Hi
Far from being an expert in this field I assume the answer is that it is
included in intercept term. It has something to do with contrasts
specification, but its usage is above my modelling knowledge.
You could try to see what
?contrasts
tells you.
Regards
Petr
you can
>
> Hello,
> I’m
Hi
It is strange. I get completely different error.
> dispersal[1] <- if (any (relativenes[1] < cost)) ((relativenes[1]-
cost)/(relativenes[1]- cost*cost)) else 0
Error: object 'relativenes' not found
> for (i in 2: time){dispersal[i] <- if (any (relativenes[i] < cost))
((relativenes[i]- cost
Hi
I am not at all an expert in this matter, so you shall copy you message to
r-help list too. However I would try na.action=na.exclude as it shall
preserve the length of variables.
Regards
Petr
>
> Hi,
>
> Sorry I didn't know the original post in this thread was not included.
I'm
> using
QQ<-diag(7)
for (j in 1:7){
for (k in 1:7){
sum=0
for (i in 1:50){
sum=sum+(Y[i,j]-mean(Y[,j]))*(Y[i,k]-mean(Y[,k]))
}
QQ[j,k]=sum/49
}
}
round(QQ,2)
this is the function I made
--
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Hi all!
I am writing because I have a small problem
I have a matrix with different variables names
I made a function to compute the covariance matrix for my first matrix, but
the variables names changed to numbers e.s [,1]
how I can explore the variable names in my covariance matrix
Can you provide a sample of code for us?
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Sent from the R help mailing list archive at Nabble.com.
__
R-he
Hello!
I am trying to get a logic estatement in R using the if function, but it is all
the time comming and error, I have tryied different methods but nothing is
working
the sentence is the following:
dispersal[1] <- if (any (relativenes[1] < cost)) ((relativenes[1]-
cost)/(relativenes[1]- co
Hi D. Isler,
It is not a very good sign if you are indeed missing the
find.package() function, this is in the base package, and if functions
from base are missing, your entire version of R is likely corrupt.
If you just do not have some packages you want, and are having
dependency issues, make su
Hi deb,
I am not clear what you want exactly, but something like:
x <- runif(1000)
y <- x^2 + rnorm(1000, mean = 0, sd = 1)
might do what you want.
See
?runif
for documentation. It generates random numbers from the uniform
distribution, and ditto for rnorm except from normal.
Cheers,
Josh
O
Data set up as one observation/subject looks like (with a total of 10 subjects)
Two treatments: shoe type with 3 categories and region with 8 categories ==>
24 "treatment" columns
Subject PHallux PMidToes PLatToe PMTH1 PMidMTH PLatMTH PMidfoot
PRearfoot LHallux LMidToes LLatToe LMTH1
I'm looking for guidance on how to implement forward stepwise regression
using lmStepAIC in Caret.
The stepwise "direction" appears to default to "backward". When I try to
use "scope" to provide a lower and upper model, Caret still seems to
default to "backward".
Any thoughts on how I can make t
Hi:
I am trying to generate data form a simple linear regression model.
The training data
T = {(x1, y1), . . . , (xn), yn}, want to sample x uniformly from the
range [0,1], find uncorrupted response y = x^2, and generate random
noise "e" from normal distribution N(0, 1). Any idea how to do in
sim
Hi all,
I am formatting some graphs and adding text to them. When using the
strwrap function in the code below, it properly cuts the main title
into sizeable chunks and pastes the second line below that of the
first. The subtitle however, is all pasted on the same line, meaning
that it reads as gi
Hi,
On Sunday, March 04, 2012 03:28:44 PM Fredrik Karlsson wrote:
> Hi Joshua,
>
> Yes, sorry - I attached an .rda file - maybe it was squashed.
> Anyway, yes, I agree with you that the function in the present state would
> not be of very much help to an end user (and your right in thinking that
>
On Mar 4, 2012, at 4:22 PM, maya wrote:
Hello,
I am a new R user, I have one R version in my Macbook and one in a
ubuntu
desktop.
I was installing some packages from a secure private source to the R
in
ubuntu and it replaced some of the existing packages. While the
version in
my macbook
Well, the only part that did not work nicely was the inside function.
So you could still probably lean on ddply() for much of the work and
just append the output of your function to the ddply results or
something like that.
You might also try to think about your problem in a different way.
Sometim
"Try this before running removePuncutation():
corpus <- tm_map(corpus, function(x) gsub("[\'\U2019]«»", " ", x))"
It will replace quotation marks with a space, and that's enough to
separate them from the rest of the word.
I try to use your solution. It's work only for characters, not for a Corpus,
Hi Joshua,
Yes, sorry - I attached an .rda file - maybe it was squashed.
