Displaying 20 results from an estimated 5000 matches similar to: "using split.screen() in Sweave"
2009 Feb 17
2
Chromatogram deconvolution and peak matching
Hi,
I'm trying to match peaks between chromatographic runs.
I'm able to match peaks when they are chromatographed with the same method,
but not when there are different methods are used and spectra comes in to
play.
While searching I found the ALS package which should be usefull for my
application, but I couldn't figure it out.
I made some dummy chroms with R, which mimic my actual
2007 Dec 06
2
Any package for deconvolution?
I want to run deconvolution of a time series by an impulse or point-spread function through Wiener filter, regularized filter, Lucy-Richardson method, or any other approaches. I searched the CRAN website and the mailing list archive, but could not find any package for such a deconvolution analysis. Does anybody know an existing R function for deconvolution?
TIA,
Gang
2003 Sep 24
1
storing objects (lm results) in an array
Hi
I have calculated lots (>1000) of linear models and would like to store
each single result as an element of a 3D matrix or a similar storage:
something like
glm[i][j][k] = lm(...)
Since I read that results are lists: Is it possible to define a matrix
of type list?
Or what do you recommend?
Many thanks
Christoph
--
Christoph Lehmann <christoph.lehmann at gmx.ch>
2007 Mar 09
2
Deconvolution of a spectrum
Dear useRs,
I have a curve which is a mixture of Gaussian curves (for example UV
emission or absorption spectrum). Do you have any suggestions how to
implement searching for optimal set of Gaussian peaks to fit the curve?
I know that it is very complex problem, but maybe it is a possibility
to do it? First supposement is to use a nls() with very large functions,
and compare AIC value, but it is
2001 Dec 08
1
plotting values "by"
I would like to produce a set of values, on the same chart, from an sql
table that is structured like...
species dbh ht expf
DF 1.2 8.9 10.0
DF 2.4 17.3 12.4342
DF 3.1 20.9 56.76
PP 2.3 16.9 100.0
PP 12.8 97.3 40.3
PP 8.2 63.0 98.34
.
.
.
SS blah, blah, blah...
is it possible to, using a single command in the plot command to plot the
different groups on the same plot or will I have to iterate
2004 Jun 25
3
sweave: graphics not at the expected location in the pdf
Hi
I use sweave for excellent pdf output (thank you- Friedrich Leisch). I
have just one problem. Quite often it happens, that the graphics are not
at the place where I expect them, but (often on a separate page) later
on in the pdf. How can I fix this, means how can I define, that I want a
graphic exactly here and now in the document?
Many thanks and best regards
Christoph
--
Christoph
2003 Dec 21
3
Sweave/LaTeX Problem with EPS PDF
Dear List:
I am unsure if my problem is with Sweave or LaTeX. Anyhow, I am using the MikTeX distribution and TexnicCenter.
I can easily create Sweave files and all goes well until I try to incorporate graphics. I use the same code as found in the examples found in the users manual.
In R, the graphics I want are created as Sweave is creating the .tex file. When I examine the .tex file
2006 Jul 11
3
least square fit with non-negativity constraints for absorption spectra fitting
I would really appreciate it if someone can give suggestions on how to
do spectra fitting in R using ordinary least square fitting and
non-negativity constraints. The lm() function works well for ordinary
least square fitting, but how to specify non-negativity constraints? It
wouldn't make sense if the fitting coefficients coming out as negative
in absorption spectra deconvolution.
Thanks.
2003 Oct 07
1
plot width in Sweave
Hi
I didn't find this in the manual: I need to change the width of a plot
while I use sweave, so which command/parameters should I insert below,
to change the width of a plot
\begin{figure}[htbp]
\begin{center}
<<echo=TRUE, fig=TRUE>>=
plot(Re(q),ylab ="",type="o",col="blue",lwd=1, sub=mystring)
@
\caption{Original stick function (stimulus
2002 Jun 04
1
how to draw two histograms in one figure
How do you draw two histograms in one figure? Two separate uses of the
"hist" function always produce two separate figures... Thanks, Roman
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Send "info", "help", or "[un]subscribe"
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2011 Apr 21
3
R CMD Sweave versus Sweave() on Windows
Dear list subscriber,
I am quite puzzled by the behaviour of processing Sweave files within an R session, i.e.
