gmane.comp.lang.r.general

distances between entities

Rory Wilson — 2013-05-18T15:44:38

Hello all, please forgive my lack of knowledge, but I am somewhat new to R and would be very grateful for some help. I have tried some searches, but am not even quite sure what to search for.  I will describe my problem:
I have a number of nodes (or entities)  and pairwise (Euclidean) distances for SOME of the pairs of nodes, but I do not know where in the space my nodes lie. Is there a package of some kind that would allow me to generate a best-possible picture of the geometry of the nodes, i.e. a guess as to where most of them are in the space (relative to each other). I know it is impossible to get the full picture (since I have incomplete information), but a kind of minimum-error guess?
Please let me know if you need more information, or if you suspect you have an answer to my query.
Thank you very much,
Rory
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[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
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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

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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
!



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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

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!

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image and color gradient

Hermann Norpois — 2013-05-17T19:58:05

Hello,

I have a nice function that makes an image of an matrix
e.g.:
qt[1:3,1:3]
             rs655246     rs943795 rs955612
rs655246           NA           NA       NA
rs943795 9.610070e-04           NA       NA
rs955612 5.555616e-05 7.915982e-07       NA


myimage <- function(x, cex.axis = 0.7, ...){
  opar <- par(mar=c(5,4,4,6),
              pty ='s')
  on.exit(par(opar))
  image(x, axes = FALSE, ...)
  ats <- 0:(nrow(x)-1)/(nrow(x)-1)
  axis(1, at=ats, lab=rownames(x), cex.axis=cex.axis, las=2)
  axis(4, at=ats, lab=colnames(x), cex.axis=cex.axis, las=2)
  box()
}

The ranges in my matrix are from 1 to 1e-08. But in my image there is no
difference between, for instance 1e-05 and 1e-06 or -07 etc.

How can I ameliorate my function myimage to do so. I guess it has something
to do with breaks but I do not understand how to handle.
Thanks
Hermann

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formatting column names of data frame

Patrick Leyshock — 2013-05-17T19:53:12

Is there any way to format the headers of data frames, for printing?

I am using Sweave to generate formatted reports.  In Sweave, I read in a
data.frame:

              result <- read.table(path.to.table);

then display it:

              print.data.frame(result);

This gives me what I expect in the eventual final output:

Record        Average          Maximum
1               34              899
2               14              15
3               433             1003
...             ...             ...

What I am hoping to do is distinguish one or more of the column headers,
for example, I want "Average" or "Maximum" to be bold, underlined, etc.,
just some way to make the column name stand out visually.

Any idea if this is possible?  Any suggestions appreciated.

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Bivariate - multivariate linear regression

Jesse Gervais — 2013-05-17T19:45:53

Hi there,



I want to do several bivariate linear regressions and, than, do a
multivariate linear regression including only variables significantly
associated *(p < 0.15)* with y in bivariate analysis, without having to
look manually to those p values.



So, here what I got for the moment.



First, I use this data set:



tolerance <- read.csv("
http://www.ats.ucla.edu/stat/r/examples/alda/data/tolerance1.txt").



Second, I used this command, allowing me to extract p-values later:



lmp <- function (modelobject) {

            if (class(modelobject) != "lm") stop("Not an object of class
'lm' ")

            f <- summary(modelobject)$fstatistic

            p <- pf(f[1],f[2],f[3],lower.tail=F)

            attributes(p) <- NULL

            return(p)}



Third, I did my bivariate linear regressions:



fit   = lm(exposure~tol11, data = tolerance)

fit_2 = lm(exposure~tol12, data= tolerance)

fit_3 = lm(exposure~tol13, data= tolerance)

fit_4 = lm(exposure~tol14, data= tolerance)

fit_5 = lm(exposure~tol1

help

masumeh akhgar — 2013-05-17T18:55:58

hi all
this command used tt function for all variables.
How can i define a different function for each variable?
 > exCox.all<-coxph(Surv(SURVT,STATUS) ~
RX+LOGWBC+SEX+tt(RX)+tt(LOGWBC)+tt(SEX),
data=rem.data,tt=function(x,t,...) log(t)*x))
thank you

help

masumeh akhgar — 2013-05-17T18:34:12

hi deer all
Estimate KM survival probabilities for each categories of RX means
‘treatment’ and ‘placebo’ separately
when write that command. it doesnt run.
what should i do?
thanks

help

masumeh akhgar — 2013-05-17T18:15:22

Hello,

I fail to tranfer data from a dataframe to a matrix.

var is from a dataframe (and belongs still to the class dataframe) and
should look like m (see below).

  vec1 vec3  d1  d2
1  172  173 223 356
structure(list(vec1 = 172L, vec3 = 173L, d1 = 223L, d2 = 356L), .Names =
c("vec1",
"vec3", "d1", "d2"), row.names = 1L, class = "data.frame")
     [,1] [,2]
[1,]  172  223
[2,]  173  356

structure(c(172, 173, 223, 356), .Dim = c(2L, 2L))

How can I transform var to m?
Thanks
akhgar

time-series aggregation of information

Chirag Maru — 2013-05-17T18:48:00

I have following data for which I need to calculate the weighted aggregate value of the parameter at each time.

Date,Parameter,Weight
2012-01-31,90,200
2012-01-31,80,400
2012-01-31,70,500
2012-01-31,60,800
2012-02-29,120,220
2012-02-29,110,410
2012-02-29,75,520
2012-02-29,65,840
2012-03-31,115,210
2012-03-31,100,405
2012-03-31,70,500
2012-03-31,60,800

So for the above sample the solution looks like:

Date,Weighted Parameter
2012-01-31,70
2012-02-29,82.96482412
2012-03-31,77.10182768

Could I potentially use tapply / aggregate for this?  Would like to avoid a for loop if possible.

