On Jun 9, 2012, at 13:16 , wl2776 wrote:
$sample.interval
[1] 0.02
$sampling.time
[1] 0.02
Apparently, it is about twise faster.
Too fast to measure, I'd say. Try system.time, and/or multiple replications.
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and provide commented, minimal, self-contained, reproducible code.
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on these two values?
That's the general mechanism, yes. (Whether the chi-square distribution holds
after variable selection is a more difficult issue. Frank Harrell might chime
in and remind us that there are books on that subject.)
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.
Opening the file in binary mode (?file) might help, but as I'm not on Windows,
I can't easily check for you.
Not sure does eveybody able to view the character, it's looks like this.
http://r.789695.n4.nabble.com/file/n4632369/arrow.png
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On Jun 2, 2012, at 15:43 , jalantho...@verizon.net wrote:
Why is igraph 0.5.5 not on the igraph sourceforgwe home page?
You'll have to ask the igraph developers... Possibly, they just ran out of
round tuits.
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.bash_profile or .bashrc.
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of
configuration trick you might be missing to get UTF-8 locales properly
installed. (Googling utf-8 ubuntu seem to come up with relevant stuff.)
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is another.
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), and some packages add their own operators like %+%.
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)$sigma^2)/2) is the bias correction
#prediction-exp(tempint+tempslope*log(lngth))*exp((summary(temppow)$sigma^2)/2)
lines(prediction)
It does help considerably to use lines(length, prediction)!
(And that bias correction looks really dubious to me, but let's not go there...)
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scheme. There have been very few
issues with 2.15.0, but some people may be waiting superstitiously for a .1
release (or expecting one come along any day now) and need to be informed about
the plans.
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.
For an incomplete design, the model is the same, but the calculations are less
simple.
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pretty obviously can't be randomized). This may or may not be
relevant, but it won't hurt to include a null effect, except for the loss of a
couple of DF.
Thanks very much,
John
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start. There should be an error
message in the output file in that case.
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gives me the date.
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an expensive
identity operation.
Also, you should probably try running Benno's exact code, just for comparison.
Some of those multicore machine are really rather slow if you only use one core
for your process.
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a is fundamentally different from the variable name a.
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, reproducible code.
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-contained, reproducible code.
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On May 18, 2012, at 09:10 , Hans W Borchers wrote:
peter dalgaard pdalgd at gmail.com writes:
On May 18, 2012, at 00:14 , Nathan Stephens wrote:
I have a very simple maximization problem where I'm solving for the vector
x:
objective function:
w'x = value to maximize
box
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. could someone please help
me with this question?
Perhaps this was what you were looking for:
d - log(c)
mean(d)
[1] 1.675003
sd(d)
[1] 0.4656469
Taking exp() of a log-normal rarely makes much sense. More commonly, you take
log() to get a normal distribution.
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, in passing, that lazy evaluation prevents the stop() from kicking in, if
a known algorithm is given). Then later on, we have
breaks - pretty(range(x), n = breaks, min.n = 1)
so by the time we need the xlim, breaks will be a numeric vector which you
_can_ take the range of.
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, or
a formula expression consisting of factors, vectors or matrices
connected by formula operators.
Not the most informative documentation. But Peter Dalgaard is a most
authoritative source!
And also I have checked:
Any more thoughts?
Data frames are odd things; a column need
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, yes. One might quibble over the use of large because, but it's
not important for the main point.
-pd
John
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Sent: Monday, May 7, 2012 11:07 PM
Subject: Re: [R] low R
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properties and interpretation, but not in the general case. In
a suitable sense, the orthogonal contrast matrices are their own inverses.
However, for the generic case, you need to find a solution to the equation
A'C=I given A or C.
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to
sapply(x.split,`[`,1)
[1] 12/31/11 1/1/12
sapply(x.split,`[`,2)
[1] 23:45 2:15
It's a bit inefficient, though. Other ideas:
sub( .*$, , x)
[1] 12/31/11 1/1/12
sub(^.* , , x)
[1] 23:45 2:15
read.table(text=x)
V1V2
1 12/31/11 23:45
2 1/1/12 2:15
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sense to me.
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.
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On May 1, 2012, at 09:06 , Ulfa Hasanah wrote:
what is nb2listw in index moran?
Did you try help(nb2listw) ?
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(read.table(pipe(pbpaste)))
Ozone Solar.R Wind Temp Month Day
141 190 7.4 67 5 1
236 118 8.0 72 5 2
312 149 12.6 74 5 3
418 313 11.5 62 5 4
5NA NA 14.3 56 5 5
628 NA 14.9 66 5 6
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to terms() what kind of variable x is. So if
you want different behavior if x is a factor than if it is continuous, you're
out of luck...
Not quite sure the last three lines of my note below make sense, though.
-pd
On Jan 29, 2012, at 11:24 , peter dalgaard wrote:
On Jan 29, 2012, at 02:42
.
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]
Yes, I was wondering why people were avoiding the obvious use of grepl(). I'm
not too happy about the nmuta technique though: What about deletionmutation
and such? Might as well do the safe(r) thing:
i2 - grepl(unmuta, tmp) | grepl(nonmuta, tmp)
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).
Could you please document that? There are many misconceptions about when
eigenvectors are correct and platform dependencies too. As far as I can tell,
both R and Matlab use the same LAPACK routines.
