date:20120730

On Mon, Jul 30, 2012 at 01:09:34PM +0200, Peppe Ricci wrote:
 Hi,
 
 thank Petr for your help.
 I have implemented you code suggestion but there is another problem.
 It seems that the code:
 
 for (i in 1:m){
  Z[i,,]=table(occ, data_matrix[,i])
 }
 
 don't charge any values in Z.
 Is there some error?

Hi.

I already deleted your previous emails from my folder and you
start a new thread. Can you recall, what is exactly the problem
and which code you use? Without a context, the reason for the
above problem cannot be understood.

Petr Savicky.

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Re: [R] On Reproducible Code

2012-07-30 Thread peter dalgaard


On Jul 30, 2012, at 13:05 , Thomas Adams wrote:

 
 Nothing is gained by punishing people over their internet 'manners'∑
 
 Tom

On the contrary, everything can be lost by allowing abusers to persevere!

(And yes, there are people who no longer attempt to help, because of ungrateful 
and downright arrogant behavior they have experienced on the lists.)

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

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Re: [R] On Reproducible Code

2012-07-30 Thread Thomas Adams

Peter,

You may have misunderstood me (I did not define correctly whose internet
manners I was referring to) â I was referring to the internet manners of
those seeking help, that, yes, they may not have adequately researched
things before asking a question, or may not have supplied a reproducible
example, etc., but they don't deserve to be treated disrespectfully â I
think we are agreeing on thisâ¦

Tom

On Mon, Jul 30, 2012 at 8:02 AM, peter dalgaard pda...@gmail.com wrote:


 On Jul 30, 2012, at 13:05 , Thomas Adams wrote:

 
  Nothing is gained by punishing people over their internet 'manners'â
 
  Tom

 On the contrary, everything can be lost by allowing abusers to persevere!

 (And yes, there are people who no longer attempt to help, because of
 ungrateful and downright arrogant behavior they have experienced on the
 lists.)

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











-- 

Thomas E Adams
National Weather Service
Ohio River Forecast Center
1901 South State Route 134
Wilmington, OH 45177

EMAIL:  thomas.ad...@noaa.gov
VOICE:  937-383-0528
FAX:937-383-0033

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[R] cluster of points

2012-07-30 Thread Frederico Mestre

Hello:

 

What I want to do is quite simple, but I can't find a way.

 

I have a data frame with several points (x and y coords). I want to add
another column with cluster membership. For example aggregate all the points
that stand within a distance of 40 from each other. 

 

I've tried using nncluster from the package nnclust, but the results are
not correct, for some reason (probably my mistake).

 

This is what I did:

 

x - nncluster(as.matrix(dframe[,1:2]), threshold=35, fill = 1, maxclust =
NULL, give.up = 500,verbose=FALSE,start=NULL)#avaliar as clusters 

 

Thanks,

 

Frederico

 


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Re: [R] Convert variable to STring

2012-07-30 Thread John Kane

Can you give us an example of what you mean by the functions listed below.

Otherwise something like this seems to work
xx -  1 :4
x1  -  as.character(xx)

mat  -  matrix(1:12, nrow=3)

John Kane
Kingston ON Canada


 -Original Message-
 From: ala...@yahoo.com
 Sent: Mon, 30 Jul 2012 03:47:45 -0700 (PDT)
 To: r-help@r-project.org
 Subject: [R] Convert variable to STring
 
 Dear all,
 I have a variable that I would like also to use it as a string. The
 reasons is that I want to collect results from different function to one
 table.. So when I use the
  colnames(mymatrix) -c(function1.function2,function3)
 the function1, function2, function3 to be converted to simple strings
 so as
 
  colnames(mymatrix) -c(function1,function2,function3)
 
 Could you please help me understand how I can do that in R?
 
 Regards
 
   [[alternative HTML version deleted]]
 
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Re: [R] Turning off continuation prompt?

2012-07-30 Thread John Kane

If I understand the question correctly the answer is probably not to type 
anything at the R command line. A good editor or interface is a better approach
http://en.wikipedia.org/wiki/R_%28programming_language%29#Interfaces

John Kane
Kingston ON Canada


 -Original Message-
 From: ted.hard...@wlandres.net
 Sent: Mon, 30 Jul 2012 09:58:09 +0100 (BST)
 To: r-help@r-project.org
 Subject: [R] Turning off continuation prompt?
 
 Greetings All.
 My apologies for a question whose answer is probably
 readily available somewhere (for some interpetation
 of somewhere) ...
 
 Say I have just typed (from a sheet of paper) several
 lines into the R command-line, and what I see is:
 
 chisq.test(matrix(c(3,6,3,4,4,
 + 4,1,4,6,5,
 + 2,7,4,2,5,
 + 8,2,4,4,2,
 + 3,4,5,4,4),ncol=5))
 
 Later, I find that would like to re-input the data part
 of this command (matrix(c(...)...)). Without the +
 continuation prompts, it would be easy to do this by
 copypaste with the mouse in one operation. With the +
 marks there, I have to do the copypaste for each separate line.
 
 So is there a way to suppress the output of the + at the
 beginning of each continuation line?
 
 (The above is one of the smaller examples of this situation;
 sometimes I have wished to do this for commands extending over,
 say, 15-20 lines).
 
 With thanks,
 Ted.
 
 -
 E-Mail: (Ted Harding) ted.hard...@wlandres.net
 Date: 30-Jul-2012  Time: 09:58:02
 This message was sent by XFMail
 
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Re: [R] Convert variable to STring

2012-07-30 Thread Michael Weylandt

You may be searching for the deparse(substitute(x)) idiom. 

Michael

On Jul 30, 2012, at 5:47 AM, Alaios ala...@yahoo.com wrote:

 Dear all,
 I have a variable that I would like also to use it as a string. The reasons 
 is that I want to collect results from different function to one table.. So 
 when I use the
 �colnames(mymatrix) -c(function1.function2,function3)
 the function1, function2, function3 to be converted to simple strings so as
 
 �colnames(mymatrix) -c(function1,function2,function3)
 
 Could you please help me understand how I can do that in R?
 
 Regards
 
[[alternative HTML version deleted]]
 
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 https://stat.ethz.ch/mailman/listinfo/r-help
 PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
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Re: [R] distance matrix and hclustering

2012-07-30 Thread Jean V Adams

Look at the examples for the functions dist() and hclust(),
?dist
?hclust

Jean


eliza botto eliza_bo...@hotmail.com wrote on 07/30/2012 01:43:55 AM:
 
 Dear R Users,i am very new to R. I want your help on an issue 
 regarding distance matrix and cluster analysis
 i had discharge data of 4 rivers(a,b,c,d) in 4 vectors each having 364 
values
  dput(qmu)structure(list(a = c(0.26, 0.25, 0.25, 0.25, 0.24, 0.23, 
 0.22, 0.21, 0.21, 0.21, 0.2, 0.19, 0.19, 0.19, 0.19, 0.18, 0.18, 0.
 18, 0.17, 0.17, 0.17, 0.17, 0.17, 0.17, 0.17, 0.17, 0.17, 0.17, 0.
 17, 0.17, 0.17, 0.17, 0.17, 0.17, 0.18, 0.19, 0.19, 0.19, 0.2, 0.21,
 0.21, 0.21, 0.21, 0.21, 0.21, 0.19, 0.19, 0.18, 0.17, 0.17, 0.15, 0.
 18, 0.2, 0.21, 0.2, 0.19, 0.19, 0.19, 0.2, 0.24, 0.3, 0.3, 0.3, 0.
 32, 0.34, 0.42, 0.46, 0.48, 0.67, 0.82, 0.79, 0.73, 0.69, 0.67, 0.
 67, 0.66, 0.64, 0.61, 0.58, 0.56, 0.55, 0.55, 0.55, 0.52, 0.49, 0.
 48, 0.51, 0.53, 0.52, 0.49, 0.48, 0.48, 0.46, 0.46, 0.44, 0.43, 0.
 43, 0.41, 0.48, 0.55, 0.57, 0.55, 0.56, 0.6, 0.64, 0.67, 0.73, 0.84,
 0.94, 1.09, 1.24, 1.28, 1.19, 1.11, 1, 0.92, 0.86, 0.79, 0.76, 0.76,
 0.76, 0.76, 0.92, 0.98, 1.03, 1.03, 1.03, 1.03, 1.07, 1.11, 1.24, 1.
 44, 2.12, 3.26, 15, 9.45, 5.07, 4.59, 3.5, 2.84, 2.54, 2.57, 3.01, 
 2.32, 2.32, 2.97, 2.92, 3.88, 4.76, 5.99, 3.74, 2.92, 2.65, 2.57, 2.
 97, 3.4, 4.13, 4.31, 3.89, 3.45, 3.01, 2.88, 2!
  .5, 2.29, 2.39, 2.25, 2.02, 1.87, 1.87, 2.54, 2.69, 2.76, 3.18, 3.
 74, 4.59, 4.76, 4.36, 6.56, 5.07, 3.84, 3.55, 3.84, 3.84, 5.49, 5.
 32, 3.74, 3.31, 3.4, 3.26, 3.09, 2.69, 2.54, 2.46, 2.39, 2.25, 2.22,
 2.22, 2.25, 2.29, 2.22, 2.18, 2.05, 2.18, 2.39, 2.18, 2.29, 2.11, 1.
 81, 1.6, 1.44, 1.41, 1.32, 1.37, 1.37, 1.65, 2.31, 2.25, 1.68, 1.41,
 1.26, 1.15, 3.28, 1.93, 1.6, 1.53, 1.28, 1.13, 1.03, 1.03, 1.03, 1.
 03, 1, 0.96, 0.92, 0.87, 0.82, 0.79, 0.76, 0.73, 0.7, 0.67, 0.64, 0.
 64, 0.61, 0.61, 0.61, 1.76, 1.19, 1.24, 1.37, 1.68, 2.39, 2.05, 1.
 78, 1.58, 1.41, 1.39, 1.5, 1.41, 1.32, 1.19, 1.11, 1.02, 1.07, 4.57,
 1.96, 1.68, 1.5, 1.37, 1.24, 1.11, 1.03, 0.96, 0.94, 2.93, 2.88, 2.
 92, 2.76, 2.02, 1.71, 1.5, 1.37, 1.22, 1.09, 1, 0.94, 0.87, 0.81, 0.
 76, 0.73, 0.7, 0.67, 0.61, 0.58, 0.57, 0.55, 0.53, 0.51, 0.48, 0.47,
 0.44, 0.43, 0.43, 0.41, 0.41, 0.38, 0.4, 0.4, 0.42, 0.42, 0.41, 0.
 46, 0.53, 0.55, 0.52, 0.49, 0.51, 0.53, 0.55, 0.7, 1.03, 1.03, 1.17,
 1.24, 1.19, 1.11, 1.03, 0.98, 0.92, 0.84,!
   0.79, 0.75, 0.7, 0.67, 0.61, 0.58, 0.56, 0.56, 0.55, 0.53, 0.51, 0.48
 , 0.46, 0.43, 0.41, 0.38, 0.37, 0.35, 0.34, 0.32, 0.31, 0.3, 0.29, 
 0.28, 0.27, 0.25, 0.26, 0.24, 0.23, 0.22, 0.22, 0.21, 0.21, 0.21), b
 = c(0.19, 0.19, 0.19, 0.18, 0.18, 0.18, 0.17, 0.17, 0.17, 0.17, 0.
 16, 0.17, 0.17, 0.15, 0.15, 0.15, 0.15, 0.15, 0.15, 0.15, 0.15, 0.
 15, 0.15, 0.15, 0.15, 0.15, 0.15, 0.15, 0.15, 0.15, 0.14, 0.14, 0.
 14, 0.14, 0.14, 0.14, 0.14, 0.14, 0.14, 0.14, 0.14, 0.14, 0.14, 0.
 14, 0.14, 0.15, 0.15, 0.15, 0.15, 0.15, 0.15, 0.15, 0.16, 0.16, 0.
 17, 0.17, 0.18, 0.18, 0.18, 0.18, 0.18, 0.18, 0.18, 0.18, 0.19, 0.
 21, 0.21, 0.21, 0.22, 0.23, 0.24, 0.24, 0.23, 0.24, 0.24, 0.25, 0.
 25, 0.25, 0.28, 0.29, 0.29, 0.3, 0.31, 0.31, 0.34, 0.41, 0.46, 0.51,
 0.57, 0.61, 0.64, 0.67, 0.7, 0.76, 0.82, 0.86, 1.05, 1.24, 1.05, 0.
 94, 0.92, 0.9, 0.86, 0.82, 0.76, 0.76, 0.76, 0.78, 0.82, 0.9, 1.07, 
 1.76, 3.13, 3.64, 3.45, 3.01, 2.39, 2.02, 1.87, 2.11, 2.02, 1.78, 1.
 63, 1.53, 1.63, 4.84, 12.5, 8.11, 3.89, 2.73, 2.11, 1.96, 3.17, 2.
 65, 2.54, 3.01, 3.31, 3.6, 3.36, 2.76, 2.39, 2.11, 2.!
  25, 2.08, 1.99, 2.11, 2.36, 3.13, 7.16, 5.39, 5.52, 5.32, 4.25, 3.
 45, 3.26, 3.18, 3.74, 4.35, 5.79, 5.45, 4.42, 3.84, 3.36, 2.84, 2.
 39, 3.84, 3.18, 3.22, 2.97, 2.73, 2.65, 2.92, 4.33, 3.01, 3.01, 3.
 26, 3.09, 3.6, 3.64, 4.05, 4.25, 4.48, 3.69, 3.74, 3.6, 3.18, 2.76, 
 4.11, 2.92, 2.69, 2.73, 2.69, 3.93, 2.69, 2.18, 2.52, 2.69, 1.99, 2.
 57, 1.81, 1.55, 1.44, 1.37, 1.28, 1.19, 1.19, 1.03, 1.03, 1, 0.94, 
 0.89, 0.87, 0.86, 0.86, 2.3, 1.55, 1.19, 1.11, 1.5, 1.39, 1.22, 1.
 24, 1.07, 1.02, 0.96, 0.92, 1.34, 1.15, 1.03, 2.06, 1.76, 1.3, 1.15,
 1.05, 0.98, 0.92, 0.89, 0.84, 0.81, 0.76, 0.73, 1.59, 5.2, 3.01, 2.
 05, 1.65, 1.68, 5.29, 2.73, 1.96, 1.6, 1.41, 1.28, 1.15, 1.11, 1.13,
 1.09, 1.03, 6.99, 10.6, 5.39, 3.45, 2.5, 1.87, 1.68, 1.78, 1.53, 1.
 41, 1.3, 1.17, 1.05, 0.98, 0.92, 0.9, 0.87, 0.86, 0.82, 0.78, 0.75, 
 0.72, 0.67, 0.82, 1.6, 0.89, 0.94, 0.96, 0.92, 0.87, 0.82, 0.79, 0.
 75, 0.7, 0.67, 0.64, 0.61, 0.58, 0.56, 0.53, 0.51, 0.48, 0.47, 0.46,
 0.43, 0.41, 0.41, 0.68, 16.3, 17.2, 6.05, 3.6!
  9, 2.92, 2.25, 1.87, 1.63, 1.46, 1.32, 1.19, 1.07, 1, 0.94, 0.89, 0.87
 , 0.86, 0.81, 0.76, 0.73, 0.7, 0.7, 0.7, 0.7, 0.7, 0.67, 0.67, 0.66,
 0.64, 0.61, 0.58, 0.56, 0.55, 0.53, 0.51, 0.48, 0.46, 0.44, 0.43, 0.
 43, 0.41, 0.4, 0.38, 0.37, 0.36, 0.35, 0.34, 0.33, 0.32, 0.31, 0.31,
 0.3, 0.3, 0.29, 0.29, 0.28, 0.27, 0.27), c = c(0.27, 0.25, 0.25, 0.
 25, 0.24, 0.24, 0.23, 0.22, 0.22, 0.21, 0.21, 0.21, 0.21, 0.21, 0.2,
 0.19, 0.19, 0.19, 0.19, 0.19, 0.19, 0.19, 0.19, 0.19, 0.19, 0.19, 0.
 19, 0.19, 0.19, 0.18,

