Displaying 20 results from an estimated 8162 matches for "factorized".

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factorize

2011 Nov 08

2

why NA coefficients

Hi, I am trying to run ANOVA with an interaction term on 2 factors (treat has 7 levels, group has 2 levels). I found the coefficient for the last interaction term is always 0, see attached dataset and the code below:
> test<-read.table("test.txt",sep='\t',header=T,row.names=NULL)
> lm(y~factor(treat)*factor(group),test)
Call:
lm(formula = y ~ factor(treat) *

2010 Jan 13

0

need a clarification on logistic regression

hello
I need a clarification.
in logistic regression, saturated model having all combinations and
interactions of variables should be constructed in this way ? :
> rsat2=glm(cbind(landp,landa) ~
as.factor(rlito)*as.factor(rslp)*as.factor(rasp)*as.factor(rplc)*as.factor(rwi),family=
binomial(link=logit),data=rdf46)
then
using stepAIC to eliminate some models having high AIC values.
>

2016 Apr 19

0

Indicator Species analysis; trouble with multipatt

Hi Ansely,
As Jim points out we really need some sample data to go with the code.
Have a look at ?dput which is the best way to supply sample data here or have a look at http://stackoverflow.com/questions/5963269/how-to-make-a-great-r-reproducible-example and/or http://adv-r.had.co.nz/Reproducibility.html for some general suggestions on asking questions

2016 Apr 18

3

Indicator Species analysis; trouble with multipatt

Hello,
*Error in tx %*% comb : non-conformable arguments*
Suggestions greatly appreciated. I am a beginner and this is my first time
posting.
I would like to get the summary for indicator species analysis, using
package indicspecies with multipatt. I am getting errors, I believe, do to
my data organization. After reorganizing and reorganizing, nothing has
helped.
> data<-

2007 Jun 05

4

Refactor all factors in a data frame

Hi all,
Assume I have a data frame with numerical and factor variables that I
got through merging various other data frames and subsetting the
resulting data frame afterwards. The number levels of the factors seem
to be the same as in the original data frames, probably because subset()
calls [.factor without drop = TRUE (that's what I gather from scanning
the mailing lists).
I wonder if

2008 Sep 02

0

Error in .local(object, ...) : test vector does not match model !

I am getting a really strange error when I am using predict on an ksvm model. The error is "Error in .local(object, ...) : test vector does not match model !". I do understand that this happens when the test vectors do not match the
Model. But in this case it is not so. I am attaching a portion of both the test data used for prediction and the data used to build the model. I could

2006 Feb 07

1

post-hoc comparisons following glmm

Dear R community,
I performed a generalized linear mixed model using glmmPQL (MASS
library) to analyse my data i.e : y is the response with a poisson
distribution, t and Trait are the independent variables which are
continuous and categorical (3 categories C, M and F) respectively, ind
is the random variable.
mydata<-glmmPQL(y~t+Trait,random=~1|ind,family=poisson,data=tab)
Do you think it

2013 Jan 31

2

use name (not values!) of a dataframe inside a funktion

Dear Listers,
can anyone help me, please.
Since several days i try to figure out, how to assign values, vectors,
functions etc to variables with dynamically generated names inside of
functions.
Sometimes I succeed, but the success is rather arbitrary, it seems. up to
now i don't fully understand, why things like get, assign, <<- etc do
sometimes work, and sometimes not.
here's one

2015 Apr 30

2

predict nlme

Estimado Oliver Nuñez
Envío un ejemplo reproducible.
Javier Marcuzzi
# de donde tomo datos, y tiene el modelo (en el pdf)
library(MCMCglmm)
# librería con las funciónes que voy a usar
library(nlme)
datos0<-ChickWeight
# creo algunos datos que agrego a los origonales
Factor<-as.numeric(datos0$Chick)
Factor[Factor > 0 & Factor <= 10] <- 'A'
Factor[Factor > 10

2005 Feb 13

0

corrupt data frame: columns will be truncated or padded with NAs in: format.data.frame(x, digits = digits)

