gmane.comp.lang.r.general

Re: formatting column names of data frame

John Kane — 2013-05-18T16:49:39

As Rainer Schuermann says, use the xtable package. Have a look at the documentation entitled xtable gallary --available at your friendly neigbourhood CRAN site and have a look at section 3.1. and it's probably a good idea to look at ?print.xtable as well

You might also what to have a look at the latex function in Hmisc.

Here is an example with "Average" in bold. It ran for me using LyX and knitr rather than Sweave.
###==================================
library( xtable ) 
x <- structure(list(Record = 1:3, Average = c(34L, 14L, 433L), Maximum = c(899L, 15L, 1003L)), .Names = c("Record", "Average", "Maximum"), class = "data.frame", row.names = c(NA, -3L)) 
names(x)  <-  c("Record", "\\textbf{Average}", "Maximum")
x <- xtable( x ) 
print( x,  sanitize.text.function = function(x){x} )
###==================================






John Kane
Kingston ON Canada



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Re: exporting data into STATA

John Kane — 2013-05-18T14:53:49

Presumably it means what it says : You have  a non-ASCII character in the dataset. Any accented character, for example, is going to be outside the basic ASCII character set.

Examples
  mydata  <-  data.frame(étt=1:10) # error
  newdata  <- data.frame(ntt=1:10) # no error

See http://www.theasciicode.com.ar/ for a list 

  
  write.dta(mydata, file = "stata.dta")

John Kane
Kingston ON Canada



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

[R-pkgs] Probabilistic neural network (PNN)

Pierre-Olivier Chasset — 2013-05-17T12:07:47

Dear useRs,

I am pleased to announce the release of the new package PNN.

PNN implements the algorithm proposed by Specht (1990). It is written in the R statistical language. It solves a common problem in automatic learning. Knowing a set of observations described by a vector of quantitative variables, we classify them in a given number of groups. Then, the algorithm is trained with this datasets and should guess afterwards the group of any new observation. This neural network has the main advantage to begin generalization instantaneously even with a small set of known observations. It is delivered with four functions - learn, smooth, perf and guess - and a dataset. The functions are documented with examples and provided with unit tests.

Continue reading at http://flow.chasset.net/r-pnn/

Have a good week-end,

Pierre-Olivier Chasset
_______________________________________________
R-packages mailing list
R-packages< at >r-project.org
https://stat.ethz.ch/mailman/listinfo/r-packages

Re: exporting data into STATA

Prof Brian Ripley — 2013-05-18T13:29:26

ASCII is an encoding, not a file format.


Use ASCII characters: that *was* on the help page!

exporting data into STATA

Niklas Fischer — 2013-05-18T13:05:57

Dear All,

I am not very familiar with ASCII file format, and now I am trying to
export R data frame into STATA
When I run write.dta command, there is a warning below appear.

5: In abbreviate(ll, 80L) : abbreviate used with non-ASCII chars

Do you have any suggestion to fix it?

All the bests

[[alternative HTML version deleted]]

Re: glmer.nb: function not in downloaded lme4 package?

Ben Bolker — 2013-05-18T12:41:57

negative binomial distribution please.

  This question would probably be better on the
r-sig-mixed-models< at >r-project.org list.

   glmer.nb() is _only_ in development versions of lme4,
not in the stable version on CRAN, and furthermore it is
still quite new and poorly tested.  If you want to try it
out you are probably best of installing from github via
install_github in the devtools package.  Alternatively
you could try the glmmADMB package, on r-forge.
   