[R] APIs and R

2012-07-30 Thread Ajay Ohri

Dear List,

I am compiling a list of R packages that work with social media APIs. Uptil
now I have found that

1- Most of these are based on RCurl, XML, rjson

2- Even though the API format is standardized for REST- there seems to be
many differences within the APIs, leadinbg to multiple packages that do the
required API call -

3- I am specific issues with the following

Google +API access through R
Linkedin API access through R
Twitter +API access through ROAuth
and Datasift API (basically a firehose for all social media) within R

Again most of the APIs I find have libraries supported by java,
ruby,python,curl- but there seems to be some overlooking of R within
various API vendors.

Has anybody been working with R for social media- I am specifically looking
for use cases that avoid rate limiting of free api- and specifically OAuth
and APis for LinkedIn, Google + and Datasift using R.

Many Thanks,

Ajay ohri

Websites-
Technology
http://decisionstats.com

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[R] If/then statement, if in a list then

I need to write an if/then statement saying something along the lines of:
if (VALUE is in list) {...
How do I write that in R's language.
Thanks!



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Re: [R] If/then statement, if in a list then

2012-07-30 Thread John Kane


if(cond) expr

if(cond) cons.expr  else  alt.expr

Also see ?ifelse
John Kane
Kingston ON Canada


 -Original Message-
 From: bunnylove...@optonline.net
 Sent: Mon, 30 Jul 2012 05:55:34 -0700 (PDT)
 To: r-help@r-project.org
 Subject: [R] If/then statement, if in a list then
 
 I need to write an if/then statement saying something along the lines of:
 if (VALUE is in list) {...
 How do I write that in R's language.
 Thanks!
 
 
 
 --
 View this message in context:
 http://r.789695.n4.nabble.com/If-then-statement-if-in-a-list-then-tp4638346.html
 Sent from the R help mailing list archive at Nabble.com.
 
 __
 R-help@r-project.org mailing list
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Re: [R] If/then statement, if in a list then

2012-07-30 Thread Jean V Adams

?is.element

Jean


cm bunnylove...@optonline.net wrote on 07/30/2012 07:55:34 AM:
 
 I need to write an if/then statement saying something along the lines 
of:
 if (VALUE is in list) {...
 How do I write that in R's language.
 Thanks!

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Re: [R] If/then statement, if in a list then

Thank you so much Jean!



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Re: [R] MANOVA polynomial contrasts

2012-07-30 Thread John Fox

Dear Gian Mauro,

On Mon, 30 Jul 2012 14:44:44 +0200
 Manzoni, GianMauro gm.manz...@auxologico.it wrote:
 Dear Prof. John Fox,
 thank you very much for your suggestions.
 However, I still do not know how to use the contrasts after generating them.
 Once I generate the matrix with the polynomial contrasts, what are the
 following steps toward the statistical test?

Here's a contrived example, which uses the Anova() and linearHypothesis() 
functions in the car package:

- snip --

 Y - matrix(rnorm(300), 100, 3)
 colnames(Y) - c(y1, y2, y3)
 f - ordered(sample(letters[1:4], 100, replace=TRUE))
 (mod - lm(Y ~ f))

Call:
lm(formula = Y ~ f)

Coefficients:
 y1y2y3  
(Intercept)   0.06514  -0.01683  -0.13787
f.L  -0.37837   0.18309   0.29736
f.Q  -0.02102  -0.39894   0.08455
f.C   0.05898   0.09358  -0.17634

 Anova(mod)

Type II MANOVA Tests: Pillai test statistic
  Df test stat approx F num Df den Df Pr(F)
f  3   0.11395   1.2634  9288 0.2566

 linearHypothesis(mod, f.L)

Sum of squares and products for the hypothesis:
  y1y2y3
y1  3.607260 -1.745560 -2.834953
y2 -1.745560  0.844680  1.371839
y3 -2.834953  1.371839  2.227995

Sum of squares and products for error:
  y1y2y3
y1 86.343376 -8.054928 -3.711756
y2 -8.054928 95.473020  2.429151
y3 -3.711756  2.429151 89.593163

Multivariate Tests: 
 Df test stat approx F num Df den Df   Pr(F)  
Pillai1 0.0648520 2.172951  3 94 0.096362 .
Wilks 1 0.9351480 2.172951  3 94 0.096362 .
Hotelling-Lawley  1 0.0693495 2.172951  3 94 0.096362 .
Roy   1 0.0693495 2.172951  3 94 0.096362 .
---
Signif. codes:  0 *** 0.001 ** 0.01 * 0.05 . 0.1   1 

- snip --

You could do similar tests for the quadratic and cubic contrasts.

I hope this helps,
 John


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

 
 A whole example would be very useful.
 
 Thank you very much in advance!
 
 Best regards,
 Gian Mauro Manzoni
 
 
 
 2012/7/25 John Fox j...@mcmaster.ca
 
  Dear Gian,
 
  How contrasts are created by default is controlled by the contrasts option:
 
   getOption(contrasts)
  unordered   ordered
  contr.treatment  contr.poly
 
  So, unless you've changed this option, contr.poly() will be used to
  generate orthogonal polynomial contrasts for an ordered factor, and you
  therefore need do nothing special to get this result. For example:
 
   (f - ordered(sample(letters[1:3], 10, replace=TRUE)))
   [1] c c a a c c b c a c
  Levels: a  b  c
 
   round(contrasts(f), 4)
.L  .Q
  [1,] -0.7071  0.4082
  [2,]  0. -0.8165
  [3,]  0.7071  0.4082
 
  For more information, see section 11 on statistical models in the manual
  An Introduction to R, which is part of the standard R distribution, and
  in particular sections 11.1 and 11.1.1.
 
  I hope that this clarifies the issue.
 
  Best,
   John
 
  
  John Fox
  Sen. William McMaster Prof. of Social Statistics
  Department of Sociology
  McMaster University
  Hamilton, Ontario, Canada
  http://socserv.mcmaster.ca/jfox/
 
  On Wed, 25 Jul 2012 11:58:30 +0200
   Manzoni, GianMauro gm.manz...@auxologico.it wrote:
   Dear Greg Snow,
   thank you very much for your suggestions. However, I need an example in
   order to understand fully.
   I was told that, given the ordinal factor, I do not need to specify the
   contr.poly function because R does it automatically.
   However, I don not know if I have to add an argument into the
  manova/anova
   function or something else.
   Please write me an illustrative example.
   Many thanks.
  
   Best regards,
   Gian Mauro Manzoni
  
   2012/7/25 Greg Snow 538...@gmail.com
  
You should not need to write them yourself. Look at the contr.poly
function along with the C function (Note uppercase C) or the contrasts
function.
   
   
On Monday, July 23, 2012, Manzoni, GianMauro wrote:
   
Dear all,
I am quite new to R and I am having trouble writing the polynomial
contrasts for an ordinal factor in MANOVA.
# I have a model such as this
fit-manova(cbind(Y1,Y2,Y3)~Groups,data=Events) # where groups is an
ordinal factor with 4 levels
# how to set polynomial contrasts for the Groups factor ?
   
Thank you very much in advance for any help!
   
Best regards,
Mauro
   
--
Dr. Gian Mauro Manzoni
PhD, PsyD
Psychology Research Laboratory
San Giuseppe Hospital
Istituto Auxologico Italiano
Verbania - Italy
e-mail: gm.manz...@auxologico.it
cell. phone +39 338 4451207
Tel. +39 0323 514278
   
[[alternative HTML version deleted]]

Re: [R] Turning off continuation prompt?

2012-07-30 Thread John Fox

Dear Ted,

I don't think that it's really advisable to do so, because it might lead to 
confusion, but you could set options(continue= ).

Best,
 John


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

On Mon, 30 Jul 2012 09:58:09 +0100 (BST)
 (Ted Harding) ted.hard...@wlandres.net wrote:
 Greetings All.
 My apologies for a question whose answer is probably
 readily available somewhere (for some interpetation
 of somewhere) ...
 
 Say I have just typed (from a sheet of paper) several
 lines into the R command-line, and what I see is:
 
  chisq.test(matrix(c(3,6,3,4,4,
 + 4,1,4,6,5,
 + 2,7,4,2,5,
 + 8,2,4,4,2,
 + 3,4,5,4,4),ncol=5))
 
 Later, I find that would like to re-input the data part
 of this command (matrix(c(...)...)). Without the +
 continuation prompts, it would be easy to do this by
 copypaste with the mouse in one operation. With the +
 marks there, I have to do the copypaste for each separate line.
 
 So is there a way to suppress the output of the + at the
 beginning of each continuation line?
 
 (The above is one of the smaller examples of this situation;
 sometimes I have wished to do this for commands extending over,
 say, 15-20 lines).
 
 With thanks,
 Ted.
 
 -
 E-Mail: (Ted Harding) ted.hard...@wlandres.net
 Date: 30-Jul-2012  Time: 09:58:02
 This message was sent by XFMail
 
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Re: [R] Turning off continuation prompt?

2012-07-30 Thread Robert Baer

On Mon, 30 Jul 2012 09:58:09 +0100 (BST) (Ted Harding) 
ted.hard...@wlandres.net wrote:

Greetings All.
My apologies for a question whose answer is probably
readily available somewhere (for some interpetation
of somewhere) ...

Say I have just typed (from a sheet of paper) several
lines into the R command-line, and what I see is:


chisq.test(matrix(c(3,6,3,4,4,

+ 4,1,4,6,5,
+ 2,7,4,2,5,
+ 8,2,4,4,2,
+ 3,4,5,4,4),ncol=5))

Later, I find that would like to re-input the data part
of this command (matrix(c(...)...)). Without the +
continuation prompts, it would be easy to do this by
copypaste with the mouse in one operation. With the +
marks there, I have to do the copypaste for each separate line.

So is there a way to suppress the output of the + at the
beginning of each continuation line?
At some level of complexity it is worth thinking of using a programming 
front end to R rather than the basic GUI.  I have really benefited from 
downloading RStudio myself, but there are any number of other choices 
that might be suited to your needs as well. Knowing how long you have 
been around this list, you probably can name more than I.

Rob

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Re: [R] If/then statement, if in a list then

2012-07-30 Thread Bert Gunter

Not necessarily. If the OP really meant the R list() structure, then
is.element does not apply. If a vector was what was intended, then it
does -- provide all elements are of the same mode. With such a vague
post, it's hard to know.


-- Bert

On Mon, Jul 30, 2012 at 6:06 AM, Jean V Adams jvad...@usgs.gov wrote:
 ?is.element

 Jean


 cm bunnylove...@optonline.net wrote on 07/30/2012 07:55:34 AM:

 I need to write an if/then statement saying something along the lines
 of:
 if (VALUE is in list) {...
 How do I write that in R's language.
 Thanks!

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Bert Gunter
Genentech Nonclinical Biostatistics

Internal Contact Info:
Phone: 467-7374
Website:
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Re: [R] using save() to work with objects that exceed memory capacity

2012-07-30 Thread David Romano

On Sun, Jul 29, 2012 at 7:08 AM, R. Michael Weylandt 
michael.weyla...@gmail.com wrote:

 On Sat, Jul 28, 2012 at 10:48 AM, David Romano roma...@grinnell.edu
 wrote:
  Context:  I'm relatively new to R and am working with very large
 datasets.
 
  General problem:  If working on a dataset requires that I produce more
 than
  two objects of roughly the size of the dataset, R quickly uses up its
  available memory and slows to a virtual halt.
 
  My tentative solution:  To save and remove objects as they're created,
 and
  load them when I need them.  To do this I'm trying to automatically
  generate file names derived from these objects, and use these in save().
 
  My specific question to the list:  How do I capture the string that names
  an object I want to save, in such a way that I can use it in a function
  that calls save()?
 
  For example, suppose I create a matrix and then save it follows:
  mat-matrix(1:9,3,3)
  save(mat, file=matfile)
  Then I get a file of the kind I'd like: the command 'load(matfile)'
  retrieves the correct matrix, with the original name 'mat'.
 
  Further, if I instead save it this way:
  objectname-mat
  save(list=ls(pattern=objectname), file=matfile)
  then I get the same positive result.
 
  But now suppose I create a function
  saveobj - function(objectname,objectfile)
  +   {
  + save(list=ls(pattern=objectname),file=objectfile);
  + return()};
  Then if I now try to save 'mat' by
  matname-mat
  saveobj(matname,matfile)
  I do not get the same result; namely, the command 'load(mat)' retrieves
  no objects.  Why is this?

 load(matfile) no?


Yes.


