Duncan Murdoch <[EMAIL PROTECTED]> [Sat, Nov 17, 2007 at 11:56:52PM CET]:
> On 17/11/2007 5:04 PM, Johannes Hüsing wrote:
> >I just realized that the "..." argument in a function cannot be used
> >without
> >taking precautions sometimes. The following behaviour is what I
> >stumbled upon:
>
> Th
One way of doing I think is to do it by 'hand'. In ls we minimizes SSE
w.r.t. mu and beta. However in your case, you can concentrate only on
minimizing SSE w.r.t. beta
Regards,
-Original Message-
From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] On
Behalf Of Tom La Bone
Sent: Sunday, Nove
Greetings
Am newbie using MacOS10.4.10
R version 2.6.0 Patched (2007-11-09 r43408)
Gostats 2.4.0
RGraphviz 1.16.0
Xcode X11 1.1.3 - XFree86 4.4.0
I want to make an induced GO tree from 1019 genes so I did
> x <- scan(file="GeneIDsToBioconductor2.txt", sep=",")
Read 1019 items
> xc <- as.characte
or simply insert the offset term into the formula
lm(y~offset(x))
Washington
On 17/11/2007 8:29 PM, Tom La Bone wrote:
> Is there a way to do a linear regression with lm (having one predictor
> variable) and constrain the slope of the line to equal 1?
>
You could use the offset parameter to do
On 17/11/2007 8:29 PM, Tom La Bone wrote:
> Is there a way to do a linear regression with lm (having one predictor
> variable) and constrain the slope of the line to equal 1?
>
You could use the offset parameter to do this, e.g.
x <- 1:20
y <- x + rnorm(20)
lm(y ~ 1, offset=x)
Duncan Murdoch
_
Is there a way to do a linear regression with lm (having one predictor
variable) and constrain the slope of the line to equal 1?
Tom
--
View this message in context:
http://www.nabble.com/Linear-Regression-with-lm-Forcing-the-Slope-to-Equal-1-tf4828804.html#a13815432
Sent from the R help maili
On 17/11/2007 5:04 PM, Johannes Hüsing wrote:
> Dear expRts,
> I just realized that the "..." argument in a function cannot be used without
> taking precautions sometimes. The following behaviour is what I stumbled
> upon:
The problem is the way replicate is implemented. It takes the
expression
Dear expRts,
I just realized that the "..." argument in a function cannot be used without
taking precautions sometimes. The following behaviour is what I stumbled
upon:
> myrepl <- function(length, fun, ...) {
+ replicate(length, fun(...))}
> myrepl(20, sample, 1:5)
[1] 0 0 0 0 0 0 0 0 0 0 0 0 0
On 17/11/2007 3:45 PM, [EMAIL PROTECTED] wrote:
> Dear R-helpers,
>
> I hope someone can help me with the following problem:
>
> I derived a variable from many others and produced data.frame: toktempo
> When I look at the variable name I get:
>
> names(toktempo)
> [1] "otok.V5"
>
> But I wa
Dear R-helpers,
I hope someone can help me with the following problem:
I derived a variable from many others and produced data.frame: toktempo
When I look at the variable name I get:
names(toktempo)
[1] "otok.V5"
But I want the variable name to be Tempo so I googled around to find info
abo
Ross Boylan wrote:
>
> I have a data frame "reading" that includes a logical variable "OLT"
> along with response variable "Reading" and predictor "True" (BOTH are
> numeric variables; it's "True" as in the true value).
>
> When I suppress the intercept, model.matrix gives me OLTTRUE and
> OLT
On Fri, 2007-11-16 at 18:31 -0500, Daniel Malter wrote:
> Hi, I have seen a discussion in the R-help asking whether nonparametric
> manova has been implemented in R yet. This discussion is form March 2006 and
> there seemed to be no package or function implemented at the time. Has this
> changed?
Jiao Yang wrote:
>
> Dear all,
>
> In MATLAB, to generate multivariate F with specified correlation matrix Pn
> I can use the code such as
>
> Z = mvnrnd([0 0 0 0 0], Pn, N);
> U = normcdf(Z,0,1);
> X = [finv(U(:,1),5,15) finv(U(:,2),5,15) finv(U(:,3),5,15)
> finv(U(:,4),5,15) finv(U(:,5),5,1
Hi,
I know this topic has come up several times, but I still haven't
found a satisfactory solution to it.
