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
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-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
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-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
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
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
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,
>
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
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
:
>
> ## 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
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
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
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
= 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
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
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
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
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:
>
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
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
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
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
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 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
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
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
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
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
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)
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
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
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
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
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 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
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("
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
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 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
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-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
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
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:
>
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"
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")
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
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";)
48 matches
Mail list logo