[R] [R-pkgs] stringr: version 0.4
Strings are not glamorous, high-profile components of R, but they do play a big role in many data cleaning and preparations tasks. R provides a solid set of string operations, but because they have grown organically over time, they can be inconsistent and a little hard to learn. Additionally, they lag behind the string operations in other programming languages, so that some things that are easy to do in languages like Ruby or Python are rather hard to do in R. The `stringr` package aims to remedy these problems by providing a clean, modern interface to common string operations. More concretely, `stringr`: * Processes factors and characters in the same way. * Gives functions consistent names and arguments. * Simplifies string operations by eliminating options that you don't need 95% of the time. * Produces outputs than can easily be used as inputs. This includes ensuring that missing inputs result in missing outputs, and zero length inputs result in zero length outputs. * Completes R's string handling functions with useful functions from other programming languages. New in stringr 0.4: * all functions now vectorised with respect to string, pattern (and where appropriate) replacement parameters * fixed() function now tells stringr functions to use fixed matching, rather than escaping the regular expression. Should improve performance for large vectors. * new ignore.case() modifier tells stringr functions to ignore case of pattern. * str_replace renamed to str_replace_all and new str_replace function added. This makes str_replace consistent with all functions. * new str_sub- function (analogous to substring-) for substring replacement * str_sub now understands negative positions as a position from the end of the string. -1 replaces Inf as indicator for string end. * str_pad side argument can be left, right, or both (instead of center) * str_trim gains side argument to better match str_pad * stringr now has a namespace and imports plyr (rather than requiring it) -- Assistant Professor / Dobelman Family Junior Chair Department of Statistics / Rice University http://had.co.nz/ ___ R-packages mailing list r-packa...@r-project.org https://stat.ethz.ch/mailman/listinfo/r-packages __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] SEM : Warning : Could not compute QR decomposition of Hessian
Dear Anne,
I started to diagram your model but stopped when I noticed some problems:
(1) Some variables, such as pays_alti, are clearly endogenous, since they
have arrows pointing to them, yet are declared as fixed exogenous variables;
that clearly doesn't make sense. (2) You've placed conflicting constraints
on factor loadings and the variances of latent variables, for example
setting both the path from the factor type_paysage to the indicator
pays_alti and the variance of type_paysage to 1; again, that doesn't make
sense.
Do you have a path diagram for the model that you're trying to fit? It would
in particular be helpful to have a diagram of the structural part of the
model.
Regards,
John
John Fox
Senator William McMaster
Professor of Social Statistics
Department of Sociology
McMaster University
Hamilton, Ontario, Canada
web: socserv.mcmaster.ca/jfox
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org]
On
Behalf Of Anne Mimet
Sent: August-25-10 10:27 AM
To: r-help@r-project.org
Subject: [R] SEM : Warning : Could not compute QR decomposition of Hessian
Hi useRs,
I'm trying for the first time to use a sem. The model finally runs,
but gives a warning saying :
In sem.default(ram = ram, S = S, N = N, param.names = pars, var.names
= vars, : Could not compute QR decomposition of Hessian.
Optimization probably did not converge.
I found in R-help some posts on this warning, but my attemps to modify
the code didn't change the warning message (i tried to give an error
of 1 to the latente variables). I can't figure what the problem is.
Here is the code :
tab-read.table(F:/Mes documents/stats/sem/donnees_corr.txt,
header=T, sep=,na.strings = NA)
tab[,46]-as.factor(tab[,46])
tab[,24]-as.factor(tab[,24])
tab[,40]-as.factor(tab[,40])
fct_cor-hetcor(tab, ML=T)
cor_tab- fct_cor$correlations
moment_tab-read.moments(diag=F, names=c('c1','c2', 'c3','c4','c5',
'c6','c7', 'c8', 'c9', 'ind_plando', 'long_sup15', 'long_inf15',
'pente', 'est', 'sud','ouest', 'nord' ,'reg_hydriq', 'prof_sol',
'pierro', 'efferv', 'struct','drainage','texture', 'route1_pond',
'route2_pond',
'pourcactif', 'tx_chomage', 'pourcagric', 'pourc_jeunes', 'pop99',
'rev_imp_foyer','eq_CONC', 'eq_sante', 'eq_edu', 'sold_nat',
'sold_mig', 'tps_dom_emp','TXEMPLOI','ORIECO','dist_paris','axe1',
'axe2', 'axe3', 'dist_protect','urbanisation','pays_incli','pays_alti'))
# after comes the moment matrix (triangular)
ram_tab-specify.model()
type_paysage-pays_alti,NA,1
type_paysage-pays_incli, pays2, NA
pedo-reg_hydriq, NA, 1
pedo-prof_sol, ped8, NA
pedo-pierro, ped9, NA
pedo-efferv, ped10, NA
pedo-struct, ped11, NA
pedo-drainage, ped12, NA
pedo-texture, ped13, NA
adj_99-c1, NA,1
adj_99-c2, adj2,NA
adj_99-c3, adj3,NA
adj_99-c4, adj4,NA
adj_99-c5, adj5,NA
adj_99-c6, adj6,NA
adj_99-c7, adj7,NA
adj_99-c8, adj8,NA
adj_99-c9, adj9,NA
etat_hexa-axe1, NA, 1
etat_hexa-axe2, et2, NA
etat_hexa-axe3, et3, NA
socioBV-sold_mig, BV1, NA
socioBV-sold_nat, BV2, NA
socioBV-TXEMPLOI, BV3, NA
socioBV-ORIECO, BV4, NA
socioBV-tps_dom_emp, NA, 1
eqBV-eq_CONC, NA, 1
eqBV-eq_sante, eq2, NA
eqBV-eq_edu, eq3, NA
socio_com-pourcactif , NA, 1
socio_com-tx_chomage, com2, NA
socio_com-pourcagric, com3, NA
socio_com-pourc_jeunes, com4, NA
socio_com-pop99, com5, NA
socio_com-rev_imp_foyer, com7, NA
access_hexa-route1_pond, NA, 1
access_hexa-route2_pond, acc2, NA
hydro-ind_plando, NA, 1
hydro-long_sup15, eau2, NA
hydro-long_inf15, eau3, NA
topog-pente, NA, 1
topog-est, top2, NA
topog-sud, top3, NA
topog-nord, top4, NA
topog-ouest, top5, NA
dist_protect- urbanisation, cor1,NA
dist_protect- adj_99, cor2, NA
dist_protect- etat_hexa, cor3, NA
topog- urbanisation, cor4, NA
topog- adj_99, cor5, NA
topog- etat_hexa, cor6, NA
topog- access_hexa, cor7, NA
topog-hydro, cor8, NA
topog-pedo, cor9, NA
pedo- urbanisation, cor10, NA
pedo- adj_99, cor11, NA
pedo- etat_hexa, cor12, NA
pedo-hydro, cor1, NA
hydro- urbanisation, cor13, NA
hydro- adj_99, cor14, NA
hydro- etat_hexa, cor15, NA
access_hexa- urbanisation, cor16, NA
access_hexa- etat_hexa, cor17, NA
socio_com- etat_hexa, cor18, NA
socio_com- adj_99, cor19, NA
socio_com- urbanisation, cor20, NA
dist_paris- socio_com, cor21, NA
dist_paris- access_hexa, cor22, NA
dist_paris- adj_99, cor23, NA
dist_paris- etat_hexa, cor24, NA
dist_paris- urbanisation, cor25, NA
dist_paris- socioBV, cor26, NA
socioBV- eqBV, cor27, NA
socioBV- urbanisation, cor28, NA
socioBV- adj_99, cor29, NA
socioBV- etat_hexa, cor30, NA
eqBV- etat_hexa, cor31, NA
eqBV- adj_99, cor32, NA
eqBV- urbanisation, cor33, NA
etat_hexa- urbanisation, cor34, NA
etat_hexa- adj_99, cor35, NA
adj_99- urbanisation, cor36, NA
type_paysage- urbanisation, cor37, NA
type_paysage- adj_99, cor38, NA
type_paysage- etat_hexa, cor39, NA
dist_paris-dist_paris, auto1, NA
Re: [R] output values from within a function
Put the line:
cat(z1,'\n')
in your function. You may also want to put the flush.console() command right
after that.
Hope this helps,
--
Gregory (Greg) L. Snow Ph.D.
Statistical Data Center
Intermountain Healthcare
greg.s...@imail.org
801.408.8111
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-
project.org] On Behalf Of Maas James Dr (MED)
Sent: Friday, August 20, 2010 7:41 AM
To: r-help@r-project.org
Subject: [R] output values from within a function
Is it possible to get R to output the value of an expression, that is
being calculated within a function? I've attached a very simple
example but for more complicated ones would like to be able to debug by
seeing what the value of specific expressions are each time it cycles
through a loop that executes the expression. I'm relatively new to
this so there may be much simpler more elegant ways to do it.
For example, is there a command I can put within the function funct01
that will output the value of z1 to the screen?
Thanks
Jim
==
## Practice file to try out evaluations
y - 5
(x - y^2)
funct01 - function (x,y) {
z1 - x + y
z2 - x * y
z3 - x^y
results - data.frame (z1=z1, z2=z2, z3=z3)
return(results)
}
funct01(7,9)
===
Dr. Jim Maas
University of East Anglia
Norwich, UK
NR4 7TJ
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and provide commented, minimal, self-contained, reproducible code.
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and provide commented, minimal, self-contained, reproducible code.
Re: [R] Removing inter-bar spaces in barchart
Using the barplot function in base graphics you just set space=0, but that function does not have a box.ratio argument which would imply that you are using something else. If you let us know which function (and which package it is in) then it is easier (possible) for us to help you, even better is to give the information asked for at the bottom of every post and the posting guide. -- Gregory (Greg) L. Snow Ph.D. Statistical Data Center Intermountain Healthcare greg.s...@imail.org 801.408.8111 -Original Message- From: r-help-boun...@r-project.org [mailto:r-help-boun...@r- project.org] On Behalf Of Jonathan Greenberg Sent: Tuesday, August 24, 2010 8:21 PM To: r-help Subject: [R] Removing inter-bar spaces in barchart Rhelpers: I'm trying to make a barchart of a 2-group dataset (barchart(x~y,data=data,groups=z,horizontal=FALSE)). My problem is that I can't, for the life of me, seem to get rid of the inter-bar space -- box.ratio set to 1 doesn't do much. Any ideas? I'd ideally want zero space between the bars. Thanks! --j -- Jonathan A. Greenberg, PhD Assistant Project Scientist Center for Spatial Technologies and Remote Sensing (CSTARS) Department of Land, Air and Water Resources University of California, Davis One Shields Avenue Davis, CA 95616 Phone: 415-763-5476 AIM: jgrn307, MSN: jgrn...@hotmail.com, Gchat: jgrn307 __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting- guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] What does this warning message (from optim function) mean?
You mean 'TRUE': 'T' is a variable in R, with initial value TRUE.
On Wed, 25 Aug 2010, Sally Luo wrote:
Hi R users,
I am trying to use the optim function to maximize a likelihood funciton, and
I got the following warning messages.
Could anyone explain to me what messege 31 means exactly? Is it a cause for
concern?
Since the value of convergence turns out to be zero, it means that the
converging is successful, right?
So can I assume that the parameter estimates generated thereafter are
reliable MLE estimates?
Thanks a lot for your help.
