Dear Newmiller,
Thank you for your reply. I’ve just posted the same question in the
another forum for stats as you suggested.
Meanwhile, I would like to keep the question submitted to learn from R
users, if it is available .
Park
On Sun, May 9, 2021 at 6:01 AM Jeff Newmiller
wrote:
> This no
: Saturday, May 08, 2021 at 5:05 AM
From: "Hyun Soo Park"
To: "r-help@r-project.org"
Subject: [R] factor analysis of dynamic structure (FADS) for a huge time-series
data
Dear R users,
I want to find the latent factors from a kind of time-series data
describing temporal changes of c
This not being a question about R, but rather about statistics, or possibly
about a contributed package, means (per the Posting Guide) that you should be
asking in a statistics forum like stats.stackexchange.com or corresponding with
the author of the package in question. If you are lucky someon
Dear R users,
I want to find the latent factors from a kind of time-series data
describing temporal changes of concentration using a factor analysis
technique called 'factor analysis of dynamic structure (FADS).' I learned
how to form the data for the analysis using a proper package embedding
FADS
to R3.20
I have installed R Factor v2.4.2
This package requires 'polycor' library
Unfortunately, 'polycor' does not exist in R3.20
That's not really correct. There is an archived version that should be
compatible with your out-of-date version of R. See:
ftp://cran.r-pr
> On Aug 5, 2017, at 7:02 PM, Gavin Brown wrote:
>
> I am not an R-Head, hence I use nice utilities that integrate R into SPSS
> I have SPSS v24, R3.20 and R3.40
> I have run IBM SPSS R Integration which requires linking to R3.20
> I have installed R Factor v2.4.2
>
.20 and R3.40
> I have run IBM SPSS R Integration which requires linking to R3.20
> I have installed R Factor v2.4.2
> This package requires 'polycor' library
> Unfortunately, 'polycor' does not exist in R3.20
>
> DATASET ACTIVATE DataSet1.
> *Mário Basto, José
I am not an R-Head, hence I use nice utilities that integrate R into SPSS
I have SPSS v24, R3.20 and R3.40
I have run IBM SPSS R Integration which requires linking to R3.20
I have installed R Factor v2.4.2
This package requires 'polycor' library
Unfortunately, 'polycor' doe
Clearly I have been doing something weird. Thanks >It is possible that
somewhere along the way, you set options(stringsAsFactors = FALSE)No not in a
long time. I found that it was great for any personal work but any time I tried
to use someone' else's raw data, say from a text file, it would mes
Thanks Marc. It never occurred to me that I would need a ""stringsAsFactors"
expression in a data.frame. I could have sworn I never did before when mocking
up some data but clearly I was wrong or there has been a change in R v. 3.4.1
which seems unlikely.
On Friday, July 7, 2017, 10:37:29 AM
> On Jul 7, 2017, at 7:03 PM, John Kane wrote:
>
> Thanks Marc.
> It never occurred to me that I would need a ""stringsAsFactors" expression in
> a data.frame. I could have sworn I never did before when mocking up some
> data but clearly I was wrong or there has been a change in R v. 3.4.1 wh
> On Jul 7, 2017, at 6:03 AM, John Kane via R-help wrote:
>
> This is not serious problem but I just wonder if someone can explain what is
> happening.
> The same command within a dataframe is giving me a factor and as a plain
> vector is giving me a character. It's probably something simple
This is not serious problem but I just wonder if someone can explain what is
happening.
The same command within a dataframe is giving me a factor and as a plain vector
is giving me a character. It's probably something simple that I have read and
forgotten but I thought I'd ask.
Thanks
#==
Hello,
I am running a factor analysis with a multiply imputed dataset in zelig
(specifying model = factor.bayes). I cannot find a way to get fit
statistics for the specified model (a 2-factor model in this particular
analysis).
The summary() function only returns lambda (factor loadings) and psi
Thank you for your reply. I am grouping citation and some social media
indictors ( number of tweets, mendeley readers, etc). The number of
citaions a paper recievs or the number of social media indicators that a
papers receives depends on time. For example a paper published in 2009 has
more time to
That step is easy, but context is hard. You really need to provide a
reproducible example. There are many models, many analysis tools, and many
timescales to choose from. In fact, this could easily be mistaken for a
question about statistics (not really on-topic here) since you have failed to
Hi,
Is it possible to use time as an offset (exposure variable) in factor
analysis? If yes, would you please advise how?
