[R] [R-SIG-Finance] regarding bootstrapping... REVISITED
hi Thomas/All,
I went through the thread(
https://stat.ethz.ch/pipermail/r-sig-finance/2006q1/000682.html which
concerns with swaps). Yeah it is correct that i would like to quote both
David and Krishna that the curve interpolation may vary considerably (for
e.g. any polynomial/parametric fit is very different from and curve
fitting whether it is free hand or by NURBS ( complex version of Basis
Splines ZZZzzz). My problem is that i want to know how can i generate spot
curve using bootstrap methodin R.Further, even if you do not have fixed
maturity bonds i.e. when you need to create fictitious or virtual paper of
varied fixed maturities like 1 month, 6 month, 1 year, 5 year, 10 year
. so that you can create a spot curve from the traded points which may
be like as follows for e.g.
Price,Residual Maturity, Coupon, Frequency, Redemption, Basis
98.45,0.53,5%,2,100,4
100.15,1.54,8%,2,100,4
99.56,8.5,4%,1,100,4
and
97.65,20.6,10%,2,100,4
thanks to all
with warm regards
-gaurav
Thomas Steiner [EMAIL PROTECTED]
09-10-06 09:18 PM
To
[EMAIL PROTECTED] [EMAIL PROTECTED]
cc
[EMAIL PROTECTED]
Subject
Re: [R-SIG-Finance] regarding bootstrapping
Gaurav,
some time ago I asked a very similar question. I got some very helpful
answers and some lines of code. Perhaps you want to read this (after
consulting Hull and the others):
https://stat.ethz.ch/pipermail/r-sig-finance/2006q1/000682.html
If you want to see some of my present code, just let me know.
Yours,
Thomas
DISCLAIMER AND CONFIDENTIALITY CAUTION:\ \ This message and ...{{dropped}}
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[R] Goodness of fit measures for model with autocorrelation?
I wonder if anyone can help me with this query? I am undertaking textual content analyses which result in time series data (i.e., relative frequency counts of word categories in successive segments of a larger text or in successive individual texts in a temporal order). In regard to the content category that interests me, previous research has suggested an underlying five-stage psychological process and has proposed a fifth-degree polynomial as a general model for this in texts: cat_freq ~ I(time) + I(time^2) + I(time^3) + I(time^4) + I(time^5) I want to test the adequacy of this model on new data. If I fit the model using lm, I get two measures of goodness of fit: RSq and the significance of the ANOVA. However, autocorrelation is normally present in these data, so I should instead be using gls with an appropriate autocorrelation process such as cor(AR1). However, gls in R does not offer the RSq and ANOVA measures. Is there any way of getting these measures for a gls fit, or is there perhaps even a better way of testing the theoretical model on new data? Many thanks, Andrew Wilson Email: [EMAIL PROTECTED] __ R-help@stat.math.ethz.ch 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] Problem building a DLL from Fortran 90 source under Windows (with solution)
On Mon, 9 Oct 2006, Mstislav Elagin wrote: Hi, All, I have come across a problem building a DLL from .f90 source (R 2.3.1, Windows XP). When using the R CMD SHLIB procedure, the DLL itself was being built, but its export table was empty. Among the output from R CMD SHLIB the following message appeared: c:\mingw\bin\nm.exe: 'a.out': No such file The reason is the empty list of dependencies in the pattern rule for DLLs contained in the file MkRules: %.dll: @$(ECHO) EXPORTS $*.def @$(NM) $^ | $(SED) -n 's/^ [BCDRT] _/ /p' $*.def The dependencies list is constructed in the MakeDll file that contains the following: CFSOURCES=$(wildcard -f *.c *.f *.f90 *.f95) CSOURCES=$(wildcard -f *.c) CXXSOURCES=$(wildcard -f *.cc *.cpp *.C) FSOURCES=$(wildcard -f *.f) FCSOURCES=$(wildcard -f *.f90 *.f95) OBJSA=$(foreach i,$(CSOURCES) $(FSOURCES) $(CXXSOURCES),$(basename $i).o) Observe that the neither CFSOURCES nor FCSOURCES in included in OBJSA, i.e. an object file made from a f90 source never becomes part of dependencies. If I add $(FCSOURCES) to OBJSA, the DLL gets built correctly. I wonder whether there is a reason for not including .f90 (and .f95) sources in the list of dependencies or this can be considered a bug in R. Please use a current version of R (as the posting guide asks). From the CHANGES file for 2.3.1 patched: Objects from Fortran 9X sources no longer need to be declared to R CMD SHLIB/INSTALL via OBJS in Makevars.win. -- Brian D. Ripley, [EMAIL PROTECTED] 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 __ R-help@stat.math.ethz.ch 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] File Encoding Warnings during package installs
On Fri, 29 Sep 2006, Sarosh Jamal wrote: I've been getting the following warning on each package update/install (Rv2.4 on SunOS9): Each? This only applies to a few packages, and is only a warning. Warning message: 'DESCRIPTION' file has 'Encoding' field and re-encoding is not possible. This was the case on R2.3 and now on R2.4 as well - any insight would be much appreciated. Most likely you need to install libiconv and rebuild R: Solaris's (I think you mean Solaris 9 or SunOS 5.9) iconv is not adequate for recent versions of R (in particular, for UTF-8 locales or packages). -- Brian D. Ripley, [EMAIL PROTECTED] 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 __ R-help@stat.math.ethz.ch 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] Rank Function
y-round(c(0.68,0.95,b,c,d),2) rank(y) [1] 3.5 5.0 1.0 2.0 3.5 On 10/10/06, Li Zhang [EMAIL PROTECTED] wrote: Does anyone know why the two rank functions gives different results? I need to use the rank function in a for loop, so the sequence to be ranked is given values in the form of part (1). How can I use assignment like in part (1) to get correct ranks as in part (2)? Thank You Part (1) i-1.94 b-0.95-i c-1.73-i d-2.62-i y-c(0.68,0.95,b,c,d) y 0.68 0.95 -0.99 -0.21 0.68 rank(y) 3 5 1 2 4 Part(2) rank(c(0.68,0.95,-0.99,-0.21,0.68)) 3.5 5.0 1.0 2.0 3.5 __ R-help@stat.math.ethz.ch 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. -- = David Barron Said Business School University of Oxford Park End Street Oxford OX1 1HP [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch 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] Rank Function
Because y[1] and y[5] are not the same in Part1 but are in Part2: # using y from Part1 y[5] - y[1] [1] 1.110223e-16 You could round your numbers to 2 digits, say: rank(round(100*y)) # y is from Part1 [1] 3.5 5.0 1.0 2.0 3.5 On 10/10/06, Li Zhang [EMAIL PROTECTED] wrote: Does anyone know why the two rank functions gives different results? I need to use the rank function in a for loop, so the sequence to be ranked is given values in the form of part (1). How can I use assignment like in part (1) to get correct ranks as in part (2)? Thank You Part (1) i-1.94 b-0.95-i c-1.73-i d-2.62-i y-c(0.68,0.95,b,c,d) y 0.68 0.95 -0.99 -0.21 0.68 rank(y) 3 5 1 2 4 Part(2) rank(c(0.68,0.95,-0.99,-0.21,0.68)) 3.5 5.0 1.0 2.0 3.5 __ R-help@stat.math.ethz.ch 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@stat.math.ethz.ch 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] Rank Function
On Tue, 10 Oct 2006, Gabor Grothendieck wrote: Because y[1] and y[5] are not the same in Part1 but are in Part2: # using y from Part1 y[5] - y[1] [1] 1.110223e-16 Yes, this is FAQ 7.31: Why doesn't R think these numbers are equal? i-1.94 d-2.62-i print(0.68, digits=16) [1] 0.68 print(d, digits=16) [1] 0.6802 identical(d, 0.68) [1] FALSE all.equal(d, 0.68) [1] TRUE with the internal rank function ignoring numeric fuzz. You could round your numbers to 2 digits, say: rank(round(100*y)) # y is from Part1 [1] 3.5 5.0 1.0 2.0 3.5 On 10/10/06, Li Zhang [EMAIL PROTECTED] wrote: Does anyone know why the two rank functions gives different results? I need to use the rank function in a for loop, so the sequence to be ranked is given values in the form of part (1). How can I use assignment like in part (1) to get correct ranks as in part (2)? Thank You Part (1) i-1.94 b-0.95-i c-1.73-i d-2.62-i y-c(0.68,0.95,b,c,d) y 0.68 0.95 -0.99 -0.21 0.68 rank(y) 3 5 1 2 4 Part(2) rank(c(0.68,0.95,-0.99,-0.21,0.68)) 3.5 5.0 1.0 2.0 3.5 __ R-help@stat.math.ethz.ch 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@stat.math.ethz.ch 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. -- Roger Bivand Economic Geography Section, Department of Economics, Norwegian School of Economics and Business Administration, Helleveien 30, N-5045 Bergen, Norway. voice: +47 55 95 93 55; fax +47 55 95 95 43 e-mail: [EMAIL PROTECTED] __ R-help@stat.math.ethz.ch 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] compiling error R-2.4.0
Hi there,
I am trying to install from the source R-2.4.0 on my mac (osx 10.4.8
G5 DP)
The error imply Tcl/Tk.
I install it by all the way I know: darwinport, the Tcl/Tk package
from the dmg available from CRAN but without success.