Maomao
p-optim(c(0,0,0), f, method =BFGS, hessian =T, y=y,X=X,W=W)
There were 31 warnings (use warnings() to see them)
warnings()
Warning messages:
1: In log(det(I_N - pd * wd - po * wo - pw * ww)) : NaNs produced
2: In log(det(I_N - pd * wd - po * wo - pw * ww)) : NaNs produced
3: In log(det(I_N - pd * wd - po * wo - pw * ww)) : NaNs produced
4: In log(det(I_N - pd * wd - po * wo - pw * ww)) : NaNs produced
5: In log(det(I_N - pd * wd - po * wo - pw * ww)) : NaNs produced
6: In log(det(I_N - pd * wd - po * wo - pw * ww)) : NaNs produced
7: In log(det(I_N - pd * wd - po * wo - pw * ww)) : NaNs produced
8: In log(det(I_N - pd * wd - po * wo - pw * ww)) : NaNs produced
9: In log(det(I_N - pd * wd - po * wo - pw * ww)) : NaNs produced
10: In log(det(I_N - pd * wd - po * wo - pw * ww)) : NaNs produced
11: In log(det(I_N - pd * wd - po * wo - pw * ww)) : NaNs produced
12: In log(det(I_N - pd * wd - po * wo - pw * ww)) : NaNs produced
13: In log(det(I_N - pd * wd - po * wo - pw * ww)) : NaNs produced
14: In log(det(I_N - pd * wd - po * wo - pw * ww)) : NaNs produced
15: In log(det(I_N - pd * wd - po * wo - pw * ww)) : NaNs produced
16: In log(det(I_N - pd * wd - po * wo - pw * ww)) : NaNs produced
17: In log(det(I_N - pd * wd - po * wo - pw * ww)) : NaNs produced
18: In log(det(I_N - pd * wd - po * wo - pw * ww)) : NaNs produced
19: In log(det(I_N - pd * wd - po * wo - pw * ww)) : NaNs produced
20: In log(det(I_N - pd * wd - po * wo - pw * ww)) : NaNs produced
21: In log(det(I_N - pd * wd - po * wo - pw * ww)) : NaNs produced
22: In log(det(I_N - pd * wd - po * wo - pw * ww)) : NaNs produced
23: In log(det(I_N - pd * wd - po * wo - pw * ww)) : NaNs produced
24: In log(det(I_N - pd * wd - po * wo - pw * ww)) : NaNs produced
25: In log(det(I_N - pd * wd - po * wo - pw * ww)) : NaNs produced
26: In log(det(I_N - pd * wd - po * wo - pw * ww)) : NaNs produced
27: In log(det(I_N - pd * wd - po * wo - pw * ww)) : NaNs produced
28: In log(det(I_N - pd * wd - po * wo - pw * ww)) : NaNs produced
29: In log(det(I_N - pd * wd - po * wo - pw * ww)) : NaNs produced
30: In log(det(I_N - pd * wd - po * wo - pw * ww)) : NaNs produced
31: In if (hessian) { ... :
the condition has length 1 and only the first element will be used
p$counts
function gradient
148 17
p$convergence
[1] 0
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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.
--
Brian D. Ripley, rip...@stats.ox.ac.uk
Professor of Applied Statistics, http://www.stats.ox.ac.uk/~ripley/
University of Oxford, Tel: +44 1865 272861 (self)
1 South Parks Road, +44 1865 272866 (PA)
Oxford OX1 3TG, UKFax: +44 1865 272595
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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.
Re: [R] approxfun-problems (yleft and yright ignored)
The plots did not come through, see the posting guide for which attachments are
allowed. It will be easier for us to help if you can send reproducible code
(we can copy and paste to run, then examine, edit, etc.). Try finding a subset
of your data for which the problem still occurs, then send the data if
possible, or similar simulated data if you cannot send original data.
--
Gregory (Greg) L. Snow Ph.D.
Statistical Data Center
Intermountain Healthcare
greg.s...@imail.org
801.408.8111
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-
project.org] On Behalf Of Samuel Wuest
Sent: Wednesday, August 25, 2010 8:20 AM
To: r-help@r-project.org
Subject: [R] approxfun-problems (yleft and yright ignored)
Dear all,
I have run into a problem when running some code implemented in the
Bioconductor panp-package (applied to my own expression data), whereby
gene
expression values of known true negative probesets (x) are interpolated
onto
present/absent p-values (y) between 0 and 1 using the *approxfun -
function*{stats}; when I have used R version 2.8, everything had
worked fine,
however, after updating to R 2.11.1., I got unexpected output
(explained
below).
Please correct me here, but as far as I understand, the yleft and
yright
arguments set the extreme values of the interpolated y-values in case
the
input x-values (on whose approxfun is applied) fall outside range(x).
So if
I run approxfun with yleft=1 and yright=0 with y-values between 0 and
1,
then I should never get any values higher than 1. However, this is not
the
case, as this code-example illustrates:
### define the x-values used to construct the approxfun, basically
these
are 2000 expression values ranging from ~ 3 to 7:
xNeg - NegExprs[, 1]
xNeg - sort(xNeg, decreasing = TRUE)
### generate 2000 y-values between 0 and 1:
yNeg - seq(0, 1, 1/(length(xNeg) - 1))
### define yleft and yright as well as the rule to clarify what
should
happen if input x-values lie outside range(x):
interp - approxfun(xNeg, yNeg, yleft = 1, yright = 0, rule=2)
Warning message:
In approxfun(xNeg, yNeg, yleft = 1, yright = 0, rule = 2) :
collapsing to unique 'x' values
### apply the approxfun to expression data that range from ~2.9 to
13.9
and can therefore lie outside range(xNeg):
PV - sapply(AllExprs[, 1], interp)
range(PV)
[1]0.000 6208.932
summary(PV)
Min. 1st Qu.Median Mean 3rd Qu. Max.
0.000e+00 0.000e+00 2.774e-03 1.299e+00 3.164e-01 6.209e+03
So the resulting output PV object contains data ranging from 0 to 6208,
the
latter of which lies outside yleft and is not anywhere close to extreme
y-values that were used to set up the interp-function. This seems wrong
to
me, and from what I understand, yleft and yright are simply ignored?
I have attached a few histograms that visualize the data distributions
of
the objects I xNeg, yNeg, AllExprs[,1] (== input x-values) and PV (the
output), so that it is easier to make sense of the data structures...
Does anyone have an explanation for this or can tell me how to fix the
problem?
Thanks a million for any help, best, Sam
sessionInfo()
R version 2.11.1 (2010-05-31)
x86_64-apple-darwin9.8.0
locale:
[1] en_IE.UTF-8/en_IE.UTF-8/C/C/en_IE.UTF-8/en_IE.UTF-8
attached base packages:
[1] stats graphics grDevices utils datasets methods base
other attached packages:
[1] panp_1.18.0 affy_1.26.1 Biobase_2.8.0
loaded via a namespace (and not attached):
[1] affyio_1.16.0 preprocessCore_1.10.0
--
-
Samuel Wuest
Smurfit Institute of Genetics
Trinity College Dublin
Dublin 2, Ireland
Phone: +353-1-896 2444
Web: http://www.tcd.ie/Genetics/wellmer-2/index.html
Email: wue...@tcd.ie
--
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
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Re: [R] lattice help required
.. thanks again, richard. and you swiftly saw the next problem comming up - when using par.settings = list(layout.widths = list(axis.panel = c(1, 0))) getting rid of the double tick labeling would be natural - but i'll leave it at that for today. many thanks, kay Richard M. Heiberger wrote: The multiple y axes are protecting you in this situation. z - cbind(rnorm(100,c(1,10),1), rnorm(100,c(20,30),1)) dotplot(z[,1]+z[,2] ~ facs$Treatment|facs$Sites, outer=TRUE, scales = list( y = list( relation=free)), ylab=c(y1, y2), xlab=c(Site 1, Site 2), strip=FALSE, main=problem) dotplot(z[,1]+z[,2] ~ facs$Treatment|facs$Sites, outer=TRUE, scales = list( y = list( relation=free, limits=list(c(-5,13),c(-5,13),c(18,32),c(18,32, ylab=c(y1, y2), xlab=c(Site 1, Site 2), strip=FALSE, main=protected) For more control (such as suppressing the y-tick labels in the right-hand column, I recommend Deepayan Sarkar's book. Rich [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. - Kay Cichini Postgraduate student Institute of Botany Univ. of Innsbruck -- View this message in context: http://r.789695.n4.nabble.com/lattice-help-required-tp2338382p2338707.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] GLM outputs in condensed versus expanded table
On 08/25/2010 05:17 PM, francogrex wrote: Hi I'm having different outputs from GLM when using a condensed table V1V2 V3 Present Absent 0 0 0 3 12 0 0 1 0 0 0 1 0 0 0 0 1 1 1 0 1 0 0 7 20 1 0 1 0 0 1 1 0 3 0 1 1 1 6 0 resp=cbind(Present, Absent) glm(resp~V1+V2+V3+I(V1*V2*V3),family=binomial) Deviance Residuals: [1] 0 0 0 0 0 0 0 0 etc and also coefficients... And when using the same but expanded table V1 V2 V3 condition (1 present 0 abscent) Id1 1 0 0 1 id2 1 1 1 1 ... etc glm(condition~V1+V2+V3+I(V1*V2*V3),family=binomial) Deviance Residuals: Min 1Q Median 3Q Max -0.7747317 -0.7747317 -0.6680472 0.0001315 1.7941226 and also coefficients are different from above. What could I be doing wrong? Not necessarily anything. Anything technical, that is. You have 3 uninformative combinations where the total is zero. The model has 5 parameters. This is quite likely to generate a perfect fit to the aggregated data. With the groups having zeros in the absent category, the fit probably diverged so some coefficients are numerically large. Refitting with individual data will likely give slightly different coefficients, since it sort of depends on how far you came on the way to infinity. With the aggregated data, a perfect fit gives residuals of zero, but with individual data, the 0's and 1's give negative and positive residuals. Try z - rep(0:1,5) zz - cbind(5,5) summary(glm(z~1, binomial)) summary(glm(zz~1, binomial)) -- Peter Dalgaard Center for Statistics, Copenhagen Business School Phone: (+45)38153501 Email: pd@cbs.dk Priv: pda...@gmail.com __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] several odfWeave questions
Ben,
1a. am I right in believing that odfWeave does not respect the
'keep.source' option? Am I missing something obvious?
I believe it does, since this gets passed directly to Sweave.
1b. is there a way to set global options analogous to \SweaveOpts{}
directives in Sweave? (I looked at odfWeaveControl, it doesn't seem to
do it.)
Yes. There are examples of this in the 'examples' package directory.
2. I tried to write a Makefile directive to process files from the
command line:
%.odt: %_in.odt
$(RSCRIPT) -e library(odfWeave);
odfWeave(\$*_in.odt\,\$*.odt\);
This works, *but* the resulting output file gives a warning (The file
'odftest2.odt' is corrupt and therefore cannot be opened.
OpenOffice.org can try to repair the file ...). Based on looking at
the contents, it seems that a spurious/unnecessary 'Rplots.pdf' file is
getting
created and zipped in with the rest of the archive; when I unzip, delete
the Rplots.pdf file and re-zip, the ODT file opens without a warning.
Obviously I could post-process but it would be nicer to find a
workaround within R ...
Get the latest version form R-Forge. I haven't gotten this fix onto
CRAN yet (I've been on a caret streak lately).
3. I find the requirement that all file paths be specified as absolute
rather than relative paths somewhat annoying -- I understand the reason,
but it goes against one practice that I try to encourage for
reproducibility, which is *not* to use absolute file paths -- when
moving a same set of data and analysis files across computers, it's hard
to enforce them all ending up in the same absolute location, which then
means that the recipient has to edit the ODT file. It would be nice if
there were hooks for read.table() and load() as there are for plotting
and package/namespace loading -- then one could just copy them into the
working directory on the fly.
has anyone experienced this/thought of any workarounds?
(I guess one solution is to zip any necessary source files into the archive
beforehand,
as illustrated in the vignette.)
You can set the working directory with the (wait for it...) 'workDir'
argument. Using 'workDir = getwd()' will pack and unpack the files in
the current location and you wouldn't need to worry about setting the
path. I use the temp directory because I started over-wrting files.
Max
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Re: [R] package MuMI
Marino Taussig De Bodonia, Agnese agnese.marino09 at imperial.ac.uk writes: I am using the package MuMI to run all the possible combinations of variables in my full model, and select my best models. When I enter my variables in the original model I write them like this lm(y~ a +b +c +a:b) However, MuMI will also use the variable b:a, which I do not want in my model. How do I stop that from happening? (1) I think you mean MuMIn. (2) Please send a reproducible example! (Hint: see the posting guide that is referred to at the bottom of all R-help messages ...) (3) [now proceeding to try to guess what you mean ...] a:b and b:a have identical meanings in the R formula syntax. I tried to do something like what I thought you might have meant and got what seemed to be reasonable answers. library(MuMIn) dat - data.frame(y=runif(100),a=factor(sample(1:5,replace=TRUE,size=100)), +b= factor(sample(1:5,replace=TRUE,size=100)), c=runif(100)) m1 - lm(y~a+b+c+a:b,data=dat) dredge(m1) Global model: lm(formula = y ~ a + b + c + a:b, data = dat) --- Model selection table (Int) a b c a:b k R.sqAdj.R.sq RSS AIC AICc delta weight 4 0.6385 -0.1643 3 0.03019 0.020290 7.386 29.23 29.48 0. 0.420 1 0.5530 2 0.0 0.00 7.616 30.29 30.42 0.9392 0.262 7 0.6283 + -0.1461 7 0.09397 0.045770 6.900 30.42 31.64 2.1650 0.142 3 0.5533 + 6 0.07067 0.031540 7.077 30.96 31.87 2.3890 0.127 6 0.6134 + -0.2199 7 0.06700 0.017370 7.105 33.36 34.57 5.0980 0.033 2 0.5202 +6 0.01989 -0.021380 7.464 36.28 37.19 7.7100 0.009 8 0.6232 + + -0.1941 11 0.12050 0.032540 6.698 35.45 38.45 8.9760 0.005 5 0.5352 + + 10 0.08490 0.004455 6.969 37.42 39.89 10.4100 0.002 10 0.7433 + + -0.2504 + 27 0.27800 0.034140 5.498 47.71 68.71 39.2400 0.000 9 0.6193 + + + 26 0.23150 -0.014420 5.853 51.96 71.19 41.7200 0.000 __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Change value of a slot of an S4 object within a method.
