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= \n,s, \n)
out-list(s=s,d=d); sink()
return(out)
}
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-contained, reproducible code.
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of the
variables has the influence on the outcome in the anorexia data?
Please don't shout!! happy to be pointed to a reference but would
prefer one in common english not some stats mumbo jumbo!
Calum
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at the answer? You are the
one who know what f is and you are the one who has the option of
increasing maxit. If the question is how to increase maxit, then the
answer is perhaps as easy as:
?optim
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On Sep 5, 2010, at 6:06 AM, st...@wittongilbert.free-online.co.uk wrote:
David Winsemius wrote:
1. is glm the right thing to use before I waste my time
Yes, but if your outcome variable is binomial then the family
argument should be binomial. (And if you thought it should
to
decimal ASCII
[1] 49 50 51 32 116 104 105 115 32 105 115 32 97 32 115 116
114 105 110 103
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possible type identifiers? We need
an example that has enough complexity to allow testing.
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On Sun, Sep 5, 2010 at 6:44 PM, David Winsemius dwinsem...@comcast.net
wrote:
On Sep 5, 2010, at 8:48 AM, rajesh j wrote:
Hi,
Is it possible to convert a string vector to integer
. The number of items in the vector is
unknown.
here's an example,
a list has vectors
INT
2
3
4
NUM
2.37
4.56
On Sun, Sep 5, 2010 at 6:56 PM, David Winsemius dwinsem...@comcast.net
wrote:
On Sep 5, 2010, at 9:22 AM, rajesh j wrote:
for e.g., I get the following as a string vector
int
-list(INT=c(1,2,3),NUM=c(2.34,4.56,6.78))
cc - lapply(cc, as.numeric)
cc
$INT
[1] 1 2 3
$NUM
[1] 2.34 4.56 6.78
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On Sun, Sep 5, 2010 at 7:57 PM, David Winsemius dwinsem...@comcast.net
wrote:
So there is one item per line and the task is to recognize the
strings INT and NUM
://finzi.psych.upenn.edu/R/library/VGAM/html/dirichlet.html
(I'm looking for something like this: http://repec.org/bocode/d/dirifit.html
,
that allows for both dependent variables summing to 1 predictive
variables
of any sort.)
Don
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On Sep 5, 2010, at 1:21 PM, Aks Ism wrote:
Hi,
I've looked at previous discussions and did not get anything. I want
to be
able to append to a list in a loop. Is this possible?
Of course:
?c
?[[
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have entered:
defaults write com.apple.Finder AppleShowAllFiles YES
. and then pt-click-hold on Dock-Finder-icon, choose relaunch
Or you could open a Terminal window which will by default open in /
Users/ctu/ and type:
rm .RData
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.
Perhaps you mean something like this:
ll - list();
for (i in 1:20) {
my.element - scan();
ll - c(ll, my.element);
if (is.na(my.element)){return(ll);break}}
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On Mon, Sep 6, 2010 at 1:01 AM, David Winsemius dwinsem...@comcast.net
wrote:
On Sep 5, 2010, at 1:21 PM
330 ...
$ :List of 3
..$ level: num 160
..$ x: num [1:31] 270 263 262 260 260 ...
..$ y: num [1:31] 310 320 330 340 350 ...
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Sergey
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that package, sometimes reads rhelp but I'm not sure
on what schedule. You might see if he makes his email address
available on those pages.
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On Sep 6, 2010, at 1:13 PM, tamas barjak wrote:
Hello!
I need some help.
How I know it to draw the formula of the poisson distribution?
expr-expression(P(xi == k) == frac(lambda^k, factorial(k))*e^-
lambda) ---
not good
?plotmath
(Do not see factorial as a plotmath function
Try:
On Sep 6, 2010, at 12:15 PM, Dimitri Shvorob wrote:
I have a (very big - 1.5 rows) dataframe with a (POSIXt POSIXlt)
column h
(hour). Surprisingly, I cannot calculate a simple aggregate over the
dataframe.
n.h1 = sqldf(select distinct h, count(*) from x group by h)
Error in
On Sep 6, 2010, at 2:07 PM, tooblue wrote:
I simply put, plot(density(), main=, + xlab = XXX), it
says that
I have an unexpected = in it.
