On 28/05/2012 02:20, Hunsicker, Lawrence wrote:
Greetings, all:
I am using factanal in R.
When I enter a matrix or a formula, the print method winds up with something
like this:
Test of the hypothesis that 6 factors are sufficient.
The chi square statistic is 28.1 on 22 degrees of freedom.
Th
It is isn't easy to write code that works with column names that have spaces.
You could rewrite rpart, or just rename the columns in your data frame to work
around the "bug". See ?names.
---
Jeff Newmiller
On 28/05/2012 08:27, Jeff Newmiller wrote:
It is isn't easy to write code that works with column names that have spaces. You could
rewrite rpart, or just rename the columns in your data frame to work around the
"bug". See ?names.
In any case, rpart pre-dates the `` notation that made this pos
On 05/28/2012 02:44 AM, Manish Gupta wrote:
Thanks it works!
How can i make horizontal bar graph using barp?
Hi Manish,
The sad fact is that I haven't gotten around to coding it yet.
Jim
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mail
hello everyone,
i often work on contingency table that I create from data.frame (with
table() function)
but a friend sent me an excel sheet wich *already is* a contingency
table (just a simple 2 way table !...)
any clue on how to import it in R (keeping row names and col names) ?
any tuto I com
Hello,
Try function read.xls in library gdata
Also, a good way of avoiding such doubts is
library(sos)
findFn('xls')
It returns read.xls as the first line.
Hope this helps,
Rui Barradas
Em 28-05-2012 11:32, sylvain willart escreveu:
hello everyone,
i often work on contingency table that I
Thanks Rui,
but my problem is not to read an xls file, I converted already to csv,
but rather to read a contingency table into R, and telling R it is
astually a contingency table, and not a data.frame...
file below, if it helps...
Sylv
,AUC,Alin,BLG,BrDep,CRF,CrfMkt,CAS,Casto,Confo,ElecDep,Geant
no,
the problem is that the lines in my file do not correspond to
individuals, but are variables, just like are the columns,
my file is already a contingency table, with each cell being a frequency:
here is a sample of it:
***
,AUC,Alin,BLG,B
I have fitted a proportional odds model, but would like to compare it to
a continuation ratio model. However, I am unable to fit the CR model
_including_ imputated data.
I guess my troubles start with settuping the data for the CR model.
Any hint is appreciated!
Christian
library(Hmisc)
libra
Le lundi 28 mai 2012 à 15:19 +0200, sylvain willart a écrit :
> no,
> the problem is that the lines in my file do not correspond to
> individuals, but are variables, just like are the columns,
> my file is already a contingency table, with each cell being a frequency:
>
> here is a sample of it:
>
Hello everyone,
I tried to understand the relationship between temperature and the
death of an organism by using logistic regression.
glm(formula = Death ~ Temperature, family = binomial(link = "logit"),
data = mydata)
Coefficients:
Estimate Std. Error z value Pr(>|z|)
Ok, try the following.
df2table <- function(x, Var1="Var1", Var2="Var2"){
tbl <- as.matrix(x)
dnames <- list(rownames(x), colnames(x))
names(dnames) <- c(Var1, Var2)
attr(tbl, "dimnames") <- dnames
attr(tbl, "class") <- "table"
tbl
}
df2table(xls_contingency) # using de
How about this?
exdf <- read.table("clipboard", sep=",", header=T, row.names=1)
extbl <- as.table(as.matrix(exdf))
--
David L Carlson
Associate Professor of Anthropology
Texas A&M University
College Station, TX 77843-4352
> -Original Message-
It works.
There goes my clever function.
Rui Barradas
Em 28-05-2012 16:31, David L Carlson escreveu:
How about this?
exdf<- read.table("clipboard", sep=",", header=T, row.names=1)
extbl<- as.table(as.matrix(exdf))
--
David L Carlson
Associate Profes
OK, a couple of things (I only looked through quickly):
1. R doesn't allow variable names to begin with a number. Be sure you don't
try that.
2. What's the overall goal here? Read them in, change the name, then write
them out? Let us know and it will be easier to help you.
3. Regardless of
Look at
?predict.glm
mydata.glm <- glm(formula = Death ~ Temperature, family = binomial(link =
"logit"),
data = mydata)
and see that predict(mydata.glm, type="response") gives the predictions on
the probability scale.