Howdy, On Wed, Aug 25, 2010 at 1:21 PM, Joris Meys jorism...@gmail.com wrote: Hi Steve, thanks for the tip. I'll definitely take a closer look at your solution for implementation for future use. But right now I don't have the time to start rewriting my class definitions. Luckily, I found where exactly things were going wrong. After reading into the documentation about the evaluation in R, I figured out I have to specify the environment where substitute should look explicitly as parent.frame(1). I still don't understand completely why exactly, but it does the job. Thus : eval(eval(substitute(expression(obj...@extra[[name]] - value should become : eval( eval( substitute( expression(obj...@extra[[name]] - value) ,env=parent.frame(1) ) ) ) My eyes start to gloss over on their first encounter of `substitute` ... add enough `eval`s and (even) an `expression` (!) to that, and you'll probably see me running for the hills ... but hey, if it makes sense to you, more power to you ;-) Glad you found a fix that works, -steve -- Steve Lianoglou Graduate Student: Computational Systems Biology | Memorial Sloan-Kettering Cancer Center | Weill Medical College of Cornell University Contact Info: http://cbio.mskcc.org/~lianos/contact __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Draw a perpendicular line?
Hi:
The function below plots the line segment between two points A and B as well
as the normal from C to AB, with a dot as the intersection point (D), which
is returned with the function call. The aspect ratio is kept at one so that
the orthogonality between the two lines is not distorted by the axis
scaling.
The input is a 3 x 2 matrix with point A in the first row, B in the second
and C in the third.
normalLine - function(pts) {
# A function that takes points A, B, C (rows of matrix pts)
# as input
# find slope of line between A and B
# find equation of normal that passes through C
# find the intersection point D between AB and its normal
# through C
if(nrow(pts) != 3 || ncol(pts) != 2) stop('input matrix must be 3 x 2')
A - pts[1, , drop = TRUE]
B - pts[2, , drop = TRUE]
C - pts[3, , drop = TRUE]
slopeAB - (B[2] - A[2])/(B[1] - A[1])
slopeNorm - -1/slopeAB
a - A[2] - slopeAB * A[1]
b - C[2] - slopeNorm * C[1]
xintersect - (b - a)/(slopeAB - slopeNorm)
yintersect - b + slopeNorm * xintersect
mxlim - max(pts)
mnlim - min(pts)
plot(pts[, 1], pts[, 2], pch = c('A', 'B', 'C'), xlab = , ylab = ,
xlim = c(mnlim, mxlim), ylim = c(mnlim, mxlim))
segments(A[1], A[2], B[1], B[2])
segments(xintersect, yintersect, C[1], C[2])
points(xintersect, yintersect, pch = 16, cex = 1.2)
c(xintersect, yintersect)
}
## Test:
pts - matrix(c(1, 2, 5, 9, 3, 4), ncol = 2, byrow = TRUE)
normalLine(pts)
HTH,
Dennis
On Wed, Aug 25, 2010 at 8:54 AM, William Revelle li...@revelle.net wrote:
At 3:04 PM -0700 8/23/10, CZ wrote:
Hi,
I am trying to draw a perpendicular line from a point to two points.
Mathematically I know how to do it, but to program it, I encounter some
problem and hope can get help. Thanks.
I have points, A, B and C. I calculate the slope and intercept for line
drawn between A and B.
I am trying to check whether I can draw a perpendicular line from C to
line
AB and get the x,y value for the point D at the intersection.
Assume I get the slope of the perpendicular line, I will have my point (D)
using variable x and y which is potentially on line AB. My idea was
using
|AC|*|AC| = |AD|*|AD|+ |CD|*|CD|. I don't know what function I may need
to
call to calculate the values for point D (uniroot?).
This is easier than you think.
Think of the x,y coordinates of each point :
Then, the slope is slope = rise/run = (By- Ay)/(Bx- Ax)
The Dx coordinate = Cx and the Dy = (Dx - Ax) * slope
Then, to draw the line segment from C to D
lines(C,D)
In R:
A - c(2,4)
B - c(4,1)
C - c(8,10)
slope -( C[2]- A[2])/(C[1]-A[1]) #rise/run
D - c(B[1],(B[1]-A[1])*slope + A[2]) # find D
my.data - rbind(A,B,C,D)
colnames(my.data) - c(X,Y)
my.data#show it
plot(my.data,type=n) #graph it without the points
text(my.data,rownames(my.data)) #put the points in
segments(A[1],A[2],C[1],C[2]) #draw the line segments
segments(B[1],B[2],D[1],D[2])
Bill
Thank you.
--
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Re: [R] What does this warning message (from optim function) mean?
Hi,
You did not give us any information about your likelihood function, f, nor
did you provide a reproducible example. So, I cannot tell for sure whether
the parameter estimates are reliable.
Ravi.
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
Behalf Of Sally Luo
Sent: Wednesday, August 25, 2010 11:26 AM
To: r-help@r-project.org
Subject: [R] What does this warning message (from optim function) mean?
Hi R users,
I am trying to use the optim function to maximize a likelihood funciton, and
I got the following warning messages.
Could anyone explain to me what messege 31 means exactly? Is it a cause for
concern?
Since the value of convergence turns out to be zero, it means that the
converging is successful, right?
So can I assume that the parameter estimates generated thereafter are
reliable MLE estimates?
Thanks a lot for your help.
Maomao
p-optim(c(0,0,0), f, method =BFGS, hessian =T, y=y,X=X,W=W)
There were 31 warnings (use warnings() to see them)
warnings()
Warning messages:
1: In log(det(I_N - pd * wd - po * wo - pw * ww)) : NaNs produced
2: In log(det(I_N - pd * wd - po * wo - pw * ww)) : NaNs produced
3: In log(det(I_N - pd * wd - po * wo - pw * ww)) : NaNs produced
4: In log(det(I_N - pd * wd - po * wo - pw * ww)) : NaNs produced
5: In log(det(I_N - pd * wd - po * wo - pw * ww)) : NaNs produced
6: In log(det(I_N - pd * wd - po * wo - pw * ww)) : NaNs produced
7: In log(det(I_N - pd * wd - po * wo - pw * ww)) : NaNs produced
8: In log(det(I_N - pd * wd - po * wo - pw * ww)) : NaNs produced
9: In log(det(I_N - pd * wd - po * wo - pw * ww)) : NaNs produced
10: In log(det(I_N - pd * wd - po * wo - pw * ww)) : NaNs produced
11: In log(det(I_N - pd * wd - po * wo - pw * ww)) : NaNs produced
12: In log(det(I_N - pd * wd - po * wo - pw * ww)) : NaNs produced
13: In log(det(I_N - pd * wd - po * wo - pw * ww)) : NaNs produced
14: In log(det(I_N - pd * wd - po * wo - pw * ww)) : NaNs produced
15: In log(det(I_N - pd * wd - po * wo - pw * ww)) : NaNs produced
16: In log(det(I_N - pd * wd - po * wo - pw * ww)) : NaNs produced
17: In log(det(I_N - pd * wd - po * wo - pw * ww)) : NaNs produced
18: In log(det(I_N - pd * wd - po * wo - pw * ww)) : NaNs produced
19: In log(det(I_N - pd * wd - po * wo - pw * ww)) : NaNs produced
20: In log(det(I_N - pd * wd - po * wo - pw * ww)) : NaNs produced
21: In log(det(I_N - pd * wd - po * wo - pw * ww)) : NaNs produced
22: In log(det(I_N - pd * wd - po * wo - pw * ww)) : NaNs produced
23: In log(det(I_N - pd * wd - po * wo - pw * ww)) : NaNs produced
24: In log(det(I_N - pd * wd - po * wo - pw * ww)) : NaNs produced
25: In log(det(I_N - pd * wd - po * wo - pw * ww)) : NaNs produced
26: In log(det(I_N - pd * wd - po * wo - pw * ww)) : NaNs produced
27: In log(det(I_N - pd * wd - po * wo - pw * ww)) : NaNs produced
28: In log(det(I_N - pd * wd - po * wo - pw * ww)) : NaNs produced
29: In log(det(I_N - pd * wd - po * wo - pw * ww)) : NaNs produced
30: In log(det(I_N - pd * wd - po * wo - pw * ww)) : NaNs produced
31: In if (hessian) { ... :
the condition has length 1 and only the first element will be used
p$counts
function gradient
148 17
p$convergence
[1] 0
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[R] Checking a point behind two polygons
Hi, I am working on a problem of checking whether a point is behind two convex hulls. By 'behind', I mean a point which is outside of two convex hulls can't reach either of these two hulls without reaching the other. My idea is to find the segments between the point and the vertices of a hull first. And then check whether there is any intersection between the segments and the edges of the other hull. If for all the edges to be checked, there is no intersection point found outside of the edges -- the point is said to be behind that hull. Basically, I am doing line-line intersection. I know if I use the line-line intersection checking, I will have to deal with lots of edges and segments. Besides, the line-line intersection checking may not be working...I wonder someone knows a simply way and more accurate way to check it, e.g.line-polygon intersection. Thank you. -- View this message in context: http://r.789695.n4.nabble.com/Checking-a-point-behind-two-polygons-tp2338823p2338823.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] What does this warning message (from optim function) mean?
the deteminant is a nonpositive value. log(det(...)) produce NaNs... -- View this message in context: http://r.789695.n4.nabble.com/What-does-this-warning-message-from-optim-function-mean-tp2338689p2338719.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] how do R calculates the number of intervals between tick-marks
Hi, With axTicks I did what I wanted. Thank you all for the attention. Below is the code I did to have a plot with three y axes. Suggestions to improve the routine are welcome. All the best, Antonio Olinto Data in a spreadsheet YEARL NFB NFT 200526,352158 150025,014 200632,514789 210035,024 200727,458126 180030,254 200828,568971 150043,254 200932,564789 300060,32 # comma is a decimal symbol dat.lnbnt - read.delim(clipboard,dec=,,header=T) attach(dat.lnbnt) summary(dat.lnbnt) YEARL NFBNFT Min. :2005 Min. :26.35 Min. :1500 Min. :25.01 1st Qu.:2006 1st Qu.:27.46 1st Qu.:1500 1st Qu.:30.25 Median :2007 Median :28.57 Median :1800 Median :35.02 Mean :2007 Mean :29.49 Mean :1980 Mean :38.77 3rd Qu.:2008 3rd Qu.:32.51 3rd Qu.:2100 3rd Qu.:43.25 Max. :2009 Max. :32.56 Max. :3000 Max. :60.32 y1min - 26 y1max - 33 y1range - y1max-y1min y2min - 1500 y2max - 3000 y2range - y2max-y2min y3min - 25 y3max - 61 y3range - y3max-y3min par(mar=c(6, 6, 2,12)) plot(L, ylim=c(y1min,y1max), ylab=landings (×1000 t), type=l,col=red,las=0, cex.axis=1.2,cex.lab=1.4,xaxt=n,xlab=) points(y1range*((NFB-y2min)/y2range)+y1min,type=l,col=blue) points(y1range*((NFT-y3min)/y3range)+y1min,type=l,col=darkgreen) yticks - axTicks(2) interval - yticks[2]-yticks[1] divisions - length(yticks)-1 axis(1,at=axTicks(1),labels=as.character(YEAR),las=2,cex.axis=1.2) axis(4,at=seq(y1min,y1max,interval),labels=as.integer(seq(y2min,y2max,y2range/divisions)),las=0,cex.axis=1.2) axis(4,at=seq(y1min,y1max,interval),labels=as.integer(seq(y3min,y3max,y3range/divisions)),cex.axis=1.2, las=0,line=5) mtext(nº of fishing boats,4,3,cex=1.4) mtext(nº of fishing trips (×1000),4,8, cex=1.4) legend(4,33,c(L,Nº FB,Nº FT),bty=n,lty=c(1,1,1),col=c(red,blue,darkgreen),cex=1.4) Citando David Winsemius dwinsem...@comcast.net: On Aug 24, 2010, at 7:05 PM, Antonio Olinto wrote: Hello, I want to know how do R calculates the number of intervals between tick-marks in the y axis in a plot. ?axTicks # and then look at the other citations and the code as needed I'm making a three y-axes plot and this information would help me a lot. Thanks in advance. Antonio Olinto -- David Winsemius, MD West Hartford, CT Webmail - iBCMG Internet http://www.ibcmg.com.br __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Removing inter-bar spaces in barchart
Hi: On Wed, Aug 25, 2010 at 11:28 AM, Greg Snow greg.s...@imail.org wrote: Using the barplot function in base graphics you just set space=0, but that function does not have a box.ratio argument which would imply that you are using something else. If you let us know which function (and which package it is in) then it is easier (possible) for us to help you, even better is to give the information asked for at the bottom of every post and the posting guide. It looks like he's using barchart() in lattice, which does have a box.ratio argument. -- Gregory (Greg) L. Snow Ph.D. Statistical Data Center Intermountain Healthcare greg.s...@imail.org 801.408.8111 -Original Message- From: r-help-boun...@r-project.org [mailto:r-help-boun...@r- project.org] On Behalf Of Jonathan Greenberg Sent: Tuesday, August 24, 2010 8:21 PM To: r-help Subject: [R] Removing inter-bar spaces in barchart Rhelpers: I'm trying to make a barchart of a 2-group dataset (barchart(x~y,data=data,groups=z,horizontal=FALSE)). My problem is that I can't, for the life of me, seem to get rid of the inter-bar space -- box.ratio set to 1 doesn't do much. Any ideas? I'd ideally want zero space between the bars. Thanks! I'm more interested in why you'd 'ideally want zero space between the bars'. By default, bar chart functions in R have spaces between bars because the bars represent different categories of a discrete/factor variable and the space is a visual device meant to connote to the viewer that the categories are not to be interpreted as an underlying continuum. In contrast, there are no spaces in between bars of a histogram precisely because the x-scale is meant to be continuous. Since you're using barchart(), why do you want to remove the space between bars? Cheers, Dennis --j -- Jonathan A. Greenberg, PhD Assistant Project Scientist Center for Spatial Technologies and Remote Sensing (CSTARS) Department of Land, Air and Water Resources University of California, Davis One Shields Avenue Davis, CA 95616 Phone: 415-763-5476 AIM: jgrn307, MSN: jgrn...@hotmail.com, Gchat: jgrn307 __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting- guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Draw a perpendicular line?
