date:20091029

On Wed, 28 Oct 2009 14:48:34 -0700 Bert Gunter gunter.ber...@gene.com 
wrote:
 If generalities -- with the attendant risk of occasional specific caveats
 and violations -- can be tolerated, then George Box's (paraphrased) comments
 of circa 40-50 years ago seem apropos: why do statisticians obsess over
 normality, to which most analyses -- i.e. inference (especially from
 balanced designs)-- are robust, when lack of independence of the
 observations is the violation of assumptions that can reek the greatest
 havoc on the statistical analysis?

While I very much agree with the sentiment, ordinary linear regression 
is *not* very robust when dealing with very heavy-tailed errors, such as 
the t-distribution with low degrees of freedom. Of course, one could use 
'rlm' instead of 'lm' (and I often prefer to do so), but with 
proper/better modelling of the error distribution, you will better 
confidence and prediction intervals.

-- 
Karl Ove Hufthammer

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[R] Parameter Estimation of Inverse Gamma distribution

2009-10-29 Thread Alla Bulashevska

Dear R users,
i am searching for the solution of this problem in R: 
how to estimate the shape and scale parameters of Inverse
Gamma distribution.
Your views and help will be appreciated.

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[R] Re trieving comments from PostgreSQL tables with RODBC and psqlodbc

2009-10-29 Thread Jim Cser


I don't know whether this is an RODBC issue, a psqlodbc issue or a PostgreSQL
issue; I've searched the documentation and forums for each, with no success.

I want to list all the tables in a PostgreSQL schema, and return the table
comments as well as the table names.  As an example, one table is defined as
follows:

CREATE TABLE scen990.ezone_locationprice_year1
(
  ezone integer NOT NULL,
  lprice_man_year1 real,
  lprice_war_year1 real,
  lprice_ret_year1 real,
  lprice_gen_year1 real,
  lprice_med_year1 real,
  lprice_gov_year1 real,
  CONSTRAINT ezone_locationprice_year1_pkey PRIMARY KEY (ezone)
) 
WITHOUT OIDS;
ALTER TABLE scen990.ezone_locationprice_year1 OWNER TO mscope;
COMMENT ON TABLE scen990.ezone_locationprice_year1 IS 'calculated';

The RODBC function sqlTables() returns all the tables, but the records look
like this:

TABLE_QUALIFIER TABLE_OWNER TABLE_NAME TABLE_TYPE REMARKS
mscope scen990 ezone_locationprice_year1 TABLE 

All the REMARKS values are empty strings.  Can I get it to return the
COMMENT ON values instead?  (the table_owner and tablet_qualifier are
switched but that doesn't worry me)   I'm using  R 2.8.1, RODBC 1.2-4,
psqlodbc 8.0.2, and PostgreSQL 8.1.5 .  

Regards,
Jim Cser
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[R] roc plot with zero length labels error

2009-10-29 Thread Chris Anderson

I am trying to create the roc plot bootstrap method from library(verification), 
and when I set the plot =both or emp then I get the following error. The 
roc.plot works fine when the plot is set to binorm. This is my first time using 
this function in R and am not sure what this error means or how to resolve the 
issue. It seems to work ok with the example dataset. Is there an option or data 
formatting I need to due to resolve this error? 
 
roc.plot((scicomp5.model1[164]==High Price), pred, CI = T, n.boot = 100, plot 
= 'both',binormal = TRUE,
+  main = paste('AREA UNDER CURVE (AUC) =', round(auc$A, 2)))
Error in text.default(DAT[id, 3, i], DAT[id, 2, i], plot.thres[id2], pos = 4,  
: 
  zero length 'labels'



Chris Anderson

http://www.seocodebreaker.com/?thankyou-page=429

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[R] help on paramter optimization

2009-10-29 Thread Jiefeng Xu

Hi,
  I am new to R and I need to solve the following problem: I have an ODE model 
with variables x(t) and y(t) that can be solved using the deSolve package. That 
model uses several parameters. Now I have a set of observations for y(t), 
denoted as Y(t). I need to optimize the value of these parameters such that 
error between the model output y(t) and the observation Y(t) is minimized. In 
another word, y(t) is fitted to Y(t) as much as possible. What R package is 
available to optimize these parameter? Is there any sample R codes available?  
Thank you very much for your help!

Jiefeng

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[R] GLMMPQL and negbinomial: trouble with the X-axis in PREDICT

2009-10-29 Thread Erin Latham

Ahh, yes. The level made all the difference. I originally thought the
level only applied to binomial response variables, so I disregarded
it.
And I'll just add that plot(e) below still produced an indexed x
axis, so I just combined the vectors and then was able to plot the
scaled variable properly.

 MyData-data.frame(BERRIES = seq(from =20, to=3600, by=100))
 e-predict(m1,MyData,type=response, level=0)
 q2-cbind(MyData,e)
 plot(q2$BERRIES,q2$e)

Thanks for the fast responses!

~erin

Erin Latham, MGIS Candidate
Geography Dept, University of Calgary, AB, Canada

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[R] Re ading and Creating Shape Files

2009-10-29 Thread PDXRugger


Hello R Community,
   I have imported a dataset which contain X Y coordinates and would like to
recreate a shape file after some data analysis.  What i have done is to
import some taxlot data and join them based on some criteria.  I want to
check to see how well the joining went by reviewing the results in GIS.  

A couple things.  I cant seem to import a shape file correctly using the
maptools package and the readShapeSpatial.  I have tried 

Building=file(data/input/BuildingShape/Building.shp)
Bldg-readShapeSpatial(fn=data/input/BuildingShape/Building,proj4string=NAD83)
#--
Bldg-readShapeSpatial(data/input/BuildingShape/Building,proj4string=NAD83)
#---
Building=file(data/input/BuildingShape/Building.shp)
Bldg-readShapeSpatial(Building,proj4string=NAD83)

I know i am mis interpreting the documentation but it doesnt seem like it is
very complicated so i am of course confused.


Also, i am wondering if i can create a shape file by simply using XY
coordinates from a data frame.
So for:


Ycoord=c( 865296.4, 865151.5, 865457.0 ,865363.4 ,865311.0, 865260.9
,865210.7 ,865173.3,
865123.6 ,865038.2 ,864841.1 ,864745.4 ,864429.1 ,864795.6 ,864334.9
,864882.0)

Xcoord=c( 4227640 ,4227816 ,4228929 ,4228508 ,4229569 ,4229498 ,4226747,
4226781, 4229597,
4229204, 4228910, 4228959 ,4229465 ,4229794 ,4229596 ,4229082)

Lot-c(1900 , 2000,  2100  , 100   ,200  , 300,   400 ,  500 ,  600 ,  701 , 
900 , 1000 , 1100,
  300   ,100,   200)   

XYcoord-spCbind(Ycoord,Xcoord) #doesnt work so

XYcoord=c(Ycoord,Xcoord)

TaxLots-cbind(Ycoord,Xcoord,Lot)

writeSpatialShape(XYcoord, TaxLots..,
file=data/input/test/Taxlots,strictFilename=FALSE)



So help reading in shape files and then creating them using XY coordinates
if possible
Any help would be appreciated.  Thank you.




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Re: [R] GGPLOT2 Different Layers Different X Values

2009-10-29 Thread Bleyhl, Jonathan

Hello Hadley,

I'm not sure how easily I'd be able to come up with a good example. My data 
contain thousands of points, and I can't really divulge this data outside my 
company. But I can tell you that I found a solution (though somewhat hackish). 
I just created another matrix with each date having a single value. I then 
added that as a separate histogram scaled to something ridiculously small (like 
0.0001). It acts as a sort of place holder for all the other layers. It seems 
to do that job.

Thanks,
Jon

-Original Message-
From: Hadley Wickham [mailto:h.wick...@gmail.com] 
Sent: Wednesday, October 28, 2009 6:59 AM
To: Bleyhl, Jonathan
Cc: r-help@r-project.org
Subject: Re: [R] GGPLOT2 Different Layers Different X Values

Hi John,

Could you please provide a small reproducible example?

Thanks,

Hadley

Sent from my iPhone

On 26/10/2009, at 6:50 PM, Jonathan Bleyhl jonathan_ble...@affymetrix.com 
  wrote:


 I'm trying to plot values based on a date and then overlay a  
 histogram also
 by date. The problem is that both data sets don't have exactly the  
 same
 dates. So when I add the histogram, any dates that aren't already  
 there from
 from the scatter plot just get tacked on at the end and are out of  
 order. I
 tried using scale_x_discrete, but to no avail.

 Thanks,
 Jon
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[R] multiple pages with ggplot2 facet_wrap?

2009-10-29 Thread Bill Gillespie

I currently use lattice functions to produce multiple pages of plots  
using the layout argument to specify the number of rows and columns  
of panels, e.g.,


xyplot(price ~ carat | clarity, diamonds, layout = c(2, 2))

This results in 2 pages of 4 panels each. diamonds is a data.frame  
distributed with ggplot2.


I would like to do the same with ggplot2 but have been unsuccessful.  
The following sequence of statements seemed like a logical way to do it:

p - ggplot(diamonds, aes(carat, price))
p + geom_point() + facet_wrap(~clarity, ncol = 2, nrow = 2)
But they result in the error statement: Error in nrow * ncol : non- 
numeric argument to binary operator.


Is facet_wrap or facet_grid capable of producing multiple pages of  
plots and, if so, how?


Thanks,
Bill

William R Gillespie, VP Strategic Modeling  Simulation
Metrum Research Group LLC
2 Tunxis Road, Suite 112, Tariffville, CT 06081
Direct  FAX:919-371-2786, Main:860-735-7043
bi...@metrumrg.com
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[R] write to RTF format - possible?

2009-10-29 Thread Mark Heckmann


Hi everybody,

To write a .txt file I use write.table()

Is there a way to write to a .rtf file as well?

TIA
Mark

–––
Mark Heckmann
Dipl. Wirt.-Ing. cand. Psych.
Vorstraße 93 B01
28359 Bremen
Blog: www.markheckmann.de
R-Blog: http://ryouready.wordpress.com

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[R] Basic simple question

2009-10-29 Thread Jim Burke


I have a statement like below.

my_num - 
c(101,102,103,104,105,107,108,112,113,114,115)


my_df has 8,000 different STREPS in it.

I have a statement where I can select ONE item from data frame my_df
txhd_df - prec_dist_df[ my_df$STREP==101, c(COUNTY, FIPS, 
STSEN, STREP, PREC)]


I would like to make R compare and deliver only the STREPS that are in 
my_num.


A comparative SQL command WHERE STREP 
IN(101,102,103,104,105,107,108,112,113,114,115)


Thanks for any help,
Jim

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Re: [R] Labelling individual points on 3D PCA scatterplot

2009-10-29 Thread trz


Thank you Duncan! That is very helpful, after posting this i did figure out
how to convert the 2d coordinates into 3d. I was initially hesitant about
using rgl because i thought it looked messy. With the modifications you
suggested for the plot, it actually looks very clear now in rgl! And it
rotates :-D

Thanks again!

Tom Rzeszutek
MSc. Candidate
McMaster University



Duncan Murdoch-2 wrote:
 
 On 10/28/2009 11:21 AM, trz wrote:
 Hi There,
 I'm attempting to plot 10 values on a three-dimensional PCA with text
 labels
 next to each point. While i have no trouble doing this on 2D plots using
 the
 'text' or 'textxy' function, I cannot find a function to do this on a 3D
 plot.
 
 I am using princomp for my PCA:
 
PCA-princomp(eucdata, cor=TRUE)
PCA$scores [,1:3]# the three principal components i want to plot
 
 Then i am using 'scatterplot3d' to plot my first 3 principal components:
 
scatterplot3d(PCA$scores [,1:3],xlab=Component 1 (26.9%),main=My 3D
 PCA,ylab=Component 2 
 + (17.9%), zlab=Component 3
 (12.4%),type=h,box=FALSE,pch=21,bg=color)
 
 
 And i get this:
 
 http://www.nabble.com/file/p26096592/myPCA.jpeg 
 
 'text' and 'textxy' only accept 2D coordinates so they do not label my
 points in a way that makes sense.
 I'm open to other 3D plotting functions in R, i just think this one is
 easy
 to visually understand. In addition, If there is any way to move the grid
 up
 to z=0 please let me know. To be clear i am trying to make my PCA look
 more
 like this one i found in a journal article (Trejaut et al., 2005):
 
 http://www.nabble.com/file/p26096592/journal.pbio.0030247.g003.png 
 
 scatterplot3d will return a function that can transform 3D coordinates 
 to 2D so you can plot labels in the right place; see its man page.
 
 Alternatively, you could use the rgl package.  Using the 
 example(princomp) results, the call is similar to what you had, with 
 some extra work for embellishment:
 
 plot3d(pc.cr$scores [,1:3],xlab=Component 1,main=My 3D 
 PCA,ylab=Component 2, zlab=Component 3, type=h, box=F, axes=F)
 spheres3d(pc.cr$scores[,1:3], radius=0.1, col=pink)
 grid3d(side=z, at=list(z=0))
 text3d(pc.cr$scores[,1:3], text=rownames(pc.cr$scores), adj=1.3)
 
 This looks too crowded, but maybe you have fewer points to plot.
 
 Duncan Murdoch
 
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[R] output of for

2009-10-29 Thread Matteo Mattiuzzi

Hello!
 
I've got troubles on following thing:
 
As output I need a string like: 
INPUT_FILENAMES = 
H:\\AKK_temp\\Modis\\MOD13Q1\\austria\\MOD09Q1.A2008073.h18v04.005.2008083134952.hdf|H:\\AKK_temp\\Modis\\MOD13Q1\\austria\\MOD09Q1.A2008073.h19v04.005.2008082182221.hdf
The filenames (dates) are with pathname (wrkdr) and separated with |. The 
number of input filenames changes (minimum 1, maximum n)
 
wrkdr  - H:\\akk_temp\\Modis\\MOD09
dates  is a data.frame where each row is a date and each column is a 
Modis-image (the name of it)
 
I did this as following:
paste('INPUT_FILENAMES = ', for(g in 1:(ntiles-1)) 
{paste(wrkdr,'\\',dates[i,g+1],'|',sep='')} , for(g in ntiles){ 
paste(wrkdr,'\\',dates[i,g+1],sep='') } ,sep='')
 
and this did work (in an other wrkdr and other dates) but after 
changing, for gave no more output. And I don't know why!!! 
 
Why i use for twice? Because after the last file there is no |...
 
 
Thanks for help 
 
Matteo
 
 
 

 
 
 

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Re: [R] superscript in ylab

2009-10-29 Thread Walmes Zeviani


Try

ylab=expression(Temperature*degree*C))

type demo(mathplot) at R prompt for more customizations.

Walmes Zeviani
Lavras - MG, Brasil.



Lathouri, Maria wrote:
 
 Dear all
 
 I am doing some plots in R.
 
 I want to have as label in y-axis Temperature (oC). I have used
 ylab=expression(paste({Temperature} ^o*C)) but what I get is
 TemperatureoC.
 How can I have a space between Temperature and the units and also the
 units to be in brackets?
 
 Many thanks
 Maria
 
 
 
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[R] sample weight for box plot?

2009-10-29 Thread Sean Parks


Hi,

I would like to make a box and whisker plot but use a sample weight for each
observation.  I've searched around a bit and have not found a method of
doing this.  

Anyone have any advice?

Thanks much,
Sean Parks
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[R] operation with if/else on a dataframe

2009-10-29 Thread Fran100681


Hi to all,
I have this dataframe (I show the first six rows)

head(table)

 A   RFold.Change P.Value 
Count1   Count2
1 ENSRNOE002_at 0 1.13 0.601  
1
2 ENSRNOE009_at 0-1.04 0.733  
3
3 ENSRNOE020_at 0-1.08 0.680  
0
4 ENSRNOE021_at 0-1.31 0.201  
2
5 ENSRNOE023_at 0-1.06 0.643  
3
6 ENSRNOE024_at 0-1.14 0.403  
3

I would like to generate a function that determine for each row a new value
(resulting in a new vector of values to add to the dataframe). 
The function should give for every row the same value showed in column R.
However,I need of this because not all the R-values reported in this table
are correctly determined following the criteria mentioned below.


These new values calculated by the function must be:

1)UP 
if all the following conditions are verified in a certain row:
Fold.Change is = +1.5, 
P.Value is  0.05 
Count1 = 2

2)DOWN
if all the following conditions are verified in a certain row:
Fold.Change is = -1.5, 
P.Value is  0.05 
Count2 = 2

3) 0 if both of previous conditions are not verified

So I have set these fllowing parameters (new objects) because I'll have to
repeat this procedure to different dataframes in which the order of columns
of interest might change (So I can change these parameters depending on the
order of the columns in any different table)

Fold.change -  3 #(because in this table, Fold.Change value is the third
column and so on...)
P.Value -  4
Count1 -  5
Count2 -  6

The function is:

func_UP_0_DOWN - function(x) {
if (x[Fold.Change] = 1.5  x[P.Value] = 0.05  x[Count1] = 2) {
return (UP)} else { 
if (x[Fold.Change] = -1.5  x[P.Value] = 0.05  x[Count2] = 2) { 
return (DOWN)} else {
return(0)} 
   } }

and then I have decided to apply the function for every row:

values.vector - apply(table,1,func_UP_0_DOWN)

But when I check the resulting vector of calculated values:

 table(values.vector)
values.vector
 0 
192279 

I have all zero but this is wrong since I expect some UP and DOWN

However, if I apply this function on a single row in which I just know that
there is a UP-value...
(for example row 66):

table[66,]
RFold.Change  
P.Value  Count1  Count2
66 ENSRNOE162_atUP 1.84 0.013  
3

the function seems to work correctly:

 func_UP_0_DOWN(table[66,])
[1] UP

This is my problem, I cannot use the function to recalculate values in
R-column for all rows in my dataframe. I don't understand  where is the
problem, can someone help me?
Thanks a lot!!

