On Sat, 25 Jun 2005, Faheem Mitha wrote:
>
> Hi,
>
> In src/nmath/dgamma.c the comment at the top says
>
> * DESCRIPTION
> *
> * Computes the density of the gamma distribution,
> *
> * 1/s (x/s)^{a-1} exp(-x/s)
> *p(x;a,s) = ---
> *
Gregory Gentlemen wrote:
> Spencer: Thank you for the helpful suggestions. I have another
> question following some code I wrote. The function below gives a
> crude approximation for the x of interest (that value of x such that
> g(x,n) is less than 0 for all n).
>
> # // btilda optimize g(n,x) f
Dear List,
If any of observed and/or expected data has less than
5 frequencies, then chisq.test (Pearson's Chi-squared
Test for Count Data from package:stats) gives warning
messages. For example,
x<-c(10, 14, 10, 11, 11, 7, 8, 4, 1, 4, 4, 2, 1, 1, 2,
1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1)
y<-c(9.1
Hi All,
I am currently conducting a mixed model. I have 7 repeated measures on a
simulated clinical trial. If I understand the model correctly, the
outcome is the measure (as a factor) the predictors are clinical group
and trial (1-7). The fixed factors are the measure and group. The rando
Ma première question: puis-je écrire en français, mon anglais est
pire que pire. J'essaie en français; sinon, j'essaierai une traduction!
Je sduis débutant. Ma question est donc simple, j'imagine.
Je travaille sur des tableaux de statistiques. Je prends un exemple:
J'ai un "frame" que j'appelle
Thomas,
I usually use the aggregate() function, proceeding as follows.
Construct a vector of 1s that is the same length as your entire data.
Then use aggregate() with sum as a function on this vector and your
grouping variable as a by argument (note the use of na.rm in sum). This
will create a
John,
In addition to 'VarCorr(nlme) and VarCorr(Matrix)', you could also try
'varcomp' function in the 'ape' package.This requires an 'lme' class of
file, from the 'nlme package as input.It additionally gives the options to
scale the components by normalizing them (both cumulatively and in total)
Dear everybody,
In our game-theory lesson we have run several classroom-experiments where the
students had to decide for a natural number between one and seven. I have
troubles now to visualize the results: be a the vector of answers.
hist(a) will not assume natural numbers as answers, but rati
On 6/26/05, Mag. Ferri Leberl <[EMAIL PROTECTED]> wrote:
> Dear everybody,
>
> In our game-theory lesson we have run several classroom-experiments where the
> students had to decide for a natural number between one and seven. I have
> troubles now to visualize the results: be a the vector of answe
Mag. Ferri Leberl wrote:
> Dear everybody,
>
> In our game-theory lesson we have run several classroom-experiments where the
> students had to decide for a natural number between one and seven. I have
> troubles now to visualize the results: be a the vector of answers.
>
> hist(a) will not assu
On Fri, Jun 24, 2005 at 11:27:28AM -0400, [EMAIL PROTECTED] wrote:
> Hi,
> I was wondering if someone can help me
> interpret the results of running
> weibreg.
>
> I run the following and get the
> following R output.
> > weibreg(Surv(time, censor)~covar)
> fit$fail = 0
> Call:
> weibreg(formu
On 6/26/05, Stephen <[EMAIL PROTECTED]> wrote:
>
>
>
> Hi All,
>
>
>
> I am currently conducting a mixed model. I have 7 repeated measures on a
> simulated clinical trial. If I understand the model correctly, the
> outcome is the measure (as a factor) the predictors are clinical group
> and t
Hi,
The description of all.equal states "is.all.equal should be used for
programming, typically in if expressions. It is a simple wrapper using
identical as shown in the documentation there.", but is.all.equal is
not explicitly defined there (although there is a hint in the comments
that is.all.eq
Hi, Renaud:
Thanks for the link to Doug Bates' recent comment on this.
Unfortunately, the information contained therein is not enough for me to
easily obtain all the required standard errors and write code to make
confidence intervals. To be more specific, I'm able to produce the same
a
On 6/26/05, Mohammad Ehsanul Karim <[EMAIL PROTECTED]> wrote:
> Dear List,
>
>
> If any of observed and/or expected data has less than
> 5 frequencies, then chisq.test (Pearson's Chi-squared
> Test for Count Data from package:stats) gives warning
> messages. For example,
>
> x<-c(10, 14, 10, 11
Hi, Laura:
How about the following:
> sapply(x1.df, class)
d x y
"factor" "numeric" "numeric"
> (byLH <- by(x1.df[-1], x1.df$d, var))
x1.df$d: 1/1/2005
x y
x 1.272688 -0.783719
y -0.783719 0.884866
---
Have you studied "An Introduction to R" [available via help.start()]?
Note especially the sections on "Factors", "Arrays and matrices", and
"Lists and data frames". There are also a couple of French-language
introductions to R available via "www.r-project.org" -> Documentation:
The precision is not a problem, only the display, as Uwe indicated.