Thank you all for your suggestions and especially the codes. Now I am able to solve it by finding the intercepts and slopes for both lines first, and then I can find the coordinates of x and y at the intersection. At first I forgot how to use C to get the intercept of line CD... Thanks again. -- View this message in context: http://r.789695.n4.nabble.com/Draw-a-perpendicular-line-tp2335882p2338864.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] accessing the attr(*,label.table) after importing from spss
Dear all, I just received a file from a colleague in spss. The read.spss could not finish the file due to an error (Unrecognized record type 7, subtype 18 encountered in system file) so instead I converted the file using stat-transfer. Looking at my data I see that most labels are in the attributes and I´d love to access them and assign the pertinent variables to factors without doing the whole factor(levels,labels) thing manually. Is there any remedy to this ?? regards, M str(dat) 'data.frame': 860 obs. of 19 variables: $ Ag: int 15 15 15 15 15 15 15 15 15 15 ... $ G: int 2 2 2 1 1 1 1 1 1 2 ... $ GCQ: int 15 15 15 15 15 15 15 15 15 15 ... $ Amn: int 2 2 2 2 2 2 1 1 1 1 ... $ HI : int 1 1 1 1 1 1 2 2 2 2 ... $ Hos: int 2 2 2 2 2 2 2 2 2 2 ... $ Risk : int 2 2 2 2 2 2 2 2 2 2 ... $ CTO : int 2 2 2 2 2 1 1 1 1 1 ... $ pat : int NA NA NA NA NA 2 2 2 2 2 ... $ Day: int 7 7 7 5 4 6 5 7 7 5 ... $ Ho : int NA NA NA NA NA NA NA NA NA NA ... $ coh : int 1 1 1 1 1 1 1 1 1 1 ... $ Comp : int 1 1 1 1 1 1 1 1 1 1 ... $ Ethan: int 2 2 2 2 2 2 2 2 2 2 ... $ Pro : num NA NA NA NA NA NA NA NA NA NA ... $ Ye : int 1 1 1 3 3 3 1 1 1 3 ... $ Ageg: int 1 1 1 1 1 1 1 1 1 1 ... $ BAC: int 0 0 0 0 0 0 0 0 0 0 ... - attr(*, val.labels)= chr VL_Gender VL_Amnesia ... - attr(*, var.labels)= chr Age (years) Gender GCQSSA Amnesty ... - attr(*, label.table)=List of 19 ..$ : NULL ..$ : Named num 1 2 .. ..- attr(*, names)= chr Male Female ..$ : NULL ..$ : Named num 1 2 .. ..- attr(*, names)= chr Yes No ..$ : NULL ..$ : Named num 1 2 .. ..- attr(*, names)= chr Yes No ..$ : Named num 1 2 .. ..- attr(*, names)= chr Yes No ..$ : Named num 1 2 .. ..- attr(*, names)= chr Yes No ..$ : Named num 1 2 .. ..- attr(*, names)= chr Yes No ..$ : Named num 1 2 3 4 5 6 7 .. ..- attr(*, names)= chr Monday Tuesday Wednesday Thursday ... ..$ : Named num 1 2 .. ..- attr(*, names)= chr Yes No ..$ : Named num 1 2 .. ..- attr(*, names)= chr Yes No ..$ : Named num 1 2 3 4 5 6 7 .. ..- attr(*, names)= chr Yes Overtriage Undertriage with admission Overtriage with pos ... ..$ : Named num 1 2 .. ..- attr(*, names)= chr Yes No ..$ : NULL ..$ : NULL ..$ : Named num 1 2 3 4 .. ..- attr(*, names)= chr 15 - 24 25-39 40-59 60- ..$ : Named num 1 2 3 .. ..- attr(*, names)= chr 0.10-0.99 1.00-1.99 2.00- ..$ : Named num 0 1 2 3 4 .. ..- attr(*, names)= chr sorry undeweight 0.10-0.99 1.00-1.99 ... __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Removing inter-bar spaces in barchart
On Aug 24, 2010, at 10:20 PM, Jonathan Greenberg wrote: Rhelpers: I'm trying to make a barchart of a 2-group dataset (barchart(x~y,data=data,groups=z,horizontal=FALSE)). My problem is that I can't, for the life of me, seem to get rid of the inter-bar space -- box.ratio set to 1 doesn't do much. Any ideas? I'd ideally want zero space between the bars. Thanks! You didn't provide any data (nor did you illustrate with one of the available datasets that are used in examples.) Compare these two outputs: barchart(yield ~ year, data = barley, groups = variety, ylab = Barley Yield (bushels/acre), ) barchart(yield ~ variety, data = barley, groups = year, ylab = Barley Yield (bushels/acre), ) ... and notice that the variables in the group have no separation of their bars whereas the rhs variables do. This is the opposite of what I expected, so perhaps you think as I did and reversing the roles ofy and z might help? -- David Winsemius, MD West Hartford, CT __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] package MuMIn
[cc'ing back to r-help: this is good etiquette so that the responses
will be seen by others/
archived for future reference.]
On 10-08-25 04:35 PM, Marino Taussig De Bodonia, Agnese wrote:
Yes, I meant MuMIn
the global formula I introduced was:
rc4.mod-lm(central$hunting~ central$year + central$gender +
central$hunter + central$k.score + central$seen.in.wild +
central$captivity + central$had.damage +
central$importance.of.being.informed + central$RC1+ central$RC2 +
central$year:central$hunter + central$year:central$had.damage +
central$year:central$seen.in.wild +central$year:central$RC1 +
central$year:central$RC2)
In general it would be much clearer and more generalizable if you
instead used:
rc4.mod-lm(hunting~ year + gender + hunter + k.score + seen.in.wild +
captivity +
had.damage + importance.of.being.informed + RC1+ RC2 + year:hunter +
year:had.damage +
year:seen.in.wild +year:RC1 + year:RC2, data=central)
or even
rc4.mod - lm(hunting ~ gender + k.score + captivity +
importance.of.being.informed+
year*(RC1 + RC2 + hunter + had.damage + seen.in.wild), data=central)
I don't know why those spurious interactions are showing up. It
*might* be a bug in MuMIn.
So far, I can't reproduce this behavior with my simple example:
=
library(MuMIn)
dat - data.frame(y=runif(100),a=factor(sample(1:5,replace=TRUE,size=100)),
b= factor(sample(1:5,replace=TRUE,size=100)),
c=runif(100))
m1 - lm(y~a+b+c+a:b,data=dat)
gm - function(x) {
model.avg(get.models(dredge(x),subset=deltaInf))$relative.importance
}
gm(m1)
m1 - lm(y~dat$a+dat$b+dat$c+dat$a:dat$b,data=dat)
gm(m1)
==
You have to keep working to simplify your example until you can get to
a pair
of examples, one simpler and one more complicated, that 'bracket' the
problem --
the simpler example doesn't show the problem, and the more complicated
(closer
to your original example) does.
Alternatively, you may simplify the problem out of existence and
decide that you don't need to spend any more
time figuring out what it was ...
Two more quick notes to consider:
* why are you not considering models with delta4 ... ?
* it is quite tricky to compare relative importance of 'main effect'
parameters in models with and without interactions.
There are 15 explanatory variables in this model.
my code was:
rc4.mod.Mu - dredge(rc4.mod, rank = AICc)
rc4.mod.Mu
rc4.mod.Mu.avg-model.avg(get.models(rc4.mod.Mu, subset = delta 4))
the output for the command rc4.mod.Mu.avg$relative.importance was:
central$hunter1
central$seen.in.wild1
central$year0.986960449
central$had.damage0.89670109
central$RC20.83866013
central$k.score0.656654734
central$had.damage:central$year0.517130185
central$year:central$hunter0.305988097
central$hunter:central$year0.212045101
central$year:central$seen.in.wild0.190520501
central$seen.in.captivity0.148263242
central$gender0.119314202
central$RC10.098234445
central$importance.of.being.informed0.091842088
central$year:central$RC20.069501158
central$seen.in.wild:central$year0.065788243
central$RC2:central$year0.024221603
There are 17 variables above: central$year:central$RC2 and
central$RC2:central$year are both present, as are
central$hunter:central$year and central$year:central$hunter .
Can you tell me why, if they are the same thing, they are present twice
and have different values?
Thanks a lot,
Agnese
From: r-help-boun...@r-project.org [r-help-boun...@r-project.org] On
Behalf Of Ben Bolker [bbol...@gmail.com]
Sent: 25 August 2010 20:36
To: r-h...@stat.math.ethz.ch
Subject: Re: [R] package MuMI
Marino Taussig De Bodonia, Agnese agnese.marino09 at imperial.ac.uk
writes:
I am using the package MuMI to run all the possible combinations
of variables in my full model, and select
my best models. When I enter my variables in the original model I
write them like this
lm(y~ a +b +c +a:b)
However, MuMI will also use the variable b:a, which I do not want
in my model.
How do I stop that from happening?
(1) I think you mean MuMIn.
(2) Please send a reproducible example! (Hint: see the posting
guide that is referred to at the bottom of all R-help messages ...)
(3) [now proceeding to try to guess what you mean ...]
a:b and b:a have identical meanings in the R formula syntax.
I tried to do something like what I thought you might have meant
and got what seemed to be reasonable answers.
library(MuMIn)
dat -
data.frame(y=runif(100),a=factor(sample(1:5,replace=TRUE,size=100)),
+b= factor(sample(1:5,replace=TRUE,size=100)), c=runif(100))
m1 - lm(y~a+b+c+a:b,data=dat)
dredge(m1)
Global model: lm(formula = y ~ a + b + c + a:b, data = dat)
---
Model selection table
(Int) a b c a:b k R.sqAdj.R.sq RSS AIC AICc
delta weight
4 0.6385 -0.1643 3 0.03019
Re: [R] Secant Method Convergence (Method to replicate Excel XIRR/IRR)
Hi Adrian,
Your code for the secant method is correct.
I just tweaked it a bit w/o the calendar time feature (assuming that the
cash flows will be available on an annual basis) as follows:
ANXIRR - function (cashFlow, guess, tol=1.e-04){
npv - function (cashFlow, irr) {
n - length(cashFlow)
sum(cashFlow / (1 + irr)^{0: (n-1)})
}
irrprev - c(0)
irr - guess
pvPrev - sum(cashFlow)
pv - npv(cashFlow, irr)
eps - abs(pv-pvPrev)
while (eps = tol) {
tmp - irrprev
irrprev - irr
irr - irr - ((irr - tmp) * pv / (pv - pvPrev))
pvPrev - pv
pv - npv(cashFlow, irr)
eps - abs(pv - pvPrev)
}
list(irr = irr, npv = pv)
}
# example from Wikipedia enntry
cashflow - c(-4000, 1200, 1410, 1875, 1050)
ANXIRR(cashflow, guess=0.05)
ANXIRR(cashflow, guess=0.05)
$irr
[1] 0.1429934
$npv
[1] 1.705303e-12
Hope this helps,
Ravi.
