date:20110810

Dear all
It is difficult to use round(..., digits=2) on a data frame since one
has to first take care to remove non-numeric variables such as
'character' or 'factor':
 head(round(iris, 2))
Error in Math.data.frame(list(Sepal.Length = c(5.1, 4.9, 4.7, 4.6, 5,  :
  non-numeric variable in data frame: Species
 head(round(iris[1:4], 2))
  Sepal.Length Sepal.Width Petal.Length Petal.Width
1  5.1 3.5  1.4 0.2
2  4.9 3.0  1.4 0.2
3  4.7 3.2  1.3 0.2
4  4.6 3.1  1.5 0.2
5  5.0 3.6  1.4 0.2
6  5.4 3.9  1.7 0.4

Is there an elegant way to use round() on a data frame containing
'character' variables without removing them? Thank you
Liviu



-- 
Do you know how to read?
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Re: [R] round() a data frame containing 'character' variables?

2011-08-10 Thread Dimitris Rizopoulos


One approach is the following:

numVars - sapply(iris, is.numeric)
iris[numVars] - lapply(iris[numVars], round, digits = 2)
head(iris)

You can also put it in one lapply() call if you like.

I hope it helps.

Best,
Dimitris


On 8/10/2011 11:34 AM, Liviu Andronic wrote:

Dear all
It is difficult to use round(..., digits=2) on a data frame since one
has to first take care to remove non-numeric variables such as
'character' or 'factor':

head(round(iris, 2))

Error in Math.data.frame(list(Sepal.Length = c(5.1, 4.9, 4.7, 4.6, 5,  :
   non-numeric variable in data frame: Species

head(round(iris[1:4], 2))

   Sepal.Length Sepal.Width Petal.Length Petal.Width
1  5.1 3.5  1.4 0.2
2  4.9 3.0  1.4 0.2
3  4.7 3.2  1.3 0.2
4  4.6 3.1  1.5 0.2
5  5.0 3.6  1.4 0.2
6  5.4 3.9  1.7 0.4

Is there an elegant way to use round() on a data frame containing
'character' variables without removing them? Thank you
Liviu





--
Dimitris Rizopoulos
Assistant Professor
Department of Biostatistics
Erasmus University Medical Center

Address: PO Box 2040, 3000 CA Rotterdam, the Netherlands
Tel: +31/(0)10/7043478
Fax: +31/(0)10/7043014
Web: http://www.erasmusmc.nl/biostatistiek/

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Re: [R] round() a data frame containing 'character' variables?

2011-08-10 Thread R. Michael Weylandt michael.weyla...@gmail.com

Hello

On Wed, Aug 10, 2011 at 11:41 AM, Dimitris Rizopoulos
d.rizopou...@erasmusmc.nl wrote:
 One approach is the following:

 numVars - sapply(iris, is.numeric)
 iris[numVars] - lapply(iris[numVars], round, digits = 2)
 head(iris)

That's interesting, but still doesn't do what I need. Since it's a
read-only View() operation I would like to avoid at all cost modifying
the original data frame. And since my data frames are relatively big,
I would like to avoid generating unnecessary copies. Basically I would
need round() to ignore objects that it knows it cannot handle.

Any other ideas? Thanks
Liviu

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[R] plot 3d info in 2d

2011-08-10 Thread Andrew Halford

Hi Listers,

Is it possible to produce an ordination plot in 2d, where bubbles represent
the location of sites (this part is easy enough) and the size of the bubbles
is proportional to the sites location in 3d space (I am stuck on this
option). So sites that are very near the 2d plane of the xy axes would be
larger while sites that are actually further away in 3 d space would be
proportionally smaller.

any help/advice appreciated

Andy

-- 
Andrew Halford Ph.D
Associate Research Scientist
Marine Laboratory
University of Guam
Ph: +1 671 734 2948

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Re: [R] plot 3d info in 2d

2011-08-10 Thread Paul Hiemstra

 On 08/10/2011 10:02 AM, Andrew Halford wrote:
 Hi Listers,

 Is it possible to produce an ordination plot in 2d, where bubbles represent
 the location of sites (this part is easy enough) and the size of the bubbles
 is proportional to the sites location in 3d space (I am stuck on this
 option). So sites that are very near the 2d plane of the xy axes would be
 larger while sites that are actually further away in 3 d space would be
 proportionally smaller.

 any help/advice appreciated

 Andy

Hi Andy!

I think ggplot2 would be the package I would use to do this kind of
plots. However, without commented, minimal, self-contained,
reproducible code I cannot provide an example of how to do it.

cheers,
Paul

-- 
Paul Hiemstra, Ph.D.
Global Climate Division
Royal Netherlands Meteorological Institute (KNMI)
Wilhelminalaan 10 | 3732 GK | De Bilt | Kamer B 3.39
P.O. Box 201 | 3730 AE | De Bilt
tel: +31 30 2206 494

http://intamap.geo.uu.nl/~paul
http://nl.linkedin.com/pub/paul-hiemstra/20/30b/770

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Re: [R] plot 3d info in 2d

2011-08-10 Thread S Ellison

Very easy if you note that cex in plot can be a vector.

example:

x - runif(100)
y-runif(100)
z-runif(100)

#shift and scale z for convenience 9the scaling is based on range 'cos we know 
this is in [0,1] 
#your mileage may vary but the principle is )
 z.scaled - 0.05 + (z-min(z))/diff(range(z))

plot(x, y, cex=2*z.scaled)
#Symbol size increases linearly with z

You can add a key by giving legend() a list of three or four cex values and 
corresponding distances in z, if yo like.

But ggplot (as a previous poster indicated) is also a natural way to do this, 
and adds a nice key for you if you map a variable to an aesthetic.

 -Original Message-
 From: r-help-boun...@r-project.org 
 [mailto:r-help-boun...@r-project.org] On Behalf Of Andrew Halford
 Sent: 10 August 2011 11:03
 To: r-help@r-project.org
 Subject: [R] plot 3d info in 2d
 
 Hi Listers,
 
 Is it possible to produce an ordination plot in 2d, where 
 bubbles represent the location of sites (this part is easy 
 enough) and the size of the bubbles is proportional to the 
 sites location in 3d space (I am stuck on this option). So 
 sites that are very near the 2d plane of the xy axes would be 
 larger while sites that are actually further away in 3 d 
 space would be proportionally smaller.
 
 any help/advice appreciated
 
 Andy
 
 --
 Andrew Halford Ph.D
 Associate Research Scientist
 Marine Laboratory
 University of Guam
 Ph: +1 671 734 2948
 
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Re: [R] subqueries in sqlQuery function (package RODBC)

2011-08-10 Thread andrija djurovic

I thought subqueries in sense of commad inside the command (in my example
two select commands). It works as you proposed an I thought in this case
(subqueries) that I need different syntax for sqlQury function combining SQL
query and paste.

But now I have another problem and again sorry if it is to general and basic
but I just can't find the right option to set up.
Namely, when I import table from SQL server into R, columns that are defined
in SQL table as char (with leading zeros as 001, 002,...) are imported as
integers.

Could you, please,  guide me on some options that should be set up to solve
this problem?
On Wed, Aug 10, 2011 at 10:16 AM, Prof Brian Ripley
rip...@stats.ox.ac.ukwrote:

 In what sense is this a 'subquery'?  It is just an SQL command (write it on
 one line, no terminating ;, which is not part of the query).


 On Wed, 10 Aug 2011, andrija djurovic wrote:

   Hi R users.

 sorry for missing example and if question is to general but I am wondering
 if it is possible to execute subqueries in function sqlQuery (package
 RODBC)
 with opened connection with Excel or SQL server 2000. I couldn't find any
 example of this.

 And if it is possible what should be a correct syntax for this query:

 SELECT ct,COUNT(*) as n
 FROM (SELECT COUNT(*) AS ct FROM children
 GROUP BY family_id) AS x
 GROUP BY ct;

 sqlQuery(connecton,  CORRECT SYNTAX )

 (This query is an example from book Data Manipulation with R, Phil
 Spector,
 page 47)

 Thanks for any help

 Andrija

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 posting-guide.html http://www.r-project.org/posting-guide.html
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 --
 Brian D. Ripley,  rip...@stats.ox.ac.uk
 Professor of Applied Statistics,  
 http://www.stats.ox.ac.uk/~**ripley/http://www.stats.ox.ac.uk/~ripley/
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Re: [R] How to get the date of specific value within a zoo object?

Xts is an extension of zoo that has some other nice features: character
subsetting, periodic apply functions, good built in time conversions, etc.

More importantly for my work, the authors put a lot of work into making sure it
plays well with all of R's many ts classes so I almost always start any project
by changing input from any source to xts so I don't have to think about the
occasional inconsistencies.

If you arent working with multiple data sources and don't need the extended
functionality, there's absolutely nothing wrong with zoo - it's also a great
package.

Michael Weylandt

PS - more than anything, I was somewhat distracted when answering and couldn't
get your code to work so I falsely assumed it was a zoo problem: so I just
wrote an example I knew worked in xts - when I thought for a moment and saw
your bug, I realized the same trick would work in zoo.

On Aug 10, 2011, at 1:53 AM, Richard Ma xuanlong...@uts.edu.au wrote:

Hi Michael,

Thanks for your kindly help. Problem solved!

Just curious why you prefer xts rather than zoo? Is xts more powerful?

BTW, It's my mistake that incorrectly type the code. ;-)

Cheers,
Richard

R. Michael Weylandt lt;michael.weyla...@gmail.comgt; wrote:

I'd suggest you look into the xts class and write

require(xts)
xts = as.xts(1:5,Sys.Date()+1:5)
time(xts)[xts==3]

By the way, your code isn't pastable for me: not sure why.

Michael Weylandt

-
Richard Ma
PhD student, Ecology Remote Sensing
Climate Change Cluster, Department of Environment Science
University of Technology, Sydney
http://everydropr.wordpress.com
--
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Re: [R] How to quickly convert a data.frame into a structure of lists

I would use the tapply function (which is designed for the case in which 
data exists for most pairs of the levels of A and B) or the 
reshape::sparseby function, or something else in the reshape package. 
These won't give you exactly the structure you were asking for, but they 
will separate the data properly.


By the way, it's a good idea when posting a question to post a simple 
example; then other solutions can be illustrated on the same example. 
It doesn't need to contain millions of rows.


Duncan Murdoch

On 11-08-09 8:58 PM, Frederic F wrote:
 Hello,

 This is my first project in R, so I'm trying to work 'the R way', but it
 still feels awkward sometimes.

 The problem that I'm facing right now is that I need to convert a 
data.frame
 into a structure of lists. The data.frame has columns in the order of 
tens
 (I need to focus on only three of them) and rows in the order of 
millions.

 So it's quite a big dataset.
 Let say that the columns of interest are A, B and C. I need to take the
 data.frame and construct a structure of list where I have a list for 
every

 level of A, those list all contain lists for every levels of B, and the
 'b-lists' contains all the values of C that match the corresponding 
levels

 of A and B.
 So, I should be able to write something like this:
 MyData@list_structure$x_level_of_A$y_level_of_B
 and get a vector of the values of C that were on rows where 
A=x_level_of_A

 and B=y_level_of_B.

 My first attempt was to use two imbricated lapply functions running
 something like this:

 list_structure-lapply(levels(A) function(x) {
as.character(x) = lapply( levels(B), function(y) {
  as.character(y) = C[A==x  B==y]
})
 })

 The real code was not quite as simple, but I managed to have it work, 
and it
 worked well on my first dataset (where A and B had only few levels). 
I was
 quite happy... but the imbricated loops killed me on a second dataset 
where

 A had several thousand levels. So I tried something else.

 My second attempt was to go through every row of the data.frame and 
append

 the value to the appropriate vector.

 I first initialized a structure of lists ending with NULL vector, 
then I did

 something like this:

 for (i in 1:nrow(DataFrame)) {
eval(
  substitute(
append(MyData@list_structure$a_value$b_value, c_value),
list(a_value=as.character(DF$A[i]), b_value=as.character(DF$B[i]),
 c_value=as.character(DF$C[i]))
  )
)
 }

 This works... but way too slowly for my purpose.

 I would like to know if there is a better road to take to do this
 transformation. Or, if there is a way of speeding one of the two 
solutions

 that I have tried.

 Thank you very much for your help!

 (And in your replies, please remember that this is my first project 
in R, so

 don't hesitate to state the obvious if it seems like I am missing it!)

 Frederic

 --
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Re: [R] plot 3d info in 2d

2011-08-10 Thread Paul Hiemstra

 On 08/10/2011 10:02 AM, Andrew Halford wrote:
 Hi Listers,

 Is it possible to produce an ordination plot in 2d, where bubbles represent
 the location of sites (this part is easy enough) and the size of the bubbles
 is proportional to the sites location in 3d space (I am stuck on this
 option). So sites that are very near the 2d plane of the xy axes would be
 larger while sites that are actually further away in 3 d space would be
 proportionally smaller.

 any help/advice appreciated

 Andy


Plotting the dataset which was proposed by S. Ellison using ggplot2 is
done in this fashion:

library(ggplot2)
theme_set(theme_bw())

dat = data.frame(x - runif(100),
 y-runif(100),
 z-runif(100))

ggplot(aes(x = x, y = y, size = z), data = dat) + geom_point(color =
'lightblue')
# Using log(z) in stead of z
ggplot(aes(x = x, y = y, size = z), data = dat) + geom_point(color =
'lightblue') +
scale_size_continuous(trans = 'log')
# Alternatively, making the color of the point dependend on the value of z
ggplot(aes(x = x, y = y, color = z), data = dat) + geom_point(size = 6) +
scale_color_gradient(low = 'white', high = 'blue')

cheers,
Paul

-- 
Paul Hiemstra, Ph.D.
Global Climate Division
Royal Netherlands Meteorological Institute (KNMI)
Wilhelminalaan 10 | 3732 GK | De Bilt | Kamer B 3.39
P.O. Box 201 | 3730 AE | De Bilt
tel: +31 30 2206 494

http://intamap.geo.uu.nl/~paul
http://nl.linkedin.com/pub/paul-hiemstra/20/30b/770

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[R] Source Code glm() question

2011-08-10 Thread Axel Urbiz

Dear List,

I'm fairly new in R. I'd like to see how glm() uses the argument family in
fitting a model. Specifically, I'd like to see how a glm with a gamma family
is fitted.

Thanks for any help,
Axel.

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Re: [R] Source Code glm() question

2011-08-10 Thread ONKELINX, Thierry

Just type glm at the prompt.

 glm
function (formula, family = gaussian, data, weights, subset, 
na.action, start = NULL, etastart, mustart, offset, control = list(...), 
model = TRUE, method = glm.fit, x = FALSE, y = TRUE, contrasts = NULL, 
...) 
{
call - match.call()
if (is.character(family)) 
family - get(family, mode = function, envir = parent.frame())
if (is.function(family)) 
family - family()

and so on

 -Oorspronkelijk bericht-
 Van: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org]
 Namens Axel Urbiz
 Verzonden: woensdag 10 augustus 2011 13:16
 Aan: R-help@r-project.org
 Onderwerp: [R] Source Code glm() question
 
 Dear List,
 
 I'm fairly new in R. I'd like to see how glm() uses the argument family in 
 fitting
 a model. Specifically, I'd like to see how a glm with a gamma family is 
 fitted.
 
 Thanks for any help,
 Axel.
 
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Re: [R] glmnet

2011-08-10 Thread Patrick Breheny


On 08/10/2011 03:00 AM, Nick Sabbe wrote:

Finally, to avoid downward bias, you could run a normal glm with only the
variables selected in the previous step.


At the cost, of course, of introducing upward bias

--
Patrick Breheny
Assistant Professor
Department of Biostatistics
Department of Statistics
University of Kentucky

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Re: [R] reflecting a PCA biplot

2011-08-10 Thread Gavin Simpson

On Tue, 2011-08-09 at 22:57 +1000, Andrew Halford wrote:
 Hi Listers,
 
 I am trying to reflect a PCA biplot in the x-axis (i.e. PC1) but am not
 having much success. In theory I believe all I need to do is multiply the
 site and species scores for the PC1 by -1, which would effectively flip the
 biplot.
 
 I am creating a blank plot using the plot command and accessing the results
 from a call to rda. I then use the calls to scores to obtain separate site
 and species coordinates and I have worked out how to multiply the
 appropriate PC1 scores by -1 to create the site and species scores I want.
 However I am not sure how to change the call to plot which accesses the
 results of the call to rda to draw the blank plot. The coordinates it is
 accessing are for the unreflected ordination and this does not match the new
 site and species scores that I have.
 
 
  fish.pca - rda(fish.hel)
  fish.site - scores(fish.pca,display=sites,scaling=3)
  fish.spp - scores(fish.pca,display=species,scaling=3)
 
  fish.site[,PC1] - -1*(fish.site[,PC1])
  fish.spp[,PC1] - -1*(fish.spp[,PC1])
 
  graph - plot(fish.pca,display=c(sites,species),type=n,scaling=3) #
 how do I get the plot to draw up the blank display based on the reversed
 site and species scores?