On Mon, May 28, 2012 at 10:16 AM, eddie smith wrote:
> Hello everyone,
>
> I
Hi
I'm working in the excellent ICEinfer package which calculates bootstrapped
cost-effectiveness ratios
Obenchain, B. (2011). ICEinfer: #Incremental Cost-Effectiveness (ICE)
Statistical Inference from Two Unbiased Samples. R package version 1.0-0.
http://CRAN.R-#project.org/package=ICEinfer
thanks rui
that's what I was looking for
I have another related question:
- why of the difference between the max distance D calculated with ks.test()
and the max distance D “manually” calculated as in (2)?
I guess it has something to do with the fact that KS is obtained with a
maximisation th
Fantastic, thanks very much Richard. The addition of 'tz="GMT"' worked a
treat. Best wishes, Des
From: Richard M. Heiberger [mailto:r...@temple.edu]
Sent: 28 May 2012 02:09
To: Des Callaghan
Cc: r-help@r-project.org
Subject: Re: [R] Problem with strptime
Some of your dates are displayed BS
Dear list,
this is perhaps more of a statistics question than an R question, but perhaps
someone could help me out anyway.
I'm doing sociological research and am currently in the process of
familiarizing myself with the basic concepts of multiple imputation.
Eventually, my goal is to perform q
I'm doing:
> alyL32007z <- zoo(alyL32007,alyL32007$time)
> range(time(alyL32007z))
[1] "2007-01-01 00:00:00 UTC" "2007-12-31 23:30:00 UTC"
But then, while the original variable is:
> summary(alyL32007$NEE_st)
Min. 1st Qu. MedianMean 3rd Qu.Max.NA's
-15.340 -1.615 -0.054 -0.814
He folks=)
I am trying to compute a supWald Test in R, trying to show that a subset of
my regressors is significantly different from 0 but I am not able to compute
this test. My sample looks as follows: I am regressing fit1 <-
(Y~X1+X2+X3,tau=tau). I know that if I want to show that e.g. X2 is
sign
He folks=)
I want to check whether a coefficient has an impact on a quantile regression
(by applying the sup-wald test for a given quantile range [0.05,0.95].
Therefore I am doing the following calculations:
a=0;
for (i in 5:95/100){
fitrestricted=rq(Y~X1+X2,tau=i)
tifunrestrited=rq(Y~X1+X2+X3,tau=
Hello,
That's a very difficult question. See
Marsaglia, Tsang, Wang (2003)
http://www.jstatsoft.org/v08/i18/
Simard, L'Ecuyer (2011)
http://www.jstatsoft.org/v39/i11
R's ks functions are a port of Marsaglia et al. to the .C interface.
Rui Barradas
maxbre wrote
>
> thanks rui
>
> that's what
hello,
I try to run simulation of levene's test to find the p-value but the error
of replacement has length zero occur, could anyone help me to fix this
problem?
asim <- 1000
pv<-rep(NA,asim)
for(i in 1:asim)
{print(i)
set.seed(i)
g1 <- rnorm(20,0,2)
g2 <- rnorm(20,0,2)
g3 <- rnorm(20,0,2)
x
thanks for the help: I'll have a look at the papers
max
Il 28/05/2012 12:31, Rui Barradas [via R] ha scritto:
> Hello,
>
> That's a very difficult question. See
>
> Marsaglia, Tsang, Wang (2003)
> http://www.jstatsoft.org/v08/i18/
>
> Simard, L'Ecuyer (2011)
> http://www.jstatsoft.org/v39/i11
>
>
Hello -
I am using glmnet to generate a model for multiple cohorts i. For each i, I
run 5 separate models, each with a different x variable. I want to compare
the fit statistic for each i and x combination.
When I use auc, the output is in some cases is < .5 (.49). In addition, if
I compare mean
Dear R-list members,
I have a problem of estimation of parameters represented in a covariance matrix
using maximum likelihood function. The problem is essentially a multivariate
Gaussian random field model. The maximum likelihood function is
L(m, *S2 , *N2 ; F1) =1/ (2* sqrt(det(*
Dear jim and rest of R users,
I initially had the following data set
01.01.1967 0.87
02.01.1967 0.87
03.01.1968 0.87
04.01.1968 0.87
05.01.1969 0.87
06.01.1969 0.87
07.01.1970 0.87
08.01.1970 0.87
09.01.1971 0.87
10.01.1971 0.87
11.01.1972 0.87
12.01.1972 0.87
13.01.1973 0.69
14.01.1973 0.70
15.0
Dear all,
I want to determine the standard error or the mean squared error for the
parameter estimate for beta and eta base on the real data.