Thanks,
Tahereh
Tahereh Dehdarirad
PhD Student of Library and Information Science
University of Barcelona, Spain
[[alternative HTML version deleted]]
__
upposed to be numeric you can check if they really are by
str(df)
Cheers
Petr
> -Original Message-
> From: R-help [mailto:r-help-boun...@r-project.org] On Behalf Of ch.elahe
> via R-help
> Sent: Tuesday, June 14, 2016 5:29 PM
> To: R-help Mailing List
> Subject: [R
Hi all,
I want to use Supervised Self organizing Maps from Kohonen package for my data.
I need to divide my df into training set and test set, but a part of my df
contains column with factor levels and I don't know how to bring them into my
training set. Currently I use the following command fo
Hi all,
I want to use Supervised Self organizing Maps from Kohonen package for my data.
I need to divide my df into training set and test set, but a part of my df
contains column with factor levels and I don't know how to bring them into my
training set. Currently I use the following command
Hi R-users,
I have 1020 time series ( each of length 10,000), say, X1,X2,..,X1020
and I want to perform Factor Analysis using 50 factors on their correlation
matrix.
The issue is: for every series, I have a weight, i.e. *the series X_i has a
pre-defined weight of w_i* ( i = 1,2,, 1020). I
> ruipbarra...@sapo.pt
> Maybe the following (untested).
>
> table(df$Protocol[df$Speed == "SLOW"])
Could also use which.max to get the particular item: ...
tprot <- table(df$Protocol[df$Speed == "SLOW"])
tprot[which.max(tprot)]
S Ellison
*
Hello,
Maybe the following (untested).
table(df$Protocol[df$Speed == "SLOW"])
Hope this helps,
Rui Barradas
Citando ch.elahe via R-help :
> Hi all,
> I have the following df:
>
> $ Protocol : Factor w/ 48 levels "DP FS QTSE SAG",..: 2 3
> 43 42 31 36 37 30 28 5 ...
>
> $ Speed
Hi all,
I have the following df:
$ Protocol : Factor w/ 48 levels "DP FS QTSE SAG",..: 2 3 43 42 31 36
37 30 28 5 ...
$ Speed : chr "SLOW" "SLOW" "SLOW" "VerySLOW" ...
How can I get the most frequent Protocol when Speed is "SLOW"?
Thanks for any help!
Elahe
_
On 05/11/2016 08:00 AM, ch.elahe via R-help wrote:
Hi all,
I have a plot for TSTMean vs. SNRMean and both of these variables are factors.
How can I use Logistic Regression for factor variables?
Currently I use model=lm(TSTMean~SNRMean,data=df) but when I check
summary(model) I get this error:
Hi all,
I have a plot for TSTMean vs. SNRMean and both of these variables are factors.
How can I use Logistic Regression for factor variables?
Currently I use model=lm(TSTMean~SNRMean,data=df) but when I check
summary(model) I get this error: r error in quartile.default (resid) factors
are not
Best advice I can give: Find a local statistical expert to work with.
You appear to be asking for help understanding statistical methodology
when you do not have the necessary background to do so.
Cheers,
Bert
Bert Gunter
"Data is not information. Information is not knowledge. And knowledge
is c
Hello everyone,
I am going to ask this certainly tricky question here not (yet) with the
intention of getting a definitive answer, as I need to deepen my questions much
more, but just to have an approximate idea of which direction taking next.
I have a dataset where the potential respo
Thanks to Marc Schwartz I was able to order a "two factors"
interaction boxplot with median associated to one factor alone.
I tried further to generate facets plot (3x2 boxplots in ggplot2) for the
dataframe reported at bottom and I'm not able to reach a correct plot.
The dataframe is a simulation
I would to visualize in boxplot a data frame with two factors ordering one
factor with the median.
