On Sat, 19 Jun 2004, Shin wrote:
I am curious if there is any way to list only variables or functions in
current environment, rather than listing all objects? Thanks.
Not really. What you can do is list the objects and then get the objects
and look at the modes, as ls.str does. That could
Dear r-helpers,
does anyone knows how to constrain the value of parameters in a non-linear
mixed model:
I've got something like that:
nlme(log(response)~z+y*log(var1)+x*log(var2)+log((b*var3+a)/(b*var4+a)^x)
where 0x1, a1 and (b*(var3 or var4: range of values similar)+a) between 0
and 1.
or even
WHY THIS FUNCTIONAL FORM?
If this were my problem, I would first want to know why we needed
this particular functional form and why (b*var3+a) and (b*var4+a) had to
be between 0 and 1. When I see things like this, my first impulse is
that these may only be approximations to something
These two functions will list the functions and variables
respectively:
ls.funs -
function(env=sys.frame(-1))unlist(lapply(ls(env=env),function(x)if(is.function(get(x)))x))
ls.vars -
function(env=sys.frame(-1))unlist(lapply(ls(env=env),function(x)if(!is.function(get(x)))x))
To use:
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I have code that will crash R every time on an OS X machine, but runs
fine on other platforms. The developer I've been working with suspects
it's a compiler issue, but I only have the same ones he does (g77).
Obviously, the way to test this would be to compile R using another
compiler. XLF is
Neat! Thanks.
How about incorporating this support into standard commands, ls() or
objects()?
Daehyok Shin (Peter)
-Original Message-
From: Gabor Grothendieck [mailto:[EMAIL PROTECTED]
Sent: Sunday, June 20, 2004 AM 10:06
To: [EMAIL PROTECTED]; [EMAIL PROTECTED]
Subject: RE: [R] A
Interesting, but it seems too complex.
In my opinion, just listing functions or non-functions meets my need.
Anyway, thanks.
Daehyok Shin (Peter)
-Original Message-
From: S?en Merser [mailto:[EMAIL PROTECTED]
Sent: Sunday, June 20, 2004 AM 10:51
To: Shin
Subject: Re: [R] A way to
I ran into an interesting oddity of R,
if (0) { print(1); }
else { print(2); }
is a syntax error, while
if (0) { print(1); } else { print(2); }
or
if (0) { print(1);
} else { print(2); }
is not. I presume it has to do with the duality of the newline
functioning as an end of
On Sun, 20 Jun 2004, Shin, Daehyok wrote:
Neat! Thanks.
Note that these are not correct, as the get is not done from the
correct environment. The function ls.str I pointed you to is correct.
How about incorporating this support into standard commands, ls() or
objects()?
Well, there
This has been discussed several times on this list. Note that line 2 of
paragraph 2 of help(if) says the following :
In particular, you should not have a newline between '}' and 'else' to
avoid a syntax error in entering a 'if ... else' construct at the
keyboard or via 'source'.
On Sun,
try
?if
Best,
Matthias
[EMAIL PROTECTED] wrote:
I ran into an interesting oddity of R,
if (0) { print(1); }
else { print(2); }
is a syntax error, while
if (0) { print(1); } else { print(2); }
or
if (0) { print(1);
} else { print(2); }
is not. I presume it has to do with the
[EMAIL PROTECTED] wrote:
I ran into an interesting oddity of R,
if (0) { print(1); }
else { print(2); }
In both cases the ; is superflously .
If if(){}else{} is within an expression (e.g. a function's body), both
ways work.
Do you really want to use it outside a function's body? If so,
Hi, friends!
Has R estimation (library, for example) to do estimation in HMM?
Thanks in advance,
Cezar Freitas
Estatistico - Comissao Permanente para os Vestibulares / UNICAMP
Probabilidade e Estatistica Aplicadas - IME / USP | IMECC / UNICAMP
Campinas |
Hi Frank. I am (somewhat) new to R as well, but almost a 10 yr SAS veteran.
I work for a very large US Bank and have spent a considerable part of my
career in Corp Mktg leveraging data for, arguably, data mining, next
purchase, attrition, balance diminishment and the like. I am now managing
an
On Sun, Jun 20, 2004 at 09:00:39AM -0700, [EMAIL PROTECTED] wrote:
I wanted to look up the internal R documentation for if via ?if, but
this does not work. making the latter work would be a good idea.
FYI, it also works inside of (X)Emacs where the (highly recommend) ESS mode
is kind enough
Charles and Kimberly Maner wrote:
Hi Frank. I am (somewhat) new to R as well, but almost a 10 yr SAS veteran.
I work for a very large US Bank and have spent a considerable part of my
career in Corp Mktg leveraging data for, arguably, data mining, next
purchase, attrition, balance diminishment and
Here they are again modified to remove that bug:
ls.funs -
function(env=sys.frame(-1))unlist(lapply(ls(env=env),function(x)if(is.function(get(x,env=env)))x))
ls.var -
function(env=sys.frame(-1))unlist(lapply(ls(env=env),function(x)if(!is.function(get(x,env=env)))x))
Date: Sun, 20 Jun
Is there any way to echo comments from an R source file into an
SWeave-LaTeX document?