 It seems to work for me:

 R x - matrix(1:9, ncol = 3)
 R saveobj - function(obj, file){
 + save(list = obj, file = file)
 + }
 R exists(x)
 [1] FALSE
 R saveobj(x, amatrix.rdat)
 R rm(x)
 R load(amatrix.rdat)
 R x
  [,1] [,2] [,3]
 [1,]147
 [2,]258
 [3,]369

 Cheers,
 Michael


Thanks, Michael, for locating the trouble in the unessary call to ls(), and
thanks to Duncan Murdoch, too, for pointing out how ls() was causing the
observed behavior: without including an argument like envir=parent.frame(),
ls() only returns local objects created after the call to saveobj.   Very
helpful -- thanks to you both!

Best,
David




  
 
  I'd be grateful for any help on either my specific questions, or
  suggestions of a better ways to address the issue of limited memory.
 
  Thanks,
  David Romano
 
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[R] If equal statement for character objects

I have a data frame (trans)
123 P
143 z   ...ect

I want to write a statement that says:
if ((trans[i,2]=p)(trans[1+1,2]=z)){

but I get the warning message that states:
 operations are possible only for numeric, logical or complex types

Any ideas?
Thanks



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Re: [R] If equal statement for character objects

nevermind. all.equal() works!



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[R] A matching problem

2012-07-30 Thread Christofer Bogaso

Dear all, I was encountering with a typical Matching problem and was 
wondering whether R can help me to solve it directly.


Let say, I have 2 vectors of equal length:
vector1 - LETTERS[1:6]
vector2 - letters[1:6]

Now I need to match these 2 vectors with all possible ways like:

(A,B,C,D,E)  (a,b,c,d,e) is 1 match. Another match can be (A,B,C,D,E)  
(b,a,c,d,e), however there cant be any duplication.


Is there any direct way to doing that in R?

Thanks and regards,

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[R] use R to read/print the system hardware configuration

2012-07-30 Thread Jie

Dear All,

I am curious if there is any why to use R to know the specification of the
machine that runs on,
i.e. read the cpu model, memory size, those hardware info, maybe even with
thesoftware information.
Thank you for your attention.

Best wishes,
Jie

[[alternative HTML version deleted]]

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

[R] locked binding of setwd() in R 2.15.x causes .Rprofile to fail

2012-07-30 Thread Michael Friendly


[Env: Win XP, R 2.14.2, R 2.15.0]

I have a replacement function for setwd() in my .Rprofile which displays 
the current R path in the R window
title. It no longer works in R 2.15.x, giving the error below. Worse, 
the error prevents the rest of my

.Rprofile script from completing.

Is there some way to rescue this, i.e., preserve this behavior in
R 2.15? If not, how can I modify my script so it will still work under R 
2.14.2 and not fail under R 2.15?


Error in utils::assignInNamespace(setwd, function(dir) { :
locked binding of ‘setwd’ cannot be changed

##
#-- functions from .Rprofile
# modify setwd() to also show the current path in the window title
# assigns .lastdir in the global environment
local({
oldsetwd - setwd
utils::assignInNamespace(setwd,
function(dir) {
.lastdir - oldsetwd(dir)
utils::setWindowTitle( short.path(base::getwd()) )
.lastdir
}, base)
})

# setwd replacement, allowing cd() to be like 'cd -' on unix (return to 
last dir)

cd - function(dir) {
if(missing(dir)) dir - .lastdir
.lastdir - base::setwd(dir)
utils::setWindowTitle( short.path(base::getwd()) )
}

short.path - function(dir, len=2) {
np -length(parts - unlist(strsplit(dir, '/')))
parts -rev( rev(parts)[1:min(np,len)] )
dots - ifelse (nplen, '...', '')
paste(dots,paste(parts, '/', sep='', collapse=''))
}

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

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Re: [R] locked binding of setwd() in R 2.15.x causes .Rprofile to fail

2012-07-30 Thread William Dunlap

Does trace() do what you want from a .Rprofile file?  It works
from the command line:

   trace(setwd,print=FALSE,quote(options(prompt=paste0(dir, 
  Tracing function setwd in
  package base
  [1] setwd
   setwd(z:/)
  z:/ getwd()
  [1] z:/
  z:/ setwd(c:/temp)
  c:/temp pi
  [1] 3.141593
  c:/temp

Bill Dunlap
Spotfire, TIBCO Software
wdunlap tibco.com


 -Original Message-
 From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On 
 Behalf
 Of Michael Friendly
 Sent: Monday, July 30, 2012 8:11 AM
 To: R-help
 Subject: [R] locked binding of setwd() in R 2.15.x causes .Rprofile to fail
 
 [Env: Win XP, R 2.14.2, R 2.15.0]
 
 I have a replacement function for setwd() in my .Rprofile which displays
 the current R path in the R window
 title. It no longer works in R 2.15.x, giving the error below. Worse,
 the error prevents the rest of my
 .Rprofile script from completing.
 
 Is there some way to rescue this, i.e., preserve this behavior in
 R 2.15? If not, how can I modify my script so it will still work under R
 2.14.2 and not fail under R 2.15?
 
 Error in utils::assignInNamespace(setwd, function(dir) { :
 locked binding of 'setwd' cannot be changed
 
 ##
 #-- functions from .Rprofile
 # modify setwd() to also show the current path in the window title
 # assigns .lastdir in the global environment
 local({
 oldsetwd - setwd
 utils::assignInNamespace(setwd,
 function(dir) {
 .lastdir - oldsetwd(dir)
 utils::setWindowTitle( short.path(base::getwd()) )
 .lastdir
 }, base)
 })
 
 # setwd replacement, allowing cd() to be like 'cd -' on unix (return to
 last dir)
 cd - function(dir) {
 if(missing(dir)) dir - .lastdir
 .lastdir - base::setwd(dir)
 utils::setWindowTitle( short.path(base::getwd()) )
 }
 
 short.path - function(dir, len=2) {
 np -length(parts - unlist(strsplit(dir, '/')))
 parts -rev( rev(parts)[1:min(np,len)] )
 dots - ifelse (nplen, '...', '')
 paste(dots,paste(parts, '/', sep='', collapse=''))
 }
 
 --
 Michael Friendly Email: friendly AT yorku DOT ca
 Professor, Psychology Dept.
 York University  Voice: 416 736-5115 x66249 Fax: 416 736-5814
 4700 Keele StreetWeb:   http://www.datavis.ca
 Toronto, ONT  M3J 1P3 CANADA
 
 __
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 PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
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Re: [R] use R to read/print the system hardware configuration

2012-07-30 Thread Prof Brian Ripley


On 30/07/2012 16:08, Jie wrote:

Dear All,

I am curious if there is any why to use R to know the specification of the
machine that runs on,
i.e. read the cpu model, memory size, those hardware info, maybe even with
thesoftware information.
Thank you for your attention.


There are ways ... however, there are no remotely portable ways and you 
have not even told us your OS (as asked for in the posting guide).


This sort of thing depends not just on the OS but the precise version of 
the OS.  Look at paralllel::detectCores() for how hard one has to work 
to get the number of processors.




Best wishes,
Jie

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

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Re: [R] use R to read/print the system hardware configuration

2012-07-30 Thread Jie

Sorry to miss the OS information. I run it on Window 7 32 bit, 64 bit and
Unix (version unclear, it is a grid machine.)

Best wishes,
Jie

On Mon, Jul 30, 2012 at 11:39 AM, Prof Brian Ripley
rip...@stats.ox.ac.ukwrote:

On 30/07/2012 16:08, Jie wrote:

Dear All,

I am curious if there is any why to use R to know the specification of the
machine that runs on,
i.e. read the cpu model, memory size, those hardware info, maybe even with
thesoftware information.
Thank you for your attention.

There are ways ... however, there are no remotely portable ways and you
have not even told us your OS (as asked for in the posting guide).

This sort of thing depends not just on the OS but the precise version of
the OS. Look at paralllel::detectCores() for how hard one has to work to
get the number of processors.

Best wishes,
Jie

[[alternative HTML version deleted]]

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posting-guide.html http://www.r-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.

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

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Re: [R] A matching problem

Is this what you want:

 vector1 - sample(LETTERS[1:6]) # randomize
 vector2 - letters[1:6]
 # convert to lower case for matching
 vector1 - tolower(vector1)
 vector2 - tolower(vector2)
 # count the number of matches so order does not matter
 count - match(vector1, vector2)
 if (length(count[!is.na(count)]) == length(vector1)) print(match) else 
 print('no match')
[1] match
 vector1 - sample(letters, 6)
 vector1
[1] d o t z g q
 count - match(vector1, vector2)
 if (length(count[!is.na(count)]) == length(vector1)) print(match) else 
 print('no match')
[1] no match




On Mon, Jul 30, 2012 at 10:55 AM, Christofer Bogaso
bogaso.christo...@gmail.com wrote:
 Dear all, I was encountering with a typical Matching problem and was
 wondering whether R can help me to solve it directly.

 Let say, I have 2 vectors of equal length:
 vector1 - LETTERS[1:6]
 vector2 - letters[1:6]

 Now I need to match these 2 vectors with all possible ways like:

 (A,B,C,D,E)  (a,b,c,d,e) is 1 match. Another match can be (A,B,C,D,E) 
 (b,a,c,d,e), however there cant be any duplication.

 Is there any direct way to doing that in R?

 Thanks and regards,

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



-- 
Jim Holtman
Data Munger Guru

What is the problem that you are trying to solve?
Tell me what you want to do, not how you want to do it.

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[R] replace values in vector from a replacement table

2012-07-30 Thread Liviu Andronic

Dear all
I've got stuck when trying to replace values in a vector by selecting
replacements from a replacement table. I'm trying to use only base
functions. Here's a dummy example:
 (x - rep(letters,2))
 [1] a b c d e f g h i j k l m n o p
q r s t u v
[23] w x y z a b c d e f g h i j k l
m n o p q r
[45] s t u v w x y z
 values - c(aa, a, b, NA, d, zz)
 repl - c(aa, A, B, NA, D, zz)
 (repl.tab - cbind(values, repl))
 values repl
[1,] aa   aa
[2,] aA
[3,] bB
[4,] NA NA
[5,] dD
[6,] zz   zz


Now I can easily compute all four combinations of 'match' and '%in%':
 (ind - match(x, repl.tab[ ,1]))
 [1]  2  3 NA  5 NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA
NA NA NA NA  2  3 NA
[30]  5 NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA
 (ind - match(repl.tab[ ,1], x))
[1] NA  1  2 NA  4 NA
 (ind - x %in% repl.tab[ ,1])
 [1]  TRUE  TRUE FALSE  TRUE FALSE FALSE FALSE FALSE FALSE FALSE FALSE
FALSE FALSE FALSE
[15] FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE
FALSE  TRUE  TRUE
[29] FALSE  TRUE FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE
FALSE FALSE FALSE
[43] FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE
 (ind - repl.tab[ ,1] %in% x)
[1] FALSE  TRUE  TRUE FALSE  TRUE FALSE


But how do I actually proceed to obtain the following vector?  Can it
be done without an explicit apply() or loop?
 res
 [1] A B c D e f g h i j k l m n o p
q r s t u v
[23] w x y z A B c D e f g h i j k l
m n o p q r
[45] s t u v w x y z


Regards
Liviu


-- 
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Re: [R] replace values in vector from a replacement table

try this:

 (x - rep(letters,2))
 [1] a b c d e f g h i j k l m n o p
q r s t u v w
[24] x y z a b c d e f g h i j k l m
n o p q r s t
[47] u v w x y z
 values - c(aa, a, b, NA, d, zz)
 repl - c(aa, A, B, NA, D, zz)
 (repl.tab - cbind(values, repl))
 values repl
[1,] aa   aa
[2,] aA
[3,] bB
[4,] NA NA
[5,] dD
[6,] zz   zz
 indx - match(x, repl.tab[, 1], nomatch = 0)
 x[indx != 0] - repl.tab[indx, 2]
 x
 [1] A B c D e f g h i j k l m n o p
q r s t u v w
[24] x y z A B c D e f g h i j k l m
n o p q r s t
[47] u v w x y z




On Mon, Jul 30, 2012 at 11:53 AM, Liviu Andronic landronim...@gmail.com wrote:
 Dear all
 I've got stuck when trying to replace values in a vector by selecting
 replacements from a replacement table. I'm trying to use only base
 functions. Here's a dummy example:
 (x - rep(letters,2))
  [1] a b c d e f g h i j k l m n o p
 q r s t u v
 [23] w x y z a b c d e f g h i j k l
 m n o p q r
 [45] s t u v w x y z
 values - c(aa, a, b, NA, d, zz)
 repl - c(aa, A, B, NA, D, zz)
 (repl.tab - cbind(values, repl))
  values repl
 [1,] aa   aa
 [2,] aA
 [3,] bB
 [4,] NA NA
 [5,] dD
 [6,] zz   zz


 Now I can easily compute all four combinations of 'match' and '%in%':
 (ind - match(x, repl.tab[ ,1]))
  [1]  2  3 NA  5 NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA
 NA NA NA NA  2  3 NA
 [30]  5 NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA
 (ind - match(repl.tab[ ,1], x))
 [1] NA  1  2 NA  4 NA
 (ind - x %in% repl.tab[ ,1])
  [1]  TRUE  TRUE FALSE  TRUE FALSE FALSE FALSE FALSE FALSE FALSE FALSE
 FALSE FALSE FALSE
 [15] FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE
 FALSE  TRUE  TRUE
 [29] FALSE  TRUE FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE
 FALSE FALSE FALSE
 [43] FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE
 (ind - repl.tab[ ,1] %in% x)
 [1] FALSE  TRUE  TRUE FALSE  TRUE FALSE


 But how do I actually proceed to obtain the following vector?  Can it
 be done without an explicit apply() or loop?
 res
  [1] A B c D e f g h i j k l m n o p
 q r s t u v
 [23] w x y z A B c D e f g h i j k l
 m n o p q r
 [45] s t u v w x y z


 Regards
 Liviu


 --
 Do you know how to read?
 http://www.alienetworks.com/srtest.cfm
 http://goodies.xfce.org/projects/applications/xfce4-dict#speed-reader
 Do you know how to write?
 http://garbl.home.comcast.net/~garbl/stylemanual/e.htm#e-mail

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



-- 
Jim Holtman
Data Munger Guru

What is the problem that you are trying to solve?
Tell me what you want to do, not how you want to do it.