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
Behalf Of Adrian Ng
Sent: Wednesday, August 25, 2010 8:39 AM
To: r-help@r-project.org
Subject: [R] Secant Method Convergence (Method to replicate Excel XIRR/IRR)
Hi,
I am new to R, and as a first exercise, I decided to try to implement an
XIRR function using the secant method. I did a quick search and saw another
posting that used the Bisection method but wanted to see if it was possible
using the secant method.
I would input a Cash Flow and Date vector as well as an initial guess. I
hardcoded today's initial date so I could do checks in Excel. This code
seems to only converge when my initial guess is very close to the correct
IRR.
Maybe I have some basic errors in my coding/logic? Any help would be greatly
appreciated.
The Wikipedia article to secant method and IRR:
http://en.wikipedia.org/wiki/Internal_rate_of_return#Numerical_solution
Thanks!
ANXIRR - function (cashFlow, cfDate, guess){
cfDate-as.Date(cfDate,format=%m/%d/%Y)
irrprev - c(0); irr- guess
pvPrev- sum(cashFlow)
pv-
sum(cashFlow/((1+irr)^(as.numeric(difftime(cfDate,2010-08-24,units=days)
)/360)))
print(pv)
print(Hi)
while (abs(pv) = 0.001) {
t-irrprev; irrprev- irr;
irr-irr-((irr-t)*pv/(pv-pvPrev));
pvPrev-pv;
pv-sum(cashFlow/((1+irr)^(as.numeric(difftime(cfDate,2010-08-24,units=da
ys))/365)))
print(irr);print(pv)
}
}
Please consider the environment before printing this e-mail.
[[alternative HTML version deleted]]
__
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https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] Secant Method Convergence (Method to replicate Excel XIRR/IRR)
Another approach is to use `uniroot' to find the zero of the NPV function:
npv - function (cashFlow, irr) {
n - length(cashFlow)
sum(cashFlow / (1 + irr)^{0: (n-1)})
}
uniroot(f=npv, interval=c(0,1), cashFlow=cashFlow)
However, there may be situations where there are no real zeros or there are
multiple zeros of the NPV function.
Ravi.
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
Behalf Of Adrian Ng
Sent: Wednesday, August 25, 2010 8:39 AM
To: r-help@r-project.org
Subject: [R] Secant Method Convergence (Method to replicate Excel XIRR/IRR)
Hi,
I am new to R, and as a first exercise, I decided to try to implement an
XIRR function using the secant method. I did a quick search and saw another
posting that used the Bisection method but wanted to see if it was possible
using the secant method.
I would input a Cash Flow and Date vector as well as an initial guess. I
hardcoded today's initial date so I could do checks in Excel. This code
seems to only converge when my initial guess is very close to the correct
IRR.
Maybe I have some basic errors in my coding/logic? Any help would be greatly
appreciated.
The Wikipedia article to secant method and IRR:
http://en.wikipedia.org/wiki/Internal_rate_of_return#Numerical_solution
Thanks!
ANXIRR - function (cashFlow, cfDate, guess){
cfDate-as.Date(cfDate,format=%m/%d/%Y)
irrprev - c(0); irr- guess
pvPrev- sum(cashFlow)
pv-
sum(cashFlow/((1+irr)^(as.numeric(difftime(cfDate,2010-08-24,units=days)
)/360)))
print(pv)
print(Hi)
while (abs(pv) = 0.001) {
t-irrprev; irrprev- irr;
irr-irr-((irr-t)*pv/(pv-pvPrev));
pvPrev-pv;
pv-sum(cashFlow/((1+irr)^(as.numeric(difftime(cfDate,2010-08-24,units=da
ys))/365)))
print(irr);print(pv)
}
}
Please consider the environment before printing this e-mail.
[[alternative HTML version deleted]]
__
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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.
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
[R] OT: R for iPhone/iPad OS?
No, seriously: I've had more than one person at work wonder what math toolset could be loaded onto iOS. So, before Matlab, FreeMat, Mathematia, SciLab, Octave, or numpy (:-) ) produces a version for iPad, any chance someone is working on R for iPad? __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] OT: R for iPhone/iPad OS?
I believe that the iPad conditions for producing apps are incompatible with the GPL (gnuGo was removed from the apps store recently for this reason), so don't hold your breath. There may be some possibility for an app on the iPad that would access R on a server somewhere else, but that is beyond my abilities/interest. -- Gregory (Greg) L. Snow Ph.D. Statistical Data Center Intermountain Healthcare greg.s...@imail.org 801.408.8111 -Original Message- From: r-help-boun...@r-project.org [mailto:r-help-boun...@r- project.org] On Behalf Of Carl Witthoft Sent: Wednesday, August 25, 2010 3:24 PM To: r-help@r-project.org Subject: [R] OT: R for iPhone/iPad OS? No, seriously: I've had more than one person at work wonder what math toolset could be loaded onto iOS. So, before Matlab, FreeMat, Mathematia, SciLab, Octave, or numpy (:-) ) produces a version for iPad, any chance someone is working on R for iPad? __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting- guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] OT: R for iPhone/iPad OS?
Instructions for installing R on jailbroken devices: http://rwiki.sciviews.org/doku.php?id=getting-started:installation:iphone Carl Witthoft wrote: No, seriously: I've had more than one person at work wonder what math toolset could be loaded onto iOS. So, before Matlab, FreeMat, Mathematia, SciLab, Octave, or numpy (:-) ) produces a version for iPad, any chance someone is working on R for iPad? __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] OT: R for iPhone/iPad OS?
Hi, On Wed, Aug 25, 2010 at 5:37 PM, Greg Snow greg.s...@imail.org wrote: I believe that the iPad conditions for producing apps are incompatible with the GPL (gnuGo was removed from the apps store recently for this reason), so don't hold your breath. There may be some possibility for an app on the iPad that would access R on a server somewhere else, but that is beyond my abilities/interest. This has come up a few times before on R-sig-mac. Here's one: http://thread.gmane.org/gmane.comp.lang.r.mac/4912 You can search that list for others discussions that have popped up to get a feel of what was discussed about this already. If you have more to add to the conversation, I reckon you'd get the most traction if you post directly to that mailing list ... -steve -- Steve Lianoglou Graduate Student: Computational Systems Biology | Memorial Sloan-Kettering Cancer Center | Weill Medical College of Cornell University Contact Info: http://cbio.mskcc.org/~lianos/contact __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Secant Method Convergence (Method to replicate Excel XIRR/IRR)
Yes, the secant method (like Newton Raphson) is not guaranteed to converge,
unlike the bisection method, but it has a superlinear convergence (not that
this matters much!). Brent's method, which is used in `uniroot', is a
reliable and fast method, which is why I suggested it in my previous email.
Having said that, I am not sure about the convergence problem that you are
having without seeing the actual example.
Ravi.
-Original Message-
From: Adrian Ng [mailto:a...@hamiltonlane.com]
Sent: Wednesday, August 25, 2010 5:28 PM
To: Ravi Varadhan; r-help@r-project.org
Subject: RE: [R] Secant Method Convergence (Method to replicate Excel
XIRR/IRR)
Hi Ravi,
Thanks for the responses. I was actually trying to calculate IRR based on
unevenly spaced cash flows, and that's why I decided to use the secant
method. I'm not sure if my answer isn't converging because I have some
careless mistake in the code, or if it's simply because unlike the bisection
method, the secant method doesn't 'sandwich' the desired root.
-Original Message-
From: Ravi Varadhan [mailto:rvarad...@jhmi.edu]
Sent: Wednesday, August 25, 2010 5:24 PM
To: Adrian Ng; r-help@r-project.org
Subject: RE: [R] Secant Method Convergence (Method to replicate Excel
XIRR/IRR)
Another approach is to use `uniroot' to find the zero of the NPV function:
npv - function (cashFlow, irr) {
n - length(cashFlow)
sum(cashFlow / (1 + irr)^{0: (n-1)})
}
uniroot(f=npv, interval=c(0,1), cashFlow=cashFlow)
However, there may be situations where there are no real zeros or there are
multiple zeros of the NPV function.
Ravi.
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
Behalf Of Adrian Ng
Sent: Wednesday, August 25, 2010 8:39 AM
To: r-help@r-project.org
Subject: [R] Secant Method Convergence (Method to replicate Excel XIRR/IRR)
Hi,
I am new to R, and as a first exercise, I decided to try to implement an
XIRR function using the secant method. I did a quick search and saw another
posting that used the Bisection method but wanted to see if it was possible
using the secant method.
I would input a Cash Flow and Date vector as well as an initial guess. I
hardcoded today's initial date so I could do checks in Excel. This code
seems to only converge when my initial guess is very close to the correct
IRR.
Maybe I have some basic errors in my coding/logic? Any help would be greatly
appreciated.
The Wikipedia article to secant method and IRR:
http://en.wikipedia.org/wiki/Internal_rate_of_return#Numerical_solution
Thanks!
ANXIRR - function (cashFlow, cfDate, guess){
cfDate-as.Date(cfDate,format=%m/%d/%Y)
irrprev - c(0); irr- guess
pvPrev- sum(cashFlow)
pv-
sum(cashFlow/((1+irr)^(as.numeric(difftime(cfDate,2010-08-24,units=days)
)/360)))
print(pv)
print(Hi)
while (abs(pv) = 0.001) {
t-irrprev; irrprev- irr;
irr-irr-((irr-t)*pv/(pv-pvPrev));
pvPrev-pv;
pv-sum(cashFlow/((1+irr)^(as.numeric(difftime(cfDate,2010-08-24,units=da
ys))/365)))
print(irr);print(pv)
}
}
Please consider the environment before printing this e-mail.
[[alternative HTML version deleted]]
__
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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.
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] OT: R for iPhone/iPad OS?
On Aug 25, 2010, at 4:23 PM, Carl Witthoft wrote: No, seriously: I've had more than one person at work wonder what math toolset could be loaded onto iOS. So, before Matlab, FreeMat, Mathematia, SciLab, Octave, or numpy (:-) ) produces a version for iPad, any chance someone is working on R for iPad? See this coverage of recent discussions at R-Bloggers: http://www.r-bloggers.com/could-we-run-a-statistical-analysis-on-iphoneipad-using-r/ Unless you are willing to jailbreak the devices, the basic answer is not as a traditional stand alone application. However, using a client/server model based GUI application (think Wolfram|Alpha) or via a web browser connecting to a remote R server, yes you can. HTH, Marc Schwartz __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Looking for an image (R 64-bit on Linux 64-bit) on Amazon EC2
I have found an existing image on Amazon EC2 including R. But unfortunately, it is 32-bit R on 32-bit Linux. Does anybody know if there exists an mage (R 64-bit on Linux 64-bit) on Amazon EC2? Or how can I install 64-bit R on my own Linux instance there? Thanks. -- View this message in context: http://r.789695.n4.nabble.com/Looking-for-an-image-R-64-bit-on-Linux-64-bit-on-Amazon-EC2-tp2338938p2338938.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] OT: R for iPhone/iPad OS?
On 8/25/2010 2:23 PM, Carl Witthoft wrote: No, seriously: I've had more than one person at work wonder what math toolset could be loaded onto iOS. So, before Matlab, FreeMat, Mathematia, SciLab, Octave, or numpy (:-) ) produces a version for iPad, any chance someone is working on R for iPad? This has come up before and one conclusion was that there may be issues regarding GPL software and the App Store's restrictions, or at least design issues that would make this somewhere between difficult and impossible. For a much more thorough discussion see the threads: https://stat.ethz.ch/pipermail/r-help/2010-May/240669.html https://stat.ethz.ch/pipermail/r-help/2010-June/240901.html -- Brian Diggs Senior Research Associate, Department of Surgery Oregon Health Science University __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Looking for an image (R 64-bit on Linux 64-bit) on Amazon EC2
You have a 64 bit Linux? If so... Dowload the sources make sure you have all of the dependencies unpack tarball cd to de-compressed directory issue ./configure make sudo make install or maybe use your distros packages managment tool. Am I missing something? On Wed, Aug 25, 2010 at 4:39 PM, noclue_ tim@netzero.net wrote: I have found an existing image on Amazon EC2 including R. But unfortunately, it is 32-bit R on 32-bit Linux. Does anybody know if there exists an mage (R 64-bit on Linux 64-bit) on Amazon EC2? Or how can I install 64-bit R on my own Linux instance there? Thanks. -- View this message in context: http://r.789695.n4.nabble.com/Looking-for-an-image-R-64-bit-on-Linux-64-bit-on-Amazon-EC2-tp2338938p2338938.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Stephen Sefick | Auburn University | | Department of Biological Sciences | | 331 Funchess Hall | | Auburn, Alabama | | 36849 | |___| | sas0...@auburn.edu | | http://www.auburn.edu/~sas0025 | |___| Let's not spend our time and resources thinking about things that are so little or so large that all they really do for us is puff us up and make us feel like gods. We are mammals, and have not exhausted the annoying little problems of being mammals. -K. Mullis A big computer, a complex algorithm and a long time does not equal science. -Robert Gentleman __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] OT: R for iPhone/iPad OS?