  Ben Bolker

Re: inverse for formula transformations on LHS

Stephen Milborrow — 2013-05-18T12:05:19

plotmo in the plotmo package has an inverse.func argument,
so something like the following might work for you?

library(MASS)
library(plotmo)
log.brain <- log(Animals$brain)
log.body  <- log(Animals$body)
m2 <- lm(log.brain ~ log.body)

myplot <- function(...)
    plotmo(m2, do.par=F, nrug=-1, col.resp=2, pch=20, 
        se=1, main="",  xlab="log(body)", ...)

par(mfrow = c(2, 2))
myplot(ylab="log(brain)")
myplot(ylab="brain", inverse.func=exp)
termplot(m2, se=T, rug=T) #  for comparison

how to do an external validation with R

beginner — 2013-05-18T11:45:42

I would like to do external validation using R software. So far I have used
packages like "Design" and "DAAG". However they perform internal validation
rather than external one. In order to perform external validation I would
have to split my data beforehand into training and test set, leave the test
set on the site and use only the training set to select a model. I would
then test the selected model with the sample set left initially on the site.
I would like to repeat this process several times to make sure that all the
samples are included at least once in a test set. I thought that I need to
use a loop function in R to perform this process automatically. As I am new
to R I don't know how to make a loop. Could you please help me with this or
suggest an R package ? I would be very very grateful for help with this task
!



--
View this message in context: http://r.789695.n4.nabble.com/how-to-do-an-external-validation-with-R-tp4667404.html
Sent from the R help mailing list archive at Nabble.com.

glmer.nb: function not in downloaded lme4 package?

Ross Marriott — 2013-05-18T06:23:34

Dear R Help,



I would like to use the glmer.nb function for mixed modelling using negative binomial distribution please.



On the CRAN website apparently this function is called from the lme4 package (version 0.99999911-1).



I have downloaded the latest version of the lme4 package (version 0.999999-2) and have recently reinstalled the latest version of 64-bit R (version 3.0.1) but after loading the package and calling:



library(lme4)

help(glmer.nb)

# I receive the following message:



No documentation for glmer.nb in specified packages and libraries:
you could try ??glmer.nb

# I called the glmer() function successfully, i.e.



glmer(CNos ~ Year+Month+(1|Vessel), data=Occ.cr, family=poisson, offset=Occ.cr$logtrapd)



# but when I tried this:



glmer.nb(CNos ~ Year+Month+(1|Vessel), data=Occ.cr, offset=Occ.cr$logtrapd)



# I got the error message:



Error: could not find function "glmer.nb"



# Can you please provide advice on how I might use glmer.nb or alternative code on how I might

Heterogeneous negative binomial

Joseph Hilbe — 2013-05-17T21:58:51

I have seen several queries about parameterizing the negative binomial scale
parameter. This is called 

the heterogeneous negative binomial. I have written a function called
"nbinomial" which is in the 

msme package on CRAN. Type ?nbinomial to see the help file.  The default
model is a negative binomial 

for which the dispersion parameter is directly related to mu, which is how
Stata, SAS, SPSS, Limdep, and

so forth parameterize the negative binomial. The direct parameterization
make sense in that the more 

variation or correlation there is in a Poisson model,  the greater is the
value of the dispersion parameter 

which is adjusting for the excessive variation. With this parameterization
the dispersion parameter is 

directly related to both mu, as well as the dispersion statistic, or Pearson
Chi2/(residual DOF).  

A dispersion parameter of 0 is Poisson, which is equidispersed. When the
dispersion parameter for 

other mixture models such as generalized Poisson and Poisson inverse
Gaussian is zero, th

Re: Error with adehabitatHR and kernelbb

Michael Sumner — 2013-05-18T03:08:37

I presume it's related to the fact that the X coordinate is duplciated
in the two records, and the grid generated by adehabitatHR is a single
column,

Here's a test:
require(adehabitatHR)
m <- matrix(c(5419574  ,    390051, 5419490   ,  390051), ncol = 2,
byrow = TRUE)
 tt <- as.POSIXct(c("2012-05-30 14:00:00", "2012-05-30 16:00:00"))
id <- c("Ade=5", "Ade-5")

x <- kernelbb(as.ltraj(m, date = tt, id = id, typeII=TRUE), sig1 =
4.5766, sig2 = 5, grid = 1000)

This degenerate Y dimension affects downstream methods (like image),
and I'd say this is worth reporting to the package author as an issue:

dim(x[[1]])
[1] 1000    1
 dim(x[[2]])
[1] 1000    1

The clue comes from points2grid in sp:
Warning messages:
1: In points2grid(points, tolerance, round) :
  cell size from constant coordinate 2 possibly taken from other coordinate


You could override the auto-generated grid by passing in your own,
probably something you will want anyway so that your collection share
the same extent and resolution. See ?kernelbb,

Re: filter rows by value

Ye Lin — 2013-05-18T00:26:03

it works!Thanks!