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[R] lattice legen and auto.key conflict

2012-07-30 Thread GOUACHE David

Hello R-helpers,

I'm trying to customize a graphic in lattice using the 'legend' argument to add 
labels on my plot but in the process I'm losing the legend drawn by 'auto.key', 
despite the fact that I'm actually not sticking these on the same sides of the 
graphic. I worked up a quick and simple example with the iris data :

### here's the basic graph
xyplot(Sepal.Length+Sepal.Width~Petal.Length+Petal.Width,data=iris,groups=Species,auto.key=list(space=right))

### now I try to add a 'legend' argument :
xyplot(Sepal.Length+Sepal.Width~Petal.Length+Petal.Width,data=iris,groups=Species,auto.key=list(space=right),legend=list(bottom=list(fun=grid.text,args=list(label=youpi
 !

### and sadly the my initial legend has disappeared...


Any pointers ?
Many thanks in advance for your help

David Gouache
ARVALIS - Institut du Végétal
Service Génétique Physiologie et Protection des Plantes
IBP - Université Paris Sud
Rue de Noetzlin - Bât. 630
91405 - ORSAY CEDEX (Adresse Postale)
91190 - GIF SUR YVETTE (Adresse GPS, livraison)
Tél : +33.(0)1.69.93.85.60
Port : +33.(0)6.86.08.94.32
Fax : +33.(0)1.69.93.85.69


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[R] how to sort huge ( 2^31 row) dataframes quickly

2012-07-30 Thread Matthew Keller

Hello all,

I have some genetic datasets (gzipped) that contain 6 columns and
upwards of 10s of billions of rows. The largest dataset is about 16 GB
on file, gzipped (!). I need to sort them according to columns 1, 2,
and 3. The setkey() function in the data.table package does this
quickly, but of course we're limited by R not being able to index
vectors with  2^31 elements, and bringing in only the parts of the
dataset we need is not applicable here.

I'm asking for practical advice from people who've done this or who
have ideas. We'd like to be able to sort the biggest datasets in hours
rather than days (or weeks!). We cannot have any process take over 50
GB RAM max (we'd prefer smaller so we can parallelize). .

Relational databases seem too slow, but maybe I am wrong. A quick look
at the bigmemory package doesn't turn up an ability to sort like this,
but again, maybe I'm wrong. My computer programmer writes in C++, so
if you have ideas in C++, that works too.

Any help would be much appreciated... Thanks!

Matt


-- 
Matthew C Keller
Asst. Professor of Psychology
University of Colorado at Boulder
www.matthewckeller.com

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Re: [R] Histogram

2012-07-30 Thread Jim Silverton

I have the following codes:
Now, suppose I have x = runif(1000,0,1, rep(1, 250), rep(0, 100) and I want
to create a 'bin' for the 0's and the 1's and put the rest of the values in
say about 20 bins. How can this be done?

Jim





On Thu, Jul 5, 2012 at 4:08 PM, Sarah Goslee sarah.gos...@gmail.com wrote:

 Which gives Jim two ways to arrive at exactly the same result, just
 different means of specifying the probs for quantile().

 Sarah

 On Thu, Jul 5, 2012 at 4:01 PM, Rui Barradas ruipbarra...@sapo.pt wrote:
  Hello,
 
  With the confusion between bin size and width the OP started, I'll
 repost my
  answer with a final line. Sorry for the repetition.
 
 
  h - hist(x, breaks=quantile(x, probs=seq(0, 1, by=1/20)))
  h$counts
  [1] 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50
 
 
  Hope this helps,
 
  Rui Barradas
 
  Em 05-07-2012 20:47, Sarah Goslee escreveu:
 
  There's no reason you can't do that with normally-distributed data,
  though I'm not sure why you'd want to. My point was rather that you
  can't specify the bin width and size both. If you let the bin size
  vary, this will work:
 
  set.seed(1234)
  mydata - rnorm(1000, mean = 2, sd = 4)
  mydata.hist - hist(mydata, breaks=quantile(mydata, probs=seq(0, 1,
  length.out = length(mydata)/50 + 1)))
  mydata.hist$counts
 
 
  Sarah
 
  On Thu, Jul 5, 2012 at 3:37 PM, Jim Silverton jim.silver...@gmail.com
  wrote:
 
  Thanks Sarah!!
  Ok so if I have say x = runif(1000,0,1) say instead if the normal and I
  want
  a histogram with bins that have an equal number of observations. For
  example
  if I want each bin to have 50 observations, how do I do this?
 
 
 
  On Thu, Jul 5, 2012 at 3:34 PM, Sarah Goslee sarah.gos...@gmail.com
  wrote:
 
 
  Hi Jim,
 
  You can't specify both number of bins and bin size. You can specify
  breaks: either the number of bins or the location of breakpoints. A
  histogram with 20 bins of 50 observations each must by definition come
  from a uniform distribution.
 
  What are you trying to accomplish?
 
  Sarah
 
  On Thu, Jul 5, 2012 at 3:29 PM, Jim Silverton 
 jim.silver...@gmail.com
  wrote:
 
  I have a column of 1000 datapoints from the normal distribution with
  mean 2
  and variance 4. How can I get a histogram of these observations with
 20
  bins with each bin having 50 observations?
 
  --
  Thanks,
  Jim.
 
 
 
 




-- 
Thanks,
Jim.

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Re: [R] replace values in vector from a replacement table

2012-07-30 Thread Liviu Andronic

On Mon, Jul 30, 2012 at 6:00 PM, jim holtman jholt...@gmail.com wrote:
 try this:
 indx - match(x, repl.tab[, 1], nomatch = 0)
 x[indx != 0] - repl.tab[indx, 2]
 x
  [1] A B c D e f g h i j k l m n o p
 q r s t u v w
 [24] x y z A B c D e f g h i j k l m
 n o p q r s t
 [47] u v w x y z


This is excellent!

Thank you
Liviu

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Re: [R] how to sort huge ( 2^31 row) dataframes quickly

This is where the sort/merge application on the mainframe has excelled
for the last 40 years.  If you can not send it to a mainframe, you can
look at the SyncSort package that runs on UNIX machines.

On Mon, Jul 30, 2012 at 12:25 PM, Matthew Keller mckellerc...@gmail.com wrote:
 Hello all,

 I have some genetic datasets (gzipped) that contain 6 columns and
 upwards of 10s of billions of rows. The largest dataset is about 16 GB
 on file, gzipped (!). I need to sort them according to columns 1, 2,
 and 3. The setkey() function in the data.table package does this
 quickly, but of course we're limited by R not being able to index
 vectors with  2^31 elements, and bringing in only the parts of the
 dataset we need is not applicable here.

 I'm asking for practical advice from people who've done this or who
 have ideas. We'd like to be able to sort the biggest datasets in hours
 rather than days (or weeks!). We cannot have any process take over 50
 GB RAM max (we'd prefer smaller so we can parallelize). .

 Relational databases seem too slow, but maybe I am wrong. A quick look
 at the bigmemory package doesn't turn up an ability to sort like this,
 but again, maybe I'm wrong. My computer programmer writes in C++, so
 if you have ideas in C++, that works too.

 Any help would be much appreciated... Thanks!

 Matt


 --
 Matthew C Keller
 Asst. Professor of Psychology
 University of Colorado at Boulder
 www.matthewckeller.com

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



-- 
Jim Holtman
Data Munger Guru

What is the problem that you are trying to solve?
Tell me what you want to do, not how you want to do it.

__
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Re: [R] use R to read/print the system hardware configuration

2012-07-30 Thread David Winsemius



On Jul 30, 2012, at 8:42 AM, Jie wrote:

Sorry to miss the OS information. I run it on Window 7 32 bit, 64  
bit and

Unix (version unclear, it is a grid machine.)


Have you looked at:

?.Machine


--
David.


Best wishes,
Jie

On Mon, Jul 30, 2012 at 11:39 AM, Prof Brian Ripley
rip...@stats.ox.ac.ukwrote:


On 30/07/2012 16:08, Jie wrote:


Dear All,

I am curious if there is any why to use R to know the  
specification of the

machine that runs on,
i.e. read the cpu model, memory size, those hardware info, maybe  
even with

thesoftware information.
Thank you for your attention.



There are ways ... however, there are no remotely portable ways and  
you

have not even told us your OS (as asked for in the posting guide).

This sort of thing depends not just on the OS but the precise  
version of
the OS.  Look at paralllel::detectCores() for how hard one has to  
work to

get the number of processors.


Best wishes,

Jie

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




--
Brian D. Ripley,  rip...@stats.ox.ac.uk
Professor of Applied Statistics,  http://www.stats.ox.ac.uk/~**ripley/ 
http://www.stats.ox.ac.uk/~ripley/

University of Oxford, Tel:  +44 1865 272861 (self)
1 South Parks Road, +44 1865 272866 (PA)
Oxford OX1 3TG, UKFax:  +44 1865 272595



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David Winsemius, MD
Alameda, CA, USA

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Re: [R] distance matrix and hclustering

2012-07-30 Thread David L Carlson

For basic hierarchical clustering use

groups - hclust(dist(qmu))

But there are many options for function dist() and for function hclust() and
these will affect the results. For basic plotting use

plot(groups)

But as you will see, there are too many cases for the labels to be legible.
Hierarchical clustering gives you from 1 to ncases groups. You have to
decide where to cut the dendrogram to define a particular number of
clusters. Since you are clustering on a single value, you may find kmeans
clustering to be more useful You must specify the number of groups in
advance, but you can try several group sizes to see what makes the most
sense for your data.

--
David L Carlson
Associate Professor of Anthropology
Texas AM University
College Station, TX 77843-4352



 -Original Message-
 From: r-help-boun...@r-project.org [mailto:r-help-bounces@r-
 project.org] On Behalf Of eliza botto
 Sent: Monday, July 30, 2012 1:44 AM
 To: r-help@r-project.org; smartpink...@yahoo.com
 Subject: [R] distance matrix and hclustering
 
 
 Dear R Users,i am very new to R. I want your help on an issue regarding
 distance matrix and cluster analysis
 i had discharge data of 4 rivers(a,b,c,d) in 4 vectors each having 364
 values
  dput(qmu)structure(list(a = c(0.26, 0.25, 0.25, 0.25, 0.24, 0.23,
 0.22, 0.21, 0.21, 0.21, 0.2, 0.19, 0.19, 0.19, 0.19, 0.18, 0.18, 0.18,
 0.17, 0.17, 0.17, 0.17, 0.17, 0.17, 0.17, 0.17, 0.17, 0.17, 0.17, 0.17,
 0.17, 0.17, 0.17, 0.17, 0.18, 0.19, 0.19, 0.19, 0.2, 0.21, 0.21, 0.21,
 0.21, 0.21, 0.21, 0.19, 0.19, 0.18, 0.17, 0.17, 0.15, 0.18, 0.2, 0.21,
 0.2, 0.19, 0.19, 0.19, 0.2, 0.24, 0.3, 0.3, 0.3, 0.32, 0.34, 0.42,
 0.46, 0.48, 0.67, 0.82, 0.79, 0.73, 0.69, 0.67, 0.67, 0.66, 0.64, 0.61,
 0.58, 0.56, 0.55, 0.55, 0.55, 0.52, 0.49, 0.48, 0.51, 0.53, 0.52, 0.49,
 0.48, 0.48, 0.46, 0.46, 0.44, 0.43, 0.43, 0.41, 0.48, 0.55, 0.57, 0.55,
 0.56, 0.6, 0.64, 0.67, 0.73, 0.84, 0.94, 1.09, 1.24, 1.28, 1.19, 1.11,
 1, 0.92, 0.86, 0.79, 0.76, 0.76, 0.76, 0.76, 0.92, 0.98, 1.03, 1.03,
 1.03, 1.03, 1.07, 1.11, 1.24, 1.44, 2.12, 3.26, 15, 9.45, 5.07, 4.59,
 3.5, 2.84, 2.54, 2.57, 3.01, 2.32, 2.32, 2.97, 2.92, 3.88, 4.76, 5.99,
 3.74, 2.92, 2.65, 2.57, 2.97, 3.4, 4.13, 4.31, 3.89, 3.45, 3.01, 2.88,
 2!
  .5, 2.29, 2.39, 2.25, 2.02, 1.87, 1.87, 2.54, 2.69, 2.76, 3.18, 3.74,
 4.59, 4.76, 4.36, 6.56, 5.07, 3.84, 3.55, 3.84, 3.84, 5.49, 5.32, 3.74,
 3.31, 3.4, 3.26, 3.09, 2.69, 2.54, 2.46, 2.39, 2.25, 2.22, 2.22, 2.25,
 2.29, 2.22, 2.18, 2.05, 2.18, 2.39, 2.18, 2.29, 2.11, 1.81, 1.6, 1.44,
 1.41, 1.32, 1.37, 1.37, 1.65, 2.31, 2.25, 1.68, 1.41, 1.26, 1.15, 3.28,
 1.93, 1.6, 1.53, 1.28, 1.13, 1.03, 1.03, 1.03, 1.03, 1, 0.96, 0.92,
 0.87, 0.82, 0.79, 0.76, 0.73, 0.7, 0.67, 0.64, 0.64, 0.61, 0.61, 0.61,
 1.76, 1.19, 1.24, 1.37, 1.68, 2.39, 2.05, 1.78, 1.58, 1.41, 1.39, 1.5,
 1.41, 1.32, 1.19, 1.11, 1.02, 1.07, 4.57, 1.96, 1.68, 1.5, 1.37, 1.24,
 1.11, 1.03, 0.96, 0.94, 2.93, 2.88, 2.92, 2.76, 2.02, 1.71, 1.5, 1.37,
 1.22, 1.09, 1, 0.94, 0.87, 0.81, 0.76, 0.73, 0.7, 0.67, 0.61, 0.58,
 0.57, 0.55, 0.53, 0.51, 0.48, 0.47, 0.44, 0.43, 0.43, 0.41, 0.41, 0.38,
 0.4, 0.4, 0.42, 0.42, 0.41, 0.46, 0.53, 0.55, 0.52, 0.49, 0.51, 0.53,
 0.55, 0.7, 1.03, 1.03, 1.17, 1.24, 1.19, 1.11, 1.03, 0.98, 0.92, 0.84,!
   0.79, 0.75, 0.7, 0.67, 0.61, 0.58, 0.56, 0.56, 0.55, 0.53, 0.51, 0.48
 , 0.46, 0.43, 0.41, 0.38, 0.37, 0.35, 0.34, 0.32, 0.31, 0.3, 0.29,
 0.28, 0.27, 0.25, 0.26, 0.24, 0.23, 0.22, 0.22, 0.21, 0.21, 0.21), b =
 c(0.19, 0.19, 0.19, 0.18, 0.18, 0.18, 0.17, 0.17, 0.17, 0.17, 0.16,
 0.17, 0.17, 0.15, 0.15, 0.15, 0.15, 0.15, 0.15, 0.15, 0.15, 0.15, 0.15,
 0.15, 0.15, 0.15, 0.15, 0.15, 0.15, 0.15, 0.14, 0.14, 0.14, 0.14, 0.14,
 0.14, 0.14, 0.14, 0.14, 0.14, 0.14, 0.14, 0.14, 0.14, 0.14, 0.15, 0.15,
 0.15, 0.15, 0.15, 0.15, 0.15, 0.16, 0.16, 0.17, 0.17, 0.18, 0.18, 0.18,
 0.18, 0.18, 0.18, 0.18, 0.18, 0.19, 0.21, 0.21, 0.21, 0.22, 0.23, 0.24,
 0.24, 0.23, 0.24, 0.24, 0.25, 0.25, 0.25, 0.28, 0.29, 0.29, 0.3, 0.31,
 0.31, 0.34, 0.41, 0.46, 0.51, 0.57, 0.61, 0.64, 0.67, 0.7, 0.76, 0.82,
 0.86, 1.05, 1.24, 1.05, 0.94, 0.92, 0.9, 0.86, 0.82, 0.76, 0.76, 0.76,
 0.78, 0.82, 0.9, 1.07, 1.76, 3.13, 3.64, 3.45, 3.01, 2.39, 2.02, 1.87,
 2.11, 2.02, 1.78, 1.63, 1.53, 1.63, 4.84, 12.5, 8.11, 3.89, 2.73, 2.11,
 1.96, 3.17, 2.65, 2.54, 3.01, 3.31, 3.6, 3.36, 2.76, 2.39, 2.11, 2.!
  25, 2.08, 1.99, 2.11, 2.36, 3.13, 7.16, 5.39, 5.52, 5.32, 4.25, 3.45,
 3.26, 3.18, 3.74, 4.35, 5.79, 5.45, 4.42, 3.84, 3.36, 2.84, 2.39, 3.84,
 3.18, 3.22, 2.97, 2.73, 2.65, 2.92, 4.33, 3.01, 3.01, 3.26, 3.09, 3.6,
 3.64, 4.05, 4.25, 4.48, 3.69, 3.74, 3.6, 3.18, 2.76, 4.11, 2.92, 2.69,
 2.73, 2.69, 3.93, 2.69, 2.18, 2.52, 2.69, 1.99, 2.57, 1.81, 1.55, 1.44,
 1.37, 1.28, 1.19, 1.19, 1.03, 1.03, 1, 0.94, 0.89, 0.87, 0.86, 0.86,
 2.3, 1.55, 1.19, 1.11, 1.5, 1.39, 1.22, 1.24, 1.07, 1.02, 0.96, 0.92,
 1.34, 1.15, 1.03, 2.06, 1.76, 1.3, 1.15, 1.05, 0.98, 0.92, 0.89, 0.84,
 0.81, 0.76, 0.73, 1.59, 5.2, 3.01, 2.05, 1.65, 1.68, 5.29,

