Solved:
I reinstalled R 2.8.1.
B.
Birgitle wrote:
>
> It is also not possible if I use this commands
>
> install.packages("analogue","/Library/Frameworks/R.framework/Versions/2.8/Resources/library",repos="http://R-Forge.R-project.org";)
where TRUE/FALSE is necessary
Error in .readRDS(pfile) : unbekanntes Eingabeformat
Fehler in if (!noCache && file.exists(dest) && file.info(dest)$mtime > :
Fehlender Wert, wo TRUE/FALSE nötig ist
Please could somebody give me an advice. I can not do anything because I
need some packa
Hello R-User!
I am running R 2.8.1 on an Intel Mac.
I just tried to install a package using the GUI and got the following error
message:
Fehler in if (14 + nchar(dcall, type = "w") + nchar(sm[1], type = "w") > :
Fehlender Wert, wo TRUE/FALSE nötig ist
Error in (14 + nchar(dcall, type = "w")
Hello R-list members!
I tried to do the following with my dataset that contains factor and
numerics, (80columns,about 600 rows)
Dataset.afdm<-AFDM(Dataset[282:595,], type=TypeVector, ncp=3)
Fehler in svd(X) : infinite or missing values in 'x'
TypeVector
[1] "n" "n" "n" "n" "n" "n" "n" "n" "n"
A),FUN=function(x)sum(x=='1')) # '1' is
> here the level of the corresponding factor to be counted.
>
> will work. A more sophisticated version could include some "factor to
> numeric" conversion, see FAQ 7.10.
> hth.
>
> Birgitle schrieb:
>
e with 3 factor
> variables due to the way cbind processes the input variables - which is
> not intended I think.
>
> You can do sth like
>
> ABC<-data.frame(A,B,C)
> aggregate(ABC[,2:3],by=list(A),sum)
>
> hth.
>
> Birgitle schrieb:
>> Hello R
Hello R-Users!
I need a little help to build up a contingency table out of several
variables.
A<-c("F","M","M","F","F","F","F","M","F","M","F","F")
B<-c(0,0,0,0,0,0,1,1,1,1,0,1)
C<-c(0,1,1,1,1,1,1,1,1,0,0,0)
ABC<-as.data.frame(cbind(A,B,C))
ABC
A B C
1 F 0 0
2 M 0 1
3 M 0 1
4
Thanks Gavin and sorry to all for this unnecessary question.
B.
Gavin Simpson wrote:
>
> On Tue, 2008-09-16 at 10:47 -0700, Birgitle wrote:
>> Hello R-User!
>>
>> I try to do the following:
>>
>> New<-iris[c(1:7,90:97),1:5]
>> New.rp
Hello R-User!
I try to do the following:
New<-iris[c(1:7,90:97),1:5]
New.rpart<-rpart(Species~., data=New, method="class")
New.rpart
n= 15
node), split, n, loss, yval, (yprob)
* denotes terminal node
1) root 15 7 versicolor (0.467 0.533) *
Does it mean it is not possible to fi
agnes(cluster) to
perform a clustering?
Thanks again
B.
Birgitle wrote:
>
> I try to perform a clustering using an existing dissimilarity matrix that
> I calculated using distance (analogue)
> I tried two different things. One of them worked and one not and I don`t
> understand
I try to perform a clustering using an existing dissimilarity matrix that I
calculated using distance (analogue)
I tried two different things. One of them worked and one not and I don`t
understand why.
Here the code:
not working example
library(cluster)
library(analogue)
iris2<-as.data.frame(ir
Hello R-User!
When I plot my rpart-object:
plot(data.rpart); text(data.rpart,cex=0.2, use.n=T)
I get this error message
No object with name: PDE
Can somebody tell me why and what the meaning of the message is?
Many thanks in advance
B.
-
The art of living is more like wrestling tha
Thanks again.
Unfortunately I have always this missing values problem.
But the missings have also a meaning and its impossible to code it
differently or impute.
Also thanks for the explanation. Now I understand.
B.
Achim Zeileis wrote:
>
> On Wed, 13 Aug 2008, Birgitle wrote:
>
Many thanks.
Much easier than my solution
B.
Birgitle wrote:
>
> I just started to write tiny functions and therefore I appologise in
> advance if I am asking stupid question.
>
> I wrote a tiny function to give me back from the original matrix, a matrix
> showing only
I just started to write tiny functions and therefore I appologise in advance
if I am asking stupid question.
I wrote a tiny function to give me back from the original matrix, a matrix
showing only the values smaller -0.8 and bigger 0.8.
y<-c(0.1,0.2,0.3,-0.8,-0.4,0.9)
x<-c(0.5,0.3,0.9,-0.9,-0.7
ation would be should be used as partitioning variables.
> ...
>
>
What do you mean with "linear specification"? I would be very happy if you
could explain.
Thanks again
B.