On Fri, May 17, 2013 at 2:28 PM, Rui Barradas <ruipbarradas< at >sapo.pt> wrote:


[[alternative HTML version deleted]]

Re: filter rows by value

arun — 2013-05-17T22:43:11


Hi,
dat1<- read.table(text="
Var  Time
1          51
2          151
3          251
4            234
5          331
6            351
",sep="",header=TRUE)
dat1[!is.na(match(gsub(".*(\\d{2})$","\\1",dat1$Time),51)),]
#  Var Time
#1   1   51
#2   2  151
#3   3  251
#6   6  351
#or
dat1[substr(dat1$Time,nchar(dat1$Time)-1,nchar(dat1$Time))=="51",]
#  Var Time
#1   1   51
#2   2  151
#3   3  251
#6   6  351

#or 
library(stringr)

dat1[str_detect(dat1$Time,"51$"),]
#  Var Time
#1   1   51
#2   2  151
#3   3  251
#6   6  351
#or
dat1[grepl("51$",dat1$Time),]
#  Var Time
#1   1   51
#2   2  151
#3   3  251
#6   6  351
#or
 dat1[str_sub(dat1$Time, start=-2)%in%51,]
  Var Time
#1   1   51
#2   2  151
#3   3  251
#6   6  351
A.K.


----- Original Message -----
From: Ye Lin <yelin< at >lbl.gov>
To: R help <r-help< at >r-project.org>
Cc: 
Sent: Friday, May 17, 2013 5:01 PM
Subject: [R] filter rows by

Re: Bivariate - multivariate linear regression

Peter Ehlers — 2013-05-17T22:16:22

(Thanks for providing reproducible code!)

It seems to me that you're just missing two things:

1. a way to determine the names of the variables to be included
    in the multiple (not 'multivariate' to be nitpicky) regression;

2. a way to build the formula for the multiple regression once
    you know which predictors to include.

To get the variables:

   varnames <- names(tolerance)[2:6]
   pvec <- c(lmp(fit), lmp(fit_2), lmp(fit_3), lmp(fit_4), lmp(fit_5))
   use <- varnames[pvec < 0.15]
   use
   #[1] "tol14" "tol15"

To construct the formula:

   rhs <- paste(use, collapse = " + ")
   form <- paste("exposure ~", rhs)

And then use it:

   fit_multi <- lm(formula = form, data = tolerance)

Peter Ehlers

Re: filter rows by value

Rui Barradas — 2013-05-17T21:28:35

Hello,

Try the following.


dat <- read.table(text = "
Var   Time
1           51
2          151
3           251
4            234
5           331
6            351
", header = TRUE)

dat[dat$Time %% 100 == 51, ]



Em 17-05-2013 22:01, Ye Lin escreveu:

filter rows by value

Ye Lin — 2013-05-17T21:01:20

Hey All,

I want to delete rows based on the last 2 digits on the value in one column
but I dont know how to do that.

Suppose my data looks like this:

Var   Time
1           51
2          151
3           251
*4            234*
*5           331*
6            351

I want to delete the rows that the value in column "Time", the last 2 digit
is not 51, in this case the rows highlighted will be removed.

Thanks for your help!

[[alternative HTML version deleted]]

Re: #Keeping row names when using as.data.frame.matrix

arun — 2013-05-17T19:58:25


Hi,
library(plyr)
res<-dcast(dataset,Date~ScowNo,sum,value.var="EstimatedQuantity")
 rownames(res)<- res[,1]
res[,-1]
#         3002 4001 4002 BR 8
#9/7/2010 2772 3535 6763 6685
#9/8/2010    0 3305    0    0
A.K.