Francesco
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[R] please help

2009-10-29 Thread karuna m

Is there any function for finding number of groups using upper tail rule or 
moving average approach suggested by Mojena?
I would also like to find functions for other stopping rules in Hierarchical 
clustering method.
If anyone knows please kindly let me know.
Kind regards,
Ms.Karunambigai M
PhD Scholar
Dept. of Biostatistics
NIMHANS
Bangalore
India


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[R] Bezier interpolation

2009-10-29 Thread Tim Clark

Dear List,

I am trying to interpolate animal tracking data using Bezier curves.  I need a 
function similar to spline() or approx() but that has a method Bezier.  I have 
tried xspline() but it does not allow you to set the number of points to 
interpolate between a given interval (n points between min(x) and max(x)).  
Mark Hindell asked the same question in 2006 
(http://tolstoy.newcastle.edu.au/R/e2/help/06/12/7034.html).  I contacted him 
and he never found a workable function.  Has one been developed since then?  

Thanks,

Tim
 
Tim Clark
Department of Zoology 
University of Hawaii

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Re: [R] Re trieving comments from PostgreSQL tables with RODBC and psqlodbc

2009-10-29 Thread Prof Brian Ripley

You seriously need to update your way-obsolete software (see the 
posting guide), but you are seeing what your ODBC driver is returning, 
not an R issue.


R-sig-db is a better list for help on database connectivity.

On Wed, 28 Oct 2009, Jim Cser wrote:



I don't know whether this is an RODBC issue, a psqlodbc issue or a PostgreSQL
issue; I've searched the documentation and forums for each, with no success.

I want to list all the tables in a PostgreSQL schema, and return the table
comments as well as the table names.  As an example, one table is defined as
follows:

CREATE TABLE scen990.ezone_locationprice_year1
(
 ezone integer NOT NULL,
 lprice_man_year1 real,
 lprice_war_year1 real,
 lprice_ret_year1 real,
 lprice_gen_year1 real,
 lprice_med_year1 real,
 lprice_gov_year1 real,
 CONSTRAINT ezone_locationprice_year1_pkey PRIMARY KEY (ezone)
)
WITHOUT OIDS;
ALTER TABLE scen990.ezone_locationprice_year1 OWNER TO mscope;
COMMENT ON TABLE scen990.ezone_locationprice_year1 IS 'calculated';

The RODBC function sqlTables() returns all the tables, but the records look
like this:

TABLE_QUALIFIER TABLE_OWNER TABLE_NAME TABLE_TYPE REMARKS
mscope scen990 ezone_locationprice_year1 TABLE 

All the REMARKS values are empty strings.  Can I get it to return the
COMMENT ON values instead?  (the table_owner and tablet_qualifier are
switched but that doesn't worry me)   I'm using  R 2.8.1, RODBC 1.2-4,
psqlodbc 8.0.2, and PostgreSQL 8.1.5 .

Regards,
Jim Cser
--

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

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Re: [R] operation with if/else on a dataframe

2009-10-29 Thread Bernardo Rangel Tura

On Thu, 2009-10-29 at 01:47 -0700, Fran100681 wrote:
 Hi to all,
 I have this dataframe (I show the first six rows)
 
 head(table)
 
  A   RFold.Change P.Value 
 Count1   Count2
 1 ENSRNOE002_at 0 1.13 0.601  
 1
 2 ENSRNOE009_at 0-1.04 0.733  
 3
 3 ENSRNOE020_at 0-1.08 0.680  
 0
 4 ENSRNOE021_at 0-1.31 0.201  
 2
 5 ENSRNOE023_at 0-1.06 0.643  
 3
 6 ENSRNOE024_at 0-1.14 0.403  
 3
 
 I would like to generate a function that determine for each row a new value
 (resulting in a new vector of values to add to the dataframe). 
 The function should give for every row the same value showed in column R.
 However,I need of this because not all the R-values reported in this table
 are correctly determined following the criteria mentioned below.
 
 
 These new values calculated by the function must be:
 
 1)UP 
 if all the following conditions are verified in a certain row:
 Fold.Change is = +1.5, 
 P.Value is  0.05 
 Count1 = 2
 
 2)DOWN
 if all the following conditions are verified in a certain row:
 Fold.Change is = -1.5, 
 P.Value is  0.05 
 Count2 = 2
 
 3) 0 if both of previous conditions are not verified
 
 So I have set these fllowing parameters (new objects) because I'll have to
 repeat this procedure to different dataframes in which the order of columns
 of interest might change (So I can change these parameters depending on the
 order of the columns in any different table)
 
 Fold.change -  3 #(because in this table, Fold.Change value is the third
 column and so on...)
 P.Value -  4
 Count1 -  5
 Count2 -  6

[ Quote text]

 This is my problem, I cannot use the function to recalculate values in
 R-column for all rows in my dataframe. I don't understand  where is the
 problem, can someone help me?
 Thanks a lot!!
 
 Francesco

Francesco,


I think you solve this problem with a simple way.
Remember in R the most function and operations are vectorized so look
this example: 

set.seed(123)
x-rpois(20,5)
y-rpois(20,15)
z-rpois(20,10)
dta-data.frame(x,y,z)
dta
dta$NEW-ifelse(x5  y15  z10,UP,
ifelse(x5  y15  z10,DOWN,
0))
dta

First, I use ifelse command to simplify your nested conditional
situation.

Second, I know that R test this nested condition in order so the first
position will result test x[1],y[1] and z[1], the second postion will
result test x[2],y[2] and z[2] ...

The new vector result is the same order the original data.frame so I use
dta$NEW to create a new column in data.frame 



-- 
Bernardo Rangel Tura, M.D,MPH,Ph.D
National Institute of Cardiology
Brazil

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[R] correlated binary data and overall probability

2009-10-29 Thread Christian Lerch

Dear All,

I try to simulate correlated binary data for a clinical research project.
Unfortunately, I do not come to grips with bindata().

Consider
corr-data.frame(ID=as.factor(rep(c(1:10), each=5)),
task=as.factor(rep(c(1:5),10)))

[this format might be more appropriate:]
corr2-data.frame(ID=as.factor(rep(c(1:10),5)),
tablet=as.factor(rep(c(1:5),each=10)))

Now, I want to add one column 'outcome' for the binary response:
* within each 'task', the probabilty of success (1) is fixed, (say 
rbinom(x,1,0.7))
* within each 'ID', the outcomes are correlated (say 0.8)

How can I generate this column 'outcome' with the given proporties?

Many thanks for hints or even code!

Regards,
Christian
-- 

http://portal.gmx.net/de/go/dsl02

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Re: [R] boxplot, data frame and html

2009-10-29 Thread Silvano


Gregoire,

it worked very well, thanks

--
Silvano Cesar da Costa
Departamento de Estatística
Universidade Estadual de Londrina
Fone: 3371-4346
--
- Original Message - 
From: Gregoire Pau gregoire@embl.de

To: r-help@r-project.org
Cc: Silvano silv...@uel.br
Sent: Wednesday, October 28, 2009 10:08 AM
Subject: Re: [R] boxplot, data frame and html



Hello Silvano,

'hwrite' appends HTML elements in a web page. The web page 
has to be opened before adding elements in it.


The following code should work:

require(hwriter)
p = openPage('T1000.html')
hwrite(t1000[,c(1,5,6)], p, bgcolor='#ffdc98',
   row.bgcolor='#ffdc98', br=TRUE)
hwriteImage('caixa.jpg', p, br=TRUE)
hwrite('', p, br=TRUE)
closePage(p)

Greg
---
Gregoire Pau
EMBL Research Officer
http://www.ebi.ac.uk/~gpau/


Silvano wrote:

Hi,

I'm trying put in same page:

- a data frame with 3 columns and 45 lines;
- a box plot;

the code is:

require(hwriter)
hwrite(t1000[,c(1,5,6)], 'T1000.html', bgcolor='#ffdc98',
  row.bgcolor='#ffdc98', br=TRUE)

p = openPage('T1000.html')
hwriteImage('caixa.jpg', p, br=TRUE)
hwrite('',p, br=TRUE)
closePage(p)

but isn't working. What's wrong?

Thanks,

--
Silvano Cesar da Costa
Departamento de Estatística
Universidade Estadual de Londrina
Fone: 3371-4346

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Re: [R] write to RTF format - possible?

2009-10-29 Thread Liviu Andronic

Hello

On 10/28/09, Mark Heckmann mark.heckm...@gmx.de wrote:
  To write a .txt file I use write.table()

  Is there a way to write to a .rtf file as well?

With odfWeave [1] you can create OOo documents, from which the
conversion to .rtf would be trivial.

If you wan to export a table, then you could also export it in an HTML
file (say, via xtable() ) and then Open with Gnumeric. I think MS
Office would also do this, but OOo doesn't seem to.
Liviu

[1] http://cran.r-project.org/web/packages/odfWeave/index.html

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[R] changing default character size depending on settings for mfcol: How to scale font consistently?

2009-10-29 Thread Kim MILFERSTEDT


Dear R-users,

I would like to create pdf files with varying paper sizes. The paper 
size is determine by the number of plots that I squeeze on a page using 
par(mfcol).


I also add text to each of these plots. I observed that depending on the 
number of plots per page, the default character size is changing, 
leading to differing font sizes in my plot annotations.


I would like to use cex to scale the font size so that I obtain 
identical font sizes in my pdf, independent of the size of my page in 
the pdf.


My question is, how do I find out the current default for character size 
so that I can calculate an appropriate scaling factor? Or are there any 
better alternatives?


Below, I added an example code to produce three pdf files. Note that the 
red letters in the plot have different sizes, depending on the number of 
plots per page.


Thank you very much in advance!

Kim

platform   i386-pc-mingw32
arch   i386
os mingw32
system i386, mingw32
status
major  2
minor  10.0
year   2009
month  10
day26
svn rev50208
language   R
version.string R version 2.10.0 (2009-10-26)

Example code

test.data - matrix(rnorm(300, 100, 25), nrow=3, ncol = 100, byrow = TRUE)
x - seq(1,100)
how.many.to.plot - c(1,2,3)


for(i in 1:length(how.many.to.plot))
{
test.data.red - matrix(test.data[1:how.many.to.plot[i],], nrow 
=how.many.to.plot[i], ncol = 100)
pdf.name - paste(test, how.many.to.plot[i], .pdf, 
sep=,collapse=)


pdf(file= pdf.name, width = 7, height=(7/5)*nrow(test.data.red))
par(mar=c(0, 0, 0, 0.0), cex = 1, oma = c(0,0,0,0), mfcol= 
c(nrow(test.data.red),1))


for(j in 1:(nrow(test.data.red)))
{
plot(   x,
test.data.red[j,],
ylim = c(0, max(test.data.red[j,])),
type = l,
xaxt = n,
yaxt = n,
)
text(   x = 30,
y = 75,
labels = LETTERS[j],
cex = 1.6,
col =red
)
}
dev.off()
}
--

Kim Milferstedt, PhD
Postdoctoral Researcher
INRA
Laboratoire de Biotechnologie de l'Environnement
Avenue des Etangs
F-11100 Narbonne
France

phone: (+33) 04 68 42 51 87
fax: (+33) 04 68 42 51 60
email: milfe...@supagro.inra.fr

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Re: [R] HOW to determine the number of components of the mixture model stratfied by age

On Oct 28, 2009, at 10:41 PM, lybaomc wrote:

DOES NOBODY KNOW?
HELP!

You submitted a question that appeared somewhat vague and had no
reproducible example. Sometimes people will step forward and create a
particular example to demonstrate coding options but there is no
expectation that someone _should_ do so. Generally the cycle time to
an answer is less than 24 hours so during the last week you should
have been asking yourself what _you_ might have in the construction of
your question that made it answer-unworthy.

Read the Posting Guide. The answer to how to write a good question
for r-help should be there.

--
David.

lybaomc wrote:

Hi,all

with my data,there are more than 1000 quantitative results of
antibody
concentrations, there may be 2 components(positive and negative),
or 3

components (may be strong positive, positive, and negative), or 4-6
components. Could you tell me how to determine the number of
components of
the mixture model? the anova.mix in mixdist of the R software
seems not

work.

my data is a little complicated(TABLE ), there are more than 1000
quantitative results of antibody concentrations stratified by age.The
overall density of results at age j, Fj, is a mixture of the
component

densities, so if there are 5 age groups, then there will be 5 mixture
models. Do i have to analyse each stratum respectively?

TABLE

age Bin length freq

1 1 19.75 4
1 2 21.7510

11241.7536

2 1 19.75 4
2 2 21.7510

21241.7536

---

appreciated

lybao

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Heritage Laboratories
West Hartford, CT

Re: [R] Is there a faster way to do it?

2009-10-29 Thread Adaikalavan Ramasamy


You might also want to consider using na.string=9 in the scan().



jim holtman wrote:

Here is a faster way of doing the replacement: (provide reproducible
data next time)


x - matrix(sample(6:9, 64, TRUE), 8)
x

 [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8]
[1,]87767879
[2,]77867677
[3,]77769667
[4,]99768766
[5,]69988989
[6,]97697867
[7,]79897978
[8,]99699886

x.f - 1:8  # replacement values based on column
x.ind - which(x == 9, arr.ind=TRUE)
x.ind

  row col
 [1,]   4   1
 [2,]   6   1
 [3,]   8   1
 [4,]   4   2
 [5,]   5   2
 [6,]   7   2
 [7,]   8   2
 [8,]   5   3
 [9,]   6   4
[10,]   7   4
[11,]   8   4
[12,]   3   5
[13,]   8   5
[14,]   5   6
[15,]   7   6
[16,]   1   8
[17,]   5   8

x[x.ind] - x.f[x.ind[,'col']]
x

 [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8]
[1,]87767878
[2,]77867677
[3,]77765667
[4,]12768766
[5,]62388688
[6,]17647867
[7,]72847678
[8,]12645886


On Wed, Oct 28, 2009 at 12:55 PM, Marcio Resende
mresende...@yahoo.com.br wrote:

#Mdarts is a matrix 2343x788
#frequencia is a vector 2343x1
# 9 in Mdarts[fri,frj] stands for my missing values which i want to replace
by the value in the vector frequencia


Mdarts-t(matrix(scan(C:/GWS/CNB/dartg.txt),ncol=nindT,nrow=nm, byrow=T))
frequencia - matrix(scan(C:/GWS/CNB/freq.txt),ncol=1)
for (fri in 1:nindT){
for (frj in 1:nm){
Mdarts[fri,frj] - if (Mdarts[fri,frj] == 9) frequencia[frj] else
Mdarts[fri,frj]
Mdarts[fri,frj] - Mdarts[fri,frj]/1-(frequencia[frj]^2)
}
}

Is there a faster way to it?
Maybe using any apply function?
Thanks in advance
--
View this message in context: 
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Re: [R] Removing generating data by category

2009-10-29 Thread Adaikalavan Ramasamy


Here is another way based on pasting ids as hinted below:

a - data.frame(id=c(c(A1,A2,A3,A4,A5),
   c(A3,A2,A3,A4,A5)),
   loc=c(B1,B2,B3,B4,B5),
   clm=c(rep((General),6),rep(Life,4)))

a$uid - paste(a$id, ., a$loc, sep=)

out - tapply( a$clm, a$uid, paste ) # can also add collapse=,
$A1.B1
[1] General

$A2.B2
[1] General Life

$A3.B1
[1] General

$A3.B3
[1] General Life

$A4.B4
[1] General Life

$A5.B5
[1] General Life


Then here are those with single policies.

 out[ which( sapply(out, length) == 1 ) ]
$A1.B1
[1] General

$A3.B1
[1] General



David Winsemius wrote:

On Oct 28, 2009, at 9:30 PM, Steven Kang wrote:


Dear R users,


Basically, from the following arbitrary data set:

a -
data
.frame
(id
=
c
(c
(A1
,A2
,A3
,A4
,A5
),c
(A3
,A2
,A3
,A4,A5)),loc=c(B1,B2,B3,B4,B5),clm=c(rep((General), 
6),rep(Life,4)))



a

   id   loc  clm
1  A1  B1 General
2  A2  B2 General
3  A3  B3 General
4  A4  B4 General
5  A5  B5 General
6  A3  B1 General
7  A2  B2Life
8  A3  B3Life
9  A4  B4Life
10 A5  B5Life

I desire removing records (highlighted records above) with identical  
values

in each fields (id  loc) but with different value of clm (i.e
according to category)


Take a look at this merge operation on separate rows of a.

  merge( a[a$clm==Life, ], a[a$clm==General, ] , by=c(id,  
loc), all=T)

   id loc clm.x   clm.y
1 A1  B1  NA General
2 A2  B2  Life General
3 A3  B1  NA General
4 A3  B3  Life General
5 A4  B4  Life General
6 A5  B5  Life General

Assignment of that object and selection with is.na should complete the  
process.


  a2m - merge( a[a$clm==Life, ], a[a$clm==General, ] ,  
by=c(id, loc), all=T)


  a2m[ is.na(a2m$clm.x) | is.na(a2m$clm.y), ]
   id loc clm.x   clm.y
1 A1  B1  NA General
3 A3  B1  NA General

Alternate methods might include paste-ing id to loc and removing  
duplicates.




i.e

categ - table(a$id,a$clm)
categ

General Life
 A1   10
 A2   11
 A3   21
 A4   11
 A5   11

The desired output is

   id   loc  clm
1  A1  B1 General
6  A3  B1 General

Because the data set I am working on is quite big (~ 800,000 x 20)
with majority of the fields values being long strings, looping  
turned out to

be very inefficient in comapring individual rows..

Are there any alternative efficient methods in implementing this  
problem?

Steven


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Re: [R] sample weight for box plot?



On Oct 28, 2009, at 8:00 PM, Sean Parks wrote:



Hi,

I would like to make a box and whisker plot but use a sample weight  
for each
observation.  I've searched around a bit and have not found a method  
of

doing this.