Any help on how to obtain these estimated errors.
library(survival)
d <- data.frame(ob=c(149971, 70808, 133518, 145658, 175701, 50960, 126606,
82329),
Hi Michel,
More than equal to function, how can I use other function ?,
For example :-
Age between
Age Greater Than or Equal to like that â¦. How ?
From: Michael Weylandt [via R]
[mailto:ml-node+s789695n4631319...@n4.nabble.com]
Sent: Friday, May 25, 2012 7:48 PM
To: Akkara
Hi,
Here i have been an matrix like this,
*NAMEAGE PALCETRUE/FALSE*
ABC 20 INDIA
XYZ 30 FRANCE
PQR40 USA
MNO 30KENIYA
DEF25AUSTRALIA
GTY34 CANADA
BNH 38JAPAN
Here, *TRUE/F
The code given below worked well. However, the problem is that when I typed
dir1 to see the results I found that R order the files as:
[1] "data1.flt" "data10.flt" "data100.flt" "data101.flt"
[5] "data102.flt" "data103.flt" "data104.flt" "data105.flt"
[9] "data106.flt" "data107.flt" "data1
Dear Rui, Kevin, Bryan and Nutter
Thank you so much for your very helpful hints!
Now I have extracted all the file names and managed to edit them using the
code (1)-(4) below and obtained the name format as I wanted
(1) files<-list.files(path = "myworking directory", pattern = NULL,
all.files
On Mon, May 28, 2012 at 5:35 AM, Agustin Lobo
wrote:
> I'm doing:
>
>> alyL32007z <- zoo(alyL32007,alyL32007$time)
>> range(time(alyL32007z))
> [1] "2007-01-01 00:00:00 UTC" "2007-12-31 23:30:00 UTC"
>
> But then, while the original variable is:
>> summary(alyL32007$NEE_st)
> Min. 1st Qu. Media
Dear Dila,
Try the following:
library(Rcmdr)
asim <- 1000
pv<-NULL
for(i in 1:asim)
{
print(i)
set.seed(i)
g1 <- rnorm(20,0,2)
g2 <- rnorm(20,0,2)
g3 <- rnorm(20,0,2)
x <- c(g1,g2,g3)
group<-as.factor(c(rep(1,20),rep(2,20),rep(3,20)))
pv<-c(pv,leveneTest(x,group)$"Pr(>F)"[1])
}
Best
Ozgur
It's because those are character strings and they are sorted lexically
(i.e., alphabetically). I think you probably can get what you prefer
by using the mixedsort/mixedorder functions of the gtools package.
Take a look at this
x <- paste0("data",1:100, ".fit")
order(x)
sort(x)
library(gtools)
m
Greater than or equal to is simply >= like most languages.
For more complicated questions, simply combine booleans:
E.g., between 5 and 25
with(dat, (AGE < 25) & (AGE > 5))
and so on.
Michael
On Mon, May 28, 2012 at 9:34 AM, Rantony wrote:
> Hi Michel,
>
>
>
> More than equal to function, ho
On May 28, 2012, at 5:20 AM, Christopher Kelvin wrote:
Dear all,
I want to determine the standard error or the mean squared error
for the parameter estimate for beta and eta base on the real data.
Any help on how to obtain these estimated errors.
library(survival)
d <- data.frame(ob=c(14
On Mon, May 28, 2012 at 12:14 PM, Özgür Asar wrote:
> Dear Dila,
>
> Try the following:
>
> library(Rcmdr)
Or avoid the unncessary overhead of Rcmdr and use
library(car)
to provide levenTest instead.
> asim <- 1000
> pv<-NULL
It's also many orders of magnitude more efficient to preallocate "
I already answered this. Don't double post questions.
On Mon, May 28, 2012 at 10:01 AM, Rantony wrote:
> Hi,
>
> Here i have been an matrix like this,
>
> *NAME AGE PALCE TRUE/FALSE*
> ABC 20 INDIA
> XYZ 30 FRANCE
> PQR 40 USA
> MNO
On 27.05.2012 19:50, Michael wrote:
R Console closed and exited quietly after loading "RODBC"...
I suspect that's segment fault crash...
Could be.
Have you read the posting guide? How could we reproduce? WHich OS,
version of R, version of RODBC? ??
but how do I figure out what exact
Here's what I would do then, to keep it simple.
1. Put all the relevant csv files into a single directory.
2. setwd() to that directory.