The PATH is correct.
tclConfig.sh is localised in /opt/local/lib and have those permission:
-rw-r--r-- 2 root admin7K Oct 9 09:49 tclConfig.sh
Here is the error produce by ./configure CFLAGS='-O3 -faltivec -fgcse-
sm -funroll-loops -fstrict-aliasing -fsched-interblock -falign-
loops=16 -falign-jumps=16 -falign-functions=16 -malign-natural -
freorder-blocks -freorder-blocks-and-partition -mpowerpc-gfxopt -
mpowerpc-gpopt -fstrict-aliasing -ftree-vectorize -mtune=G5 -mcpu=G5'
FFLAGS='-O3 -faltivec -fgcse-sm -funroll-loops -fstrict-aliasing -
fsched-interblock -falign-loops=16 -falign-jumps=16 -falign-
functions=16 -malign-natural -freorder-blocks -freorder-blocks-and-
partition -mpowerpc-gfxopt -mpowerpc-gpopt -fstrict-aliasing -ftree-
vectorize -mtune=G5 -mcpu=G5'
[...]
checking for tclConfig.sh... /opt/local/lib/tclConfig.sh
checking for tkConfig.sh... no
checking for tkConfig.sh in library (sub)directories... /usr/local/
lib/tkConfig.sh
checking /opt/local/include/tcl8.4/generic/tcl.h usability... no
checking /opt/local/include/tcl8.4/generic/tcl.h presence... no
checking for /opt/local/include/tcl8.4/generic/tcl.h... no
checking /opt/local/include/tcl8.4/tcl.h usability... no
checking /opt/local/include/tcl8.4/tcl.h presence... no
checking for /opt/local/include/tcl8.4/tcl.h... no
checking /opt/local/include/tcl.h usability... yes
checking /opt/local/include/tcl.h presence... yes
checking for /opt/local/include/tcl.h... yes
checking /usr/local/include/tk8.4/generic/tk.h usability... no
checking /usr/local/include/tk8.4/generic/tk.h presence... no
checking for /usr/local/include/tk8.4/generic/tk.h... no
checking /usr/local/include/tk8.4/tk.h usability... no
checking /usr/local/include/tk8.4/tk.h presence... no
checking for /usr/local/include/tk8.4/tk.h... no
checking /usr/local/include/tcl8.4/tk.h usability... no
checking /usr/local/include/tcl8.4/tk.h presence... no
checking for /usr/local/include/tcl8.4/tk.h... no
checking /usr/local/include/tk.h usability... no
checking /usr/local/include/tk.h presence... yes
configure: WARNING: /usr/local/include/tk.h: present but cannot be
compiled
configure: WARNING: /usr/local/include/tk.h: check for missing
prerequisite headers?
configure: WARNING: /usr/local/include/tk.h: see the Autoconf
documentation
configure: WARNING: /usr/local/include/tk.h: section Present But
Cannot Be Compiled
configure: WARNING: /usr/local/include/tk.h: proceeding with the
preprocessor's result
configure: WARNING: /usr/local/include/tk.h: in the future, the
compiler will take precedence
configure: WARNING: ## --- ##
configure: WARNING: ## Report this to [EMAIL PROTECTED] ##
configure: WARNING: ## --- ##
checking for /usr/local/include/tk.h... yes
checking whether compiling/linking Tcl/Tk code works... no
[...]
Then after it goes all right until the end and then produce a error
in the make step:
make[4]: `libintl.a' is up to date.
gcc -I. -I../../src/include -I../../src/include -I/usr/local/include -
DHAVE_CONFIG_H -fPIC -O3 -faltivec -fgcse-sm -funroll-loops -
fstrict-aliasing -fsched-interblock -falign-loops=16 -falign-jumps=16
-falign-functions=16 -malign-natural -freorder-blocks -freorder-
blocks-and-partition -mpowerpc-gfxopt -mpowerpc-gpopt -fstrict-
aliasing -ftree-vectorize -mtune=G5 -mcpu=G5 -c cpoly.c -o cpoly.o
cpoly.c: In function 'fxshft':
cpoly.c:292: error: expected identifier or '(' before 'unsigned'
[...]
make[3]: *** [cpoly.o] Error 1
make[2]: *** [R] Error 2
make[1]: *** [R] Error 1
make: *** [R] Error 1
I try lot of things but I run out of idea.
Any idea would be really helpful...
David
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[R] Haplo.Stats: error (recursive default argument reference)
Dear colleagues, I face a problem doing haplotype analyses with haplostats: when I use the haplo.em function, the programme gives an error message because of 'recursive default argument reference.' I am not able to figure out what this means. Could you perhaps help me? The full output is the following: library(haplo.stats) datafile.dat-read.table(datafile.dat,header=TRUE) attach(datafile.dat) names(datafile.dat) [1] SEXE AGE OUTCOME1 OUTCOME2 OUTCOME3 OUTCOME4 [7] OUTCOME5 SNP1X SNP1Y SNP2X SNP2Y SNP3X [13] SNP3Y SNP4X SNP4Y SNP5X SNP5Y SNP6X [19] SNP6Y SNP7X SNP7Y block1-datafile.dat[,c(8:21)] loci-c(1,2,3,4,5,6,7) em-haplo.em(geno=block1,locus.label=loci,miss.val=c(0,NA)) Error in .Call(R_lazyLoadDBfetch, key, file, compressed, hook, PACKAGE = base) : recursive default argument reference Thank you very much in advance! Michiel Bos -- Michiel J. Bos, MD MSc PhD-student Neuro-epidemiology Erasmus MC Dept. of Epidemiology and Biostatistics Room Ee2130 PO Box 1738 3000 DR Rotterdam tel: +31 (0)10-4087478 fax: +31 (0)10-4089382 E-mail: [EMAIL PROTECTED] __ R-help@stat.math.ethz.ch 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] copula
Dear R-helper, Is there any thing that I am doing wrong in the following codes: norm.cop - normalCopula(0.5) persp(norm.cop, dcopula) The last command produces what follows Error in persp(x, y, z, xlim, ylim, zlim, theta, phi, r, d, scale, expand, : invalid 'x' argument In addition: Warning messages: 1: no non-missing arguments to min; returning Inf 2: no non-missing arguments to max; returning -Inf 3: no non-missing arguments to min; returning Inf 4: no non-missing arguments to max; returning -Inf 5: is.na() applied to non-(list or vector) in: is.na(x) 6: no non-missing arguments to min; returning Inf 7: no non-missing arguments to max; returning -Inf I remember that I tried the same two commands exactly the way they appear above and I had the surface drawn properly. Does anyone have an idea what this could possibly mean? thanks, Dominique __ R-help@stat.math.ethz.ch 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] Rank Function
Li Zhang wrote: Does anyone know why the two rank functions gives different results? I need to use the rank function in a for loop, so the sequence to be ranked is given values in the form of part (1). How can I use assignment like in part (1) to get correct ranks as in part (2)? Thank You Part (1) i-1.94 b-0.95-i c-1.73-i d-2.62-i y-c(0.68,0.95,b,c,d) y 0.68 0.95 -0.99 -0.21 0.68 rank(y) 3 5 1 2 4 Part(2) rank(c(0.68,0.95,-0.99,-0.21,0.68)) 3.5 5.0 1.0 2.0 3.5 You have specified the exact numbers in part(2). Try part(1) with the following: rank(zapsmall(y)) zapsmall removes tiny floating point errors that are not visible with the default representation of numbers. Jim __ R-help@stat.math.ethz.ch 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] Surfaceplot3D with wireframe
Hi, I want to make a surface3D plot of a landscape. I have cordinates (x, y, z) recorded with a GPS. The datapoints are not evenly distributed within the rectangular area. To do a fast 3D plot I used following. library(grid) library(lattice) v - read.table(clipboard) names(v) - c(x, y, z) wireframe(z ~ x * y, data = v) However I just get an empty box with no 3D grid. I have tried with other data and then it usually works. But I cant get it to work with my data. Here is a sample of my data (x, y, z). I think there is something in my data set that wireframe doesnt like. Any ideas?? Are there better commands to use? (I have tried to change the magnitude of the numbers but that doesnt help) y xz 6659592.078 1585162.959 59.229 6659591.348 1585164.402 59.159 6659590.494 1585163.294 59.205 6659589.466 1585158.968 59.290 6659589.229 1585159.390 59.246 6659588.963 1585158.895 59.201 6659589.475 1585158.617 59.199 6659589.808 1585159.206 59.135 6659592.361 1585161.707 59.447 6659592.387 1585161.319 59.437 6659592.590 1585161.871 59.217 6659591.989 1585162.156 59.163 6659591.934 1585161.489 59.358 6659590.340 1585158.060 59.352 6659590.228 1585157.714 59.138 6659590.957 1585157.947 59.145 6659590.422 1585158.344 59.214 6659589.904 1585158.024 59.175 6659589.877 1585162.033 59.346 6659589.967 1585161.674 59.197 6659590.109 1585162.062 59.205 6659589.778 1585162.572 59.152 6659589.591 1585161.876 59.187 6659591.783 1585156.210 59.561 6659592.164 1585156.764 59.266 6659590.352 1585156.154 59.339 6659590.890 1585154.720 59.278 6659592.520 1585155.023 59.250 6659593.127 1585166.690 59.217 Regards, Gustaf _ Gustaf Granath (PhD student) Dept of Plant Ecology Evolutionary Biology Centre (EBC) Uppsala University Villavägen 14, SE - 752 36 Uppsala, Sweden Tel: +46 (0)18-471 76 88 Fax: 018 55 34 19 Email: [EMAIL PROTECTED] __ R-help@stat.math.ethz.ch 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] read.table() and scientific notation
Dear all, I am having troubles importing values written as scientific notation using read.table(). I'm sure this is a frequent problem, as many people in my lab have this problem as well, so I'm sure that I just have troubles googling for the right solution. The problem is, that, given a file like that: a 1 2e-4 b 2 3e-8 ... the third column gets imported as a factor, or a string if I set the as.is parameter of read.table to TRUE for this column. However, I just want a simple numeric vector :-) I'm sure there is a simple trick for this. If you can point me to the right function, or manual, I think I should be able to find out the details myself. Thanks in advance, January -- January Weiner 3 -+--- Division of Bioinformatics, University of Muenster | Schloßplatz 4 (+49)(251)8321634 | D48149 Münster http://www.uni-muenster.de/Biologie.Botanik/ebb/| Germany __ R-help@stat.math.ethz.ch 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] r 2.4.0
Can someone help interprete the error message below? i was trying to load the package copula from the R command prompt. Error in loadNamespace(package, c(which.lib.loc, lib.loc), keep.source = keep.source) : in 'copula' methods specified for export, but none defined: show, summary, persp, contour Error: package/namespace load failed for 'copula' Thanks, Dominique __ R-help@stat.math.ethz.ch 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] factor levels with umlauts
On Fri, 6 Oct 2006, Christian Bieli wrote: Hi all I have to generate some test data for import in an sql database. The database is meant for web-based data entry in a study taking place in a german speaking region, so factor levels of the variables include umlauts. The variables in the dataframe t.muster are generated e.g. like this: t.muster$screening - rep(ausgefüllt,50) and exported to a .csv file by: write.table(t.muster,MakeMuster041006/MusterDaten.csv, col.names=FALSE,row.names=FALSE,na=,sep=;) After export the factor level including an umlaut of t.muster$screening look like this in the sql-database as well as in an excel spreadsheet: ausgefüllt I think the problem is rather how you imported them. That is the UTF-8 representation of the ausgefüllt viewed in a single-byte locale. R on Windows does not handle UTF-8, so something else has done the conversion. [...] -- Brian D. Ripley, [EMAIL PROTECTED] 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__ R-help@stat.math.ethz.ch 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] solaris 64 build?