----- Original Message -----
From: Tim <tfox< at >mde.state.md.us>
To: r-help< at >r-project.org
Cc: 
Sent: Friday, May 17, 2013 12:46 PM
Subject: [R] #Keeping row names when using as.data.frame.matrix

#question I have the following data set:

Date<-c("9/7/2010","9/7/2010","9/7/2010","9/7/2010","9/7/2010","9/7/2010","9/8/2010")

EstimatedQuantity<-c(3535,2772,3279,3411,3484,3274,3305)

ScowNo<-c("4001","3002","4002","BR 8","4002","BR 8","4001")

dataset<- data.frame(EstimatedQuantity,Date,ScowNo)

#I'm trying to convert the data set into a contingency table and then back
into a regular data frame:

        
xtabdata<-as.data.frame.matrix(xtabs(EstimatedQuantity~Date+ScowNo,data=dataset),
         row.names=(dataset$Date),optional=F)

#I'm trying to keep the row names (in xtabsdat

Re: formatting column names of data frame

Rainer Schuermann — 2013-05-17T20:31:43

Have you tried xtable?

library( xtable )
x <- structure(list(Record = 1:3, Average = c(34L, 14L, 433L), Maximum = c(899L, 
15L, 1003L)), .Names = c("Record", "Average", "Maximum"), class = "data.frame", row.names = c(NA, 
-3L))
x <- xtable( x )
print( x )
% latex table generated in R 2.15.2 by xtable 1.7-1 package                                                                                                                                  
% Fri May 17 22:22:00 2013                                                                                                                                                                   
\begin{table}[ht]                                                                                                                                                                            
\centering                                                                                                                                                                                   
\begin

Re: image and color gradient

David Winsemius — 2013-05-17T20:22:30

On May 17, 2013, at 12:58 PM, Hermann Norpois wrote:


If you had used dput to offer that test case I would have tested my suggestions. As it is I will simply suggest:

... ,  breaks=10^-(0:8), ...

You will need to specify the colors to match the breaks.


David Winsemius
Alameda, CA, USA

Re: Problems using lmer {lme4}

Andrea Goijman — 2013-05-17T20:12:36

Hi Patrick,

Thanks for you reply. I tried adding site fixed effect as you told me, but
the program failed again (R stopped working).
Basically, what I am trying to do is to test for auto-correlation between
sites within routes. My survey takes place in routes, but each route is
divided in segments (or sites), and the presence of the species is
monitored at a site level.
A simple table for one species only would be as follows:

  Route site Visit1 Visit2 Visit3 Visit4 Visit5 #detections (punto6)  1 1-1
1 0 0 0 1 2  1 1-2 1 1 1 1 0 4  1 1-3 0 1 1 1 1 4  1 1-4 0 1 1 0 0 2  1 1-5
1 0 0 0 1 2  2 2-1 0 0 1 1 1 3  2 2-2 1 1 1 1 1 5  2 2-3 0 0 0 1 0 1  2 2-4
0 0 1 1 1 3  2 2-5 0 0 1 0 1 2  3 3-1 1 1 0 1 0 3  3 3-2 1 0 1 1 1 4  3 3-3
1 0 1 1 1 4  3 3-4 0 1 1 1 0 3  3 3-5 0 0 0 1 1 2


On Fri, May 17, 2013 at 2:15 PM, Patrick Coulombe <
patrick.coulombe< at >gmail.com> wrote:

Re: inverse for formula transformations on LHS

Roger Koenker — 2013-05-17T19:58:00

Paul,

Inverting log(y)  is just the beginning of the problem,  after that you need to
teach predict.lm()  that E(y |x) = exp(x'betahat + .5*sigmahat^2) and then further
lessons are required to get it to understand how to adapt its confidence and
prediction bands…  and then you need to generalize all of this to other
transformations.  Quite a project!

Best,
Roger

Roger Koenker
rkoenker< at >illinois.edu




On May 17, 2013, at 12:21 PM, Paul Johnson wrote: