I was going through this example of maxent (Maximun entropy package) use:
http://cran.r-project.org/web/packages/maxent/maxent.pdf
# LOAD LIBRARY
library(maxent)
# READ THE DATA, PREPARE THE CORPUS, and CREATE THE MATRIX
data - read.csv(system.file(data/NYTimes.csv.gz,package=maxent))
corpus -
Hello,
Has any1 ever tried to parse a useragent string to get the OS and Browser
information out of it?
There are many Java, PHP and Python ways to do it, but I was wondering if
there is a way to do it with R.
Googling retrieved no help at all.
A useragent string might look like this :
Hi,
Sorry to bother you again.
I would like to estimate the effect of several categorical factors (two
between subjects and one within subjects) on two continuous dependent
variables that probably covary, with subjects as a random effect. *I want
to control for the covariance between those two
'Unicode' is a not an encoding. As the help says
fileEncoding: character string: if non-empty declares the encoding used
on a file (not a connection) so the character data can be
re-encoded. See the ‘Encoding’ section of the help for
‘file’, the ‘R Data
As a matter of principle, yes, multivariate mixed models do exist, look at the
last bit of example(manova) (in reasonably recent versions of R).
In practice, it often doesn't really buy you much. It just gives a joint test
for all the DVs, the estimates are the same as in separate analyses.
In the pairs help is an example how to plot the correlation as txt.
How do I get rid of the axes
in the upper panel?
regards
## put (absolute) correlations on the upper panels,
## with size proportional to the correlations.
panel.cor - function(x, y, digits = 3, prefix = , cex.cor,axis=F, ...)
Le mardi 08 octobre 2013 à 16:02 -0700, Ira Sharenow a écrit :
A colleague is sending me quite a few files that have been saved with MS
SQL Server 2005. I am using R 2.15.1 on Windows 7.
I am trying to read in the files using standard techniques. Although the
file has a csv extension when
Please ignore. My apologies for the noise.
cheers,
Rolf Turner
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and provide commented,
I have a problem that I am trying to resolve with no success. More than two
days searching and I didn't get a single clue. Sorry if the answer is out
there and I didn't find it.
Suppose that you have a logistic equation regression (binary model) from an
old model that you estimated some years
Hi Peter, and thank you for your quick and helpful reply !
Do you want to know whether the predictors affect the marginal
distributions of Y1, Y2,... or are you interested in conditional effects
given other DVs (aka test for additional information)?
Hmmm I think that, yes, i am looking for that
Dear useRs,
A new version of the package âTestSurvRecâ is now available on CRAN. In
this version I add new function and new data of recurrent events.
The new function called âPlot.Data.Eventsâ plots recurrent event data and
the new data is useful as example for see the
Hello,
I am trying to write the default OrchardSprays R data frame into HDFS
using the rhdfs package. I want to write this data frame directly into
HDFS without first storing it into any file in local file system.
Which rhdfs command i should use? Can some one help me? I am very new to R
and
Hi,
I have a set of parameters that I want to test for covariance using Spearman's
correlation coefficient.
cor.test(x=x,y=y,method=spearman)
These parameters are taken from plots which may be spatially correlated. How
can I use Dutilleul's method to account for spatial autocorrelation while
Hi,
I have trying to run a simple MR program using rmr2 in a single node Hadoop
cluster. Here is the environment for the setup
Ubuntu 12.04 (32 bit)
R (Ubuntu comes with 2.14.1, so updated to 3.0.2)
Installed the latest rmr2 and rhdfs from
Hi,
New to R here. Lots of fun. Still rather green though.
I'd like to select unique items from a list that looks like this (for
example):
xyz
$x
[1] 8 6 9 0 0 3 9 7 1 9
$y
[1] 1 2 9 5 1 2 0 9 2 9
$z
[1] 5 6 9 0 5 1 1 7 3 4
I'd like to select unique (x,y), while preserving association
Suppose that you have a logistic equation regression (binary model) from an
old model that you estimated some years ago. Therefore you know the
parameters âk (k = 1, 2, ..., p) because they were estimated in the
past.