Which Solaris is this? I am not seeing any problems with our Solaris 8 systems. Was the compiler built on that exact system? What gcc options were used (-std=gnu99, as recommended, for instance)? It looks like a system header problem: __builtin_isnan ought to be built in, not an external entry point. Since R does not use it, you need to trace through the headers to see why isnan is expanding to it. On Thu, 5 Oct 2006, roger koenker wrote: We have a solaris/sparc machine that has been running an old version of R-devel: Version 2.2.0 Under development (unstable) (2005-06-04 r34577) which was built as m64 from sources. Attempting to upgrade to 2.4.0 the configure step goes ok, but I'm getting early on from make: gcc -m64 -L/opt/sfw/lib/sparcv9 -L/usr/lib/sparcv9 -L/usr/openwin/lib/sparcv9 -L/usr/local/lib -o R.bin Rmain.o CConverters.o CommandLineArgs.o Rdynload.o Renviron.o RNG.o apply.o arithmetic.o apse.o array.o attrib.o base.o bind.o builtin.o character.o coerce.o colors.o complex.o connections.o context.o cov.o cum.o dcf.o datetime.o debug.o deparse.o deriv.o dotcode.o dounzip.o dstruct.o duplicate.o engine.o envir.o errors.o eval.o format.o fourier.o gevents.o gram.o gram-ex.o graphics.o identical.o internet.o iosupport.o lapack.o list.o localecharset.o logic.o main.o mapply.o match.o memory.o model.o names.o objects.o optim.o optimize.o options.o par.o paste.o pcre.o platform.o plot.o plot3d.o plotmath.o print.o printarray.o printvector.o printutils.o qsort.o random.o regex.o registration.o relop.o rlocale.o saveload.o scan.o seq.o serialize.o size.o sort.o source.o split.o sprintf.o startup.o subassign.o subscript.o subset.o summary.o sysutils.o unique.o util.o version.o vfonts.o xxxpr.o mkdtemp.o ../unix/libunix.a ../appl/libappl.a ../nmath/libnmath.a -L../../lib -lRblas -L/usr/local/encap/gf7764-3.4.3+2/lib/gcc/sparc64-sun-solaris2.9/3.4.3 -L/usr/ccs/bin/sparcv9 -L/usr/ccs/bin -L/usr/ccs/lib -L/usr/local/encap/gf7764-3.4.3+2/lib/sparcv9 -L/usr/local/encap/gf7764-3.4.3+2/lib -lg2c -lm -lgcc_s ../extra/zlib/libz.a ../extra/bzip2/libbz2.a ../extra/pcre/libpcre.a ../extra/intl/libintl.a -lreadline -ltermcap -lnsl -lsocket -ldl -lm Undefined first referenced symbol in file __builtin_isnan arithmetic.o ld: fatal: Symbol referencing errors. No output written to R.bin collect2: ld returned 1 exit status I've tried to look at the difference in outcomes in the old R-devel version -- if I touch arithmetic.c there and then type make I get a something almost the same as above except for the following bits that are new to 2.4.0 (this diff is after replacing spaces with linebreaks obviously.) ysidro.econ.uiuc.edu% diff t0 t1 54a55 localecharset.o 81a83 rlocale.o 101a104 mkdtemp.o 104a108,109 -L../../lib -lRblas Has there been some change in the way that Rblas is used, or in isnan? It didn't seem so from a look at arithmetic.c, but this is well beyond me. I hope that someone sees something suspicious, or could point me toward a better diagnostic. Thanks, Roger url:www.econ.uiuc.edu/~rogerRoger Koenker email[EMAIL PROTECTED]Department of Economics vox: 217-333-4558University of Illinois fax: 217-244-6678Champaign, IL 61820 __ R-help@stat.math.ethz.ch 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. -- Brian D. Ripley, [EMAIL PROTECTED] 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 __ R-help@stat.math.ethz.ch 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] read.table() and scientific notation
Your example does not exhibit that behavior when I try it (below). Can you provide a reproducible example following the style shown here: Lines - a 1 2e-4 + b 2 3e-8 DF - read.table(textConnection(Lines)) str(DF) 'data.frame': 2 obs. of 3 variables: $ V1: Factor w/ 2 levels a,b: 1 2 $ V2: int 1 2 $ V3: num 2e-04 3e-08 R.version.string # Windows XP [1] R version 2.4.0 (2006-10-03) On 10/10/06, January Weiner [EMAIL PROTECTED] wrote: Dear all, I am having troubles importing values written as scientific notation using read.table(). I'm sure this is a frequent problem, as many people in my lab have this problem as well, so I'm sure that I just have troubles googling for the right solution. The problem is, that, given a file like that: a 1 2e-4 b 2 3e-8 ... the third column gets imported as a factor, or a string if I set the as.is parameter of read.table to TRUE for this column. However, I just want a simple numeric vector :-) I'm sure there is a simple trick for this. If you can point me to the right function, or manual, I think I should be able to find out the details myself. Thanks in advance, January -- January Weiner 3 -+--- Division of Bioinformatics, University of Muenster | Schloßplatz 4 (+49)(251)8321634 | D48149 Münster http://www.uni-muenster.de/Biologie.Botanik/ebb/| Germany __ R-help@stat.math.ethz.ch 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@stat.math.ethz.ch 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] read.table() and scientific notation
On FC5 Linux: gannet% cat foo.dat a 1 2e-4 b 2 3e-8 gannet% R ... read.table(foo.dat) V1 V2V3 1 a 1 2e-04 2 b 2 3e-08 sapply(read.table(foo.dat), class) V1V2V3 factor integer numeric so please tell us your environment and give a reproducible example. (This is using the OS function strtod, so it might be a deficiency in your OS's implementation of ISO C.) On Tue, 10 Oct 2006, January Weiner wrote: Dear all, I am having troubles importing values written as scientific notation using read.table(). I'm sure this is a frequent problem, as many people in my lab have this problem as well, so I'm sure that I just have troubles googling for the right solution. The problem is, that, given a file like that: a 1 2e-4 b 2 3e-8 ... the third column gets imported as a factor, or a string if I set the as.is parameter of read.table to TRUE for this column. However, I just want a simple numeric vector :-) I'm sure there is a simple trick for this. If you can point me to the right function, or manual, I think I should be able to find out the details myself. Thanks in advance, January -- Brian D. Ripley, [EMAIL PROTECTED] 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 __ R-help@stat.math.ethz.ch 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] factor levels with umlauts
Thanks for your answer. I went round the problem by directly connect to the sql-database instead of generating a .csv file and then upload it. This works perfectly with the RODBC package and is much more suitable, too. Kind regards Christian Prof Brian Ripley schrieb: On Fri, 6 Oct 2006, Christian Bieli wrote: Hi all I have to generate some test data for import in an sql database. The database is meant for web-based data entry in a study taking place in a german speaking region, so factor levels of the variables include umlauts. The variables in the dataframe t.muster are generated e.g. like this: t.muster$screening - rep(ausgefüllt,50) and exported to a .csv file by: write.table(t.muster,MakeMuster041006/MusterDaten.csv, col.names=FALSE,row.names=FALSE,na=,sep=;) After export the factor level including an umlaut of t.muster$screening look like this in the sql-database as well as in an excel spreadsheet: ausgefüllt I think the problem is rather how you imported them. That is the UTF-8 representation of the ausgefüllt viewed in a single-byte locale. R on Windows does not handle UTF-8, so something else has done the conversion. [...] __ R-help@stat.math.ethz.ch 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] r 2.4.0
Dominique Katshunga wrote: Can someone help interprete the error message below? i was trying to load the package copula from the R command prompt. Error in loadNamespace(package, c(which.lib.loc, lib.loc), keep.source = keep.source) : in 'copula' methods specified for export, but none defined: show, summary, persp, contour Error: package/namespace load failed for 'copula' Thanks, Dominique You have to reinstall all packages depending on methods after upgrade to R-2.4.0. For convenience, type: update.packages(checkBuilt = TRUE) Uwe Ligges __ R-help@stat.math.ethz.ch 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@stat.math.ethz.ch 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] read.table() and scientific notation
Oh, thanks, that was hint enough :-) I see it now. I turns that R does not understand e-10 ...which stands for 1e-10 and is produced by some of the bioinformatic applications that I use (notably BLAST). However, R instead of being verbose on that just assumes that the whole column is a string. Is there a way to enforce a specific conversion in R (for example, to be able to see where the errors are?). January -- January Weiner 3 -+--- Division of Bioinformatics, University of Muenster | Schloßplatz 4 (+49)(251)8321634 | D48149 Münster http://www.uni-muenster.de/Biologie.Botanik/ebb/| Germany __ R-help@stat.math.ethz.ch 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] read.table() and scientific notation
I think the colClasses argument to read.table() is what you need. Either that, or explicitly cast columns in the data.frame that's returned by read.table(). That's how you get data types that aren't directly supported by read.table(), like various date formats. - Martin January Weiner wrote: Oh, thanks, that was hint enough :-) I see it now. I turns that R does not understand e-10 ...which stands for 1e-10 and is produced by some of the bioinformatic applications that I use (notably BLAST). However, R instead of being verbose on that just assumes that the whole column is a string. Is there a way to enforce a specific conversion in R (for example, to be able to see where the errors are?). January __ R-help@stat.math.ethz.ch 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] read.table() and scientific notation
On Tue, 10 Oct 2006, January Weiner wrote: Oh, thanks, that was hint enough :-) I see it now. I turns that R does not understand e-10 ...which stands for 1e-10 and is produced by some of the bioinformatic applications that I use (notably BLAST). And that is not standard C notation. However, R instead of being verbose on that just assumes that the whole column is a string. Is there a way to enforce a specific conversion in R (for example, to be able to see where the errors are?). Please study ?read.table, especially 'colClasses'. -- Brian D. Ripley, [EMAIL PROTECTED] 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 __ R-help@stat.math.ethz.ch 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] bimodal / trimodal
Hi, If you want parametric modes you could fit mixtures with different numbers of components and select the best model with your favourite model-selection criterion. Various packages such as mclust and fitdistr may be helpful in doing this, the cluster task view has many more: http://cran.at.r-project.org/src/contrib/Views/Cluster.html I am not aware of non-parametric mode testing routines in R, which ones were you thinking of? Hth, Ingmar From: array chip [EMAIL PROTECTED] Date: Mon, 9 Oct 2006 14:28:32 -0700 (PDT) To: r-help@stat.math.ethz.ch Subject: [R] bimodal / trimodal Hi, is there any package/function that can tell if a numeric vector (continuous data) has a bimodal or trimodal distribution and caluclate the location of the corresponding modes? Thanks __ R-help@stat.math.ethz.ch 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@stat.math.ethz.ch 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] read.table() and scientific notation
A cheeky solution by subverting the coerce mechanism and read.table:
# install a coerce function which can fix the e+10 syntax for an
imaginary class myDouble:
setAs(character, myDouble, function(from)as.double(sub('^(-?)