Re: [R] lattice legen and auto.key conflict

2012-07-30 Thread Bert Gunter

David:

I think one needs to carefully parse the xyplot help, where it says:

To use more than one legend, or to have arbitrary legends not
constrained by the structure imposed by key, use the legend argument.


So I presume that this is to be interpreted as: ONLY the legend
argument will be used when both key(including auto.key) and legend
arguments are given. However, this is not clear to me either. The
Help appears to leave the behavior when one tries to use both
unspecified.

-- Bert

On Mon, Jul 30, 2012 at 9:10 AM, GOUACHE David
d.goua...@arvalisinstitutduvegetal.fr wrote:
 Hello R-helpers,

 I'm trying to customize a graphic in lattice using the 'legend' argument to 
 add labels on my plot but in the process I'm losing the legend drawn by 
 'auto.key', despite the fact that I'm actually not sticking these on the same 
 sides of the graphic. I worked up a quick and simple example with the iris 
 data :

 ### here's the basic graph
 xyplot(Sepal.Length+Sepal.Width~Petal.Length+Petal.Width,data=iris,groups=Species,auto.key=list(space=right))

 ### now I try to add a 'legend' argument :
 xyplot(Sepal.Length+Sepal.Width~Petal.Length+Petal.Width,data=iris,groups=Species,auto.key=list(space=right),legend=list(bottom=list(fun=grid.text,args=list(label=youpi
  !

 ### and sadly the my initial legend has disappeared...


 Any pointers ?
 Many thanks in advance for your help

 David Gouache
 ARVALIS - Institut du Végétal
 Service Génétique Physiologie et Protection des Plantes
 IBP - Université Paris Sud
 Rue de Noetzlin - Bât. 630
 91405 - ORSAY CEDEX (Adresse Postale)
 91190 - GIF SUR YVETTE (Adresse GPS, livraison)
 Tél : +33.(0)1.69.93.85.60
 Port : +33.(0)6.86.08.94.32
 Fax : +33.(0)1.69.93.85.69


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Internal Contact Info:
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[R] Possible bug in class 'POSIXlt' when including microseconds?

2012-07-30 Thread Janko Thyson

Dear list,

I'm a bit puzzled by an ambiguity with respect to the representation of 
micro-/milliseconds when using 'POSIXlt' objects.

It seems that the last digit of  the 'sec' attribute sometimes seems to 
differ from the digits shown when printing the 'POSIXlt' object. You'll 
find a little SO post with some example code here: 
http://stackoverflow.com/questions/11725517/ambiguity-with-posixlt-representation-when-microseconds-are-included.

In case you don't want to have a look at that, here's another short example:

|opts-  options(digits.secs=6)
x-  2012-07-30 12:10:09.123123

posix-  as.POSIXlt(x,  tz=Europe/Berlin)

  posix
[1]  2012-07-30 12:10:09.123122 Europe/Berlin

  posix$sec
[1]  9.123123
|

# Manually changing the 'sec' attribute

|posix$sec-  9.123122
  posix
[1]  2012-07-30 12:10:09.123122 Europe/Berlin
# Still '.123122'

posix$sec-  9.123124
  posix
[1]  2012-07-30 12:10:09.123124 Europe/Berlin
# Now it's '.123124' in both 'representations'
|


Thanks a lot for any clarification on this!
Janko

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Re: [R] cannot install RSTAR, MSVAR, and MSVECM packages

2012-07-30 Thread Uwe Ligges




On 30.07.2012 11:42, peter dalgaard wrote:


On Jul 30, 2012, at 11:08 , Ario Ario wrote:


*Hi all,
I got problems installing RSTAR, MSVAR, and MSVECM packages. *


No such thing exists on CRAN (nor its Archive section), so little wonder.

Similarly named packages exist for the Ox matrix language, though. Such 
packages, however, do not readily convert themselves for use with R.




... or they have different names or have been removed from CRAN, see e.g.:

http://cran.r-project.org/web/packages/STAR/index.html

http://cran.r-project.org/web/packages/MSVAR/index.html


Best,
Uwe Ligges

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Re: [R] Possible bug in class 'POSIXlt' when including microseconds?

FAQ 7.31

In floating point, you get about 15 digits of precision.  If you look
at the value of time for today you will see:

1343668783

If you add 5 more digits after the decimal point, you will see that
there is not resolution to a microsecond level.  I usually assume that
I can get millisecond resolution from POSIXct values.

On Mon, Jul 30, 2012 at 12:52 PM, Janko Thyson
janko.thyson.rst...@googlemail.com wrote:
 Dear list,

 I'm a bit puzzled by an ambiguity with respect to the representation of
 micro-/milliseconds when using 'POSIXlt' objects.

 It seems that the last digit of  the 'sec' attribute sometimes seems to
 differ from the digits shown when printing the 'POSIXlt' object. You'll
 find a little SO post with some example code here:
 http://stackoverflow.com/questions/11725517/ambiguity-with-posixlt-representation-when-microseconds-are-included.

 In case you don't want to have a look at that, here's another short example:

 |opts-  options(digits.secs=6)
 x-  2012-07-30 12:10:09.123123

 posix-  as.POSIXlt(x,  tz=Europe/Berlin)

  posix
 [1]  2012-07-30 12:10:09.123122 Europe/Berlin

  posix$sec
 [1]  9.123123
 |

 # Manually changing the 'sec' attribute

 |posix$sec-  9.123122
  posix
 [1]  2012-07-30 12:10:09.123122 Europe/Berlin
 # Still '.123122'

 posix$sec-  9.123124
  posix
 [1]  2012-07-30 12:10:09.123124 Europe/Berlin
 # Now it's '.123124' in both 'representations'
 |


 Thanks a lot for any clarification on this!
 Janko

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What is the problem that you are trying to solve?
Tell me what you want to do, not how you want to do it.

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Re: [R] Z score in gplots

2012-07-30 Thread David Winsemius



On Jul 29, 2012, at 9:49 PM, Khush gg wrote:


Hi,

Can anyone tell me how to set Z-score according to my own  
requirement as

the below code is taking as per the file entries. Any help would
be appreciable.



That was a bit vague and hard to parse, but I'm thinking I may need to  
make allowance for a non-native speaker. The help page for heatmap.2  
has an example that demonstrates changing the color palette to  
'bluered'. If you run that that example and then look at the output,  
you should see that the 'breaks' value is returned and is on a Z-scale  
as a result of setting the scale argument ... as you may already  
understand. Since there is also a breaks argument, that would appear  
to be the setting you should be modifying.



library(gplots)
x=read.table(final.txt, header=TRUE)


However, ...  you have not provided a reproducible example, and that  
is not something I make much allowance for:


 x=read.table(final.txt, header=TRUE)
Error in file(file, rt) : cannot open the connection
In addition: Warning message:
In file(file, rt) :
  cannot open file 'final.txt': No such file or directory

So I cannot offer a tested example. If you need further commentary  
perhaps you could use the help page examples as a starting point for  
further discussion rather than offering code that will surely fail.




snipped code.




Thank you in advance

KS

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--
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David Winsemius, MD
Alameda, CA, USA

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Re: [R] How can I use IPF function correctly?

2012-07-30 Thread David L Carlson

Im not sure what SAS is doing (or if you are using it correctly).  In R you
do not create marginal totals independent of the data and try to fit them to
the data. In your first example you create a matrix called raw, but you do
not use it for anything. Your loglin() call is for the all cells 16.67 and
then you fit a model in which the row and column marginal totals are used
but not the row*column marginal. Im not really sure what you are trying to
accomplish with that. 

In your second example you create three variables and then want to fit
another set of marginal totals that seem to be roughly equal distribution
for rows/columns/pages except that race has four categories but tart.reg has
only three???

If the null hypothesis for these data is no interaction between the
variables and that each category should have the same proportion of cases:
age=c(1/3, 1/3, 1/3), gender=c(1/2, 1/2), race=c(1/4, 1/4, 1/4, 1/4), then
try this:

mytable - xtabs(~age+gender+race, rawdat) # table() loses the variable
names

loglin(mytable, margin=list(0), fit=TRUE)

If you want to preserve the marginal totals for each variable, but not any
interactions between them use

loglin(mytable, margin=list(1, 2, 3), fit=TRUE)

If you want to fit the three two-way interactions use

loglin(mytable, margin=list(c(1, 2), c(2, 3), c(1, 3)), fit=TRUE)

If you want to fit the saturated table (all interactions), use

loglin(mytable, margin=list(c(1, 2, 3)), fit=TRUE)

---
David


From: Miao Zhang [mailto:mandyzhangpub...@gmail.com] 
Sent: Monday, July 30, 2012 9:35 AM
To: dcarl...@tamu.edu
Subject: Re: [R] How can I use IPF function correctly?

Thanks David,
 
The purpose of doing this is that i am trying to weighted the data to get
the target values (yes, I am using percentage instead of counts here), I
could get what I need for 2 way tables as using loglin() codes as below, I
have the row target and column target value:
 
raw-matrix(c(28.571,14.286,23.809,4.762,9.523,19.049),3,2,byrow=TRUE)
rowmarg-c(33.4,33.3,33.3)
colmarg-c(50,50)
newmat1 - loglin( rowmarg%o%colmarg/sum(colmarg), margin=list(1,2),
start=raw, fit=TRUE, eps=1.e-05, iter=100)$fit 
 
Am I am not sure how to expending into 3 or higher dimensions(I need
expending into higher dimentions latter), that's why I am considering
Iterative proportional fitting/ipf(), SAS can use ipf call, but i am not
sure how to apply in R, here we could use counts instead of %, here is an
example, say we have age, gender and region 3 variables, by using frequency:
 
### set a rawdata and view the frequency
age    -
c(1,1,1,1,1,1,1,1,2,2,2,2,2,2,2,2,3,3,3,3,3,3,3,3,1,1,2,1,2,2,2,3,3,3,3,1,1,
1,1,2,2,2,2,2,2,2,2,3,3,3,3)
gender -
c(1,1,1,1,2,2,2,2,1,1,1,1,2,2,2,2,1,1,1,1,2,2,2,2,1,1,2,2,2,2,2,1,1,1,1,2,2,
2,2,1,1,1,1,2,2,2,2,1,1,1,1)
race   -
c(1,2,3,4,1,2,3,4,1,2,3,4,1,2,3,4,1,2,3,4,1,2,3,4,1,2,2,2,2,3,4,1,2,3,4,1,2,
3,4,1,2,3,4,1,2,3,4,1,2,3,4)
rawdat-data.frame (age,gender,race)
View (rawdat)
mytable - table (rawdat$age,rawdat$gender,rawdat$race) #generates a
cross-tab of counts
View (mytable)
 
### set target value to weight the frequency 3  dimensions, NOTE, we are
using counts here not percentage, trying to fit the frequency to the target
value 
targ.age-c(17,17,17)
targ.gen-c(24,26)
targ.reg-c(13,13,12)
 
f2-ipf(mytable, margins=c(1,2,0,1,3,0,2,3), eps = 1e-04, maxits = 50,
showits = TRUE) #no 3 way interaction
Where and how should I input/set my target value here? Any sugguestions? or
I have to write my own function?
 