It may be a case of a confused parser. You have an extraneous + in
there:
= rnorm(100)
plot(density(), main=, xlab = XXX)
talking about?
Thanks
Jason
Dr. Iasonas Lamprianou
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since you assigned a result to those
names.
(Greater clarity would occur if you offered at least str(NEVER)
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On Sep 6, 2010, at 7:54 PM, Brant Inman wrote:
R-helpers,
I am using the package drc to fit a 4 parameter logistic model.
When I
use the predict function to get prediction on a new dataset, I am not
getting the requested confidence or prediction intervals. Any idea
what
is going on?
.
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density
estimation
function (e.g. lambdahat in the spatialkernel package) with boundary
corrections to allow univariate density estimation?
When this question has been (multiply) posed in the past, the
suggested answer has been to use package logspline.
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, 4)
[,1] [,2] [,3]
[1,]1 81 125
[2,]8 16 1296
BTW, I have a 64bit R version (2.11) for Linux. Any advice would be
appreciated.
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, I'll take this moment to remind anyone interested that R
still has trouble with embedded zeros in character strings. I may be
abusing terminology, but I think that makes R 8-bit dirty.
-Matt
On Tue, 2010-09-07 at 14:01 -0400, David Winsemius wrote:
On Sep 7, 2010, at 1:35 PM, Matt Shotwell wrote
On Sep 7, 2010, at 2:53 PM, Johann Hibschman wrote:
David Winsemius dwinsem...@comcast.net writes:
On Sep 7, 2010, at 11:02 AM, Johann Hibschman wrote:
Even so, I would prefer to only save the coefficients
Have you read through the Value section of glm's help page?
...and
?coef
I have
On Sep 7, 2010, at 3:16 PM, David Winsemius wrote:
On Sep 7, 2010, at 2:53 PM, Johann Hibschman wrote:
David Winsemius dwinsem...@comcast.net writes:
On Sep 7, 2010, at 11:02 AM, Johann Hibschman wrote:
Even so, I would prefer to only save the coefficients
Have you read through
(MatchedValues[Value2,Value])
[1] 4420
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parameters some time.
*
Can anyone help me this this? Thanks so much!
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Sincerely,
Changbin
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series
when se.fit = TRUE.
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On Sep 7, 2010, at 9:33 PM, David Winsemius wrote:
On Sep 7, 2010, at 7:51 PM, lord12 wrote:
For each arima model, can you output an associated confidence
interval for
the predicted value at each time point?
?arima0
arima0 will return ... a list with components pred
])])
s - with(measure_bkf_not_zero[bkf_min:bkf_max,],
approxfun(bankfull_depths_m, measurment_num), ties=mean)
int_bkf - s(0)
On Tue, Sep 7, 2010 at 8:28 PM, David Winsemius dwinsem...@comcast.net
wrote:
On Sep 7, 2010, at 9:06 PM, stephen sefick wrote:
s - 1.00
max(s)
sprintf(%.2f, max(s
[-
which.max(vector)] ). So as Jim Holtman's tag line says: what problem
are you trying to solve?
Again, I am sorry for being vague. I get wrapped up in a problem and
forget that I need to communicate.
kindest regards,
Stephen
On Tue, Sep 7, 2010 at 8:48 PM, David Winsemius dwinsem
of the for-loop using mapply (or some
other
function)?
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-font-size-in-plot-function-tp2531127p2531161.html
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.html
and I got results like this:
Profiling...
2.5 % 97.5 %
2.393297 2.412650
What do these suggest?
please tell me about this if someone knows.
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])}
}
Works - but takes too long time.
I would appreciate alternative solutions.
Best regards, Jakob
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for your help.