e','\\11e',from)))
Warning message:
in the method signature for function 'coerce' no definition for
class: “myDouble” in: matchSignature(signature, fdef, where)
# load some data:
Lines - scan(sep=\n, what=)
a 1 3e-8
b 2 1e+10
c 3 e-10
d 4 e+3
e 5 e+1
# process it without using the imaginary class - use a real double
instead to see what happens:
# Note I've used textConnection(Lines) here, where your filename
would go
T - read.table(textConnection(Lines), colClasses=list
(character, integer, double))
Error in scan(file, what, nmax, sep, dec, quote, skip, nlines,
na.strings, :
scan() expected 'a real', got 'e-10'
# process it, specifying the imaginary class myDouble.
T - read.table(textConnection(Lines), colClasses=list
(character, integer, myDouble))
T
V1 V2V3
1 a 1 3e-08
2 b 2 1e+10
3 c 3 1e-10
4 d 4 1e+03
5 e 5 1e+01
lapply(T, class)
$V1
[1] character
$V2
[1] integer
$V3
[1] numeric
Someone's bound to shoot me down for hackery here :-)
-Alex
On 10 Oct 2006, at 11:43, January Weiner wrote:
Dear all,
I am having troubles importing values written as scientific notation
using read.table(). I'm sure this is a frequent problem, as many
people in my lab have this problem as well, so I'm sure that I just
have troubles googling for the right solution.
The problem is, that, given a file like that:
a 1 2e-4
b 2 3e-8
...
the third column gets imported as a factor, or a string if I set the
as.is parameter of read.table to TRUE for this column. However, I just
want a simple numeric vector :-) I'm sure there is a simple trick for
this. If you can point me to the right function, or manual, I think I
should be able to find out the details myself.
Thanks in advance,
January
--
January Weiner 3 -+---
Division of Bioinformatics, University of Muenster | Schloßplatz 4
(+49)(251)8321634 | D48149 Münster
http://www.uni-muenster.de/Biologie.Botanik/ebb/| Germany
__
R-help@stat.math.ethz.ch 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.
__
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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] Fwd: read.table() and scientific notation
note: this e-mail is supposed to precede my coerce hack one. As an example of the other posters mentioning colClasses, with some debugging notes: # create a pretend file for this example Lines - scan(sep=\n, what=) a 1 3e-8 b 2 1e+10 c 3 e-10 d 4 e+3 file - textConnection(Lines) # import as you would a file, and specify the column types. T - read.table(file, colClasses=list(character, integer, double)) Error in scan(file, what, nmax, sep, dec, quote, skip, nlines, na.strings, : scan() expected 'a real', got 'e-10' # decide that's not very helpful. let's just import everything as character: # restarting the file. file - textConnection(Lines) T - read.table(file, colClasses=character) lapply(T, mode) $V1 [1] character $V2 [1] character $V3 [1] character T V1 V2V3 1 a 1 3e-8 2 b 2 1e+10 3 c 3 e-10 4 d 4 e+3 # try the conversion to double: (D-as.double(T$V3)) [1] 3e-08 1e+10NANA Warning message: NAs introduced by coercion # let's see which are bad: T[is.na(D),] V1 V2 V3 3 c 3 e-10 4 d 4 e+3 -Alex On 10 Oct 2006, at 12:17, January Weiner wrote: Oh, thanks, that was hint enough :-) I see it now. I turns that R does not understand e-10 ...which stands for 1e-10 and is produced by some of the bioinformatic applications that I use (notably BLAST). However, R instead of being verbose on that just assumes that the whole column is a string. Is there a way to enforce a specific conversion in R (for example, to be able to see where the errors are?). January -- January Weiner 3 -+--- Division of Bioinformatics, University of Muenster | Schloßplatz 4 (+49)(251)8321634 | D48149 Münster http://www.uni-muenster.de/Biologie.Botanik/ebb/| Germany __ R-help@stat.math.ethz.ch 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@stat.math.ethz.ch 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] Survival Sample Size package / function
Hi, My name is Edison and this is the first time I write to this list. I have been trying without success to find a package or function in R that calculates sample sizes for multiple events survival models. Does anyone have an advice that could help me? best regards, edison _ O Windows Live Spaces é seu espaço na internet com fotos (500 por mês), blog e agora com rede social http://spaces.live.com/ __ R-help@stat.math.ethz.ch 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] cokriging
Hi, when using the script file for cokriging, is there any way to fit cross variogram other than linear, I mean what other options can I put in place of fit.lmc I ll be very appreciated if somebody who know the answer can response to me thanks __ R-help@stat.math.ethz.ch 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] Block comments in R?
On 10/9/2006 10:06 AM, hadley wickham wrote: Current .Rd documentation has some obvious problems: - the parser strips comments out of examples when it runs them - there's no way to put images into the documentation - the keywords aren't much use - there's isn't a definition anywhere of what the format really is, so it's hard to know whether something will work other than by trying it -- and it may break with the next release. - there's no way to link from a help man page to a vignette or other form of documentation. The main thing I don't like about the current system is the amount of duplication - you have to supply a lot of information in the documentation that is also encoded in the function (ie. everything in the codoc check). This makes it unnecessarily painful when updating documentation to reflect minor changes. The large distance between code and documentation also makes it easy to forget to update the documentation when changing the code. These are things that are helped by inline documentation (which I am using) Inline documentation isn't the only way to achieve what you want: better editing tools would help too. I don't know if ESS helps with this, but I could imagine a smart editor would be able to help with the updates in each direction. The big problem with this solution is that there is such a variety of editors in use, and most people are convinced that everyone else has made a big mistake in choosing theirs. However, there was talk in the summer about adding more keywords to .Rd files, to expand in various ways. A keyword that expanded to the arg list of a function might be a nice way to avoid duplication. If we had a way to put argument descriptions into the R file it could do a better job, but it would be tricky: e.g. what if the file arg in two objects to be documented in the same .Rd had different descriptions? I agree with Richard that we don't want to force separate man pages for every documented object. Duncan Murdoch __ R-help@stat.math.ethz.ch 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] eps embedded fonts again
Dear friends, I am sorry, I again rise that boring question about font embedding in EPS figure. I found some discussions on this topic but there were no strait solution. The publisher (AIP) demands submission of separate EPS file for each figure with all fonts embedded in it (even the standard 14 Adobe fonts). As I understand the R does not do this embedding. It inserts only comments what font resources are needed. Then I should use some other software to do this task. Is there a solution for problem within open source environment (without using of proprietary software)? My system is Debian GNU/Linux (testing). The second picture format that AIP journals accept is TIFF, but R has no tiff device to produce that pictures. Does someone know a strait way to convert the EPS file into the high resolution high quality TIFF figure? -- Sincerely yours, Yury Yuryev. __ R-help@stat.math.ethz.ch 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] generate random numbers that sum up to 1
I am trying to generate a vector of random numbers with the constraint that
they have to sum up to one with uniform distribution.
eg. {0.1,0.7,0.2 }
any function to do this? Thanks.
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and provide commented, minimal, self-contained, reproducible code.
Re: [R] generate random numbers that sum up to 1
As I have previously asked, in response to a similar question: Is this a homework problem? cheers, Rolf Turner [EMAIL PROTECTED] __ R-help@stat.math.ethz.ch 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] eps embedded fonts again
On Tue, 10 Oct 2006, Yury Yuryev wrote: Dear friends, I am sorry, I again rise that boring question about font embedding in EPS figure. It gives us an opportunity to point you to the solution included in R 2.4.0. I found some discussions on this topic but there were no strait solution. The publisher (AIP) demands submission of separate EPS file for each figure with all fonts embedded in it (even the standard 14 Adobe fonts). As I understand the R does not do this embedding. It inserts only comments what font resources are needed. Yes, that's the EPS standard. Then I should use some other software to do this task. Is there a solution for problem within open source environment (without using of proprietary software)? My system is Debian GNU/Linux (testing). ghostscript: see the R-News 6/2 article for an R wrapper function embedFonts() to help you with this. The second picture format that AIP journals accept is TIFF, but R has no tiff device to produce that pictures. Does someone know a strait way to convert the EPS file into the high resolution high quality TIFF figure? ghostscript, ImageMagick. -- Brian D. Ripley, [EMAIL PROTECTED] 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 __ R-help@stat.math.ethz.ch 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] generate random numbers that sum up to 1
Rolf Turner [EMAIL PROTECTED] writes: As I have previously asked, in response to a similar question: Is this a homework problem? If so, he has a pretty nasty teacher... -- O__ Peter Dalgaard Øster Farimagsgade 5, Entr.B c/ /'_ --- Dept. of Biostatistics PO Box 2099, 1014 Cph. K (*) \(*) -- University of Copenhagen Denmark Ph: (+45) 35327918 ~~ - ([EMAIL PROTECTED]) FAX: (+45) 35327907 __ R-help@stat.math.ethz.ch 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] generate random numbers that sum up to 1
sun [EMAIL PROTECTED] writes:
I am trying to generate a vector of random numbers with the constraint that
they have to sum up to one with uniform distribution.
eg. {0.1,0.7,0.2 }
any function to do this? Thanks.