But you don't have the data that were used to fit the model.
My
Hello all,
I need to infer if the slopes and intercept of each group of my lme analyze,
showed below, are different from each other. Like a Tukey test. I don't
need to compare the slope and interceptof each subject of my fixed data,
but the slope and intercept of each group of them.
Never mind -- after getting detailed help at SO, I rebooted Windows and
makeCluster() worked just fine. I have no idea what cruft in my system was
getting in the way, but all's well.
--
View this message in context:
Hi,
trees - structure(list(Girth = c(8.3, 8.6, 8.8, 10.5, 10.7, 10.8, 11,
11, 11.1, 11.2, 11.3, 11.4, 11.4, 11.7, 12, 12.9, 12.9, 13.3,
13.7, 13.8, 14, 14.2, 14.5, 16, 16.3, 17.3, 17.5, 17.9, 18, 18,
20.6), Height = c(70L, 65L, 63L, 72L, 81L, 83L, 66L, 75L, 80L,
75L, 79L, 76L, 76L, 69L, 75L,
Johannes Radinger johannesradinger at gmail.com writes:
Hi,
I'd like to use Latin Hypercube Sampling (LHC) in the the context of
uncertainty / sensitivity analysis of a complex model with approximately
10
input variables. With the LHC approach I'd like to generate parameter
combinations
Hi,
Check this link (not tested):
http://stackoverflow.com/questions/7771577/clear-startup-screen-in-r-rstudio
A.K.
I just want to remove the help information when I start R console.
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Hi,
Not sure about your expected output.
xyz-
list(x=c(8,6,9,0,0,3,9,7,1,9),y=c(1,2,9,5,1,2,0,9,2,9),z=c(5,6,9,0,5,1,1,7,3,4))
indx-sort(unique(unlist(lapply(xyz[1:2],function(u)
which(!duplicated(u))),use.names=FALSE)))
xyz[1:2]-lapply(xyz[1:2],function(u) u[!duplicated(u)])
xyz[3]$z-
I can't comment on frailtypack issues, but would like to mention that coxme will handle
nested models, contrary to the statement below that frailtypack is perhaps the only
for nested survival data.
To reprise the original post's model
cgd.nfm - coxme(Surv(Tstart, Tstop, Status) ~
You can create a glm fit with only an offset created from the coefficients
that you have, then use the regular predict function with that. For
example using the iris data (first fitting a model on the real data, then
fitting a new model using dummy data and the coefficients from the first
fit):
Thanks for everyoneâs feedback. It may be that the person who is using
SQL Server 2005 is doing something that is not helpful.
In the data there are supposed to be 10 columns with headers. The first
column is a date time that I was reading in as a string. The second is a
string. The rest are
Hi,
May be this is what you are looking for.
indx- xyz[[1]]==xyz[[2]]
indx1- which(indx)[duplicated(xyz[[1]][indx])]
lapply(xyz,function(u) u[-indx1])
$x
[1] 8 6 9 0 0 3 9 7 1
$y
[1] 1 2 9 5 1 2 0 9 2
$z
[1] 5 6 9 0 5 1 1 7 3
lapply(xyz, function(u) u[indx1])
$x
[1] 9
$y
[1] 9
$z
[1] 4
On 09-10-2013, at 13:50, Ronald Peterson r...@hub.yellowbank.com wrote:
Hi,
New to R here. Lots of fun. Still rather green though.
I'd like to select unique items from a list that looks like this (for
example):
xyz
$x
[1] 8 6 9 0 0 3 9 7 1 9
$y
[1] 1 2 9 5 1 2 0 9 2 9
$z
[1] 5
Hi,
I recall that the form for an R formula statement has a formal name and
a reference, perhaps in the computer science area. I'm referring to a
statement such as Y ~ x1 + x2 + x3. Does anyone know the name for this
form and where it came from? I would like to use this in a presentation
Hi everyone,
I'm attempting to use the bnlearn package to calculate conditional
probabilities, and I'm running into a problem when the cpquery function is
used within a loop. I've created an example, shown below, using data included
with the package. When using the cpquery function in a loop,
Hi,
I got given some code that uses the R function pbionom:
p - mut * t
sumprobs - pbinom( N, B, p ) * 1000
Which gives the output of a probability as a percentage like 5, 50, 95.