Anyway, yes, I agree with you that the function in the present state would
not be of very much help to an end user (and your right in thinking that
this is not the end implementation of the function, just an example). I too
c
Dear Kino,
As it stands this model doesn't make sense: You've *both* set a reference
indicator for each factor, by fixing a factor loading to 1, and fixed the
variances of the factors 1. You should do one or the other. You say near the
end of your message that you did try freeing the factor varian
packet.number() will retrieve the panel you are working in. With a
third conditioning variable, you now have say a list of *.err$x, one
for each panel/level of 3rd var. In that case you need
dsl[[packet.number()]] to get at the right place in the list (or
[,packet.number()] if it's a matrix etc.).
On Sun, Mar 04, 2012 at 03:18:33PM -0600, Matyas Sustik wrote:
> Hi All,
>
> I am confused by a type conversion happening against my intent.
>
> In an R script I allocate a matrix X, and I pass it to a C function
> by using
>
> tmp -< .C(..., as.double(X),...).
>
> I use as.double() because I r
[cc'ing back to r-help]
On 12-03-04 05:07 PM, Mateusz Kędzior wrote:
> Sorry, may I ask one more question?
> What does "X" exactly meand and why without it, it's invalid variable name?
Follow the documentation link to ?make.names to see the definition of
a valid variable name (hint, it can't
On Sun, 4 Mar 2012, j.straka wrote:
Hello,
I?m using zero-inflated Poisson regression via the zeroinfl() function to
analyze data on seed-set of plants, but for some reason, I don?t seem to be
getting the output for all three levels of my two categorical predictors.
More about my data and mod
Mateusz Kędzior gmail.com> writes:
> I want to prepare boxplot graphs from CSV files which containt in first row
> years.
> I've tried to use both read.table and read.CSV functions, but each of them
> add "X" to year in header,
> so I can see: X2012 instead of simply 2012.
> After calling boxplot
Hi All,
I'm not able to figure out how to perform a Jackknife test for a 2-sample
dispersion test in R. Is there a built-in function to perform this or do we
have to take a step by step approach to calculate the test statistic?
Any help would be awesome.
Thanks!
--
View this message in contex
I want to prepare boxplot graphs from CSV files which containt in first row
years.
I've tried to use both read.table and read.CSV functions, but each of them
add "X" to year in header,
so I can see: X2012 instead of simply 2012.
After calling boxplot those strange signs weren't removed.
Is there an
Hi,
I created the model below, which returns me the following warning message:
In sem.default(ram = ram, S = S, N = N, param.names = pars, var.names =
vars, :
Could not compute QR decomposition of Hessian.
Optimization probably did not converge.
# Model
mDPDF =
data.frame(mj1
Hello,
I’m using zero-inflated Poisson regression via the zeroinfl() function to
analyze data on seed-set of plants, but for some reason, I don’t seem to be
getting the output for all three levels of my two categorical predictors.
More about my data and model:
My response variable is the number o
Hello,
I am a new R user, I have one R version in my Macbook and one in a ubuntu
desktop.
I was installing some packages from a secure private source to the R in
ubuntu and it replaced some of the existing packages. While the version in
my macbook works all ok, in the Ubuntu version of R, I am mis
I would like to convert my point data to grid data
my data contains e.g
lat <- c(25.1618 , 24.8664, 21.0575, 21.1537, 21.2477, 21.3403,
20.7617, 20.8577, 20.9516, 21.0443)
lon <- c(55.7660, 55.6890, 54.7205, 54.2142, 53.7105,53.2053, 54.6470,
54.1419, 53.6389, 53.1348)
z <- c(
Hi All,
I am confused by a type conversion happening against my intent.
In an R script I allocate a matrix X, and I pass it to a C function
by using
tmp -< .C(..., as.double(X),...).
I use as.double() because I read that it makes sure that the
parameter passing is correct.
I return the matrix
Hi,
Still no data (either not attached or a format the mail server
scrubs), but I made some up. ddply seems to be a bit tricky in that
it holds off evaluating its arguments. I honestly am not sure how to
get around that---no amount of fancy footwork with eval() or quote()
is going to change the
On 03/04/2012 11:06 AM, Yadav Sapkota wrote:
Hi,
I want to merge multiple chromosomal regions based on their common
intersecting regions. I tried couple of things using while and if loops but
did not work out.
I would appreciate if anyone could provide me a small piece of code in R to
get the i
Hello,
I have a lot of data and it has a lot of NaN values. I want to compress the
data so I don't have memory issues later.
Using the Matrix package, sparseMatrix function, and some fiddling around,
I have successfully reduced the 'size' of my data (as measured by
object.size()). However, NaN va
R still doesn't seem to recognize the data.frame ... I get a [6] ERROR:
object 'depy.w' not found from the following code:
dep <- pls[,1:4]
ind <- pls[,5:8]
eqn <- data.frame(depy = dep, indx = ind)
apls <- plsr(depy.w + depy.h + depy.d + depy.s ~ indx.a + indx.i + indx.r +
indx.x, data=eqn)
Thanks Michael. I had tried to drop the I(as.matrix(...)) conversions, and
fiddled with a number of other permutations of code ... I still can't seem
to get it right.
