Hi..
I pass a lm() object to step() but after a few steps it stops with the
error message.
say:
x1<-lm(y~.,data=z)
x2<-step(x1)
i have 516 observations and 299 variables.
after a few steps i get the error message:
Error in one %*% x : requires numeric matrix/vector arguments
Hi..
I have to invert a 15000 x 15000 matrix (generalized inverse). I do run
the process on a fairly powerful computer. but still complains indufficient
memory.
Is there a way one can invert a large matrix in some other efficient manner.
Thanks
Harsh
Hi..
I a have a symmetric matrix to plot . I would like to plot only the Upper
triangle but with the diagonal as the Base of the rectangle. Is there an easy
way to do it.
Thanks.
Harsh
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Hi ..
I would like to have math symbols in perspective plots
i tried : persp(x,y,z,xlab=expression(phi))
but it plots it as phi.
Thanks.
Harsh
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Hi All,
when i run
cv.lars(x3,y3)
it runs fine. but when i run
cv.lars(x3,y3,fraction=seq(0,0.1,100))
I get the following error.
Error in apply((y[omit] - fit)^2, 2, mean) : dim(X) must have a positive
length
Any help/suggestions will be appreciated.
Th
Hi..
i was using image.plot from the library fields along with layout
i had the follwoing commands:
layout(matrix(c(1,1,2,3,4,5),ncol=2,nrow=3,byrow=TRUE))
followed by 5 image plots.
i wanted image 1 on the top row followed by the 4 images on the next two rows.
it works
Hi ..
lm() returns an effects component in its output. I read the explanation in
R but was not quite clear.
say my response is Y
and design matrix is X
say X has QR decomposition X=QR
is effects = Q (Q'Q)^-1 Q' Y ???
i am sure this is wrong as it did not match the ou
get an error saying row size changing: remove NAs.
the data does not have NA's
Thanks
Harsh
"Doran, Harold" <[EMAIL PROTECTED]> wrote: You can do this yourself with
ginv()
-Original Message-
From: [EMAIL PROTECTED]
[mailto:[EMAIL PROTECTED] On Beh
Hi,
is there a lm which will implement the moore-pensrose generalized
inverse.
Thanks.
Harsh
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R-help@stat.math.ethz.ch mailing list
Hi ,
I am using nls(formula)
where formula is y~ _expression
y is 1 character
~ is 1 character
now i have an _expression that is greater tha 502 characters. but R does not
recognize anything greater than 502 characters. (Note the _expression itself
is created using pas
Hi ,
I am using nls(formula)
where formula is y~ expression
y is 1 character
~ is 1 character
now i have an expression that is greater tha 502 characters. but R does not
recognize anything greater than 502 characters. (Note the expression itself is
created using paste
Hi ,
I am using nls(formula)
where formula is y~ expression
y is 1 character
~ is 1 character
now i have an expression that is greater tha 502 characters. but R does not
recognize anything greater than 502 characters. (Note the expression itself is
created using paste
Hi..
is there a limit on the number of explanatory variables in nls ?
i have a dataframe with the columns names x1,x2..,x300
when i run nls it gives the error: " x181 not found"
thought it does run when i have x1,x2,...,x170 variables.
Thanks
Harsh
Charles Annis, P.E.
[EMAIL PROTECTED]
phone: 561-352-9699
eFax: 614-455-3265
http://www.StatisticalEngineering.com
-----Original Message-
From: [EMAIL PROTECTED]
[mailto:[EMAIL PROTECTED] On Behalf Of Cal Stats
Sent: Saturday, March 11, 2006 3:43 PM
To: r-help@stat.math.ethz.ch
Subject: [R] nls s
Hi,
I have a large number of parameters to estimate in nls say 100:
beta1--beta100
lets say i have 100 values in a vector
is there a way where i can create the start vector for nls using a loop
instead of individually filling the 100 values.
Thanks
Harsh
Hi..
i have an expression of the form:
model1<-nls(y~beta1*(x1+(k1*x2)+(k1*k1*x3)+(k2*x4)+(k2*k1*x5)+(k2*k2*x6)+(k3*x7)+(k3*k4*x8)+(k3*k2*x9)+(k3*k3*x10)+
(k4*x11)+(k4*k1*x12)+(k4*k2*x13)+(k4*k3*x14)+(k4*k4*x15)+(k5*x16)+(k5*k1*x17)+(k5*k2*x18)+(k5*k3*x19)+
(
Hi..,
is there any funcrion in R to fit a periodic B-spline Surface
Harsh
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Brings words and photos together (easily) with
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R-help@stat.math.
Hi..
i was trying to integrate the indicator funtion but had problems when
limits where negative or equal to the indicator condition
my function is
fun1<-function(x){
as.numeric(x>=2)
}
_
which should be Ind(x>=
is there a command in R that would return the expression of an Indefinite
Integral.
i know there is integrate() command for definite integrals.
Thanks.
Harsh.
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Hi..,
is there a function in R to fit bivariate splines
?
I came across 'polymars' (POLSPLINE) and 'mars' (mda)
packages. Are these the one to use or are there other
specific commands?
Thanks.
Harsh
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