What the code currently does is find me the values of t I need, by using the
above two code lines in a loop, each
Dear All,
I am trying to model the habitat mapping of 4 sharks species and my data are
zero inflated and positively skewed. The zero percentages in my data vary
between 56-77%. I did some tests fitting GAMs with negative binomial family
(with theta between 1-10). Is this the best way to go?
Thank you very much for responding. Yes, I incorrectly stated that
frailtypack was the only widely available software for the analysis of
nested frailty models.When I initially began my search to identify
software well suited to nested frailty analysis, Frailtypack dominated the
google
Thanks. That's not quite what I'm looking for, but it's good see different
ways to slice and dice data.
In my example, the one duplicated x,y pair would 9,9, so I would want to
reduce the original list to
xyz
$x
[1] 8 6 9 0 0 3 9 7 1
$y
[1] 1 2 9 5 1 2 0 9 2
$z
[1] 5 6 9 0 5 1 1 7 3
and if
On Wed, Oct 9, 2013 at 11:49 AM, Berend Hasselman b...@xs4all.nl wrote:
On 09-10-2013, at 13:50, Ronald Peterson r...@hub.yellowbank.com wrote:
Hi,
New to R here. Lots of fun. Still rather green though.
I'd like to select unique items from a list that looks like this (for
example):
Very cool! Thanks Berend and arun.
R.
On Wed, Oct 9, 2013 at 2:49 PM, Berend Hasselman b...@xs4all.nl wrote:
On 09-10-2013, at 13:50, Ronald Peterson r...@hub.yellowbank.com wrote:
Hi,
New to R here. Lots of fun. Still rather green though.
I'd like to select unique items from a
On 13-10-09 8:11 AM, Data Analytics Corp. wrote:
Hi,
I recall that the form for an R formula statement has a formal name and
a reference, perhaps in the computer science area. I'm referring to a
statement such as Y ~ x1 + x2 + x3. Does anyone know the name for this
form and where it came
Hi everyone,
I'm attempting to use the bnlearn package to calculate conditional
probabilities, and I'm running into a problem when the cpquery function is
used within a loop. I've created an example, shown below, using data included
with the package. When using the cpquery function in a loop,
Hi,
Try:
example1-
as.matrix(read.table(example1.txt,header=TRUE,stringsAsFactors=FALSE,sep=\t,row.names=1,check.names=FALSE))
example2 - example1
example1New - cbind(example1, `MAX without
Restriction`=apply(example1[,1:4],1,max,na.rm=TRUE))
#or
library(matrixStats)
`MAX without Restriction`-
Thank you for the fast answer!
Is there any way to use the first solution way, but work with the column
names instead of the number of the column. Because in further calculation I
need to drop more columns than just the restriction ones and the real
dataset has to many columns to work with the
Dear R wizards,
Though I hate to do it after weeks of my code not working I need some help
since I cant find an example which seems to work.
I am trying to create a graph which show the probability of predation of a
nest on one side (either 1 to 0) or (0% to 100%) on one side
and grass height at
Hi,
No problem.
It depends upon how many columns you want to delete, or if they have any common
names as in Restriction
example1-
as.matrix(read.table(example1.txt,header=TRUE,stringsAsFactors=FALSE,sep=\t,row.names=1,check.names=FALSE))
indx- which(!grepl(Restriction, colnames(example1)))
Hello
I've got a matrix of the following form
BARNBCGEBCVNBEANRestriction 1 Restriction 2
Restriction 3 Restriction 4
Restriction 5
alpha.1 0.000172449 7.87E-05-0.0032710440.000921609
9.28E-192.00E-05
-0.000608211NA NA
alpha.2
Thank you again! But somehow the which(function) didn't work...