Do you mean something like...?

require(vegan)
data(dune)
mod - rda(dune)
si.scrs - scores(mod, display = sites, scaling = 3, choices = 1:2)
sp.scrs - scores(mod, display = species, scaling = 3, choices = 1:2)

si.scrs[, PC1] - -1 * si.scrs[, PC1]
sp.scrs[, PC1] - -1 * sp.scrs[, PC1]

xlim - range(0, si.scrs[, 1], sp.scrs[, 1])
ylim - range(0, si.scrs[, 2], sp.scrs[, 2])

plot(si.scrs[,1], si.scrs[,2], ylim = ylim, xlim = xlim,
 ylab = PC2, xlab = PC1, cex = 0.7, asp = 1)
abline(h = 0, lty = dotted)
abline(v = 0, lty = dotted)
points(sp.scrs[,1], sp.scrs[,2], col = red, pch = 3, cex = 0.7)
box()

For non-standard plotting of ordination objects, our advice has always
been to build the plot up from lower-level plotting functions rather
than the specific methods supplied with vegan.

A comparison: (not quite the same, I know, but close enough)

layout(matrix(1:2, ncol = 2))
plot(mod, display=c(sites,species), type = p, scaling=3,
 main = Original)
plot(si.scrs[,1], si.scrs[,2], ylim = ylim, xlim = xlim,
 ylab = PC2, xlab = PC1, cex = 0.7, asp = 1,
 main = Flipped PC1)
abline(h = 0, lty = dotted)
abline(v = 0, lty = dotted)
points(sp.scrs[,1], sp.scrs[,2], col = red, pch = 3, cex = 0.7)
box()
layout(1)

If you want type = t, the default for plot.cca, then use the same
coordinates but extract the relevant labels from the two scores objects:

plot(si.scrs[,1], si.scrs[,2], ylim = ylim, xlim = xlim,
 ylab = PC2, xlab = PC1, cex = 0.7, asp = 1,
 type = n) ## no plotting this time first
abline(h = 0, lty = dotted)
abline(v = 0, lty = dotted)
text(si.scrs[,1], si.scrs[,2], labels = rownames(si.scrs),
 cex = 0.7) ## add site and species scores as labels
text(sp.scrs[,1], sp.scrs[,2], labels = rownames(sp.scrs),
 col = red, cex = 0.7)
box()

HTH

G

 Any help appreciated.
 
 cheers
 
 Andy

-- 
%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%
 Dr. Gavin Simpson [t] +44 (0)20 7679 0522
 ECRC, UCL Geography,  [f] +44 (0)20 7679 0565
 Pearson Building, [e] gavin.simpsonATNOSPAMucl.ac.uk
 Gower Street, London  [w] http://www.ucl.ac.uk/~ucfagls/
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[R] igraph - designing graph plot by attributes

2011-08-10 Thread Gaitan, Andreea

Hi,

I'm working on some social networks and I managed to create the graphs with 
labels and edges weight, but I would also like to change the size of the 
vertices according to the age of the persons in the network and the shape 
according to the gender. Now for the age, I have people with ages between 20 
and 78, and I would like to have 4 categories (sizes): 20-35, 36-50, 50-65, 
65. I have entered the ages as attributes of the vertices from a table, so 
they are included in the graph, but how do I change the size in the plot? And 
the same for gender with different shapes (circle and square maybe).

Thanks in advance and regards,
Andreea.

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[R] Floats in Microsoft Basic format

2011-08-10 Thread taraxacum

Hi all,
I need to convert a floating point value from Microsoft Basic format to IEEE
format.
There's a simple way to achieve this in R or I have to write my own
function?
(e.g. convert the C code below)

thanks
t

#include string.h/* for strncpy  */

  int _fmsbintoieee(float *src4, float *dest4)
 {
 unsigned char *msbin = (unsigned char *)src4;
 unsigned char *ieee = (unsigned char *)dest4;
 unsigned char sign = 0x00;
 unsigned char ieee_exp = 0x00;
 int i;

 /* MS Binary Format */
 /* byte order =m3 | m2 | m1 | exponent */
 /* m1 is most significant byte = sbbb| */
 /* m3 is the least significant byte */
 /*  m = mantissa byte   */
 /*  s = sign bit*/
 /*  b = bit */

 sign = msbin[2]  0x80;  /* 1000|b  */

 /* IEEE Single Precision Float Format   */
 /*m3m2m1 exponent   */
 /* | | emmm| seee|  */
 /*  s = sign bit*/
 /*  e = exponent bit*/
 /*  m = mantissa bit*/

 for (i=0; i4; i++) ieee[i] = 0;

 /* any msbin w/ exponent of zero = zero */
 if (msbin[3] == 0) return 0;

 ieee[3] |= sign;

 /* MBF is bias 128 and IEEE is bias 127. ALSO, MBF places   */
 /* the decimal point before the assumed bit, while  */
 /* IEEE places the decimal point after the assumed bit. */

 ieee_exp = msbin[3] - 2;/* actually, msbin[3]-1-128+127 */

 /* the first 7 bits of the exponent in ieee[3] */
 ieee[3] |= ieee_exp  1;

 /* the one remaining bit in first bin of ieee[2] */
 ieee[2] |= ieee_exp  7;

 /* 0111|b : mask out the msbin sign bit */
 ieee[2] |= msbin[2]  0x7f;

 ieee[1] = msbin[1];
 ieee[0] = msbin[0];

 return 0;
 }


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[R] Using ANCOVA in R

2011-08-10 Thread January Weiner

Hello,

I have a problem with using the following design with ANCOVA in R.

There are two groups (control + treatment), each with ten subjects.
The subjects show a response that is monitored over time (four time
points). For a single given subject, the response can be analysed with
linear regression with time as the independent variable.

The question is, how does the response differ between the two groups.
It is to be expected that the slope of the response differs between
the two groups, while the intercept itself is of minor interest.

I have tried a linear model as in

g - lm( response ~ time + subject + group)

but somehow I think that this is not correct, as it would not show a
difference in the slope of the response.

Any kind of help would be greatly appreciated.

--
January

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[R] Different approach to set up Cohen Kappa

2011-08-10 Thread gavfung

Hi, I just started learning R, and one of the most frequent thing that I need
to calculate is cohen kappa in my psychology lab and I figure being able to
do inter rater reliability is a great way for me to explore R. There are
two different scenario in which I need help with. (By the way, I have
installed irr and concord.)

1) In the lab I am working in, we go through transcripts and find certain
words to code, and we usually compare the codes between two raters. In this
case, the two rater can agree (both rater coded apple with A), have a
mismatch (one rater coded apple with A but another coded it as B) or a
miss (a rater coded apple with A but the other rater did not code it at
all). There are several codes with similar procedures and as the codes are
tallied together, a chart is constructed, similar to the one attached(
http://r.789695.n4.nabble.com/file/n3732320/example1.xls example1.xls )
title example1.xls. The maroon color represents the cells with data, and
the lavender cells is just the total in each row/column.

My question in this case is How would I calculate the Cohen Kappa for the
cell shaded in maroon?

2)I helped run participants over the summer for a psych summer internship
and after coding them, I will enter the data as shown in the attachment
title example2.xls (
http://r.789695.n4.nabble.com/file/n3732320/example2.xls example2.xls )
There is also another research assistant that entered the data and I want to
find a way to check whether we are reliable or not, and want to calculate
reliability for the following:TimeI, TimeA, TriesI, and TriesA.

Once again, i would need to convert the excel file into csv, but aside from
that, I am lost as to what I need to do. Any help is appreciated! Thank
you!

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Re: [R] R-help Digest, Vol 102, Issue 10

2011-08-10 Thread fraenzi . korner

Wir sind bis am 20. August in den Ferien und werden keine e-mails beantworten. 
Bei dringenden Fällen melden Sie sich bei Stefanie von Felten 
steffi.vonfel...@oikostat.ch

We are on vacation until 20. August. In urgent cases, please contact Stefanie 
von Felten steffi.vonfel...@oikostat.ch

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[R] varpart with repeated measures

2011-08-10 Thread VG

Dear all,

I would like to do variance partitioning of community dissimilarity matrix
(Y) using 4 explanatory tables:
X1=environmental characteristics
X2=species traits related to dispersal
X3=species characteristics (abundances and richnness)
X4= xy spatial coordinates

The problem is that I have repeated measures design (longitudinal data). I
treated Time (factor with 4 levels) as fixed factor in the main analysis
(relating community dissimilarity matrix with environmental characteristics)
and checked for sphericity etc.

Now, I would like to see what proportion of variation is due to these 4
explanatory tables, but I am not sure how to deal with spatial coordinates
(X4), without having to pool all data across dates?
Should I just repeat xy data 4 times in the table?
Should I do pcnm with X4, before using it in varpart?

I would greatly appreciate any help,
Vesna

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[R] function runif in for loop

2011-08-10 Thread Johannes Radinger

Hello,

I'd like to perform a regression using MCMCregress (MCMCpack).
One variable therefore should be a function rather than a variable:

I want to use X as an input and X should be defined as a random number between 
to values. Therefore I want to use the function runif like:
X -(1, Xa, Xb) but it seems that runif doesn't allow to use vectors.
So I think I've to calculate the new vector X by using a for loop.

I tried for (i in 1:length(lT)) T-runif(1,lT,uT) but that doesn't work.
What is the correct for-loop function to create this new vector/variable?

Can I use that function then as an input for MCMCregress?

thank you

Johannes

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Re: [R] round() a data frame containing 'character' variables?

The function format() might serve your needs.

 format(head(iris), digits=1)
  Sepal.Length Sepal.Width Petal.Length Petal.Width Species
15   41 0.2  setosa
25   31 0.2  setosa
35   31 0.2  setosa
45   32 0.2  setosa
55   41 0.2  setosa
65   42 0.4  setosa

Jean

`·.,,  (((º   `·.,,  (((º   `·.,,  (((º

Jean V. Adams
Statistician
U.S. Geological Survey
Great Lakes Science Center
223 East Steinfest Road
Antigo, WI 54409  USA



From:
Liviu Andronic landronim...@gmail.com
To:
r-help@r-project.org Help r-help@r-project.org
Date:
08/10/2011 04:37 AM
Subject:
[R] round() a data frame containing 'character' variables?
Sent by:
r-help-boun...@r-project.org



Dear all
It is difficult to use round(..., digits=2) on a data frame since one
has to first take care to remove non-numeric variables such as
'character' or 'factor':
 head(round(iris, 2))
Error in Math.data.frame(list(Sepal.Length = c(5.1, 4.9, 4.7, 4.6, 5,  :
  non-numeric variable in data frame: Species
 head(round(iris[1:4], 2))
  Sepal.Length Sepal.Width Petal.Length Petal.Width
1  5.1 3.5  1.4 0.2
2  4.9 3.0  1.4 0.2
3  4.7 3.2  1.3 0.2
4  4.6 3.1  1.5 0.2
5  5.0 3.6  1.4 0.2
6  5.4 3.9  1.7 0.4

Is there an elegant way to use round() on a data frame containing
'character' variables without removing them? Thank you
Liviu



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[R] rbind/cbind

2011-08-10 Thread Anthony Ching Ho Ng

Dear list,

I wonder if there a better way to have rbind/cbind/append to create
the first element (if it is empty) instead of doing the following in a
loop?

for (i in 1:10) {
  if (i == 1) {
 aRow = SomeExpression(i)
  } else {
aRow = rbind(aRow,SomeExpression(i))
  }
}

Thanks

Anthony

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Re: [R] Plotting Ellipses and Points of Matching Colors in an Ordination

Gabe,

Since you didn't provide a small example of your data, I can't test out 
your code.  However, I used an example from the ?ordiellipse help page to 
draw different colored ellipses (using the show.groups= argument) with 
labels (using the label= argument).

Hope this helps.

library(vegan)

data(dune)
data(dune.env)
mod - cca(dune ~ Management, dune.env)
attach(dune.env)

plot(mod, type=n)
points(mod, display=sites, pch=as.numeric(Management), 
col=as.numeric(Management))
groupz - sort(unique(Management))
for(i in seq(groupz)) {
ordiellipse(mod, Management, kind=se, conf=0.95, label=T, 
font=2, cex=1.5, col=i, show.groups=groupz[i])
}

Jean


`·.,,  (((º   `·.,,  (((º   `·.,,  (((º

Jean V. Adams
Statistician
U.S. Geological Survey
Great Lakes Science Center
223 East Steinfest Road
Antigo, WI 54409  USA



From:
Gabriel Yospin yosp...@gmail.com
To:
r-help@r-project.org
Date:
08/09/2011 10:34 PM
Subject:
[R] Plotting Ellipses and Points of Matching Colors in an Ordination
Sent by:
r-help-boun...@r-project.org



Hello, R-Help -

I am trying to plot the results of an ordination from package vegan. The
tricky part for me right now is getting the colors of the ellipses 
denoting
the 95% confidence intervals of the group centroids to match the colors of
the points for those same groups.

From and earlier post, I saw the code to make a plot of the ordination 
using
different colors for my different groups.  My functional code is below:

library(vegan)
comvars - read.csv(commvars9trim2.csv)
com.mds - metaMDS(comvars, trace=FALSE)
firefactors - read.csv(commvars9factors.csv)
plot(com.mds, type = n)
points(com.mds, dis=si, pch = as.numeric(firefactors$comm4), col =
as.numeric(firefactors$comm4))

This code will generate the ordination and plot it, using the factor 
levels
from firefactors$comm4 to determine the colors and characters to use in 
the
plot.

What I would like to do next is plot the ellipses denoting the 95%
confidence intervals of the group centroids, with colors matching those of
the points. The following piece of code plots the centroids:

with(firefactors, ordiellipse(com.mds, comm4, kind = se, conf = 0.95, 
col
= as.numeric(firefactors$comm4)), label = TRUE)

But the above code makes all of the ellipses blue. Blue is the fourth 
color
in my default palette(), and the first value returned by
as.numeric(firefactors$comm4) is 4. I assume that's not a coincidence, but 
I
could be wrong.

I have also tried using:

plot(com.mds, display = sites, type = p)
with(firefactors, ordiellipse(com.mds, comm4, kind = se, conf = 0.95, 
col
= as.numeric(firefactors$comm4))

But that code also gives me blue ellipses.

Finally, I'd like to label the ellipses. The only way I've found to do 
that
is by using the ordispider() function. Is there any way to make ordispider
draw no lines? I've tried

with(firefactors, ordispider(com.mds, comm4, col = green3, label = TRUE,
lty = 3))

My two questions:

1. How do I make the ellipse color match the color of the points each
ellipse represents?

2. How do I label those ellipses, without drawing the dashed lines as per
ordispider()?

Many thanks,

Gabe

-- 
Gabriel I. Yospin

Center for Ecology  Evolutionary Biology
Bridgham Lab
University of Oregon
Eugene, OR 97403-5289

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Re: [R] Loops for repetitive task

2011-08-10 Thread Dennis Murphy

Hi:

Try this:

## Function that takes a data frame as input and outputs a data frame:
chrSumm - function(d) {   # d is a data frame
colnames(d) - c(chr,start,end,base1,base2,
 totalreads,methylation,strand)
TR - nrow(d)
RG1 - sum(d['totalreads'] = 1)
percent - TR/RG1
methylSumm - summary(d$methylation)
names(methylSumm) - c('Min', 'Q1', 'Median', 'Mean', 'Q3', 'Max')
data.frame(TR, RG1, percent, as.data.frame(as.list(methylSumm)))
  }

# Read the data files into a list and apply the function to each file
recursively,
# resulting in a data frame

# vector of file names
files - c('chr1.out.txt', 'chr2.out.txt')
# use lapply() to read files into a list
filelist - lapply(files, read.table, header = FALSE)
# Use the ldply() function from the plyr package to
# process the list and return a data frame
library('plyr')
ldply(filelist, chrSumm)

# Result from your example:
 ldply(filelist, chrSumm)
  TR RG1 percent  Min Q1 MedianMean Q3  Max
1  4   4 1.0 0.04 0.0475   0.07 0.07500 0.0975 0.12
2  3   2 1.5 0.00 0.0150   0.03 0.03667 0.0550 0.08

HTH,
Dennis

On Tue, Aug 9, 2011 at 9:31 PM, a217 aj...@case.edu wrote:
 Hello,

 I have an R script that I use as a template to perform a task for multiple
 files (in this case, multiple chromosomes).

 What I would like to do is to utilize a simple loop to parse through each
 chromosome number so that I don't have to type the same code over and over
 again in the R console.

 I've tried using:

 for(i in 1:22){
 etc..
 }

 and replacing each chromosome number with [[i]], but that did not seem to
 work.