As example,suppose to have the InsectSprays dataframe, where an "operator"
factor with two levels, op1 and op2, has been added as shown at bottom here.
How may be generated a boxplot showing boxes fo
Great! Thank you
All the best
SF
Il 05/set/2015 15:08, "Marc Schwartz" ha scritto:
>
> > On Sep 5, 2015, at 7:29 AM, Sergio Fonda
> wrote:
> >
> > I would to visualize in boxplot a data frame with two factors ordering
> one
> > factor with the median.
> > As example,suppose to have the InsectSpr
> On Sep 5, 2015, at 7:29 AM, Sergio Fonda wrote:
>
> I would to visualize in boxplot a data frame with two factors ordering one
> factor with the median.
> As example,suppose to have the InsectSprays dataframe, where an "operator"
> factor with two levels, op1 and op2, has been added as shown a
Hi,
Suppose your data frame is called data and the name of the factor column
is named tobeConverted. I have tried this and it worked. Hope this helps.
> as.numeric(as.character(data$tobeConverted))
--
View this message in context:
http://r.789695.n4.nabble.com/factor-levels-numeric-values
ent of Anthropology
>Texas A&M University
>College Station, TX 77840-4352
>
>-Original Message-
>From: r-help-boun...@r-project.org
>[mailto:r-help-boun...@r-project.org] On Behalf Of Gerrit Eichner
>Sent: Wednesday, November 12, 2014 8:06 AM
>To: David Studer
>Cc: r-he
rrit Eichner
Sent: Wednesday, November 12, 2014 8:06 AM
To: David Studer
Cc: r-help@r-project.org
Subject: Re: [R] factor levels > numeric values
Hello, David,
take a look at the beginning of the "Warning" section of ?factor.
Hth -- Gerrit
Hi everybody,
I have another que
c: r-help@r-project.org
Subject: Re: [R] factor levels > numeric values
Hello, David,
take a look at the beginning of the "Warning" section of ?factor.
Hth -- Gerrit
> Hi everybody,
>
> I have another question (to which I could not find an answer in my r-books.
> I am
Hello, David,
take a look at the beginning of the "Warning" section of ?factor.
Hth -- Gerrit
Hi everybody,
I have another question (to which I could not find an answer in my r-books.
I am sure, it's not a great issue, but I simply lack of a good idea how to
solve this:
One of my variable
Hi everybody,
I have another question (to which I could not find an answer in my r-books.
I am sure, it's not a great issue, but I simply lack of a good idea how to
solve this:
One of my variables gets imported as a factor instead of a numeric variable.
Now I have a...
Factor w/ 63 levels "0","0
integer, levels is numeric
[1]
Levels: 29 3e+05 31
Bill Dunlap
TIBCO Software
wdunlap tibco.com
On Sat, Sep 20, 2014 at 3:52 AM, Suharto Anggono Suharto Anggono
wrote:
> In R:
>
>> factor(30, levels=1:30)
> [1]
> 30 Levels: 1 2 3 4 5 6 7 8 9 10 11 12 13
#. .OO#. rocks...1k
---
Sent from my phone. Please excuse my brevity.
On September 20, 2014 3:52:15 AM PDT, Suharto Anggono Suharto Anggono
wrote:
>In R:
>
>> factor(30, levels=1:30)
>[1]
>30 L
In R:
> factor(30, levels=1:30)
[1]
30 Levels: 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 ...
30
The NA above is undesirable in my view, because 30 is in 1:30.
I have just got bitten by it.
I have figured out why it happens. The results
Thanks, Bert. It seems I got these NAs because I already had MSA population
controlled for in my model, besides the fixed effect variable, which led to
overestimation. Those NAs disappeared after I dropped the population
variable.
Gary
On Sun, Dec 1, 2013 at 10:27 AM, Bert Gunter wrote:
> You
You may wish to talk to a local statistician or read up on linear
models, as you appear to not understand some basics. Anyway, either
1. You have other covariates in your model that you haven't shown and
your model is overdetermined.