Anyone have any advice?


There are a variety of ways to get weighted quantiles. Two that have  
come up in recent r-help postings are the facilities in the quantreg  
package and wtd.quantile in Hmisc. Once you have calculated the five  
numbers that define a box-whisker plot they can be passed to bxp.


?bxp



--

David Winsemius, MD
Heritage Laboratories
West Hartford, CT

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Re: [R] Basic simple question

2009-10-29 Thread jim holtman

try:

txhd_df - prec_dist_df[ my_df$STREP %in% my_num, c(COUNTY, FIPS,
STSEN, STREP, PREC)]

On Thu, Oct 29, 2009 at 1:04 AM, Jim Burke j.bu...@earthlink.net wrote:
 I have a statement like below.

 my_num -
 c(101,102,103,104,105,107,108,112,113,114,115)

 my_df has 8,000 different STREPS in it.

 I have a statement where I can select ONE item from data frame my_df
 txhd_df - prec_dist_df[ my_df$STREP==101, c(COUNTY, FIPS, STSEN,
 STREP, PREC)]

 I would like to make R compare and deliver only the STREPS that are in
 my_num.

 A comparative SQL command WHERE STREP
 IN(101,102,103,104,105,107,108,112,113,114,115)

 Thanks for any help,
 Jim

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-- 
Jim Holtman
Cincinnati, OH
+1 513 646 9390

What is the problem that you are trying to solve?

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[R] lmer and negative binomial family

2009-10-29 Thread Patrick Giraudoux

Dear listers,

One of my former students is trying to fit a model of the negative 
binomial family with lmer. In the past (two years ago), the following 
call was working well:

m1a-lmer(mapos~ninter+saison+milieu*zone+(1|code),family=neg.bin(0.451),REML=TRUE,data=manu)

But now (R version 2.9.2 and lme4 version  0.999375-32), that gives 
(even with the library MASS loaded):

 
m1a-lmer(mapos~ninter+saison+milieu*zone+(1|code),family=neg.bin(0.451),REML=TRUE,data=manu)
Error in famType(glmFit$family) : unknown GLM family: 'Negative Binomial'

Any idea about what happens ?

Patrick



[[alternative HTML version deleted]]

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[R] lmer and negative binomial family

2009-10-29 Thread Patrick Giraudoux

Dear listers,

One of my former students is trying to fit a model of the negative 
binomial family with lmer. In the past (two years ago), the following 
call was working well:

m1a-lmer(mapos~ninter+saison+milieu*zone+(1|code),family=neg.bin(0.451),REML=TRUE,data=manu)

But now (R version 2.9.2 and lme4 version  0.999375-32), that gives 
(even with the library MASS loaded):

 
m1a-lmer(mapos~ninter+saison+milieu*zone+(1|code),family=neg.bin(0.451),REML=TRUE,data=manu)
Error in famType(glmFit$family) : unknown GLM family: 'Negative Binomial'

Any idea about what happens ?

Patrick




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Re: [R] lmer and negative binomial family

2009-10-29 Thread Patrick Giraudoux

Patrick Giraudoux a écrit :
 Dear listers,

 One of my former students is trying to fit a model of the negative 
 binomial family with lmer. In the past (two years ago), the following 
 call was working well:

 m1a-lmer(mapos~ninter+saison+milieu*zone+(1|code),family=neg.bin(0.451),REML=TRUE,data=manu)

 But now (R version 2.9.2 and lme4 version  0.999375-32), that gives 
 (even with the library MASS loaded):

  
 m1a-lmer(mapos~ninter+saison+milieu*zone+(1|code),family=neg.bin(0.451),REML=TRUE,data=manu)
 Error in famType(glmFit$family) : unknown GLM family: 'Negative Binomial'

 Any idea about what happens ?

 Patrick





Oups. Sorry to reply to myself, but the answer was here:

http://www.nabble.com/What-happen-for-Negative-binomial-link-in-Lmer-fonction--td26013041.html



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Re: [R] lmer and negative binomial family

2009-10-29 Thread John Sorkin

Patrick,
I am not sure about this but 
(1) I think you need to load the lme4 package, not MASS
(2) I don't think neg.bin is allowable. I suggest you load lme4 and then type 
?family

John

John David Sorkin M.D., Ph.D.
Chief, Biostatistics and Informatics
University of Maryland School of Medicine Division of Gerontology
Baltimore VA Medical Center
10 North Greene Street
GRECC (BT/18/GR)
Baltimore, MD 21201-1524
(Phone) 410-605-7119
(Fax) 410-605-7913 (Please call phone number above prior to faxing)

 Patrick Giraudoux patrick.giraud...@univ-fcomte.fr 10/29/2009 8:43 AM 
Dear listers,

One of my former students is trying to fit a model of the negative 
binomial family with lmer. In the past (two years ago), the following 
call was working well:

m1a-lmer(mapos~ninter+saison+milieu*zone+(1|code),family=neg.bin(0.451),REML=TRUE,data=manu)

But now (R version 2.9.2 and lme4 version  0.999375-32), that gives 
(even with the library MASS loaded):

 
m1a-lmer(mapos~ninter+saison+milieu*zone+(1|code),family=neg.bin(0.451),REML=TRUE,data=manu)
Error in famType(glmFit$family) : unknown GLM family: 'Negative Binomial'

Any idea about what happens ?

Patrick




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Re: [R] x11(title=' ')

2009-10-29 Thread Patrick Connolly

From: r-help-boun...@r-project.org on behalf of Jim Lemon
Sent: Thu 29-Oct-09 0:20
Cc: R Help
Subject: Re: [R] x11(title='  ')

On 10/28/2009 09:51 PM, Prof Brian Ripley wrote:
 ...
 Are you perchance x11() on Windows, when you should be using 
 windows()?  The posting guide asked you to tell us your OS, amongst 
 other things 
I say, my dear professor, if this MS chap who has used a Microsoft 
email client and a Microsoft email service turns out to be using Linux, 
he's covered his tracks pretty well.

Well, he wouldn't be the only one who can use Linux for serious R work, but is 
forced by institutional policy to use MS for email.

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[R] Invert the sign of a number

2009-10-29 Thread Fran100681


Hello,

just a simple question

How can I do to invert the sign of a number? (for example: -4 to 4, 2 to -2
and so on..)
I was looking for a specific function in R but I didn't found it...
thank you

Francesco
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[R] similarity measure for binary data

2009-10-29 Thread karuna m

I am doing hierarchical clustering with cluster package.  I couldnot find 
similarity measures like matching coefficient, Jaccard coefficient and sokal 
and sneath. Could anyone please tell package with similarity measures for 
binary data?
kind regards,

Ms.Karunambigai M
PhD Scholar
Dept. of Biostatistics
NIMHANS
Bangalore
India


  From cricket scores to your friends. Try the Yahoo! India Homepage! 
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[R] singular variance-covariance warning in lmer

2009-10-29 Thread Weber, Sam

Dear R Users,

I was hoping for some help with a recurrent error message in lmer. I am trying 
to model the effect of temperature on metabolic rate in animals (response = 
int.length) at different temperatures (mean.sst), with repeated measurements on 
the same individuals (random effect = female). Ideally I would make a random 
slope and intercept model where the rate can change differently with 
temperature for different individuals:

model-lmer(int.length~mean.sst+(mean.sst|female))

However, I get the following warning message:

Warning message:
Estimated variance-covariance for factor 'female' is singular in: 
`LMEoptimize-`(`*tmp*`, value = list(maxIter = 200L, tolerance = 
1.49011611938477e-08,
summary(model)

Linear mixed-effects model fit by REML
Formula: int.length ~ mean.sst + (mean.sst | female)
   AIC   BIC logLik MLdeviance REMLdeviance
 155.4 164.5  -72.7  142.8145.4
Random effects:
 Groups   NameVariance   Std.Dev.   Corr
 female   (Intercept) 6.8459e-10 2.6165e-05
  mean.sst6.8169e-10 2.6109e-05 -0.065
 Residual 1.3634e+00 1.1676e+00
number of obs: 46, groups: female, 18
Fixed effects:
Estimate Std. Error t value
(Intercept)  48.8249 6.5895   7.409
mean.sst -1.3609 0.2518  -5.406
Correlation of Fixed Effects:
 (Intr)
mean.sst -1.000





If I try and run just a random intercepts model I get similar problems:



model2-lmer(int.length~mean.sst+(1|female))

Warning message: Estimated variance for factor 'female' is effectively zero in: 
`LMEoptimize-`(`*tmp*`, value = list(maxIter = 200L, tolerance = 
1.49011611938477e-08,



I have tried disabling PQL iterations  using control = list(usePQL = FALSE, 
msVerbose=TRUE), following Douglas Bates' recommendation on the mailing list 
archives but I still get a similar message. Does this mean that the variance 
among subjects is too close to zero for estimation of the random effects? I 
compared the random effects model to a linear model with just lm(int.length ~ 
mean.sst) using a likelihood ratio test and got p = 1.0 (which is always 
suspicious). It would actually make sense for there to be negligible variation 
among subjects in their response to temperature, however I am concerned that I 
am making a fundamental error somewhere along the line.



I would greatly appreciate any suggestions you may have.



Best regards



Sam Weber



University of Exeter, UK.




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Re: [R] Invert the sign of a number

2009-10-29 Thread jim holtman

x - -x

On Thu, Oct 29, 2009 at 6:06 AM, Fran100681 franar...@hotmail.it wrote:

 Hello,

 just a simple question

 How can I do to invert the sign of a number? (for example: -4 to 4, 2 to -2
 and so on..)
 I was looking for a specific function in R but I didn't found it...
 thank you

 Francesco
 --
 View this message in context: 
 http://www.nabble.com/Invert-the-sign-of-a-number-tp26110267p26110267.html
 Sent from the R help mailing list archive at Nabble.com.

 __
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 PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
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-- 
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Cincinnati, OH
+1 513 646 9390

What is the problem that you are trying to solve?

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Re: [R] Invert the sign of a number


On 29/10/2009 6:06 AM, Fran100681 wrote:

Hello,

just a simple question

How can I do to invert the sign of a number? (for example: -4 to 4, 2 to -2
and so on..)
I was looking for a specific function in R but I didn't found it...
thank you


You want the unary - operator.  For example,

-x

has opposite signs to x.

Even

--4

will work to give 4, but it will scare C programmers, so it's probably 
best to use parentheses:


-(-4)

Duncan Murdoch

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Re: [R] operation with if/else on a dataframe

2009-10-29 Thread Fran100681


 Ok, thank you a lot!!


Bernardo Rangel tura wrote:
 
 On Thu, 2009-10-29 at 01:47 -0700, Fran100681 wrote:
 Hi to all,
 I have this dataframe (I show the first six rows)
 
 head(table)
 
  A   RFold.Change P.Value 
 Count1   Count2
 1 ENSRNOE002_at 0 1.13 0.601 
  
 1
 2 ENSRNOE009_at 0-1.04 0.733 
  
 3
 3 ENSRNOE020_at 0-1.08 0.680 
  
 0
 4 ENSRNOE021_at 0-1.31 0.201 
  
 2
 5 ENSRNOE023_at 0-1.06 0.643 
  
 3
 6 ENSRNOE024_at 0-1.14 0.403 
  
 3
 
 I would like to generate a function that determine for each row a new
 value
 (resulting in a new vector of values to add to the dataframe). 
 The function should give for every row the same value showed in column
 R.
 However,I need of this because not all the R-values reported in this
 table
 are correctly determined following the criteria mentioned below.
 
 
 These new values calculated by the function must be:
 
 1)UP 
 if all the following conditions are verified in a certain row:
 Fold.Change is = +1.5, 
 P.Value is  0.05 
 Count1 = 2
 
 2)DOWN
 if all the following conditions are verified in a certain row:
 Fold.Change is = -1.5, 
 P.Value is  0.05 
 Count2 = 2
 
 3) 0 if both of previous conditions are not verified
 
 So I have set these fllowing parameters (new objects) because I'll have
 to
 repeat this procedure to different dataframes in which the order of
 columns
 of interest might change (So I can change these parameters depending on
 the
 order of the columns in any different table)
 
 Fold.change -  3 #(because in this table, Fold.Change value is the third
 column and so on...)
 P.Value -  4
 Count1 -  5
 Count2 -  6
 
 [ Quote text]
 
 This is my problem, I cannot use the function to recalculate values in
 R-column for all rows in my dataframe. I don't understand  where is the
 problem, can someone help me?
 Thanks a lot!!
 
 Francesco
 
 Francesco,
 
 
 I think you solve this problem with a simple way.
 Remember in R the most function and operations are vectorized so look
 this example: 
 
 set.seed(123)
 x-rpois(20,5)
 y-rpois(20,15)
 z-rpois(20,10)
 dta-data.frame(x,y,z)
 dta
 dta$NEW-ifelse(x5  y15  z10,UP,
 ifelse(x5  y15  z10,DOWN,
 0))
 dta
 
 First, I use ifelse command to simplify your nested conditional
 situation.
 
 Second, I know that R test this nested condition in order so the first
 position will result test x[1],y[1] and z[1], the second postion will
 result test x[2],y[2] and z[2] ...
 
 The new vector result is the same order the original data.frame so I use
 dta$NEW to create a new column in data.frame 
 
 
 
 -- 
 Bernardo Rangel Tura, M.D,MPH,Ph.D
 National Institute of Cardiology
 Brazil
 
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[R] R crashes

2009-10-29 Thread premmad


My R crashes frequently when run  with huge data.
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Re: [R] Invert the sign of a number

2009-10-29 Thread Peter Flom

Fran100681 franar...@hotmail.it wrote

just a simple question

How can I do to invert the sign of a number? (for example: -4 to 4, 2 to -2
and so on..)
I was looking for a specific function in R but I didn't found it...
thank you


Is there something wrong with *-1 ?

Peter

Peter L. Flom, PhD
Statistical Consultant
Website: www DOT peterflomconsulting DOT com
Writing; http://www.associatedcontent.com/user/582880/peter_flom.html
Twitter:   @peterflom

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[R] The system cannot find the file specified

What is the problem here? I did an install package from the Rgui menu.
Windows Vista

SNIP

trying URL 
'http://cran.cnr.Berkeley.edu/bin/windows/contrib/2.9/mvtnorm_0.9-8.zip'
Content type 'application/zip' length 236089 bytes (230 Kb)
opened URL
downloaded 230 Kb

trying URL 
'http://cran.cnr.Berkeley.edu/bin/windows/contrib/2.9/QRMlib_1.4.4.zip'
Content type 'application/zip' length 2369023 bytes (2.3 Mb)
opened URL
downloaded 2.3 Mb

package 'robustbase' successfully unpacked and MD5 sums checked
package 'fUtilities' successfully unpacked and MD5 sums checked
package 'fEcofin' successfully unpacked and MD5 sums checked
package 'fCalendar' successfully unpacked and MD5 sums checked
package 'fSeries' successfully unpacked and MD5 sums checked
package 'mvtnorm' successfully unpacked and MD5 sums checked
package 'QRMlib' successfully unpacked and MD5 sums checked
Error in normalizePath(path) :
  path[1]=C:\Program Files\R\R-2.9.2\library/QRMlib: The system
cannot find the file specified

SNIP

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Re: [R] R crashes

2009-10-29 Thread John Kane

And

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

--- On Thu, 10/29/09, premmad mtechp...@gmail.com wrote:

 From: premmad mtechp...@gmail.com
 Subject: [R]  R crashes
 To: r-help@r-project.org
 Received: Thursday, October 29, 2009, 6:13 AM

 My R crashes frequently when run  with huge data.
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 mailing list
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 reproducible code.

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Re: [R] column names of a correlation matrix

2009-10-29 Thread Lee William

Thanks! guys for the help.

cheers!
Lee

On Wed, Oct 28, 2009 at 7:03 PM, Tony Plate tpl...@acm.org wrote:

 Here's a simple example that might help you get what you want:

  set.seed(1)
 x - matrix(rnorm(30), ncol=3, dimnames=list(NULL, letters[1:3]))
 (xc - cor(x))

  a  b  c
 a  1.000 -0.3767034 -0.7158385
 b -0.3767034  1.000  0.6040273
 c -0.7158385  0.6040273  1.000

 (cd - data.frame(elt=outer(colnames(xc), colnames(xc), paste,
 sep=:)[upper.tri(xc)], row=row(xc)[upper.tri(xc)],
 col=col(xc)[upper.tri(xc)], cor=xc[upper.tri(xc)]))

  elt row colcor
 1 a:b   1   2 -0.3767034
 2 a:c   1   3 -0.7158385
 3 b:c   2   3  0.6040273

 cd[order(-cd$cor),]

  elt row colcor
 3 b:c   2   3  0.6040273
 1 a:b   1   2 -0.3767034
 2 a:c   1   3 -0.7158385



 If you need something more efficient, try using which(..., arr.ind) to pick
 out matrix style indices, e.g.:

 (ii - which(xc  -0.4  upper.tri(xc), arr.ind=T))

  row col
 a   1   2
 b   2   3

 cbind(ii, cor=xc[ii])

  row colcor
 a   1   2 -0.3767034
 b   2   3  0.6040273



 -- Tony Plate

 Lee William wrote:

 Hi! All,
 I am working on a correlation matrix of 4217x4217 named 'cor_expN'. I wish
 to obtain pairs with highest correlation values. So, I did this

 b=matrix(data=NA,nrow=4217,ncol=1)
 rownames(b)=rownames(cor_expN)
 for(i in 1:4217){b[i,]=max(cor_expN[i,])}
 head(b)

   [,1]
 aaeA_b3241_14 0.7181912
 aaeB_b3240_15 0.7513084
 aaeR_b3243_15 0.7681684
 aaeX_b3242_12 0.5230587
 aas_b2836_14   0.6615927
 aat_b0885_140.6344144

 Now I want the corresponding columns for the above values. For that I
 tried
 this

 c=matrix(data=NA,nrow=4217,ncol=1)
 for(i in 1:4217){b[i,]=colnames(max(cor_expN[i,]))}


 And got the following error:
 Error in b[i, ] = colnames(max(cor_expN[i, ])) : number of items to
 replace
 is not a multiple of replacement length
 Any thoughts?