Consider the following:
> (seq(25.5,25.6,length=20)-25.5)[c(1, 2, 19, 20)]
[1] 0.00e+00 5.25e-07 9.50e-02 1.00e-01
> ?options
> options(digits=20)
> seq(25.5,25.6,length=20)[c(1, 2, 1
When I try the following code with the Windows
graphics window, a new window is opened for each
multiple of four images I produce.
par(layout(matrix(c(1,2,3,4), 2, 2, byrow = TRUE)),
mar = c(2, 3, 2, 3))
for (i in 1:n) {
image(... )
}
When I try to do the same with Sweave to produce a pdf
docu
On 6/26/05, Spencer Graves <[EMAIL PROTECTED]> wrote:
> Hi, Renaud:
>
> Thanks for the link to Doug Bates' recent comment on this.
> Unfortunately, the information contained therein is not enough for me to
> easily obtain all the required standard errors and write code to make
> confiden
Hi, Doug:
Thanks for your reply. I'm keenly aware of many of the issues you've
mentioned. (I've learned much of what I know about this from your
innovation research.)
Besides, "lme" provides one answer to that question, if not a great
one. Moreover, if someone is really
Is there a way in R to globally setup a default format for dates for all, or
nearly all, or just "many" R date/time-related functions?
For example, dates in library(chron) expects, by default, MDY with
forward-slash delimiters, whereas my habit and my data sets tend to
be YMD with dash delimiter
On 6/26/05, Mike R <[EMAIL PROTECTED]> wrote:
> Is there a way in R to globally setup a default format for dates for all, or
> nearly all, or just "many" R date/time-related functions?
>
> For example, dates in library(chron) expects, by default, MDY with
> forward-slash delimiters, whereas my hab
> Here gamma is the usual gamma function, see ?gamma. (I notice in the R
> documentation of the Weibull distribution that "E(X) = b Gamma(1+1/a)",
> which is an error; the G should be g (lowercase).)
It is not an error. R's function gamma() is said to implement
\eqn{\Gamma}{Gamma}: `see ?gamma'!
OK. How about the following:
R> x1.df$d. <- as.Date(as.character(x1.df$d), format="%m/%d/%Y")
R> sapply(x1.df, class)
d x yd.
"factor" "numeric" "numeric""Date"
R>
R> (byLH <- by(x1.df[2:3], x1.df$d., var))
x1.df$d.: 2005-01-01
x
Dear useRs,
Is there a way to convert a list generated by "by" command into a data frame?
Or, more generally, are there any functions that operate on "by" lists?
My "by" list consists of numeric vectors of equal length and the objective is
to calculate the sum of the same index elements (i.e., col
On 6/26/05, Tudor Bodea <[EMAIL PROTECTED]> wrote:
> Dear useRs,
>
> Is there a way to convert a list generated by "by" command into a data frame?
Here is an example which creates a by object whose elements
are the first row of each Species of iris. The next line converts
that to a data frame:
How can one convert back slashes to forward slashes, e.g, changing
"c:\a\b" to "c:/a/b"? I tried the following:
> gsub("", "/", "c:\a\b")
[1] "c:\a\b"
Since this didn't work, I eliminated the replacement as a contributor
as follows:
> gsub("X", "/", "c:XaXb")
[1] "c:
On Sun, Jun 26, 2005 at 11:28:04PM +0100, Prof Brian Ripley wrote:
> >Here gamma is the usual gamma function, see ?gamma. (I notice in the R
> >documentation of the Weibull distribution that "E(X) = b Gamma(1+1/a)",
> >which is an error; the G should be g (lowercase).)
>
> It is not an error. R's
On 26 June 2005 at 20:30, Spencer Graves wrote:
| How can one convert back slashes to forward slashes, e.g, changing
| "c:\a\b" to "c:/a/b"? I tried the following:
|
| > gsub("", "/", "c:\a\b")
| [1] "c:\a\b"
This does work, provided you remember that single backslashed "don't exi
On 6/27/05, Dirk Eddelbuettel <[EMAIL PROTECTED]> wrote:
>
> On 26 June 2005 at 20:30, Spencer Graves wrote:
> | How can one convert back slashes to forward slashes, e.g, changing
> | "c:\a\b" to "c:/a/b"? I tried the following:
> |
> | > gsub("", "/", "c:\a\b")
> | [1] "c:\a\b"
>
>
Hi
Thank you for your comments.
Yes you are correct its a very big data set.
Perhaps I am best splitting it up and then importing to R.
The reason for the loop is that I am conducting the equivalent of Split
file in SPSS.
Specifically, I am conducting the analysis for each value of on the
grouping
Hi
It is corrected in the newer R-version. The help page now says:
'all.equal(x,y)' is a utility to compare R objects 'x' and 'y'
testing "near equality". If they are different, comparison is
still made to some extent, and a report of the differences is
returned.Don't use 'all.equal' direct
hadley wickham wrote:
> Hi,
>
> The description of all.equal states "is.all.equal should be used for
> programming, typically in if expressions. It is a simple wrapper using
> identical as shown in the documentation there.", but is.all.equal is
> not explicitly defined there (although there is a
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