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messages and continue
executing?
?try
Sorry if this has been addressed already, but, even aided by the new
awesome
google i haven't been able to find it.
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https
On Sep 8, 2010, at 11:46 PM, Philippe Grosjean wrote:
On 08/09/10 19:25, David Winsemius wrote:
On Sep 8, 2010, at 1:18 PM, telm8 wrote:
Hi,
I am having some strange problem with detecting try-error. From
what I
have read so far the following statement:
try( log(a) ) == try-error
('state','Indiana')
map.axes()
?par
bounds - par(usr)
bounds
[1] -88.12964 -84.77184 37.74583 41.82082
??Something that lets me know the y-axis is from ~38 to ~42 and
store this
information into a vector
Is there some way to query what the bounds of the current graph are?
Thanks!
David
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Hmisc::summary.formula
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On Sep 9, 2010, at 11:20 AM, David Winsemius wrote:
On Sep 8, 2010, at 7:32 PM, Jonathan Finlay wrote:
Thanks David, gmodels::Crosstable partially work because can show
only 1 x 1
tablen
CrossTable(x,y,...)
I need something how can process at less 1 variable in X an 10 in Y.
A further
that attitude and there are
now quite a few worked examples of common task in which SAS, SPSS,
Stata and R are applied using the same data.
http://www.ats.ucla.edu/stat/dae/
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HTH,
Marc Schwartz
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On Sep 9, 2010, at 10:57 AM, David Winsemius wrote:
On Sep 9, 2010, at 6:50 AM, Jan private wrote:
Hello Bernardo,
-
If I understood your problem this script solve your problem:
q-0.15 + c(-.1,0,.1)
h-10 + c(-.1,0,.1)
5*q*h
[1] 2.475 7.500 12.625
-
OK, this solves
/
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. Is
there any
package can do it?
If you minimize R0 - sum(r * Xt ), you should get your answer.
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Thanks.
Peter
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On Sep 10, 2010, at 5:44 PM, David Winsemius wrote:
On Sep 10, 2010, at 5:35 PM, Chien-Pang Chin wrote:
Hi all:
I'm looking for a package that similar to solver in MS. All I need
is find a
r to satisfy R0=sum( , where t are from 1 to n and Xt are come from
another formula.
The most
of Biostatistics, Vanderbilt University
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(3,4,6),
text.col = green4, lty = c(2, -1, 1), pch = c(-1, 3, 4),
merge = TRUE, bg = 'gray90')
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in advance.
Gregory
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On Sep 10, 2010, at 10:05 PM, Peng, C wrote:
I mean to display 001,010, ..., as there are. In other words,
whether there
is a function, say func(), such that func(001,010) displays 001, 010.
Not hard to construct one, but does not behave properly in the sub-
unity decimal range. Not
On Sep 10, 2010, at 10:32 PM, Paul Johnson wrote:
Hi, everybody
On Wed, Sep 8, 2010 at 5:43 PM, Min-Han Tan minhan.scie...@gmail.com
wrote:
David said my R code text attachment got rejected by the mailing list.
Pooh. I don't think that's nice. I don't see anything in the
posting
at csdb.cn is 2.7.1 from May 2009. There have
been several revisions since that date.)
Warning message:
In getDependencies(pkgs, dependencies, available, lib) :
package gplots is not available
any suggestion will be appreciable.
Thank you
Khush
David Winsemius, MD
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that you do not have Ghostscript
installed properly. While you are at it, you should probably updated R
as well.
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Any help?
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On Sep 11, 2010, at 8:39 AM, Peng, C wrote:
or:
k=0
for (i in 1:k) if(k0) print(i)
Because of the way the : operator works, I would have tested k =1
k=0.5
for (i in 1:k) if (k0){print(i)}
[1] 1
But Gabor's suggestion to use seq_len(k) is cleaner, anyway.