 Lee

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Re: [R] R 2.10 under Windows XP: glitches and successes

2009-10-29 Thread James W. MacDonald


Hi Duncan,

Duncan Murdoch wrote:

On 28/10/2009 5:54 PM, Ross Boylan wrote:

I just installed 2.10 on XP; ESS is my primary interface.

I seem to be able to access all the help files; under 2.8.1 I was having
seemingly random problems accessing some help topics (including one day
I could and a few days later I couldn't, and vice-versa).  That's good.

I noticed a few glitches.

First, the installer asks if you want text, CHM, or HTML help, but
offers only text or HTML.  I understand CHM has been removed; I guess
this is just an old message.


Yes, already fixed in the patched version.


Second, install.packages(intervals) produced the usual line about
selecting a mirror, but no selection list popped up.  It just sat there
until I interrupted the session and reentered the command with a
pre-specified repos.  I selected custom install but took the defaults; I
noticed there was some optional tk/tcl stuff (which I did not select);
I'm guessing it's required for these selections.  Perhaps it should be
installed by default?


That's something I haven't seen before.  I do get the mirror dialog, and 
it doesn't use tcl/tk, so I'd appreciate hearing if you can spot any 
pattern here (including the pattern I never get it).  (The install 
option is about installing the tcl/tk help files:  mainly useful if 
you're developing with tcl/tk, not necessary for most users.  The Tcl/tk 
code itself is always installed.)  It could be that R is being blocked 
by your anti-virus software; what it's trying to do is to read a list of 
mirrors from


This is a longstanding problem for R under ess/emacs on Windows (and 
maybe Xemacs as well, but I don't know for sure).


This problem manifests itself in install.packages(), but the real 
problem comes from the line


 if (.Platform$OS.type == windows | .Platform$GUI == AQUA)
return(.Internal(select.list(list, preselect, multiple,
title)))

in select.list(). If you run R under ess/emacs and do something as 
simple as


select.list(letters)

it will freeze and you have to use C-g, C-c C-c to 'unfreeze'. 
Paradoxically, after doing so, select.list() will then function correctly.


You can get around this problem by first loading the tcltk package, if 
that is of any help.


This problem only occurs under emacs/ess; if you run R from a DOS prompt 
or via the GUI then select.list() works correctly the first time.


Best,

Jim




http://cran.r-project.org/CRAN_mirrors.csv

using the code in getCRANmirrors().  If that download fails or is 
delayed forever, you won't get the selection list.



Third, every time one asks for help on a new topic there is a
considerable delay (seemed like forever, probably c 15 seconds) before
the help comes up.  My first thought was that the on-demand help was
being assembled, but I didn't see much CPU use.  It might be latency in
talking to MSIE, and this latency might even be from my (Sophos)
antivirus software, which kept warning me about a program suspiciously
launching MSIE.  If you ask for help you already requested there is no
delay (good) and it does not focus on the tab with the relevant help
(which strikes me as the ideal behavior, though it may be hard to do).


When you ask for help the first time, it starts the internal server, but 
that should be fast.  Then it passes the URL to Windows to open, and 
Windows needs to open the browser to show it, which tends to be slow. 
When you ask subsequently, it will send the request to Windows again, 
but if the browser is open, that should be quick.  I'd guess this is 
your antivirus checking out the URL; when I do it (with no antivirus 
checking on such things) it's pretty slow on the first occurence 
(because it needs to start Firefox), then pretty fast after that.



Fourth, I read the previous comments about how more beta testers would
be useful :)


Yep.

Duncan Murdoch



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--
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Biostatistician
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Re: [R] Prediction Error Calculation

2009-10-29 Thread quaildoc


Any help would be appreciated.

quaildoc wrote:
 
 Hello List,
 
 I am fitting a logistic regression model for some presence/absence type
 data.  I have numerous covariates I am fitting to explain variation, and I
 am using AIC to rank models.  However, I would like to report how well my
 best model (s) do at prediction.  I have looked over the archives and the
 web and have come up with something that gives me what I think is the mean
 prediction error, BUT I am not sure of that. I am sort of unfamiliar with
 these types of statistics.  Here is my code:
 
 
 metrics.global-glm(Type~MPI+IJI+ED+PRD+class2+class3+class5,
 family=binomial, data=metrics)## ##Type is my binary response 0 or 1
 
 muhat-metrics.global$fitted.values
 ##assigns the fitted values a name muhat
 global.diag-glm.diag(metrics.global)
 ##creates a the diagnostic values
 cv.err-mean((metrics.global$y-muhat)^2/(1-global.diag$h)^2)
 ###calculates cv.err
 cv.err
 
 
 My main problem is I am unsure how to interpret what cv.err means for my
 model.  I know that h is a leverage statistic for each observation.  I
 would appreciate some interpretation clarification.
 
 Thank you.
 
 
 
 
 

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Re: [R] The system cannot find the file specified

Is 2.10 out? I'll give it a try.

Thanks,
Mark

On Thu, Oct 29, 2009 at 6:45 AM, Duncan Murdoch murd...@stats.uwo.ca wrote:
 On 29/10/2009 9:43 AM, Mark Knecht wrote:

 On Thu, Oct 29, 2009 at 6:39 AM, Duncan Murdoch murd...@stats.uwo.ca
 wrote:

 On 29/10/2009 9:11 AM, Mark Knecht wrote:

 What is the problem here? I did an install package from the Rgui menu.
 Windows Vista

 normalizePath uses the Windows functions GetFullPathName and
 GetLongPathName
 to clean up paths, and one of them returned the error you saw.  Does that
 path exist on your system?

 Duncan Murdoch

 SNIP

 Error in normalizePath(path) :
  path[1]=C:\Program Files\R\R-2.9.2\library/QRMlib: The system
 cannot find the file specified

 SNIP

 C:\Program Files\R\R-2.9.2\library does exist

 C:\Program Files\R\R-2.9.2\library/QRMlib does not.

 As it was the QRMlib package being installed it doesn't surprise me
 that it's not there, but please note the '/' instead of '\' like the
 rest of the path.

 I've seen this a number of times recently. This is the first I've posted
 here.

 Do you have write permission in C:\Program Files\R\R-2.9.2\library?  It
 could be that the installer just tried to create the QRMlib subdir, and
 failed, and that's why it doesn't exist.

 What happens in 2.10.0?

 Duncan Murdoch


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Re: [R] The system cannot find the file specified


On 29/10/2009 9:11 AM, Mark Knecht wrote:

What is the problem here? I did an install package from the Rgui menu.
Windows Vista


normalizePath uses the Windows functions GetFullPathName and 
GetLongPathName to clean up paths, and one of them returned the error 
you saw.  Does that path exist on your system?


Duncan Murdoch



SNIP

trying URL 
'http://cran.cnr.Berkeley.edu/bin/windows/contrib/2.9/mvtnorm_0.9-8.zip'
Content type 'application/zip' length 236089 bytes (230 Kb)
opened URL
downloaded 230 Kb

trying URL 
'http://cran.cnr.Berkeley.edu/bin/windows/contrib/2.9/QRMlib_1.4.4.zip'
Content type 'application/zip' length 2369023 bytes (2.3 Mb)
opened URL
downloaded 2.3 Mb

package 'robustbase' successfully unpacked and MD5 sums checked
package 'fUtilities' successfully unpacked and MD5 sums checked
package 'fEcofin' successfully unpacked and MD5 sums checked
package 'fCalendar' successfully unpacked and MD5 sums checked
package 'fSeries' successfully unpacked and MD5 sums checked
package 'mvtnorm' successfully unpacked and MD5 sums checked
package 'QRMlib' successfully unpacked and MD5 sums checked
Error in normalizePath(path) :
  path[1]=C:\Program Files\R\R-2.9.2\library/QRMlib: The system
cannot find the file specified

SNIP

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Re: [R] R 2.10 under Windows XP: glitches and successes


On 29/10/2009 9:26 AM, James W. MacDonald wrote:

Hi Duncan,

Duncan Murdoch wrote:

On 28/10/2009 5:54 PM, Ross Boylan wrote:

I just installed 2.10 on XP; ESS is my primary interface.

I seem to be able to access all the help files; under 2.8.1 I was having
seemingly random problems accessing some help topics (including one day
I could and a few days later I couldn't, and vice-versa).  That's good.

I noticed a few glitches.

First, the installer asks if you want text, CHM, or HTML help, but
offers only text or HTML.  I understand CHM has been removed; I guess
this is just an old message.

Yes, already fixed in the patched version.


Second, install.packages(intervals) produced the usual line about
selecting a mirror, but no selection list popped up.  It just sat there
until I interrupted the session and reentered the command with a
pre-specified repos.  I selected custom install but took the defaults; I
noticed there was some optional tk/tcl stuff (which I did not select);
I'm guessing it's required for these selections.  Perhaps it should be
installed by default?
That's something I haven't seen before.  I do get the mirror dialog, and 
it doesn't use tcl/tk, so I'd appreciate hearing if you can spot any 
pattern here (including the pattern I never get it).  (The install 
option is about installing the tcl/tk help files:  mainly useful if 
you're developing with tcl/tk, not necessary for most users.  The Tcl/tk 
code itself is always installed.)  It could be that R is being blocked 
by your anti-virus software; what it's trying to do is to read a list of 
mirrors from


This is a longstanding problem for R under ess/emacs on Windows (and 
maybe Xemacs as well, but I don't know for sure).


Okay, I'll leave it to those guys to debug.  If R is doing something it 
shouldn't, let me know what, and I'll try to fix it.  (I'd guess the 
problem is that Emacs makes assumptions about the OS that just aren't 
true in Windows, but it might be a bug we could work around.)


Duncan Murdoch



This problem manifests itself in install.packages(), but the real 
problem comes from the line


  if (.Platform$OS.type == windows | .Platform$GUI == AQUA)
 return(.Internal(select.list(list, preselect, multiple,
 title)))

in select.list(). If you run R under ess/emacs and do something as 
simple as


select.list(letters)

it will freeze and you have to use C-g, C-c C-c to 'unfreeze'. 
Paradoxically, after doing so, select.list() will then function correctly.


You can get around this problem by first loading the tcltk package, if 
that is of any help.


This problem only occurs under emacs/ess; if you run R from a DOS prompt 
or via the GUI then select.list() works correctly the first time.


Best,

Jim



http://cran.r-project.org/CRAN_mirrors.csv

using the code in getCRANmirrors().  If that download fails or is 
delayed forever, you won't get the selection list.



Third, every time one asks for help on a new topic there is a
considerable delay (seemed like forever, probably c 15 seconds) before
the help comes up.  My first thought was that the on-demand help was
being assembled, but I didn't see much CPU use.  It might be latency in
talking to MSIE, and this latency might even be from my (Sophos)
antivirus software, which kept warning me about a program suspiciously
launching MSIE.  If you ask for help you already requested there is no
delay (good) and it does not focus on the tab with the relevant help
(which strikes me as the ideal behavior, though it may be hard to do).
When you ask for help the first time, it starts the internal server, but 
that should be fast.  Then it passes the URL to Windows to open, and 
Windows needs to open the browser to show it, which tends to be slow. 
When you ask subsequently, it will send the request to Windows again, 
but if the browser is open, that should be quick.  I'd guess this is 
your antivirus checking out the URL; when I do it (with no antivirus 
checking on such things) it's pretty slow on the first occurence 
(because it needs to start Firefox), then pretty fast after that.



Fourth, I read the previous comments about how more beta testers would
be useful :)

Yep.

Duncan Murdoch


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Re: [R] The system cannot find the file specified

On Thu, Oct 29, 2009 at 6:39 AM, Duncan Murdoch murd...@stats.uwo.ca wrote:
 On 29/10/2009 9:11 AM, Mark Knecht wrote:

 What is the problem here? I did an install package from the Rgui menu.
 Windows Vista

 normalizePath uses the Windows functions GetFullPathName and GetLongPathName
 to clean up paths, and one of them returned the error you saw.  Does that
 path exist on your system?

 Duncan Murdoch
SNIP
 Error in normalizePath(path) :
  path[1]=C:\Program Files\R\R-2.9.2\library/QRMlib: The system
 cannot find the file specified

 SNIP

C:\Program Files\R\R-2.9.2\library does exist

C:\Program Files\R\R-2.9.2\library/QRMlib does not.

As it was the QRMlib package being installed it doesn't surprise me
that it's not there, but please note the '/' instead of '\' like the
rest of the path.

I've seen this a number of times recently. This is the first I've posted here.

Thanks,
Mark

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[R] fast cumulative matrix multiplication

2009-10-29 Thread Todd Schneider

Hi all,

I am looking for a function like cumprod() that works for matrix
multiplication.

In other words, I have matrices [M1, M2, ..., Mn], and I want to calculate
[M1, M1%*%M2, M1%*%M2%*%M3, ..., M1%*%...%*%Mn] as quickly as possible.
Right now I'm using a for() loop but it seems like there should be a faster
way.

Any help is appreciated!

Thanks,

Todd Schneider
todd.w.schnei...@gmail.com

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Re: [R] The system cannot find the file specified


On 29/10/2009 9:43 AM, Mark Knecht wrote:

On Thu, Oct 29, 2009 at 6:39 AM, Duncan Murdoch murd...@stats.uwo.ca wrote:

On 29/10/2009 9:11 AM, Mark Knecht wrote:

What is the problem here? I did an install package from the Rgui menu.
Windows Vista

normalizePath uses the Windows functions GetFullPathName and GetLongPathName
to clean up paths, and one of them returned the error you saw.  Does that
path exist on your system?

Duncan Murdoch

SNIP

Error in normalizePath(path) :
 path[1]=C:\Program Files\R\R-2.9.2\library/QRMlib: The system
cannot find the file specified

SNIP


C:\Program Files\R\R-2.9.2\library does exist

C:\Program Files\R\R-2.9.2\library/QRMlib does not.

As it was the QRMlib package being installed it doesn't surprise me
that it's not there, but please note the '/' instead of '\' like the
rest of the path.

I've seen this a number of times recently. This is the first I've posted here.


Do you have write permission in C:\Program Files\R\R-2.9.2\library?  It 
could be that the installer just tried to create the QRMlib subdir, and 
failed, and that's why it doesn't exist.


What happens in 2.10.0?

Duncan Murdoch

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[R] Error installing RSPerl.

2009-10-29 Thread Grainne Kerr

Dear list,

I have updated to version R-2.10.0. When I try to load the RSPerl
library I get the following error:

 library(RSPerl)
Error in dyn.load(file, DLLpath = DLLpath, ...) :
  unable to load shared library
'/usr/local/lib/R/library/RSPerl/libs/RSPerl.so':
  /usr/local/lib/R/library/RSPerl/libs/RSPerl.so: undefined symbol:
boot_DB_File__Glob
Error: package/namespace load failed for 'RSPerl'

I do not know how to fix this. Can anyone please help?

I'm runninn R on Ubuntu 9.04

Many thanks,
Grainne.


 sessionInfo()
R version 2.10.0 (2009-10-26)
i686-pc-linux-gnu

locale:
 [1] LC_CTYPE=en_US.UTF-8   LC_NUMERIC=C
 [3] LC_TIME=en_US.UTF-8LC_COLLATE=en_US.UTF-8
 [5] LC_MONETARY=C  LC_MESSAGES=en_US.UTF-8
 [7] LC_PAPER=en_US.UTF-8   LC_NAME=C
 [9] LC_ADDRESS=C   LC_TELEPHONE=C
[11] LC_MEASUREMENT=en_US.UTF-8 LC_IDENTIFICATION=C

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


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Re: [R] Year and Month extraction from Date object.

2009-10-29 Thread Gavin Simpson


On Tue, 2009-10-27 at 14:49 -0700, rkevinbur...@charter.net wrote:
 Hello,
 
 I have seen much discussion on Date. But I can't seem to do this
 simple operation. I can convert a string to a date:
 
 d - as.Date(DATE, format=%m/%d/%Y)
 
 But what I want to do is extract the year and month so I can construct
 an element in a ts object. Ideally I would like to see d$year but that
 doesn't seem to be available. Once I have a Date object how can I get
 an integer year?

as.numeric(format(d, format = %Y))
as.numeric(format(d, format = %m))

and so on.

Look at the formats in ?strftime for further details.

HTH

G

 
 Thank you.
 
 Kevin
 
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 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
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Re: [R] x11(title=' ')

2009-10-29 Thread ms.com


Dear all
sorry that i am late to reply for the question about my OS, I am using MS 
Windows XP 2003 OS
thanking you
Warm RegardMSNepal



 Date: Wed, 28 Oct 2009 22:20:50 +1100
 From: j...@bitwrit.com.au
 CC: r-help@r-project.org
 Subject: Re: [R] x11(title='  ')
 
 On 10/28/2009 09:51 PM, Prof Brian Ripley wrote:
  ...
  Are you perchance x11() on Windows, when you should be using 
  windows()?  The posting guide asked you to tell us your OS, amongst 
  other things 
 I say, my dear professor, if this MS chap who has used a Microsoft 
 email client and a Microsoft email service turns out to be using Linux, 
 he's covered his tracks pretty well.
 
 Jim
 
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_


WLMTAGL:ON:WL:en-US:WWL_WIN_evergreen2:102009
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[R] Trouble retrieving data (.xls) from folder on my computer.

2009-10-29 Thread tobiasfa


Using gdata on windows and im having trouble to retrieve an excel file from a
folder on my computer.