 Below is the script I have. Basically everywhere you see a '2' I would like
 there to be an 'i' so that the script can be applied in a general sense.
 Code###

 chr2.data-read.table(file=chr2.out.txt, header=F)
 colnames(chr2.data)-c(chr,start,end,base1,base2,totalreads,methylation,strand)
 splc2-split(chr2.data, paste(chr2.data$chr))
 chr2.df-as.data.frame(t(sapply(splc2, function(x)
 list(TR=NROW(x[['totalreads']]),    RG1=sum(x[['totalreads']]=1),
 percent=(NROW(x[['totalreads']]=1)/sum(x[['totalreads']]))
 chr2.df.summ-as.data.frame(t(sapply(splc2, function(x)
 summary(x$methylation
 chr2.summ-cbind(chr2.df,chr2.df.summ)

 ##


 Here are some sample input files in case you'd like to test the code:
 ##
 # chr1.out.txt
 ##
 chr1    100     159     104     104     1       0.05    +
 chr1    100     159     145     145     1       0.04    +
 chr1    200     260     205     205     1       0.12    +
 chr1    500     750     600     600     1       0.09    +

 ##
 # chr2.out.txt
 ##
 chr2    100     200     105     105     1       0.03    +
 chr2    100     200     110     110     1       0.08    +
 chr2    300     400     350     350     0       0       +


 The code works perfectly fine just typing everything out by hand, but that
 is very inefficient given that there are 24 chromosomes for each dataset. I
 am just looking for any suggestions as to how I can write a general version
 of this code.


 --
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[R] column names issue with read.csv

2011-08-10 Thread Anthony Ching Ho Ng

Dear List,

I wonder why when using read.csv(), if the column name contains a
numeric i.e. a stock symbols-0001.HK, it will automatically insert
an X character to the column names - X0001.HK.

Now I have to manually do a loop and use substring() to remove the X
character. Any advice? Thanks

Anthony

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Re: [R] function runif in for loop

Johannes,

You have the loop set up right, you just need to add indexing to refer to 
the looping variable, i.

lT - sample(1:10)
uT - sample(21:30)
X - numeric(length(lT))

for (i in 1:length(lT)) X[i] - runif(1, lT[i], uT[i])

X

Note that I changed the name of the result from T to X, because T has 
special meaning in R.

Jean

`·.,,  (((º   `·.,,  (((º   `·.,,  (((º

Jean V. Adams
Statistician
U.S. Geological Survey
Great Lakes Science Center
223 East Steinfest Road
Antigo, WI 54409  USA



From:
Johannes Radinger jradin...@gmx.at
To:
r-help@r-project.org
Date:
08/10/2011 07:23 AM
Subject:
[R] function runif in for loop
Sent by:
r-help-boun...@r-project.org



Hello,

I'd like to perform a regression using MCMCregress (MCMCpack).
One variable therefore should be a function rather than a variable:

I want to use X as an input and X should be defined as a random number 
between to values. Therefore I want to use the function runif like:
X -(1, Xa, Xb) but it seems that runif doesn't allow to use vectors.
So I think I've to calculate the new vector X by using a for loop.

I tried for (i in 1:length(lT)) T-runif(1,lT,uT) but that doesn't work.
What is the correct for-loop function to create this new vector/variable?

Can I use that function then as an input for MCMCregress?

thank you

Johannes

--

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http://www.R-project.org/posting-guide.html
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Re: [R] column names issue with read.csv

Anthony,

See ?make.names for a description of valid names.  Here's an excerpt:

A syntactically valid name consists of letters, numbers and the dot or 
underline characters and starts with a letter or the dot not followed by a 
number. ... The character X is prepended if necessary.

There is an argument in read.csv() called check.names.  Try setting this 
to FALSE and see if that works.

read.csv(file=Something.csv, check.names=F)

Jean

`·.,,  (((º   `·.,,  (((º   `·.,,  (((º

Jean V. Adams
Statistician
U.S. Geological Survey
Great Lakes Science Center
223 East Steinfest Road
Antigo, WI 54409  USA



From:
Anthony Ching Ho Ng anthony.ch...@gmail.com
To:
r-help@r-project.org
Date:
08/10/2011 08:18 AM
Subject:
[R] column names issue with read.csv
Sent by:
r-help-boun...@r-project.org



Dear List,

I wonder why when using read.csv(), if the column name contains a
numeric i.e. a stock symbols-0001.HK, it will automatically insert
an X character to the column names - X0001.HK.

Now I have to manually do a loop and use substring() to remove the X
character. Any advice? Thanks

Anthony

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

Re: [R] How to quickly convert a data.frame into a structure of lists

2011-08-10 Thread Dennis Murphy

To borrow shamelessly from one of the prominent helpers on this list:

What is the problem you're trying to solve?(attribution: Jim Holtman)

I have the sense you want to do something over many subsets of your
data frame. If so, breaking things up into lists of lists of lists is
not necessarily productive, nor may it be necessary to use loops
explicitly, depending on the nature of what you want to do. If you're
more explicit about the nature of your task, it's entirely possible
that there may be a nice 'R way' to do it. Read the posting guide and
if at all possible, provide a small, reproducible example that
demonstrates what you want to accomplish.
(See ?dput to learn how to transmit data by e-mail.)

HTH,
Dennis

On Tue, Aug 9, 2011 at 5:58 PM, Frederic F fournier.frede...@gmail.com wrote:
 Hello,

 This is my first project in R, so I'm trying to work 'the R way', but it
 still feels awkward sometimes.

 The problem that I'm facing right now is that I need to convert a data.frame
 into a structure of lists. The data.frame has columns in the order of tens
 (I need to focus on only three of them) and rows in the order of millions.
 So it's quite a big dataset.
 Let say that the columns of interest are A, B and C. I need to take the
 data.frame and construct a structure of list where I have a list for every
 level of A, those list all contain lists for every levels of B, and the
 'b-lists' contains all the values of C that match the corresponding levels
 of A and B.
 So, I should be able to write something like this:
 MyData@list_structure$x_level_of_A$y_level_of_B
 and get a vector of the values of C that were on rows where A=x_level_of_A
 and B=y_level_of_B.

 My first attempt was to use two imbricated lapply functions running
 something like this:

 list_structure-lapply(levels(A) function(x) {
  as.character(x) = lapply( levels(B), function(y) {
    as.character(y) = C[A==x  B==y]
  })
 })

 The real code was not quite as simple, but I managed to have it work, and it
 worked well on my first dataset (where A and B had only few levels). I was
 quite happy... but the imbricated loops killed me on a second dataset where
 A had several thousand levels. So I tried something else.

 My second attempt was to go through every row of the data.frame and append
 the value to the appropriate vector.

 I first initialized a structure of lists ending with NULL vector, then I did
 something like this:

 for (i in 1:nrow(DataFrame)) {
  eval(
    substitute(
      append(MyData@list_structure$a_value$b_value, c_value),
      list(a_value=as.character(DF$A[i]), b_value=as.character(DF$B[i]),
 c_value=as.character(DF$C[i]))
    )
  )
 }

 This works... but way too slowly for my purpose.

 I would like to know if there is a better road to take to do this
 transformation. Or, if there is a way of speeding one of the two solutions
 that I have tried.

 Thank you very much for your help!

 (And in your replies, please remember that this is my first project in R, so
 don't hesitate to state the obvious if it seems like I am missing it!)

 Frederic

 --
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 http://r.789695.n4.nabble.com/How-to-quickly-convert-a-data-frame-into-a-structure-of-lists-tp3731746p3731746.html
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Re: [R] Floats in Microsoft Basic format


On 10/08/2011 5:58 AM, taraxacum wrote:

Hi all,
I need to convert a floating point value from Microsoft Basic format to IEEE
format.
There's a simple way to achieve this in R or I have to write my own
function?
(e.g. convert the C code below)


You'll need to write your own function.  It can be very similar to what 
you have below, except:
 - you'll need to produce double rather than single (easy, just produce 
a single and assign to a double, C does the conversion for you)

 - you need to change the args to your function to some that R understands.

Duncan Murdoch


thanks
t

#includestring.h /* for strncpy  */

   int _fmsbintoieee(float *src4, float *dest4)
  {
  unsigned char *msbin = (unsigned char *)src4;
  unsigned char *ieee = (unsigned char *)dest4;
  unsigned char sign = 0x00;
  unsigned char ieee_exp = 0x00;
  int i;

  /* MS Binary Format */
  /* byte order = m3 | m2 | m1 | exponent */
  /* m1 is most significant byte =  sbbb| */
  /* m3 is the least significant byte */
  /*  m = mantissa byte   */
  /*  s = sign bit*/
  /*  b = bit */

  sign = msbin[2]  0x80;  /* 1000|b  */

  /* IEEE Single Precision Float Format   */
  /*m3m2m1 exponent   */
  /* | | emmm| seee|  */
  /*  s = sign bit*/
  /*  e = exponent bit*/
  /*  m = mantissa bit*/

  for (i=0; i4; i++) ieee[i] = 0;

  /* any msbin w/ exponent of zero = zero */
  if (msbin[3] == 0) return 0;

  ieee[3] |= sign;

  /* MBF is bias 128 and IEEE is bias 127. ALSO, MBF places   */
  /* the decimal point before the assumed bit, while  */
  /* IEEE places the decimal point after the assumed bit. */

  ieee_exp = msbin[3] - 2;/* actually, msbin[3]-1-128+127 */

  /* the first 7 bits of the exponent in ieee[3] */
  ieee[3] |= ieee_exp  1;

  /* the one remaining bit in first bin of ieee[2] */
  ieee[2] |= ieee_exp  7;

  /* 0111|b : mask out the msbin sign bit */
  ieee[2] |= msbin[2]  0x7f;

  ieee[1] = msbin[1];
  ieee[0] = msbin[0];

  return 0;
  }


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Re: [R] function runif in for loop


On 10/08/2011 7:28 AM, Johannes Radinger wrote:

Hello,

I'd like to perform a regression using MCMCregress (MCMCpack).
One variable therefore should be a function rather than a variable:

I want to use X as an input and X should be defined as a random number between 
to values. Therefore I want to use the function runif like:
X-(1, Xa, Xb) but it seems that runif doesn't allow to use vectors.
So I think I've to calculate the new vector X by using a for loop.


runif() does allow vectors.  Assuming Xa and Xb are vectors of length n, 
then


X - runif(n, Xa, Xb)

will work.  (Xa and Xb don't both have to be vectors; values will be 
recycled as necessary.)


Duncan Murdoch

I tried for (i in 1:length(lT)) T-runif(1,lT,uT) but that doesn't work.
What is the correct for-loop function to create this new vector/variable?

Can I use that function then as an input for MCMCregress?

thank you

Johannes

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Re: [R] rbind/cbind



On Aug 10, 2011, at 9:08 AM, Anthony Ching Ho Ng wrote:


Dear list,

I wonder if there a better way to have rbind/cbind/append to create
the first element (if it is empty) instead of doing the following in a
loop?

for (i in 1:10) {
 if (i == 1) {
aRow = SomeExpression(i)
 } else {
   aRow = rbind(aRow,SomeExpression(i))
 }
}



Generally one is advised not to use rbind in this manner but rather to  
pre-allocate aRow to the size needed and then to add information by  
rows using [.


For a matrix this might be:

aRow - matrix(NA, ncol=3, nrow=10)

for (i in 1:10) {
aRow[1,] - SomeExpression(i)
 }


--
David Winsemius, MD
West Hartford, CT

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Re: [R] rbind/cbind

2011-08-10 Thread ONKELINX, Thierry

Can't you use sapply?

sapply(seq_len(10), function(i){SomeExpression(i)})

Best regards,

Thierry

 -Oorspronkelijk bericht-
 Van: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org]
 Namens David Winsemius
 Verzonden: woensdag 10 augustus 2011 15:50
 Aan: Anthony Ching Ho Ng
 CC: r-help@r-project.org
 Onderwerp: Re: [R] rbind/cbind
 
 
 On Aug 10, 2011, at 9:08 AM, Anthony Ching Ho Ng wrote:
 
  Dear list,
 
  I wonder if there a better way to have rbind/cbind/append to create
  the first element (if it is empty) instead of doing the following in a
  loop?
 
  for (i in 1:10) {
   if (i == 1) {
  aRow = SomeExpression(i)
   } else {
 aRow = rbind(aRow,SomeExpression(i))  } }
 
 
 Generally one is advised not to use rbind in this manner but rather to pre-
 allocate aRow to the size needed and then to add information by rows using 
 [.
 
 For a matrix this might be:
 
 aRow - matrix(NA, ncol=3, nrow=10)
 
 for (i in 1:10) {
  aRow[1,] - SomeExpression(i)
   }
 
 
 --
 David Winsemius, MD
 West Hartford, CT
 
 __
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[R] studentized and standarized residuals

2011-08-10 Thread Jen

Hi,
I must be doing something silly here, because I can't get the studentised
and standardised residuals from r output of a linear model to agree with
what I think they should be from equation form. 
Thanks in advance,
Jennifer

x = seq(1,10)
y = x + rnorm(10)
mod = lm(y~x)
rstandard(mod)
residuals(mod)/(summary(mod)$sigma)
rstudent(mod)
residuals(mod)/(summary(mod)$sigma*sqrt(1-lm.influence(mod)$hat))

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Re: [R] Floats in Microsoft Basic format

2011-08-10 Thread Barry Rowlingson

On Wed, Aug 10, 2011 at 2:34 PM, Duncan Murdoch
murdoch.dun...@gmail.com wrote:
 On 10/08/2011 5:58 AM, taraxacum wrote:

 Hi all,
 I need to convert a floating point value from Microsoft Basic format to
 IEEE
 format.
 There's a simple way to achieve this in R or I have to write my own
 function?
 (e.g. convert the C code below)

 You'll need to write your own function.  It can be very similar to what you
 have below, except:
  - you'll need to produce double rather than single (easy, just produce a
 single and assign to a double, C does the conversion for you)
  - you need to change the args to your function to some that R understands.

 Wouldn't it be easier, given a working C implementation, to write a
wrapper that calls the C code? R isn't built for bit-twiddling, C very
definitely is.

 Also, I imagine you may have a large number of these to change, in
which case you could do it all with a single call to a C function that
loops over the numbers from an R vector.

 See the R docs for how to load compiled C into R.

Barry

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[R] useR! 2012 Nashville TN: Your Input Needed

2011-08-10 Thread Frank Harrell


So far we have received over 70 responses to our survey.  If you have NOT
responded already and are likely to attend useR! 2012, please take the
extremely short survey today.  The link is below.

More information about Nashville may be seen at
http://biostat.mc.vanderbilt.edu/UseR-2012
In a day or two the conference web address will be
http://www.r-project.org/useR-2012/

Thanks!

--

The 2012 R User Conference - useR! 2012 - will be held in Nashville
Tennessee USA, June 12-15, 2012 on the campus of Vanderbilt University.
We would like to begin estimating the number of attendees, their area of
interest, and the number seeking hotel vs. lower-cost housing.  If you are
likely to attend useR! 2012 please go to the following link to answer two
questions: https://redcap.vanderbilt.edu/surveys/?s=2wiqWo

There are many fun things to do in Nashville -- Music City USA -- around
the time of the meeting.  Vanderbilt is 2.2 miles (3.5 km) from the center
of the action.

I hope to see many of you at useR! 2011 at the University of Warwick in
Coventry UK in just over a week.

If anyone knows of another site to which I should send this announcement
please e-mail me your suggestion.

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

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Re: [R] Floats in Microsoft Basic format


On 10/08/2011 10:16 AM, Barry Rowlingson wrote:

On Wed, Aug 10, 2011 at 2:34 PM, Duncan Murdoch
murdoch.dun...@gmail.com  wrote:
  On 10/08/2011 5:58 AM, taraxacum wrote:

  Hi all,
  I need to convert a floating point value from Microsoft Basic format to
  IEEE
  format.
  There's a simple way to achieve this in R or I have to write my own
  function?
  (e.g. convert the C code below)

  You'll need to write your own function.  It can be very similar to what you
  have below, except:
- you'll need to produce double rather than single (easy, just produce a
  single and assign to a double, C does the conversion for you)
- you need to change the args to your function to some that R understands.

  Wouldn't it be easier, given a working C implementation, to write a
wrapper that calls the C code? R isn't built for bit-twiddling, C very
definitely is.


That's actually what I was suggesting.

Duncan Murdoch

  Also, I imagine you may have a large number of these to change, in
which case you could do it all with a single call to a C function that
loops over the numbers from an R vector.

  See the R docs for how to load compiled C into R.

Barry


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[R] Reading XML files masquerading as XL files

2011-08-10 Thread Dennis Fisher

R version 2.13.1
OS X (or Windows)

Colleagues,

I received a number of files with a .xls extension.  These files open in XL 
and, by all appearances, are XL files.  However, it appears to me that the 
files are actually XML:

 readLines(dir()[16])[1:10]
 [1] ?xml version=\1.0\?  
  