2. You have NA's in your data that causes 1) to occur.
As an e
Dear R users,
I am running a linear regression in R. My observations are Census Tracts in
several metropolitan areas (MSAs). In my data set, each MSA has at least 50
observations. I use factor(msa_code) in the lm formula to control for
metropolitan fixed effects. But I kept getting something like
How do you extend factor() without abstract data types?
The idea is to have
factor (x, TYPE)
to transform all or selected components of x to factor TYPE.
Value:
a data structure of same type as x, with factor-like components
transformed to be like TYPE.
In the Pascal family, you would
In addition to David's great answer also read
?contr.SAS
?contr.treatment
?relevel
and also sections 4 and 11.1.1 in An Introduction to R.
On Fri, May 3, 2013 at 9:47 PM, David Winsemius wrote:
>
> On May 3, 2013, at 3:32 PM, Iuri Gavronski wrote:
>
> > Hi,
> > I would like to know the criteri
On May 3, 2013, at 3:32 PM, Iuri Gavronski wrote:
> Hi,
> I would like to know the criteria by which R removes a factor in linear
> models. For example, I have a four level factor, and R creates 3 dummies to
> estimate coefficients. Which level is chosen? Can I chance it?
The default order is al
Hi,
I would like to know the criteria by which R removes a factor in linear
models. For example, I have a four level factor, and R creates 3 dummies to
estimate coefficients. Which level is chosen? Can I chance it?
Thanks,
Iuri
[[alternative HTML version deleted]]
On 2013-04-01 13:08, Matthew Lundberg wrote:
Note the edited subject line! I don't know why I typed it as it was before.
This says that as.numeric(as.character(f)) will work regardless of the
implementation, and I agree.
It's the recommendation to use as.numeric(levels(f))[f] that has me
wonde
Note the edited subject line! I don't know why I typed it as it was before.
This says that as.numeric(as.character(f)) will work regardless of the
implementation, and I agree.
It's the recommendation to use as.numeric(levels(f))[f] that has me
wondering about section 2.3.1 of the language defini
Yup. Note also:
> as.character.factor
function (x, ...)
levels(x)[x]
But of course this is OK, since this can change if the implementation
does. Which is the whole point, of course.
-- Bert
On Mon, Apr 1, 2013 at 12:16 PM, Matthew Lundberg
wrote:
>
> When used as an index, the factor is impl
When used as an index, the factor is implicitly converted to integer. In
the expression as.numeric(levels(f))[f], the vector as.numeric(levels(f))
is indexed by as.integer(f).
This appears to rely on the current implementation, as mentioned in section
2.3.1 of the language definition.
On Mon, A
On 2013-04-01 10:48, Matthew Lundberg wrote:
These two seem to be at odds. Is this the case?
From help(factor) - section Warning:
To transform a factor f to approximately its original numeric values,
as.numeric(levels(f))[f] is recommended and slightly more efficient than
as.numeric(as.chara
Hi Mathew,
In what way are they at odds?
On Mon, Apr 1, 2013 at 1:48 PM, Matthew Lundberg
wrote:
> These two seem to be at odds. Is this the case?
>
> >From help(factor) - section Warning:
>
> To transform a factor f to approximately its original numeric values,
> as.numeric(levels(f))[f] is re
These two seem to be at odds. Is this the case?
>From help(factor) - section Warning:
To transform a factor f to approximately its original numeric values,
as.numeric(levels(f))[f] is recommended and slightly more efficient than
as.numeric(as.character(f)).
>From the language definition - secti
Hi
> -Original Message-
> From: r-help-boun...@r-project.org [mailto:r-help-bounces@r-
> project.org] On Behalf Of maths123
> Sent: Monday, November 26, 2012 8:25 PM
> To: r-help@r-project.org
> Subject: Re: [R] Factor function for coded numerical values
>
> The
The question says that there is an experiement to investigate the effect on
breathing rate when doing different types of exercise with wearing more
clothes or less clothes (factor A, coded 1,2).
The duration of exercise was 10min, 20min, 30min (factor B, coded 1,2,3).
Could you give m anymore hel
Le jeudi 22 novembre 2012 à 13:12 -0800, maths123 a écrit :
> I have s data set where 2 of the columns give the coded versions of the
> factors A and B. Factor A is coded with 1, 2, 3. Factor B is coded with 1,2.