This is my dummy:
R  download.file(http://people.su.se/~lundh/data/cpi_kpix.rda;,
+ cpi_kpix.rda)

R  URL - http://people.su.se/;
R  PATH - ~lundh/data/
R  FILE - cpi_kpix.rda
R  download.file(paste(URL,PATH,FILE,sep=),
+ cpi_kpix.rda)

R  library(gdata)
R  FILE - Stock_Returns_1931_2002.xls
R  stock.xls - read.xls(paste(URL, PATH, FILE, sep = ),
+ header = TRUE)[, 3:4]
Downloading...
Done.
Converting xls file to csv file... Done.
Reading csv file... Done.  

Im not suppose to download the file though its already on my computer. How
do i type in the path så R finds my .xls??

kind regards Tobias

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Re: [R] What happen for Negative binomial link in Lmer fonction?

2009-10-29 Thread E. Robardet

Thank you for your answers,

I have an exemple of that i was using:

m1a-lmer(atpos~ninter+saison+milieu*zone+(1|code),family=neg.bin(0.429),method=Laplace,data=manu)
summary(m1a)
Generalized linear mixed model fit using Laplace
Formula: atpos ~ ninter + saison + milieu * zone + (1 | code)
Data: manu
Family: Negative Binomial(log link)
AIC BIC logLik deviance
125.1 147.6 -54.57109.1

I think It was the version lme4 0.9975-10.
Unfortunately, I have this version no more available on my computer..
I wonder if this old results are still valid..

Ben Bolker wrote:

ROBARDET Emmanuelle wrote:

Dear R users,
I'm performing some GLMMs analysis with a negative binomial link.
I already performed such analysis some months ago with the lmer()
function but when I tried it today I encountered this problem:
Erreur dans famType(glmFit$family) : unknown GLM family: 'Negative
Binomial'

Does anyone know if the negative binomial family has been removed from
this function?
I really appreciate any response.
Emmanuelle

I would be extremely surprised if this worked in the past; to
the best of my knowledge the negative binomial family has
never been implemented in lmer. One could in principle
do a glmmPQL fit with the negative binomial family
(with a fixed value of the overdispersion parameter).
glmmADMB is another option.
Can you say which version etc. you were using???

Follow-ups should probably be sent to r-sig-mixed-mod...@r-project.org

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Re: [R] Re ading and Creating Shape Files

2009-10-29 Thread cls59



PDXRugger wrote:
 
 Hello R Community,
I have imported a dataset which contain X Y coordinates and would like
 to recreate a shape file after some data analysis.  What i have done is to
 import some taxlot data and join them based on some criteria.  I want to
 check to see how well the joining went by reviewing the results in GIS.  
 
 A couple things.  I cant seem to import a shape file correctly using the
 maptools package and the readShapeSpatial.  I have tried 
 
 Building=file(data/input/BuildingShape/Building.shp)
 Bldg-readShapeSpatial(fn=data/input/BuildingShape/Building,proj4string=NAD83)
 #--
 Bldg-readShapeSpatial(data/input/BuildingShape/Building,proj4string=NAD83)
 #---
 Building=file(data/input/BuildingShape/Building.shp)
 Bldg-readShapeSpatial(Building,proj4string=NAD83)
 
 I know i am mis interpreting the documentation but it doesnt seem like it
 is very complicated so i am of course confused.
 
 

I haven't used the maptools package for this kind of operation, so I offer
any specific advice. However, the value of proj4string seems to be a little
odd-- unless NAD83 is a variable that contains a string. It should probably
be a quoted list of PROJ4 declarations such as:

proj4string = '+proj=longlat +datum=NAD83'

If maptools is leveraging the sp package, then you probably need to enclose
the string in the CRS() function:

proj4string = CRS('+proj=longlat +datum=NAD83')

Note that the above example assumes coordinates are in lat/lon using the
NAD83 datum. If your data is in a different projection, such as UTM, you
will need to change the PROJ4 string accordingly.



PDXRugger wrote:
 
 
 Also, i am wondering if i can create a shape file by simply using XY
 coordinates from a data frame.
 So for:
 
 
 Ycoord=c( 865296.4, 865151.5, 865457.0 ,865363.4 ,865311.0, 865260.9
 ,865210.7 ,865173.3,
 865123.6 ,865038.2 ,864841.1 ,864745.4 ,864429.1 ,864795.6 ,864334.9
 ,864882.0)
 
 Xcoord=c( 4227640 ,4227816 ,4228929 ,4228508 ,4229569 ,4229498 ,4226747,
 4226781, 4229597,
 4229204, 4228910, 4228959 ,4229465 ,4229794 ,4229596 ,4229082)
 
 Lot-c(1900 , 2000,  2100  , 100   ,200  , 300,   400 ,  500 ,  600 ,  701
 ,  900 , 1000 , 1100,
   300   ,100,   200)   
 
 XYcoord-spCbind(Ycoord,Xcoord) #doesnt work so
 
 XYcoord=c(Ycoord,Xcoord)
 
 TaxLots-cbind(Ycoord,Xcoord,Lot)
 
 writeSpatialShape(XYcoord, TaxLots..,
 file=data/input/test/Taxlots,strictFilename=FALSE)
 
 
 
 So help reading in shape files and then creating them using XY coordinates
 if possible
 Any help would be appreciated.  Thank you.
 
 
 

Maybe maptools provides a nice way to do this-- again I haven't used it
much. My shapefile workflow usually centers on the sp and rgdal packages.
First, sp is used to create a spatial object that holds the coordinates and
data. Assuming you have point data:

  require(rgdal)
  lots - SpatialPointsDataFrame( coords = cbind(Xcoord,Ycoord), data =
data.frame( Lot = Lot ))

*Note that SpatialPointsDataFrame also takes a proj4string argument of the
form:

proj4string = CRS( 'proj declarations' )

I have omitted it since I don't know what projection your data is in.


You can then create a shapefile using the writeOGR() routine in rgdal:

  writeOGR( lots, dsn = 'tstShapefile', layer = 'tstShapefile', driver='ESRI
Shapefile')


The readOGR() function can also be used to read a shapefile-- note that you
give it the name of the directory containing the shapefile components and
not the name of an individual component such as 'shapefile.shp'.

Hope this helps!

-Charlie

-
Charlie Sharpsteen
Undergraduate
Environmental Resources Engineering
Humboldt State University
-- 
View this message in context: 
http://www.nabble.com/Reading-and-Creating-Shape-Files-tp26098828p26114143.html
Sent from the R help mailing list archive at Nabble.com.

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

Re: [R] R 2.10 under Windows XP: glitches and successes

2009-10-29 Thread Blair Christian

On a related note, on windows xp + ESS + R 2.9.2, I have long waits when I
run:

browseURL( unitTestOutFile,  browser=C:/Program Files/Mozilla
Firefox/firefox.exe)

Firefox is always already open, and sometimes the old test results are
already open too.  Sometimes it takes minutes to open up, sometimes using
submit+step in emacs seems to be correlated with this getting this to go a
bit faster (but I can't tell what causes the huge variation in time opening
the webpage)...  I'm upgrading to 2.10.x next time I can schedule it on my
work machine.

Blair


On Thu, Oct 29, 2009 at 9:42 AM, Duncan Murdoch murd...@stats.uwo.cawrote:

 On 29/10/2009 9:26 AM, James W. MacDonald wrote:

 Hi Duncan,

 Duncan Murdoch wrote:

 On 28/10/2009 5:54 PM, Ross Boylan wrote:

 I just installed 2.10 on XP; ESS is my primary interface.

 I seem to be able to access all the help files; under 2.8.1 I was having
 seemingly random problems accessing some help topics (including one day
 I could and a few days later I couldn't, and vice-versa).  That's good.

 I noticed a few glitches.

 First, the installer asks if you want text, CHM, or HTML help, but
 offers only text or HTML.  I understand CHM has been removed; I guess
 this is just an old message.

 Yes, already fixed in the patched version.

  Second, install.packages(intervals) produced the usual line about
 selecting a mirror, but no selection list popped up.  It just sat there
 until I interrupted the session and reentered the command with a
 pre-specified repos.  I selected custom install but took the defaults; I
 noticed there was some optional tk/tcl stuff (which I did not select);
 I'm guessing it's required for these selections.  Perhaps it should be
 installed by default?

 That's something I haven't seen before.  I do get the mirror dialog, and
 it doesn't use tcl/tk, so I'd appreciate hearing if you can spot any pattern
 here (including the pattern I never get it).  (The install option is about
 installing the tcl/tk help files:  mainly useful if you're developing with
 tcl/tk, not necessary for most users.  The Tcl/tk code itself is always
 installed.)  It could be that R is being blocked by your anti-virus
 software; what it's trying to do is to read a list of mirrors from


 This is a longstanding problem for R under ess/emacs on Windows (and maybe
 Xemacs as well, but I don't know for sure).


 Okay, I'll leave it to those guys to debug.  If R is doing something it
 shouldn't, let me know what, and I'll try to fix it.  (I'd guess the problem
 is that Emacs makes assumptions about the OS that just aren't true in
 Windows, but it might be a bug we could work around.)

 Duncan Murdoch



 This problem manifests itself in install.packages(), but the real problem
 comes from the line

  if (.Platform$OS.type == windows | .Platform$GUI == AQUA)
 return(.Internal(select.list(list, preselect, multiple,
 title)))

 in select.list(). If you run R under ess/emacs and do something as simple
 as

 select.list(letters)

 it will freeze and you have to use C-g, C-c C-c to 'unfreeze'.
 Paradoxically, after doing so, select.list() will then function correctly.

 You can get around this problem by first loading the tcltk package, if
 that is of any help.

 This problem only occurs under emacs/ess; if you run R from a DOS prompt
 or via the GUI then select.list() works correctly the first time.

 Best,

 Jim


  http://cran.r-project.org/CRAN_mirrors.csv

 using the code in getCRANmirrors().  If that download fails or is delayed
 forever, you won't get the selection list.

  Third, every time one asks for help on a new topic there is a
 considerable delay (seemed like forever, probably c 15 seconds) before
 the help comes up.  My first thought was that the on-demand help was
 being assembled, but I didn't see much CPU use.  It might be latency in
 talking to MSIE, and this latency might even be from my (Sophos)
 antivirus software, which kept warning me about a program suspiciously
 launching MSIE.  If you ask for help you already requested there is no
 delay (good) and it does not focus on the tab with the relevant help
 (which strikes me as the ideal behavior, though it may be hard to do).

 When you ask for help the first time, it starts the internal server, but
 that should be fast.  Then it passes the URL to Windows to open, and Windows
 needs to open the browser to show it, which tends to be slow. When you ask
 subsequently, it will send the request to Windows again, but if the browser
 is open, that should be quick.  I'd guess this is your antivirus checking
 out the URL; when I do it (with no antivirus checking on such things) it's
 pretty slow on the first occurence (because it needs to start Firefox), then
 pretty fast after that.

  Fourth, I read the previous comments about how more beta testers would
 be useful :)

 Yep.

 Duncan Murdoch

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 R-help@r-project.org mailing list

Re: [R] The system cannot find the file specified

Thanks Duncan. 2.10 seems to install things correctly.

Cheers,
Mark

On Thu, Oct 29, 2009 at 6:52 AM, Mark Knecht markkne...@gmail.com wrote:
 Is 2.10 out? I'll give it a try.

 Thanks,
 Mark

 On Thu, Oct 29, 2009 at 6:45 AM, Duncan Murdoch murd...@stats.uwo.ca wrote:
 On 29/10/2009 9:43 AM, Mark Knecht wrote:

 On Thu, Oct 29, 2009 at 6:39 AM, Duncan Murdoch murd...@stats.uwo.ca
 wrote:

 On 29/10/2009 9:11 AM, Mark Knecht wrote:

 What is the problem here? I did an install package from the Rgui menu.
 Windows Vista

 normalizePath uses the Windows functions GetFullPathName and
 GetLongPathName
 to clean up paths, and one of them returned the error you saw.  Does that
 path exist on your system?

 Duncan Murdoch

 SNIP

 Error in normalizePath(path) :
  path[1]=C:\Program Files\R\R-2.9.2\library/QRMlib: The system
 cannot find the file specified

 SNIP

 C:\Program Files\R\R-2.9.2\library does exist

 C:\Program Files\R\R-2.9.2\library/QRMlib does not.

 As it was the QRMlib package being installed it doesn't surprise me
 that it's not there, but please note the '/' instead of '\' like the
 rest of the path.

 I've seen this a number of times recently. This is the first I've posted
 here.

 Do you have write permission in C:\Program Files\R\R-2.9.2\library?  It
 could be that the installer just tried to create the QRMlib subdir, and
 failed, and that's why it doesn't exist.

 What happens in 2.10.0?

 Duncan Murdoch



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

Re: [R] Trouble retrieving data (.xls) from folder on my computer.



On Oct 29, 2009, at 9:36 AM, tobiasfa wrote:



Using gdata on windows and im having trouble to retrieve an excel  
file from a

folder on my computer.

This is my dummy:
R  download.file(http://people.su.se/~lundh/data/cpi_kpix.rda;,
+ cpi_kpix.rda)

R  URL - http://people.su.se/;
R  PATH - ~lundh/data/
R  FILE - cpi_kpix.rda
R  download.file(paste(URL,PATH,FILE,sep=),
+ cpi_kpix.rda)

R  library(gdata)
R  FILE - Stock_Returns_1931_2002.xls
R 
Downloading...
Done.
Converting xls file to csv file... Done.
Reading csv file... Done.

Im not suppose to download the file though its already on my  
computer. How

do i type in the path så R finds my .xls??


It's probably not where you think it is. Try using:

stock.xls - read.xls(file.choose(),  header = TRUE)[, 3:4]

... and navigating your directory tree.



--

David Winsemius, MD
Heritage Laboratories
West Hartford, CT

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Re: [R] singular variance-covariance warning in lmer

2009-10-29 Thread Ista Zahn

Hi Sam,
Just a stab in the dark here, but is your grouping variable really
female? What does

str(data.frame(mean.sst, female)

look like? How many levels does female have?

-Ista

On Thu, Oct 29, 2009 at 7:10 AM, Weber, Sam sam.we...@exeter.ac.uk wrote:
 Dear R Users,

 I was hoping for some help with a recurrent error message in lmer. I am 
 trying to model the effect of temperature on metabolic rate in animals 
 (response = int.length) at different temperatures (mean.sst), with repeated 
 measurements on the same individuals (random effect = female). Ideally I 
 would make a random slope and intercept model where the rate can change 
 differently with temperature for different individuals:

 model-lmer(int.length~mean.sst+(mean.sst|female))

 However, I get the following warning message:

 Warning message:
 Estimated variance-covariance for factor 'female' is singular in: 
 `LMEoptimize-`(`*tmp*`, value = list(maxIter = 200L, tolerance = 
 1.49011611938477e-08,
 summary(model)

 Linear mixed-effects model fit by REML
 Formula: int.length ~ mean.sst + (mean.sst | female)
   AIC   BIC logLik MLdeviance REMLdeviance
  155.4 164.5  -72.7      142.8        145.4
 Random effects:
  Groups   Name        Variance   Std.Dev.   Corr
  female   (Intercept) 6.8459e-10 2.6165e-05
          mean.sst    6.8169e-10 2.6109e-05 -0.065
  Residual             1.3634e+00 1.1676e+00
 number of obs: 46, groups: female, 18
 Fixed effects:
            Estimate Std. Error t value
 (Intercept)  48.8249     6.5895   7.409
 mean.sst     -1.3609     0.2518  -5.406
 Correlation of Fixed Effects:
         (Intr)
 mean.sst -1.000





 If I try and run just a random intercepts model I get similar problems:



 model2-lmer(int.length~mean.sst+(1|female))

 Warning message: Estimated variance for factor 'female' is effectively zero 
 in: `LMEoptimize-`(`*tmp*`, value = list(maxIter = 200L, tolerance = 
 1.49011611938477e-08,



 I have tried disabling PQL iterations  using control = list(usePQL = FALSE, 
 msVerbose=TRUE), following Douglas Bates' recommendation on the mailing list 
 archives but I still get a similar message. Does this mean that the variance 
 among subjects is too close to zero for estimation of the random effects? I 
 compared the random effects model to a linear model with just lm(int.length ~ 
 mean.sst) using a likelihood ratio test and got p = 1.0 (which is always 
 suspicious). It would actually make sense for there to be negligible 
 variation among subjects in their response to temperature, however I am 
 concerned that I am making a fundamental error somewhere along the line.



 I would greatly appreciate any suggestions you may have.



 Best regards



 Sam Weber



 University of Exeter, UK.




        [[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.




-- 
Ista Zahn
Graduate student
University of Rochester
Department of Clinical and Social Psychology
http://yourpsyche.org

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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] Trouble retrieving data (.xls) from folder on my computer.

2009-10-29 Thread Gabor Grothendieck

If its already on your computer try one of these:

xls - read.xls(myfile.xls)   # if its in current directory

xls - read.xls(/path/to/myfile.xls)   # to specify path

xls - read.xls(file.choose())   # use explorer style interface to choose




On Thu, Oct 29, 2009 at 9:36 AM, tobiasfa tobias.farnly...@gmail.com wrote:

 Using gdata on windows and im having trouble to retrieve an excel file from a
 folder on my computer.

 This is my dummy:
 R  download.file(http://people.su.se/~lundh/data/cpi_kpix.rda;,
 + cpi_kpix.rda)

 R  URL - http://people.su.se/;
 R  PATH - ~lundh/data/
 R  FILE - cpi_kpix.rda
 R  download.file(paste(URL,PATH,FILE,sep=),
 + cpi_kpix.rda)

 R  library(gdata)
 R  FILE - Stock_Returns_1931_2002.xls
 R  stock.xls - read.xls(paste(URL, PATH, FILE, sep = ),
 + header = TRUE)[, 3:4]
 Downloading...
 Done.
 Converting xls file to csv file... Done.
 Reading csv file... Done.