Achim Zeileis wrote:
>
> On Wed, 13 Aug 2008, Birgitle wrote:
>
>> I try tu use
= Classe72)
then "party with the mob":
library(party)
Test.mob<-mob(X1~1|X34+X45+X73, data=DFExample, model=glinearModel,
family=binomial())
Fehler in `[.data.frame`(x, r, vars, drop = drop) :
undefined columns selected
B.
Birgitle wrote:
>
> I try tu use mob() with my
I try tu use mob() with my data.frame ('data.frame':288 obs. of 81
variables; factors, numerics and ordered factors)
My response is a binary variable and I should use for modelling a logistic
regression (family=binomial).
I read in the "MOB" Vignette that I could use a formula like this if I
You could perhaps do it like that
a<-a[c(1,2,4,3),]
B.
Zhang Yanwei - Princeton-MRAm wrote:
>
> Hi all,
> I have a 4 by 4 matrix, and I want to switch row 2 and row 3 first, then
> switch column 2 and column 3. Is there an easy way to do it?
> The following is a tedious way to get what I wa
sorry here the right thing
a<-a[c(1,3,2,4),c(1,3,2,4)]
B.
Birgitle wrote:
>
> You could perhaps do it like that
>
> a<-a[c(1,2,4,3),]
>
> B.
>
>
> Zhang Yanwei - Princeton-MRAm wrote:
>>
>> Hi all,
>> I have a 4 by 4 matrix, and I w
:
>
> ## The first fails; the second works
> hetcor(TestPart[,c(1:11,13:22,24:43,45:60)], pd=T, std.err=F)
> hetcor(TestPart[,c(1:72)], pd=F, std.err=F)
>
>
Mark Difford wrote:
>
> Hi Birgitle,
>
>>> It seems than, that it is not possible to use all variabl
Many Thanks Mark for your answer.
It seems than, that it is not possible to use all variables without somehow
imputing missing values.
But I will try which variables I can finally use.
Many thanks again.
B.
Mark Difford wrote:
>
> Hi Birgitle,
>
> It seems to be failing on t
t;of",
12))
TestPart<-read.table("TestPart.txt", header=TRUE,row.names=1,
na.strings="NA" ,colClasses = Classe72)
library(polycor)
TestPart.hetcor<-hetcor(TestPart, use="complete.obs")
B.
Birgitle wrote:
>
> Thanks Mark and I am sorry that I
eric",6), rep ("factor",
12))
TestPart<-read.table("TestPart.txt", header=TRUE,row.names=1,
na.strings="NA" ,colClasses = Classe72)
library(polycor)
TestPart.hetcor<-hetcor(TestPart, use="complete.obs")
Mark Difford wrote:
>
> Hi Birgitle,
>
Sorry if this post should be long but I tried to give you a piece of my data
to reproduce my error message using hetcor:
Fehler in result$rho : $ operator is invalid for atomic vectors
Zusätzlich: Warning messages:
1: In polychor(x, y, ML = ML, std.err = std.err) :
1 row with zero marginal remo
Many, many thanks that was fast and exactly what I was looking for.
B.
Mark Difford wrote:
>
> Hi Birgitle,
>
>>> ... my variables are dichotomous factors, continuous (numerical) and
>>> ordered factors. ...
>>> Now I am confused what I should use to cal
Hello R-User!
I appologise in advance if this should also go into statistics but I am
presently puzzled.
I have a data.frame (about 300 rows and about 80 variables) and my variables
are dichotomous factors, continuous (numerical) and ordered factors.
I would like to calculate the linear correlat
Hello R-User!
I have a data.frame with 82 variables (columns) and 290 rows.
The variables are set to classes factor, ordered factor and numeric.
I used the following code
Matrix.My.data<-as.matrix(Df.My.Data[2:82])
Matrix.My.data.rcorr<-rcorr(Matrix.My.data, type="spearman")
and got the foll
Hello R-Users, again me!
I have a data.frame with 291 rows, 82 columns.
Tha variables in the columns are factors, numerics and ordered factors.
The response variable is a factor with two levels.
I would like to find the best model by trying every possible variable
combination using a logistic
I am sorry I just found the stupid mistake.
I did not specify dec="," because I usually use .
Anyway thanks for having the opportunity to get help in this list.
B.
Birgitle wrote:
>
> Hello R-User
>
> I have a table as tab-delimited textfile (291 rows, 83 columns
Hello R-User
I have a table as tab-delimited textfile (291 rows, 83 columns).
The first row are labels and the first line the variable names.
I used the following code several times with different similar tables and it
always worked.
But now:
setClass("of")
setAs("character", "of", functi
Still the same question:
Birgitle wrote:
>
> I try to use ?randomForest to find variables that are the most important
> to divide my dataset (continuous, categorical variables) in two given
> groups.
>
> But when I plot the outlier:
>
> plot(outlier(rfObject, cls=grou
Hello RUser!