Hey, Many thanks to all who responded! I'll pass along the links to my pals. Personally I can't imagine doing R on a virtual keyboard in the first place, but to each his own. Carl __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] how to plot y-axis on the right of x-axis
Thank you. The answers provides the right direction and a code was written accordingly. One more request, if the label of axis X wants to be drawn from 5 to 1 (left to right) rather than 1 to 5, is it fine to change axis (4, at = NULL) ? If so, which value should be input ? Thanks again. Elaine code plot(1:5,axes=FALSE) axis(1) axis(4) box() On Wed, Aug 25, 2010 at 5:39 PM, Jim Lemon j...@bitwrit.com.au wrote: On 08/25/2010 09:12 AM, elaine kuo wrote: Dear List, I have a richness data distributing across 20 N to 20 S latitude. (120 E-140 E longitude). I would like to draw the richness in the north hemisphere and a regression line in the plot (x-axis: latitude, y-axis: richness in the north hemisphere). The above demand is done using plot. Then, south hemisphere richness and regression are required to be generated using the same y-axis above but an x-axis on the left side of the y-axis. (The higher latitude in the south hemisphere, the left it would move) Please kindly share how to design the south plot and regression line for richness. Also, please advise if any more info is in need. Hi Elaine, Changing the position of the axes is easily done by changing the side and pos arguments of the axis function. If you want to move the y-axis to the right (or as you state it, the x axis to the left): # y axis on the left plot(1:5,axes=FALSE) axis(1) axis(2) # add a y axis one third of the way to the right xylim-par(usr) axis(2,pos=xylim[1]+diff(xylim[1:2])/3) # add another y axis two thirds of the way axis(4,pos=xylim[2]-diff(xylim[1:2])/3) # add one more on the right axis(4) You can move the x axis up and down using the same tricks. Jim [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] how to plot y-axis on the right of x-axis
Yes, I appreciated your answers which well hit my questions. (esp the perfect model parts). About the plot part, one more question. Is it possible to make the two plots (northern and southern richness) sharing the same Y axis (latitude = 0, the equator) ? In other words, southern richness would be on the left side of the Y axis, while northern one on the right side. Elaine Two plot panels on the same device like: layout(1:2) plot(1:100) plot(1:100) layout(1) = Yes that's what I want The second panel for southern species richness, do you mean you want the plot to go like this: plot(1:100, xlim = c(100,1)) = Yes, too. [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Secant Method Convergence (Method to replicate Excel XIRR/IRR)
The secant method converges just fine. Your problem might have occurred due
to improper conversion of dates to elapsed time. You want to calculate IRR
using year as the time unit, not days.
Here is the secant function (modified to account for irregular times) and
the results for your example:
ANXIRR - function (cashFlow, dates, guess, tol=1.e-04){
npv - function (cashFlow, times, irr) {
n - length(cashFlow)
sum(cashFlow / (1 + irr)^times)
}
if (guess == 0)stop(Initial guess must be strictly greater than 0)
times - as.numeric(difftime(dates, dates[1], units=days) /
365.24)
irrprev - c(0)
irr - guess
pvPrev - sum(cashFlow)
pv - npv(cashFlow, times, irr)
eps - abs(pv-pvPrev)
while (eps = tol) {
tmp - irrprev
irrprev - irr
irr - irr - ((irr - tmp) * pv / (pv - pvPrev))
pvPrev - pv
pv - npv(cashFlow, times, irr)
eps - abs(pv - pvPrev)
}
list(irr = irr, npv = pv)
}
CF - c(-1000,500,500,500,500,500)
dates -
c(1/1/2001,2/1/2002,3/1/2003,4/1/2004,5/1/2005,6/1/2006)
ANXIRR(CF, dates, guess=0.1)
ANXIRR(CF, dates, guess=0.1)
$irr
[1] 0.4106115
$npv
[1] 2.984279e-13
Ravi.
-Original Message-
From: Adrian Ng [mailto:a...@hamiltonlane.com]
Sent: Wednesday, August 25, 2010 6:23 PM
To: Ravi Varadhan
Subject: RE: [R] Secant Method Convergence (Method to replicate Excel
XIRR/IRR)
The forum is kind of slow so I'm just re-sending you the message here:
Hi Ravi,
I'm just trying a fairly simple example:
CFs: -1000,500,500,500,500,500
dates-c(1/1/2001,2/1/2002,3/1/2003,4/1/2004,5/1/2005,6/1/2006)
Thanks a lot for your help.
Adrian
-Original Message-
From: Ravi Varadhan [mailto:rvarad...@jhmi.edu]
Sent: Wednesday, August 25, 2010 5:44 PM
To: Adrian Ng; r-help@r-project.org
Subject: RE: [R] Secant Method Convergence (Method to replicate Excel
XIRR/IRR)
Yes, the secant method (like Newton Raphson) is not guaranteed to converge,
unlike the bisection method, but it has a superlinear convergence (not that
this matters much!). Brent's method, which is used in `uniroot', is a
reliable and fast method, which is why I suggested it in my previous email.
Having said that, I am not sure about the convergence problem that you are
having without seeing the actual example.
Ravi.
-Original Message-
From: Adrian Ng [mailto:a...@hamiltonlane.com]
Sent: Wednesday, August 25, 2010 5:28 PM
To: Ravi Varadhan; r-help@r-project.org
Subject: RE: [R] Secant Method Convergence (Method to replicate Excel
XIRR/IRR)
Hi Ravi,
Thanks for the responses. I was actually trying to calculate IRR based on
unevenly spaced cash flows, and that's why I decided to use the secant
method. I'm not sure if my answer isn't converging because I have some
careless mistake in the code, or if it's simply because unlike the bisection
method, the secant method doesn't 'sandwich' the desired root.
-Original Message-
From: Ravi Varadhan [mailto:rvarad...@jhmi.edu]
Sent: Wednesday, August 25, 2010 5:24 PM
To: Adrian Ng; r-help@r-project.org
Subject: RE: [R] Secant Method Convergence (Method to replicate Excel
XIRR/IRR)
Another approach is to use `uniroot' to find the zero of the NPV function:
npv - function (cashFlow, irr) {
n - length(cashFlow)
sum(cashFlow / (1 + irr)^{0: (n-1)})
}
uniroot(f=npv, interval=c(0,1), cashFlow=cashFlow)
However, there may be situations where there are no real zeros or there are
multiple zeros of the NPV function.
Ravi.
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
Behalf Of Adrian Ng
Sent: Wednesday, August 25, 2010 8:39 AM
To: r-help@r-project.org
Subject: [R] Secant Method Convergence (Method to replicate Excel XIRR/IRR)
Hi,
I am new to R, and as a first exercise, I decided to try to implement an
XIRR function using the secant method. I did a quick search and saw another
posting that used the Bisection method but wanted to see if it was possible
using the secant method.
I would input a Cash Flow and Date vector as well as an initial guess. I
hardcoded today's initial date so I could do checks in Excel. This code
seems to only converge when my initial guess is very close to the correct
IRR.
Maybe I have some basic errors in my coding/logic? Any help would be greatly
appreciated.
The Wikipedia article to secant method and IRR:
http://en.wikipedia.org/wiki/Internal_rate_of_return#Numerical_solution
Thanks!
ANXIRR - function (cashFlow, cfDate, guess){
cfDate-as.Date(cfDate,format=%m/%d/%Y)
irrprev - c(0); irr- guess
pvPrev- sum(cashFlow)
pv-
sum(cashFlow/((1+irr)^(as.numeric(difftime(cfDate,2010-08-24,units=days)
)/360)))
print(pv)
print(Hi)
while (abs(pv) = 0.001) {
t-irrprev; irrprev- irr;
irr-irr-((irr-t)*pv/(pv-pvPrev));
pvPrev-pv;
Re: [R] lattice help required
Deepayan Sarkar has a function combineLimits() in the development version of the latticeExtra package (i.e. the version on r-forge.r-project.org) which will set common scales in each row or column of your layout. It can also remove the internal axes. # Felix On 26 August 2010 04:43, Kay Cichini kay.cich...@uibk.ac.at wrote: .. thanks again, richard. and you swiftly saw the next problem comming up - when using par.settings = list(layout.widths = list(axis.panel = c(1, 0))) getting rid of the double tick labeling would be natural - but i'll leave it at that for today. many thanks, kay Richard M. Heiberger wrote: The multiple y axes are protecting you in this situation. z - cbind(rnorm(100,c(1,10),1), rnorm(100,c(20,30),1)) dotplot(z[,1]+z[,2] ~ facs$Treatment|facs$Sites, outer=TRUE, scales = list( y = list( relation=free)), ylab=c(y1, y2), xlab=c(Site 1, Site 2), strip=FALSE, main=problem) dotplot(z[,1]+z[,2] ~ facs$Treatment|facs$Sites, outer=TRUE, scales = list( y = list( relation=free, limits=list(c(-5,13),c(-5,13),c(18,32),c(18,32, ylab=c(y1, y2), xlab=c(Site 1, Site 2), strip=FALSE, main=protected) For more control (such as suppressing the y-tick labels in the right-hand column, I recommend Deepayan Sarkar's book. Rich [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. - Kay Cichini Postgraduate student Institute of Botany Univ. of Innsbruck -- View this message in context: http://r.789695.n4.nabble.com/lattice-help-required-tp2338382p2338707.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Felix Andrews / 安福立 http://www.neurofractal.org/felix/ __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Adding column to dataframe
To update on this. I ran the same command on a grid of computers with 32gb ram, and it completed in 15 seconds, compared to the ~20 minutes on my desktop. Simply a ram issue as suspected by a few on the list here. Thanks -- View this message in context: http://r.789695.n4.nabble.com/Adding-column-to-dataframe-tp2330556p2339076.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Rotate x-axis label on log scale
Hi Jim
Thanks so much for this. The updated staxlab function works perfectly!
Tim
-Original Message-
From: Jim Lemon [mailto:j...@bitwrit.com.au]
Sent: Wednesday, August 25, 2010 6:36 PM
To: tesutton
Cc: r-help@r-project.org
Subject: Re: [R] Rotate x-axis label on log scale
On 08/25/2010 01:37 PM, Tim Elwell-Sutton wrote:
Hi Jim
Thanks for this. The staxlab function seems very useful. Unfortunately,
the
rotation option doesn't seem to work for me when the y-axis is on a log
scale. It will stagger the labels but not rotate them. There's no error
message. On a linear axis the rotation works nicely. Any ideas?
The example below works if you omit log='y' or srt=45
Hi Tim,
Thanks for letting me know about this problem. I never did get around to
making staxlab work with log axes, but I think this new version:
staxlab-function(side=1,at,labels,nlines=2,top.line=0.5,
line.spacing=0.8,srt=NA,ticklen=0.03,...) {
if(missing(labels)) labels-at
nlabels-length(labels)
if(missing(at)) at-1:nlabels
if(is.na(srt)) {
linepos-rep(top.line,nlines)
for(i in 2:nlines) linepos[i]-linepos[i-1]+line.spacing
linepos-rep(linepos,ceiling(nlabels/nlines))[1:nlabels]
axis(side=side,at=at,labels=rep(,nlabels))
mtext(text=labels,side=side,line=linepos,at=at,...)
}
else {
xylim-par(usr)
if(side == 1) {
xpos-at
if(par(ylog)) ypos-10^(xylim[3]-ticklen*(xylim[4]-xylim[3]))
else ypos-xylim[3]-ticklen*(xylim[4]-xylim[3])
}
else {
ypos-at
if(par(xlog)) xpos-10^(xylim[1]-ticklen*(xylim[2]-xylim[1]))
else xpos-xylim[1]-ticklen*(xylim[2]-xylim[1])
}
par(xpd=TRUE)
text(xpos,ypos,labels,srt=srt,adj=1,...)
par(xpd=FALSE)
}
}
will do what you want.
Jim
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
[R] How to obtain the graph of fitted values against one variable after estimation?
Hi there, I have a question regarding the plot command after estimation. Specifically, I estimate a model, say regressing y on x and z. And after estimation, I would like to plot the fitted values against x, but I don't need that for z. The following command always gives two graphs, for both variables x and z. plot.np-plot(model, plot.errors.method = asymptotic) My question is, what option should I specify in order to get the graph for x only? I know this is probably a very simple question, but I searched around for a while without any luck. Thank you for your time. Best, Le __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Secant Method Convergence (Method to replicate Excel XIRR/IRR)
You should use cfDate[1] as the time origin. You cannot use 08-24-2010 as the
time origin, since that will yield negative times.
Here is the correct solution.