3. Use the approach I suggested before:
files <- list.files(pattern = "\\.(csv|CSV)$")
for (i in 1:length(files)) {
temp <- read.csv(files[i],
Dear Researchers,
sorry for this simple question. I have a point plot with mean values and i
wish to plot line with Standard Deviation as "Whiskers". I calculate the
mean+sd and mean-sd, but i can not figure out the way to add the line.
mydata <-
data.frame(mean=c(0.42,0.41,0.41,0.43,0.45,0.43,0
Hi all
I did an estimation of a simple regression model (ror_xxx~ror_spi_xxx) and
assessed the quality of this estimation. After having detected that there
are indications of autocorrelatio and an AR(1) process, I used an arima
model:
absi.arima=arima(ror_absi, order=c(1,0,0), xreg=ror_spi_absi)
Stefan,
You could try this: make a private version of anova.rqlist and change the
call to lapply
that computes summaries so that se = "ker" instead of se = "nid". Please let
me know
if this does what you would like to do. This is about 20 lines into the code.
Could you also explain what yo
Many thanks to Michael Weylandt and Prof. Ripley for answers to yesterday's
query.
1. The response to
methods(print)
is that the print.princomp method is "non-visible," not "suppressed," as I
misquoted.
The method can be located by either
getAnywhere(print.princomp)
as suggested by Michae
Hello,
I am trying the postForm() function on a very simple webpage:
http://www.colby.edu/chemistry/PChem/Hartree.html
I am simply trying to fill the "Hartrees" text form with the value "100".
But running this:
url = "http://www.colby.edu/chemistry/PChem/Hartree.html";
test = postForm(url, H="
Dear Bryan
Thank you so much for your prompt reply!
Please see my responds below under = in your reply...
Many thanks again!
HJ
On Mon, May 28, 2012 at 4:45 PM, Bryan Hanson wrote:
> OK, a couple of things (I only looked through quickly):
>
> 1. R doesn't allow variable names to begin w
Hi,
I guess this is what you are looking for,
dat[dat$AGE<=40 & da$AGE>=30,"TRUE/FALSE"]<-TRUE
A.K.
- Original Message -
From: Rantony
To: r-help@r-project.org
Cc:
Sent: Monday, May 28, 2012 10:01 AM
Subject: [R] Hash Table - Select and Change Data iniside Matrix Using "Between"
H
On May 28, 2012, at 9:55 AM, gianni lavaredo wrote:
Dear Researchers,
sorry for this simple question. I have a point plot with mean
values and i
wish to plot line with Standard Deviation as "Whiskers". I calculate
the
mean+sd and mean-sd, but i can not figure out the way to add the line.
On Mon, 28 May 2012, and_mue wrote:
Hi all
I did an estimation of a simple regression model (ror_xxx~ror_spi_xxx) and
assessed the quality of this estimation. After having detected that there
are indications of autocorrelatio and an AR(1) process, I used an arima
model:
absi.arima=arima(ror_ab
Hi, Troels:
I'm still trying to understand the structure of your data.
Please check the discussion below. If what I suggest is correct, it
should make the analysis much more routine and therefore easier
requiring less time to analyze.
On 5/21/2012 1:33 PM, Troels Ring wrote:
Dear f
The function i am looking is a bars from the mean points of the plot in
boxplot style. I tryed several forum but I have no clear the way to create
these bars.
Gianni
On Mon, May 28, 2012 at 7:13 PM, David Winsemius wrote:
>
> On May 28, 2012, at 9:55 AM, gianni lavaredo wrote:
>
> Dear Research
This post are useful.
http://myowelt.blogspot.com.br/2008/03/beautiful-error-bars-in-r.html
http://mapas.mma.gov.br/i3geo/pacotes/rlib/win/gplots/html/plotCI.html
Walmes.
==
Walmes Marques Zeviani
LEG (Laboratório de Estatís
Hello,
The function 'arrows' with angle=90 can do the job.
mydata <-
data.frame(mean=c(0.42,0.41,0.41,0.43,0.45,0.43,0.43,0.42,0.44,0.45,0.45,0.45,0.46,0.43,0.42,0.37,0.44,0.46,0.46,0.39,0.40),
sdUP=c(0.58,0.56,0.55,0.57,0.61,0.55,0.57,0.59,0.61,0.60,0.57,0.60,0.62,0.57,0.59,0.56,0.57,0.61,0.61,
hello,
Is there any function in R to transform a numerical continuos variable in a
ranked variable?