 [2] Workbook xmlns=\urn:schemas-microsoft-com:office:spreadsheet\ 
  
 [3]  xmlns:o=\urn:schemas-microsoft-com:office:office\ 
  
 [4]  xmlns:x=\urn:schemas-microsoft-com:office:excel\  
  
 [5]  xmlns:ss=\urn:schemas-microsoft-com:office:spreadsheet\   
  
 [6]  xmlns:html=\http://www.w3.org/TR/REC-html40\; 
  
 [7]  DocumentProperties xmlns=\urn:schemas-microsoft-com:office:office\  
  
 [8]   Version12.0/Version
  
 [9]  /DocumentProperties   
  
[10]  OfficeDocumentSettings 
xmlns=\urn:schemas-microsoft-com:office:office\

 I had initially tried to read the files using read.xls (gdata) but that failed 
(not surprisingly).  I could open each Excel file, then save as csv, then use 
read.csv.  However, there are many files so I would love to have a solution 
that does not require this brute force approach.

Are there any packages that would allow me to read these files without the 
additional steps?

Dennis


Dennis Fisher MD
P  (The P Less Than Company)
Phone: 1-866-PLessThan (1-866-753-7784)
Fax: 1-866-PLessThan (1-866-753-7784)
www.PLessThan.com

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Re: [R] studentized and standarized residuals

2011-08-10 Thread Patrick Breheny


On 08/10/2011 10:03 AM, Jen wrote:

Hi,
I must be doing something silly here, because I can't get the studentised
and standardised residuals from r output of a linear model to agree with
what I think they should be from equation form.

x = seq(1,10)
y = x + rnorm(10)
mod = lm(y~x)
rstandard(mod)
residuals(mod)/(summary(mod)$sigma)
rstudent(mod)
residuals(mod)/(summary(mod)$sigma*sqrt(1-lm.influence(mod)$hat))


The terms studentized and standardized are sometimes used 
differently by different authors and software packages.  In R, the 
standardized residuals are based on your second calculation above. 
The studentized residuals are similar, but involve estimating sigma in 
a way that leaves out the ith data point when calculating the ith 
residual (some authors call these the studentized deleted residuals or 
the externally studentized residuals).  There is a closed form 
expression, but it is somewhat bulky.


--
Patrick Breheny
Assistant Professor
Department of Biostatistics
Department of Statistics
University of Kentucky

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Re: [R] function runif in for loop

Following the suggestion by Duncan Murdoch, this should work for you.

X - runif(length(lT), lT, uT)

Jean



From:
Johannes Radinger jradin...@gmx.at
To:
Jean V Adams jvad...@usgs.gov
Cc:
r-help@r-project.org
Date:
08/10/2011 08:40 AM
Subject:
Re: [R] function runif in for loop



Jean,

thank you for your answer.
especially the line X - numeric(length(lT)) helped me a lot.

Anyway, in my case I'd like to get a dynamic variable or better a
function for X. I mean if i try to call X I'd like that this drawing of 
random number is performed. In the case now if I call X several times I'll 
always get the same random numbers.
 
I thought about something like:
X - for (i in 1:length(lT)) runif(1, lT[i], uT[i])

So that I can use X as a variable for multiple runs and
each run new random values are used.

thank you
Johannes

 Original-Nachricht 
 Datum: Wed, 10 Aug 2011 08:19:07 -0500
 Von: Jean V Adams jvad...@usgs.gov
 An: Johannes Radinger jradin...@gmx.at
 CC: r-help@r-project.org
 Betreff: Re: [R] function runif in for loop

 Johannes,
 
 You have the loop set up right, you just need to add indexing to refer 
to 
 the looping variable, i.
 
 lT - sample(1:10)
 uT - sample(21:30)
 X - numeric(length(lT))
 
 for (i in 1:length(lT)) X[i] - runif(1, lT[i], uT[i])
 
 X
 
 Note that I changed the name of the result from T to X, because T has 
 special meaning in R.
 
 Jean
 
 `·.,,  (((º   `·.,,  (((º   `·.,,  (((º
 
 Jean V. Adams
 Statistician
 U.S. Geological Survey
 Great Lakes Science Center
 223 East Steinfest Road
 Antigo, WI 54409  USA
 
 
 
 From:
 Johannes Radinger jradin...@gmx.at
 To:
 r-help@r-project.org
 Date:
 08/10/2011 07:23 AM
 Subject:
 [R] function runif in for loop
 Sent by:
 r-help-boun...@r-project.org
 
 
 
 Hello,
 
 I'd like to perform a regression using MCMCregress (MCMCpack).
 One variable therefore should be a function rather than a variable:
 
 I want to use X as an input and X should be defined as a random number 
 between to values. Therefore I want to use the function runif like:
 X -(1, Xa, Xb) but it seems that runif doesn't allow to use vectors.
 So I think I've to calculate the new vector X by using a for loop.
 
 I tried for (i in 1:length(lT)) T-runif(1,lT,uT) but that doesn't 
work.
 What is the correct for-loop function to create this new 
vector/variable?
 
 Can I use that function then as an input for MCMCregress?
 
 thank you
 
 Johannes
 
 --
 
 __
 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.
 
 

-- 
NEU: FreePhone - 0ct/min Handyspartarif mit Geld-zurück-Garantie!   
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Re: [R] studentized and standarized residuals

2011-08-10 Thread Jen

Thanks Patrick - at least I know I wasn't being too silly :-)
Jen

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[R] join columns

2011-08-10 Thread Anthony Ching Ho Ng

Dear R-help,

I wonder if you could give me some suggestions in how to do a union
join of two data frames as follow:
- union join the common column, and insert a 0 if one is missing.

I made a function to perform the following, and I know it may not that
quite welly written, but it works.

Any suggestions are welcome, many thanks.

Anthony

 q1 = data.frame(a=1,b=2,c=3,row.names=q1)
 a b c
q1 1 2 3

 q2 = data.frame(d=4,b=1,a=4, row.names=q2)
 d b a
q2 4 1 4

-  myJoinColumns(q1,q2)
 a b c d
q1 1 2 3 0
q2 4 1 0 4


myJoinColumns - function(q1,q2){
  allNames = sort(union(colnames(q1),colnames(q2)))
  for (i in 1:length(allNames)){
t1 = which(colnames(q1) == allNames[i])
t2 = which(colnames(q2) == allNames[i])

if (length(t1) == 1){
  sec1 = q1[,t1]
} else {
  sec1 = 0
}

if (length(t2) == 1){
  sec2 = q2[,t2]
} else {
  sec2 = 0
}

if (i == 1){
  qTable = matrix(c(sec1,sec2))
}else{
  qTable = cbind(qTable,c(sec1,sec2))
}
  }
  colnames(qTable) = allNames
  rownames(qTable) = c(q1,q2)
  qTable
}

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Re: [R] join columns



On Aug 10, 2011, at 11:04 AM, Anthony Ching Ho Ng wrote:


Dear R-help,

I wonder if you could give me some suggestions in how to do a union
join of two data frames as follow:
- union join the common column, and insert a 0 if one is missing.

I made a function to perform the following, and I know it may not that
quite welly written, but it works.

Any suggestions are welcome, many thanks.

Anthony


q1 = data.frame(a=1,b=2,c=3,row.names=q1)

a b c
q1 1 2 3


q2 = data.frame(d=4,b=1,a=4, row.names=q2)

d b a
q2 4 1 4

-  myJoinColumns(q1,q2)
a b c d
q1 1 2 3 0
q2 4 1 0 4


 temp - merge(q1,q2, all=TRUE)
 temp[is.na(temp)] - 0
 temp
  a b c d
1 1 2 3 0
2 4 1 0 4



myJoinColumns - function(q1,q2){
 allNames = sort(union(colnames(q1),colnames(q2)))
 for (i in 1:length(allNames)){
   t1 = which(colnames(q1) == allNames[i])
   t2 = which(colnames(q2) == allNames[i])

   if (length(t1) == 1){
 sec1 = q1[,t1]
   } else {
 sec1 = 0
   }

   if (length(t2) == 1){
 sec2 = q2[,t2]
   } else {
 sec2 = 0
   }

   if (i == 1){
 qTable = matrix(c(sec1,sec2))
   }else{
 qTable = cbind(qTable,c(sec1,sec2))
   }
 }
 colnames(qTable) = allNames
 rownames(qTable) = c(q1,q2)
 qTable
}

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Re: [R] Using ANCOVA in R

2011-08-10 Thread Richard M. Heiberger

January,

It looks like you will need an interaction effect, perhaps
   g - lm( response ~ subject + group*time)
Please see the ancova function in the HH package.
install.packages(HH) ## if necessary
library(HH)
?ancova

Rich




On Wed, Aug 10, 2011 at 5:15 AM, January Weiner january.wei...@gmail.comwrote:

 Hello,

 I have a problem with using the following design with ANCOVA in R.

 There are two groups (control + treatment), each with ten subjects.
 The subjects show a response that is monitored over time (four time
 points). For a single given subject, the response can be analysed with
 linear regression with time as the independent variable.

 The question is, how does the response differ between the two groups.
 It is to be expected that the slope of the response differs between
 the two groups, while the intercept itself is of minor interest.

 I have tried a linear model as in

 g - lm( response ~ time + subject + group)

 but somehow I think that this is not correct, as it would not show a
 difference in the slope of the response.

 Any kind of help would be greatly appreciated.

 --
 January

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Re: [R] round() a data frame containing 'character' variables?

Hello

On Wed, Aug 10, 2011 at 2:31 PM, Jean V Adams jvad...@usgs.gov wrote:
 The function format() might serve your needs.

This looks very promising, but yields some strange results. See below:

 x - data.frame(a=rnorm(10), b=rnorm(10), c=rnorm(10), d=letters[1:10])
 x
 a   bc d
1   0.54114449 -0.11195580  1.526279364 a
2   3.27109063  0.50848249 -0.215760332 b
3  -0.27064475 -1.04749725  0.082319811 c
4  -0.06638611 -0.58600572  0.004148253 d
5  -0.06170739 -0.37885203  0.689125494 e
6   0.53211363 -0.09150913 -0.463972307 f
7  -0.43314431 -0.28981614 -0.973410994 g
8   0.52137857 -1.15077343  0.163120205 h
9  -1.39581552  1.27378389  0.136708313 i
10  0.06348058 -0.00369746 -0.570214119 j
 format(x, digits=2)  ##it displays 3 or 4 digits, instead of the required 2
a   b   c d
1   0.541 -0.1120  1.5263 a
2   3.271  0.5085 -0.2158 b
3  -0.271 -1.0475  0.0823 c
4  -0.066 -0.5860  0.0041 d
5  -0.062 -0.3789  0.6891 e
6   0.532 -0.0915 -0.4640 f
7  -0.433 -0.2898 -0.9734 g
8   0.521 -1.1508  0.1631 h
9  -1.396  1.2738  0.1367 i
10  0.063 -0.0037 -0.5702 j
 format(x, digits=2, nsmall=1, scientific=7)  ##no change when setting related 
 arguments
a   b   c d
1   0.541 -0.1120  1.5263 a
2   3.271  0.5085 -0.2158 b
3  -0.271 -1.0475  0.0823 c
4  -0.066 -0.5860  0.0041 d
5  -0.062 -0.3789  0.6891 e
6   0.532 -0.0915 -0.4640 f
7  -0.433 -0.2898 -0.9734 g
8   0.521 -1.1508  0.1631 h
9  -1.396  1.2738  0.1367 i
10  0.063 -0.0037 -0.5702 j
 round(x[1:3], digits=2)  ##works as expected
   a b c
1   0.54 -0.11  1.53
2   3.27  0.51 -0.22
3  -0.27 -1.05  0.08
4  -0.07 -0.59  0.00
5  -0.06 -0.38  0.69
6   0.53 -0.09 -0.46
7  -0.43 -0.29 -0.97
8   0.52 -1.15  0.16
9  -1.40  1.27  0.14
10  0.06  0.00 -0.57

Any ideas why format() and round() give so different results? Can
format() be set to behave similarly to round? Regards
Liviu

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[R] [R-pkgs] New Matrix and lme4: Must reinstall lme4 if got new Matrix

2011-08-10 Thread Martin Maechler

We have released to CRAN a new version of the (recommended)
package Matrix, and of package lme4 yesterday.

Anyone who gets the new version of Matrix *MUST* re-install lme4
-- if (s)he is using lme4 at all.  Technical details about that further below.

The fact that yesterday's version number of Matrix is 0.9995875-2, 
indicates that Matrix' version is indeed approaching 1.0 (*),
and I'd declare this version as  release candidate for 1.0.
As a recommended package, Matrix is part of every R distribution,
our aim is to release Matrix_1.0-0 (or higher) with the next
non-patch release of R, i.e., R-2.14.0 somewhere in October.

For this reason, we are asking R useRs, programmeRs and
provideRs, to ``hash at'' the package, trying to find problems /
bugs, badly lacking (or wrong) documentation, etc,
and report it to us (to 'Matrix-authors@...' or possibly R-devel@...),
so we can prepare a  shiny sparkling  Matrix_1.0-0 in time.
Thank you in advance for such a contribution to the Free
Software universe.

Martin Maechler and Doug Bates
(and Ben Bolker for lme4).

-- -- --

- Why must lme4 be re-installed as well ?  [Technical !]

 This is because lme4 has a 'LinkingTo: Matrix' in its
 DESCRIPTION and indeed, lme4's C code is using part of Matrix' C
 code.
 The change in Matrix: Part of the C-level interface in Matrix is
 now (again, after several years) using the standard CHOLMOD
 typedef of UFlong.  This means that the C API of Matrix should
 behave conformly with what other instances of CHOLMOD export.

(*) For some, it may be amusing to read
https://stat.ethz.ch/pipermail/r-packages/2008/000911.html
which was the last announcement of Matrix on R-packages.

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Re: [R] studentized and standarized residuals

2011-08-10 Thread Dénes TÓTH


Dear Jen,

Actually you can check out what R does by looking at the source.

# first type the name of the function
 rstandard
function (model, ...)
UseMethod(rstandard)
environment: namespace:stats

# ?methods will list you the corresponding functions
 methods(rstandard)
[1] rstandard.glm rstandard.lm

# choose rstandard.lm
 rstandard.lm
function (model, infl = lm.influence(model, do.coef = FALSE),
sd = sqrt(deviance(model)/df.residual(model)), ...)
{
res - infl$wt.res/(sd * sqrt(1 - infl$hat))
res[is.infinite(res)] - NaN
res
}

# in case the function is not visible,
# you can use package-name:::function-name to display it
stats:::rstandard.lm


Best,
  Denes






 Thanks Patrick - at least I know I wasn't being too silly :-)
 Jen

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 http://r.789695.n4.nabble.com/studentized-and-standarized-residuals-tp3732997p3733173.html
 Sent from the R help mailing list archive at Nabble.com.

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Re: [R] join columns

2011-08-10 Thread Henrique Dallazuanna

Try this:

merge(q1, q2, all = TRUE)

On Wed, Aug 10, 2011 at 12:04 PM, Anthony Ching Ho Ng
anthony.ch...@gmail.com wrote:
 Dear R-help,

 I wonder if you could give me some suggestions in how to do a union
 join of two data frames as follow:
 - union join the common column, and insert a 0 if one is missing.

 I made a function to perform the following, and I know it may not that
 quite welly written, but it works.

 Any suggestions are welcome, many thanks.

 Anthony

 q1 = data.frame(a=1,b=2,c=3,row.names=q1)
     a b c
 q1 1 2 3

 q2 = data.frame(d=4,b=1,a=4, row.names=q2)
     d b a
 q2 4 1 4

 -  myJoinColumns(q1,q2)
     a b c d
 q1 1 2 3 0
 q2 4 1 0 4


 myJoinColumns - function(q1,q2){
  allNames = sort(union(colnames(q1),colnames(q2)))
  for (i in 1:length(allNames)){
    t1 = which(colnames(q1) == allNames[i])
    t2 = which(colnames(q2) == allNames[i])

    if (length(t1) == 1){
      sec1 = q1[,t1]
    } else {
      sec1 = 0
    }

    if (length(t2) == 1){
      sec2 = q2[,t2]
    } else {
      sec2 = 0
    }

    if (i == 1){
      qTable = matrix(c(sec1,sec2))
    }else{
      qTable = cbind(qTable,c(sec1,sec2))
    }
  }
  colnames(qTable) = allNames
  rownames(qTable) = c(q1,q2)
  qTable
 }

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

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Re: [R] round() a data frame containing 'character' variables?

As it says in

?format

the digits argument specifies ... how many significant digits are to be 
used ... enough decimal places will be used so that the smallest (in 
magnitude) number has this many significant digits ...

In your example, the last value in column a is 0.06348058 which is 
written as 0.063 to two significant digits.

There is no way (that I know of) to make the format() function do the same 
sort of thing as round().

If digits won't meet your needs, you could try something like this

data.frame(lapply(x, function(y) if(is.numeric(y)) round(y, 2) else y))

Jean

`·.,,  (((º   `·.,,  (((º   `·.,,  (((º

Jean V. Adams
Statistician
U.S. Geological Survey
Great Lakes Science Center
223 East Steinfest Road
Antigo, WI 54409  USA

Liviu Andronic landronim...@gmail.com wrote on 08/10/2011 10:26:43 AM:

 [image removed] 
 
 Re: [R] round() a data frame containing 'character' variables?
 