 Im not suppose to download the file though its already on my computer. How
 do i type in the path så R finds my .xls??

 kind regards Tobias

 --
 View this message in context: 
 http://www.nabble.com/Trouble-retrieving-data-%28.xls%29-from-folder-on-my-computer.-tp26113172p26113172.html
 Sent from the R help mailing list archive at Nabble.com.

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


__
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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] R 2.10 under Windows XP: glitches and successes


On 29/10/2009 10:23 AM, Blair Christian wrote:

On a related note, on windows xp + ESS + R 2.9.2, I have long waits when I
run:

browseURL( unitTestOutFile,  browser=C:/Program Files/Mozilla
Firefox/firefox.exe)

Firefox is always already open, and sometimes the old test results are
already open too.  Sometimes it takes minutes to open up, sometimes using
submit+step in emacs seems to be correlated with this getting this to go a
bit faster (but I can't tell what causes the huge variation in time opening
the webpage)...  I'm upgrading to 2.10.x next time I can schedule it on my
work machine.



You should probably report ESS problems to their mailing list.  I don't 
use it or support it, and I don't know if they'll read this message 
since it doesn't mention ESS in the subject line.


Duncan Murdoch



Blair


On Thu, Oct 29, 2009 at 9:42 AM, Duncan Murdoch murd...@stats.uwo.cawrote:


On 29/10/2009 9:26 AM, James W. MacDonald wrote:


Hi Duncan,

Duncan Murdoch wrote:


On 28/10/2009 5:54 PM, Ross Boylan wrote:


I just installed 2.10 on XP; ESS is my primary interface.

I seem to be able to access all the help files; under 2.8.1 I was having
seemingly random problems accessing some help topics (including one day
I could and a few days later I couldn't, and vice-versa).  That's good.

I noticed a few glitches.

First, the installer asks if you want text, CHM, or HTML help, but
offers only text or HTML.  I understand CHM has been removed; I guess
this is just an old message.


Yes, already fixed in the patched version.

 Second, install.packages(intervals) produced the usual line about

selecting a mirror, but no selection list popped up.  It just sat there
until I interrupted the session and reentered the command with a
pre-specified repos.  I selected custom install but took the defaults; I
noticed there was some optional tk/tcl stuff (which I did not select);
I'm guessing it's required for these selections.  Perhaps it should be
installed by default?


That's something I haven't seen before.  I do get the mirror dialog, and
it doesn't use tcl/tk, so I'd appreciate hearing if you can spot any pattern
here (including the pattern I never get it).  (The install option is about
installing the tcl/tk help files:  mainly useful if you're developing with
tcl/tk, not necessary for most users.  The Tcl/tk code itself is always
installed.)  It could be that R is being blocked by your anti-virus
software; what it's trying to do is to read a list of mirrors from


This is a longstanding problem for R under ess/emacs on Windows (and maybe
Xemacs as well, but I don't know for sure).


Okay, I'll leave it to those guys to debug.  If R is doing something it
shouldn't, let me know what, and I'll try to fix it.  (I'd guess the problem
is that Emacs makes assumptions about the OS that just aren't true in
Windows, but it might be a bug we could work around.)

Duncan Murdoch




This problem manifests itself in install.packages(), but the real problem
comes from the line

 if (.Platform$OS.type == windows | .Platform$GUI == AQUA)
return(.Internal(select.list(list, preselect, multiple,
title)))

in select.list(). If you run R under ess/emacs and do something as simple
as

select.list(letters)

it will freeze and you have to use C-g, C-c C-c to 'unfreeze'.
Paradoxically, after doing so, select.list() will then function correctly.

You can get around this problem by first loading the tcltk package, if
that is of any help.

This problem only occurs under emacs/ess; if you run R from a DOS prompt
or via the GUI then select.list() works correctly the first time.

Best,

Jim


 http://cran.r-project.org/CRAN_mirrors.csv

using the code in getCRANmirrors().  If that download fails or is delayed
forever, you won't get the selection list.

 Third, every time one asks for help on a new topic there is a

considerable delay (seemed like forever, probably c 15 seconds) before
the help comes up.  My first thought was that the on-demand help was
being assembled, but I didn't see much CPU use.  It might be latency in
talking to MSIE, and this latency might even be from my (Sophos)
antivirus software, which kept warning me about a program suspiciously
launching MSIE.  If you ask for help you already requested there is no
delay (good) and it does not focus on the tab with the relevant help
(which strikes me as the ideal behavior, though it may be hard to do).


When you ask for help the first time, it starts the internal server, but
that should be fast.  Then it passes the URL to Windows to open, and Windows
needs to open the browser to show it, which tends to be slow. When you ask
subsequently, it will send the request to Windows again, but if the browser
is open, that should be quick.  I'd guess this is your antivirus checking
out the URL; when I do it (with no antivirus checking on such things) it's
pretty slow on the first occurence (because it needs to start Firefox), then
pretty fast after that.

 Fourth, I read the

Re: [R] Error installing RSPerl.

2009-10-29 Thread Duncan Temple Lang

Hi Grainne

There is one likely cause. But before getting into the explanation,
can you send me the output from when you installed the package, e.g. the output 
from

 R CMD INSTALL   RSPerl

and any configuration arguments you specified.

You can send this to me off-list and we can summarize at the end.

 Thanks,
   D.

Grainne Kerr wrote:
 Dear list,
 
 I have updated to version R-2.10.0. When I try to load the RSPerl
 library I get the following error:
 
 library(RSPerl)
 Error in dyn.load(file, DLLpath = DLLpath, ...) :
   unable to load shared library
 '/usr/local/lib/R/library/RSPerl/libs/RSPerl.so':
   /usr/local/lib/R/library/RSPerl/libs/RSPerl.so: undefined symbol:
 boot_DB_File__Glob
 Error: package/namespace load failed for 'RSPerl'
 
 I do not know how to fix this. Can anyone please help?
 
 I'm runninn R on Ubuntu 9.04
 
 Many thanks,
 Grainne.
 
 sessionInfo()
 R version 2.10.0 (2009-10-26)
 i686-pc-linux-gnu
 
 locale:
  [1] LC_CTYPE=en_US.UTF-8   LC_NUMERIC=C
  [3] LC_TIME=en_US.UTF-8LC_COLLATE=en_US.UTF-8
  [5] LC_MONETARY=C  LC_MESSAGES=en_US.UTF-8
  [7] LC_PAPER=en_US.UTF-8   LC_NAME=C
  [9] LC_ADDRESS=C   LC_TELEPHONE=C
 [11] LC_MEASUREMENT=en_US.UTF-8 LC_IDENTIFICATION=C
 
 attached base packages:
 [1] stats graphics  grDevices utils datasets  methods   base
 
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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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and provide commented, minimal, self-contained, reproducible code.

[R] The 'subset matching' challenge

2009-10-29 Thread Yvonnick Noël


Dear all,

The following problem just has been submitted to me by an accountant.

In his new job, he has to close some old accounts. He has yearly 
amounts, and a list of products that have been bought over the years, at 
certain prices for which he has an exhaustive record. The problem is: He 
does not know what product was bought this or that year (don't ask). He 
does not want to find back the real story, but just write realistic 
accounts, for which the sum of a subset of product prices will give the 
exact yearly amount.


Here is a real example from his data:

# A list of 64 product prices
products = 
c(30500,30500,30500,30500,42000,42000,42000,42000,42000,42000,42000,42000,42000,42000,71040,90900,76950,35100,71190,

53730,456000,70740,70740,533600,83800,59500,27465,28000,28000,28000,28000,28000,26140,49600,77000,123289,27000,27000,27000,
27000,27000,27000,8,33000,33000,55000,77382,48048,51186,4,35000,21716,63051,15025,15025,15025,15025,80,111,
59700,25908,829350,1198000,1031655)

# Global amount
amount = 4748652

Now he wants to find all subsets of the 'product' vector which sums to 
'amount'.


I wrote the following code, which is clearly not optimal:

# Create a matrix of subsets of size r among the integer set 1:n
subsets - function(n, r, v = 1:n) {
 if(r = 0) vector(mode(v), 0)
 else if(r = n) v[1:n]
 else rbind(cbind(v[1], Recall(n-1, r-1, v[-1])),Recall(n-1, r, v[-1]))
}

# Main function
find.amount = function(amount,products) {

 if(sum(products)amount) {
   cat(There is no solution.\n)
   return()
 }

 l = length(products)
 cat(\nThere are,l,product prices\n\n)
 names(products) = paste(Product,1:l,sep=)
 products = sort(products)

 for(i in 2:l) {

   # If the sum of the i smallest prices is greater than amount, then stop
   if(sum(products[1:i])amount) break

   # Look for matching subsets only in the case when the sum of i 
largest prices is greater than amount

   if(sum(rev(products)[1:i])=amount) {
 # Generates all subsets of i indicies in 1:l
 subs = subsets(l,i)
 nl = nrow(subs)
 nc = ncol(subs)

 # Compute sums of corresponding price subsets
 sums = rowSums(matrix(products[subs],nl,nc))

 # Which ones match the global amount ?
 w = which(sums == amount)
 how.many = length(w)
 if(how.many0) {
   cat(\n-- There are,how.many,solutions with,nc,products :\n)
   for(j in 1:how.many) {
 print(products[subs[w[j],]])
   }
 }
 else cat(\n-- There is no solution with,nc,products.\n)
   }
   else cat(\n-- There is no solution with,i,products.\n)
 }
}


Then I can use these functions on a smaller example:

 find.amount(4,c(1,1,1,1,2,2))

and a number of matching subsets are provided. The problem is: This 
approach creates a whole matrix of subsets of r integers among 1:n, 
which rapidly gives huge matrices, and this is clearly not optimal for 
the real data provided above.


Would anyone have a suggestion as to an alternative and more efficient 
strategy?


Good luck,

Yvonnick Noel
University of Brittany, Rennes 2
France

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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] Invert the sign of a number

On Thu, 29 Oct 2009 09:05:22 -0400 Duncan Murdoch murd...@stats.uwo.ca 
wrote:
 You want the unary - operator.  For example,
 
 -x

It even has a nice help page:

?-

-- 
Karl Ove Hufthammer

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Re: [R] The 'subset matching' challenge

2009-10-29 Thread Ben Bolker




yvonnick noel wrote:
 
 Dear all,
 
 The following problem just has been submitted to me by an accountant.
 
 In his new job, he has to close some old accounts. He has yearly 
 amounts, and a list of products that have been bought over the years, at 
 certain prices for which he has an exhaustive record. The problem is: He 
 does not know what product was bought this or that year (don't ask). He 
 does not want to find back the real story, but just write realistic 
 accounts, for which the sum of a subset of product prices will give the 
 exact yearly amount.
 
 Here is a real example from his data:
 
 # A list of 64 product prices
 products = 
 c(30500,30500,30500,30500,42000,42000,42000,42000,42000,42000,42000,42000,42000,42000,71040,90900,76950,35100,71190,
 53730,456000,70740,70740,533600,83800,59500,27465,28000,28000,28000,28000,28000,26140,49600,77000,123289,27000,27000,27000,
 27000,27000,27000,8,33000,33000,55000,77382,48048,51186,4,35000,21716,63051,15025,15025,15025,15025,80,111,
 59700,25908,829350,1198000,1031655)
 
 # Global amount
 amount = 4748652
 
 Now he wants to find all subsets of the 'product' vector which sums to 
 'amount'.
 
 I wrote the following code, which is clearly not optimal:
 
 # Create a matrix of subsets of size r among the integer set 1:n
 subsets - function(n, r, v = 1:n) {
   if(r = 0) vector(mode(v), 0)
   else if(r = n) v[1:n]
   else rbind(cbind(v[1], Recall(n-1, r-1, v[-1])),Recall(n-1, r, v[-1]))
 }
 
 # Main function
 find.amount = function(amount,products) {
  
   if(sum(products)amount) {
 cat(There is no solution.\n)
 return()
   }
  
   l = length(products)
   cat(\nThere are,l,product prices\n\n)
   names(products) = paste(Product,1:l,sep=)
   products = sort(products)
 
   for(i in 2:l) {
 
 # If the sum of the i smallest prices is greater than amount, then
 stop
 if(sum(products[1:i])amount) break
 
 # Look for matching subsets only in the case when the sum of i 
 largest prices is greater than amount
 if(sum(rev(products)[1:i])=amount) {
   # Generates all subsets of i indicies in 1:l
   subs = subsets(l,i)
   nl = nrow(subs)
   nc = ncol(subs)
 
   # Compute sums of corresponding price subsets
   sums = rowSums(matrix(products[subs],nl,nc))
 
   # Which ones match the global amount ?
   w = which(sums == amount)
   how.many = length(w)
   if(how.many0) {
 cat(\n-- There are,how.many,solutions with,nc,products
 :\n)
 for(j in 1:how.many) {
   print(products[subs[w[j],]])
 }
   }
   else cat(\n-- There is no solution with,nc,products.\n)
 }
 else cat(\n-- There is no solution with,i,products.\n)
   }
 }
 
 
 Then I can use these functions on a smaller example:
 
   find.amount(4,c(1,1,1,1,2,2))
 
 and a number of matching subsets are provided. The problem is: This 
 approach creates a whole matrix of subsets of r integers among 1:n, 
 which rapidly gives huge matrices, and this is clearly not optimal for 
 the real data provided above.
 
 Would anyone have a suggestion as to an alternative and more efficient 
 strategy?
 
 Good luck,
 
 Yvonnick Noel
 University of Brittany, Rennes 2
 France
 
 

  I believe this is the knapsack problem.
RSiteSearch(knapsack) might help a little bit.

http://xkcd.com/287/

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[R] change in default behavior of ?functionname in R 2.10?

2009-10-29 Thread Jenny Drnevich

Hi everyone,

I just installed R 2.10 using the Windows binary. At some point in 
the past using the Windows binary installation, the default behavior 
of ?functionname changed from plain text to a nice html interface. 
Now with R 2.10, the default behavior has changed back to the plain 
text help. I assume this has some thing to do with these changes 
listed for R 2.10, but I don't quite understand what it's saying or 
how to change my system back to the html-style help:


Package help is now converted from Rd by the R-based converters
 that were first introduced in 2.9.0.  This means

 - Packages that were installed by R-devel after 2009-08-09
   should not be used with earlier versions of R, and most
   aspects of package help (including the runnable examples)
   will be missing if they are so used.

 - Text, HTML and latex help and examples for packages
   installed under the new system are converted on-demand from
   stored parsed Rd files.  (Conversions stored in packages
   installed under R  2.10.0 are used if no parsed Rd files
   are found.  It is recommended that such packages be
   re-installed.)


 o   HTML help is now generated dynamically using an HTTP server
 running in the R process and listening on the loopback
 interface.

 - Those worried about security implications of such a server
   can disable it by setting the environment variable
   R_DISABLE_HTTPD to a non-empty value.  This disables
   help.start() and HTML help (so text help is shown instead).

 - The Java/Javascript search engine has been replaced by an
   HTML interface to help.search().  help.start() no longer has
   an argument 'searchEngine' as it is no longer needed.

 - The HTML help can now locate cross-references of the form
   \link[pkg]{foo} and \link[pkg:foo]{bar} where 'foo' is an
   alias in the package, rather than the documented (basename
   of a) filename (since the documentation has been much
   ignored).


I'd appreciate any help, as the html-style help is so much more convenient!

Thanks,
Jenny

  sessionInfo()
R version 2.10.0 (2009-10-26)
i386-pc-mingw32

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

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

other attached packages:
  [1] ath1121501cdf_2.5.0  limma_3.2.1  affyQCReport_1.24.0
  [4] lattice_0.17-26  xtable_1.5-5 simpleaffy_2.22.0
  [7] genefilter_1.28.0made4_1.20.0 scatterplot3d_0.3-29
[10] gplots_2.7.3 caTools_1.10 bitops_1.0-4.1
[13] gdata_2.6.1  gtools_2.6.1 RColorBrewer_1.0-2
[16] ade4_1.4-11  affyPLM_1.22.0   preprocessCore_1.8.0
[19] gcrma_2.18.0 affycoretools_1.18.0 KEGG.db_2.3.5
[22] GO.db_2.3.5  RSQLite_0.7-3DBI_0.2-4
[25] AnnotationDbi_1.8.0  affy_1.24.0  Biobase_2.6.0
[28] RWinEdt_1.8-1

loaded via a namespace (and not attached):
  [1] affyio_1.14.0 annaffy_1.18.0annotate_1.24.0   biomaRt_2.2.0
  [5] Biostrings_2.14.0 Category_2.12.0   GOstats_2.12.0graph_1.24.0
  [9] GSEABase_1.8.0IRanges_1.4.0 RBGL_1.20.0   RCurl_1.2-1
[13] splines_2.10.0survival_2.35-7   tools_2.10.0  XML_2.6-0
 



Jenny Drnevich, Ph.D.

Functional Genomics Bioinformatics Specialist
W.M. Keck Center for Comparative and Functional Genomics
Roy J. Carver Biotechnology Center
University of Illinois, Urbana-Champaign

330 ERML
1201 W. Gregory Dr.
Urbana, IL 61801
USA

ph: 217-244-7355
fax: 217-265-5066
e-mail: drnev...@illinois.edu 
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[R] Keeping NAs in model.matrix for glm.fit

2009-10-29 Thread Ted


Hi,

For efficiency purposes I'm using glm.fit instead of glm, and I'm 
pre-constructing a model design matrix using model.matrix().