Thanks
-
Mario Garrido Escudero
PhD student
Dpto. de Biología Animal, Ecología, Parasitología, Edafología y Qca. Agrícola
Universidad de Salamanca
--
View this message in context:
http://r.78
Just a final correction.
I was wrong, stats::ks.test doesn't use only Marsaglia et al.
It's even clearly written in the help page.
Read the documentation before stating!
Rui Barradas
Em 28-05-2012 11:51, maxbre escreveu:
> thanks for the help: I'll have a look at the papers
> max
>
> Il 28/05/20
I was wondering if there was a quick way to extract out certain rows from a
data set in R?
I have a data.frame, LOG, where in one column, sample_data_tx, there is a list
of 62 different types of treatment. I've sub-selected the rows that contain the
names, "PLO" and "NOY" to make a new vector
? rank
Michael
On Mon, May 28, 2012 at 2:17 PM, gaiarrido wrote:
> hello,
> Is there any function in R to transform a numerical continuos variable in a
> ranked variable?
>
> Thanks
>
> -
> Mario Garrido Escudero
> PhD student
> Dpto. de Biología Animal, Ecología, Parasitología, Edafología y
hi,
read an Introduction to R.
?rank is what you are looking for?
kd
2012.05.28. 20:17 keltezéssel, gaiarrido írta:
hello,
Is there any function in R to transform a numerical continuos variable in a
ranked variable?
Thanks
-
Mario Garrido Escudero
PhD student
Dpto. de Biología Animal, E
On Mon, May 28, 2012 at 3:16 PM, Kelly Cool wrote:
> I was wondering if there was a quick way to extract out certain rows from a
> data set in R?
> I have a data.frame, LOG, where in one column, sample_data_tx, there is a
> list of 62 different types of treatment. I've sub-selected the rows tha
Hi Kelly,
Check
?subset
in the R console. Here is a piece of code (untested):
subset(LOG, sample_data %in% c("Noy", "PLO"))
HTH,
Jorge.-
On Mon, May 28, 2012 at 3:16 PM, Kelly Cool <> wrote:
> I was wondering if there was a quick way to extract out certain rows from
> a data set in R?
> I
Dear colleagues,
Could I get some assistance using postForm() to scrape the business names and
addresses at this website:
http://www.brantford.ca/business/LocalBusinessCommunity/Pages/BusinessDirectorySearch.aspx
I've read through (http://www.omegahat.org/RCurl/RCurlJSS.pdf) and scoured the
web
Hi all,
I have a set of files (which is growing) in a folder. The files are text
files...
The form of files is such :
...with numbers for Length (m) going up to 2000 ...
Anyway...i just need the data from first two columns (length (m) and
Temperature (C)), and no data before that...
This Len
On 28/05/12 20:46, Simon Kiss wrote:
Dear colleagues,
Could I get some assistance using postForm() to scrape the business names and
addresses at this website:
http://www.brantford.ca/business/LocalBusinessCommunity/Pages/BusinessDirectorySearch.aspx
I've read through (http://www.omegahat.org/RC
I managed to sort something out with a for loop, but it's till not working
ok...
What it does is it loops through all files in the folder, it imports each
file from line 763 on. Than it just takes the second column (Temprerature)
and binds the columns (cbind).
BUT it just binds the values of th
Sure, I have lots of ideas, mostly involving you overwriting your
results with each iteration. But unless you post your code to the the
list, I'll never know if my ideas are right.
Please read the posting guide. Using the Nabble interface does not
exempt you from posting manners.
Sarah
On Mon, M
Thanks Peter for your clarifications.
Yes, the definition I'm looking for is:
- I have 0.1 observations identical to this one,
i.e. this row and nine others similar (but not identical) to it together
represent a single observation.
> in lm/glm ... the weights are really only relative
This i
Hi,
if you're on a mac, I would recommend Automator. If you're on unix I would
recommend a handy bash script with regex. And on windows.. I don't know.. you
could do regex in R, couldn't you?
Am 28.05.2012 um 21:02 schrieb mpavlic:
> Hi all,
>
> I have a set of files (which is growing) in a
Dear R-users,
In advance excuse me for this basic question. I´m trying to compare the
coloration patterns on three spider species. In order to do that, I was
trying to convert the image on a pixel matrix and compare them.
Because of that, I´d like to know how to convert an image to a pixel matrix
I want to just convert from float32 to 16b with scale factor of 10. I wonder
why some files were converted correctly while some were not converted
correctly. By means, the results of some files are weird. the original files
are all ok!.
dir1<- list.files("C:\\New folder (13)", "*.img", full.nam
Thanks for your interest.