The col names appear to be depy and indx ... here is the output (and the
rows are just line numbers)
> colnames(eqn)
[1] "dep
The attach(eqn) seems to change some things in the format, but doesn't solve
the problem ... here is my script
> dep <- pls[,1:4]
> ind <- pls[,5:8]
> eqn <- data.frame(depy = dep, indx = ind)
> attach(eqn)
The following object(s) are masked from 'eqn (position 3)':
depy.d, depy.h, depy.
Joshua,
Thank you for explaining this. I will think of a work-around.
/Sergey
On Sun, Mar 4, 2012 at 3:36 PM, Joshua Ulrich wrote:
> Hi Sergey,
>
> Internally, xts objects are a matrix with an index attribute. While
> you *can* make a matrix of lists, that is not supported in zoo or xts.
>
> B
Hi,
I want to merge multiple chromosomal regions based on their common
intersecting regions. I tried couple of things using while and if loops but
did not work out.
I would appreciate if anyone could provide me a small piece of code in R to
get the intersection of following example:
chr1: 100-15
It's nice to cc the list for archival reasons -- it also usually gets
you a faster response as more folks can see how the thread develops.
The problem is that the colnames aren't ctually depy and indx: they
are depy.w, depy.h, etc. If you want to model, you need to use those
as is: e.g., with your
Dear R Helpers,
I am still having trouble with the getOptionChain command in quantmod. I
have the latest version of quantmod, etc. so I was under the impression
that the problem was solved with updates to the package.
If someone could let me know what I need to install in order to make this
work
No. Do not do that. While it looks nice at first, it quickly becomes
the source of innumerable errors.
Michael
On Sun, Mar 4, 2012 at 1:56 AM, Kazuo Ishii wrote:
> 2012/3/4, westland :
>> I am still/again having trouble getting PLSR to recognize the input data
>> frames. Here is what I have do
Hi,
Sorry all - I will provide a reproducable version of this. I am still
seeing the same problem - maybe it is due to me having to use summarize?
Anyway, here is an example (using the data set attached):
Two test functions:
insidefun <- function(x){
return(x+1)
}
testfun <- function(data,fa
On Mar 4, 2012, at 4:51 AM, Ajay Askoolum wrote:
I know how to do this when creating the time series variable in one
expression, e.g.
valuesTS1<-ts(values,start=2000); # Frequency naturally defaults to 1
How can I specify the frequency of a time series?
I do not see a `frequency<-` usag
On Mar 4, 2012, at 12:20 AM, Fredrik Karlsson wrote:
Hi Michael,
No, sorry - that is neither the problem or the solution.
suspicious.vowels(pb,c("Type","Sex","Vowel"),F1,F2)
Error in mean(y, na.rm = na.rm) : object 'f1' not found
Obviously you have still failed to offer reproducible code
Hi Sergey,
Internally, xts objects are a matrix with an index attribute. While
you *can* make a matrix of lists, that is not supported in zoo or xts.
Best,
--
Joshua Ulrich | FOSS Trading: www.fosstrading.com
R/Finance 2012: Applied Finance with R
www.RinFinance.com
On Sun, Mar 4, 2012 at
On Mar 4, 2012, at 12:21 , Dunia Scheid wrote:
> Dear all,
>
> I am fitting a GLM similar to
>
> library(MASS)
> anorex.1 <- glm(Treat~Postwt+Prewt,family = binomial, data = anorexia)
I hope that's just for an example. The actual analysis makes zero sense to me...
>
> I have found two ways o
Le samedi 03 mars 2012 à 16:56 -0800, Mickael R problem a écrit :
> Hello everybody,
> I don't give up the fight, but it's hard. I have finded a solution for the
> ligature with a best converter wich tranlated more precisely PDF to plain
> text. But a new problem has occured. In french particulary,
Dear all,
I am fitting a GLM similar to
library(MASS)
anorex.1 <- glm(Treat~Postwt+Prewt,family = binomial, data = anorexia)
I have found two ways of computing the p-value of the fitted model:
pval1 <- 1-pchisq(anorex.1$deviance,anorex.1$df.residual)
pval2 <- 1-pchisq(anorex.1$null.deviance - an
Hi R programmers,
I have stumbled across what seems a very simple problem. My goal is to
create a xts time series object which contains vectors as values. In
other words, I try to create something like this:
2009-01-01 => c('aa', 'bb', 'dd')
...
2010-02-01 => c('mm')
I have figured out parts of
Thanks, a lot!!
On Fri, 2012-03-02 at 18:00 -0500, Gabor Grothendieck wrote:
> On Fri, Mar 2, 2012 at 5:15 PM, sluedtke wrote:
> > Dear List,
> >
> > I am struggling with the trellis graphic. A similar problem was mentioned
> > here:
> >
> > http://r.789695.n4.nabble.com/R-How-can-you-get-N-rep
I know how to do this when creating the time series variable in one expression,
e.g.
valuesTS1<-ts(values,start=2000); # Frequency naturally defaults to 1
How can I specify the frequency of a time series?
> values=c(12,34,65,12);
> values<-ts(values);
> frequency(values);
[1] 1
> start(values)
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