I'm finding that the 'NAs' in the x are being removed by model.matrix by 
default, and would like to know how could I keep them (or remove the 
corresponding NAs in the Y) so that the length for both response and 
predictor would be the same.  Thanks for your reply.



myformula = ~ offset(Bgrid[i,1] * x)

MM - model.matrix(myformula)

alpha[i] - coef(glm.fit(MM, Y, family=binomial()))


Ted

-
Ted Chiang
 Bioinformatics Analyst
 Centre for Computational Biology
 Hospital for Sick Children, Toronto
 416.813.7028
 tchi...@sickkids.ca

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

Re: [R] lmer and negative binomial family

2009-10-29 Thread Ben Bolker




John Sorkin wrote:
 
 Patrick,
 I am not sure about this but 
 (1) I think you need to load the lme4 package, not MASS
 (2) I don't think neg.bin is allowable. I suggest you load lme4 and then
 type ?family
 
 John
 
 

This conversation would probably go better on
r-sig-mixed-mod...@r-project.org ...
In the meanwhile --
   for a long time (I thought always, but I guess that's not true) the
allowable
families have been hard-coded within lme4, for purposes of efficiency.
The bottom line is that you can only use the following families:

famNms - c(binomial, gaussian, Gamma, inverse.gaussian,
poisson, quasibinomial, quasipoisson, quasi)

'quasipoisson' might be good, if you're willing to settle for a quasi-
answer,
although you should be careful with random effects variances -- in earlier
versions there are some problems with the scaling there

   If you really need NB I would suggest glmmADMB ...


-- 
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http://www.nabble.com/lmer-and-negative-binomial-family-tp26112313p26115443.html
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Re: [R] output of for

2009-10-29 Thread Greg Snow

Look at the collapse argument to the paste function.  You should be able to do 
what you want much simpler without any explicit for loops.

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


 -Original Message-
 From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-
 project.org] On Behalf Of Matteo Mattiuzzi
 Sent: Wednesday, October 28, 2009 11:28 AM
 To: R-help@r-project.org
 Subject: [R] output of for
 
 Hello!
 
 I've got troubles on following thing:
 
 As output I need a string like:
 INPUT_FILENAMES =
 H:\\AKK_temp\\Modis\\MOD13Q1\\austria\\MOD09Q1.A2008073.h18v04.005.2008
 083134952.hdf|H:\\AKK_temp\\Modis\\MOD13Q1\\austria\\MOD09Q1.A2008073.h
 19v04.005.2008082182221.hdf
 The filenames (dates) are with pathname (wrkdr) and separated with |.
 The number of input filenames changes (minimum 1, maximum n)
 
 wrkdr  - H:\\akk_temp\\Modis\\MOD09
 dates  is a data.frame where each row is a date and each column is a
 Modis-image (the name of it)
 
 I did this as following:
 paste('INPUT_FILENAMES = ', for(g in 1:(ntiles-1))
 {paste(wrkdr,'\\',dates[i,g+1],'|',sep='')} , for(g in ntiles){
 paste(wrkdr,'\\',dates[i,g+1],sep='') } ,sep='')
 
 and this did work (in an other wrkdr and other dates) but after
 changing, for gave no more output. And I don't know why!!!
 
 Why i use for twice? Because after the last file there is no |...
 
 
 Thanks for help
 
 Matteo
 
 
 
 
 
 
 
 
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 guide.html
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Re: [R] The 'subset matching' challenge

2009-10-29 Thread Detlef Steuer


If you look for subset sum problem, you will find relevant 
information.

A starter:
http://en.wikipedia.org/wiki/Subset_sum_problem

Detlef

On Thu, 29 Oct 2009 15:47:22 +0100
Yvonnick Noël yvonnick.n...@uhb.fr wrote:

 Dear all,
 
 The following problem just has been submitted to me by an accountant.
 
 In his new job, he has to close some old accounts. He has yearly 
 amounts, and a list of products that have been bought over the years, at 
 certain prices for which he has an exhaustive record. The problem is: He 
 does not know what product was bought this or that year (don't ask). He 
 does not want to find back the real story, but just write realistic 
 accounts, for which the sum of a subset of product prices will give the 
 exact yearly amount.
 
 Here is a real example from his data:
 
 # A list of 64 product prices
 products = 
 c(30500,30500,30500,30500,42000,42000,42000,42000,42000,42000,42000,42000,42000,42000,71040,90900,76950,35100,71190,
 53730,456000,70740,70740,533600,83800,59500,27465,28000,28000,28000,28000,28000,26140,49600,77000,123289,27000,27000,27000,
 27000,27000,27000,8,33000,33000,55000,77382,48048,51186,4,35000,21716,63051,15025,15025,15025,15025,80,111,
 59700,25908,829350,1198000,1031655)
 
 # Global amount
 amount = 4748652
 
 Now he wants to find all subsets of the 'product' vector which sums to 
 'amount'.
 
 I wrote the following code, which is clearly not optimal:
 
 # Create a matrix of subsets of size r among the integer set 1:n
 subsets - function(n, r, v = 1:n) {
   if(r = 0) vector(mode(v), 0)
   else if(r = n) v[1:n]
   else rbind(cbind(v[1], Recall(n-1, r-1, v[-1])),Recall(n-1, r, v[-1]))
 }
 
 # Main function
 find.amount = function(amount,products) {
  
   if(sum(products)amount) {
 cat(There is no solution.\n)
 return()
   }
  
   l = length(products)
   cat(\nThere are,l,product prices\n\n)
   names(products) = paste(Product,1:l,sep=)
   products = sort(products)
 
   for(i in 2:l) {
 
 # If the sum of the i smallest prices is greater than amount, then stop
 if(sum(products[1:i])amount) break
 
 # Look for matching subsets only in the case when the sum of i 
 largest prices is greater than amount
 if(sum(rev(products)[1:i])=amount) {
   # Generates all subsets of i indicies in 1:l
   subs = subsets(l,i)
   nl = nrow(subs)
   nc = ncol(subs)
 
   # Compute sums of corresponding price subsets
   sums = rowSums(matrix(products[subs],nl,nc))
 
   # Which ones match the global amount ?
   w = which(sums == amount)
   how.many = length(w)
   if(how.many0) {
 cat(\n-- There are,how.many,solutions with,nc,products :\n)
 for(j in 1:how.many) {
   print(products[subs[w[j],]])
 }
   }
   else cat(\n-- There is no solution with,nc,products.\n)
 }
 else cat(\n-- There is no solution with,i,products.\n)
   }
 }
 
 
 Then I can use these functions on a smaller example:
 
   find.amount(4,c(1,1,1,1,2,2))
 
 and a number of matching subsets are provided. The problem is: This 
 approach creates a whole matrix of subsets of r integers among 1:n, 
 which rapidly gives huge matrices, and this is clearly not optimal for 
 the real data provided above.
 
 Would anyone have a suggestion as to an alternative and more efficient 
 strategy?
 
 Good luck,
 
 Yvonnick Noel
 University of Brittany, Rennes 2
 France
 
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Re: [R] New variables remember how they were created?

2009-10-29 Thread Skipper Seabold

On Wed, Oct 28, 2009 at 12:40 PM, Adaikalavan Ramasamy
a.ramas...@imperial.ac.uk wrote:
 Your example is too complicated for me. But few points:

 1) What do you mean by instrument? Do you mean variable?


By instruments, I mean instrumental variables.  Very common in
econometrics: http://en.wikipedia.org/wiki/Instrumental_variable

Doing two-stage least squares in this example requires using
instrumental variables.

 2) diff(demand) is identical to demand[-1] - demand[-204]


Just trying to be explicit with the use of the lag and then including
it as an instrument.

 3) system() is a built-in R function, so avoid using it as variable name


Ok thanks.

 4) The variable yd is in the eqInvest formula and subsequently to the
 system formula. The variable y.1 is in the instruments formula. Both
 formulas are passed onto systemfit() call. Thus I see no surprises here.

 Try simplifying and rephrasing please if you want further help.


Ok, here it goes.

This is an introductory example, so I'm not sure how much more I can
simplify it, and the details of the estimator aren't that important.
Everything here works, and I understand what's going on, but I just
wonder how R knows that yd was creating using y.1.

In the second stage of the fit, the endogenous regressors on the RHS
are replaced by the fitted values from the first stage found from
regressing these endogenous regressors on all of the instrumental
variables (these are exogenous and commonly called instruments) in the
system.

So I have yd which is partially an endogenous variable and partially
an exogenous variable y - y.1.  In the second stage of the estimation
it seems that this yd is replaced by (z - y.1), where z is the
instrumental variable (the result of fitting y against all of the
instruments).  So how does R know that yd should be replace by (z -
y.1) unless yd carries some information that it was originally created
as (y - y.1).

Maybe this question is best asked on the devel list?

Cheers,

Skipper

 Regards, Adai




 Skipper Seabold wrote:

 Hello all,

 I hope this question is appropriate for this ML.

 Basically, I am wondering if when you create a new variable, if the
 variable holds some information about how it was created.

 Let me explain, I have the following code to replicate an example in a
 textbook (Greene's Econometric Analysis), using the systemfit package.

 dta -
 read.table('http://pages.stern.nyu.edu/~wgreene/Text/Edition6/TableF5-1.txt',
 header = TRUE)
 attach(dta)
 library(systemfit)
 demand - realcons + realinvs + realgovt
 c.1 - realcons[-204]
 y.1 - demand[-204]
 yd - demand[-1] - y.1
 eqConsump - realcons[-1] ~ demand[-1] + c.1
 eqInvest - realinvs[-1] ~ tbilrate[-1] + yd
 system - list( Consumption = eqConsump, Investment = eqInvest)
 instruments - ~ realgovt[-1] + tbilrate[-1] + c.1 + y.1
 # 2SLS
 greene2sls - systemfit( system, 2SLS, inst = instruments,
 methodResidCov = noDfCor )

 When I do the 2SLS fit, it seems that even though I declared y.1 as an
 instrument that the estimator knows that yd was created using y1, so
 it (correctly) transforms yd to use the instrument in the final
 estimation.

 So I'm wondering if yd somehow carries knowledge of how it was created.

 Thanks,

 Skipper


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Re: [R] singular variance-covariance warning in lmer

2009-10-29 Thread Weber, Sam


Hi Ista,

The command looks like:

$ female  : Factor w/ 18 levels 2,4,5,8,..: 1 1 1 1 2 2 2 3 4 4 ...

Female is a factor with 18 levels, so I assume this is how the analysis is 
being grouped.

Best

Sam 

From: Ista Zahn [istaz...@gmail.com]
Sent: 29 October 2009 14:35
To: Weber, Sam
Cc: r-help@R-project.org
Subject: Re: [R] singular variance-covariance warning in lmer

Hi Sam,
Just a stab in the dark here, but is your grouping variable really
female? What does

str(data.frame(mean.sst, female)

look like? How many levels does female have?

-Ista

On Thu, Oct 29, 2009 at 7:10 AM, Weber, Sam sam.we...@exeter.ac.uk wrote:
 Dear R Users,

 I was hoping for some help with a recurrent error message in lmer. I am 
 trying to model the effect of temperature on metabolic rate in animals 
 (response = int.length) at different temperatures (mean.sst), with repeated 
 measurements on the same individuals (random effect = female). Ideally I 
 would make a random slope and intercept model where the rate can change 
 differently with temperature for different individuals:

 model-lmer(int.length~mean.sst+(mean.sst|female))

 However, I get the following warning message:

 Warning message:
 Estimated variance-covariance for factor 'female' is singular in: 
 `LMEoptimize-`(`*tmp*`, value = list(maxIter = 200L, tolerance = 
 1.49011611938477e-08,
 summary(model)

 Linear mixed-effects model fit by REML
 Formula: int.length ~ mean.sst + (mean.sst | female)
   AIC   BIC logLik MLdeviance REMLdeviance
  155.4 164.5  -72.7  142.8145.4
 Random effects:
  Groups   NameVariance   Std.Dev.   Corr
  female   (Intercept) 6.8459e-10 2.6165e-05
  mean.sst6.8169e-10 2.6109e-05 -0.065
  Residual 1.3634e+00 1.1676e+00
 number of obs: 46, groups: female, 18
 Fixed effects:
Estimate Std. Error t value
 (Intercept)  48.8249 6.5895   7.409
 mean.sst -1.3609 0.2518  -5.406
 Correlation of Fixed Effects:
 (Intr)
 mean.sst -1.000





 If I try and run just a random intercepts model I get similar problems:



 model2-lmer(int.length~mean.sst+(1|female))

 Warning message: Estimated variance for factor 'female' is effectively zero 
 in: `LMEoptimize-`(`*tmp*`, value = list(maxIter = 200L, tolerance = 
 1.49011611938477e-08,



 I have tried disabling PQL iterations  using control = list(usePQL = FALSE, 
 msVerbose=TRUE), following Douglas Bates' recommendation on the mailing list 
 archives but I still get a similar message. Does this mean that the variance 
 among subjects is too close to zero for estimation of the random effects? I 
 compared the random effects model to a linear model with just lm(int.length ~ 
 mean.sst) using a likelihood ratio test and got p = 1.0 (which is always 
 suspicious). It would actually make sense for there to be negligible 
 variation among subjects in their response to temperature, however I am 
 concerned that I am making a fundamental error somewhere along the line.



 I would greatly appreciate any suggestions you may have.



 Best regards



 Sam Weber



 University of Exeter, UK.




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--
Ista Zahn
Graduate student
University of Rochester
Department of Clinical and Social Psychology
http://yourpsyche.org
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Re: [R] Trouble retrieving data (.xls) from folder on my computer.

2009-10-29 Thread tobiasfa


This is what R says:

 xls - read.xls(file.choose())
Converting xls file to csv file... Error in system(cmd, intern = !verbose) :
perl not found
Error in file.exists(tfn) : invalid 'file' argument
 xls - read.xls(file.choose(), perl=C:/perl/bin/perl.exe)
Converting xls file to csv file... Error in system(cmd, intern = !verbose) :
perl not found
Error in file.exists(tfn) : invalid 'file' argument

(I have installed Perl!) 


Gabor Grothendieck wrote:
 
 If its already on your computer try one of these:
 
 xls - read.xls(myfile.xls)   # if its in current directory
 
 xls - read.xls(/path/to/myfile.xls)   # to specify path
 
 xls - read.xls(file.choose())   # use explorer style interface to choose
 
 
 
 
 On Thu, Oct 29, 2009 at 9:36 AM, tobiasfa tobias.farnly...@gmail.com
 wrote:

 Using gdata on windows and im having trouble to retrieve an excel file
 from a
 folder on my computer.

 This is my dummy:
 R  download.file(http://people.su.se/~lundh/data/cpi_kpix.rda;,
 + cpi_kpix.rda)

 R  URL - http://people.su.se/;
 R  PATH - ~lundh/data/
 R  FILE - cpi_kpix.rda
 R  download.file(paste(URL,PATH,FILE,sep=),
 + cpi_kpix.rda)

 R  library(gdata)
 R  FILE - Stock_Returns_1931_2002.xls
 R  stock.xls - read.xls(paste(URL, PATH, FILE, sep = ),
 + header = TRUE)[, 3:4]
 Downloading...
 Done.
 Converting xls file to csv file... Done.
 Reading csv file... Done.

 Im not suppose to download the file though its already on my computer.
 How
 do i type in the path så R finds my .xls??

 kind regards Tobias

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 http://www.nabble.com/Trouble-retrieving-data-%28.xls%29-from-folder-on-my-computer.-tp26113172p26113172.html
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[R] Weird error: Error in xj[i] : invalid subscript type 'list'

2009-10-29 Thread Peng Yu

I got the error. I haven't been able to get a stand along case so that
I can show it here. But could somebody give some clue on what could
cause this error? Since I never defined xj[i], I don't understand
where this error come from.

Error in xj[i] : invalid subscript type 'list'

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[R] Rounding and printing

2009-10-29 Thread Alan Cohen

Hello,

I am trying to print a table with numbers all rounded to the same number of 
digits (one after the decimal), but R seems to want to not print .0 for 
integers.  I can go in and fix it one number at a time, but I'd like to 
understand the principle.  Here's an example of the code.  The problem is the 
13th element, 21 or 21.0:
nvb_deaths - round(ss[,10]/100,digits=1)   
 nvb_deaths
 [1] 56.5  1.6  0.2  3.9  0.1  2.2  0.2  2.6  1.5  4.1  1.1  6.1 21.0
nvb_dths - paste(nvb_deaths, 
(,round(100*nvb_deaths/nvb_deaths[1],digits=1),%),sep=)
 nvb_dths
 [1] 56.5 (100%) 1.6 (2.8%)  0.2 (0.4%)  3.9 (6.9%)  0.1 (0.2%)  2.2 
(3.9%) 
 [7] 0.2 (0.4%)  2.6 (4.6%)  1.5 (2.7%)  4.1 (7.3%)  1.1 (1.9%)  6.1 
(10.8%)
[13] 21 (37.2%) 
 print(nvb_deaths,digits=1)
 [1] 56.5  1.6  0.2  3.9  0.1  2.2  0.2  2.6  1.5  4.1  1.1  6.1 21.0
 paste(print(nvb_deaths,digits=1), 
 (,round(100*nvb_deaths/nvb_deaths[1],digits=1),%),sep=)
 [1] 56.5  1.6  0.2  3.9  0.1  2.2  0.2  2.6  1.5  4.1  1.1  6.1 21.0
 [1] 56.5 (100%) 1.6 (2.8%)  0.2 (0.4%)  3.9 (6.9%)  0.1 (0.2%)  2.2 
(3.9%) 
 [7] 0.2 (0.4%)  2.6 (4.6%)  1.5 (2.7%)  4.1 (7.3%)  1.1 (1.9%)  6.1 
(10.8%)
[13] 21 (37.2%) 

I'm running R v2.8.1 on Windows.  Any help is much appreciated.

Cheers,
Alan Cohen

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Re: [R] how to fetch rows with certain characteristics

2009-10-29 Thread Waverley @ Palo Alto

Thanks.  That works.

However, in my own case, there are more columns of other kinds of
data.  So to me, it is more important to get the row index of those
that has the min values of particular column in particular class
(which is another column).

Can you help more as how to get those row index? One issue is that for
some class they may share the same min value so that using %in% does
not work.  My goal is to reduce the original matrix size and get the
result back in the original matrix format.


Thanks.