 Liviu Andronic 
 
 to:
 
 Jean V Adams
 
 08/10/2011 10:27 AM
 
 Cc:
 
 r-help@r-project.org Help
 
 Hello
 
 On Wed, Aug 10, 2011 at 2:31 PM, Jean V Adams jvad...@usgs.gov wrote:
  The function format() might serve your needs.
 
 This looks very promising, but yields some strange results. See below:
 
  x - data.frame(a=rnorm(10), b=rnorm(10), c=rnorm(10), 
d=letters[1:10])
  x
  a   bc d
 1   0.54114449 -0.11195580  1.526279364 a
 2   3.27109063  0.50848249 -0.215760332 b
 3  -0.27064475 -1.04749725  0.082319811 c
 4  -0.06638611 -0.58600572  0.004148253 d
 5  -0.06170739 -0.37885203  0.689125494 e
 6   0.53211363 -0.09150913 -0.463972307 f
 7  -0.43314431 -0.28981614 -0.973410994 g
 8   0.52137857 -1.15077343  0.163120205 h
 9  -1.39581552  1.27378389  0.136708313 i
 10  0.06348058 -0.00369746 -0.570214119 j
  format(x, digits=2)  ##it displays 3 or 4 digits, instead of the 
required 2
 a   b   c d
 1   0.541 -0.1120  1.5263 a
 2   3.271  0.5085 -0.2158 b
 3  -0.271 -1.0475  0.0823 c
 4  -0.066 -0.5860  0.0041 d
 5  -0.062 -0.3789  0.6891 e
 6   0.532 -0.0915 -0.4640 f
 7  -0.433 -0.2898 -0.9734 g
 8   0.521 -1.1508  0.1631 h
 9  -1.396  1.2738  0.1367 i
 10  0.063 -0.0037 -0.5702 j
  format(x, digits=2, nsmall=1, scientific=7)  ##no change when 
 setting related arguments
 a   b   c d
 1   0.541 -0.1120  1.5263 a
 2   3.271  0.5085 -0.2158 b
 3  -0.271 -1.0475  0.0823 c
 4  -0.066 -0.5860  0.0041 d
 5  -0.062 -0.3789  0.6891 e
 6   0.532 -0.0915 -0.4640 f
 7  -0.433 -0.2898 -0.9734 g
 8   0.521 -1.1508  0.1631 h
 9  -1.396  1.2738  0.1367 i
 10  0.063 -0.0037 -0.5702 j
  round(x[1:3], digits=2)  ##works as expected
a b c
 1   0.54 -0.11  1.53
 2   3.27  0.51 -0.22
 3  -0.27 -1.05  0.08
 4  -0.07 -0.59  0.00
 5  -0.06 -0.38  0.69
 6   0.53 -0.09 -0.46
 7  -0.43 -0.29 -0.97
 8   0.52 -1.15  0.16
 9  -1.40  1.27  0.14
 10  0.06  0.00 -0.57
 
 Any ideas why format() and round() give so different results? Can
 format() be set to behave similarly to round? Regards
 Liviu

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[R] RExcel

2011-08-10 Thread Dr. Alireza Zolfaghari

Hi list,
I used to work with RExcel in excel 2003. Now in 2007, I tried the same
RExcel, but it did not work. I got R version 12. I downloaded/installed the
latest version of RExcel 3.2.0 from http://sunsite.univie.ac.at/rcom/. It
has added the RExcel add-ins, but when I click on starting R in add-ins, I
get the following sequentional errors:

Errors:
SCtools not available
SCTools can not be loaded.
could not start Rserver
There seems to be no R proecess conneceted to Excel


I used to install Rsrv200.exe, and do not know if I still need to install it
in this version.

Any help, please,


thanks,
Alireza

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[R] spfeml error message

2011-08-10 Thread Darshana Udayanganie

Hello


I am trying to learn the spatial panel data analysis (newbie). I have the R
version 2.13.1 and I did download the spml package required for the spatial
panel data analysis. However, when I tried the analysis, I get the following
error message. could not find function spfeml.  Can somebody help me
with this.

Thanks in advance.


-Dar

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Re: [R] How to quickly convert a data.frame into a structure of lists

2011-08-10 Thread Frederic F

Hello Duncan,  

Here is a small example to illustrate what I am trying to do.

# Example data.frame
df=data.frame(A=c(a,a,b,b), B=c(X,X,Y,Z), C=c(1,2,3,4)) 
#   A B C
# 1 a X 1
# 2 a X 2
# 3 b Y 3
# 4 b Z 4

### First way of getting the list structure (ls1) using imbricated lapply
loops:
# Get the structure and populate it:
ls1-lapply(levels(df$A), function(levelA) { 
  lapply(levels(df$B), function(levelB) {df$C[df$A==levelA 
df$B==levelB]})
})
# Apply the names:
names(list_structure)-levels(df$A)
for (i in 1:length(list_structure))
{names(list_structure[[i]])-levels(df$B)}

# Result:
ls1$a$X
# [1] 1 2
ls1$b$Z
# [1] 4

The data.frame will always be 'complete', i.e., there will be a value in
every row for every column. 
I want to produce a structure like this one quickly (I aim at something
below 10 seconds) for a dataset containing between 1 and 2 millions of rows. 

I hope that this helps clarify things.

Thanks for your help,

Frederic 

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Re: [R] instal tar.gz package on windows

2011-08-10 Thread Lippel, Anna

Hi Duncan,

I have tried to install a tar.gz package following your instructions 
(https://stat.ethz.ch/pipermail/r-help/2008-August/169599.html) but without 
success. Here are the steps I followed:

I installed the last version of Rtools and ran Rcmd INSTALL rJava_0.8-8.tar.gz 
and got the error message attach in errorLog.txt. It seems that there is a 
problem with JRI but I´m really stuck. Do you have any idea of what it could be?

 

Thanks!

 


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P:\DesktopRcmd INSTALL rJava_0.8-8.tar.gz
* installing to library 'C:/Program Files/R/R-2.13.1/library'
* installing *source* package 'rJava' ...
Generate Windows-specific files (src/jvm-w32) ...
cygwin warning:
  MS-DOS style path detected: C:/PROGRA~1/R/R-213~1.1/etc/i386/Makeconf
  Preferred POSIX equivalent is: /cygdrive/c/PROGRA~1/R/R-213~1.1/etc/i386/Makec
onf
  CYGWIN environment variable option nodosfilewarning turns off this warning.
  Consult the user's guide for more details about POSIX paths:
http://cygwin.com/cygwin-ug-net/using.html#using-pathnames
make: Entering directory `/cygdrive/c/DOCUME~1/lippelan/LOCALS~1/Temp/RtmpHz0yZg
/R.INSTALL6d282489/rJava/src/jvm-w32'
dlltool --input-def jvm.def --kill-at --dllname jvm.dll --output-lib libjvm.dll.
a
gcc -O2 -c -o findjava.o findjava.c
gcc -s -o findjava.exe findjava.o
make: Leaving directory `/cygdrive/c/DOCUME~1/lippelan/LOCALS~1/Temp/RtmpHz0yZg/
R.INSTALL6d282489/rJava/src/jvm-w32'
Find Java...
  JAVA_HOME=C:/PROGRA~1/Java/JRE16~1.0_1
=== Building JRI ===
  JAVA_HOME=C:/PROGRA~1/Java/JRE16~1.0_1
  R_HOME=C:/PROGRA~1/R/R-213~1.1
Creating Makefiles ...
Configuration done.
make -C src JRI.jar
make[1]: Entering directory `/cygdrive/c/DOCUME~1/lippelan/LOCALS~1/Temp/RtmpHz0
yZg/R.INSTALL6d282489/rJava/jri/src'
C:/PROGRA~1/Java/JRE16~1.0_1/bin/javac -target 1.4 -source 1.4 -d . ../Mutex.jav
a ../RBool.java ../RConsoleOutputStream.java ../REXP.java ../RFactor.java ../RLi
st.java ../RMainLoopCallbacks.java ../RVector.java ../Rengine.java ../package-in
fo.java
make[1]: C:/PROGRA~1/Java/JRE16~1.0_1/bin/javac: Command not found
make[1]: *** [org/rosuda/JRI/Rengine.class] Error 127
make[1]: Leaving directory `/cygdrive/c/DOCUME~1/lippelan/LOCALS~1/Temp/RtmpHz0y
Zg/R.INSTALL6d282489/rJava/jri/src'
make: *** [src/JRI.jar] Error 2
 WARNING: JRI could NOT be built
Set IGNORE=1 if you want to build rJava anyway.
ERROR: configuration failed for package 'rJava'
* removing 'C:/Program Files/R/R-2.13.1/library/rJava'
* restoring previous 'C:/Program Files/R/R-2.13.1/library/rJava'
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Re: [R] function runif in for loop

2011-08-10 Thread Johannes Radinger

Jean,

thank you for your answer.
especially the line X - numeric(length(lT)) helped me a lot.

Anyway, in my case I'd like to get a dynamic variable or better a
function for X. I mean if i try to call X I'd like that this drawing of random 
number is performed. In the case now if I call X several times I'll always get 
the same random numbers.
 
I thought about something like:
X - for (i in 1:length(lT)) runif(1, lT[i], uT[i])

So that I can use X as a variable for multiple runs and
each run new random values are used.

thank you
Johannes

 Original-Nachricht 
 Datum: Wed, 10 Aug 2011 08:19:07 -0500
 Von: Jean V Adams jvad...@usgs.gov
 An: Johannes Radinger jradin...@gmx.at
 CC: r-help@r-project.org
 Betreff: Re: [R] function runif in for loop

 Johannes,
 
 You have the loop set up right, you just need to add indexing to refer to 
 the looping variable, i.
 
 lT - sample(1:10)
 uT - sample(21:30)
 X - numeric(length(lT))
 
 for (i in 1:length(lT)) X[i] - runif(1, lT[i], uT[i])
 
 X
 
 Note that I changed the name of the result from T to X, because T has 
 special meaning in R.
 
 Jean
 
 `·.,,  (((º   `·.,,  (((º   `·.,,  (((º
 
 Jean V. Adams
 Statistician
 U.S. Geological Survey
 Great Lakes Science Center
 223 East Steinfest Road
 Antigo, WI 54409  USA
 
 
 
 From:
 Johannes Radinger jradin...@gmx.at
 To:
 r-help@r-project.org
 Date:
 08/10/2011 07:23 AM
 Subject:
 [R] function runif in for loop
 Sent by:
 r-help-boun...@r-project.org
 
 
 
 Hello,
 
 I'd like to perform a regression using MCMCregress (MCMCpack).
 One variable therefore should be a function rather than a variable:
 
 I want to use X as an input and X should be defined as a random number 
 between to values. Therefore I want to use the function runif like:
 X -(1, Xa, Xb) but it seems that runif doesn't allow to use vectors.
 So I think I've to calculate the new vector X by using a for loop.
 
 I tried for (i in 1:length(lT)) T-runif(1,lT,uT) but that doesn't work.
 What is the correct for-loop function to create this new vector/variable?
 
 Can I use that function then as an input for MCMCregress?
 
 thank you
 
 Johannes
 
 --
 
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Re: [R] function runif in for loop

2011-08-10 Thread Johannes Radinger



 Original-Nachricht 
 Datum: Wed, 10 Aug 2011 09:38:38 -0400
 Von: Duncan Murdoch murdoch.dun...@gmail.com
 An: Johannes Radinger jradin...@gmx.at
 CC: r-help@r-project.org
 Betreff: Re: [R] function runif in for loop

 On 10/08/2011 7:28 AM, Johannes Radinger wrote:
  Hello,
 
  I'd like to perform a regression using MCMCregress (MCMCpack).
  One variable therefore should be a function rather than a variable:
 
  I want to use X as an input and X should be defined as a random number
 between to values. Therefore I want to use the function runif like:
  X-(1, Xa, Xb) but it seems that runif doesn't allow to use vectors.
  So I think I've to calculate the new vector X by using a for loop.
 
 runif() does allow vectors.  Assuming Xa and Xb are vectors of length n, 
 then
 
 X - runif(n, Xa, Xb)

Thank you very much, I just got confused with the n, as I wanted always only 
one value between Xa and Xb.

So far my solution based on your suggestion is:
X - runif(length(Xa), Xa, Xb)

/johannes


 
 will work.  (Xa and Xb don't both have to be vectors; values will be 
 recycled as necessary.)
 
 Duncan Murdoch
  I tried for (i in 1:length(lT)) T-runif(1,lT,uT) but that doesn't
 work.
  What is the correct for-loop function to create this new
 vector/variable?
 
  Can I use that function then as an input for MCMCregress?
 
  thank you
 
  Johannes
 
  --
 
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Re: [R] How to quickly convert a data.frame into a structure of lists

2011-08-10 Thread Frederic F

Hello Denis,

 To borrow shamelessly from one of the prominent helpers on this list:
 What is the problem you're trying to solve? Â  Â (attribution: Jim Holtman)

I'm trying to connect two sets of legacy R tools: the output of the
first one can be transformed in a data.frame without loss of
information, the input of the second one takes the form of a structure
of list.

  it's entirely possible
 that there may be a nice 'R way' to do it. Read the posting guide and
 if at all possible, provide a small, reproducible example that
 demonstrates what you want to accomplish.

Here is the first way attacked the problems illustrated on a tiny
dataset (this way does not work quickly enough on a real dataset
unfortunately):

df=data.frame(A=c(a,a,b,b), B=c(X,X,Y,Z), C=c(1,2,3,4))

# Get the structure and populate it:
ls1-lapply(levels(df$A), function(levelA) {
  lapply(levels(df$B), function(levelB) {df$C[df$A==levelA  df$B==levelB]})
})
# Get the names:
names(list_structure)-levels(df$A)
for (i in 1:length(list_structure)) {names(list_structure[[i]])-levels(df$B)}

# Results:
ls1$a$X
# [1] 1 2
ls1$b$Z
# [1] 4

Thanks for your help,

Frederic


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Re: [R] scan(file, encoding=?)

2011-08-10 Thread tibi3000

Hi,

this command gives you all possible encoding options on your platform:
iconvlist()


hope it answers your question,

T

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Re: [R] RExcel

Don't they have their own support mailing list? You should review  
their documentation for specifics.


--
David.
On Aug 10, 2011, at 12:24 PM, Dr. Alireza Zolfaghari wrote:


Hi list,
I used to work with RExcel in excel 2003. Now in 2007, I tried the  
same
RExcel, but it did not work. I got R version 12. I downloaded/ 
installed the
latest version of RExcel 3.2.0 from http://sunsite.univie.ac.at/ 
rcom/. It
has added the RExcel add-ins, but when I click on starting R in add- 
ins, I

get the following sequentional errors:

Errors:
SCtools not available
SCTools can not be loaded.
could not start Rserver
There seems to be no R proecess conneceted to Excel

I used to install Rsrv200.exe, and do not know if I still need to  
install it

in this version.

Any help, please,


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Re: [R] How to quickly convert a data.frame into a structure of lists


On 10/08/2011 10:30 AM, Frederic F wrote:

Hello Duncan,

Here is a small example to illustrate what I am trying to do.

# Example data.frame
df=data.frame(A=c(a,a,b,b), B=c(X,X,Y,Z), C=c(1,2,3,4))
#   A B C
# 1 a X 1
# 2 a X 2
# 3 b Y 3
# 4 b Z 4

### First way of getting the list structure (ls1) using imbricated lapply
loops:
# Get the structure and populate it:
ls1-lapply(levels(df$A), function(levelA) {
   lapply(levels(df$B), function(levelB) {df$C[df$A==levelA
df$B==levelB]})
})
# Apply the names:
names(list_structure)-levels(df$A)
for (i in 1:length(list_structure))
{names(list_structure[[i]])-levels(df$B)}

# Result:
ls1$a$X
# [1] 1 2
ls1$b$Z
# [1] 4

The data.frame will always be 'complete', i.e., there will be a value in
every row for every column.
I want to produce a structure like this one quickly (I aim at something
below 10 seconds) for a dataset containing between 1 and 2 millions of rows.



I don't know what the timing would be like for your real data, but this 
does look like by() would work:


ls1 - by(df$C, df[,1:2], identity)

When I repeat the rows of df a million times each, this finishes in a 
few seconds.  It would definitely be slower if there were more levels of 
A or B.


Now ls1 will be a matrix whose entries are the subsets of C that you 
want, so you can see your two results with slightly different syntax:


 ls1[[a, X]]
[1] 1 2
 ls1[[b,Z]]
[1] 4

Duncan Murdoch

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Re: [R] instal tar.gz package on windows


On 10/08/2011 11:59 AM, Lippel, Anna wrote:

Hi Duncan,

I have tried to install a tar.gz package following your instructions 
(https://stat.ethz.ch/pipermail/r-help/2008-August/169599.html) but without 
success. Here are the steps I followed:

I installed the last version of Rtools and ran Rcmd INSTALL rJava_0.8-8.tar.gz 
and got the error message attach in errorLog.txt. It seems that there is a 
problem with JRI but I´m really stuck. Do you have any idea of what it could be?