Manythanks,
 
Mandy
 
On Fri, Jul 27, 2012 at 5:11 PM, David L Carlson dcarl...@tamu.edu wrote:
It is not clear what you are trying to do. The ipf() function you are using
seems to be the one included in package cat for imputing missing values for
categorical variables. For ipf() you have not read the instructions
carefully because you have entered the marginal values, not their dimensions
and you have given ipf() a 2 way table but miss-specified a three way model.
No wonder it is confused. Function loglin() which is part of the included
stats package also does iterative proportional fitting.

Iterative proportional fitting (ipf) is used for fitting models for
categorical data when there are three or more variables. There is no need
for ipf on a table with two variables since, the values can be directly
calculated.

Your example data does not include the raw data counts (as it should), but
percentages for each of the 3 x 2 cells (I assume, since they sum to 100).
The marginal values you list (again percentages) are for a model assuming
equal margins. That is easily computed as 1/3*1/2*100 (one third in each row
by one half in each column times 100). So each cell should be 16.667 percent
of the total. Using loglin() that would be specified as follows:

 loglin(raw, margin=list(0), fit=TRUE)
0 iterations: deviation
$lrt
[1] 25.87661

$pearson
[1] 23.80933

$df
[1] 5

$margin
[1] 0

$fit
         [,1]     [,2]
[1,] 16.7 16.7
[2,] 16.7 16.7
[3,]

Re: [R] Hmisc improveProb() and PredictABEL reclassification () function and continuous NRI

2012-07-30 Thread Frank Harrell

If you type improveProb at the command line you'll see all the code, which is
quite simple. I'm glad you are not attempting to classify risk as this is
almost always futile.

Frank

Svingen, Gard Frodahl wrote

Dear Sirs.

I am working with the R packages Hmisc and PredictABEL to make NRI
estimates from my Cox models with and without a specific biomarker.
According to Pencina et al (Statistics in Medicine 2010, DOI:
0.1002/sim.4085 ), a continuous/non-categorical NRI (NRI0) is to be used
when there are no obvious reason to categorize risk, such as the risk of
future cardiovascular events in patients with established cardiovascular
disease.

My question is therefore: Which value(s) are to be used in the calculation
of continuous NRI from the output in Hmisc or in PredictABEL? Does
continuous NRI equal total NRI in the output?

Yours sincerely

Gard Frodahl T. Svingen
PhD student
the University of Bergen
Bergen, Norway

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[R] length of variable in mlogit

2012-07-30 Thread Lee van Cleef

Dear all,

does anybody have experience with building logits in Mlogit?

I want to test the use of a couple of alternative specific variables with a
generic regression coefficient. However, one of them simply does not work. R
says the length of this variable is different.

Problem: If I check the length of this special variable, I get a value,
which also other variables have – and with those, the logit model works! I
also checked the class of the explaining variables. The problem-laden
variable is a matrix – but the model is working with other variables which
are of the matrix and the data frame class. I also see no problems because
of wide/long formats. 

Does anybody happen to know what else I can check?

Any help would be much appreciated.

Best regards!




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Re: [R] cell blocking in log-linear models with a survey sample

2012-07-30 Thread Robert Arnold

Hello.  I hope someone can suggest a way to test log-linear models with 
cells
blocked when working with survey data. 

The tables  I am working with come from a multi-stage stratified sample. 
The 
command 'svyloglin' in the package 'survey' works well when I want to 
choose between models with two or three-way effects, but I haven't been 
able to figure out how to get proper p-values to compare models with and 
without blocked cells.

Thanks in advance for any suggestions.

Bob Arnold 
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[R] Marginal effects in mlogit

2012-07-30 Thread Lee van Cleef

Dear all,

does anybody have experience with the calculation of marginal effects
(“effects”) in Mlogit (see Croissant, Package ‘mlogit’, p.8)?

1)  Is there a good qualitative explanation available for the listed options
for the argument “type” (“aa”, …)? When do I have to choose aa, ar…? And
when can I leave it out?
2)  The table in the output – how is it to read? Does it refer to relative
changes (percentage) in the probability or absolute values (percentage
points)?

Any help including hints for literature is welcome.

Best regards!





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[R] pairwise comparisons in accordance with regression fit

2012-07-30 Thread Andres LaCortadora

Hello,

I have a unconventional question arising from my current master thesis on
regression modeling. Suppose we have fitted a (linear) relationship between
a dependent variable y and an independent variable x. Now we choose two
points on the x-axis, i.e. according to percentiles x10 and x90. These two
points are chosen to select the data points for two groups in order to
perform pairwise comparisons. In the first run, we choose 20 data pairs
around the x-values of x10 and x90 and run a statistical test in order to
infere if their y-values differ significantly. In a subsequent step, we
choose, say 30 data points around x90 and test against x10; in the next step
40 data points around x90 and test against x10 and so on. The intention for
this is to determine the group size and consequently the corresponding
lowest x-value where the comparison turns out to be significant.
How would you approach this problem. I thought of a many-to-one procedure
like the Dunnett test where multiple groups are tested against the same
control (which could be x10 in our example) (package multcomp or others).
However, these multiple groups would contain partly the same subjects. Or is
some sort of adaptive design the right choice. Or something completely
different?
I'd be very happy for any kind of help since I'm completely stuck with this
question and have no idea how to solve it
Kind regards
Andres

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[R] Problem with Bootres dcc and Dendroclim match

2012-07-30 Thread Rasmus_Jorgensen

Dear all,

I hope someone can assist with soliving my problems with the DCC function in
the BOOTRES package.
 
Initially did my analyses in Dendroclim, but when I tried to repeat the
analyses in dcc I encountered some problems, that I think I should report.
 
First I encountered that that one variable (April precipitation) that was
determined as significant in Dendroclim's response function analysis was not
significant when using the corresponding dcc function, using exactly the
same data and level of significans.
 
Moreover the values in the output-files of dcc does not reflect completely
the values from Dendroclim (with all decimals).
 
While Dendroclim returns exactly the same values each time the same analysis
is run, dcc values seem to change a bit each time. This seems a bit odd, but
I do not know the analysis good enough to say it is wrong.
 
After more testing I found out that 'April precipitation' is significant
first time I run the dcc analysis after starting R, but in all succeeding
runs, the variable is claimed NS. And this despite the confidence intervals
do not always include the zero value in each run, (which according to my
knowledge is the way the significant variables are determined).
 
One last question: I would like to know if values from correlation analysis
or response function analysis fits the data best. Is there a way to to
obtain a value for model performance from dcc like e.g. a AIC value?
 
I use :
R version 2.15.1 (2012-06-22) -- Roasted Marshmallows, Copyright (C) 2012
The R Foundation for Statistical Computing ISBN 3-900051-07-0, Platform:
i386-pc-mingw32/i386 (32-bit). For Windows XP.
But I also encountered the same problems in R version 2.15.0.
 
My script:
response - dcc(kronologiprewhite, list(temp,precip), method = response,
start = -6, end = 7, timespan=c(1963,2008), ci = 0.05, boot=TRUE)
 
I very much hope someone can find the time to answer my questions.

Rasmus Halfdan Jørgensen



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[R] barplot question

2012-07-30 Thread Michael Eisenring

 Dear r-help members.

I would like to:

a) control the margin around my legend box. Unfortunately I did not find an 
appropriate command under ?legend. The margin around the actual legend is way 
too wide. There is a lot of unnecessary empty space on the right side.

b) increase the width of the individual barplots. I saw that this can be 
obtained with the command width and xlim.However, since I have 3 Barplots 
next to each other beside=T, I could not figure it out how to do it (2 
Barplots are disappearing) 

c) I wanted to label each bar (names.arg) unfortunately nothing happens if I 
use this command (in my code i just used 1:10 for the names, in the original 
plot I would replace the numbers with real names)

Thank you very much:

Input data (dput)

structure(list(total = c(28L, 17L, 11L, 6L, 6L, 5L, 4L, 3L, 3L, 
2L), young = c(29L, 22L, 15L, 8L, 5L, 3L, 2L, 2L, 2L, 2L), old = c(36L, 
11L, 11L, 8L, 6L, 5L, 4L, 3L, 2L, 2L)), .Names = c(total, young, 
old), class = data.frame, row.names = c(NA, -10L))




R-code:
barplot(as.matrix(plants_herbs_input_top10),xaxt=n,
space=c(0.5,12),ylab= Relative frequencies (%),beside=TRUE,

col=c(rep(black,10),rep(chartreuse1,10),rep(chartreuse4,10)),ylim=c(0,50),cex.names=0.8,names.arg=c(1,
 2, 3,4,5,6,7,8,9,10,1, 2, 
3,4,5,6,7,8,9,10,1, 2, 3,4,5,6,7,8,9,10))
legend(topright,c(Total,Young secondary forest,Old secondary 
forest),cex=0.9,pt.cex=1,y.intersp=0.4,pch=15,col=c(black,chartreuse1,chartreuse4))


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Re: [R] Problem creation tensor

2012-07-30 Thread GiuseppeRicci

Hi,

thank Petr for your help.
I have implemented you code suggestion but there is another problem.
It seems that the code:

for (i in 1:m){
 Z[i,,]=table(occ, data_matrix[,i])
}

don't charge any values in Z.
Is there some error?
Thanks.
Giuseppe



--
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Re: [R] MANOVA polynomial contrasts

2012-07-30 Thread Manzoni, GianMauro

Dear Prof. John Fox,
thus all I should do to test quadratic and cubic effects is to change the
second argument of the linearHypothesis() function, right?
So, for testing the cubic effect:
  linearHypothesis (mod, f.C)

Is there a chapter or paragragh about contrasts in your book An R
companion for applied regression?

Best regards,
GMM

2012/7/30 John Fox j...@mcmaster.ca

 Dear Gian Mauro,

 On Mon, 30 Jul 2012 14:44:44 +0200
  Manzoni, GianMauro gm.manz...@auxologico.it wrote:
  Dear Prof. John Fox,
  thank you very much for your suggestions.
  However, I still do not know how to use the contrasts after generating
 them.
  Once I generate the matrix with the polynomial contrasts, what are the
  following steps toward the statistical test?

 Here's a contrived example, which uses the Anova() and linearHypothesis()
 functions in the car package:

 - snip --

  Y - matrix(rnorm(300), 100, 3)
  colnames(Y) - c(y1, y2, y3)
  f - ordered(sample(letters[1:4], 100, replace=TRUE))
  (mod - lm(Y ~ f))

 Call:
 lm(formula = Y ~ f)

 Coefficients:
  y1y2y3
 (Intercept)   0.06514  -0.01683  -0.13787
 f.L  -0.37837   0.18309   0.29736
 f.Q  -0.02102  -0.39894   0.08455
 f.C   0.05898   0.09358  -0.17634

  Anova(mod)

 Type II MANOVA Tests: Pillai test statistic
   Df test stat approx F num Df den Df Pr(F)
 f  3   0.11395   1.2634  9288 0.2566

  linearHypothesis(mod, f.L)

 Sum of squares and products for the hypothesis:
   y1y2y3
 y1  3.607260 -1.745560 -2.834953
 y2 -1.745560  0.844680  1.371839
 y3 -2.834953  1.371839  2.227995

 Sum of squares and products for error:
   y1y2y3
 y1 86.343376 -8.054928 -3.711756
 y2 -8.054928 95.473020  2.429151
 y3 -3.711756  2.429151 89.593163

 Multivariate Tests:
  Df test stat approx F num Df den Df   Pr(F)
 Pillai1 0.0648520 2.172951  3 94 0.096362 .
 Wilks 1 0.9351480 2.172951  3 94 0.096362 .
 Hotelling-Lawley  1 0.0693495 2.172951  3 94 0.096362 .
 Roy   1 0.0693495 2.172951  3 94 0.096362 .
 ---
 Signif. codes:  0 *** 0.001 ** 0.01 * 0.05 . 0.1   1

 - snip --

 You could do similar tests for the quadratic and cubic contrasts.

 I hope this helps,
  John

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

 
  A whole example would be very useful.
 
  Thank you very much in advance!
 
  Best regards,
  Gian Mauro Manzoni
 
 
 
  2012/7/25 John Fox j...@mcmaster.ca
 
   Dear Gian,
  
   How contrasts are created by default is controlled by the contrasts
 option:
  
getOption(contrasts)
   unordered   ordered
   contr.treatment  contr.poly
  
   So, unless you've changed this option, contr.poly() will be used to
   generate orthogonal polynomial contrasts for an ordered factor, and you
   therefore need do nothing special to get this result. For example:
  
(f - ordered(sample(letters[1:3], 10, replace=TRUE)))
[1] c c a a c c b c a c
   Levels: a  b  c
  
round(contrasts(f), 4)
 .L  .Q
   [1,] -0.7071  0.4082
   [2,]  0. -0.8165
   [3,]  0.7071  0.4082
  
   For more information, see section 11 on statistical models in the
 manual
   An Introduction to R, which is part of the standard R distribution,
 and
   in particular sections 11.1 and 11.1.1.
  
   I hope that this clarifies the issue.
  
   Best,
John
  
   
   John Fox
   Sen. William McMaster Prof. of Social Statistics
   Department of Sociology
   McMaster University
   Hamilton, Ontario, Canada
   http://socserv.mcmaster.ca/jfox/
  
   On Wed, 25 Jul 2012 11:58:30 +0200
Manzoni, GianMauro gm.manz...@auxologico.it wrote:
Dear Greg Snow,
thank you very much for your suggestions. However, I need an example
 in
order to understand fully.
I was told that, given the ordinal factor, I do not need to specify
 the
contr.poly function because R does it automatically.
However, I don not know if I have to add an argument into the
   manova/anova
function or something else.
Please write me an illustrative example.
Many thanks.
   
Best regards,
Gian Mauro Manzoni
   
2012/7/25 Greg Snow 538...@gmail.com
   
 You should not need to write them yourself. Look at the contr.poly
 function along with the C function (Note uppercase C) or the
 contrasts
 function.