Depending on what you mean by uniform, this may be a solution
x-runif(3)
x/sum(x)
[1] 0.1130642 0.4098608 0.4770750
What you can't have is the individual components uniform and
identically distributed because the sum of the means would be 1.5 and
equal to the mean of the sum which is 1...
--
O__ Peter Dalgaard Øster Farimagsgade 5, Entr.B
c/ /'_ --- Dept. of Biostatistics PO Box 2099, 1014 Cph. K
(*) \(*) -- University of Copenhagen Denmark Ph: (+45) 35327918
~~ - ([EMAIL PROTECTED]) FAX: (+45) 35327907
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Re: [R] generate random numbers that sum up to 1
The trick is in defining random in the face of the three apparently
incompatible constraints that the numbers be random, that they sum to 1,
and that they be uniformly distributed. Tell us more. Perhaps extend
Peter D's solution by first deciding (at random) how many components
each solution will have.
Ben Fairbank
-Original Message-
From: [EMAIL PROTECTED]
[mailto:[EMAIL PROTECTED] On Behalf Of sun
Sent: Tuesday, October 10, 2006 8:28 AM
To: r-help@stat.math.ethz.ch
Subject: [R] generate random numbers that sum up to 1
I am trying to generate a vector of random numbers with the constraint
that they have to sum up to one with uniform distribution.
eg. {0.1,0.7,0.2 }
any function to do this? Thanks.
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[R] 2 linux/R environment questions
I was working on windows xp for a while and now i am back to linux ( not sure which kind but I can find out if that helps ) and i just have two questions that are really environment R questions rather than R code questions. 1) i use emacs ( or xemacs if need be ) and i vaguely being able to cut a line or a region in an emacs window and then paste it at my R prompt so that it ran. Could I be mistaken or is there an option I need to set in my environment to be able to do this. It doesn't seem to work and I would prefer to avoid learning ESS. 2) I can make my linux window ( where i am working in R ) quite wide but if i do a print or a head on a dataframe of say 8 columns , it doesn't know that there is all this extra space in the window and it ends up putting the columns on top of each other instead of straight across and it's hard to read the data this way ? Is there a way to make it know that it has a lot more room than it is using. Thanks a lot. Mark This is not an offer (or solicitation of an offer) to buy/sell the securities/instruments mentioned or an official confirmation. Morgan Stanley may deal as principal in or own or act as market maker for securities/instruments mentioned or may advise the issuers. This is not research and is not from MS Research but it may refer to a research analyst/research report. Unless indicated, these views are the author's and may differ from those of Morgan Stanley research or others in the Firm. We do not represent this is accurate or complete and we may not update this. Past performance is not indicative of future returns. For additional information, research reports and important disclosures, contact me or see https://secure.ms.com/servlet/cls. You should not use e-mail to request, authorize or effect the purchase or sale of any security or instrument, to send transfer instructions, or to effect any other transactions. We cannot guarantee that any such requests received via ! e-mail will be processed in a timely manner. This communication is solely for the addressee(s) and may contain confidential information. We do not waive confidentiality by mistransmission. Contact me if you do not wish to receive these communications. In the UK, this communication is directed in the UK to those persons who are market counterparties or intermediate customers (as defined in the UK Financial Services Authority's rules). [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch 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] find weighted group mean
HI, I am trying to figure out an efficient way to calculate group means and associate each entry with it. I made up an example: A = rep(rep(0:1,each=2),3) B = rep(rep(0:1,4),3) C = rep(rep(c(0,0,1,1),2),3) X =cbind(rnorm(24,0,1),runif(24,0,1),A,B,C) A B C [1,] -1.92926469 0.32213127 0 0 0 [2,] -0.83935617 0.77794096 0 1 0 [3,] -1.27799751 0.26276934 1 0 1 Suppose I want to compute a weighted mean of X[,1] by for each group, which is defined by unique vector (A,B,C) with weights are X[,2]. And then add a column for the weighted group mean. How can I do this ? My matrix is fairly large (few thousands) but, luckily, I have only a few factors (10). Thanks a lot, Young. [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch 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] 2 linux/R environment questions
On 10 October 2006 at 10:51, Leeds, Mark \(IED\) wrote: | 1) i use emacs ( or xemacs if need be ) and i vaguely being able to cut | a line or a region in an emacs window and then paste it at my R prompt | so that it ran. Could I be mistaken | or is there an option I need to set in my environment to be able to do | this. It doesn't seem to work and I would prefer to avoid learning ESS. All three questions in section 6 in the R FAQ details R and (X)Emacs. And yes, you do want to learn ESS as that is how (X)Emacs and R play together. | 2) I can make my linux window ( where i am working in R ) quite wide | but if i do a print or a head on a dataframe of say 8 columns , it | doesn't know that there is all this extra space | in the window and it ends up putting the columns on top of each other | instead of straight across and it's hard to read the data this way ? Is | there a way to make it know that it has a lot more room than it is | using. Put options(width=160) # or whatever nb of columns you want into ~/.Rprofile. You may want to set other preferences too, see help(options) help(Startup) Good luck, Dirk -- Hell, there are no rules here - we're trying to accomplish something. -- Thomas A. Edison __ R-help@stat.math.ethz.ch 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] generate random numbers that sum up to 1
On 10/10/2006 9:53 AM, Peter Dalgaard wrote:
sun [EMAIL PROTECTED] writes:
I am trying to generate a vector of random numbers with the constraint that
they have to sum up to one with uniform distribution.
eg. {0.1,0.7,0.2 }
any function to do this? Thanks.
Depending on what you mean by uniform, this may be a solution
x-runif(3)
x/sum(x)
[1] 0.1130642 0.4098608 0.4770750
What you can't have is the individual components uniform and
identically distributed because the sum of the means would be 1.5 and
equal to the mean of the sum which is 1...
Another definition of uniform is to have equal density for all possible
vectors; the Dirichlet distribution with parameters (1,1,1) would give
you that. RSiteSearch finds at least a couple of packages (VGAM,
MCMCpack) that provide simulations from the Dirichlet.
Duncan Murdoch
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Re: [R] find weighted group mean
You can do it directly from the X matrix like so: by(X, as.list(as.data.frame(X[,3:5])), function(R)weighted.mean(R [1], R[2])) A: 0 B: 0 C: 0 [1] 0.4912458 A: 1 B: 0 C: 0 [1] NA A: 0 B: 1 C: 0 [1] -0.2694550 A: 1 B: 1 C: 0 [1] NA A: 0 B: 0 C: 1 [1] NA A: 1 B: 0 C: 1 [1] 0.01102718 A: 0 B: 1 C: 1 [1] NA A: 1 B: 1 C: 1 [1] -0.5622041 Or you can poke about a bit with it first: Z = as.data.frame(X) names(Z) = c(x, w, A, B, C) by(Z[,c(x, w)], list(paste(A, Z$A, B, Z$B, C, Z$C)), function(R)weighted.mean(R$x, R$w)) : A 0 B 0 C 0 [1] 0.4912458 : A 0 B 1 C 0 [1] -0.2694550 : A 1 B 0 C 1 [1] 0.01102718 : A 1 B 1 C 1 [1] -0.5622041 -Alex On 10 Oct 2006, at 16:13, Young Cho wrote: HI, I am trying to figure out an efficient way to calculate group means and associate each entry with it. I made up an example: A = rep(rep(0:1,each=2),3) B = rep(rep(0:1,4),3) C = rep(rep(c(0,0,1,1),2),3) X =cbind(rnorm(24,0,1),runif(24,0,1),A,B,C) A B C [1,] -1.92926469 0.32213127 0 0 0 [2,] -0.83935617 0.77794096 0 1 0 [3,] -1.27799751 0.26276934 1 0 1 Suppose I want to compute a weighted mean of X[,1] by for each group, which is defined by unique vector (A,B,C) with weights are X[,2]. And then add a column for the weighted group mean. How can I do this ? My matrix is fairly large (few thousands) but, luckily, I have only a few factors (10). Thanks a lot, Young. [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch 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@stat.math.ethz.ch 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] find weighted group mean
Here are two ways: f1 - function(i) weighted.mean(X[i,1], X[i,2]) aggregate(list(wmean = 1:nrow(X)), as.data.frame(X[,3:5]), f1) f2 - function(x) data.frame(wmean = weighted.mean(x[,1], x[,2]), x[1, 3:5]) do.call(rbind, by(X, as.data.frame(X[,3:5]), f2)) Also you check out the na.rm= argument in ?weighted.mean which may or may not be relevant to you. On 10/10/06, Young Cho [EMAIL PROTECTED] wrote: HI, I am trying to figure out an efficient way to calculate group means and associate each entry with it. I made up an example: A = rep(rep(0:1,each=2),3) B = rep(rep(0:1,4),3) C = rep(rep(c(0,0,1,1),2),3) X =cbind(rnorm(24,0,1),runif(24,0,1),A,B,C) A B C [1,] -1.92926469 0.32213127 0 0 0 [2,] -0.83935617 0.77794096 0 1 0 [3,] -1.27799751 0.26276934 1 0 1 Suppose I want to compute a weighted mean of X[,1] by for each group, which is defined by unique vector (A,B,C) with weights are X[,2]. And then add a column for the weighted group mean. How can I do this ? My matrix is fairly large (few thousands) but, luckily, I have only a few factors (10). Thanks a lot, Young. [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch 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@stat.math.ethz.ch 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] update.default evaluating in wrong environment?