The important error message is:

make[1]: C:/PROGRA~1/Java/JRE16~1.0_1/bin/javac: Command not found
make[1]: *** [org/rosuda/JRI/Rengine.class] Error 127

So you don't have the Java compiler, or don't have it where that package 
was looking for it.  But I don't know why you wouldn't just install the 
binary version; why do you want to compile it yourself?


Duncan Murdoch

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Re: [R] function runif in for loop


On 10/08/2011 9:40 AM, Johannes Radinger wrote:

Jean,

thank you for your answer.
especially the line X- numeric(length(lT)) helped me a lot.

Anyway, in my case I'd like to get a dynamic variable or better a
function for X. I mean if i try to call X I'd like that this drawing of random 
number is performed. In the case now if I call X several times I'll always get 
the same random numbers.


Such things do exist in R, but they aren't easy to set up.  Why not just 
make X be a function explicitly?  That is,


X - function() runif(length(lT), lT, uT)

Then use X() to call the function where you were previously using X.

Duncan Murdoch


I thought about something like:
X- for (i in 1:length(lT)) runif(1, lT[i], uT[i])

So that I can use X as a variable for multiple runs and
each run new random values are used.

thank you
Johannes

 Original-Nachricht 
  Datum: Wed, 10 Aug 2011 08:19:07 -0500
  Von: Jean V Adamsjvad...@usgs.gov
  An: Johannes Radingerjradin...@gmx.at
  CC: r-help@r-project.org
  Betreff: Re: [R] function runif in for loop

  Johannes,

  You have the loop set up right, you just need to add indexing to refer to
  the looping variable, i.

  lT- sample(1:10)
  uT- sample(21:30)
  X- numeric(length(lT))

  for (i in 1:length(lT)) X[i]- runif(1, lT[i], uT[i])

  X

  Note that I changed the name of the result from T to X, because T has
  special meaning in R.

  Jean

  `·.,,(((º`·.,,(((º`·.,,(((º

  Jean V. Adams
  Statistician
  U.S. Geological Survey
  Great Lakes Science Center
  223 East Steinfest Road
  Antigo, WI 54409  USA



  From:
  Johannes Radingerjradin...@gmx.at
  To:
  r-help@r-project.org
  Date:
  08/10/2011 07:23 AM
  Subject:
  [R] function runif in for loop
  Sent by:
  r-help-boun...@r-project.org



  Hello,

  I'd like to perform a regression using MCMCregress (MCMCpack).
  One variable therefore should be a function rather than a variable:

  I want to use X as an input and X should be defined as a random number
  between to values. Therefore I want to use the function runif like:
  X-(1, Xa, Xb) but it seems that runif doesn't allow to use vectors.
  So I think I've to calculate the new vector X by using a for loop.

  I tried for (i in 1:length(lT)) T-runif(1,lT,uT) but that doesn't work.
  What is the correct for-loop function to create this new vector/variable?

  Can I use that function then as an input for MCMCregress?

  thank you

  Johannes

  --

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



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Re: [R] Can R handle a matrix with 8 billion entries?

2011-08-10 Thread Corey Dow-Hygelund

You might want to look into the packages bigmemory and biganalytics.

Corey

On Tue, Aug 9, 2011 at 8:38 PM, Chris Howden
ch...@trickysolutions.com.auwrote:

 Hi,

 Im trying to do a hierarchical cluster analysis in R with a Big Data set.
 Im running into problems using the dist() function.

 Ive been looking at a few threads about Rs memory and have read the
 memory limits section in R help. However Im no computer expert so Im
 hoping Ive misunderstood something and R can handle my Big Data set,
 somehow. Although at the moment I think my dataset is simply too big and
 there is no way around it, but Id like to be proved wrong!

 My data set has 90523 rows of data and 24 columns.

 My understanding is that this means the distance matrix has a min of
 90523^2 elements which is 8194413529. Which roughly translates as 8GB of
 memory being required (if I assume each entry requires 1 bit). I only have
 4GB on a 32bit build of windows and R. So there is no way thats going to
 work.

 So then I thought of getting access to a more powerful computer, and maybe
 using cloud computing.

 However the R memory limit help mentions  On all builds of R, the maximum
 length (number of elements) of a vector is 2^31 - 1 ~ 2*10^9. Now as the
 distance matrix I require has more elements than this does this mean its
 too big for R no matter what I do?

 Any ideas would be welcome.

 Thanks.


 Chris Howden
 Founding Partner
 Tricky Solutions
 Tricky Solutions 4 Tricky Problems
 Evidence Based Strategic Development, IP Commercialisation and Innovation,
 Data Analysis, Modelling and Training
 (mobile) 0410 689 945
 (fax / office)
 ch...@trickysolutions.com.au

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Re: [R] How to quickly convert a data.frame into a structure of lists

2011-08-10 Thread Eik Vettorazzi

Hi Frederic,
shouldn't there be an result for the 3rd row as well, eg ls1$b$Y?

Maybe this will do what you want?

dtf-within(dtf,index-factor(A:B))
tapply(dtf$C,dtf$index,list)

Hth.

Am 10.08.2011 16:30, schrieb Frederic F:
 Hello Duncan,  
 
 Here is a small example to illustrate what I am trying to do.
 
 # Example data.frame
 df=data.frame(A=c(a,a,b,b), B=c(X,X,Y,Z), C=c(1,2,3,4)) 
 #   A B C
 # 1 a X 1
 # 2 a X 2
 # 3 b Y 3
 # 4 b Z 4
 
 ### First way of getting the list structure (ls1) using imbricated lapply
 loops:
 # Get the structure and populate it:
 ls1-lapply(levels(df$A), function(levelA) { 
   lapply(levels(df$B), function(levelB) {df$C[df$A==levelA 
 df$B==levelB]})
 })
 # Apply the names:
 names(list_structure)-levels(df$A)
 for (i in 1:length(list_structure))
 {names(list_structure[[i]])-levels(df$B)}
 
 # Result:
 ls1$a$X
 # [1] 1 2
 ls1$b$Z
 # [1] 4
 
 The data.frame will always be 'complete', i.e., there will be a value in
 every row for every column. 
 I want to produce a structure like this one quickly (I aim at something
 below 10 seconds) for a dataset containing between 1 and 2 millions of rows. 
 
 I hope that this helps clarify things.
 
 Thanks for your help,
 
 Frederic 
 
 --
 View this message in context: 
 http://r.789695.n4.nabble.com/How-to-quickly-convert-a-data-frame-into-a-structure-of-lists-tp3731746p3733073.html
 Sent from the R help mailing list archive at Nabble.com.
 
 __
 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.

-- 
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Institut für Medizinische Biometrie und Epidemiologie
Universitätsklinikum Hamburg-Eppendorf

Martinistr. 52
20246 Hamburg

T ++49/40/7410-58243
F ++49/40/7410-57790

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Re: [R] How to quickly convert a data.frame into a structure of lists

2011-08-10 Thread William Dunlap

I was going to suggest
   AB - df[c(A,B)]
   ls2 - array(split(df$C, AB), dim=sapply(AB, nlevels), dimnames=sapply(AB, 
levels))
which produces a matrix very similar to what Duncan's by() call produces
   ls1 - by(df$C, df[,1:2], identity)
E.g.,
   ls2[[a,X]]
  [1] 1 2
   ls1[[a,X]]
  [1] 1 2
   ls1[[a,Y]] # by assigns NULL to unoccupied slots
  NULL
   ls2[[a,Y]] # split gives the same type to all slots, copied from input
  numeric(0)

They both are quick because they use split() to avoid the repeated
evaluations of
  bigVector[ anotherBigVector == scalar ]
that your nested (not imbricated) loops do.  If you really need to convert
the matrix to a list of lists that will probably be a quick transformation.

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 Duncan Murdoch
 Sent: Wednesday, August 10, 2011 9:43 AM
 To: Frederic F
 Cc: r-help@r-project.org
 Subject: Re: [R] How to quickly convert a data.frame into a structure of lists
 
 On 10/08/2011 10:30 AM, Frederic F wrote:
  Hello Duncan,
 
  Here is a small example to illustrate what I am trying to do.
 
  # Example data.frame
  df=data.frame(A=c(a,a,b,b), B=c(X,X,Y,Z), C=c(1,2,3,4))
  #   A B C
  # 1 a X 1
  # 2 a X 2
  # 3 b Y 3
  # 4 b Z 4
 
  ### First way of getting the list structure (ls1) using imbricated lapply
  loops:
  # Get the structure and populate it:
  ls1-lapply(levels(df$A), function(levelA) {
 lapply(levels(df$B), function(levelB) {df$C[df$A==levelA
  df$B==levelB]})
  })
  # Apply the names:
  names(list_structure)-levels(df$A)
  for (i in 1:length(list_structure))
  {names(list_structure[[i]])-levels(df$B)}
 
  # Result:
  ls1$a$X
  # [1] 1 2
  ls1$b$Z
  # [1] 4
 
  The data.frame will always be 'complete', i.e., there will be a value in
  every row for every column.
  I want to produce a structure like this one quickly (I aim at something
  below 10 seconds) for a dataset containing between 1 and 2 millions of rows.
 
 
 I don't know what the timing would be like for your real data, but this
 does look like by() would work:
 
 ls1 - by(df$C, df[,1:2], identity)
 
 When I repeat the rows of df a million times each, this finishes in a
 few seconds.  It would definitely be slower if there were more levels of
 A or B.
 
 Now ls1 will be a matrix whose entries are the subsets of C that you
 want, so you can see your two results with slightly different syntax:
 
   ls1[[a, X]]
 [1] 1 2
   ls1[[b,Z]]
 [1] 4
 
 Duncan Murdoch
 
 __
 R-help@r-project.org mailing list
 https://stat.ethz.ch/mailman/listinfo/r-help
 PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
 and provide commented, minimal, self-contained, reproducible code.

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Re: [R] function runif in for loop

2011-08-10 Thread Bert Gunter

Duncan et. al:

Inline below.

On Wed, Aug 10, 2011 at 9:48 AM, Duncan Murdoch
murdoch.dun...@gmail.com wrote:
 On 10/08/2011 9:40 AM, Johannes Radinger wrote:

 Jean,

 thank you for your answer.
 especially the line X- numeric(length(lT)) helped me a lot.

 Anyway, in my case I'd like to get a dynamic variable or better a
 function for X. I mean if i try to call X I'd like that this drawing of
 random number is performed. In the case now if I call X several times I'll
 always get the same random numbers.

 Such things do exist in R, but they aren't easy to set up.

Well, how about:

Well... How about:

X - function()runif(1)
class(X) - c(wizz,class(X))
print.wizz - function(x){y -x(); print(y);y }

 X
[1] 0.875768
 X
[1] 0.955208
 X
[1] 0.1150938
 z - X
 z
[1] 0.3760085
 z - X
 z
[1] 0.1506062


Cheers,
Bert


 Why not just
 make X be a function explicitly?  That is,

 X - function() runif(length(lT), lT, uT)

 Then use X() to call the function where you were previously using X.

 Duncan Murdoch

 I thought about something like:
 X- for (i in 1:length(lT)) runif(1, lT[i], uT[i])

 So that I can use X as a variable for multiple runs and
 each run new random values are used.

 thank you
 Johannes

  Original-Nachricht 
   Datum: Wed, 10 Aug 2011 08:19:07 -0500
   Von: Jean V Adamsjvad...@usgs.gov
   An: Johannes Radingerjradin...@gmx.at
   CC: r-help@r-project.org
   Betreff: Re: [R] function runif in for loop

   Johannes,
 
   You have the loop set up right, you just need to add indexing to refer
  to
   the looping variable, i.
 
   lT- sample(1:10)
   uT- sample(21:30)
   X- numeric(length(lT))
 
   for (i in 1:length(lT)) X[i]- runif(1, lT[i], uT[i])
 
   X
 
   Note that I changed the name of the result from T to X, because T has
   special meaning in R.
 
   Jean
 
   `·.,,(((º    `·.,,(((º    `·.,,(((º
 
   Jean V. Adams
   Statistician
   U.S. Geological Survey
   Great Lakes Science Center
   223 East Steinfest Road
   Antigo, WI 54409  USA
 
 
 
   From:
   Johannes Radingerjradin...@gmx.at
   To:
   r-help@r-project.org
   Date:
   08/10/2011 07:23 AM
   Subject:
   [R] function runif in for loop
   Sent by:
   r-help-boun...@r-project.org
 
 
 
   Hello,
 
   I'd like to perform a regression using MCMCregress (MCMCpack).
   One variable therefore should be a function rather than a variable:
 
   I want to use X as an input and X should be defined as a random number
   between to values. Therefore I want to use the function runif like:
   X-(1, Xa, Xb) but it seems that runif doesn't allow to use vectors.
   So I think I've to calculate the new vector X by using a for loop.
 
   I tried for (i in 1:length(lT)) T-runif(1,lT,uT) but that doesn't
  work.
   What is the correct for-loop function to create this new
  vector/variable?
 
   Can I use that function then as an input for MCMCregress?
 
   thank you
 
   Johannes
 
   --
 
   __
   R-help@r-project.org mailing list
   https://stat.ethz.ch/mailman/listinfo/r-help
   PLEASE do read the posting guide
   http://www.R-project.org/posting-guide.html
   and provide commented, minimal, self-contained, reproducible code.
 
 

 --

 __
 R-help@r-project.org mailing list
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 PLEASE do read the posting guide
 http://www.R-project.org/posting-guide.html
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 __
 R-help@r-project.org mailing list
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-- 
Men by nature long to get on to the ultimate truths, and will often
be impatient with elementary studies or fight shy of them. If it were
possible to reach the ultimate truths without the elementary studies
usually prefixed to them, these would not be preparatory studies but
superfluous diversions.

-- Maimonides (1135-1204)

Bert Gunter
Genentech Nonclinical Biostatistics

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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
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[R] Histograms in R

2011-08-10 Thread lt2

HI everyone,
I'm plotting a histogram in R and within that histogram i need to
demonstrate the percentage of another variable (Percentage of MutStatus)
within the bins plotted inthe histogramI don't know how to do that!

Data:Validation_Status  Mutation_Status TvarRatio
WildtypeNone0.08
WildtypeNone0.08
WildtypeNone0.08
WildtypeNone0.08
WildtypeNone0.080139373
WildtypeNone0.080152672
WildtypeNone0.080213904
WildtypeNone0.080357143
WildtypeNone0.080357143
WildtypeNone0.080357143
WildtypeNone0.08045977
WildtypeNone0.1
WildtypeLOH 0.1
WildtypeNone0.1
WildtypeNone0.1
WildtypeNone0.1
WildtypeNone0.1
WildtypeNone0.1
WildtypeSomtatic0.1
WildtypeNone0.100558659
WildtypeNone0.100591716
WildtypeNone0.101010101
WildtypeNone0.101123596
WildtypeGline   0.10133
WildtypeNone0.101369863
WildtypeNone0.101449275
WildtypeNone0.101522843
WildtypeNone0.101604278
WildtypeNone0.102040816
WildtypeGline   0.102040816
WildtypeNone0.102362205
WildtypeNone0.102459016
WildtypeNone0.102564103
WildtypeNone0.102702703
WildtypeNone0.102739726
WildtypeNone0.102803738
Valid   Somatic 0.102941176
WildtypeNone0.102941176

--
View this message in context: 
http://r.789695.n4.nabble.com/Histograms-in-R-tp3733644p3733644.html
Sent from the R help mailing list archive at Nabble.com.

__
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Re: [R] Histograms in R

Perhaps you could shade the bars as appropriate?

I'm not going to use your data because it's not an easily paste-able but how
about this:

x = rnorm(100)
y = sample(c(A,B),100,replace=T,prob=c(0.7,0.3))

d = data.frame(level = x, status = y)
n = 10 # Number of bins
breaks = quantile(d$level, (0:n)/n)
#breaks = with(d,hist(level,breaks=n,plot=F)$breaks)
breaksAssign = findInterval(d$level,breaks)

percentB = unique(ave(d$status==B,breaksAssign)[order(breaksAssign)])
percentB = gray(percentB/max(percentB)) # Many other color functions
available as well.

with(d, hist(level, breaks = breaks, col = percentB))

If you want to use hist()'s smart choice of bins (which I'd recommend), you
can call this hist command once with plot=F and get the breaks from there.
I.e., breaks = with(d,hist(level,breaks=n,plot=F)$breaks)

There's probably a smarter way to do all this, but this does seem to work...

Hope this helps,

Michael Weylandt

On Wed, Aug 10, 2011 at 1:16 PM, lt2 l...@bcm.edu wrote:

 HI everyone,
 I'm plotting a histogram in R and within that histogram i need to
 demonstrate the percentage of another variable (Percentage of MutStatus)
 within the bins plotted inthe histogramI don't know how to do that!