 On Monday, July 23, 2012, Manzoni, GianMauro wrote:

 Dear all,
 I am quite new to R and I am having trouble writing the polynomial
 contrasts for an ordinal factor in MANOVA.
 # I have a model such as this
 fit-manova(cbind(Y1,Y2,Y3)~Groups,data=Events) # where groups is
 an
 ordinal

Re: [R] Convert variable to STring

2012-07-30 Thread arun

Hello,

Not sure whether this is what you need.

colnames(mymatrix)-c(function1,function2,function3)
 toString(colnames(mymatrix))
#[1] function1, function2, function3

A.K.



- Original Message -
From: Alaios ala...@yahoo.com
To: R help R-help@r-project.org
Cc: 
Sent: Monday, July 30, 2012 6:47 AM
Subject: [R] Convert variable to STring

Dear all,
I have a variable that I would like also to use it as a string. The reasons is 
that I want to collect results from different function to one table.. So when I 
use the
 colnames(mymatrix) -c(function1.function2,function3)
the function1, function2, function3 to be converted to simple strings so as

 colnames(mymatrix) -c(function1,function2,function3)

Could you please help me understand how I can do that in R?

Regards

    [[alternative HTML version deleted]]


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Re: [R] MANOVA polynomial contrasts

2012-07-30 Thread jesse.fox

Dear GMM,

 -Original Message-
 From: Manzoni, GianMauro [mailto:gm.manz...@auxologico.it]
 Sent: July-30-12 9:49 AM
 To: John Fox
 Cc: r-help@r-project.org; Greg Snow
 Subject: Re: [R] MANOVA polynomial contrasts

 Dear Prof. John Fox,
 thus all I should do to test quadratic and cubic effects is to change the
second
 argument of the linearHypothesis() function, right?
 So, for testing the cubic effect:
   linearHypothesis (mod, f.C)

Yes, but wouldn't it have been faster simply to try it? Also see
?linearHypothesis.

 Is there a chapter or paragragh about contrasts in your book An R
 companion for applied regression?

There are discussions of contrasts and of linear hypotheses about
coefficients, though not in the context of *multivariate* linear models;
that's the subject of an on-line appendix, at 
http://socserv.socsci.mcmaster.ca/jfox/Books/Companion/appendix/Appendix-Mul
tivariate-Linear-Models.pdf.

Best,
 John

 Best regards,
 GMM

 2012/7/30 John Fox j...@mcmaster.ca

   Dear Gian Mauro,

   On Mon, 30 Jul 2012 14:44:44 +0200
Manzoni, GianMauro gm.manz...@auxologico.it wrote:
Dear Prof. John Fox,
thank you very much for your suggestions.
However, I still do not know how to use the contrasts after
 generating them.
Once I generate the matrix with the polynomial contrasts, what are
 the
following steps toward the statistical test?

   Here's a contrived example, which uses the Anova() and
 linearHypothesis() functions in the car package:

   - snip --

Y - matrix(rnorm(300), 100, 3)
colnames(Y) - c(y1, y2, y3)
f - ordered(sample(letters[1:4], 100, replace=TRUE))
(mod - lm(Y ~ f))

   Call:
   lm(formula = Y ~ f)

   Coefficients:
y1y2y3
   (Intercept)   0.06514  -0.01683  -0.13787
   f.L  -0.37837   0.18309   0.29736
   f.Q  -0.02102  -0.39894   0.08455
   f.C   0.05898   0.09358  -0.17634

Anova(mod)

   Type II MANOVA Tests: Pillai test statistic
 Df test stat approx F num Df den Df Pr(F)
   f  3   0.11395   1.2634  9288 0.2566

linearHypothesis(mod, f.L)

   Sum of squares and products for the hypothesis:
 y1y2y3
   y1  3.607260 -1.745560 -2.834953
   y2 -1.745560  0.844680  1.371839
   y3 -2.834953  1.371839  2.227995

   Sum of squares and products for error:
 y1y2y3
   y1 86.343376 -8.054928 -3.711756
   y2 -8.054928 95.473020  2.429151
   y3 -3.711756  2.429151 89.593163

   Multivariate Tests:
Df test stat approx F num Df den Df   Pr(F)
   Pillai1 0.0648520 2.172951  3 94 0.096362 .
   Wilks 1 0.9351480 2.172951  3 94 0.096362 .
   Hotelling-Lawley  1 0.0693495 2.172951  3 94 0.096362 .
   Roy   1 0.0693495 2.172951  3 94 0.096362 .
   ---
   Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1

   - snip --

   You could do similar tests for the quadratic and cubic contrasts.

   I hope this helps,

John

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

A whole example would be very useful.

Thank you very much in advance!

Best regards,
Gian Mauro Manzoni

2012/7/25 John Fox j...@mcmaster.ca

 Dear Gian,

 How contrasts are created by default is controlled by the
 contrasts option:

  getOption(contrasts)
 unordered   ordered
 contr.treatment  contr.poly

 So, unless you've changed this option, contr.poly() will be used
to
 generate orthogonal polynomial contrasts for an ordered factor,
 and you
 therefore need do nothing special to get this result. For
example:

  (f - ordered(sample(letters[1:3], 10, replace=TRUE)))
  [1] c c a a c c b c a c
 Levels: a  b  c

  round(contrasts(f), 4)
   .L  .Q
 [1,] -0.7071  0.4082
 [2,]  0. -0.8165
 [3,]  0.7071  0.4082

 For more information, see section 11 on statistical models in
the
 manual
 An Introduction to R, which is part of the standard R
 distribution, and
 in particular sections 11.1 and 11.1.1.

 I hope that this clarifies the issue.

 Best,
  John

 John Fox
 Sen. William

Re: [R] curve comparison

2012-07-30 Thread Will Shadish

I'm very much a novice with R, and not a statistician. But as my group 
has been working with generalized additive models (semi-parametric 
regression), we have followed Wood's advice about using the R anova 
function to do model comparison for different regressions. I would 
imagine at least some of yours might be nested, e.g., adding polynomial 
functions to some? If so, model comparisons like this might be justified.

Will Shadish

On 7/29/2012 10:38 PM, Luis Fernando García Hernández wrote:
 Dear R users,

 I have seven regression lines I´d like to compare, in order to find out if
 these are significatively different. The main problem is that these are
 curves,  non normal, non homogeneous data, I´ve tried to linearize them but
 it has not worked. So I´d like to know if you know any command or source in
 R which explains how to perform this kind of comparison.

 Thanks in advance for your help!

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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.

-- 
William R. Shadish
Distinguished Professor
Founding Faculty
Chair, Psychological Sciences

Mailing Address:
William R. Shadish
University of California
School of Social Sciences, Humanities and Arts
5200 North Lake Rd
Merced CA  95343

Physical/Delivery Address:
University of California Merced
ATTN: William Shadish
School of Social Sciences, Humanities and Arts
Facilities Services Building A
5200 North Lake Rd.
Merced, CA 95343

209-228-4372 voice
209-228-4390 fax
wshad...@ucmerced.edu
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and provide commented, minimal, self-contained, reproducible code.

Re: [R] MANOVA polynomial contrasts

2012-07-30 Thread Manzoni, GianMauro

Dear Prof. John Fox,
thank you very much for your suggestions.
However, I still do not know how to use the contrasts after generating them.
Once I generate the matrix with the polynomial contrasts, what are the
following steps toward the statistical test?

A whole example would be very useful.

Thank you very much in advance!

Best regards,
Gian Mauro Manzoni



2012/7/25 John Fox j...@mcmaster.ca

 Dear Gian,

 How contrasts are created by default is controlled by the contrasts option:

  getOption(contrasts)
 unordered   ordered
 contr.treatment  contr.poly

 So, unless you've changed this option, contr.poly() will be used to
 generate orthogonal polynomial contrasts for an ordered factor, and you
 therefore need do nothing special to get this result. For example:

  (f - ordered(sample(letters[1:3], 10, replace=TRUE)))
  [1] c c a a c c b c a c
 Levels: a  b  c

  round(contrasts(f), 4)
   .L  .Q
 [1,] -0.7071  0.4082
 [2,]  0. -0.8165
 [3,]  0.7071  0.4082

 For more information, see section 11 on statistical models in the manual
 An Introduction to R, which is part of the standard R distribution, and
 in particular sections 11.1 and 11.1.1.

 I hope that this clarifies the issue.

 Best,
  John

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

 On Wed, 25 Jul 2012 11:58:30 +0200
  Manzoni, GianMauro gm.manz...@auxologico.it wrote:
  Dear Greg Snow,
  thank you very much for your suggestions. However, I need an example in
  order to understand fully.
  I was told that, given the ordinal factor, I do not need to specify the
  contr.poly function because R does it automatically.
  However, I don not know if I have to add an argument into the
 manova/anova
  function or something else.
  Please write me an illustrative example.
  Many thanks.
 
  Best regards,
  Gian Mauro Manzoni
 
  2012/7/25 Greg Snow 538...@gmail.com
 
   You should not need to write them yourself. Look at the contr.poly
   function along with the C function (Note uppercase C) or the contrasts
   function.
  
  
   On Monday, July 23, 2012, Manzoni, GianMauro wrote:
  
   Dear all,
   I am quite new to R and I am having trouble writing the polynomial
   contrasts for an ordinal factor in MANOVA.
   # I have a model such as this
   fit-manova(cbind(Y1,Y2,Y3)~Groups,data=Events) # where groups is an
   ordinal factor with 4 levels
   # how to set polynomial contrasts for the Groups factor ?
  
   Thank you very much in advance for any help!
  
   Best regards,
   Mauro
  
   --
   Dr. Gian Mauro Manzoni
   PhD, PsyD
   Psychology Research Laboratory
   San Giuseppe Hospital
   Istituto Auxologico Italiano
   Verbania - Italy
   e-mail: gm.manz...@auxologico.it
   cell. phone +39 338 4451207
   Tel. +39 0323 514278
  
   [[alternative HTML version deleted]]
  
   __
   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.
  
  
  
   --
   Gregory (Greg) L. Snow Ph.D.
   538...@gmail.com
  
 
 
 
  --
  Dr. Gian Mauro Manzoni
  PhD, PsyD
  Psychology Research Laboratory
  San Giuseppe Hospital
  Istituto Auxologico Italiano
  Verbania - Italy
  e-mail: gm.manz...@auxologico.it
  cell. phone +39 338 4451207
  Tel. +39 0323 514278
 
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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. Gian Mauro Manzoni
PhD, PsyD
Psychology Research Laboratory
San Giuseppe Hospital
Istituto Auxologico Italiano
Verbania - Italy
e-mail: gm.manz...@auxologico.it
cell. phone +39 338 4451207
Tel. +39 0323 514278

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[R] Unable to install gsl package

2012-07-30 Thread Michael Hughes


I'm attempting to install the gsl package using the following command:

install.packages(c:/users/mike/documents/R/win-library/gsl_1.9-9.tar.gz,repos=NULL,type=source)

My attempt is failing.  Please help.

The message supplied is:

Installing package(s) into C:/Users/Mike/Documents/R/win-library/2.15
(as lib is unspecified)
* installing *source* package 'gsl' ...
** package 'gsl' successfully unpacked and MD5 sums checked

   **
   WARNING: this package has a configure script
 It probably needs manual configuration
   **


** libs
gcc -m64 -IC:/PROGRA~1/R/R-215~1.0/include -DNDEBUG -I/include
-Id:/RCompile/CRANpkg/extralibs64/local/include -O2 -Wall  -std=gnu99 
-mtune=core2 -c airy.c -o airy.o
airy.c:1: sorry, unimplemented: 64-bit mode not compiled in
airy.c:1: error: bad value (core2) for -mtune= switch
C:/PROGRA~1/R/R-215~1.0/etc/x64/Makeconf:172: recipe for target `airy.o' failed
make: *** [airy.o] Error 1
ERROR: compilation failed for package 'gsl'
* removing 'C:/Users/Mike/Documents/R/win-library/2.15/gsl'
Warning messages:
1: running command 'C:/PROGRA~1/R/R-215~1.0/bin/x64/R CMD INSTALL -l 
C:/Users/Mike/Documents/R/win-library/2.15   
c:/users/mike/documents/R/win-library/gsl_1.9-9.tar.gz' had status 1 
2: In 
install.packages(c:/users/mike/documents/R/win-library/gsl_1.9-9.tar.gz,  :
  installation of package 
c:/users/mike/documents/R/win-library/gsl_1.9-9.tar.gz had non-zero exit 
status
  
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[R] multiple comparison tests using medians

2012-07-30 Thread Data Analytics Corp.


Hi,

A client has asked for a series of pair-wise tests similar to Tukey's 
but using medians as opposed to means.  The reason is that the 
distributions are highly right skewed.  Are there such tests (I'm 
finding nothing) and if so, is there anything in R?


Thanks,

Walt



Walter R. Paczkowski, Ph.D.
Data Analytics Corp.
44 Hamilton Lane
Plainsboro, NJ 08536

(V) 609-936-8999
(F) 609-936-3733
w...@dataanalyticscorp.com
www.dataanalyticscorp.com

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[R] Optimization help

2012-07-30 Thread Megh Dal

Hi, I have following optimization problem:

Min: x1 + x2 +...+ x7
subject to:

x1 + x2 = 80
x2 + x3 = 65
x3 + x4 = 40

all xi are ***positive integer***.

Can somebody help me in this optimization problem?

Thanks for your help

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Re: [R] If equal statement for character objects

2012-07-30 Thread Bert Gunter

1. Please read an Intro to R before posting further. In your previous
post you confused = with ==. These and other basic matters are
discussed there, so you can save yourself -- and this list -- a lot of
time and aggravation by familiarizing yourself with it.(Unless this
was just a typo).

2. I presume that you read the ?all.equal Help -- the part about not
using all.equal directly in if() expressions -- and adjusted your code
accordingly?

3. Please post directly to this list, not on Nabble, as context is
lost for those who access R-help directly, which means also that you
reduce the likelihood of a helpful reply.

-- Bert

On Mon, Jul 30, 2012 at 7:41 AM, cm bunnylove...@optonline.net wrote:
nevermind. all.equal() works!

--
View this message in context:
http://r.789695.n4.nabble.com/If-equal-statement-for-character-objects-tp4638359p4638361.html
Sent from the R help mailing list archive at Nabble.com.