ANXIRR - function (cashFlow, dates, guess, tol=1.e-04){
npv - function (cashFlow, times, irr) {
n - length(cashFlow)
sum(cashFlow / (1 + irr)^times)
}
if (guess == 0) stop(Initial guess must be strictly greater than 0)
cfDate - as.Date(cfDate,format=%m/%d/%Y)
times - as.numeric(difftime(cfDate, cfDate[1], units=days) /
365.24)
irrprev - c(0)
irr - guess
pvPrev - sum(cashFlow)
pv - npv(cashFlow, times, irr)
eps - abs(pv-pvPrev)
while (eps = tol) {
tmp - irrprev
irrprev - irr
irr - irr - ((irr - tmp) * pv / (pv - pvPrev))
pvPrev - pv
pv - npv(cashFlow, times, irr)
eps - abs(pv - pvPrev)
}
list(irr = irr, npv = pv)
}
CF - c(-1000,500,500,500,500,500)
dates - c(1/1/2001,2/1/2002,3/1/2003,4/1/2004,5/1/2005,6/1/2006)
ANXIRR(CF, dates, guess=0.1)
ANXIRR(CF, dates, guess=0.1)
$irr
[1] 0.3740656
$npv
[1] 2.102695e-09
Hope this helps,
Ravi.
Ravi Varadhan, Ph.D.
Assistant Professor,
Division of Geriatric Medicine and Gerontology
School of Medicine
Johns Hopkins University
Ph. (410) 502-2619
email: rvarad...@jhmi.edu
- Original Message -
From: Adrian Ng a...@hamiltonlane.com
Date: Wednesday, August 25, 2010 8:33 pm
Subject: RE: [R] Secant Method Convergence (Method to replicate Excel XIRR/IRR)
To: Ravi Varadhan rvarad...@jhmi.edu
Cc: r-help@r-project.org r-help@r-project.org
Hi Ravi,
Using days and dividing it by 365 effectively converts the number to
years anyway and allows for the irregular times to be specific to the
days.
Also, when I replace dates[1] in your line:
times - as.numeric(difftime(dates, dates[1], units=days) /
365.24) with 2010-08-24 I think I am getting some irregular
results.
Effectively, what I was trying to do was match what Excel produced
with its XIRR function. With the example I gave excel returned an IRR
of ~0.37 (or 37%)
I am still in the process of debugging it...
-Original Message-
From: Ravi Varadhan [
Sent: Wednesday, August 25, 2010 7:24 PM
To: Adrian Ng
Cc: r-help@r-project.org
Subject: RE: [R] Secant Method Convergence (Method to replicate Excel
XIRR/IRR)
The secant method converges just fine. Your problem might have
occurred due
to improper conversion of dates to elapsed time. You want to
calculate IRR
using year as the time unit, not days.
Here is the secant function (modified to account for irregular times)
and
the results for your example:
ANXIRR - function (cashFlow, dates, guess, tol=1.e-04){
npv - function (cashFlow, times, irr) {
n - length(cashFlow)
sum(cashFlow / (1 + irr)^times)
}
if (guess == 0)stop(Initial guess must be strictly greater than 0)
times - as.numeric(difftime(dates, dates[1], units=days) /
365.24)
irrprev - c(0)
irr - guess
pvPrev - sum(cashFlow)
pv - npv(cashFlow, times, irr)
eps - abs(pv-pvPrev)
while (eps = tol) {
tmp - irrprev
irrprev - irr
irr - irr - ((irr - tmp) * pv / (pv - pvPrev))
pvPrev - pv
pv - npv(cashFlow, times, irr)
eps - abs(pv - pvPrev)
}
list(irr = irr, npv = pv)
}
CF - c(-1000,500,500,500,500,500)
dates -
c(1/1/2001,2/1/2002,3/1/2003,4/1/2004,5/1/2005,6/1/2006)
ANXIRR(CF, dates, guess=0.1)
ANXIRR(CF, dates, guess=0.1)
$irr
[1] 0.4106115
$npv
[1] 2.984279e-13
Ravi.
-Original Message-
From: Adrian Ng [
Sent: Wednesday, August 25, 2010 6:23 PM
To: Ravi Varadhan
Subject: RE: [R] Secant Method Convergence (Method to replicate Excel
XIRR/IRR)
The forum is kind of slow so I'm just re-sending you the message here:
Hi Ravi,
I'm just trying a fairly simple example:
CFs: -1000,500,500,500,500,500
dates-c(1/1/2001,2/1/2002,3/1/2003,4/1/2004,5/1/2005,6/1/2006)
Thanks a lot for your help.
Adrian
-Original Message-
From: Ravi Varadhan [
Sent: Wednesday, August 25, 2010 5:44 PM
To: Adrian Ng; r-help@r-project.org
Subject: RE: [R] Secant Method Convergence (Method to replicate Excel
XIRR/IRR)
Yes, the secant method (like Newton Raphson) is not guaranteed to converge,
unlike the bisection method, but it has a superlinear convergence
(not that
this matters much!). Brent's method, which is used in `uniroot', is
a
reliable and fast method, which is why I suggested it in my previous
email.
Having said that, I am not sure about the convergence problem that
you are
having without seeing the actual example.
Ravi.
-Original Message-
From: Adrian Ng [
Sent: Wednesday, August 25, 2010 5:28 PM
To: Ravi Varadhan; r-help@r-project.org
Re: [R] How to obtain the graph of fitted values against one variable after estimation?
On Aug 25, 2010, at 10:46 PM, Le Wang wrote: Hi there, I have a question regarding the plot command after estimation. Specifically, I estimate a model, say regressing y on x and z. And after estimation, I would like to plot the fitted values against x, but I don't need that for z. The following command always gives two graphs, for both variables x and z. plot.np-plot(model, plot.errors.method = asymptotic) My question is, what option should I specify in order to get the graph for x only? Pick a constant value for z and vary x in a dataframe that you offer to the newdata argument of predict. ?predict Then plot those values versus x. I know this is probably a very simple question, but I searched around for a while without any luck. Thank you for your time. -- David Winsemius, MD West Hartford, CT __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Secant Method Convergence (Method to replicate Excel XIRR/IRR)
I have written a general root-finder using the secant method.
secant - function(par, fn, tol=1.e-07, itmax = 100, trace=TRUE, ...) {
# par = a starting vector with 2 starting values
# fn = a function whose first argument is the variable of interest
#
if (length(par) != 2) stop(You must specify a starting parameter vector of
length 2)
p.2 - par[1]
p.1 - par[2]
f - rep(NA, length(par))
f[1] - fn(p.1, ...)
f[2] - fn(p.2, ...)
iter - 1
pchg - abs(p.2 - p.1)
fval - f[2]
if (trace) cat(par: , par, fval: , f, \n)
while (pchg = tol abs(fval) tol iter = itmax) {
p.new - p.2 - (p.2 - p.1) * f[2] / (f[2] - f[1])
pchg - abs(p.new - p.2)
fval - fn(p.new, ...)
p.1 - p.2
p.2 - p.new
f[1] - f[2]
f[2] - fval
iter - iter + 1
if (trace) cat(par: , p.new, fval: , fval, \n)
}
list(par = p.new, value = fval, iter=iter)
}
Now we can use this function to find the zero of your NPV function as follows:
npv - function (irr, cashFlow, times) sum(cashFlow / (1 + irr)^times)
CF - c(-1000,500,500,500,500,500)
dates - c(1/1/2001,2/1/2002,3/1/2003,4/1/2004,5/1/2005,6/1/2006)
cfDate - as.Date(cfDate,format=%m/%d/%Y)
times - as.numeric(difftime(cfDate, cfDate[1], units=days))/365.24
secant(par=c(0,0.1), fn=npv, cashFlow=CF, times=times)
secant(par=c(0,0.1), fn=npv, cashFlow=CF, times=times)
par: 0 0.1 fval: 854.2388 1500
par: 0.232284 fval: 334.7318
par: 0.2990093 fval: 156.9595
par: 0.3579227 fval: 30.59229
par: 0.3721850 fval: 3.483669
par: 0.3740179 fval: 0.08815743
par: 0.3740655 fval: 0.0002613245
par: 0.3740656 fval: 1.966778e-08
$par
[1] 0.3740656
$value
[1] 1.966778e-08
$iter
[1] 8
Hope this helps,
Ravi.
Ravi Varadhan, Ph.D.
Assistant Professor,
Division of Geriatric Medicine and Gerontology
School of Medicine
Johns Hopkins University
Ph. (410) 502-2619
email: rvarad...@jhmi.edu
- Original Message -
From: Ravi Varadhan rvarad...@jhmi.edu
Date: Wednesday, August 25, 2010 10:51 pm
Subject: Re: [R] Secant Method Convergence (Method to replicate Excel XIRR/IRR)
To: Adrian Ng a...@hamiltonlane.com
Cc: r-help@r-project.org r-help@r-project.org
You should use cfDate[1] as the time origin. You cannot use
08-24-2010 as the time origin, since that will yield negative times.
Here is the correct solution.
ANXIRR - function (cashFlow, dates, guess, tol=1.e-04){
npv - function (cashFlow, times, irr) {
n - length(cashFlow)
sum(cashFlow / (1 + irr)^times)
}
if (guess == 0) stop(Initial guess must be strictly greater than 0)
cfDate - as.Date(cfDate,format=%m/%d/%Y)
times - as.numeric(difftime(cfDate, cfDate[1], units=days) /
365.24)
irrprev - c(0)
irr - guess
pvPrev - sum(cashFlow)
pv - npv(cashFlow, times, irr)
eps - abs(pv-pvPrev)
while (eps = tol) {
tmp - irrprev
irrprev - irr
irr - irr - ((irr - tmp) * pv / (pv - pvPrev))
pvPrev - pv
pv - npv(cashFlow, times, irr)
eps - abs(pv - pvPrev)
}
list(irr = irr, npv = pv)
}
CF - c(-1000,500,500,500,500,500)
dates -
c(1/1/2001,2/1/2002,3/1/2003,4/1/2004,5/1/2005,6/1/2006)
ANXIRR(CF, dates, guess=0.1)
ANXIRR(CF, dates, guess=0.1)
$irr
[1] 0.3740656
$npv
[1] 2.102695e-09
Hope this helps,
Ravi.
Ravi Varadhan, Ph.D.
Assistant Professor,
Division of Geriatric Medicine and Gerontology
School of Medicine
Johns Hopkins University
Ph. (410) 502-2619
email: rvarad...@jhmi.edu
- Original Message -
From: Adrian Ng a...@hamiltonlane.com
Date: Wednesday, August 25, 2010 8:33 pm
Subject: RE: [R] Secant Method Convergence (Method to replicate Excel
XIRR/IRR)
To: Ravi Varadhan rvarad...@jhmi.edu
Cc: r-help@r-project.org r-help@r-project.org
Hi Ravi,
Using days and dividing it by 365 effectively converts the number
to
years anyway and allows for the irregular times to be specific to
the
days.
Also, when I replace dates[1] in your line:
times - as.numeric(difftime(dates, dates[1], units=days) /
365.24) with 2010-08-24 I think I am getting some irregular
results.
Effectively, what I was trying to do was match what Excel produced
with its XIRR function. With the example I gave excel returned an
IRR
of ~0.37 (or 37%)
I am still in the process of debugging it...
-Original Message-
From: Ravi Varadhan [
Sent: Wednesday, August 25, 2010 7:24 PM
To: Adrian Ng
Cc: r-help@r-project.org
Subject: RE: [R] Secant Method Convergence (Method to replicate
Excel
XIRR/IRR)
The secant method converges just fine. Your problem might have
occurred due
to improper conversion of dates to elapsed time. You want to
calculate IRR
using year
[R] list of closures
Hi, I wanted to create a list of closures. When I use Map(), mapply(), lapply(), etc., to create this list, it appears that the wrong arguments are being passed to the main function. For example: Main function: adder - function(x) function(y) x + y Creating list of closures with Map(): plus - Map(adder,c(one=1,two=2)) plus$one(1)[1] 3 plus$two(1)[1] 3 Examining what value was bound to x: Map(function(fn) get(x,environment(fn)),plus)$one[1] 2$two[1] 2 This is what I had expected: plus - list(one=adder(1),two=adder(2)) plus$one(1)[1] 2 plus$two(1)[1] 3 Map(function(fn) get(x,environment(fn)),plus)$one[1] 1$two[1] 2 Anyone know what's going on? Thanks much! Stephen [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] list of closures
On Thu, Aug 26, 2010 at 12:04 AM, Stephen T. obsessiv...@hotmail.com wrote:
Hi, I wanted to create a list of closures. When I use Map(), mapply(),
lapply(), etc., to create this list, it appears that the wrong arguments are
being passed to the main function. For example:
Main function:
adder - function(x) function(y) x + y
Creating list of closures with Map():
plus - Map(adder,c(one=1,two=2)) plus$one(1)[1] 3 plus$two(1)[1] 3
Examining what value was bound to x:
Map(function(fn) get(x,environment(fn)),plus)$one[1] 2$two[1] 2
This is what I had expected:
plus - list(one=adder(1),two=adder(2)) plus$one(1)[1] 2 plus$two(1)[1] 3
Map(function(fn) get(x,environment(fn)),plus)$one[1] 1$two[1] 2
Anyone know what's going on? Thanks much!