 Data:Validation_Status  Mutation_Status TvarRatio
 WildtypeNone0.08
 WildtypeNone0.08
 WildtypeNone0.08
 WildtypeNone0.08
 WildtypeNone0.080139373
 WildtypeNone0.080152672
 WildtypeNone0.080213904
 WildtypeNone0.080357143
 WildtypeNone0.080357143
 WildtypeNone0.080357143
 WildtypeNone0.08045977
 WildtypeNone0.1
 WildtypeLOH 0.1
 WildtypeNone0.1
 WildtypeNone0.1
 WildtypeNone0.1
 WildtypeNone0.1
 WildtypeNone0.1
 WildtypeSomtatic0.1
 WildtypeNone0.100558659
 WildtypeNone0.100591716
 WildtypeNone0.101010101
 WildtypeNone0.101123596
 WildtypeGline   0.10133
 WildtypeNone0.101369863
 WildtypeNone0.101449275
 WildtypeNone0.101522843
 WildtypeNone0.101604278
 WildtypeNone0.102040816
 WildtypeGline   0.102040816
 WildtypeNone0.102362205
 WildtypeNone0.102459016
 WildtypeNone0.102564103
 WildtypeNone0.102702703
 WildtypeNone0.102739726
 WildtypeNone0.102803738
 Valid   Somatic 0.102941176
 WildtypeNone0.102941176

 --
 View this message in context:
 http://r.789695.n4.nabble.com/Histograms-in-R-tp3733644p3733644.html
 Sent from the R help mailing list archive at Nabble.com.

 __
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 https://stat.ethz.ch/mailman/listinfo/r-help
 PLEASE do read the posting guide
 http://www.R-project.org/posting-guide.html
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[[alternative HTML version deleted]]

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Re: [R] function runif in for loop


On 10/08/2011 1:16 PM, Bert Gunter wrote:

Duncan et. al:

Inline below.

On Wed, Aug 10, 2011 at 9:48 AM, Duncan Murdoch
murdoch.dun...@gmail.com  wrote:
  On 10/08/2011 9:40 AM, Johannes Radinger wrote:

  Jean,

  thank you for your answer.
  especially the line X- numeric(length(lT)) helped me a lot.

  Anyway, in my case I'd like to get a dynamic variable or better a
  function for X. I mean if i try to call X I'd like that this drawing of
  random number is performed. In the case now if I call X several times I'll
  always get the same random numbers.

  Such things do exist in R, but they aren't easy to set up.

Well, how about:

Well... How about:

X- function()runif(1)
class(X)- c(wizz,class(X))
print.wizz- function(x){y-x(); print(y);y }

  X
[1] 0.875768
  X
[1] 0.955208
  X
[1] 0.1150938
  z- X
  z
[1] 0.3760085
  z- X
  z
[1] 0.1506062



That sort of looks as if it works, but it doesn't:

  for (i in 1:3) {
 r - X + 1
 print(r)
  }

Duncan Murdoch


Cheers,
Bert


  Why not just
  make X be a function explicitly?  That is,

  X- function() runif(length(lT), lT, uT)

  Then use X() to call the function where you were previously using X.

  Duncan Murdoch

  I thought about something like:
  X- for (i in 1:length(lT)) runif(1, lT[i], uT[i])

  So that I can use X as a variable for multiple runs and
  each run new random values are used.

  thank you
  Johannes

   Original-Nachricht 
  Datum: Wed, 10 Aug 2011 08:19:07 -0500
  Von: Jean V Adamsjvad...@usgs.gov
  An: Johannes Radingerjradin...@gmx.at
  CC: r-help@r-project.org
  Betreff: Re: [R] function runif in for loop

  Johannes,
  
  You have the loop set up right, you just need to add indexing to refer
to
  the looping variable, i.
  
  lT- sample(1:10)
  uT- sample(21:30)
  X- numeric(length(lT))
  
  for (i in 1:length(lT)) X[i]- runif(1, lT[i], uT[i])
  
  X
  
  Note that I changed the name of the result from T to X, because T has
  special meaning in R.
  
  Jean
  
  `·.,,(((º  `·.,,(((º  `·.,,(((º
  
  Jean V. Adams
  Statistician
  U.S. Geological Survey
  Great Lakes Science Center
  223 East Steinfest Road
  Antigo, WI 54409  USA
  
  
  
  From:
  Johannes Radingerjradin...@gmx.at
  To:
  r-help@r-project.org
  Date:
  08/10/2011 07:23 AM
  Subject:
  [R] function runif in for loop
  Sent by:
  r-help-boun...@r-project.org
  
  
  
  Hello,
  
  I'd like to perform a regression using MCMCregress (MCMCpack).
  One variable therefore should be a function rather than a variable:
  
  I want to use X as an input and X should be defined as a random number
  between to values. Therefore I want to use the function runif like:
  X-(1, Xa, Xb) but it seems that runif doesn't allow to use vectors.
  So I think I've to calculate the new vector X by using a for loop.
  
  I tried for (i in 1:length(lT)) T-runif(1,lT,uT) but that doesn't
work.
  What is the correct for-loop function to create this new
vector/variable?
  
  Can I use that function then as an input for MCMCregress?
  
  thank you
  
  Johannes
  
  --
  
  __
  R-help@r-project.org mailing list
  https://stat.ethz.ch/mailman/listinfo/r-help
  PLEASE do read the posting guide
  http://www.R-project.org/posting-guide.html
  and provide commented, minimal, self-contained, reproducible code.
  
  

  --

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

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






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

Re: [R] Histograms in R

Assuming your data is in a data.frame called df, try this:

attach(df)
TR.groups - cut(TvarRatio, seq(0.07, 0.11, 0.01))
m - table(Mutation_Status, TR.groups)
mut.no - dim(m)[1]
barplot(m, col=seq(mut.no), xlab=TvarRatio, ylab=Frequency)
legend(topleft, dimnames(m)[[1]], fill=seq(mut.no), 
title=Mutation_Status)

Jean

`·.,,  (((º   `·.,,  (((º   `·.,,  (((º

Jean V. Adams
Statistician
U.S. Geological Survey
Great Lakes Science Center
223 East Steinfest Road
Antigo, WI 54409  USA

r-help-boun...@r-project.org wrote on 08/10/2011 12:16:51 PM:

 [image removed] 
 
 [R] Histograms in R
 
 lt2 
 
 to:
 
 r-help
 
 08/10/2011 12:18 PM
 
 Sent by:
 
 r-help-boun...@r-project.org
 
 HI everyone,
 I'm plotting a histogram in R and within that histogram i need to
 demonstrate the percentage of another variable (Percentage of MutStatus)
 within the bins plotted inthe histogramI don't know how to do that!
 
 Data:Validation_Status   Mutation_Status   TvarRatio
 Wildtype   None   0.08
 Wildtype   None   0.08
 Wildtype   None   0.08
 Wildtype   None   0.08
 Wildtype   None   0.080139373
 Wildtype   None   0.080152672
 Wildtype   None   0.080213904
 Wildtype   None   0.080357143
 Wildtype   None   0.080357143
 Wildtype   None   0.080357143
 Wildtype   None   0.08045977
 Wildtype   None   0.1
 Wildtype   LOH   0.1
 Wildtype   None   0.1
 Wildtype   None   0.1
 Wildtype   None   0.1
 Wildtype   None   0.1
 Wildtype   None   0.1
 Wildtype   Somtatic   0.1
 Wildtype   None   0.100558659
 Wildtype   None   0.100591716
 Wildtype   None   0.101010101
 Wildtype   None   0.101123596
 Wildtype   Gline   0.10133
 Wildtype   None   0.101369863
 Wildtype   None   0.101449275
 Wildtype   None   0.101522843
 Wildtype   None   0.101604278
 Wildtype   None   0.102040816
 Wildtype   Gline   0.102040816
 Wildtype   None   0.102362205
 Wildtype   None   0.102459016
 Wildtype   None   0.102564103
 Wildtype   None   0.102702703
 Wildtype   None   0.102739726
 Wildtype   None   0.102803738
 Valid   Somatic   0.102941176
 Wildtype   None   0.102941176
 
 --
 View this message in context: http://r.789695.n4.nabble.com/
 Histograms-in-R-tp3733644p3733644.html
 Sent from the R help mailing list archive at Nabble.com.
 
 __
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 https://stat.ethz.ch/mailman/listinfo/r-help
 PLEASE do read the posting guide 
http://www.R-project.org/posting-guide.html
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[[alternative HTML version deleted]]

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[R] Opposite of paste function

2011-08-10 Thread Soyeon Kim

Dear All,

I have vn variable
 vn
[1] V300 V376
What I want to get is
300 376

without V and  from vn variable.

Could you help me about this issue?

Thank you,
Soyeon

[[alternative HTML version deleted]]

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

Re: [R] Opposite of paste function

2011-08-10 Thread Peter Langfelder

On Wed, Aug 10, 2011 at 11:22 AM, Soyeon Kim yunni0...@gmail.com wrote:
 Dear All,

 I have vn variable
 vn
 [1] V300 V376
 What I want to get is
 300 376

as.numeric(substring(vn, 2))

HTH

Peter

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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
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Re: [R] Opposite of paste function

2011-08-10 Thread Abhijit Dasgupta

or,

gsub('V','',vn)

On 8/10/2011 2:23 PM, Peter Langfelder wrote:
 On Wed, Aug 10, 2011 at 11:22 AM, Soyeon Kimyunni0...@gmail.com  wrote:
 Dear All,

 I have vn variable
 vn
 [1] V300 V376
 What I want to get is
 300 376
 as.numeric(substring(vn, 2))

 HTH

 Peter

 __
 R-help@r-project.org mailing list
 https://stat.ethz.ch/mailman/listinfo/r-help
 PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
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[[alternative HTML version deleted]]

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

Re: [R] Opposite of paste function

2011-08-10 Thread Doran, Harold

The see also potion of paste gives you the functions you can use for this

 -Original Message-
 From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
 Behalf Of Soyeon Kim
 Sent: Wednesday, August 10, 2011 2:22 PM
 To: r-help@r-project.org
 Subject: [R] Opposite of paste function

 Dear All,

 I have vn variable
  vn
 [1] V300 V376
 What I want to get is
 300 376

 without V and  from vn variable.

 Could you help me about this issue?

 Thank you,
 Soyeon

   [[alternative HTML version deleted]]

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

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

Re: [R] function runif in for loop

2011-08-10 Thread Bert Gunter

Duncan:

Yup, you're right. Can't assign, just print.

-- Bert

On Wed, Aug 10, 2011 at 11:02 AM, Duncan Murdoch
murdoch.dun...@gmail.com wrote:
 On 10/08/2011 1:16 PM, Bert Gunter wrote:

 Duncan et. al:

 Inline below.

 On Wed, Aug 10, 2011 at 9:48 AM, Duncan Murdoch
 murdoch.dun...@gmail.com  wrote:
   On 10/08/2011 9:40 AM, Johannes Radinger wrote:
 
   Jean,
 
   thank you for your answer.
   especially the line X- numeric(length(lT)) helped me a lot.
 
   Anyway, in my case I'd like to get a dynamic variable or better a
   function for X. I mean if i try to call X I'd like that this drawing
  of
   random number is performed. In the case now if I call X several times
  I'll
   always get the same random numbers.
 
   Such things do exist in R, but they aren't easy to set up.

 Well, how about:

 Well... How about:

 X- function()runif(1)
 class(X)- c(wizz,class(X))
 print.wizz- function(x){y-x(); print(y);y }

   X
 [1] 0.875768
   X
 [1] 0.955208
   X
 [1] 0.1150938
   z- X
   z
 [1] 0.3760085
   z- X
   z
 [1] 0.1506062
 

 That sort of looks as if it works, but it doesn't:

  for (i in 1:3) {
     r - X + 1
     print(r)
  }

 Duncan Murdoch

 Cheers,
 Bert


  Why not just
   make X be a function explicitly?  That is,
 
   X- function() runif(length(lT), lT, uT)
 
   Then use X() to call the function where you were previously using X.
 
   Duncan Murdoch
 
   I thought about something like:
   X- for (i in 1:length(lT)) runif(1, lT[i], uT[i])
 
   So that I can use X as a variable for multiple runs and
   each run new random values are used.
 
   thank you
   Johannes
 
    Original-Nachricht 
       Datum: Wed, 10 Aug 2011 08:19:07 -0500
       Von: Jean V Adamsjvad...@usgs.gov
       An: Johannes Radingerjradin...@gmx.at
       CC: r-help@r-project.org
       Betreff: Re: [R] function runif in for loop
 
       Johannes,
   
       You have the loop set up right, you just need to add indexing to
  refer
     to
       the looping variable, i.
   
       lT- sample(1:10)
       uT- sample(21:30)
       X- numeric(length(lT))
   
       for (i in 1:length(lT)) X[i]- runif(1, lT[i], uT[i])
   
       X
   
       Note that I changed the name of the result from T to X, because T
  has
       special meaning in R.
   
       Jean
   
       `·.,,(((º      `·.,,(((º      `·.,,(((º
   
       Jean V. Adams
       Statistician
       U.S. Geological Survey
       Great Lakes Science Center
       223 East Steinfest Road
       Antigo, WI 54409  USA
   
   
   
       From:
       Johannes Radingerjradin...@gmx.at
       To:
       r-help@r-project.org
       Date:
       08/10/2011 07:23 AM
       Subject:
       [R] function runif in for loop
       Sent by:
       r-help-boun...@r-project.org
   
   
   
       Hello,
   
       I'd like to perform a regression using MCMCregress (MCMCpack).
       One variable therefore should be a function rather than a
  variable:
   
       I want to use X as an input and X should be defined as a random
  number
       between to values. Therefore I want to use the function runif
  like:
       X-(1, Xa, Xb) but it seems that runif doesn't allow to use
  vectors.
       So I think I've to calculate the new vector X by using a for
  loop.
   
       I tried for (i in 1:length(lT)) T-runif(1,lT,uT) but that
  doesn't
     work.
       What is the correct for-loop function to create this new
     vector/variable?
   
       Can I use that function then as an input for MCMCregress?
   
       thank you
   
       Johannes
   
       --
   
       __
       R-help@r-project.org mailing list
       https://stat.ethz.ch/mailman/listinfo/r-help
       PLEASE do read the posting guide
       http://www.R-project.org/posting-guide.html
       and provide commented, minimal, self-contained, reproducible
  code.
   
   
 
   --
 
   __
   R-help@r-project.org mailing list
   https://stat.ethz.ch/mailman/listinfo/r-help
   PLEASE do read the posting guide
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   and provide commented, minimal, self-contained, reproducible code.
 
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-- 
Men by nature long to get on to the ultimate truths, and will often
be impatient with elementary studies or fight shy of them. If it were
possible to reach the ultimate truths without the elementary studies
usually prefixed to them, these would not be preparatory studies but
superfluous diversions.

-- Maimonides (1135-1204)

Bert Gunter
Genentech Nonclinical Biostatistics
467-7374
http://pharmadevelopment.roche.com/index/pdb/pdb-functional-groups/pdb-biostatistics/pdb-ncb-home.htm

[R] convert 'list' to 'vector'?

Dear all
How does one convert a non-symmetric list to a vector? See below:

 x - list()
 x[[1]] - letters[1:5]
 x[[2]] - letters[6:10]
 x[[3]] - letters[11:12]
 x
[[1]]
[1] a b c d e

[[2]]
[1] f g h i j

[[3]]
[1] k l

 paste(x)
[1] c(\a\, \b\, \c\, \d\, \e\) c(\f\, \g\, \h\,
\i\, \j\)
[3] c(\k\, \l\)
 as.vector(x)
[[1]]
[1] a b c d e

[[2]]
[1] f g h i j

[[3]]
[1] k l

 simplify2array(x)
[[1]]
[1] a b c d e

[[2]]
[1] f g h i j

[[3]]
[1] k l


What I would need to get instead is:
 letters[1:12]
 [1] a b c d e f g h i j k l

Any ideas? 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?
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[R] choosing selective data with permutations

2011-08-10 Thread Vikram Chhatre

Hello,

I am a R beginner and hoping to obtain some hints or suggestions about
using permutations to sort a data set I have.

Here is an example dataset:


Ind1   11  00   12  15  28
Ind2  21  33   22  67  52
Ind3  22  45   21  22  56
Ind4  11  25   74  77  42
Ind5  41  32   67  45   22

This will be read into a variable using read.table.  What I want to do
is permute these individuals and every time pick 3 individuals and
write them to a new variable.  I want to do this 100 times so that in
the end I will have 100 tables containing data for 3 individuals each.
 The data (for individuals) itself is not to be permuted, rather the
selection of individuals.

I am guessing this is probably trivial to do.  But I would appreciate
any advice on this matter.

