gu wrote:
 Hi Team,

   Can you please tell me the rules of assignment in R, by value or
by reference.

 From my about 3 months of experience of part time job of R, it seems most
times it is by value, especially in function parameter and return values
assignment; and it is by reference when referencing container sub-objects of
container objects, such as elements of List objects and row/column objects of
DataFrame objectes; but it is by value when referencing the smallest unit of
element of a container object, such as cell of data frame objects.





 Xiaobo.Gu




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Re: [R] About choosing file

2010-09-15 Thread Uwe Ligges
You can specify an absolute or relative path (relative to the current 
directory) of course.


Uwe

On 15.09.2010 16:03, Stephen Liu wrote:

Hi Uwe,


(assuming R-2.11.1)

Yes


Test01=read.table(file.choose(), header=TRUE)

Enter file name:

My problem here is ;
the file is NOT on /home/user directory.  I need to browse for it.


list.files()

[1] boxplot.pdf Desktop FourPlots01.png hist.png
[5] jranke_cran.asc M-x R   M-x R~


B.R.
Stephen L






- Original Message 
From: Uwe Liggeslig...@statistik.tu-dortmund.de
To: Stephen Liusati...@yahoo.com
Cc: R-help@r-project.org
Sent: Wed, September 15, 2010 9:20:48 PM
Subject: Re: [R] About choosing file



On 26.08.2010 09:50, Stephen Liu wrote:

Hi folks,


Following command only works on Windows

Test01=read.table(file.choose(), header=TRUE)


It popup open a dialog box for choosing file.


But it doesn't work on Linux (Ubuntu);

Test01=read.table(file.choose(), header=TRUE)

Enter file name:


Entering file name doesn't work.  Please advise.  TIA



(assuming R-2.11.1)

Why not? What happens? What is the error message? Which file did you
specify and does
list.files()
lists the file? Please show the output.

Best,
Uwe Ligges



B.R.
Stephen L




__
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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.





__
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Re: [R] About choosing file

2010-09-15 Thread Henrique Dallazuanna
You can try tcltk:

library(tcltk)
tk_choose.files()

On Wed, Sep 15, 2010 at 11:03 AM, Stephen Liu sati...@yahoo.com wrote:

 Hi Uwe,

  (assuming R-2.11.1)
 Yes

  Test01=read.table(file.choose(), header=TRUE)
 Enter file name:

 My problem here is ;
 the file is NOT on /home/user directory.  I need to browse for it.

  list.files()
 [1] boxplot.pdf Desktop FourPlots01.png hist.png
 [5] jranke_cran.asc M-x R   M-x R~


 B.R.
 Stephen L






 - Original Message 
 From: Uwe Ligges lig...@statistik.tu-dortmund.de
 To: Stephen Liu sati...@yahoo.com
 Cc: R-help@r-project.org
 Sent: Wed, September 15, 2010 9:20:48 PM
 Subject: Re: [R] About choosing file



 On 26.08.2010 09:50, Stephen Liu wrote:
  Hi folks,
 
 
  Following command only works on Windows
  Test01=read.table(file.choose(), header=TRUE)
 
  It popup open a dialog box for choosing file.
 
 
  But it doesn't work on Linux (Ubuntu);
  Test01=read.table(file.choose(), header=TRUE)
  Enter file name:
 
 
  Entering file name doesn't work.  Please advise.  TIA


 (assuming R-2.11.1)

 Why not? What happens? What is the error message? Which file did you
 specify and does
 list.files()
 lists the file? Please show the output.

 Best,
 Uwe Ligges

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




 __
 R-help@r-project.org mailing list
 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.




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

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[R] Odp: Programming: loop versus vector oriented

2010-09-15 Thread Petr PIKAL
Hi

I do not want to go too much deep to internals of your function. What do 
you suppose to get as the result.

If you want to get results of your function for a vector of reynolds and 
dk you can use function outer and probably get rid of for cycle in the 
function.

outer(c(100, 530,2410), c(10, 150,200),lambda_wall)
  [,1]   [,2]   [,3]
[1,] 0.640 0.6400 0.6400
[2,] 0.1207547 0.12075472 0.12075472
[3,] 0.1081338 0.04515774 0.04515774


If you want to feed vector(s) to your function it would not be so easy as 
you need to check some predefined conditions for laminar or turbulent or 
any other flow.

In this case you cold check function

?switch

together with separate function definitions for each type of flow. But 
three if's could be as good as switch.

Basically your function probably works for vectors but only in certain 
conditions.

Laminar OK
 lambda_wall(10:20)
 [1] 6.40 5.818182 5.33 4.923077 4.571429 4.27 4.00 
3.764706
 [9] 3.56 3.368421 3.20

Turbulent needs dk 
 lambda_wall(2400:2410)
Error: argument dk is missing, with no default

But not any dk
 lambda_wall(2400:2410, 10)
Error in if (Re  65 * dk[z]) { : missing value where TRUE/FALSE needed

only a vector of the same length as reynolds

 lambda_wall(2400:2410, 10:21)
 [1] 0.10815993 0.10325278 0.09912072 0.09558754 0.09252750 0.08984834
 [7] 0.08748069 0.08537137 0.08347884 0.08177018 0.08021892

So before trying to elaborate your function further what shall be inputs 
and what is desired output?

Regards
Petr


r-help-boun...@r-project.org napsal dne 15.09.2010 11:57:28:

 Dear all,
 
 I am new to R and to it's programming philosophy. The following function
 is supposed to work on a vector, but I can't figure out how to do that
 without looping through every element of it. Is there a more elegant
 way?
 
 Note: I have shortened it, so it is NOT correct from the pipe hydraulics
 point of view
 
 # Calculate wall friction factor
 # reynolds: Reynolds number
 # dk: relative roughness of pipe
 lambda_wall - function (reynolds, dk) {
   z - 1
   result - 0
 
   for (Re in reynolds) {
 if (Re = 2320) {
   # Laminar flow
   lambda - 64/Re
 } else if (Re  65 * dk[z]) {
   # Turbulent flow
   if (Re  1e+5) {
lambda - 0.3164 / sqrt(sqrt(Re))
   } else {
lambda - 0.309/(log10(Re/7))^2
   }
 } else {
   # Intermediate area
   lambdanew - 1 / (2 * log10(3.71 * dk[z]))^2 # Start value
   iter - 0
 
   repeat {
lambda - lambdanew
lambdanew - 1 / (2 * log10(2.51/(Re * sqrt(lambda)) + 0.27/dk[z]))^2
iter - iter + 1
if ((abs(lambdanew - lambda)  0.001) || (iter  100)) break
   }
 
   lambda = lambdanew
 }
 
 result[z] - lambda
 z - z + 1
   }
 
   result
 } # lambda_wall()
 
 __
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[R] Creating publication-quality plots for use in Microsoft Word

2010-09-15 Thread dadrivr

Hi everyone,

I am trying to make some publication-quality plots for use in Microsoft
Word, but I am having trouble creating high-quality plots that are supported
by Microsoft Word.

If I use the R plot function to create the figure, the lines are jagged, and
the picture is not of high quality (same with JPEG(), TIFF(), and PNG()
functions).  I have tried using the Cairo package, but it distorts my dashed
lines, and the win.metafile results in a picture of terrible quality.  The
only way I have succeeded in getting a high quality picture in a file is by
using the pdf() function to save the plot as a pdf file, but all my attempts
to convert the image in the pdf file to a TIFF or other file type accepted
by Word result in considerably degraded quality.  Do you have any
suggestions for creating publication-quality plots in R that can be placed
in Word documents?  What packages, functions (along with options), and/or
conversions would you use?  Thanks so much for your help!
-- 
View this message in context: 
http://r.789695.n4.nabble.com/Creating-publication-quality-plots-for-use-in-Microsoft-Word-tp2540676p2540676.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
and provide commented, minimal, self-contained, reproducible code.


Re: [R] Creating publication-quality plots for use in Microsoft Word

2010-09-15 Thread Bryan Hanson
There's many ways to solve this, but you are close to one already:  Make the
pdf, put the cursor where you want it in the document, then on the menu bar
Insert -- Picture -- From File... And navigate to the file.  This works on
the Mac, and seems to store the picture internally in a different way that
selecting the graphic in a viewer and cutting and pasting.  Quality is
top-notch and the graphic is clickable to be resized (and retains it's
quality).