Hi all, update.default, which is the method used to update lm objects (among others), extracts the call element from it's first argument, updates it, then evaluates it in the parent.frame(). Shouldn't it be evaluated in environment(formula(object)), if that's non-NULL? I ask because I call lm from within a function, and the data argument is a local variable of that function. After that, I can't update the model any more, since the new lm() call (the one evaled in parent.frame()) can't find the data. Best, Martin __ R-help@stat.math.ethz.ch 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] .arch.uni in function call in arch test of vars package
I have been reviewing the arch test of vars package that is based on Engle's paper. R-metrics has this as a wants/todo item. I need an arch test and thought I might accomplish two things at once or at least try. The arch test requires a varest object and I am trying to write one that will use R-metrics arima, arch, Garch objects, or at least a vector. Also the arch function has the following line of code that I can not find the function anywhere archs.resids - apply(resids, 2, function(x) .arch.uni(x, lags.single = lags.single)) does any one know about the function .arch.uni? Thank you Joe __ R-help@stat.math.ethz.ch 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] Surfaceplot3D with wireframe
On 10/10/06, Gustaf Granath [EMAIL PROTECTED] wrote: Hi, I want to make a surface3D plot of a landscape. I have cordinates (x, y, z) recorded with a GPS. The datapoints are not evenly distributed within the rectangular area. To do a fast 3D plot I used following. library(grid) library(lattice) v - read.table(clipboard) names(v) - c(x, y, z) wireframe(z ~ x * y, data = v) However I just get an empty box with no 3D grid. I have tried with other data and then it usually works. But I cant get it to work with my data. Here is a sample of my data (x, y, z). I think there is something in my data set that wireframe doesnt like. Yes, and you have already stated what that is, namely that The datapoints are not evenly distributed within the rectangular area. As it says in the first multiple-sentence paragraph in help page for wireframe: ... In the case of 'wireframe', calculations are based on the assumption that the 'x' and 'y' values are evaluated on a rectangular grid defined by their unique values. The grid points need not be equally spaced. You need to use your favorite smoothing method to evaluate z values on a grid. ?levelplot has an example using loess (the akima package is also popular, although I'm not personally familiar with it). Another alternative is using some form of triangulation (e.g. voronoi tesselation). This should be possible, but is not trivial. -Deepayan Any ideas?? Are there better commands to use? (I have tried to change the magnitude of the numbers but that doesnt help) y xz 6659592.078 1585162.959 59.229 6659591.348 1585164.402 59.159 6659590.494 1585163.294 59.205 6659589.466 1585158.968 59.290 6659589.229 1585159.390 59.246 6659588.963 1585158.895 59.201 6659589.475 1585158.617 59.199 6659589.808 1585159.206 59.135 6659592.361 1585161.707 59.447 6659592.387 1585161.319 59.437 6659592.590 1585161.871 59.217 6659591.989 1585162.156 59.163 6659591.934 1585161.489 59.358 6659590.340 1585158.060 59.352 6659590.228 1585157.714 59.138 6659590.957 1585157.947 59.145 6659590.422 1585158.344 59.214 6659589.904 1585158.024 59.175 6659589.877 1585162.033 59.346 6659589.967 1585161.674 59.197 6659590.109 1585162.062 59.205 6659589.778 1585162.572 59.152 6659589.591 1585161.876 59.187 6659591.783 1585156.210 59.561 6659592.164 1585156.764 59.266 6659590.352 1585156.154 59.339 6659590.890 1585154.720 59.278 6659592.520 1585155.023 59.250 6659593.127 1585166.690 59.217 Regards, Gustaf __ R-help@stat.math.ethz.ch 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] update.default evaluating in wrong environment?
As a workaround use evaluate=FALSE argument to update and
evaluate it yourself fetching the environment from the innards
of the lm structure:
f - function() {
DF - data.frame(y = 1:12, x1 = gl(2, 1, 12), x2 = gl(2,6))
lm(y ~ x1, DF)
}
f.lm - f()
e - attr(terms(f.lm), .Environment)
eval(update(f.lm, formula = y ~ x2, evaluate = FALSE), e)
On 10/10/06, Martin C. Martin [EMAIL PROTECTED] wrote:
Hi all,
update.default, which is the method used to update lm objects (among
others), extracts the call element from it's first argument, updates
it, then evaluates it in the parent.frame(). Shouldn't it be evaluated
in environment(formula(object)), if that's non-NULL?
I ask because I call lm from within a function, and the data argument
is a local variable of that function. After that, I can't update the
model any more, since the new lm() call (the one evaled in
parent.frame()) can't find the data.
Best,
Martin
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Re: [R] update.default evaluating in wrong environment?
Gabor Grothendieck wrote:
As a workaround use evaluate=FALSE argument to update and
evaluate it yourself fetching the environment from the innards
of the lm structure:
f - function() {
DF - data.frame(y = 1:12, x1 = gl(2, 1, 12), x2 = gl(2,6))
lm(y ~ x1, DF)
}
f.lm - f()
e - attr(terms(f.lm), .Environment)
eval(update(f.lm, formula = y ~ x2, evaluate = FALSE), e)
Thanks, or even just:
e - environment(formula(f.lm))
But this was more of a bug report. Is update.default wrong? Should it
be changed? I don't see how evaluating in update's parent environment
would ever be better default behavior than the formula's environment.
- Martin
On 10/10/06, Martin C. Martin [EMAIL PROTECTED] wrote:
Hi all,
update.default, which is the method used to update lm objects (among
others), extracts the call element from it's first argument, updates
it, then evaluates it in the parent.frame(). Shouldn't it be evaluated
in environment(formula(object)), if that's non-NULL?
I ask because I call lm from within a function, and the data argument
is a local variable of that function. After that, I can't update the
model any more, since the new lm() call (the one evaled in
parent.frame()) can't find the data.
Best,
Martin
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and provide commented, minimal, self-contained, reproducible code.
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Re: [R] 2 linux/R environment questions
Mark, you could also try rkward a gui for R from http://rkward.sourceforge.net/ or JGR. I have experimented with both because like you I have been avoiding learning ESS, and emacs just confuses me. rkward seems fine with a few quirks that I need to report as problems, like the object browser not showing the r session objects, and a couple of package functions not working correctly. I hope these are just because I have a RHEL 4 linux system. JGR, when I manually compiled it for my system by specifying my location of java 1.5.0_06, worked fairly nicely. The autoinstall does not find my installation of this java version since Redhat EL4 does not support it. Redhat EL5 will in the near future. Maybe JGR will autoinstall then. These are just GUI/programming options that I have tried. Good Luck Joe Leeds, Mark (IED) wrote: I was working on windows xp for a while and now i am back to linux ( not sure which kind but I can find out if that helps ) and i just have two questions that are really environment R questions rather than R code questions. 1) i use emacs ( or xemacs if need be ) and i vaguely being able to cut a line or a region in an emacs window and then paste it at my R prompt so that it ran. Could I be mistaken or is there an option I need to set in my environment to be able to do this. It doesn't seem to work and I would prefer to avoid learning ESS. 2) I can make my linux window ( where i am working in R ) quite wide but if i do a print or a head on a dataframe of say 8 columns , it doesn't know that there is all this extra space in the window and it ends up putting the columns on top of each other instead of straight across and it's hard to read the data this way ? Is there a way to make it know that it has a lot more room than it is using. Thanks a lot. Mark This is not an offer (or solicitation of an offer) to buy/sell the securities/instruments mentioned or an official confirmation. Morgan Stanley may deal as principal in or own or act as market maker for securities/instruments mentioned or may advise the issuers. This is not research and is not from MS Research but it may refer to a research analyst/research report. Unless indicated, these views are the author's and may differ from those of Morgan Stanley research or others in the Firm. We do not represent this is accurate or complete and we may not update this. Past performance is not indicative of future returns. For additional information, research reports and important disclosures, contact me or see https://secure.ms.com/servlet/cls. You should not use e-mail to request, authorize or effect the purchase or sale of any security or instrument, to send transfer instructions, or to effect any other transactions. We cannot guarantee that any such requests received via ! e-mail will be processed in a timely manner. This communication is solely for the addressee(s) and may contain confidential information. We do not waive confidentiality by mistransmission. Contact me if you do not wish to receive these communications. In the UK, this communication is directed in the UK to those persons who are market counterparties or intermediate customers (as defined in the UK Financial Services Authority's rules). [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch 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@stat.math.ethz.ch 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] update.default evaluating in wrong environment?
It does seem strange to me too. I guess what one could say
is that from an oo viewpoint the internal environment within f
is the object space and update needs to be a method of it which
could be arranged by including update in your lm object:
f - function() {
DF - data.frame(y = 1:12, x1 = gl(2, 1, 12), x2 = gl(2,6))
f.lm - lm(y ~ x1, DF)
f.lm$update - function(object = f.lm, ...) update(object, ...)
f.lm
}
f.lm - f()
f.lm$update(formula = y ~ x2)
On 10/10/06, Martin C. Martin [EMAIL PROTECTED] wrote:
Gabor Grothendieck wrote:
As a workaround use evaluate=FALSE argument to update and
evaluate it yourself fetching the environment from the innards
of the lm structure:
f - function() {
DF - data.frame(y = 1:12, x1 = gl(2, 1, 12), x2 = gl(2,6))
lm(y ~ x1, DF)
}
f.lm - f()
e - attr(terms(f.lm), .Environment)
eval(update(f.lm, formula = y ~ x2, evaluate = FALSE), e)
Thanks, or even just:
e - environment(formula(f.lm))
But this was more of a bug report. Is update.default wrong? Should it
be changed? I don't see how evaluating in update's parent environment
would ever be better default behavior than the formula's environment.
- Martin
On 10/10/06, Martin C. Martin [EMAIL PROTECTED] wrote:
Hi all,
update.default, which is the method used to update lm objects (among
others), extracts the call element from it's first argument, updates
it, then evaluates it in the parent.frame(). Shouldn't it be evaluated
in environment(formula(object)), if that's non-NULL?
I ask because I call lm from within a function, and the data argument
is a local variable of that function. After that, I can't update the
model any more, since the new lm() call (the one evaled in
parent.frame()) can't find the data.
Best,
Martin
__
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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.
__
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[R] Interval pls
Hey R-helpers, Does anybody know of an implementation of interval PLS in R? Thx, Dirk -- Dirk De Becker Work: Kasteelpark Arenberg 30 3001 Heverlee phone: ++32(0)16/32.14.44 fax: ++32(0)16/32.85.90 Home: Waversebaan 90 3001 Heverlee phone: ++32(0)16/23.36.65 [EMAIL PROTECTED] mobile phone: ++32(0)498/51.19.86 Disclaimer: http://www.kuleuven.be/cwis/email_disclaimer.htm __ R-help@stat.math.ethz.ch 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] update.default evaluating in wrong environment?