Thank you.
Vikram

__
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Re: [R] convert 'list' to 'vector'?

unlist()

Michael Weylandt

On Wed, Aug 10, 2011 at 2:58 PM, Liviu Andronic landronim...@gmail.comwrote:

 Dear all
 How does one convert a non-symmetric list to a vector? See below:

  x - list()
  x[[1]] - letters[1:5]
  x[[2]] - letters[6:10]
  x[[3]] - letters[11:12]
  x
 [[1]]
 [1] a b c d e

 [[2]]
 [1] f g h i j

 [[3]]
 [1] k l

  paste(x)
 [1] c(\a\, \b\, \c\, \d\, \e\) c(\f\, \g\, \h\,
 \i\, \j\)
 [3] c(\k\, \l\)
  as.vector(x)
 [[1]]
 [1] a b c d e

 [[2]]
 [1] f g h i j

 [[3]]
 [1] k l

  simplify2array(x)
 [[1]]
 [1] a b c d e

 [[2]]
 [1] f g h i j

 [[3]]
 [1] k l


 What I would need to get instead is:
  letters[1:12]
  [1] a b c d e f g h i j k l

 Any ideas? 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

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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.


[[alternative HTML version deleted]]

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[R] xtable - caption missing with float=FALSE

2011-08-10 Thread Zev Ross


Hi,

For some reason I'm finding that my table caption is disappearing if I 
print xtable output with the floating argument set to FALSE. Below is a 
very simple Sweave file that produces two tables the first has no 
caption and the second has a caption (if you want to see it 
http://www.zevross.com/temp/test.pdf).


Does anyone know what I can do to fix this?

Zev

(I'm using Windows 7, 64 bit, R 2.12.2)

% begin Rnw file

\documentclass[a4paper]{article}

\begin{document}

results=tex, echo=FALSE=
library(xtable)

atable-data.frame(a=1:10, b=rnorm(10))
print(xtable(atable, caption=FLOAT), floating=FALSE)
print(xtable(atable, caption=NO FLOAT), floating=TRUE)
@

\end{document}


Sweave(test.Rnw)
texi2dvi(test.tex, pdf=TRUE, clean=TRUE)

--
Zev Ross
ZevRoss Spatial Analysis
120 N Aurora, Suite 3A
Ithaca, NY 14850
607-277-0004 (phone)
866-877-3690 (fax, toll-free)
z...@zevross.com

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Re: [R] convert 'list' to 'vector'?

2011-08-10 Thread Dimitris Rizopoulos


Check function unlist().


Best,
Dimitris


On 8/10/2011 8:58 PM, Liviu Andronic wrote:

Dear all
How does one convert a non-symmetric list to a vector? See below:


x- list()
x[[1]]- letters[1:5]
x[[2]]- letters[6:10]
x[[3]]- letters[11:12]
x

[[1]]
[1] a b c d e

[[2]]
[1] f g h i j

[[3]]
[1] k l


paste(x)

[1] c(\a\, \b\, \c\, \d\, \e\) c(\f\, \g\, \h\,
\i\, \j\)
[3] c(\k\, \l\)

as.vector(x)

[[1]]
[1] a b c d e

[[2]]
[1] f g h i j

[[3]]
[1] k l


simplify2array(x)

[[1]]
[1] a b c d e

[[2]]
[1] f g h i j

[[3]]
[1] k l


What I would need to get instead is:

letters[1:12]

  [1] a b c d e f g h i j k l

Any ideas? Regards
Liviu




--
Dimitris Rizopoulos
Assistant Professor
Department of Biostatistics
Erasmus University Medical Center

Address: PO Box 2040, 3000 CA Rotterdam, the Netherlands
Tel: +31/(0)10/7043478
Fax: +31/(0)10/7043014
Web: http://www.erasmusmc.nl/biostatistiek/

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Re: [R] convert 'list' to 'vector'?

unlist(x)

r-help-boun...@r-project.org wrote on 08/10/2011 01:58:57 PM:

 [image removed] 
 
 [R] convert 'list' to 'vector'?
 
 Liviu Andronic 
 
 to:
 
 r-help@r-project.org Help
 
 08/10/2011 02:02 PM
 
 Sent by:
 
 r-help-boun...@r-project.org
 
 Dear all
 How does one convert a non-symmetric list to a vector? See below:
 
  x - list()
  x[[1]] - letters[1:5]
  x[[2]] - letters[6:10]
  x[[3]] - letters[11:12]
  x
 [[1]]
 [1] a b c d e
 
 [[2]]
 [1] f g h i j
 
 [[3]]
 [1] k l
 
  paste(x)
 [1] c(\a\, \b\, \c\, \d\, \e\) c(\f\, \g\, \h\,
 \i\, \j\)
 [3] c(\k\, \l\)
  as.vector(x)
 [[1]]
 [1] a b c d e
 
 [[2]]
 [1] f g h i j
 
 [[3]]
 [1] k l
 
  simplify2array(x)
 [[1]]
 [1] a b c d e
 
 [[2]]
 [1] f g h i j
 
 [[3]]
 [1] k l
 
 
 What I would need to get instead is:
  letters[1:12]
  [1] a b c d e f g h i j k l
 
 Any ideas? 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.

[[alternative HTML version deleted]]

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Re: [R] choosing selective data with permutations

To pick random elements to sample, you can just use the sample function

sample(1:5,3,replace=T/F) # pick true or false as needed for your data.

If you replicate this, you should have no problem.

replicate(100,function() return(sample(1:5,3,replace=T/F)))

This will be plenty fast, but if you get into very large scale boot
strapping, you might want to vectorize the whole thing. That's easily done
with replace = T (just take 3*100 samples and then convert the output vector
to a matrix) but I'm not sure its quite as easy with replace = F.

Hope this helps,

Michael

On Wed, Aug 10, 2011 at 2:37 PM, Vikram Chhatre crypticline...@gmail.comwrote:

 Hello,

 I am a R beginner and hoping to obtain some hints or suggestions about
 using permutations to sort a data set I have.

 Here is an example dataset:


 Ind1   11  00   12  15  28
 Ind2  21 33 22 67 52
 Ind3  22 45 21 22 56
 Ind4  11  25   74  77  42
 Ind5  41 32 67 45 22

 This will be read into a variable using read.table.  What I want to do
 is permute these individuals and every time pick 3 individuals and
 write them to a new variable.  I want to do this 100 times so that in
 the end I will have 100 tables containing data for 3 individuals each.
  The data (for individuals) itself is not to be permuted, rather the
 selection of individuals.

 I am guessing this is probably trivial to do.  But I would appreciate
 any advice on this matter.

 Thank you.
 Vikram

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


[[alternative HTML version deleted]]

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

[R] Clustering Large Applications..sort of

2011-08-10 Thread Ken Hutchison

Hello all,
   I am using the clustering functions in R in order to work with large
masses of binary time series data, however the clustering functions do not
seem able to fit this size of practical problem. Library 'hclust' is good
(though it may be sub par for this size of problem, thus doubly poor for
this application) in that I do not want to make assumptions about the number
of clusters present, also due to computational resources and time hclust is
not functionally good enough; furthermore k-means works fine assuming the
number of clusters within the data, which is not realistic. The silhouette
functions in 'Pam' and 'Clara' and (if I remember correctly) 'cluster' seem
to be really bad through very thorough experimentation of data generation
with known clusters. I am left then with either theoretical abstractions
such as pruning hclust trees with minimal spanning trees or perhaps
hand-rolling a hierarchical k-medoids which works extremely efficiently and
without cluster number assumptions. Anybody have any suggestions as to
possible libraries which I have missed or suggestions in general? Note: this
is not a question for 'Bigkmeans' unless there exists a
'findbigkmeansnumberofclusters' function also.
Thank you in advance for your
assistance,
 Ken

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Re: [R] convert 'list' to 'vector'?

On Wed, Aug 10, 2011 at 9:02 PM, R. Michael Weylandt
michael.weyla...@gmail.com wrote:
 unlist()

Thanks all! This is perfect, and very R-ish: never where a novice
would expect it to be.

Cheers
Liviu

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Re: [R] choosing selective data with permutations

Sorry, that second line of code won't work: do it in 2.

f - function() {return(sample(1:5,3,replace=T/F))}
replicate(100,f())

Michael


On Wed, Aug 10, 2011 at 3:06 PM, R. Michael Weylandt 
michael.weyla...@gmail.com wrote:

 To pick random elements to sample, you can just use the sample function

 sample(1:5,3,replace=T/F) # pick true or false as needed for your data.

 If you replicate this, you should have no problem.

 replicate(100,function() return(sample(1:5,3,replace=T/F)))

 This will be plenty fast, but if you get into very large scale boot
 strapping, you might want to vectorize the whole thing. That's easily done
 with replace = T (just take 3*100 samples and then convert the output vector
 to a matrix) but I'm not sure its quite as easy with replace = F.

 Hope this helps,

 Michael


 On Wed, Aug 10, 2011 at 2:37 PM, Vikram Chhatre 
 crypticline...@gmail.comwrote:

 Hello,

 I am a R beginner and hoping to obtain some hints or suggestions about
 using permutations to sort a data set I have.

 Here is an example dataset:


 Ind1   11  00   12  15  28
 Ind2  21 33 22 67 52
 Ind3  22 45 21 22 56
 Ind4  11  25   74  77  42
 Ind5  41 32 67 45 22

 This will be read into a variable using read.table.  What I want to do
 is permute these individuals and every time pick 3 individuals and
 write them to a new variable.  I want to do this 100 times so that in
 the end I will have 100 tables containing data for 3 individuals each.
  The data (for individuals) itself is not to be permuted, rather the
 selection of individuals.

 I am guessing this is probably trivial to do.  But I would appreciate
 any advice on this matter.

 Thank you.
 Vikram

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




[[alternative HTML version deleted]]

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

Re: [R] xtable - caption missing with float=FALSE

2011-08-10 Thread Marc Schwartz

On Aug 10, 2011, at 2:02 PM, Zev Ross wrote:

 Hi,
 
 For some reason I'm finding that my table caption is disappearing if I print 
 xtable output with the floating argument set to FALSE. Below is a very simple 
 Sweave file that produces two tables the first has no caption and the second 
 has a caption (if you want to see it http://www.zevross.com/temp/test.pdf).
 
 Does anyone know what I can do to fix this?
 
 Zev
 
 (I'm using Windows 7, 64 bit, R 2.12.2)
 
 % begin Rnw file
 
 \documentclass[a4paper]{article}
 
 \begin{document}
 
 results=tex, echo=FALSE=
 library(xtable)
 
 atable-data.frame(a=1:10, b=rnorm(10))
 print(xtable(atable, caption=FLOAT), floating=FALSE)
 print(xtable(atable, caption=NO FLOAT), floating=TRUE)
 @
 
 \end{document}
 
 
 Sweave(test.Rnw)
 texi2dvi(test.tex, pdf=TRUE, clean=TRUE)
 

Hi,

If you compare the output of the two commands, the 'floating = TRUE' variant 
places the tabular environment within a table environment. The tabular 
environment does not support the \caption command, table does, hence no caption 
if the tabular is not contained within a float (floating = FALSE).

HTH,

Marc Schwartz

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Re: [R] convert 'list' to 'vector'?



On Aug 10, 2011, at 3:10 PM, Liviu Andronic wrote:


On Wed, Aug 10, 2011 at 9:02 PM, R. Michael Weylandt
michael.weyla...@gmail.com wrote:

unlist()


Thanks all! This is perfect, and very R-ish: never where a novice
would expect it to be.


Well, since `unlist` is linked in the See Also on the help page for  
`list`, I can only hope you meant that in complete jest.


As the Posting Guide says: ... sometimes `read the manual' is the  
appropriate response.


--
David Winsemius, MD
West Hartford, CT

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Re: [R] How to quickly convert a data.frame into a structure of lists

2011-08-10 Thread William Dunlap

Here is code to transform the matrix that by() or array(split())
produces, along with an example of the speed of the various
approaches.  Using split(), either directly or via by() or tapply(),
saves a lot of time.

f0 - function(df) {
# original code with typos fixed.
list_structure - lapply(levels(df$A), function(levelA) {
lapply(levels(df$B), function(levelB) {df$C[df$A==levelA  
df$B==levelB]})
})
# Apply the names:
names(list_structure)-levels(df$A)
for (i in 1:length(list_structure)) {
names(list_structure[[i]])-levels(df$B)
}
list_structure
}

f0a - function(df) {
# slightly faster version of f0, taking repeated
# calculations out of loops.
A - df$A
B - df$B
C - df$C
levelsA - structure(levels(A), names=levels(A))
levelsB - structure(levels(B), names=levels(B))
lapply(levelsA, function(levelA) {
tmpA - A == levelA # this is responsible for most of the savings
lapply(levelsB, function(levelB) {C[tmpA  B==levelB]})
})
}

f1 - function(df) {
# DM's code
by(df$C, df[,1:2], identity)
}

f2 - function(df) {
# WD's code
AB- df[c(A,B)]
array(split(df$C, AB), dim=sapply(AB, nlevels), dimnames=sapply(AB, levels))
}

matrix2ListOfRows - function(mat) {
# convert a matrix to a list of its rows, converting dimnames to names.
retval - structure(as.vector(mat), names=rep(colnames(mat), 
each=nrow(mat)))
retval - split(retval, row(mat))
names(retval) - rownames(mat)
retval
}

The test involves 10^5 rows of data with 26 levels for A and 200 for B.

 r200 - as.character(as.roman(1:200))
 set.seed(1)
 df - data.frame(A=factor(sample(letters, size=1e5, replace=TRUE), 
 levels=letters),
+  B=factor(sample(r200, size=1e5, replace=TRUE), levels=r200),
+  C=1:1e5)
 system.time(ls0 - f0(df))
   user  system elapsed 
  74.082.34   76.60 
 system.time(ls0a - f0a(df))
   user  system elapsed 
  43.090.44   43.73 
 all.equal(ls0, ls0a)
[1] TRUE
 system.time(ls2 - matrix2ListOfRows(f2(df)))
   user  system elapsed 
   0.090.020.11 
 all.equal(ls0, ls2)
[1] TRUE
 system.time(ls1 - matrix2ListOfRows(f1(df)))
   user  system elapsed 
   0.690.000.69 
 all.equal(ls0, ls1)
[1] TRUE


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 William Dunlap
 Sent: Wednesday, August 10, 2011 10:05 AM
 To: Duncan Murdoch; Frederic F
 Cc: r-help@r-project.org
 Subject: Re: [R] How to quickly convert a data.frame into a structure of lists
 
 I was going to suggest
AB - df[c(A,B)]
ls2 - array(split(df$C, AB), dim=sapply(AB, nlevels), 
 dimnames=sapply(AB, levels))
 which produces a matrix very similar to what Duncan's by() call produces
ls1 - by(df$C, df[,1:2], identity)
 E.g.,
ls2[[a,X]]
   [1] 1 2
ls1[[a,X]]
   [1] 1 2
ls1[[a,Y]] # by assigns NULL to unoccupied slots
   NULL
ls2[[a,Y]] # split gives the same type to all slots, copied from input
   numeric(0)
 
 They both are quick because they use split() to avoid the repeated
 evaluations of
   bigVector[ anotherBigVector == scalar ]
 that your nested (not imbricated) loops do.  If you really need to convert
 the matrix to a list of lists that will probably be a quick transformation.
 
 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 Duncan Murdoch
  Sent: Wednesday, August 10, 2011 9:43 AM
  To: Frederic F
  Cc: r-help@r-project.org
  Subject: Re: [R] How to quickly convert a data.frame into a structure of 
  lists
 
  On 10/08/2011 10:30 AM, Frederic F wrote:
   Hello Duncan,
  
   Here is a small example to illustrate what I am trying to do.
  
   # Example data.frame
   df=data.frame(A=c(a,a,b,b), B=c(X,X,Y,Z), C=c(1,2,3,4))
   #   A B C
   # 1 a X 1
   # 2 a X 2
   # 3 b Y 3
   # 4 b Z 4
  
   ### First way of getting the list structure (ls1) using imbricated lapply
   loops:
   # Get the structure and populate it:
   ls1-lapply(levels(df$A), function(levelA) {
  lapply(levels(df$B), function(levelB) {df$C[df$A==levelA
   df$B==levelB]})
   })
   # Apply the names:
   names(list_structure)-levels(df$A)
   for (i in 1:length(list_structure))
   {names(list_structure[[i]])-levels(df$B)}
  
   # Result:
   ls1$a$X
   # [1] 1 2
   ls1$b$Z
   # [1] 4
  
   The data.frame will always be 'complete', i.e., there will be a value in
   every row for every column.
   I want to produce a structure like this one quickly (I aim at something
   below 10 seconds) for a dataset containing between 1 and 2 millions of 
   rows.
  
 
  I don't know what the timing would be like for your real data, but this
  does look like by() would work:
 
  ls1 - by(df$C, df[,1:2], identity)
 
  When I repeat the rows of df a million times

Re: [R] convert 'list' to 'vector'?