I try to use ace for an ancestral state reconstruction but got back an error
message.
ace(FacVar,Tree, type="discrete")
Warning messages:
1: In nlm(function(p) dev(p), p = rep(ip, length.out = np), hessian = TRUE)
:
NA/Inf durch größte positive Zahl ersetzt (NA/Inf repla
R 2.7.2
PPC Mac OS X 10.4.11
library mice 1.13.1
I try to use mice for multivariate data imputation.
My variables are numeric, factors, count data, ordered factors.
First I created a vector for the methods to use with each variable
ImpMethMice<-c(rep("logreg", 62), rep("polyreg",1), rep("
I tried to use ctree but am not sure about the meaning of the plot.
My.data.cf<-ctree(Resp~., data=My.data)
plot(FemMalSex_NAavoid88.ct)
My data.frame contains 88 explanatory variables (continous,ordered/unordered
multistate,count data) and one response with two groups.
In the plot are only two
You could have a look at library(analogue) , function ?distance
and library (cluster), function ?agnes
B.
Chua Siang Li wrote:
>
>
>Hello there. Is there any function in R that can do cluster on a set
> of
>data that has both categorical and numerical variables? thanks.
>siangl
I try to use ?randomForest to find variables that are the most important to
divide my dataset (continuous, categorical variables) in two given groups.
But when I plot the outliers:
plot(outlier(FemMalSex_NAavoid88.rf33, cls=FemMalSex_NAavoid88$Sex),
type="h",col=c("red","green")[as.numeric(FemMa
I have an additional question concerning to this topic.
I usually use something liek that:
read.table(, colClasses=c("numeric", "factor", "character",
"my.funny.class"))
but why can I not implement "ordered.factor" in there?
Birgit
Kenn Konstabel wrote:
>
> Conversion to factor may hap
n`t know what this means)
operator
Thanks
Birgit
mel-10 wrote:
>
> Birgitle a écrit :
>> Many thanks.
>> Is there a way to give me the number of the row, if I have the row name?
>> B.
>
> a= object
> 'w' = name
>
> > match('w', name
Additonally I got this error message
TestDist = Dist.HalbDisGow88[-147,]
Fehler in Dist.HalbDisGow88[-147, ] : falsche Anzahl von Dimensionen
(Error in Dist.HalbDisGow88[-147, ] : wrong number of dimensions
Birgit
mel-10 wrote:
>
> Birgitle a écrit :
>
>> I have a dist obje
Many thanks.
Is there a way to give me the number of the row, if I have the row name?
B.
mel-10 wrote:
>
> Birgitle a écrit :
>
>> I have a dist object containing 1 row that is only NA (not very
>> intelligent
>> to have bas dataset with one NA speciesanyway)
I have a dist object containing 1 row that is only NA (not very intelligent
to have bas dataset with one NA speciesanyway).
I would like to delete this row from this object.
It may be not a difficult problem but I can not find a solution presently.
So I would be very happy if somebody could
I think you should specify your grouping factor:
g a vector or factor object giving the group for the corresponding
elements of x. Ignored if x is a list.
batlett.test(xx, groupingfactor)
Hope this helps.
Birgit
hanen wrote:
>
> i'm trying to test the homogeneity of variance of 92 samples
Many thenks to both of you:
Will have a look.
Birgit
Chuck Cleland wrote:
>
> On 6/4/2008 5:32 AM, Birgitle wrote:
>> My dataset contains missing data and I would like to do something like an
>> EM
>> algorithm or a Markov Chain Monte Carlo approach to get rid o
My dataset contains missing data and I would like to do something like an EM
algorithm or a Markov Chain Monte Carlo approach to get rid of the missing
data.
Is there a function for imputation or simulation of missing data apart from
those in the randomForest library?
Thanks in advance
Birgit
Thanks might be easier in my case because I have so many variables.
Could have found this solution on my own.
Birgit
Rogers, James A [PGRD Groton] wrote:
>
>
> Birgit Lemcke wrote:
>
>> I have a dataframe and two of my variables are in the wrong position
>> and I would like to swap those va
That works perfect.
Thanks a lot Paul!
Greets
Birgit
Paul Smith wrote:
>
> On Mon, Jun 2, 2008 at 1:04 PM, Birgitle <[EMAIL PROTECTED]>
> wrote:
>>
>> Thanks Paul.
>>
>> I am not sure if I understood well, but when I do it then I have only two
&g
Thanks Paul.
I am not sure if I understood well, but when I do it then I have only two
columns left:
> L3 <- LETTERS[1:3]
> (d <- data.frame(cbind(x=1, y=1:10, z=11:20), fac=sample(L3, 10,
> replace=TRUE)))
x y z fac
1 1 1 11 C
2 1 2 12 B
3 1 3 13 B
4 1 4 14 C
5 1 5 15 C
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