On Tue, 10 Oct 2006, Martin C. Martin wrote:
Thanks, or even just:
e - environment(formula(f.lm))
But this was more of a bug report. Is update.default wrong? Should it
be changed? I don't see how evaluating in update's parent environment
would ever be better default behavior than the formula's environment.
This is deliberate (although that doesn't necessarily imply that it is
optimal). There already is a bug report (PR#1861) on this issue, so
another one certainly isn't appropriate.
The comments for this bug report give an example taken from the scripts
for MASS chapter 6, doing a bootstrap of a linear model
ph.fun - function(data, i) {
d - data
d$calls - d$fitted + d$res[i]
coef(update(fit, data=d))
}
This is not uncommon for resampling and other perturbations of a model and
requires looking in the calling frame of update().
For other uses such as yours the right place would be the frame where the
model was created. Neither the environment of the formula nor the calling
frame of update() is guaranteed to be the right place and it is easy to
come up with examples where each works and the other doesn't.
There really isn't a trivial solution to this.
-thomas
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Re: [R] update.default evaluating in wrong environment?
Great; thanks for the detailed explanation!
- Martin
Thomas Lumley wrote:
On Tue, 10 Oct 2006, Martin C. Martin wrote:
Thanks, or even just:
e - environment(formula(f.lm))
But this was more of a bug report. Is update.default wrong? Should it
be changed? I don't see how evaluating in update's parent environment
would ever be better default behavior than the formula's environment.
This is deliberate (although that doesn't necessarily imply that it is
optimal). There already is a bug report (PR#1861) on this issue, so
another one certainly isn't appropriate.
The comments for this bug report give an example taken from the scripts
for MASS chapter 6, doing a bootstrap of a linear model
ph.fun - function(data, i) {
d - data
d$calls - d$fitted + d$res[i]
coef(update(fit, data=d))
}
This is not uncommon for resampling and other perturbations of a model
and requires looking in the calling frame of update().
For other uses such as yours the right place would be the frame where
the model was created. Neither the environment of the formula nor the
calling frame of update() is guaranteed to be the right place and it is
easy to come up with examples where each works and the other doesn't.
There really isn't a trivial solution to this.
-thomas
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[R] How to assign a rank to a range of values..
From the following: basin.map - readAsciiGrid(c:/temp/area.asc, colname=area) I have a SpatialGridDataFrame which has the x and y cordinate of a cell, and the drainage area of that cell. There are many cells with a low drainage area (in my case, 33000 with an area of 37.16) and one cell with the highest drainage area (again, in my case, a drainage area of of 80). What I'd like to do, is to rank the drainage area cells based upon the number of times they occur, with a rank of 100 going to the cells with area=37.16, and 1 going to the cell(s) with area=80). There are 6,000 different drainage areas out of 180,000 cells in this grid, so the ranks would have values like 100, 99.01, 57.34, 20, 1.08, 1 and so forth. I have been struggeling with the split, length and rank commands in R, but can't seem to figure out how to attach a new column (or make a new dataset) that has a colums of ranks, or how to calculate the rank. Thanks for any help. Thomas Colson North Carolina State University Department of Forestry and Environmental Resources (919)624-6329 (919)515 3434 [EMAIL PROTECTED] Schedule: www4.ncsu.edu/~tpcolson __ R-help@stat.math.ethz.ch 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] Function that operates on functions: it's ok, but the display isn't
The following code works fine: # g is the function that returns the square of a number g - function(y) y^2 # f1 is a function that takes one function # as argument, and returns another function f1 - function(f) function(x) f(x+1) - f(x) # h(x) is g(x+1) - g(x) or 2x + 1 h - f1(g) # h(1) = 3 # h(2) = 5 # h(3) = 7 So far, so good. But why: h shows: function(x) f(x+1)-f(x) environment: 0264BE84 I don't get it. h should show function(x) g(x+1)-g(x) or something like that. Alberto Monteiro __ R-help@stat.math.ethz.ch 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] update.default evaluating in wrong environment?
On 10/10/2006 1:32 PM, Martin C. Martin wrote:
Gabor Grothendieck wrote:
As a workaround use evaluate=FALSE argument to update and
evaluate it yourself fetching the environment from the innards
of the lm structure:
f - function() {
DF - data.frame(y = 1:12, x1 = gl(2, 1, 12), x2 = gl(2,6))
lm(y ~ x1, DF)
}
f.lm - f()
e - attr(terms(f.lm), .Environment)
eval(update(f.lm, formula = y ~ x2, evaluate = FALSE), e)
Thanks, or even just:
e - environment(formula(f.lm))
But this was more of a bug report. Is update.default wrong? Should it
be changed? I don't see how evaluating in update's parent environment
would ever be better default behavior than the formula's environment.
This is pretty old code: I think the particular line causing the
problem you see was put there in 2000, but it was a workalike for code
that was there already in 1998. This isn't to say that old code is
necessarily right, but changing it could have lots of unwanted side
effects. It's also likely that whatever was in the mind of whoever
wrote the original shows up in other places, so tracking things down to
make a consistent change would be a lot of work. You'd probably want to
read the referenced 1992 Statistical Models definition of the expected
behaviour here pretty closely (and recognize that the 1992 book is about
S, not R, so we might not want to follow it exactly), etc., etc., etc.
All of which is to say that I wouldn't pick up this bug fix unless I was
personally bitten by it: it looks like an awful lot of work. But there
are others who may want to tackle it.
Duncan Murdoch
- Martin
On 10/10/06, Martin C. Martin [EMAIL PROTECTED] wrote:
Hi all,
update.default, which is the method used to update lm objects (among
others), extracts the call element from it's first argument, updates
it, then evaluates it in the parent.frame(). Shouldn't it be evaluated
in environment(formula(object)), if that's non-NULL?
I ask because I call lm from within a function, and the data argument
is a local variable of that function. After that, I can't update the
model any more, since the new lm() call (the one evaled in
parent.frame()) can't find the data.
Best,
Martin
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Re: [R] update.default evaluating in wrong environment?
Perhaps the evaluate= argument could be extended to
allow an environment or an object for which
environment(object) yields an environment, e.g.
update(y ~ x2, evaluate = formula(f.lm))
On 10/10/06, Duncan Murdoch [EMAIL PROTECTED] wrote:
On 10/10/2006 1:32 PM, Martin C. Martin wrote:
Gabor Grothendieck wrote:
As a workaround use evaluate=FALSE argument to update and
evaluate it yourself fetching the environment from the innards
of the lm structure:
f - function() {
DF - data.frame(y = 1:12, x1 = gl(2, 1, 12), x2 = gl(2,6))
lm(y ~ x1, DF)
}
f.lm - f()
e - attr(terms(f.lm), .Environment)
eval(update(f.lm, formula = y ~ x2, evaluate = FALSE), e)
Thanks, or even just:
e - environment(formula(f.lm))
But this was more of a bug report. Is update.default wrong? Should it
be changed? I don't see how evaluating in update's parent environment
would ever be better default behavior than the formula's environment.
This is pretty old code: I think the particular line causing the
problem you see was put there in 2000, but it was a workalike for code
that was there already in 1998. This isn't to say that old code is
necessarily right, but changing it could have lots of unwanted side
effects. It's also likely that whatever was in the mind of whoever
wrote the original shows up in other places, so tracking things down to
make a consistent change would be a lot of work. You'd probably want to
read the referenced 1992 Statistical Models definition of the expected
behaviour here pretty closely (and recognize that the 1992 book is about
S, not R, so we might not want to follow it exactly), etc., etc., etc.
All of which is to say that I wouldn't pick up this bug fix unless I was
personally bitten by it: it looks like an awful lot of work. But there
are others who may want to tackle it.
Duncan Murdoch
- Martin
On 10/10/06, Martin C. Martin [EMAIL PROTECTED] wrote:
Hi all,
update.default, which is the method used to update lm objects (among
others), extracts the call element from it's first argument, updates
it, then evaluates it in the parent.frame(). Shouldn't it be evaluated
in environment(formula(object)), if that's non-NULL?
I ask because I call lm from within a function, and the data argument
is a local variable of that function. After that, I can't update the
model any more, since the new lm() call (the one evaled in
parent.frame()) can't find the data.
Best,
Martin
__
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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.