The problem is, that the columns I want to drop in further calculations will
not have the same name, that's why I'm looking for a solution to make a
vector of the columns that should be omitted, like
omit-c(Restriction 1,Restriction
Dear r genii,
I hope you can help.
I have vector 'b':
b=c((1:10),sort(1:9,decreasing=TRUE),(2:12),sort(6:11,decreasing=TRUE),(7:13))
and, from 'b' I wish to create vector 'c':
c=c(
NA,NA,NA,NA,NA,NA,1,1,1,1,1,1,1,NA,NA,NA,NA,NA,NA,NA,NA,NA,NA,NA,2,2,2,2,2,2,2,2,2,2,2,NA,3,3,3,3,3,3,3)
On Oct 9, 2013, at 3:31 PM, laro wrote:
Thank you again! But somehow the which(function) didn't work...
The problem is, that the columns I want to drop in further calculations will
not have the same name, that's why I'm looking for a solution to make a
vector of the columns that should be
Hi,
Try:
!colnames(example1)%in%omit
#[1] TRUE TRUE TRUE TRUE FALSE FALSE TRUE TRUE TRUE
cbind(example1, `MAX without
Restriction`=apply(example1[,!colnames(example1)%in% omit],1,max,na.rm=TRUE))
A.K.
Thank you again! But somehow the which(function) didn't work...
The problem is,
Hello everyone,
I am trying to test some biclustering algorithms, and I am using the package
biclust.
I tried to bicluster a very simple matrix, and it seems that I cannot obtain
what I was expecting, even though the cluster, to me, seem pretty obvious.
The code is the following:
mat1-
Hi,
Try:
b1- b
b1[!b1=7]- NA
lst1 - split(b1,cumsum(c(0,abs(diff(b=7)
indx - as.logical(((seq_along(lst1)-1)%%2))
lst1[indx]- lapply(seq_along(lst1[indx]),function(i) {lst1[indx][[i]]-
rep(i,length(lst1[indx][[i]]))})
C2 - unlist(lst1,use.names=FALSE)
all.equal(c1,C2)
#[1] TRUE
A.K.
Uwe,
Good news. I installed 3.0.2, and the parallel package examples ran
successfully. This time a firewall window popped up. Probably the
firewall was the problem with the snow package too, but for some reason the
window didn't pop up with the snow package.
Thanks for the suggestion to use
Hi
Is there any implementation for AES encryption algorithm in R?
Thanks
Bander
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and provide
Hi,
I have what I think should be a basic question on knitr. I am just
moving to knitr, and previously I had created functions which
automatically created latex wrappers for many (100s) figures. I also
have other functions which automatically create pages worth of latex tables.
The knitr
On 10/10/2013 08:35 AM, Rebecca Stirnemann wrote:
Dear R wizards,
Though I hate to do it after weeks of my code not working I need some help
since I cant find an example which seems to work.
I am trying to create a graph which show the probability of predation of a
nest on one side (either 1 to
I think in most cases you are probably on the wrong track if you have
to generate LaTeX code for figures from R code (LaTeX tables are
another story), but your case might be different. You did not give a
specific example on why you had to do that, so I cannot give any
advice for now.
My best
Thanks Jim for helping,
Your sample data actually looks like my dataset. The one I put up looks
strange for some reason so please ignore that.
I have three landusenumb variables 1 2 and 3. is rep (1,2,3) correct?
When I run the following code I am getting:
mod1 - glmer(frat ~ flandusenumb +
But with three lines for the three habitat types and grass length at the
bottom
On Thu, Oct 10, 2013 at 5:52 PM, Rebecca Stirnemann
rstirnem...@gmail.comwrote:
Thanks Jim for helping,
Your sample data actually looks like my dataset. The one I put up looks
strange for some reason so please
On 10/10/2013 03:52 PM, Rebecca Stirnemann wrote:
Thanks Jim for helping,
Your sample data actually looks like my dataset. The one I put up looks
strange for some reason so please ignore that.
I have three landusenumb variables 1 2 and 3. is rep (1,2,3) correct?
When I run the following code I
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