>
> How do use the factor function to convert these variables into factors, and
> a
I have s data set where 2 of the columns give the coded versions of the
factors A and B. Factor A is coded with 1, 2, 3. Factor B is coded with 1,2.
How do use the factor function to convert these variables into factors, and
also use the labels= command to give them a more informative name?
Hi,
still doesn't work.
> Hello,
>
> I've just seen the error, you are _not_ searching for colnames in
> mod5.sig$snps. Corrected:
>
> Selec = todos[ , colnames(todos) %in% mod5.sig$snps]
>
> Hope this helps,
>
> Rui Barradas
> Em 23-10-2012 21:17, Silvano Cesar da Costa escreveu:
>> Hi Rui,
>>
Hello,
Sorry but, what doesn't work? What is the error message or result?
As for stringsAsFactors not keeping character strings as character
strings maybe this is from calling data.frame from within the
environment used by with(), but you can use other ways of converting to
character strings,
Hi Rui,
it doesn't work:
> (mod5.sig = with(mod5, data.frame(snps = SNP[pvalor<5e-8],
stringsAsFactors=FALSE)))
> str(mod5.sig)
'data.frame': 76 obs. of 1 variable:
$ snps: Factor w/ 220 levels "rs10058955_A",..: 89 59 88 73 40 35 97 55
87 204 ...
> Selec = todos[ , colnames(todos) %in% mod5
Hello,
I've just seen the error, you are _not_ searching for colnames in
mod5.sig$snps. Corrected:
Selec = todos[ , colnames(todos) %in% mod5.sig$snps]
Hope this helps,
Rui Barradas
Em 23-10-2012 21:17, Silvano Cesar da Costa escreveu:
Hi Rui,
it doesn't work:
(mod5.sig = with(mod5, dat
Hello,
When creating the data.frame of snps use the option stringsAsFactors = FALSE
I believe your error comes from the fact that you are trying to find
colnames in a variable coded as integers (a factor, like str shows).
Hope this helps,
Rui Barradas
Em 23-10-2012 19:02, Silvano Cesar da Cos
Hi,
The program below work very well.
(snps = c('rs621782_G', 'rs8087639_G', 'rs8094221_T', 'rs7227515_A',
'rs537202_C'))
Selec = todos[ , colnames(todos) %in% snps]
head(Selec)
But, I have a data set with 1.000 columns and I need extract 70 to use
(like snps in command above).
This 70 snps ar
>From ?princomp
Arguments:
scores: a logical value indicating whether the score on each
principal component should be calculated.
Value:
scores: if ‘scores = TRUE’, the scores of the supplied data on the
principal components. These are non-null only if ‘x’ was
supp
Hi everyone,
I am trying to get the factor score for each individual case from a principal
component analysis, as I understand, both princomp() and prcomp() can not
produce this factor score, the principal() in psych package has this option:
scores=T, but after running the code, I could not
lish (my first tonge is spanish). Perhaps, there's a function in R,
> but I have not idea.
> Hope it helps.
> Thanks in advance.
> José
>
>
>
>
> De: PIKAL Petr
>
> l...@r-project.org>
> Enviado: Miércoles 19 de septiembre d
: Wednesday, September 26, 2012 6:39 PM
To: PIKAL Petr; r-help@r-project.org
Subject: Re: [R] factor expansion
Thank you PIKAL Petr,
This is used when you have a big data base of a national sample. So, this
factor is a variable that uses the sample in order to obtain information about
all the people
12 9:19
Asunto: RE: [R] factor expansion
Hi
I did not see any answer yet but can you explain what you mean by factor
expansion? Something like expand.grid?
Petr
> -Original Message-
> From: r-help-boun...@r-project.org [mailto:r-help-bounces@r-
> project.org] On Behalf Of Jose Bu
ber 12, 2012 4:16 PM
> To: r-help@r-project.org
> Subject: [R] factor expansion
>
> I'm trying to use a data base from SPSS and get all data representing
> the true data. I would like to use a variable as expansion data. In our
> data base we have the variable with the facto
I'm trying to use a data base from SPSS and get all data
representing the true data. I would like to use a variable as expansion
data. In our data base we have the variable with the factor expansion,
but we have not idea how to set it in R.