On Wed, Oct 28, 2009 at 11:55 PM, Ista Zahn istaz...@gmail.com wrote:
 There are various ways, including

 x - read.table(textConnection(1.2 1
 + 1.3 1
 + 1.3 1
 + 1.5 1
 + 2.1 2
 + 2.0 2
 + 9.9 2
 + 1.4 3
 + 1.8 3
 + 1.9 3) )

 x - as.matrix(x)

 x.min - cbind(tapply(x[,1], x[,2], min), unique(x[,V2]))

 Most of that is just formatting it in the way you requested. All you
 need to compute the values is

 tapply(x[,1], x[,2], min)

 -Ista

 On Thu, Oct 29, 2009 at 1:47 AM, Waverley @ Palo Alto
 waverley.paloa...@gmail.com wrote:
 Hi,

 I have a matrix, first column is of certain values, second column is
 the class labels or a factor.
 e.g.

 1.2 1
 1.3 1
 1.3 1
 1.5 1
 2.1 2
 2.0 2
 9.9 2
 1.4 3
 1.8 3
 1.9 3

 I want to find out what is the min values of column 1 for each
 corresponding class (column 2).  For the above example, I want to
 return a matrix of
 1.2 1
 2.0 2
 1.3 3

 Can someone suggest how to code for that?  The second column can be of
 characters.

 Thanks much.


 --
 Waverley @ Palo Alto

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 --
 Ista Zahn
 Graduate student
 University of Rochester
 Department of Clinical and Social Psychology
 http://yourpsyche.org




-- 
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Re: [R] Trouble retrieving data (.xls) from folder on my computer.

2009-10-29 Thread Gabor Grothendieck

Either make sure that perl is on your PATH or if the path/filename you
specified is correct then try with backslashes just in case:

C:\\perl\\bin\\perl.exe

Also check the upper/lower case of each component of your path just in case.

On Thu, Oct 29, 2009 at 11:52 AM, tobiasfa tobias.farnly...@gmail.com wrote:

 This is what R says:

 xls - read.xls(file.choose())
 Converting xls file to csv file... Error in system(cmd, intern = !verbose) :
 perl not found
 Error in file.exists(tfn) : invalid 'file' argument
 xls - read.xls(file.choose(), perl=C:/perl/bin/perl.exe)
 Converting xls file to csv file... Error in system(cmd, intern = !verbose) :
 perl not found
 Error in file.exists(tfn) : invalid 'file' argument

 (I have installed Perl!)


 Gabor Grothendieck wrote:

 If its already on your computer try one of these:

 xls - read.xls(myfile.xls)   # if its in current directory

 xls - read.xls(/path/to/myfile.xls)   # to specify path

 xls - read.xls(file.choose())   # use explorer style interface to choose




 On Thu, Oct 29, 2009 at 9:36 AM, tobiasfa tobias.farnly...@gmail.com
 wrote:

 Using gdata on windows and im having trouble to retrieve an excel file
 from a
 folder on my computer.

 This is my dummy:
 R  download.file(http://people.su.se/~lundh/data/cpi_kpix.rda;,
 + cpi_kpix.rda)

 R  URL - http://people.su.se/;
 R  PATH - ~lundh/data/
 R  FILE - cpi_kpix.rda
 R  download.file(paste(URL,PATH,FILE,sep=),
 + cpi_kpix.rda)

 R  library(gdata)
 R  FILE - Stock_Returns_1931_2002.xls
 R  stock.xls - read.xls(paste(URL, PATH, FILE, sep = ),
 + header = TRUE)[, 3:4]
 Downloading...
 Done.
 Converting xls file to csv file... Done.
 Reading csv file... Done.

 Im not suppose to download the file though its already on my computer.
 How
 do i type in the path så R finds my .xls??

 kind regards Tobias

 --
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Re: [R] The system cannot find the file specified

2009-10-29 Thread Jim Burke


I think your are using R 2.10.0.

I get similar messages when I have too many resources in use. Like FireFox
eating memory and CPU.

When you get this message go look in your R installation library 
subdirectory.
You will see a weird number as a folder name. Under that folder is the 
package you
were trying to install. Simply copy that package name folder to the 
correct folder
location for the other packages in library. Then clean up by removing 
that weird

folder name.

Hope this helps,
Jim

Mark Knecht wrote:

What is the problem here? I did an install package from the Rgui menu.
Windows Vista

SNIP

trying URL 
'http://cran.cnr.Berkeley.edu/bin/windows/contrib/2.9/mvtnorm_0.9-8.zip'
Content type 'application/zip' length 236089 bytes (230 Kb)
opened URL
downloaded 230 Kb

trying URL 
'http://cran.cnr.Berkeley.edu/bin/windows/contrib/2.9/QRMlib_1.4.4.zip'
Content type 'application/zip' length 2369023 bytes (2.3 Mb)
opened URL
downloaded 2.3 Mb

package 'robustbase' successfully unpacked and MD5 sums checked
package 'fUtilities' successfully unpacked and MD5 sums checked
package 'fEcofin' successfully unpacked and MD5 sums checked
package 'fCalendar' successfully unpacked and MD5 sums checked
package 'fSeries' successfully unpacked and MD5 sums checked
package 'mvtnorm' successfully unpacked and MD5 sums checked
package 'QRMlib' successfully unpacked and MD5 sums checked
Error in normalizePath(path) :
  path[1]=C:\Program Files\R\R-2.9.2\library/QRMlib: The system
cannot find the file specified

SNIP

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Re: [R] Weird error: Error in xj[i] : invalid subscript type 'list'

On Thu, 29 Oct 2009 11:28:44 -0500 Peng Yu pengyu...@gmail.com wrote:
 
 I got the error. I haven't been able to get a stand along case so that
 I can show it here. But could somebody give some clue on what could
 cause this error? Since I never defined xj[i], I don't understand
 where this error come from.
 
 Error in xj[i] : invalid subscript type 'list'

You use a function which extracts subsets of an object, but the indices 
happens to be a list. Example:

x=1:5
z=as.list(1:2)
x[z]

-- 
Karl Ove Hufthammer

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Re: [R] Weird error: Error in xj[i] : invalid subscript type 'list'



On Oct 29, 2009, at 12:28 PM, Peng Yu wrote:


I got the error. I haven't been able to get a stand along case so that
I can show it here. But could somebody give some clue on what could
cause this error? Since I never defined xj[i], I don't understand
where this error come from.

Error in xj[i] : invalid subscript type 'list'



This reminds me of the stories told by guys coming into emergency  
rooms who have been involved in fights:


I was just sitting on the corner, minding my own business, when I was  
attacked by this error.


 xj - 1:4
 xj[list(1,2)]
Error in xj[list(1, 2)] : invalid subscript type 'list'

--

David Winsemius, MD
Heritage Laboratories
West Hartford, CT

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Re: [R] Rounding and printing

On Thu, 29 Oct 2009 12:29:42 -0400 Alan Cohen coh...@smh.toronto.on.ca 
wrote:
 I am trying to print a table with numbers all rounded to the same 
 number of digits (one after the decimal), but R seems to want to not 
 print .0 for integers.

'round' only rounds numbers; it doesn't format them. Use 'formatC' 
instead. Example:

 x=c(1.234, 20.31, 25, 0.7)
 formatC(x, digits=1, format=f)
[1] 1.2  20.3 25.0 0.7 

'prettyNum' may also be useful.

-- 
Karl Ove Hufthammer

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[R] Question about is.na for missing data

2009-10-29 Thread David Kaplan


Greetings,

I am trying to a simple mean imputation using the following command 
(yes, I know it is not an ideal way to impute missing data)


asd[is.na(asd)] - mean(asd,na.rm=TRUE)

This has worked quite well in the past.  Now, however, I am getting the 
following error messsage.


Error in `[-.data.frame`(`*tmp*`, is.na(asd), value = 
c(2433.64150943396,  :

 rhs is the wrong length for indexing by a logical matrix

An inspection of the data set reveals no value of the sort give above.  
Any thoughts?


Thanks in advance,

David

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Re: [R] The system cannot find the file specified

2009-10-29 Thread hadley wickham

 Do you have write permission in C:\Program Files\R\R-2.9.2\library?  It
 could be that the installer just tried to create the QRMlib subdir, and
 failed, and that's why it doesn't exist.

One possible reason for failure is that your virus checker prevented
the R installer from creating a new directory.

Hadley

-- 
http://had.co.nz/

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Re: [R] Rounding and printing



On Oct 29, 2009, at 12:29 PM, Alan Cohen wrote:


Hello,

I am trying to print a table with numbers all rounded to the same  
number of digits (one after the decimal), but R seems to want to not  
print .0 for integers.  I can go in and fix it one number at a  
time, but I'd like to understand the principle.  Here's an example  
of the code.  The problem is the 13th element, 21 or 21.0:

nvb_deaths - round(ss[,10]/100,digits=1)
nvb_deaths

[1] 56.5  1.6  0.2  3.9  0.1  2.2  0.2  2.6  1.5  4.1  1.1  6.1 21.0


?sprintf  # fmt = %1.1f

 nvb_dths - paste(sprintf(%1.1f,  
nvb_deaths), (,round(100*nvb_deaths/ 
nvb_deaths[1],digits=1),%),sep=)


 nvb_dths
 [1] 56.5 (100%)  1.6 (2.8%)   0.2 (0.4%)   3.9 (6.9%)   0.1  
(0.2%)   2.2 (3.9%)   0.2 (0.4%)
 [8] 2.6 (4.6%)   1.5 (2.7%)   4.1 (7.3%)   1.1 (1.9%)   6.1  
(10.8%)  21.0 (37.2%)



nvb_dths - paste(nvb_deaths, (,round(100*nvb_deaths/ 
nvb_deaths[1],digits=1),%),sep=)

nvb_dths
[1] 56.5 (100%) 1.6 (2.8%)  0.2 (0.4%)  3.9 (6.9%)  0.1  
(0.2%)  2.2 (3.9%)
[7] 0.2 (0.4%)  2.6 (4.6%)  1.5 (2.7%)  4.1 (7.3%)  1.1  
(1.9%)  6.1 (10.8%)

[13] 21 (37.2%)

print(nvb_deaths,digits=1)

[1] 56.5  1.6  0.2  3.9  0.1  2.2  0.2  2.6  1.5  4.1  1.1  6.1 21.0
paste(print(nvb_deaths,digits=1), (,round(100*nvb_deaths/ 
nvb_deaths[1],digits=1),%),sep=)

[1] 56.5  1.6  0.2  3.9  0.1  2.2  0.2  2.6  1.5  4.1  1.1  6.1 21.0
[1] 56.5 (100%) 1.6 (2.8%)  0.2 (0.4%)  3.9 (6.9%)  0.1  
(0.2%)  2.2 (3.9%)
[7] 0.2 (0.4%)  2.6 (4.6%)  1.5 (2.7%)  4.1 (7.3%)  1.1  
(1.9%)  6.1 (10.8%)

[13] 21 (37.2%)

I'm running R v2.8.1 on Windows.  Any help is much appreciated.

Cheers,
Alan Cohen

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David Winsemius, MD
Heritage Laboratories
West Hartford, CT

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[R] a minor bug in venn from gplots?

2009-10-29 Thread Tao Shi



Hi list,

I found this one when I was trying to output the Venn diagram to a .pdf file.  
When there are 4 sets of groups to draw, the .pdf file automatically has 3 
pages and the figure only appears on the 3rd page in the .pdf file with the 
first 2 pages being blank.  Try the following: (I'm using R-2.9.0 on WinXP, 
gplots 2.7.1).  gplots 2.7.3 has the same problem.

###==
pdf(test.pdf)
A- 1:20

B- 1:20

C- 2:20

D- 3:21

input-list(A,B,C,D)

venn(input)

dev.off()
###==

By looking at the code of drawVennDiagram, I think the problem comes from the 
fact there are one grid.newpage() call and two plot function calls when 
numCircles==4 (see below).  I wonder the grid.newpage() and the second plot 
call are necessary?

Thanks a lot.

...Tao


##=

    else if (4 = numCircles  numCircles = 5  !simplify) {
    grid.newpage() 
===
    relocate_elp - function(e, alpha, x, y) {
    phi = (alpha/180) * pi
    xr = e[, 1] * cos(phi) + e[, 2] * sin(phi)
    yr = -e[, 1] * sin(phi) + e[, 2] * cos(phi)
    xr = x + xr
    yr = y + yr
    return(cbind(xr, yr))
    }
    lab - function(identifier, data, showLabel = showSetLogicLabel) {
    r - data[identifier, 1]
    if (showLabel) {
    return(paste(identifier, r, sep = \n))
    }
    else {
    return(r)
    }
    }
    plot(c(0, 400), c(0, 400), type = n, axes = F, ylab = ,   

    xlab = )
    if (4 == numCircles) {
    elps = cbind(162 * cos(seq(0, 2 * pi, len = 1000)), 
    108 * sin(seq(0, 2 * pi, len = 1000)))
    plot(c(0, 400), c(0, 400), type = n, axes = F, 
===
    ylab = , xlab = )
    polygon(relocate_elp(elps, 45, 130, 170))
    polygon(relocate_elp(elps, 45, 200, 200))

...
###===

  
_
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Re: [R] The system cannot find the file specified


On 10/29/2009 12:58 PM, hadley wickham wrote:

Do you have write permission in C:\Program Files\R\R-2.9.2\library?  It
could be that the installer just tried to create the QRMlib subdir, and
failed, and that's why it doesn't exist.


One possible reason for failure is that your virus checker prevented
the R installer from creating a new directory.


Yes, virus checkers cause a lot of trouble for R.  Some of them open 
files for exclusive access while checking, and when R tries to read or 
write the same file while the virus checker has it locked, it fails.


Duncan Murdoch

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Re: [R] Question about is.na for missing data

On Thu, 29 Oct 2009 11:45:19 -0500 David Kaplan 
dkap...@education.wisc.edu wrote:
 asd[is.na(asd)] - mean(asd,na.rm=TRUE)
 
 This has worked quite well in the past.  Now, however, I am getting the 
 following error messsage.
 
 Error in `[-.data.frame`(`*tmp*`, is.na(asd), value = 
 c(2433.64150943396,  :
   rhs is the wrong length for indexing by a logical matrix
 
 An inspection of the data set reveals no value of the sort give above.  
 Any thoughts?

I would venture to guess that your 'asd' object is not of the type you 
think it is. What is the value of

str(asd)

-- 
Karl Ove Hufthammer

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Re: [R] Weird error: Error in xj[i] : invalid subscript type 'list'


On 10/29/2009 12:28 PM, Peng Yu wrote:

I got the error. I haven't been able to get a stand along case so that
I can show it here. But could somebody give some clue on what could
cause this error? Since I never defined xj[i], I don't understand
where this error come from.

Error in xj[i] : invalid subscript type 'list'


Next time you see it, run traceback() afterwards to see which function 
triggered it.  That might help you to track it down and fix it.


Duncan Murdoch

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Re: [R] The system cannot find the file specified

2009-10-29 Thread Prof Brian Ripley


On Thu, 29 Oct 2009, hadley wickham wrote:

(quoting Duncan Murdoch without attribution, tut tut)


Do you have write permission in C:\Program Files\R\R-2.9.2\library?  It
could be that the installer just tried to create the QRMlib subdir, and
failed, and that's why it doesn't exist.


One possible reason for failure is that your virus checker prevented
the R installer from creating a new directory.


Another (more likely, and mentioned in the rw-FAQ) is that a file 
indexing service prevented the file being renamed.


As I understand this thread, the OP was able to install several 
packages but the last one failed.  That's more consistent with file 
indexing (a concurrent process that will interfere apparently 
randomly) than with permission issues.  Anti-virus software normally 
intervenes before the file gets written, not on renaming.


--
Brian D. Ripley,  rip...@stats.ox.ac.uk
Professor of Applied Statistics,  http://www.stats.ox.ac.uk/~ripley/
University of Oxford, Tel:  +44 1865 272861 (self)
1 South Parks Road, +44 1865 272866 (PA)
Oxford OX1 3TG, UKFax:  +44 1865 272595__
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
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[R] How to turn individual consecutive information into survival objects?

2009-10-29 Thread Guido van Steen

Dear R List, 

I have a dataset with the following structure:  

personal_id, p_0, p_1, p_2,  , p_36, p_37 
1, NA, 1, 4,  , 1, NA 
2, NA, NA, NA,  , 4, NA 
.
.
.
6020, NA, 3, 3, , NA, NA
6021, NA, 2, 2, , 4, NA


I used some made-up data. It is just meant to show the structure of the 
dataset. 

The variables of interest are p_0, ... p_37. They represent types of activity 
of the 6021 persons interviewed in 38 consecutive period. 

The values for p_0 and p_37 are coded as follows: 
1 = self-employed 
2 = employed 
3 = in training 
4 = unemployed 
NA = no information available  

p_0 is the period before p_1. This period was just before the survey, so that 
none of the individuals were interviewed in this period. 

p_37 is the period after p_36. This period was just after the survey, so that 
none of the individuals were interviewed in this period. 

I would like to transform this dataset into information on the lenght of the 
first spell of unemployment. (If there are multiple spells I would just like to 
use the first one.) 

My question is how can I convert the records in this dataset into survivul 
objects - Surv() - so that they can be used with a function like coxph().

Could anyone give me a pointer to a function that performs this specific 
conversion? 

Thanks in advance! 

Guido 



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Re: [R] Question about is.na for missing data

On Thu, 29 Oct 2009 18:12:34 +0100 Karl Ove Hufthammer k...@huftis.org 
wrote:
rhs is the wrong length for indexing by a logical matrix
  
  An inspection of the data set reveals no value of the sort give above.  
  Any thoughts?
 
 I would venture to guess that your 'asd' object is not of the type you 
 think it is.

More specifically, I think it is a data frame with more than one column. 
Then 'mean' will calculate the average value for each column, instead of 
a single value.

-- 
Karl Ove Hufthammer

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Re: [R] How to turn individual consecutive information into survival objects?