Sweave("foo.Rnw") versus R CMD Sweave foo.Rnw
In the former the environmental variable 'SWEAVE_STYLEPATH_DEFAULT = TRUE' is obeyed (this is set in etc/Renviron.site as well as under the users home directory in .Renviron). That is the hard-coded path to Sweave.sty is
2003 Jun 01
6
compositional data: percent values sum up to 1
again, under another subject:
sorry, maybe an all too trivial question. But we have power data from J
frequency spectra and to have the same range for the data of all our
subjects, we just transformed them into % values, pseudo-code:
power[i,j]=power[i,j]/sum(power[i,1:J])
of course, now we have a perfect linear relationship in our x design-matrix,
since all power-values for each subject sum up
2023 Dec 11
1
Base R wilcox.test gives incorrect answers, has been fixed in DescTools, solution can likely be ported to Base R
While using the Hodges Lehmann Mean in DescTools (DescTools::HodgesLehmann),
I found that it generated incorrect answers (see
<https://github.com/AndriSignorell/DescTools/issues/97>
https://github.com/AndriSignorell/DescTools/issues/97). The error is driven
by the existence of tied values forcing wilcox.test in Base R to switch to
an approximate algorithm that returns incorrect results - see
2004 Mar 29
2
c() question
Hi
I need to define the following
c("one group" = class.weight[2], "other group" = class.weight[1])
#class.weight = c(1,2)
but I don't like the hard-coded way and would like to use
my.group <- array(c("one group", "other group"))
but now
c(my.group[1] = class.weight[2], my.group[2] = class.weight[1])
gives an error
how can I solve this
2003 Jun 03
3
lda: how to get the eigenvalues
Dear R-users
How can I get the eigenvalues out of an lda analysis?
thanks a lot
christoph
--
Christoph Lehmann <christoph.lehmann at gmx.ch>
2003 Sep 26
2
overlay two pixmap
Hi
I need to overlay two pixmaps (library (pixmap)). One, a pixmapGrey, is
the basis, and on this I need to overlay a pixmapIndexed, BUT: the
pixmapIndexed has set only some of its "pixels" to an indexed color,
many of its pixels should not cover the basis pixmapGrey pixel, means,
for this "in pixmapIndexed not defined pixels" it should be transparent.
What would you
2003 Sep 09
2
logistic regression for a data set with perfect separation
Dear R experts
I have the follwoing data
V1 V2
1 -5.8000000 0
2 -4.8000000 0
3 -2.8666667 0
4 -0.8666667 0
5 -0.7333333 0
6 -1.6666667 0
7 -0.1333333 1
8 1.2000000 1
9 1.3333333 1
and I want to know, whether V1 can predict V2: of course it can, since
there is a perfect separation between cases 1..6 and 7..9
How can I test, whether this conclusion (being able to assign an
2003 Dec 10
2
OT: BibTex year-only citation in text?
Sorry for the off-topic question, but I know there are some talented
LaTeX users out there. Which bibliography style gives only the year in
text citations (e.g "for further details, see Anderson (1992)" )?
Thanks
Jason
--
Indigo Industrial Controls Ltd.
http://www.indigoindustrial.co.nz
64-21-343-545
jasont at indigoindustrial.co.nz
2008 Mar 26
1
deconv
I'm translating a matlab routine to R and I need some equivalent to deconv():
Description: deconv()
[q,r] = deconv(v,u) deconvolves vector u out of vector v, using long
division. The quotient is returned in vector q and the remainder in
vector r such that v = conv(u,q)+r .
If u and v are vectors of polynomial coefficients, convolving them is
equivalent to multiplying the two polynomials, and
2003 Feb 26
1
calculationg condition numbers
am I right in the assumption, that for calculation of the condition
numbers I have to use the correlation matrix of X, and not t(x) %*% x?
> e <- eigen(t(x) %*% x)
better (x must not have a first column of ones):
> e <- eigen(cor(x))
> e$val
[1] 6.6653e+07 2.0907e+05 1.0536e+05 1.8040e+04 2.4557e+01 2.0151e+00
> sqrt(e$val[1]/e$val)
[1] 1.000 17.855 25.153 60.785 1647.478