HTH.  Bryan
*
Bryan Hanson
Professor of Chemistry  Biochemistry
DePauw University, Greencastle IN USA



On 9/15/10 10:38 AM, dadrivr dadr...@gmail.com wrote:

 
 Hi everyone,
 
 I am trying to make some publication-quality plots for use in Microsoft
 Word, but I am having trouble creating high-quality plots that are supported
 by Microsoft Word.
 
 If I use the R plot function to create the figure, the lines are jagged, and
 the picture is not of high quality (same with JPEG(), TIFF(), and PNG()
 functions).  I have tried using the Cairo package, but it distorts my dashed
 lines, and the win.metafile results in a picture of terrible quality.  The
 only way I have succeeded in getting a high quality picture in a file is by
 using the pdf() function to save the plot as a pdf file, but all my attempts
 to convert the image in the pdf file to a TIFF or other file type accepted
 by Word result in considerably degraded quality.  Do you have any
 suggestions for creating publication-quality plots in R that can be placed
 in Word documents?  What packages, functions (along with options), and/or
 conversions would you use?  Thanks so much for your help!

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


Re: [R] Creating publication-quality plots for use in Microsoft Word

2010-09-15 Thread Gabor Grothendieck
On Wed, Sep 15, 2010 at 10:38 AM, dadrivr dadr...@gmail.com wrote:

 Hi everyone,

 I am trying to make some publication-quality plots for use in Microsoft
 Word, but I am having trouble creating high-quality plots that are supported
 by Microsoft Word.

 If I use the R plot function to create the figure, the lines are jagged, and
 the picture is not of high quality (same with JPEG(), TIFF(), and PNG()
 functions).  I have tried using the Cairo package, but it distorts my dashed
 lines, and the win.metafile results in a picture of terrible quality.  The
 only way I have succeeded in getting a high quality picture in a file is by
 using the pdf() function to save the plot as a pdf file, but all my attempts
 to convert the image in the pdf file to a TIFF or other file type accepted
 by Word result in considerably degraded quality.  Do you have any
 suggestions for creating publication-quality plots in R that can be placed
 in Word documents?  What packages, functions (along with options), and/or
 conversions would you use?  Thanks so much for your help!


Those are all bitmapped formats. For best quality you want a
vector-based format. This link here discusses the difference:
http://web.archive.org/web/20070221152152/http://www.stc-saz.org/resources/0203_graphics.pdf

Microsoft's metafile formats are vector formats that work well in
Word.  Try savePlot with type = wmf or type = emf as the argument
or right click a graphic in R and choose one of the metafile options.


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Re: [R] Creating publication-quality plots for use in Microsoft Word

2010-09-15 Thread Marc Schwartz
That approach will be unique to OSX, upon which PDF is a default format. You 
can copy and paste from a PDF document using Preview into Office or iWork or 
similar apps. However, when subsequently displaying that content on a non-OSX 
system, the content will be shown as a bitmap not as the vector based PDF. You 
would have to save the new document to a new PDF file via the printer dialogs, 
in order to retain the full high quality PDF content.

The ultimate question for the OP is what do you intend to do with the Word 
document relative to displaying/printing the content in the Word document?

Using EMF/WMF is one way to get vector based images from R into other Windows 
apps and typically that provides suitable quality.

However, if you truly want publication quality, then you would need to use 
encapsulated postscript (EPS) for which you can see ?postscript and pay 
attention to the Details section for proper creation.

However, after importing an EPS image into Word on Windows, you will not see 
the EPS based image, but a bitmapped preview. You would then have to print the 
document to a PS compatible printer to see the vector based content in the 
imported image in the output.

So it all depends upon how you plan to use the document.

HTH,

Marc Schwartz

On Sep 15, 2010, at 9:50 AM, Bryan Hanson wrote:

 There's many ways to solve this, but you are close to one already:  Make the
 pdf, put the cursor where you want it in the document, then on the menu bar
 Insert -- Picture -- From File... And navigate to the file.  This works on
 the Mac, and seems to store the picture internally in a different way that
 selecting the graphic in a viewer and cutting and pasting.  Quality is
 top-notch and the graphic is clickable to be resized (and retains it's
 quality).
 
 HTH.  Bryan
 *
 
 On 9/15/10 10:38 AM, dadrivr dadr...@gmail.com wrote:
 
 
 Hi everyone,
 
 I am trying to make some publication-quality plots for use in Microsoft
 Word, but I am having trouble creating high-quality plots that are supported
 by Microsoft Word.
 
 If I use the R plot function to create the figure, the lines are jagged, and
 the picture is not of high quality (same with JPEG(), TIFF(), and PNG()
 functions).  I have tried using the Cairo package, but it distorts my dashed
 lines, and the win.metafile results in a picture of terrible quality.  The
 only way I have succeeded in getting a high quality picture in a file is by
 using the pdf() function to save the plot as a pdf file, but all my attempts
 to convert the image in the pdf file to a TIFF or other file type accepted
 by Word result in considerably degraded quality.  Do you have any
 suggestions for creating publication-quality plots in R that can be placed
 in Word documents?  What packages, functions (along with options), and/or
 conversions would you use?  Thanks so much for your help!


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[R] Contour line coordinates

2010-09-15 Thread Tonja Krueger
Hi all,
I used contour() to add contour lines to a plot. Now I’m wondering if there is 
a way to get an output of the calculated x- and y- coordinates of the contour 
lines?
Tonja

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Re: [R] About choosing file

2010-09-15 Thread Stephen Liu
Hi Henrique,

Thanks for your advice which works for me.

B.R.
Stephen L





From: Henrique Dallazuanna www...@gmail.com

Cc: R-help@r-project.org
Sent: Wed, September 15, 2010 10:24:01 PM
Subject: Re: [R] About choosing file

You can try tcltk:

library(tcltk)
tk_choose.files()




Hi Uwe,


 (assuming R-2.11.1)
Yes


 Test01=read.table(file.choose(), header=TRUE)
Enter file name:

My problem here is ;
the file is NOT on /home/user directory.  I need to browse for it.

 list.files()
[1] boxplot.pdf Desktop FourPlots01.png hist.png
[5] jranke_cran.asc M-x R   M-x R~


B.R.
Stephen L







- Original Message 
From: Uwe Ligges lig...@statistik.tu-dortmund.de

Cc: R-help@r-project.org
Sent: Wed, September 15, 2010 9:20:48 PM
Subject: Re: [R] About choosing file



On 26.08.2010 09:50, Stephen Liu wrote:
 Hi folks,


 Following command only works on Windows
 Test01=read.table(file.choose(), header=TRUE)

 It popup open a dialog box for choosing file.


 But it doesn't work on Linux (Ubuntu);
 Test01=read.table(file.choose(), header=TRUE)
 Enter file name:


 Entering file name doesn't work.  Please advise.  TIA


(assuming R-2.11.1)

Why not? What happens? What is the error message? Which file did you
specify and does
list.files()
lists the file? Please show the output.

Best,
Uwe Ligges


 B.R.
 Stephen L




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Re: [R] Package build and install under Windows

2010-09-15 Thread Tao

Thanks. It seems that both Rtools210.exe and Rtools211.exe support R 1.10.x,
which one is better for R 2.10.1. 
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Re: [R] Contour line coordinates

2010-09-15 Thread David Winsemius


On Sep 15, 2010, at 11:13 AM, Tonja Krueger wrote:


Hi all,
I used contour() to add contour lines to a plot. Now I’m wondering  
if there is a way to get an output of the calculated x- and y-  
coordinates of the contour lines?


?contourLines   # as was suggested to be the first  See(n) Also in  
contour.




Tonja

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David Winsemius, MD
West Hartford, CT

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[R] characters in a string

2010-09-15 Thread raje...@cse.iitm.ac.in

Hi,

I need to check if a string rha,b,c,drh is delimited by two rh 's as 
efficiently as possible(I need to do this a lot of times) and return TRUE. Can 
someone suggest a good technique? 
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Re: [R] Contour line coordinates

2010-09-15 Thread Duncan Murdoch

 On 15/09/2010 11:13 AM, Tonja Krueger wrote:

Hi all,
I used contour() to add contour lines to a plot. Now I’m wondering if there is 
a way to get an output of the calculated x- and y- coordinates of the contour 
lines?