Example:
# Npop is population total
# Npoph0..Npoph2 are stratum totals
# Npoph is vector of stratum totals
Npop-sum(to2000)
Npoph0-sum(to2000[bg==0])
Npoph1-sum(to2000[bg==1])
Npoph2-sum(to2000[bg==2])
I have a problem with the use of locfit with censured data, when I carry out locfit
by:
fitbmt-locfit(~recur,data=BMTAGE11,cens=df.status,family=hazard,alpha=0.5)
it does not give me any message, but if I want to obtain the graph or even if I ask
for (fitbmt) made it gives me the following
Hello:
You are damn right! I haven't noticed because I rarely use the html help.
I will be working on. I'll let you know.
Regards.
On Fri, 18 Jun 2004, Emilio A. Laca wrote:
I recently upgraded from R 1.8 to 1.9. I removed 1.8 following the
instructions. Html help has not worked since.
Dear R,
I am student at University of new haven, CT.I am trying to run my R scripts where I
don't have
X11() authentication to my account. I run those R scripts and when I generate any
graphics I get
error and it comes out from system.
if possible , please let me know how can i run R scripts
- Original Message -
From: Robin Gruna
To: [EMAIL PROTECTED]
Sent: Sunday, June 20, 2004 5:42 PM
Subject: Evaluating strings as variables
Hello,
I have the following problem: I have a list as follows,
values - list(red = 1, yellow = 2, blue = 3)
values
$red
[1] 1
$yellow
[1] 2
Robin Gruna [EMAIL PROTECTED] writes:
I have the following problem: I have a list as follows,
values - list(red = 1, yellow = 2, blue = 3)
values
$red
[1] 1
$yellow
[1] 2
$blue
[1] 3
There is also a vector containing the diffrent colors as character strings:
colors -
On Sunday 20 June 2004 15:15, SAURIN wrote:
Dear R,
I am student at University of new haven, CT.I am trying to run my R
scripts where I don't have X11() authentication to my account. I run
those R scripts and when I generate any graphics I get error and it
comes out from system.
if
Hello,
And thanks again for your answers, perspectives and more...
So, as I understood, R can (nearly) do anything. So, also because it's free,
it is worth a try ;-).
I then next will start with reading some introductory texts. And, wow, I'm
quite 'overloaded', because there is so much stuff
Robin Gruna [EMAIL PROTECTED] writes:
I have the following problem: I have a list as follows,
values - list(red = 1, yellow = 2, blue = 3)
values
$red
[1] 1
$yellow
[1] 2
$blue
[1] 3
There is also a vector containing the diffrent colors as character
strings:
colors - c(red,
Charles H. Franklin wrote:
Is there any way to echo comments from an R source file into an
SWeave-LaTeX document?
Example:
# Npop is population total
# Npoph0..Npoph2 are stratum totals
# Npoph is vector of stratum totals
Npop-sum(to2000)
Npoph0-sum(to2000[bg==0])
Npoph1-sum(to2000[bg==1])
Dear Frank,
First, thank you for your kind remarks about the Rcmdr package.
Please note that the Rcmdr package is meant to be a basic statistics GUI for
R, to be used, for example, in an introductory statistics course. It covers
only a very small fraction of what's available in R.
Almost all of
I have problem evaluating the expression h = exp(x)/(exp(exp(x))-1) for
large negative x. This expression is actually the probability
that y = 1 when y is a Poisson random variable truncated at 0, hence
must satisfy 0 = h = 1. However, when
x -18, I may get an h value that is larger than 1
[EMAIL PROTECTED] writes:
sapply(colors, get)
or?
unlist(mget(colors, envir = as.environment(-1), inherits = TRUE))
Ah, yes, forgot about mget. Doesn't seem to make things much cleaner
though... (although get() also needs a bit more care about
environments).
--
O__ Peter
Zhen Chen [EMAIL PROTECTED] writes:
I have problem evaluating the expression h = exp(x)/(exp(exp(x))-1)
for large negative x. This expression is actually the probability
that y = 1 when y is a Poisson random variable truncated at 0, hence
must satisfy 0 = h = 1. However, when
x -18, I may
On Sun, 20 Jun 2004, Zhen Chen wrote:
I have problem evaluating the expression h = exp(x)/(exp(exp(x))-1) for
large negative x. This expression is actually the probability
that y = 1 when y is a Poisson random variable truncated at 0, hence
must satisfy 0 = h = 1. However, when
You would be
Date: Sun, 20 Jun 2004 18:32:46 -0400
From: Zhen Chen [EMAIL PROTECTED]
I have problem evaluating the expression h = exp(x)/(exp(exp(x))-1) for
large negative x. This expression is actually the probability
that y = 1 when y is a Poisson random variable truncated at 0, hence
must satisfy
Hi Cezar,
Download and install the repeated package by Jim Lindsey and type ?hidden
bye, ingmar
Hi, friends!
Has R estimation (library, for example) to do estimation in HMM?
Thanks in advance,
Cezar Freitas
Estatistico - Comissao Permanente para
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