I want to draw complicated figures (preferably in X11) and display
them only once they are completed.
Obviously, the main purpose for doing this is so that a sequence
of complicated plots can
There is a difference between importing a namespace and loading a package:
If you import functions from a namespace into your package, these
functions are known within that namespace, i.e. functions from your
package's namespace know the other functions and can use them. The user
still won't se
Hmmm...
Isn't this :
Deepayan Sarkar a écrit :
> On Nov 16, 2007 5:26 AM, Bob Farmer <[EMAIL PROTECTED]> wrote:
>> Yep, that did it, thanks.
>
> Great.
>
> [...]
>
>>> The obvious approach would be to use a relative unit like "cm" or
>>> "inch" instead of an absolute one like "native". Does th
Geertje Van der Heijden wrote:
>
> Hi,
>
> I am trying to fit a ANCOVA model using lmRob. The P-values of the
> variables in the model differ hugely between the summary() function and
> the anova() function (from >0.8 in the summary to <0.001in the anova for
> the same variable). I understand
On Nov 17, 2007 8:55 AM, Emmanuel Charpentier
<[EMAIL PROTECTED]> wrote:
> Dear Lucia,
>
> lucia a écrit :
> > Hello,
> > I'm very new to R, and so my question is simple.
> >
> > I have data record with 80 years of daily temperatures in one long
> > string. The dates are also recorded, in YYMMDD f
On Nov 17, 2007 8:55 AM, Emmanuel Charpentier
<[EMAIL PROTECTED]> wrote:
> Dear Lucia,
>
> lucia a écrit :
> > Hello,
> > I'm very new to R, and so my question is simple.
> >
> > I have data record with 80 years of daily temperatures in one long
> > string. The dates are also recorded, in YYMMDD f
Dear Lucia,
lucia a écrit :
> Hello,
> I'm very new to R, and so my question is simple.
>
> I have data record with 80 years of daily temperatures in one long
> string. The dates are also recorded, in YYMMDD format. I'd like to
> learn an elegant simple way to pull out the annual averages.
Assuming we use zoo, we define a test series, z, and then aggregate it:
> library(zoo)
> z <- zooreg(1000:2000, start = as.Date("2000-01-01"))
> aggregate(z, as.numeric(format(time(z), "%Y")), mean)
2000 2001 2002
1182.5 1548.0 1865.5
See:
vignette("zoo")
vignette("zoo-quickref")
?zoo
Hello,
I'm very new to R, and so my question is simple.
I have data record with 80 years of daily temperatures in one long
string. The dates are also recorded, in YYMMDD format. I'd like to
learn an elegant simple way to pull out the annual averages.
(Obviously, every 4th year has 366 days
Hi, Kalin, maybe you need this：
A=diag(4)+1
prod(diag(chol(A))^2) # it can help you get det(A)
2007/11/17, kalin lagno <[EMAIL PROTECTED]>:
> Hi,
> Which R function I should use to obtain determinant of a matrix with real(and
> complex) numbers?
>
> Kalin
>
> -
Shoaaib Mehmood wrote:
> using an example from r online help
>
>
>>state <- c("tas", "sa", "qld", "nsw", "nsw", "nt", "wa", "wa",
>
> "qld", "vic", "nsw", "vic", "qld", "qld", "sa", "tas",
> "sa", "nt", "wa", "vic", "qld", "nsw", "nsw", "wa",
> "sa", "act", "nsw", "vic", "vic", "act"
Prof Brian Ripley a écrit :
[ ... ]
> What is strange is that no one has ever thanked us for finding out
> (despite most Microsoft documentation) that you can get up to 3.5Gb in a
> 32-bit process on certain 64-bit versions of Windows and enabling you to
> use it in recent versions of R.
Maybe
Steve Powell a écrit :
> hi list members
> I am trying to use odfWeave with R 2.5.1 on Windows XP.
>
> however when running e.g.
> odfWeave(demoFile, outputFile)
>
> I get:
> Error in odfWeave(demoFile, outputFile) : Error unzipping file
> In addition: Warning message:
> unzip not found in: sy
26 matches
Mail list logo