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On Sep 11, 2010, at 8:02 AM, David Winsemius wrote:
On Sep 11, 2010, at 6:29 AM, khush wrote:
Dear all,
I have installed R using yum install R-2.9. I am able to use R for
general
functions but when I installed some library say gplots I am getting
the
below error
the elements that were not included in the list? I
realize i can do this with a for loop by going through each element
and checking if it's in the list but I am wondering if there is a
faster way.
?setdiff
?%in%
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survival_2.35-8 plyr_1.1
[25] lattice_0.18-8
loaded via a namespace (and not attached):
[1] tools_2.11.1
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I get completely different (and correct) results,
by the way the *same* you have in the bug report you've
submitted
(https://bugs.r-project.org/bugzilla3/show_bug.cgi?id=14377
=expression(abc~x = 1))
I want proper sign of weak inequality not just '='
will appreciate!
robert
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Best regards
Maik
David Winsemius, MD
West Hartford, CT
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a workaround,
grid.text(expression(bgroup(,atop(x,y),)))
Error in bgroup(, atop(x, y),) : invalid group delimiter
Regards,
baptiste
sessionInfo()
R version 2.11.1 (2010-05-31)
x86_64-apple-darwin9.8.0
David Winsemius, MD
West Hartford, CT
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achieve this
using R,
Karl
--
Karl Brand
--
David Winsemius, MD
West Hartford, CT
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and provide commented, minimal, self-contained, reproducible code.
David Winsemius, MD
West Hartford, CT
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,] -2
-4-3 -20 -18
for( i in 1:nrow(A) ) { cat(sprintf(%4.0f, A[i, ]), paste(
,if( i==3 ){+}else{ }, , sep=),sprintf(%4.0f,B[i, ]),
paste( ,if( i==3 ){=}else{ }, , sep=), sprintf(%4.0f, (A
+B)[i, ]), \n )}
--
David Winsemius, MD
West Hartford, CT
understand these codes might
correspond to extended ascii characters whose boundaries and positions
we want to borrow. Then again, maybe it's something else entirely.
Any hints?
Best wishes,
baptiste
On 12 September 2010 03:27, David Winsemius dwinsem...@comcast.net
wrote:
On Sep 11, 2010
Regards,
baptiste
sessionInfo()
R version 2.11.1 (2010-05-31)
x86_64-apple-darwin9.8.0
David Winsemius, MD
West Hartford, CT
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On Sep 12, 2010, at 3:34 PM, Dennis Murphy wrote:
Hi:
Natasha said:
I changed it so i hope it will look better now
the matrix is like this:
AgeNo. Age No. AgeNo.
Center1 52 8 7
Center210 720 9 4
On Sep 12, 2010, at 12:24 PM, David Winsemius wrote:
On Sep 12, 2010, at 11:27 AM, Cuckovic Paik wrote:
I appreciate all you help. This is only for instructional purpose:
A = matrix(c(0,1,1,-2,-3,1,2,-1,0,2,2,4,1,-3,-2,1,-4,-7,-1,-19),
ncol=5,
byrow=T)
B
=
matrix(sample(c(0,1,1,-2
On Sep 12, 2010, at 10:23 PM, Aaditya Nanduri wrote:
Hello All,
I cant seem to do a trig regression in R.
The equation is as follows : y = a+b*(sin((2*pi*x/360) - c))^2
a, b, c are coefs that I want.
y, x are input vectors.
The equation I put into R: lm(y ~ sin(2*pi*x/360)^2)
This equation
worked examples in the archives, as well as canned
solutions in widely used packages.
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David Winsemius, MD
West Hartford, CT
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, reproducible code.
David Winsemius, MD
West Hartford, CT
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at Nabble.com.
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David Winsemius
:
..., scales=list(x=list(font=3)),
In this case (and I suppose many others) lattice is much more flexible
than base. I initially tried what appeared to be valid strategies with
boxplot and ended up tied in knots. Then I noticed your subject line
and it was trivial.
David Winsemius, MD
West
, reproducible code.
David Winsemius, MD
West Hartford, CT
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