Hello R users!
I have written one function (look at the end), which will ease my
work with analysis of data in another programme, for which I need
sometimes a special data structure. However I encountered several
problems with a created data frame.
---------------------------------------------------------------
The data frame (produced from the example at the end) looks like
the way I want

2008 Jul 13

3

initialize a factor vector

What is the least surprising way of initializing a factor with
predefined levels and with length 0?
as.factor(c("eins", "zwei", "drei"))[FALSE]
does the job but looks a bit weird.
--
Johannes H?sing There is something fascinating about science.
One gets such wholesale returns of conjecture
mailto:johannes at

2011 Nov 10

4

Error in Summary.factor(c(1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 4L, : max not meaningful for factors

Hello,
Beginner, sorry if this is wasting anyone's time, but have been working on
this for a couple of days now, think it should have take a few hours!
The Problem:
Error in Summary.factor(c(1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 4L, :
max not meaningful for factors
I have tried to re-arrange and check formula. I am working in SIAR, and
cannot get model1 to run because of the error

2007 Dec 05

2

converting factors to dummy variables

Hi all -
I'm trying to find a way to create dummy variables from factors in a
regression. I have been using biglm along the lines of
ff <- log(Price) ~ factor(Colour):factor(Store) +
factor(DummyVar):factor(Colour):factor(Store)
lm1 <- biglm(ff, data=my.dataset)
but because there are lots of colours (>100) and lots of stores
(>250), I run it to memory problems. Now, not every

2010 Sep 21

1

Prime Factorization

Hi everyone, I have a very quick question:
Is there a ready-made function in R or any R packages to find the prime
factorization of an integer?
--
View this message in context: http://r.789695.n4.nabble.com/Prime-Factorization-tp2548877p2548877.html
Sent from the R help mailing list archive at Nabble.com.

2018 Apr 18

3

How to replace numeric value in the column contains Text (Factor)?

Hi R user,
Would you mind to help me on how I can change a value in a specific column
and row in a big table? but the column of the table is a factor (not
numeric).
Here is an example. I want to change dat[4:5,3]<-"20" but it generated NA>
do you have any suggestions for me?
dat<-structure(list(Sites = structure(1:5, .Label = c("Site1", "Site2",

2011 May 24

3

Beginner Question: List value without Levels

Hey folks,
I'm new to the R Project so I'm facing a great problem. I've read a file
into R:
>myVal
I V L F C M A G T W S Y P H Q D
N E K R
1 4,5 4,2 3,8 2,8 2,5 1,9 1,8 -0,4 -0,7 -0,9 -0,8 -1,3 -1,6 -3,2 -3,5 -3,5
-3,5 -3,5 -3,9 -4,5
> mode(myVal)
[1] list
Now I want to multiplicate each of this values with this one:

2009 Nov 08

1

ordered factor and unordered factor

I don't understand under what situation ordered factor rather than
unordered factor should be used. Could somebody give me some examples?
What are the implications of order vs. unordered factors? Could
somebody recommend a textbook to me?

2017 Jun 26

1

How to code the factor for ANOVA

Hi there,
I have a experimental design related question. I have done a experiment
with 3 factors. The design matrix is similar to:
> data.frame(Factor.1 = rep(rep(0:2, each = 3),3), Factor.2 =
rep(c("U","S","N"), 9), Factor.3 = rep(0:2, each = 9))
Factor.1 Factor.2 Factor.3
1 0 U 0
2 0 S 0
3 0 N

2012 May 04

7

Breaking up a Row in R (transpose)

I have the following:
Time A1 A1 B1 B1 C1 C2
x y x y x y
0 5 6 6 7 7 9
1 3 4 4 3 9 9
2 5 2 6 4 7 4
I want to change it to the following:
0 1 2
x y x y x y
A1 5 6 3 4 5 2
B1

2008 Jul 19

1

replicate matrix blocks different numbers of times into new matrix

Hi,
I am trying to replicate blocks of a matrix (defined by factors) into another matrix, but an unequal, consecutive number of times for each factor.
I need to find an elegant and fast way to do this, so loops will not work.
An example of what I am trying to do is the following:
# the data - first column entries are both data and the two factors
x<-matrix(c(1,2,3,4),2,2)
> x
[,1]