I've put the dataframe alyL32007 in
http://dl.dropbox.com/u/3180464/alyL32007.rda
ready to be used with load()
Agus
On Mon, May 28, 2012 at 6:11 PM, Gabor Grothendieck
wrote:
> On Mon, May 28, 2012 at 5:35 AM, Agustin Lobo
> wrote:
>> I'm doing:
>>
>>> alyL32007z <- z
Hello,
I've named your file 'file1.txt' and with
readLines("file1.txt")
saw 25 lines, then a header, then a table of tab separated values.
The header is full of blanks, such as the ones in 'length (m)' and
'temperature (°C)', making it impratical.
So if 'flist' is your list of files, try the
On Mon, May 28, 2012 at 5:35 AM, Agustin Lobo
wrote:
> I'm doing:
>
>> alyL32007z <- zoo(alyL32007,alyL32007$time)
The POSIXct time is erroneously being used twice: once as part of the
data and once as the index. It should be:
alyL32007z <- zoo(alyL32007[-1], alyL32007$time)
or
alyL32007z <-
Most likely you have a bug in your program. Have you looked at the
results of your calculations before writing them out? Since you have
provided no data, we can not reproduce what you are doing to show
where the error might be, or the correct way of doing it. In almost
all cases, if you think yo
Hi Roger,
thank you very much for your fast response. First of all, the mistakes I
mentioned are all of the sam type:" In summary.rq(x, se = "nid",
covariance = TRUE) : 22 non-positive fis". That is the reason why I want
to change the se-procedure as I made the experience that there problem
di
Hi there
We have 25m XY pairs to be correlated.
Data is bank financial data and smooth over Time ; observations for x and y
are 32 quarters each.
Testing 25m rships exhaustively will take forever ; this task is easily
over-engineered.
We'll use the best of XY relationships to predict. *
I believe my already posted solution works.
I've just tried it with your examples.
url <-
c("http://r.789695.n4.nabble.com/file/n4631640/channel_1_20120509_153744_1.ddf";,
"http://r.789695.n4.nabble.com/file/n4631640/channel_1_20120509_154744_1.ddf";)
flist <- url
And the rest is
Forgive me if this is a trivial question, but I couldn't find it an answer
in former forums. I'm trying to reproduce some SAS results where they set
two parameters equal. For example:
y = b1X1 + b2X2 + b1X3
Notice that the variables X1 and X3 both have the same slope and the
intercept has been re
I don't know how it ties into the tools car gives you, but one (quick
and dirty) way to do this is to simply regress on
Y ~ aX2 + b(X1+X3)
or in R code something like:
lm(Y ~ X2 + I(X1+X3), data = data.set)
which gives a linear model you can play around with. Note the I()
function [that's the c
I have know how to sort and filter and group.
can anyone answer my another question?
Is there any function in R like *lead *and *lag * in SQL. They are relative
position function. We can use them to solve problem such as : on
year-on-year basis, link relative ratio
can anyone give a tips?
--
Vie
type
??lag
at the R command line.
---
Jeff NewmillerThe . . Go Live...
DCN:Basics: ##.#. ##.#. Live Go...
Live: OO#.. Dead: OO#
Kel,
in addition, and depending on how you define "similarity", you might want to
look into the RV coefficient as a measure of it (it is actually related to a
correlation, so similarity would rather mean "similar information" though not
necessarily small Euclidean distance); coeffRV in FactoMine
Hi everyone,
I am using the fastgreedy.community function to get the $merges matrix and the
$modularity vector. This serves my purpose of testing modularity of my graph.
But I am "greedy" to plot the heat map and dendrrogram based on the $merges
dendogram matrix. I know that heatplot does the
i want to find type I error rates using modified F statistic..below is my R
code..
h=0
g=0
n=15
alpha=0.1
k=floor(alpha*n)+1
r=k-(alpha*n)
i=k+1
m=n-k
J=4
trim1<-rep(NA,asim)
trim2<-rep(NA,asim)
trim3<-rep(NA,asim)
trim4<-rep(NA,asim)
pw<-rep(NA,asim)
qw<-rep(NA,asim)
px<-rep(NA,asi
Hello everyone, I want to calculate type I error rate for modified F
statistic for one way robust anova. I need to find the j group trimmed
mean and winsorized sum of squared deviations. Here I attached my code for
j=2 to make it simple. Originally I have j=4. Hope you can help. I need to
run it f
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