Any idea?
Thanks in advance!
José
[[alternati
388-4170 | http://www.wvbauer.com
>
> > -Original Message-
> > From: [hidden email]
> [mailto:[hidden
> email] ]
> > On Behalf Of John Hodgson
> > Sent: Thursday, August 09, 2012 12:56
> > To: [hidden email]
> > Subject: [R] Factor moderators
+31 (43) 388-4170 | http://www.wvbauer.com
> -Original Message-
> From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org]
> On Behalf Of John Hodgson
> Sent: Thursday, August 09, 2012 12:56
> To: r-help@r-project.org
> Subject: [R] Factor moderators i
I'm puzzled by the behaviour of factors in rma models, see example and
comments below. I'm sure there's a simple explanation but can't see it...
Thanks for any input
John Hodgson
- code/selected output -
library(metafor)
##Set up data
On Jun 7, 2012, at 09:25 , Rui Barradas wrote:
> Hello,
>
> Try option stringsAsFactors, see ?read.csv or ?read.table
> As for the thousands separator, see ?format
help(as.Date) should also help. (Hint: there's no dateFormat= argument)
>
> Hope this helps,
>
> Rui Barradas
>
> Em 07-06-2012
Hello,
Try option stringsAsFactors, see ?read.csv or ?read.table
As for the thousands separator, see ?format
Hope this helps,
Rui Barradas
Em 07-06-2012 03:09, eric escreveu:
How do I fix this error ? I tried coercion to a vector but that didn't work.
msci <-read.csv("..MSCIexUS.csv", header
How do I fix this error ? I tried coercion to a vector but that didn't work.
msci <-read.csv("..MSCIexUS.csv", header=TRUE)
head(msci)
Date index
1 Dec 31, 1969100
2 Jan 30, 1970 97.655
3 Feb 27, 1970 96.154
4 Mar 31, 1970 95.857
5 Apr 30, 1970 85.564
6 May 29, 1970 79.005
> str(m
Please don't double post. (And see my reply to your other thread)
Michael
On Tue, May 29, 2012 at 1:50 PM, Raúl Bajo wrote:
> HI!
>
> I am working with a table with quantitative and qualitative data; one of
> the columns is "Level of education", which is an ordered factor. I want to
> get rid of
HI!
I am working with a table with quantitative and qualitative data; one of
the columns is "Level of education", which is an ordered factor. I want to
get rid of the ordered factor, and set my own scale. How to do that? All
the time I get the following error:
In `[<-.factor`(`*tmp*`, educa = "Le
ge-
> From: mayo_j...@hotmail.com
> Sent: Wed, 2 May 2012 14:08:26 +0200
> To: r-help@r-project.org
> Subject: [R] factor conversion to date/time
>
> Hi, I've been trying to convert numbers from an online temperature
> database
> into dates and time that R recognizes. I've
Please don't triple post.
Michael
On Wed, May 2, 2012 at 8:08 AM, marjolein post wrote:
>
>
>
>
> Hi, I've been trying to convert numbers from an online temperature database
> into dates and time that R recognizes. I've tried as.Date, as.POSIXlt and
> strptime the problem is that the database
On 02/05/2012 8:08 AM, marjolein post wrote:
Hi, I've been trying to convert numbers from an online temperature database
into dates and time that R recognizes. I've tried as.Date, as.POSIXlt and
strptime the problem is that the database has put a T between the numbers and
R will not accept
Hi, I've been trying to convert numbers from an online temperature database
into dates and time that R recognizes. I've tried as.Date, as.POSIXlt and
strptime the problem is that the database has put a T between the numbers and
R will not accept any conversions. currently it sees the date as
Hi David,
On Thursday, February 09, 2012 01:21:48 PM David Studer wrote:
> Hello everybody!
>
> Let's assume I have the following factor with it's levels:
>
> a<-factor(c(2,3,3,2,4,2,3,2,2,2,3,2,3))
> mydata<-data.frame(a)
#You need to specify levels and labels here, like this:
a <- factor(c(2
Hi
>
> Hello everybody!