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Genentech Nonclinical Biostatistics

Internal Contact Info:
Phone: 467-7374
Website:
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[R] te( ) interactions and AIC model selection with GAM

2012-07-30 Thread Ricardogg

Hello R users,

I'm working with a time-series of several years and to analyze it, I’m using
GAM smoothers from the package mgcv. I’m constructing models where
zooplankton biomass (bm) is the dependent variable and the continuous
explanatory variables are:
-time in Julian days (t), to creat a long-term linear trend
-Julian days of the year (t_year) to create an annual cycle 
- Mean temperature of Winter (temp_W), Temperature of September (temp_sept)
or Chla.
Questions:
1) To introduce a tensor product modifying the annual cycle in my model, I
tried 2 different approaches:
- a) gam ( bm ~ t + te (t_year, temp_W, temp_sept, k = c( 5,30), d= ( 1,2),
bs = c( “cc”,”cr”)), data = data)
-b) gam ( bm ~ t + te (t_year, temp_W, temp_sept, k = 5, bs = c(
“cc”,”cr”,”cr”)), data = data)
Here is my problem: when I’m using just 2 variables (e.g., t_year and
temp_W) for the tensor product, I can understand pretty well how the
interpolation works and visualize it with vis.gam() as a 3d plot or a
contour one. But with 3 variables is difficult to me to understand how it
works. Besides, I don’t which one is the proper way to construct it, a) or
b). Finally, when I plot a) or b) as vis.gam (model_name , view= c(“t_year”,
“temp_W”)), How should I interpret the plot? The effect of temp_W on the
annual cycle after considering already the effect of temp_sept or just the
individual effect of Temp_W on the annual cycle?
2) I’m trying to do a model selection using AIC criteria. I have several
questions about it:
- Should I use always the same type of smoothing basis (bs), the same type
of smoother ( e.g te) and the same dimension of the basis (k)? Example:
Option 1:
a) mod1 - gam (bm ~ t, data = data)
b) mod2 - gam (bm ~ te (t, k = 5, bs = “cr”), data = data)
c) mod3 - gam (bm ~ te (t_year, k = 5, bs = “cc”), data = data)
d) mod4 - gam (bm ~ te (t_year, temp_W, k = 5, bs = c(“cc”,”cr”)), data =
data)
e) mod5 - gam (bm ~ te (t_year, temp_W, temp_sept, k = 5, bs =
c(“cc”,”cr”,”cr”)), data = data).
Here the limitation for k = 5, is due to mod5, I don’t use s () because in
mod4 and mod5 te () is used and finally, I always use “cr” and “cc”.
Option 2: 
a) mod1 - gam (bm ~ t, data = data)
b) mod2 - gam (bm ~ s (t, k = 13, bs = “cr”), data = data)
c) mod3 - gam (bm ~ s (t_year, k = 13, bs = “cc”), data = data)
d) mod4 - gam (bm ~ te (t_year, temp_W, k = 11, bs = c(“cc”,”cr”)), data =
data)
e) mod5 - gam (bm ~ te (t_year, temp_W, temp_sept, k = 5, bs =
c(“cc”,”cr”,”cr”)), data = data).
I can get lower AIC for each of the models with Option 2, but are they
comparable when I use AIC criteria? Is it therefore the proper way to do it
as in Option 1? AIC (mod1, mod2, mod3, mod4, mod5).

Thank you in advance,
Best regards,
Ricardo González-Gil




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Re: [R] A matching problem

On Mon, Jul 30, 2012 at 08:40:59PM +0545, Christofer Bogaso wrote:
 Dear all, I was encountering with a typical Matching problem and was 
 wondering whether R can help me to solve it directly.
 
 Let say, I have 2 vectors of equal length:
 vector1 - LETTERS[1:6]
 vector2 - letters[1:6]
 
 Now I need to match these 2 vectors with all possible ways like:
 
 (A,B,C,D,E)  (a,b,c,d,e) is 1 match. Another match can be (A,B,C,D,E)  
 (b,a,c,d,e), however there cant be any duplication.

Hi.

If i understand correctly, all matches are obtained by taking all
permutations of (a,b,c,d,e) and relating them to unchanged (A,B,C,D,E).
Try the following.

  library(permute)
  vector2 - letters[1:3]
  p - allPerms(length(vector2), observed=TRUE)
  matrix(vector2[p], nrow=nrow(p), ncol=ncol(p))
  
   [,1] [,2] [,3]
  [1,] a  b  c 
  [2,] a  c  b 
  [3,] b  a  c 
  [4,] b  c  a 
  [5,] c  a  b 
  [6,] c  b  a 

The rows of the resulting matrix are all permutations of vector2.

Hope this helps.

Petr Savicky.

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[R] Are there any package for sub-sampling data

2012-07-30 Thread Bruce Miller


Hi all,

I need to sub sample some large data sets.
The problem is using complete data samples for bats that are ± data per 
1-minute time blocks using acoustic data collection and a method I 
published some time back. Miller, B. W. 2001. A method for determining 
relative activity of free flying bats using a new activity index for 
acoustic monitoring. Acta Chiropterologica. 3: 93-105.


I am convinced after analyzing complete data for each night to derive 
relative abundance and species compositions that the consultants who are 
doing wind power monitoring and ONLY using 5 to 10 minute sub samples 
for each hour like the first 5 or 10 minutes are missing especially eh 
rare species that have only a few occurrences so misleading results.


Anyone have any suggestions to a field ecologists who is not a dyed in 
the wool stats person on how to sub sample the data sets and robustly 
test this hypothesis?


Happy to co-author a paper if some one is interested as I have the data.

Cheers,

Bruce

--
Bruce W. Miller, Ph.D.
Conservation Ecologist
Neotropical Bat Projects


office details
11384 Alpine Road
Stanwood, Mi. 49346
Phone (231) 679-6059

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Re: [R] Calculating percentiles for multiple dates

2012-07-30 Thread guzzom

I was hoping to calculate the percentile for each date. So group all of one
date together, calculating the value of the 2.5 and 97.5 percentile.



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[R] sprintf doesn't care of escape characters

2012-07-30 Thread Edwin Helbert Aponte Angarita

Hi.

I am having trouble with something that should be simple. I am unable
to get sprintf using escape sequences:

 sprintf(a\nb)
[1] a\nb

 sprintf(a\bc\d)
[1] a\bc\d

But it seems to need them any way:

 sprintf(a\bcd)
Error: unexpected symbol in sprintf(a\bcd

Any suggestion on how to solve this issue?

My R system:
 version
   _
platform   x86_64-suse-linux-gnu
arch   x86_64
os linux-gnu
system x86_64, linux-gnu
status
major  2
minor  15.0
year   2012
month  03
day30
svn rev58871
language   R
version.string R version 2.15.0 (2012-03-30)

Thanks.

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Re: [R] Histogram

2012-07-30 Thread David Winsemius



On Jul 30, 2012, at 9:29 AM, Jim Silverton wrote:


I have the following codes:
Now, suppose I have x = runif(1000,0,1, rep(1, 250), rep(0, 100) and  
I want
to create a 'bin' for the 0's and the 1's and put the rest of the  
values in

say about 20 bins. How can this be done?


?cut

--
David.


Jim





On Thu, Jul 5, 2012 at 4:08 PM, Sarah Goslee  
sarah.gos...@gmail.com wrote:



Which gives Jim two ways to arrive at exactly the same result, just
different means of specifying the probs for quantile().

Sarah

On Thu, Jul 5, 2012 at 4:01 PM, Rui Barradas ruipbarra...@sapo.pt  
wrote:

Hello,

With the confusion between bin size and width the OP started, I'll

repost my

answer with a final line. Sorry for the repetition.


h - hist(x, breaks=quantile(x, probs=seq(0, 1, by=1/20)))
h$counts
[1] 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50


Hope this helps,

Rui Barradas

Em 05-07-2012 20:47, Sarah Goslee escreveu:


There's no reason you can't do that with normally-distributed data,
though I'm not sure why you'd want to. My point was rather that you
can't specify the bin width and size both. If you let the bin size
vary, this will work:

set.seed(1234)
mydata - rnorm(1000, mean = 2, sd = 4)
mydata.hist - hist(mydata, breaks=quantile(mydata, probs=seq(0, 1,
length.out = length(mydata)/50 + 1)))
mydata.hist$counts


Sarah

On Thu, Jul 5, 2012 at 3:37 PM, Jim Silverton jim.silver...@gmail.com 


wrote:


Thanks Sarah!!
Ok so if I have say x = runif(1000,0,1) say instead if the  
normal and I

want
a histogram with bins that have an equal number of observations.  
For

example
if I want each bin to have 50 observations, how do I do this?



On Thu, Jul 5, 2012 at 3:34 PM, Sarah Goslee sarah.gos...@gmail.com 


wrote:



Hi Jim,

You can't specify both number of bins and bin size. You can  
specify
breaks: either the number of bins or the location of  
breakpoints. A
histogram with 20 bins of 50 observations each must by  
definition come

from a uniform distribution.

What are you trying to accomplish?

Sarah

On Thu, Jul 5, 2012 at 3:29 PM, Jim Silverton 

jim.silver...@gmail.com

wrote:


I have a column of 1000 datapoints from the normal  
distribution with

mean 2
and variance 4. How can I get a histogram of these  
observations with

20

bins with each bin having 50 observations?

--
Thanks,
Jim.













--
Thanks,
Jim.

[[alternative HTML version deleted]]

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David Winsemius, MD
Alameda, CA, USA

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Re: [R] A matching problem

2012-07-30 Thread MacQueen, Don

Does this do what you want?
(using just three letters to keep the list of results short)


 tmp - expand.grid(v=letters[1:3],V=LETTERS[1:3])
 tmp
  v V
1 a A
2 b A
3 c A
4 a B
5 b B
6 c B
7 a C
8 b C
9 c C
 
 subset(tmp, tolower(tmp$V) != tmp$v)
  v V
2 b A
3 c A
4 a B
6 c B
7 a C
8 b C

-Don


-- 
Don MacQueen

Lawrence Livermore National Laboratory
7000 East Ave., L-627
Livermore, CA 94550
925-423-1062





On 7/30/12 7:55 AM, Christofer Bogaso bogaso.christo...@gmail.com
wrote:

Dear all, I was encountering with a typical Matching problem and was
wondering whether R can help me to solve it directly.

Let say, I have 2 vectors of equal length:
vector1 - LETTERS[1:6]
vector2 - letters[1:6]

Now I need to match these 2 vectors with all possible ways like:

(A,B,C,D,E)  (a,b,c,d,e) is 1 match. Another match can be (A,B,C,D,E) 
(b,a,c,d,e), however there cant be any duplication.

Is there any direct way to doing that in R?

Thanks and regards,

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Re: [R] sprintf doesn't care of escape characters

sprintf is working just fine.  Your problem is the interpretation of
what the results are.  Displaying the object will show the escape
characters, but if you use cat, or output to a file, you will see
that the result is correct:


 cat( sprintf(a\nb))
a
b
 cat(sprintf(a\bc\d))
abcd




On Mon, Jul 30, 2012 at 1:29 PM, Edwin Helbert Aponte Angarita
helber...@gmail.com wrote:
 Hi.

 I am having trouble with something that should be simple. I am unable
 to get sprintf using escape sequences:

 sprintf(a\nb)
 [1] a\nb

 sprintf(a\bc\d)
 [1] a\bc\d

 But it seems to need them any way:

 sprintf(a\bcd)
 Error: unexpected symbol in sprintf(a\bcd

 Any suggestion on how to solve this issue?

 My R system:
 version
_
 platform   x86_64-suse-linux-gnu
 arch   x86_64
 os linux-gnu
 system x86_64, linux-gnu
 status
 major  2
 minor  15.0
 year   2012
 month  03
 day30
 svn rev58871
 language   R
 version.string R version 2.15.0 (2012-03-30)

 Thanks.

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



-- 
Jim Holtman
Data Munger Guru

What is the problem that you are trying to solve?
Tell me what you want to do, not how you want to do it.

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Re: [R] Optimization help

On Mon, Jul 30, 2012 at 06:51:47AM -0700, Megh Dal wrote:
 Hi, I have following optimization problem:
 
 Min: x1 + x2 +...+ x7
 subject to:
 
 x1 + x2 = 80
 x2 + x3 = 65
 x3 + x4 = 40
 
 all xi are ***positive integer***.
 
 Can somebody help me in this optimization problem?

Hi.

As stated, there are no constraints on x5, x6, x7. So, these will be 0.
Try the following

  library(lpSolve)
  mat - rbind(
c(1, 1, 0, 0),
c(0, 1, 1, 0),
c(0, 0, 1, 1))
  obj - rep(1, times=ncol(mat))
  dir - rep(=, times=nrow(mat))
  rhs - c(80, 65, 40)
  out - lp(min, objective.in=obj, const.mat=mat, const.dir=dir, 
const.rhs=rhs, all.int=TRUE)
  out$solution

  [1] 55 25 40  0

Hope this helps.

Petr Savicky.

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Re: [R] sprintf doesn't care of escape characters

2012-07-30 Thread Jeff Newmiller

I am unable to determine what you think the problem is from your description.

A wild guess is that you just don't know what you are looking at in the output. 
Do you understand the difference between the output of the print function and 
of the cat function? Try

cat( sprintf(a\bc\d) )

and

print ( sprintf(a\bc\d) )

In a nutshell, print makes the string suitable for inclusion in source code, 
while cat sends the string directly as-is.
---
Jeff NewmillerThe .   .  Go Live...
DCN:jdnew...@dcn.davis.ca.usBasics: ##.#.   ##.#.  Live Go...
  Live:   OO#.. Dead: OO#..  Playing
Research Engineer (Solar/BatteriesO.O#.   #.O#.  with
/Software/Embedded Controllers)   .OO#.   .OO#.  rocks...1k
--- 
Sent from my phone. Please excuse my brevity.

Edwin Helbert Aponte Angarita helber...@gmail.com wrote:

Hi.

I am having trouble with something that should be simple. I am unable
to get sprintf using escape sequences:

 sprintf(a\nb)
[1] a\nb

 sprintf(a\bc\d)
[1] a\bc\d

But it seems to need them any way:

 sprintf(a\bcd)
Error: unexpected symbol in sprintf(a\bcd

Any suggestion on how to solve this issue?

My R system:
 version
   _
platform   x86_64-suse-linux-gnu
arch   x86_64
os linux-gnu
system x86_64, linux-gnu
status
major  2
minor  15.0
year   2012
month  03
day30
svn rev58871
language   R
version.string R version 2.15.0 (2012-03-30)

Thanks.

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Re: [R] Problem creation tensor