R uses lazy evaluation of function arguments. Try forcing x:
adder - function(x) { force(x); function(y) x + y }
plus - Map(adder,c(one=1,two=2))
plus$one(1)
[1] 2
plus$two(1)
[1] 3
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] list of closures
Unless for learning R, you should really consider R.oo or proto packages
that may be more convenient for you (but you don't provide enough
context to tell).
Best,
Philippe
On 26/08/10 06:28, Gabor Grothendieck wrote:
On Thu, Aug 26, 2010 at 12:04 AM, Stephen T.obsessiv...@hotmail.com wrote:
Hi, I wanted to create a list of closures. When I use Map(), mapply(),
lapply(), etc., to create this list, it appears that the wrong arguments are
being passed to the main function. For example:
Main function:
adder- function(x) function(y) x + y
Creating list of closures with Map():
plus- Map(adder,c(one=1,two=2)) plus$one(1)[1] 3 plus$two(1)[1] 3
Examining what value was bound to x:
Map(function(fn) get(x,environment(fn)),plus)$one[1] 2$two[1] 2
This is what I had expected:
plus- list(one=adder(1),two=adder(2)) plus$one(1)[1] 2 plus$two(1)[1] 3
Map(function(fn) get(x,environment(fn)),plus)$one[1] 1$two[1] 2
Anyone know what's going on? Thanks much!
R uses lazy evaluation of function arguments. Try forcing x:
adder- function(x) { force(x); function(y) x + y }
plus- Map(adder,c(one=1,two=2))
plus$one(1)
[1] 2
plus$two(1)
[1] 3
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] Secant Method Convergence (Method to replicate Excel XIRR/IRR)
Hi Ravi,
Using days and dividing it by 365 effectively converts the number to years
anyway and allows for the irregular times to be specific to the days.
Also, when I replace dates[1] in your line:
times - as.numeric(difftime(dates, dates[1], units=days) /
365.24) with 2010-08-24 I think I am getting some irregular results.
Effectively, what I was trying to do was match what Excel produced with its
XIRR function. With the example I gave excel returned an IRR of ~0.37 (or 37%)
I am still in the process of debugging it...
-Original Message-
From: Ravi Varadhan [mailto:rvarad...@jhmi.edu]
Sent: Wednesday, August 25, 2010 7:24 PM
To: Adrian Ng
Cc: r-help@r-project.org
Subject: RE: [R] Secant Method Convergence (Method to replicate Excel XIRR/IRR)
The secant method converges just fine. Your problem might have occurred due
to improper conversion of dates to elapsed time. You want to calculate IRR
using year as the time unit, not days.
Here is the secant function (modified to account for irregular times) and
the results for your example:
ANXIRR - function (cashFlow, dates, guess, tol=1.e-04){
npv - function (cashFlow, times, irr) {
n - length(cashFlow)
sum(cashFlow / (1 + irr)^times)
}
if (guess == 0)stop(Initial guess must be strictly greater than 0)
times - as.numeric(difftime(dates, dates[1], units=days) /
365.24)
irrprev - c(0)
irr - guess
pvPrev - sum(cashFlow)
pv - npv(cashFlow, times, irr)
eps - abs(pv-pvPrev)
while (eps = tol) {
tmp - irrprev
irrprev - irr
irr - irr - ((irr - tmp) * pv / (pv - pvPrev))
pvPrev - pv
pv - npv(cashFlow, times, irr)
eps - abs(pv - pvPrev)
}
list(irr = irr, npv = pv)
}
CF - c(-1000,500,500,500,500,500)
dates -
c(1/1/2001,2/1/2002,3/1/2003,4/1/2004,5/1/2005,6/1/2006)
ANXIRR(CF, dates, guess=0.1)
ANXIRR(CF, dates, guess=0.1)
$irr
[1] 0.4106115
$npv
[1] 2.984279e-13
Ravi.
-Original Message-
From: Adrian Ng [mailto:a...@hamiltonlane.com]
Sent: Wednesday, August 25, 2010 6:23 PM
To: Ravi Varadhan
Subject: RE: [R] Secant Method Convergence (Method to replicate Excel
XIRR/IRR)
The forum is kind of slow so I'm just re-sending you the message here:
Hi Ravi,
I'm just trying a fairly simple example:
CFs: -1000,500,500,500,500,500
dates-c(1/1/2001,2/1/2002,3/1/2003,4/1/2004,5/1/2005,6/1/2006)
Thanks a lot for your help.
Adrian
-Original Message-
From: Ravi Varadhan [mailto:rvarad...@jhmi.edu]
Sent: Wednesday, August 25, 2010 5:44 PM
To: Adrian Ng; r-help@r-project.org
Subject: RE: [R] Secant Method Convergence (Method to replicate Excel
XIRR/IRR)
Yes, the secant method (like Newton Raphson) is not guaranteed to converge,
unlike the bisection method, but it has a superlinear convergence (not that
this matters much!). Brent's method, which is used in `uniroot', is a
reliable and fast method, which is why I suggested it in my previous email.
Having said that, I am not sure about the convergence problem that you are
having without seeing the actual example.
Ravi.
-Original Message-
From: Adrian Ng [mailto:a...@hamiltonlane.com]
Sent: Wednesday, August 25, 2010 5:28 PM
To: Ravi Varadhan; r-help@r-project.org
Subject: RE: [R] Secant Method Convergence (Method to replicate Excel
XIRR/IRR)
Hi Ravi,
Thanks for the responses. I was actually trying to calculate IRR based on
unevenly spaced cash flows, and that's why I decided to use the secant
method. I'm not sure if my answer isn't converging because I have some
careless mistake in the code, or if it's simply because unlike the bisection
method, the secant method doesn't 'sandwich' the desired root.
-Original Message-
From: Ravi Varadhan [mailto:rvarad...@jhmi.edu]
Sent: Wednesday, August 25, 2010 5:24 PM
To: Adrian Ng; r-help@r-project.org
Subject: RE: [R] Secant Method Convergence (Method to replicate Excel
XIRR/IRR)
Another approach is to use `uniroot' to find the zero of the NPV function:
npv - function (cashFlow, irr) {
n - length(cashFlow)
sum(cashFlow / (1 + irr)^{0: (n-1)})
}
uniroot(f=npv, interval=c(0,1), cashFlow=cashFlow)
However, there may be situations where there are no real zeros or there are
multiple zeros of the NPV function.
Ravi.
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
Behalf Of Adrian Ng
Sent: Wednesday, August 25, 2010 8:39 AM
To: r-help@r-project.org
Subject: [R] Secant Method Convergence (Method to replicate Excel XIRR/IRR)
Hi,
I am new to R, and as a first exercise, I decided to try to implement an
XIRR function using the secant method. I did a quick search and saw another
posting that used the Bisection method but wanted to see if it was possible
using the secant method.
I would input a Cash Flow and Date vector as well as an initial guess. I
hardcoded today's initial date so I could
Re: [R] creation package
Dear r-help, I took your advice into consideration and i tried to solve my errors. But , when I use the command R CMD check namepackage, I find an error in this file C:/Rpack/namepackage.Rcheck/00check.txt : * using log directory 'C:/Rpack/namepackge.Rcheck' * using R version 2.10.1 (2009-12-14) * using session charset: ISO8859-1 * checking for file 'namepackge/DESCRIPTION' ... OK * checking extension type ... Package * this is package 'namepackge' version '1.0' * checking package name space information ... OK * checking package dependencies ... OK * checking if this is a source package ... OK * checking for executable files ... OK * checking whether package 'namepackge' can be installed ... OK * checking package directory ... OK * checking for portable file names ... OK * checking DESCRIPTION meta-information ... OK * checking top-level files ... OK * checking index information ... OK * checking package subdirectories ... OK * checking R files for non-ASCII characters ... OK * checking R files for syntax errors ... OK * checking whether the package can be loaded ... OK * checking whether the package can be loaded with stated dependencies ... OK * checking whether the name space can be loaded with stated dependencies ... OK * checking for unstated dependencies in R code ... OK * checking S3 generic/method consistency ... OK * checking replacement functions ... OK * checking foreign function calls ... OK * checking R code for possible problems ... OK * checking Rd files ... OK * checking Rd metadata ... OK * checking Rd cross-references ... OK * checking for missing documentation entries ... OK * checking for code/documentation mismatches ... OK * checking Rd \usage sections ... OK * checking data for non-ASCII characters ... OK * checking examples ... ERROR Running examples in 'namepackge-Ex.R' failed. The error most likely occurred in: ### * RT flush(stderr()); flush(stdout()) ### Name: RT ### Title: modeling ### Aliases: RT ### Keywords: programming models ### ** Examples data(d) data - d RT(t(d), c(2,2,2), c(1,2,3),c(V1,V2,V3)) Error in text.default(x, y, labels = labels, col = label.color, family = label.family, : family 'serif' not included in PostScript device Calls: RT- plot - plot.igraph - text - text.default Execution halted Best Regards, 2010/8/19, Michael Dewey i...@aghmed.fsnet.co.uk: At 13:57 19/08/2010, anderson nuel wrote: Dear r-help, I don't use namespace. Well, as I said in my original reply, it would be a good idea to do so. How can I make asia available? Without knowing where asia is that is quite a tough call. How do you access it when you test your code before you try to package it? I think my problem in creating the package in this: I have a singleglobal function (RT) in my package, but inside RT I need to call several other function( comb l,earn_comb, nchoo). When I used package.sekeleton, Iput in lists all the functions(comb l,earn_comb, nchoo,RT),but in the 'man' I left only namepackage-package.Rd and RT.Rd. When I did do this, is it true?? Best Regards 2010/8/18, Michael Dewey mailto:i...@aghmed.fsnet.co.uk i...@aghmed.fsnet.co.uk: At 10:27 18/08/2010, anderson nuel wrote: Dear r-help, No, I find errors in the file C:/Rp/namepackage.Rcheck/00check.txt : * using log directory 'C:/Rpackage/namepackage.Rcheck' * using R version 2.10.1 (2009-12-14) * using session charset: ISO8859-1 * checking for file 'namepackage/DESCRIPTION' ... OK * checking extension type ... Package * this is package 'namepackage' version '1.0' * checking package dependencies ... OK * checking if this is a source package ... OK * checking for executable files ... OK * checking whether package 'namepackage' can be installed ... OK * checking package directory ... OK * checking for portable file names ... OK * checking DESCRIPTION meta-information ... OK * checking top-level files ... OK * checking index information ... OK * checking package subdirectories ... OK * checking R files for non-ASCII characters ... OK * checking R files for syntax errors ... OK * checking whether the package can be loaded ... OK * checking whether the package can be loaded with stated dependencies ... OK * checking for unstated dependencies in R code ... OK * checking S3 generic/method consistency ... OK * checking replacement functions ... OK * checking foreign function calls ... OK * checking R code for possible problems ... NOTE comb: no visible global function definition for 'copy_bloc' comb: no visible global function definition for 'copy_interl' * checking Rd files ... OK * checking Rd metadata ... OK * checking Rd cross-references ... OK * checking for missing documentation entries ... WARNING Undocumented code objects: comb learn_comb nchoo All user-level objects in a package should have documentation entries. See the chapter 'Writing R documentation files' in manual 'Writing R Extensions'. * checking for
Re: [R] Looking for an image (R 64-bit on Linux 64-bit) on Amazon EC2
You have a 64 bit Linux? If so... Dowload the sources Do you mean download Linux kernel source code and then compile it on Amazon EC2? -- View this message in context: http://r.789695.n4.nabble.com/Looking-for-an-image-R-64-bit-on-Linux-64-bit-on-Amazon-EC2-tp2338938p2339072.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Specify non-default libiconv, readline, libpng, tiff etc during R compilation
Hi guys, I am trying to compile 64-bit R 2.11.1 on Solaris 10. I mainly follow the guide in here https://www.initworks.com/wiki/pages/viewpage.action?pageId=6521038 The guide suggests that install the customised libiconv, readline under the designated R installation folder and become the private libraries that are exclusive to that R only. This seems to be a great solution, as I need to compile several versions of R in the same environment, libiconv, readline, libpng etc would be different version too. This concept will allow me to do so without contaminating the global default libraries. But I found that, R ./configure does not locate the customised, provate libiconv, readline etc in the designated location. It still looks from the default one like /usr/lib, /usr/local/lib etc (typically, is older version). I am wondering if any R user has done the similar work and can share some experience. Regards, Derrick [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