>
> Let's assume I have the following factor with it's levels:
>
> a<-factor(c(2,3,3,2,4,2,3,2,2,2,3,2,3))
> mydata<-data.frame(a)
>
> When I plot the vector "a" using
>
> barplot(table(mydata$a)
>
> unfortunately the value "1" does not
> show up, as it does not appea
Hello everybody!
Let's assume I have the following factor with it's levels:
a<-factor(c(2,3,3,2,4,2,3,2,2,2,3,2,3))
mydata<-data.frame(a)
When I plot the vector "a" using
barplot(table(mydata$a)
unfortunately the value "1" does not
show up, as it does not appear in my data.
But still, it theor
ey man!!
you need to change the format of your data as matrix
data <- as.matrix(correlation_data)
solucion <- factanal(covmat = data, factors=2)
that´s all!!!
enjoy
Rafael Ch.
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Thanks for the fast response and your comments!
That works perfect!
Another little mystery solved ;)
Stefan
From: Felipe Carrillo [mailto:mazatlanmex...@yahoo.com]
Sent: Tuesday, November 01, 2011 3:54 PM
To: Schreiber, Stefan; r-help@r-project.org
Subject: Re: [R] factor level
e Interior
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From: "Schreiber, Stefan"
>To: r-help@r-project.org
>Sent: Tuesday, November 1, 2011 2:28 PM
>Subject: [R] factor level issue after subsetting
>
>Dear list,
>
>I cannot
first of all, the subsetting line is overly complicated.
dat.sub<-dat[dat$treat!='cont',]
will work just fine. R does exactly what you're describing. It knows
the levels of the factor. Once you remove 'cont' from the data, that
doesn't mean that the level is removed from the factor:
> df<-dat
> -Original Message-
> From: r-help-boun...@r-project.org [mailto:r-help-bounces@r-
> project.org] On Behalf Of Schreiber, Stefan
> Sent: Tuesday, November 01, 2011 2:29 PM
> To: r-help@r-project.org
> Subject: [R] factor level issue after subsetting
>
> Dear li
Dear list,
I cannot figure out why, after sub-setting my data, that particular item
which I don't want to plot is still in the newly created subset (please
see example below). R somehow remembers what was in the original data
set. A work around is exporting and importing the new subset. Then it's
Varimax is orthogonal, promax is oblique. Varimax is generally not
recommended. See: Preacher, K. J., & MacCallum, R. C. (2003).
Repairing Tom Swift's electric factor analysis machine. Understanding
Statistics, 2(1), 13-43. (Google the title and you'll find a PDF).
The fa() function in the psy
Hello
I seem to find only two types of rotation for the factanal function in R, the
Varimax and Promax, but is it possible to run a orthogonal and oblique
rotations in R?
Thanks in advance
Rosario
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On Fri, Jun 17, 2011 at 6:32 AM, Hugo Loyola wrote:
> Please, can you help me with the following? I want to include a Greek letter
> as part of a factor label that I need to use as factor on a xyplot. Probably
> this was already solved so can I get the place where I can look for?
>
> Th
Please, can you help me with the following? I want to include a Greek letter
as part of a factor label that I need to use as factor on a xyplot. Probably
this was already solved so can I get the place where I can look for?
Thanks,
Hugo
[[alternative HTML version deleted]]
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At 6:19 AM -0700 6/14/11, Jay wrote:
Hi,
are there readily available R packages that are able to perform FA on
ordinal and/or nominal data?
If not, what other approaches and helpful packages would you suggest?
If by ordinal and nominal you mean just a few categories (e.g., a
mood scale or per
Hi,
are there readily available R packages that are able to perform FA on
ordinal and/or nominal data?
If not, what other approaches and helpful packages would you suggest?
BR,
Jay
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Hi Matt,
Did you try reading the documentation for factanal()? You can pull it
up by typing: help("factanal")
These give basically identical results using the raw data, the
covariance matrix, and the correlation matrix.
factanal(x = mtcars, factors = 3)
factanal(factors = 3, covmat = cov(mtcars
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