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On Apr 21, 2012, at 19:13 , Berend Hasselman wrote:
On 21-04-2012, at 16:40, peter dalgaard wrote:
On Apr 21, 2012, at 15:22 , Luke Hartigan wrote:
Hi all,
In my experience, using eigen to solve generalized eigenvalue / eigenvector
problems only gives correct looking eigenvalues
, reproducible code.
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, self-contained, reproducible code.
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to the cell integral if you just multiply by the cell area. If
that isn't good enough, there are finite-difference formulas that could improve
the approximation (usually by several orders of magnitude in dx and dy) using
the nearby values on the grid.
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)
lines(x0,predict(ratelogis,newdata=data.frame(X=x0)))
lines(x0,evalq(a*(exp((b-X)/c)*(1/c))/(1 + exp((b-X)/c))^2,envir=list(a =
21.16322, b = 8.83669, c = 2.957765, X=x0)), lty=dashed)
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(or 10, maybe), but it seems
easier to divide and conquer (pardon the pun).
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), rep(NA,2), rep(2,4))
df1 - data.frame(c1 = c(1,2), c2 = c(5,6))
I would like to get vector x2:
x2
[1] 5 5 5 NA NA 6 6 6 6
Thanks a lot, OV
You mean like this?
df1$c2[match(x1, df1$c1)]
[1] 5 5 5 NA NA 6 6 6 6
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and defeat that:
solve(crossprod(x), crossprod(x,y), tol=0)
[,1]
-2.959385e+09
x1 2.218414e+00
Which, as you'll notice is, er, somewhat off the mark.
/lecture on numerical linear algebra
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* x),data = penetrationks28, start =
+ list(a=3,b = exp(4),c=.67), trace = T)
53.23186 : 3.0 54.59815 0.67000
2.188135 : 2.7131685 71.6770898 0.7187123
1.970939 : 2.6654559 71.6169034 0.7057594
...
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.)
The direct cause of your problem seems to be that if x is a vector, so is y,
and your temp.fn returns a vector, so int.fn returns a matrix, and integrate()
is unhappy with that.
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to some sort of preprocessing that removes
low-frequency components?
At any rate, this isn't really about R, is it?
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at each
fragment.
Apologies to Mr. Kamakshaiah, whose inquisitiveness really doesn't deserve
being poked fun at, but David's remark definitely calls for a nomination to the
fortunes package!
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-2.15.0.tar.gz
or wait for it to be mirrored at a CRAN site nearer to you.
Binaries for various platforms will appear in due course.
For the R Core Team
Peter Dalgaard
These are the md5sums for the freshly created files, in case you wish
to check that they are uncorrupted:
MD5 (AUTHORS
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-contained, reproducible code.
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the [[-assignment inside the anonymous
function, but the assignment is to a local copy which is deleted on exit, and
the return value is the rhs of the assignment.)
Please:
x - lapply(1:10, function(z) 1:z)
or even
x - lapply(1:10, seq_len)
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, rather like a piece
of an SPSS syntax file.
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)/(1+exp(-x*sigma)),a,Inf,rel.tol=1e-10)$value
I1+I2+I3
[1] 0.4999626
(I'm still not quite happy with the fact that I1 gets essentially set to zero
in the above, but the essential point is that you need to be prepared to do a
little analytical work to deal with tricky integrands.)
--
Peter
read the posting guide http://www.R-project.org/posting-guide.html
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Center for Statistics, Copenhagen Business School
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Phone: (+45)38153501
Email: pd
),each=4).
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Center for Statistics, Copenhagen Business School
Solbjerg Plads 3, 2000 Frederiksberg, Denmark
Phone: (+45)38153501
Email: pd@cbs.dk Priv: pda...@gmail.com
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gain a few
denominator DF. It does, however, imply that the F tests with a common
denominator are not independent, as opposed to the proper successive F tests.
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Peter Dalgaard, Professor
Center for Statistics, Copenhagen Business School
Solbjerg Plads 3, 2000 Frederiksberg, Denmark
Phone: (+45
]]
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Peter Dalgaard
,]
Many thanks,
Dominic C.
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Peter Dalgaard, Professor,
Center for Statistics, Copenhagen Business School
Solbjerg Plads 3, 2000 Frederiksberg, Denmark
Phone: (+45)38153501
Email: pd@cbs.dk Priv: pda...@gmail.com
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Peter Dalgaard, Professor,
Center for Statistics, Copenhagen Business School
-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
--
Peter Dalgaard, Professor
Center for Statistics, Copenhagen Business School
Solbjerg Plads 3, 2000 Frederiksberg, Denmark
Phone: (+45)38153501
.
2012/3/19 peter dalgaard pda...@gmail.com
There's no trivial way since you need the covariance of pred2 and pred1 to
calculate the variance of the difference.
I think you can proceed somewhat like as follows (I can't be bothered to test
it without a reproducible example to start from. You
/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
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Peter Dalgaard, Professor,
Center for Statistics, Copenhagen Business School
Solbjerg Plads 3, 2000 Frederiksberg, Denmark
Phone: (+45)38153501
Email: pd@cbs.dk Priv: pda...@gmail.com
, reproducible code.
--
Peter Dalgaard, Professor,
Center for Statistics, Copenhagen Business School
Solbjerg Plads 3, 2000 Frederiksberg, Denmark
Phone: (+45)38153501
Email: pd@cbs.dk Priv: pda...@gmail.com
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R-help@r-project.org mailing list
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list
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
--
Peter Dalgaard, Professor,
Center for Statistics, Copenhagen Business School
Solbjerg Plads 3, 2000
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