Thank you!




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

Tim — 2013-05-17T16:46:29

#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 xtabsdata) as the dates.
#But the row names keep coming up as integers.
#How can I preserve the row names as dates when
#the table is converted back to a data frame?




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inverse for formula transformations on LHS

Paul Johnson — 2013-05-17T17:21:59

This is an R formula handling question. It arose in class. We were working
on the Animals data in the MASS package. In order to see a relationship,
you need to log brain and body weight.  It's a fun one for teaching
regression, if you did not try it yet.  There are outliers too!

Students wanted to make a predicted value plot in the non-logged values of
y, for comparison, and I wondered if I couldn't automate this somehow for
them.

It made me wonder how R manages formulae and if a transformation like
log(y) can be be mechanically inverted.

So we have something concrete to talk about, suppose x and y are variables
in dat, a person fits

m1 <- lm(log(y) ~ log(x), data = dat)

termplot shows log(y) on the vertical.  What if I want y on the vertical?
Similarly, predict gives values on the log(y) scale, there's no argument
like type = "untransformed".

I want my solution to be a bit general, so that it would give back
predicted y for formulae like

sqrt(y)

or

exp(y)

or

log(y + d)

or whatever other math peo

Problems using lmer {lme4}

Andrea Goijman — 2013-05-17T17:02:38

Dear R list,

I'm attaching a sample of my data which consists on the presence/absence
("punto6", binomial n=5 occasions)
of different species ("sp"), on different sites ("site") within routes
('route").

First, I want to be able to find if there is autocorrelation of the
response variable between
the sites within each route. For this I start testing 2 models, but when I
try to run the second model,
to test for the random effects of sites within routes R stops working!

I'm not being able to find out why r is crashing... and whay am I doing
wrong.

Thanks!

Andrea

#######################################################
#dput(d)

d<-structure(list(site = structure(c(55L, 56L, 57L, 58L, 59L, 60L,
55L, 56L, 57L, 58L, 59L, 60L, 55L, 56L, 57L, 58L, 59L, 60L, 55L,
56L, 57L, 58L, 59L, 60L, 55L, 56L, 57L, 58L, 59L, 60L, 55L, 56L,
57L, 58L, 59L, 60L, 55L, 56L, 57L, 58L, 59L, 60L, 229L, 230L,
231L, 232L, 233L, 234L, 229L, 230L, 231L, 232L, 233L, 234L, 229L,
230L, 231L, 232L, 233L, 234L, 229L, 230L, 231L, 232L, 233L,

Comma separated vector

Manta — 2013-05-17T15:46:59

Hi all,

I have a vector of numbers, and to be able to pass it to RMySQL and use the
IN clause I need to have this vector to be a single list numeric and comma
separated.

I saw the post below but it is about strings, which I do not need (I cannot
pass strings in this SQL query, I need something like ' where ASSETT in
(1,2,3,4,5)'

http://stackoverflow.com/questions/6347356/creating-a-comma-separated-vector


Any clue?



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R and libre office base

Johan Lassen — 2013-05-17T14:47:02

Dear community . I would like to connect r to libre office base. Does
anyone know if and how this can be done? I think of the pendant to rodbc
for libre office. I am using windows 7.
Thanks in advance and best regards johan

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Error with adehabitatHR and kernelbb

Rémi Lesmerises — 2013-05-17T14:44:52

Dear all,

I'm trying to get a Brownian bridge kernel (kernelbb) for each combination of two consecutive animal locations (see commands below) and put them, with a loop, inside a list. It works well at the beginning but after 42 runs, it appears the following warning :



I looked at the coordinates, at the id, at the time of the run 43 and it's all good...

I looked on the net and it happened to only one person and there was no answer to his post. 

Someone could help me?

## commands

BBtraj <- list()
for (i in 1:(nrow(loc< at >data)-1)) {
BBtraj[[i]] <- kernelbb(as.ltraj(loc< at >coords[i:(i+1),], date=loc< at >data$time[i:(i+1)], id = as.character(loc< at >data$id[i:(i+1)]),
typeII = TRUE), sig1=as.numeric(as.character(loc< at >data$sig1[i])), sig2= 5, grid = 1000)
}

 
Rémi Lesmerises, biol. M.Sc.,
Candidat Ph.D. en Biologie
Université du Québec à Rimouski
remilesmerises< at >yahoo.ca

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mirt package "error in ESTIMATION..."

kende jan — 2013-05-17T14:17:59

Hello everyone,

I am trying to undertake an item bifactor analysis of graded response data from a questionnaire. I am using the mirt package, especially the bfactor function.My dataset is called "data.items", it contains about 2000 observations and 31 variables (variables represent the items in the questionnaire). The items follow a Likert scale format and represent a level of satisfaction, missing values are coded as NA). I am having trouble at the beginning of my analysis, during the exploratory model fitting. The syntax and the error message are as follows:

Error in ESTIMATION(data = data, model = model, group = rep("all", nrow(data)),  : 
  index out of bounds

I get the same error when i try with: bfactor(data.items,9,itemtype="graded") or  bfactor(data.items,9,prev.cor=cor)
where cor is the correlationmatrix that i compute without the missing values.

If someone have an idea to suggest, do not hesitate.

Thank you.

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