?contourLines

Duncan Murdoch

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Re: [R] Creating publication-quality plots for use in Microsoft Word

2010-09-15 Thread dadrivr

Thanks for your help, guys.  I'm looking to produce a high-quality plot (no
jagged lines or other distortions) with a filetype that is accepted by
Microsoft Word on a PC and that most journals will accept.  That's why I'd
prefer to stick with JPEG, TIFF, PNG, or the like.  I'm not sure EPS would
fly.

I tried inserting the PDF directly into Word, but I am on a PC and there is
a loss of quality in the transfer.  I'm not sure I know how to use the
approach that Marc suggested in reference to saving a new PDF for use in
Word.

I also tried Gabor's suggestion to save in Microsoft's metafile format
(savePlot with type = wmf and emf), but the images contain lines that are as
jagged as those created from the regular R plot output.

Is there a way to enable anti-aliasing on all regular R plot output to clean
up the jaggies and then save it in another format?  Or should I try
something else?  Thanks again!


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[R] symmetric dotplot? (Wilkinson 1999, Am Stat 53 (3) 276-281

2010-09-15 Thread Paul Artes

DeaR all,

The stripchart function (graphics) is provides jittered and stacked
univariate scatterplots, but I wonder if anyone has implemented a
*symmetric* version of this - as in the lower panel of Wilkinson's paper:

http://www.jstor.org/stable/2686111

I have looked through several functions in contributed packages that do
similar things, but not found one that can do this. What am I missing?

Thanks!

Paul

 
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Re: [R] multiple cores/building fails

2010-09-15 Thread Gregory Ryslik
Hi,

That seems to be the case. Is there a way that I can put an object in some 
global place where all the workers can access it?

Doing the following at the start (before I make multiple workers) does not work:
Assign('global.control', control, globalenv())

Thanks for your help!

Kind regards,
Greg

Sent via BlackBerry by ATT

-Original Message-
From: Uwe Ligges lig...@statistik.tu-dortmund.de
Date: Wed, 15 Sep 2010 10:56:54 
To: Gregory Ryslikrsa...@comcast.net
Cc: r-h...@stat.math.ethz.ch
Subject: Re: [R] multiple cores/building fails



On 14.09.2010 21:50, Gregory Ryslik wrote:
 Hi Everyone,

 I have written debugged and tested my code and it works (yay!). I recently 
 tried to parallelize it (1 core/per tree) and when I run the code I get an 
 error saying the object control was not found. I do have an object control 
 in my code but it always works fine as long as I am not running it on 
 multiple cores (which I am doing via the help of sleigh and the 'nws' 
 package).

 Think that this could be an error in my source statements, I tried to build 
 my package (with the hope of replace source with library()) and I get the 
 following (where XXX is my package name that I've removed for privacy 
 reasons):

 * creating vignettes ... ERROR

 Error: processing vignette '.Rnw' failed with diagnostics:
   chunk 4
 Error : object control not found

 Again it gives me that this control object is not found.

 Does anyone have an idea how I can isolate what's going on. It's slightly 
 tough to do because it all works fine in the single processor case!


Perhaps you forgot to populate the object to the client nodes?
Hard to say without any reproducible example code.

Uwe Ligges


 Thanks!

 Kind regards,
 Greg
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Re: [R] characters in a string

2010-09-15 Thread Marc Schwartz
On Sep 15, 2010, at 10:16 AM, raje...@cse.iitm.ac.in wrote:

 
 Hi,
 
 I need to check if a string rha,b,c,drh is delimited by two rh 's 
 as efficiently as possible(I need to do this a lot of times) and return TRUE. 
 Can someone suggest a good technique? 


See ?grep and ?regex

 grepl(^rh.*rh$, rha,b,c,drh)
[1] TRUE


You can pass the entire source vector to grepl():

Vec - c(rha,b,c,drh, 1, 2, 3, 4, a, b, c, drh, rh1, 2, 3, 
4rh)

 grepl(^rh.*rh$, Vec)
[1]  TRUE FALSE FALSE  TRUE


HTH,

Marc Schwartz

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Re: [R] characters in a string

2010-09-15 Thread Romain Francois

Le 15/09/10 17:16, raje...@cse.iitm.ac.in a écrit :



Hi,

I need to check if a string rha,b,c,drh is delimited by tworh 's as 
efficiently as possible(I need to do this a lot of times) and return TRUE. Can someone suggest a good technique?


Hi Rajesh,

 f - function( x ) grepl( ^rh.*rh$, x )
 f( rha,b,c,drh )
[1] TRUE

See ?grepl for details.

Romain

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Re: [R] Creating publication-quality plots for use in Microsoft Word

2010-09-15 Thread Gabor Grothendieck
On Wed, Sep 15, 2010 at 11:25 AM, dadrivr dadr...@gmail.com wrote:

 Thanks for your help, guys.  I'm looking to produce a high-quality plot (no
 jagged lines or other distortions) with a filetype that is accepted by
 Microsoft Word on a PC and that most journals will accept.  That's why I'd
 prefer to stick with JPEG, TIFF, PNG, or the like.  I'm not sure EPS would
 fly.

 I tried inserting the PDF directly into Word, but I am on a PC and there is
 a loss of quality in the transfer.  I'm not sure I know how to use the
 approach that Marc suggested in reference to saving a new PDF for use in
 Word.

 I also tried Gabor's suggestion to save in Microsoft's metafile format
 (savePlot with type = wmf and emf), but the images contain lines that are as
 jagged as those created from the regular R plot output.

You may simply be viewing the document at such a small resolution that
the resolution is not there to display smooth lines regardless of the
approach.  Try viewing the document at increased zoom.  With vector
based images they automatically adapt to the resolution available.

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Re: [R] characters in a string

2010-09-15 Thread David Winsemius


On Sep 15, 2010, at 11:16 AM, raje...@cse.iitm.ac.in wrote:



Hi,

I need to check if a string rha,b,c,drh is delimited by two  
rh 's as efficiently as possible(I need to do this a lot of  
times) and return TRUE. Can someone suggest a good technique?


 txt - rha,b,c,drh

 grep(^rh.+rh$, txt)
[1] 1


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Re: [R] Creating publication-quality plots for use in Microsoft Word

2010-09-15 Thread Marc Schwartz
On Sep 15, 2010, at 10:25 AM, dadrivr wrote:

 
 Thanks for your help, guys.  I'm looking to produce a high-quality plot (no
 jagged lines or other distortions) with a filetype that is accepted by
 Microsoft Word on a PC and that most journals will accept.  That's why I'd
 prefer to stick with JPEG, TIFF, PNG, or the like.  I'm not sure EPS would
 fly.
 
 I tried inserting the PDF directly into Word, but I am on a PC and there is
 a loss of quality in the transfer.  I'm not sure I know how to use the
 approach that Marc suggested in reference to saving a new PDF for use in
 Word.
 
 I also tried Gabor's suggestion to save in Microsoft's metafile format
 (savePlot with type = wmf and emf), but the images contain lines that are as
 jagged as those created from the regular R plot output.
 
 Is there a way to enable anti-aliasing on all regular R plot output to clean
 up the jaggies and then save it in another format?  Or should I try
 something else?  Thanks again!


The comments that I had vis-a-vis saving to PDF were specific to operating on 
an OSX platform. They won't apply to Windows/Linux.

The issue is that the notion of most journals is problematic. Many journals 
will have requirements for specific formats, including plot/graphic output, 
which I would note includes the use of LaTeX/EPS/PDF as the source content, not 
Word or similar word processing formats.

If you need to stick with bitmapped formats, then you need to generate the plot 
file with a specific size/dpi in mind, so that the content will not be resized 
during editing. It is the resizing of plots and images that gets you into 
trouble with bitmapped formats, resulting in the oft seen pixelation.

I would recommend that you check specifically with the journal(s) that you are 
targeting to find out exactly what they want. They typically publish guidelines 
for authors, which you should seek out.

Once you know exactly what the specific journals require, then you can target 
the content creation accordingly.

You can also search the archives via rseek.org (using publication quality 
plots) to review past discussions on this same topic.

HTH,

Marc

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Re: [R] Creating publication-quality plots for use in Microsoft Word

2010-09-15 Thread Thomas Lumley

On Wed, 15 Sep 2010, dadrivr wrote:



Thanks for your help, guys.  I'm looking to produce a high-quality plot (no
jagged lines or other distortions) with a filetype that is accepted by
Microsoft Word on a PC and that most journals will accept.  That's why I'd
prefer to stick with JPEG, TIFF, PNG, or the like.  I'm not sure EPS would
fly.


One simple approach, which I use when I have to create graphics for MS Office 
while on a non-Windows platform is to use PNG and set the resolution and file 
size large enough.  At 300dpi or so the physics of ink on paper does all the 
antialiasing you need.

Work out how big you want the graph to be, and use PNG with enough pixels to 
get at least 300dpi at that final size. You'll need to set the pointsize 
argument and it will help to set the resolution argument.

 -thomas

Thomas Lumley
Professor of Biostatistics
University of Washington, Seattle

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Re: [R] relative risk regression with survey data

2010-09-15 Thread Thomas Lumley

On Wed, 15 Sep 2010, Ravi Varadhan wrote:


Dear Thomas,

You said, the log-binomial model is very non-robust when the fitted values
get close to 1, and there is some controversy over the best approach.
Could you please point me to a paper that discusses the issues?

I have written some code to do maximum likelihood estimation for relative,
additive, and mixed risk regression models with binomial model.  I have been
able to obtain good convergence.  I have used bootstrap to get standard
errors.  However, I am not sure if these standard errors are valid when
fitted values were close to 0 or 1. It seems to me that when the fitted
probabilities are close to 0 or 1, there is not a good way to estimate
standard errors.


There's a technical report at 
http://www.bepress.com/uwbiostat/paper293/

with simulations, some theory, and references.  It's under review at the 
moment, after being forgotten for a few years.

The distribution of the parameter estimates when the true parameter is on the 
boundary of the parameter space is a separate mess.
 Theoretically it is the intersection of the the multivariate Normal with the 
parameter space, and if the parameter space has a piecewise linear boundary the 
log likelihood ratio has a chi-squared mixture distribution.  In practice, if 
there isn't a hard edge to the covariate distribution it's not going to be easy 
to get a good approximation to the distribution of parameter estimates. As an 
example of the complications, the sampling distributions for fixed and random 
design matrices can be very different, because a random design matrix means 
that the estimated edge of the parameter space moves from one realization to 
another.

-thomas

Thomas Lumley
Professor of Biostatistics
University of Washington, Seattle

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[R] Difficulty creating Julian day in data frame

2010-09-15 Thread Sean Parks
Hi,

I'm attempting to add a Julian Day column to a data frame.

Here is my code and the resulting data frame:

vic.data - read.table(C:/VIC/data/vic.data.csv, header=F)
names(vic.data) - c(year, month, day, precip, evap, 
runoff, baseflow, Tsup, SM1, SM2, SM3, SWE)
 
vic.data$temp.date - as.Date(paste(vic.data$year, /, 
vic.data$month, /, vic.data$day, sep=))
vic.data$julian.day - julian(vic.data$temp.date, origin = 
as.Date(1900-01-01))[1]
head(vic.data)

  year month day precip   evap runoff baseflow Tsup SM1 SM2  
SM3  SWE  temp.date julian.day
1 1916 1   1   0.00 0.0789  0   0.5037 -15.9907 20.2285 63.4011 
296.9437 214.7537 1916-01-01   5843
2 1916 1   2   0.00 0.1135  0   0.5030 -14.1946 20.1619 63.3425 
296.5659 214.7009 1916-01-02   5843
3 1916 1   3  29.74 0.1584  0   0.5024 -11.1894 20.0998 63.2817 
296.1864 244.0971 1916-01-03   5843
4 1916 1   4  11.37 0.2056  0   0.5017 -12.9745 20.0416 63.2194 
295.8052 255.3145 1916-01-04   5843
5 1916 1   5   0.00 0.1467  0   0.5011 -15.1542 19.9866 63.1561 
295.4224 255.2361 1916-01-05   5843
6 1916 1   6   0.00 0.0939  0   0.5004 -12.5409 19.9345 63.0922 
295.0379 255.2084 1916-01-06   5843


As you can see, the temp.date column behaves as expected, but the 
julian.day column is populated with the Julian day value from 1/1/1916.

Please help me out if you have some ideas.

Thanks,
Sean




*
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Ecologist | Geospatial Analyst
Aldo Leopold Wilderness Research Institute
Rocky Mountain Research Station
(406) 542-4182
http://leopold.wilderness.net/
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[R] Scripting help

2010-09-15 Thread Ayyappa Chaturvedula
Dear all, I am new to R and this group. I have good experience in S  
scripts.  I need some orientation on data imports, general plotting  
functions. Can you please direct me?


Regards,Ayyappa Chaturvedula

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Re: [R] Difficulty creating Julian day in data frame

2010-09-15 Thread David Winsemius


On Sep 15, 2010, at 12:05 PM, Sean Parks wrote:


Hi,

I'm attempting to add a Julian Day column to a data frame.

Here is my code and the resulting data frame:

   vic.data - read.table(C:/VIC/data/vic.data.csv, header=F)
   names(vic.data) - c(year, month, day, precip, evap,
runoff, baseflow, Tsup, SM1, SM2, SM3, SWE)






   vic.data$julian.day - julian(vic.data$temp.date, origin =
as.Date(1900-01-01))[1]


There's one problem. Why are you appending the [1]? to the as.Date  
function. It's just going to give you the first date.

--
David.


   head(vic.data)

 year month day precip   evap runoff baseflow Tsup SM1 SM2
SM3  SWE  temp.date julian.day
1 1916 1   1   0.00 0.0789  0   0.5037 -15.9907 20.2285  
63.4011

296.9437 214.7537 1916-01-01   5843
2 1916 1   2   0.00 0.1135  0   0.5030 -14.1946 20.1619  
63.3425

296.5659 214.7009 1916-01-02   5843
3 1916 1   3  29.74 0.1584  0   0.5024 -11.1894 20.0998  
63.2817

296.1864 244.0971 1916-01-03   5843
4 1916 1   4  11.37 0.2056  0   0.5017 -12.9745 20.0416  
63.2194

295.8052 255.3145 1916-01-04   5843
5 1916 1   5   0.00 0.1467  0   0.5011 -15.1542 19.9866  
63.1561

295.4224 255.2361 1916-01-05   5843
6 1916 1   6   0.00 0.0939  0   0.5004 -12.5409 19.9345  
63.0922

295.0379 255.2084 1916-01-06   5843


As you can see, the temp.date column behaves as expected, but the
julian.day column is populated with the Julian day value from  
1/1/1916.


Please help me out if you have some ideas.

Thanks,
Sean




*
Sean Parks
Ecologist | Geospatial Analyst
Aldo Leopold Wilderness Research Institute
Rocky Mountain Research Station
(406) 542-4182
http://leopold.wilderness.net/
*
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David Winsemius, MD
West Hartford, CT

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Re: [R] characters in a string

2010-09-15 Thread raje...@cse.iitm.ac.in
great! Thanks

- Original Message -
From: David Winsemius dwinsem...@comcast.net
To: raje...@cse.iitm.ac.in
Cc: r-help r-help@r-project.org
Sent: Wed, 15 Sep 2010 21:12:27 +0530 (IST)
Subject: Re: [R] characters in a string


On Sep 15, 2010, at 11:16 AM, raje...@cse.iitm.ac.in wrote:


 Hi,

 I need to check if a string a,b,c,d is delimited by two  
  's as efficiently as possible(I need to do this a lot of  
 times) and return TRUE. Can someone suggest a good technique?

  txt - a,b,c,d
 
  grep(^.+$, txt)
[1] 1

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David Winsemius, MD
West Hartford, CT



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[R] Computing effect sizes based on mixed models

2010-09-15 Thread Andrea Meyer
In an intervention study with subjects randomly allocated to two treatments 
(treat A and B) and three time points (time) plus an additional baseline 
measurement (dv_base), I've set up the following model to test for differences 
in temporal courses of treatments for the outcome (dv), thereby allowing for 
individual intercepts and slopes:

lmer(dv ~ dv_base+treat*time+(1+time|subject))

Fixed effects:
Estimate  Std. Error DF t value
(Intercept)-1.080041   0.126665  58  -8.527
dv_base-0.888656   0.090617  53  -9.807
treatB  0.645455   0.190541  53   3.387
time   -0.001726   0.163044  58  -0.011
treatB:time 0.377888   0.271972  58   1.389

I'm interested now in comparing estimated treatment means for let's say the 
last time point and I've centered 'time' accordingly. The term 'treatB' shows 
the difference which is relatively high and significantly different from 0 
(t(53)=3.387). Now I want to compute the effect size of this difference, based 
on the model, not on the observed values. From my understanding I could obtain 
raw effect sizes by simply reporting the value of the term 'treatB' (=0.645). 
When it comes to standardized effect sizes (comparable to Cohen's d) I could 
simply take the t-value and the degrees of freedom and use the formula 
2*t/sqrt(DF)=0.930. 

My question: Is this a correct procedure? I'm somewhat unsure as I've never 
encountered it in the literature.

And help on this is greatly appreciated.

Andrea


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Re: [R] Difficulty creating Julian day in data frame

2010-09-15 Thread David Winsemius


On Sep 15, 2010, at 12:23 PM, David Winsemius wrote:



On Sep 15, 2010, at 12:05 PM, Sean Parks wrote:


Hi,

I'm attempting to add a Julian Day column to a data frame.

Here is my code and the resulting data frame:

  vic.data - read.table(C:/VIC/data/vic.data.csv, header=F)
  names(vic.data) - c(year, month, day, precip, evap,
runoff, baseflow, Tsup, SM1, SM2, SM3, SWE)






  vic.data$julian.day - julian(vic.data$temp.date, origin =
as.Date(1900-01-01))[1]


There's one problem. Why are you appending the [1]? to the as.Date  
function.


Er, ... that would be the julian function. (But same import.)
-- David

It's just going to give you the first date.
--
David.


  head(vic.data)

year month day precip   evap runoff baseflow Tsup SM1 SM2
SM3  SWE  temp.date julian.day
1 1916 1   1   0.00 0.0789  0   0.5037 -15.9907 20.2285  
63.4011

296.9437 214.7537 1916-01-01   5843
2 1916 1   2   0.00 0.1135  0   0.5030 -14.1946 20.1619  
63.3425

296.5659 214.7009 1916-01-02   5843
3 1916 1   3  29.74 0.1584  0   0.5024 -11.1894 20.0998  
63.2817

296.1864 244.0971 1916-01-03   5843
4 1916 1   4  11.37 0.2056  0   0.5017 -12.9745 20.0416  
63.2194

295.8052 255.3145 1916-01-04   5843
5 1916 1   5   0.00 0.1467  0   0.5011 -15.1542 19.9866  
63.1561

295.4224 255.2361 1916-01-05   5843
6 1916 1   6   0.00 0.0939  0   0.5004 -12.5409 19.9345  
63.0922

295.0379 255.2084 1916-01-06   5843


As you can see, the temp.date column behaves as expected, but the
julian.day column is populated with the Julian day value from  
1/1/1916.


Please help me out if you have some ideas.

Thanks,
Sean




*
Sean Parks
Ecologist | Geospatial Analyst
Aldo Leopold Wilderness Research Institute
Rocky Mountain Research Station
(406) 542-4182
http://leopold.wilderness.net/
*
[[alternative HTML version deleted]]

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


David Winsemius, MD
West Hartford, CT

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David Winsemius, MD
West Hartford, CT

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Re: [R] How to uncompress a gz file in R

2010-09-15 Thread Wonsang You
Dear Henrik,

Thank you so much for your kind help. Unfortunately, I could not find out
any function such as 'gunzip' in R.utils package. Instead, I could be
successful by using the following command.

system(gunzip filename)

On the other hand, the function 'gzfile' supports the compression as gz
format, but I still do not know how to decompress gz file by using the
function 'gzfile'.

Best Regards,
Wonsang


On 14 September 2010 15:23, Henrik Bengtsson h...@stat.berkeley.edu wrote:

 To uncompress an *.gz file into another file on disk, see also ?gunzip
 in the R.utils package.

 /Henrik

 2010/9/14 Uwe Ligges lig...@statistik.tu-dortmund.de:
  See ?gzfile
 
  Uwe Ligges
 
 
  On 14.09.2010 11:02, Wonsang You wrote:
 
  Dear Fellows,
 
  I would like to know how to uncompress a gz file at the R console. I
 could
  not find out any help from the R-help archive.
  Thanks for your great help.
 
  Best Regards,
  Wonsang You
 
 
  -
  --
  Wonsang You
  Special Lab Non-Invasive Brain Imaging
  Leibniz Institute for Neurobiology
  http://www.ifn-magdeburg.de
 
  __
  R-help@r-project.org mailing list
  https://stat.ethz.ch/mailman/listinfo/r-help
  PLEASE do read the posting guide
 http://www.R-project.org/posting-guide.html
  and provide commented, minimal, self-contained, reproducible code.
 


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Re: [R] Creating publication-quality plots for use in Microsoft Word

2010-09-15 Thread Max Kuhn
You might want to check out the Reproducible Research task view:

   http://cran.r-project.org/web/views/ReproducibleResearch.html

There is a section on Microsoft formats, as well as other formats that
can be converted.

Max



On Wed, Sep 15, 2010 at 11:49 AM, Thomas Lumley
tlum...@u.washington.edu wrote:
 On Wed, 15 Sep 2010, dadrivr wrote:


 Thanks for your help, guys.  I'm looking to produce a high-quality plot
 (no
 jagged lines or other distortions) with a filetype that is accepted by
 Microsoft Word on a PC and that most journals will accept.  That's why I'd
 prefer to stick with JPEG, TIFF, PNG, or the like.  I'm not sure EPS would
 fly.

 One simple approach, which I use when I have to create graphics for MS
 Office while on a non-Windows platform is to use PNG and set the resolution
 and file size large enough.  At 300dpi or so the physics of ink on paper
 does all the antialiasing you need.

 Work out how big you want the graph to be, and use PNG with enough pixels to
 get at least 300dpi at that final size. You'll need to set the pointsize
 argument and it will help to set the resolution argument.

     -thomas

 Thomas Lumley
 Professor of Biostatistics
 University of Washington, Seattle

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-- 

Max

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Re: [R] Creating publication-quality plots for use in Microsoft Word

2010-09-15 Thread Mark Difford

 I'd prefer to stick with JPEG, TIFF, PNG, or the like.  I'm not sure EPS
would fly.

Preferring to stick with bitmap formats (like JPEG, TIFF, PNG) is likely to
give you the jagged lines and other distortions you profess to want to
avoid.

EPS (encapsulated postscript, which handles vector+bitmap) is one of the
graphic file formats preferred by most quality journals. Surprisingly, not
too many people seem to be aware of the fact that PDF really is a crippled
form of postscript.

Regards,
Mark.



-- 
View this message in context: 
http://r.789695.n4.nabble.com/Creating-publication-quality-plots-for-use-in-Microsoft-Word-tp2540676p2540858.html
Sent from the R help mailing list archive at Nabble.com.

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[R] Format Data Issue??

2010-09-15 Thread Randy Cass
R Users,

I am new to R and have tried to figure out how to automate this
process instead of using excel.  I have read in this dataframe into r
with read.table.  I need to reshape the data from the first table into
the format of the second table.

TractID StandID Species CruiseDate  DBHClassTreesPerAcre
Carbon  Stand 1 Loblolly Pine  5/20/201010  1.2
Carbon  Stand 1 Loblolly Pine  5/20/201012  1.2
Carbon  Stand 1 Loblolly Pine  5/20/201014  0.61
Carbon  Stand 1 Loblolly Pine  5/20/201016  0.75
Carbon  Stand 1 Hard Hardwoods  5/20/2010   6   27.4
Carbon  Stand 1 Hard Hardwoods  5/20/2010   8   16.4
Carbon  Stand 1 Hard Hardwoods  5/20/2010   10  16.1
Carbon  Stand 1 Hard Hardwoods  5/20/2010   12  18.3
Carbon  Stand 1 Hard Hardwoods  5/20/2010   14  9.9

I want the TreesPerAcre number to be divided evenly (1.2/20) over the
tenth inch DBHClasses.  For example, A 10 inch DBHClass would range
from 9.1 to 11.0, the 12 inch class would be from 11.1 to 13.0, etc.
Any help will be greatly appreciated.

TractID StandID Species CruiseDate  DBHClassTreesPerAcre
Carbon  Stand 1 Loblolly Pine   5/20/2010   9.1 0.06
Carbon  Stand 1 Loblolly Pine   5/20/2010   9.2 0.06
Carbon  Stand 1 Loblolly Pine   5/20/2010   9.3 0.06
Carbon  Stand 1 Loblolly Pine   5/20/2010   9.4 0.06
Carbon  Stand 1 Loblolly Pine   5/20/2010   9.5 0.06
Carbon  Stand 1 Loblolly Pine   5/20/2010   9.6 0.06
Carbon  Stand 1 Loblolly Pine   5/20/2010   9.7 0.06
Carbon  Stand 1 Loblolly Pine   5/20/2010   9.8 0.06
Carbon  Stand 1 Loblolly Pine   5/20/2010   9.9 0.06
Carbon  Stand 1 Loblolly Pine   5/20/2010   10.00.06
Carbon  Stand 1 Loblolly Pine   5/20/2010   10.10.06
Carbon  Stand 1 Loblolly Pine   5/20/2010   10.20.06
Carbon  Stand 1 Loblolly Pine   5/20/2010   10.30.06
Carbon  Stand 1 Loblolly Pine   5/20/2010   10.40.06
Carbon  Stand 1 Loblolly Pine   5/20/2010   10.50.06
Carbon  Stand 1 Loblolly Pine   5/20/2010   10.60.06
Carbon  Stand 1 Loblolly Pine   5/20/2010   10.70.06
Carbon  Stand 1 Loblolly Pine   5/20/2010   10.80.06
Carbon  Stand 1 Loblolly Pine   5/20/2010   10.90.06
Carbon  Stand 1 Loblolly Pine   5/20/2010   11.00.06
etc.

Information:
R  2.11.1
Windows 7

Thanks in advance,

Randy

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Re: [R] How to uncompress a gz file in R

2010-09-15 Thread Henrik Bengtsson
Hi.

On Wed, Sep 15, 2010 at 9:30 AM, Wonsang You y...@ifn-magdeburg.de wrote:
 Dear Henrik,
 Thank you so much for your kind help. Unfortunately, I could not find out
 any function such as 'gunzip' in R.utils package.

Hmm... it's been there since at least 2005.

library(R.utils);
help(gunzip)

See the example.

gzip() and gunzip() works on any OS.

/H

 Instead, I could be
 successful by using the following command.
 system(gunzip filename)
 On the other hand, the function 'gzfile' supports the compression as gz
 format, but I still do not know how to decompress gz file by using the
 function 'gzfile'.
 Best Regards,
 Wonsang

 On 14 September 2010 15:23, Henrik Bengtsson h...@stat.berkeley.edu wrote:

 To uncompress an *.gz file into another file on disk, see also ?gunzip
 in the R.utils package.

 /Henrik

 2010/9/14 Uwe Ligges lig...@statistik.tu-dortmund.de:
  See ?gzfile
 
  Uwe Ligges
 
 
  On 14.09.2010 11:02, Wonsang You wrote:
 
  Dear Fellows,
 
  I would like to know how to uncompress a gz file at the R console. I
  could
  not find out any help from the R-help archive.
  Thanks for your great help.
 
  Best Regards,
  Wonsang You
 
 
  -
  --
  Wonsang You
  Special Lab Non-Invasive Brain Imaging
  Leibniz Institute for Neurobiology
  http://www.ifn-magdeburg.de
 
  __
  R-help@r-project.org mailing list
  https://stat.ethz.ch/mailman/listinfo/r-help
  PLEASE do read the posting guide
  http://www.R-project.org/posting-guide.html
  and provide commented, minimal, self-contained, reproducible code.
 



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[R] Sas to R

2010-09-15 Thread Sarah Jilani
Hi,

I need to call an R program from Sas. I have tried using the following code
in Sas using the x command but it just calls up dos and says
'D:\Program' is not recognized as an internal or external command, operable
program or batch file.
]
SAS CODE:

options xwait xsync;

%let Rsource=S:\EPI_Data\Communicable Diseases\DR1355 Historic Communicable
Disease Analaysis\R graphics;

options xwait xsync;

x 'D:\Program Files\R\R-2.9.2\bin\Rgui.exe --no-save --quiet
rsource.\Rtest.R rsource.\Rtest.log';

What I need it to do is call R, run an R program and then go back to running
sas. Any ideas on how to do that?

Thanks!

Sarah

-- 
---
Sarah Jilani
(317) 979 2830

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Re: [R] Scripting help

2010-09-15 Thread Philipp Pagel
On Wed, Sep 15, 2010 at 12:22:15PM -0400, Ayyappa Chaturvedula wrote:
 Dear all, I am new to R and this group. I have good experience in S
 scripts.  I need some orientation on data imports, general plotting
 functions. Can you please direct me?

Welcome to R. Coming from an S background you should have no problems
to adjust quickly. For data import have a look at the R Data
Import/Export manual:

http://cran.r-project.org/doc/manuals/R-data.html

Plotting is not to different from S-Plus. As far as my knowledge goes
there are 3 differnet plotting frameworsk in R:

1) Basic plot functions like plot or hist some of which are covered in 
the Introduction to R

2) lattice (the R aequivalent of trellis graphics

covered in many manual pages, many tutorials and talks google will
quickly find and last but not least the book
Lattice: Multivariate Data Visualization with R by Deepayan
Sarkar who implemented lattice

3) ggplot2

See http://had.co.nz/ggplot2/ for documentation
and consider the book ggplot2: Elegant Graphics for Data Analysis by
Hadley Wickham (the author of ggplot2)


I use all three frameworks on a regular basis - choosing the
respective functions depending on the complexity of the task.

cu
Philipp

-- 
Dr. Philipp Pagel
Lehrstuhl für Genomorientierte Bioinformatik
Technische Universität München
Wissenschaftszentrum Weihenstephan
Maximus-von-Imhof-Forum 3
85354 Freising, Germany
http://webclu.bio.wzw.tum.de/~pagel/

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Re: [R] Sas to R

2010-09-15 Thread Erik Iverson

Sarah,

This is a SAS question, not R.  However, it seems clear that it
has something to do with the fact that there are spaces in the
command that you're sending to Windows.

Maybe try calling with the 'short directory name' notation,
I forget what that's called in Windows.

Or else follow-up on a SAS list?

Sarah Jilani wrote:

Hi,

I need to call an R program from Sas. I have tried using the following code
in Sas using the x command but it just calls up dos and says
'D:\Program' is not recognized as an internal or external command, operable
program or batch file.
]
SAS CODE:

options xwait xsync;

%let Rsource=S:\EPI_Data\Communicable Diseases\DR1355 Historic Communicable
Disease Analaysis\R graphics;

options xwait xsync;

x 'D:\Program Files\R\R-2.9.2\bin\Rgui.exe --no-save --quiet
rsource.\Rtest.R rsource.\Rtest.log';

What I need it to do is call R, run an R program and then go back to running
sas. Any ideas on how to do that?

Thanks!

Sarah



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Re: [R] Sas to R

2010-09-15 Thread Erik Iverson



David Winsemius wrote:


On Sep 15, 2010, at 1:44 PM, Sarah Jilani wrote:


Hi,

I need to call an R program from Sas. I have tried using the following 
code

in Sas using the x command but it just calls up dos and says


I went searching for a worked example and found this:

http://www.nesug.org/proceedings/nesug08/sa/sa07.pdf

Those of us with a preference for R are likely to be either amused (or 
mildly offended) by statements like ...Data manipulation is much easier 
in SAS... 


Which is just another way of saying, I am used to manipulating data
in SAS or I've never been taught to manipulate data in R.

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Re: [R] Sas to R

2010-09-15 Thread Nordlund, Dan (DSHS/RDA)
 -Original Message-
 From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-
 project.org] On Behalf Of Sarah Jilani
 Sent: Wednesday, September 15, 2010 10:44 AM
 To: r-help@r-project.org
 Subject: [R] Sas to R
 
 Hi,
 
 I need to call an R program from Sas. I have tried using the following
 code
 in Sas using the x command but it just calls up dos and says
 'D:\Program' is not recognized as an internal or external command,
 operable
 program or batch file.
 ]
 SAS CODE:
 
 options xwait xsync;
 
 %let Rsource=S:\EPI_Data\Communicable Diseases\DR1355 Historic
 Communicable
 Disease Analaysis\R graphics;
 
 options xwait xsync;
 
 x 'D:\Program Files\R\R-2.9.2\bin\Rgui.exe --no-save --quiet
 rsource.\Rtest.R rsource.\Rtest.log';
 
 What I need it to do is call R, run an R program and then go back to
 running
 sas. Any ideas on how to do that?
 
 Thanks!
 
 Sarah
 

Sarah,

I have found that making sure the command you wish to run will actually run at 
the command line is a big help.  I suspect that the command, as you have 
specified it, will not run at the command line.  Try copying everything inside 
the single quotes, open a command prompt, then paste the command and run it.  I 
don't think it will do what you expect or want.

That being said, try something like

x 'D:\Program Files\R\R-2.9.2\bin\R.exe --no-save --quiet  
rsource.\Rtest.R  rsource.\Rtest.log';

You don't want to run Rgui.exe, use R.exe.  In addition, you don't want to 
double your double-quotes, because they are inside single-quotes.  Doubling 
them will cause problems.  If this doesn't work, then the R-help related issue 
is figuring out how to run R from the command line with input and output 
redirection.  The syntax I provide above should run just fine in Windows 
XP/Vista/7.

Hope this is helpful,

Dan 

Daniel J. Nordlund
Washington State Department of Social and Health Services
Planning, Performance, and Accountability
Research and Data Analysis Division
Olympia, WA 98504-5204


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Re: [R] Sas to R

2010-09-15 Thread Joshua Wiley
Hi Sarah,

Just a couple additional notes to what's been said:

1) It seems like R CMD BATCH might be easier
2) If the space in Program Files is causing issues, you can use the
Windows environment variable %PROGRAMFILES% (which also means you do
not need the drive letter.
3) I find it handy to define a new Windows environment variable with
the path to R, then I can just reference that rather than the entire
path (a google search will show you lots of ways to do this via the
GUI and the command prompt).  For instance: %R%/Rterm.exe starts R
in the command prompt for me.

Here is an example run from the Windows command prompt:

%PROGRAMFILES%/R/R-2.11.1/bin/R.exe CMD BATCH C:/myscript.R C:/results.txt

Cheers,

Josh

On Wed, Sep 15, 2010 at 10:44 AM, Sarah Jilani sarahjil...@gmail.com wrote:
 Hi,

 I need to call an R program from Sas. I have tried using the following code
 in Sas using the x command but it just calls up dos and says
 'D:\Program' is not recognized as an internal or external command, operable
 program or batch file.
 ]
 SAS CODE:

 options xwait xsync;

 %let Rsource=S:\EPI_Data\Communicable Diseases\DR1355 Historic Communicable
 Disease Analaysis\R graphics;

 options xwait xsync;

 x 'D:\Program Files\R\R-2.9.2\bin\Rgui.exe --no-save --quiet
 rsource.\Rtest.R rsource.\Rtest.log';

 What I need it to do is call R, run an R program and then go back to running
 sas. Any ideas on how to do that?

 Thanks!

 Sarah

 --
 ---
 Sarah Jilani
 (317) 979 2830

        [[alternative HTML version deleted]]

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 R-help@r-project.org mailing list
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-- 
Joshua Wiley
Ph.D. Student, Health Psychology
University of California, Los Angeles
http://www.joshuawiley.com/

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Re: [R] Sas to R

2010-09-15 Thread Sarah Jilani
Thanks Josh and Dan!

I got it to work using the following code that Dan had suggested.

x D:\Program Files\R\R-2.9.2\bin\R.exe --no-save --quiet 
rsource.\Rtest.R  rsource.\Rtest.log;

Thanks for your help!! :)

Sarah



On Wed, Sep 15, 2010 at 2:58 PM, Joshua Wiley jwiley.ps...@gmail.comwrote:

 Hi Sarah,

 Just a couple additional notes to what's been said:

 1) It seems like R CMD BATCH might be easier
 2) If the space in Program Files is causing issues, you can use the
 Windows environment variable %PROGRAMFILES% (which also means you do
 not need the drive letter.
 3) I find it handy to define a new Windows environment variable with
 the path to R, then I can just reference that rather than the entire
 path (a google search will show you lots of ways to do this via the
 GUI and the command prompt).  For instance: %R%/Rterm.exe starts R
 in the command prompt for me.

 Here is an example run from the Windows command prompt:

 %PROGRAMFILES%/R/R-2.11.1/bin/R.exe CMD BATCH C:/myscript.R
 C:/results.txt

 Cheers,

 Josh

 On Wed, Sep 15, 2010 at 10:44 AM, Sarah Jilani sarahjil...@gmail.com
 wrote:
  Hi,
 
  I need to call an R program from Sas. I have tried using the following
 code
  in Sas using the x command but it just calls up dos and says
  'D:\Program' is not recognized as an internal or external command,
 operable
  program or batch file.
  ]
  SAS CODE:
 
  options xwait xsync;
 
  %let Rsource=S:\EPI_Data\Communicable Diseases\DR1355 Historic
 Communicable
  Disease Analaysis\R graphics;
 
  options xwait xsync;
 
  x 'D:\Program Files\R\R-2.9.2\bin\Rgui.exe --no-save --quiet
  rsource.\Rtest.R rsource.\Rtest.log';
 
  What I need it to do is call R, run an R program and then go back to
 running
  sas. Any ideas on how to do that?
 
  Thanks!
 
  Sarah
 
  --
  ---
  Sarah Jilani
  (317) 979 2830
 
 [[alternative HTML version deleted]]
 
  __
  R-help@r-project.org mailing list
  https://stat.ethz.ch/mailman/listinfo/r-help
  PLEASE do read the posting guide
 http://www.R-project.org/posting-guide.html
  and provide commented, minimal, self-contained, reproducible code.
 



 --
 Joshua Wiley
 Ph.D. Student, Health Psychology
 University of California, Los Angeles
 http://www.joshuawiley.com/




-- 
---
Sarah Jilani
(317) 979 2830

[[alternative HTML version deleted]]

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Re: [R] Homogeneity of regression slopes

2010-09-15 Thread Doug Adams
That's good insight, and gives me some good ideas for what direction
to this.  Thanks everyone !

Doug

P.S. - I guess if you have a significant interaction, that implies the
slopes of the individual regression lines are significantly different
anyway, doesn't it...



On Tue, Sep 14, 2010 at 11:33 AM, Thomas Stewart tgstew...@gmail.com wrote:
 If you are interested in exploring the homogeneity of variance assumption,
 I would suggest you model the variance explicitly.  Doing so allows you to
 compare the homogeneous variance model to the heterogeneous variance model
 within a nested model framework.  In that framework, you'll have likelihood
 ratio tests, etc.
 This is why I suggested the nlme package and the gls function.  The gls
 function allows you to model the variance.
 -tgs
 P.S. WLS is a type of GLS.
 P.P.S It isn't clear to me how a variance stabilizing transformation would
 help in this case.

 On Tue, Sep 14, 2010 at 6:53 AM, Clifford Long gnolff...@gmail.com wrote:

 Hi Thomas,

 Thanks for the additional information.

 Just wondering, and hoping to learn ... would any lack of homogeneity of
 variance (which is what I believe you mean by different stddev estimates) be
 found when performing standard regression diagnostics, such as residual
 plots, Levene's test (or equivalent), etc.?  If so, then would a WLS routine
 or some type of variance stabilizing transformation be useful?

 Again, hoping to learn.  I'll check out the gls() routine in the nlme
 package, as you mentioned.

 Thanks.

 Cliff


 On Mon, Sep 13, 2010 at 10:02 PM, Thomas Stewart tgstew...@gmail.com
 wrote:

 Allow me to add to Michael's and Clifford's responses.

 If you fit the same regression model for each group, then you are also
 fitting a standard deviation parameter for each model.  The solution
 proposed by Michael and Clifford is a good one, but the solution assumes
 that the standard deviation parameter is the same for all three models.

 You may want to consider the degree by which the standard deviation
 estimates differ for the three separate models.  If they differ wildly,
 the
 method described by Michael and Clifford may not be the best.  Rather,
 you
 may want to consider gls() in the nlme package to explicitly allow the
 variance parameters to vary.

 -tgs

 On Mon, Sep 13, 2010 at 4:52 PM, Doug Adams f...@gmx.com wrote:

  Hello,
 
  We've got a dataset with several variables, one of which we're using
  to split the data into 3 smaller subsets.  (as the variable takes 1 of
  3 possible values).
 
  There are several more variables too, many of which we're using to fit
  regression models using lm.  So I have 3 models fitted (one for each
  subset of course), each having slope estimates for the predictor
  variables.
 
  What we want to find out, though, is whether or not the overall slopes
  for the 3 regression lines are significantly different from each
  other.  Is there a way, in R, to calculate the overall slope of each
  line, and test whether there's homogeneity of regression slopes?  (Am
  I using that phrase in the right context -- comparing the slopes of
  more than one regression line rather than the slopes of the predictors
  within the same fit.)
 
  I hope that makes sense.  We really wanted to see if the predicted
  values at the ends of the 3 regression lines are significantly
  different... But I'm not sure how to do the Johnson-Neyman procedure
  in R, so I think testing for slope differences will suffice!
 
  Thanks to any who may be able to help!
 
  Doug Adams
 
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[R] test to see if a s4 property is assigned?

2010-09-15 Thread darckeen

Class(person,representation(age=numeric,weight=numeric))
[1] person
 bob - new(person,age=30)
 is.null(b...@weight)
[1] FALSE
 b...@weight
numeric(0)
 b...@weight == numeric(0)
logical(0)

How can i test to see if a s4 property is assigned?

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Re: [R] Format Data Issue??

2010-09-15 Thread Dennis Murphy
Hi:

This isn't the most elegant way, I'm sure, but here's one approach.

library(reshape)
# read in your data...I had to surround the text strings with quotes because
of the spaces
df - read.table(textConnection(
TractID StandID Species CruiseDate  DBHClassTreesPerAcre
'Carbon  Stand' 1 'Loblolly Pine'  5/20/201010  1.2
'Carbon  Stand' 1 'Loblolly Pine'  5/20/201012  1.2
'Carbon  Stand' 1 'Loblolly Pine'  5/20/201014  0.61
'Carbon  Stand' 1 'Loblolly Pine'  5/20/201016  0.75
'Carbon  Stand' 1 'Hard Hardwoods'  5/20/2010   6   27.4
'Carbon  Stand' 1 'Hard Hardwoods'  5/20/2010   8   16.4
'Carbon  Stand' 1 'Hard Hardwoods'  5/20/2010   10  16.1
'Carbon  Stand' 1 'Hard Hardwoods'  5/20/2010   12  18.3
'Carbon  Stand' 1 'Hard Hardwoods'  5/20/2010   14  9.9), header =
TRUE)

# Generate a matrix that expands the DBHs in each row and column concatenate
DBH - t(apply(df, 1, function(x) as.numeric(x[5]) + seq(-0.9, 1, by =
0.1)))
df2 - cbind(df, DBH)
# Divide the TreesPerAcre by 20
df2$TreesPerAcre - df2$TreesPerAcre/20
# reshape from wide to long using the reshape function melt()
df3 - melt(df2, id = 1:6) # a 180 x 8 data frame
df3 - df3[order(df3$DBHClass, df3$value), ]
 head(df3)
 TractID StandIDSpecies CruiseDate DBHClass TreesPerAcre
5  Carbon  Stand   1 Hard Hardwoods  5/20/20106 1.37
14 Carbon  Stand   1 Hard Hardwoods  5/20/20106 1.37
23 Carbon  Stand   1 Hard Hardwoods  5/20/20106 1.37
32 Carbon  Stand   1 Hard Hardwoods  5/20/20106 1.37
41 Carbon  Stand   1 Hard Hardwoods  5/20/20106 1.37
50 Carbon  Stand   1 Hard Hardwoods  5/20/20106 1.37
   variable value
5 1   5.1
142   5.2
233   5.3
324   5.4
415   5.5
506   5.6

You'll likely need to convert the CruiseDate to a date variable at some
point; see ?as.Date, especially the format argument.

If you need this as a function to be applied multiple times,

DBHspread - function(df) {

# Function to spread out the DBHClass values over an interval of
# length 2 in increments of 0.1, and divide TreesPerAcre by 20 to
# compensate for change in class width.

require(reshape)

# Generate a matrix that expands the DBHs in each row and column
concatenate
# Since DBHClass (variable 5) is integer, it needs to be converted to
numeric
# In this context, x is the i-th row of the data frame.
DBH - t(apply(df, 1, function(x) as.numeric(x[5]) + seq(-0.9, 1, by =
0.1)))

# Concatenate new DBHClass values to existing data frame
df2 - cbind(df, DBH)

# Divide the TreesPerAcre by 20
df2$TreesPerAcre - df2$TreesPerAcre/20

# reshape from wide to long using the reshape function melt()
df3 - melt(df2, id = 1:6)
# Sort by DBHClass and value within it, output
with(df3, df3[order(DBHClass, value), ])
  }

HTH,
Dennis

On Wed, Sep 15, 2010 at 10:17 AM, Randy Cass randynewru...@gmail.comwrote:

 R Users,

 I am new to R and have tried to figure out how to automate this
 process instead of using excel.  I have read in this dataframe into r
 with read.table.  I need to reshape the data from the first table into
 the format of the second table.

 TractID StandID Species CruiseDate  DBHClassTreesPerAcre
 Carbon  Stand 1 Loblolly Pine  5/20/201010  1.2
 Carbon  Stand 1 Loblolly Pine  5/20/201012  1.2
 Carbon  Stand 1 Loblolly Pine  5/20/201014  0.61
 Carbon  Stand 1 Loblolly Pine  5/20/201016  0.75
 Carbon  Stand 1 Hard Hardwoods  5/20/2010   6   27.4
 Carbon  Stand 1 Hard Hardwoods  5/20/2010   8   16.4
 Carbon  Stand 1 Hard Hardwoods  5/20/2010   10  16.1
 Carbon  Stand 1 Hard Hardwoods  5/20/2010   12  18.3
 Carbon  Stand 1 Hard Hardwoods  5/20/2010   14  9.9

 I want the TreesPerAcre number to be divided evenly (1.2/20) over the
 tenth inch DBHClasses.  For example, A 10 inch DBHClass would range
 from 9.1 to 11.0, the 12 inch class would be from 11.1 to 13.0, etc.
 Any help will be greatly appreciated.

 TractID StandID Species CruiseDate  DBHClassTreesPerAcre
 Carbon  Stand 1 Loblolly Pine   5/20/2010   9.1 0.06
 Carbon  Stand 1 Loblolly Pine   5/20/2010   9.2 0.06
 Carbon  Stand 1 Loblolly Pine   5/20/2010   9.3 0.06
 Carbon  Stand 1 Loblolly Pine   5/20/2010   9.4 0.06
 Carbon  Stand 1 Loblolly Pine   5/20/2010   9.5 0.06
 Carbon  Stand 1 Loblolly Pine   5/20/2010   9.6 0.06
 Carbon  Stand 1 Loblolly Pine   5/20/2010   9.7 0.06
 Carbon  Stand 1 Loblolly Pine   5/20/2010   9.8 0.06
 Carbon  Stand 1 Loblolly Pine   5/20/2010   9.9 0.06
 Carbon  Stand 1 Loblolly Pine   5/20/2010   

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