__
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Re: [R] Function that operates on functions: it's ok, but the display isn't
On 10/10/2006 3:53 PM, Alberto Monteiro wrote: The following code works fine: # g is the function that returns the square of a number g - function(y) y^2 # f1 is a function that takes one function # as argument, and returns another function f1 - function(f) function(x) f(x+1) - f(x) # h(x) is g(x+1) - g(x) or 2x + 1 h - f1(g) # h(1) = 3 # h(2) = 5 # h(3) = 7 So far, so good. But why: h shows: function(x) f(x+1)-f(x) environment: 0264BE84 I don't get it. h should show function(x) g(x+1)-g(x) or something like that. But you defined the result of f1 to be function(x) f(x+1) - f(x), so what it is showing you is correct. This will be evaluated in environment: 0264BE84, where f is defined to be function(y) y^2, which is why h gives the right answers. Remember that R isn't a macro language, function calls aren't text substitutions. Duncan Murdoch Alberto Monteiro __ R-help@stat.math.ethz.ch 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@stat.math.ethz.ch 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] Function that operates on functions: it's ok, but the display isn't
On 10/10/06, Alberto Monteiro [EMAIL PROTECTED] wrote: The following code works fine: # g is the function that returns the square of a number g - function(y) y^2 # f1 is a function that takes one function # as argument, and returns another function f1 - function(f) function(x) f(x+1) - f(x) # h(x) is g(x+1) - g(x) or 2x + 1 h - f1(g) # h(1) = 3 # h(2) = 5 # h(3) = 7 So far, so good. But why: h shows: function(x) f(x+1)-f(x) environment: 0264BE84 Presumably, 'f' takes the value 'g' in this environment, and indeed environment(h)$f function(y) y^2 I don't get it. h should show function(x) g(x+1)-g(x) or something like that. Not that I know much about functions and environments, but printing is ultimately a matter of aesthetics. What do you think should happen when h - f1(function(y) y^2) and you try to print h ? -Deepayan __ R-help@stat.math.ethz.ch 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] Including time in a zoo
dear list, I have these hourly price data over a 20 year period. Among other things, I want to plot them with time on the x-axis. The data is in a data frame, head(OWI.SE.all) Record_Number Name Condition Time_PeriodPrice 1312 WECC-OWI SouthEast All 1/1/2012 Hour: 1 41.21383 2827 WECC-OWI SouthEast All 1/1/2012 Hour: 2 38.38091 4342 WECC-OWI SouthEast All 1/1/2012 Hour: 3 39.97879 5857 WECC-OWI SouthEast All 1/1/2012 Hour: 4 40.14156 7372 WECC-OWI SouthEast All 1/1/2012 Hour: 5 38.21092 8887 WECC-OWI SouthEast All 1/1/2012 Hour: 6 40.09152 And the date-times are in a POSIX object dte1, class(dte1) [1] POSIXt POSIXlt When I tried to create a zoo with dte1 as the index, I got an error, zoo(OWI.SE.all, dte1) Error in order(x, ..., na.last = na.last, decreasing = decreasing) : unimplemented type 'list' in 'orderVector1' I guess that's saying the index of a zoo can't have a time component. Am I correct? If so, is there a way to include an hour component in a zoo object. Second question: If I just attach this dte1 to my data frame as a column named Date and try plotting it as, plot(OWI.SE.all$Date, OWI.SE.all$Price, type=l) I got non-sensical result. How do I plot hourly data in general? TIA. Horace W. Tso __ R-help@stat.math.ethz.ch 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] rarefy a matrix of counts
Hi all, I have a matrix of counts for objects (rows) by samples (columns). I aimed for about 500 counts in each sample (I have about 80 samples) and would now like to rarefy these down to 100 counts in each sample using simple random sampling without replacement. I plan on rarefying several times for each sample. I could do the tedious looping task of making a list of all objects (with its associated identifier) in each sample and then use the wonderful sampling package to select a sub-sample of 100 for each sample and thereby get a logical vector of inclusions. I would then regroup the resulting logical vector into a vector of counts by object, rinse and repeat several times for each sample. Alternately, using the same list, I could create a random index of integers between 1 and the number of objects for a sample (without repeats) and then select those objects from the list. Again, rinse and repeat several time for each sample. Is there a way to directly rarefy a matrix of counts without having to create a list of objects first? I am trying to switch to R from Matlab and am trying to pick up good programming habits from the start. Much appreciation! [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch 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] Data manipulation in columns (with apply?)
R Users, I have written a small simulation model in R which outputs a datafile consisting of ending population sizes for each simulation run (year). The data (see short data example below) is labeled by NUM (simulation run), sim (year) and N (yearly count). After searching the help files and coming up empty (probably because I used the wrong terms) I am appealing for some help for working with the output dataset. What I want to do is for each of the i simulation runs (NUM) I want to 1) take N(t+1)/N(t)=lambda(t) for each year (where in the below example t=1,...,10--total years of the simulation) 2) Sum lambda(t) and divide by t (e.g., output both the mean/se of lambda for each simulation run) 3) Take the mean of the mean(lambda's) (and associated stddev, min, max) over all NUM I think I have to write a function for use within an apply statement, but I am not quite there yet on the learning curve so most of my recent attempts in R have been useful learning experiences of what not to do... Any suggestions/direction is greatly appreciated. Bret Collier TX AM NUM sim N 1 1 466 1 2 450 1 3 473 1 4 531 1 5 515 1 6 502 1 7 471 1 8 460 1 9 458 1 10 434 2 1 289 2 2 356 2 3 387 2 4 440 2 5 457 2 6 466 2 7 467 2 8 449 2 9 387 2 10 394 3 1 367 3 2 400 3 3 476 3 4 508 3 5 478 3 6 501 3 7 513 3 8 505 3 9 492 3 10 465 platform i386-pc-mingw32 arch i386 os mingw32 system i386, mingw32 status major 2 minor 3.0 year 2006 month 04 day24 svn rev37909 language R version.string Version 2.3.0 (2006-04-24) (yeah, I need to update) __ R-help@stat.math.ethz.ch 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] r 2.4.0
Uwe Ligges wrote: Dominique Katshunga wrote: Can someone help interprete the error message below? i was trying to load the package copula from the R command prompt. Error in loadNamespace(package, c(which.lib.loc, lib.loc), keep.source = keep.source) : in 'copula' methods specified for export, but none defined: show, summary, persp, contour Error: package/namespace load failed for 'copula' Thanks, Dominique You have to reinstall all packages depending on methods after upgrade to R-2.4.0. For convenience, type: update.packages(checkBuilt = TRUE) Uwe Ligges Quite similar(?) problem occured after updating to 2.4.0 in a case of flexmix package. The error message: library(flexmix) Error in lazyLoadDBfetch(key, datafile, compressed, envhook) : internal error in R_decompress1 Error: package/namespace load failed for 'flexmix' I was using as you recommended 'update.packages(checkBuilt = TRUE)' and the error still occures. Have any an idea what to do in this case? Thanks! victor __ R-help@stat.math.ethz.ch 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@stat.math.ethz.ch 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@stat.math.ethz.ch 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] Including time in a zoo
Its saying you are trying to pass a list to zoo (a data frame is a list); however, from ?zoo we see zoo takes a first argument of: a numeric vector, matrix or a factor. On 10/10/06, Horace Tso [EMAIL PROTECTED] wrote: dear list, I have these hourly price data over a 20 year period. Among other things, I want to plot them with time on the x-axis. The data is in a data frame, head(OWI.SE.all) Record_Number Name Condition Time_PeriodPrice 1312 WECC-OWI SouthEast All 1/1/2012 Hour: 1 41.21383 2827 WECC-OWI SouthEast All 1/1/2012 Hour: 2 38.38091 4342 WECC-OWI SouthEast All 1/1/2012 Hour: 3 39.97879 5857 WECC-OWI SouthEast All 1/1/2012 Hour: 4 40.14156 7372 WECC-OWI SouthEast All 1/1/2012 Hour: 5 38.21092 8887 WECC-OWI SouthEast All 1/1/2012 Hour: 6 40.09152 And the date-times are in a POSIX object dte1, class(dte1) [1] POSIXt POSIXlt When I tried to create a zoo with dte1 as the index, I got an error, zoo(OWI.SE.all, dte1) Error in order(x, ..., na.last = na.last, decreasing = decreasing) : unimplemented type 'list' in 'orderVector1' I guess that's saying the index of a zoo can't have a time component. Am I correct? If so, is there a way to include an hour component in a zoo object. Second question: If I just attach this dte1 to my data frame as a column named Date and try plotting it as, plot(OWI.SE.all$Date, OWI.SE.all$Price, type=l) I got non-sensical result. How do I plot hourly data in general? TIA. Horace W. Tso __ R-help@stat.math.ethz.ch 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@stat.math.ethz.ch 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] Data manipulation in columns (with apply?)
Does this start to do what you want? x - NUM sim N + 1 1 466 + 1 2 450 + 1 3 473 + 1 4 531 + 1 5 515 + 1 6 502 + 1 7 471 + 1 8 460 + 1 9 458 + 1 10 434 + 2 1 289 + 2 2 356 + 2 3 387 + 2 4 440 + 2 5 457 + 2 6 466 + 2 7 467 + 2 8 449 + 2 9 387 + 2 10 394 + 3 1 367 + 3 2 400 + 3 3 476 + 3 4 508 + 3 5 478 + 3 6 501 + 3 7 513 + 3 8 505 + 3 9 492 + 3 10 465 a - read.table(textConnection(x), header=T) lambda - by(a, a$NUM, function(x) x$N[-1] / x$N[-length(x$N)]) lambda a$NUM: 1 [1] 0.9656652 1.051 1.1226216 0.9698682 0.9747573 0.9382470 0.9766454 0.9956522 0.9475983 -- a$NUM: 2 [1] 1.2318339 1.0870787 1.1369509 1.0386364 1.0196937 1.0021459 0.9614561 0.8619154 1.0180879 -- a$NUM: 3 [1] 1.0899183 1.190 1.0672269 0.9409449 1.0481172 1.0239521 0.9844055 0.9742574 0.9451220 # sum of lambdas sapply(lambda, sum) 123 8.942166 9.357799 9.263944 # mean sapply(lambda, mean) 123 0.993574 1.039755 1.029327 # sd sapply(lambda, sd) 1 2 3 0.05822850 0.10525335 0.08004527 On 10/10/06, Bret Collier [EMAIL PROTECTED] wrote: R Users, I have written a small simulation model in R which outputs a datafile consisting of ending population sizes for each simulation run (year). The data (see short data example below) is labeled by NUM (simulation run), sim (year) and N (yearly count). After searching the help files and coming up empty (probably because I used the wrong terms) I am appealing for some help for working with the output dataset. What I want to do is for each of the i simulation runs (NUM) I want to 1) take N(t+1)/N(t)=lambda(t) for each year (where in the below example t=1,...,10--total years of the simulation) 2) Sum lambda(t) and divide by t (e.g., output both the mean/se of lambda for each simulation run) 3) Take the mean of the mean(lambda's) (and associated stddev, min, max) over all NUM I think I have to write a function for use within an apply statement, but I am not quite there yet on the learning curve so most of my recent attempts in R have been useful learning experiences of what not to do... Any suggestions/direction is greatly appreciated. Bret Collier TX AM NUM sim N 1 1 466 1 2 450 1 3 473 1 4 531 1 5 515 1 6 502 1 7 471 1 8 460 1 9 458 1 10 434 2 1 289 2 2 356 2 3 387 2 4 440 2 5 457 2 6 466 2 7 467 2 8 449 2 9 387 2 10 394 3 1 367 3 2 400 3 3 476 3 4 508 3 5 478 3 6 501 3 7 513 3 8 505 3 9 492 3 10 465 platform i386-pc-mingw32 arch i386 os mingw32 system i386, mingw32 status major 2 minor 3.0 year 2006 month 04 day24 svn rev37909 language R version.string Version 2.3.0 (2006-04-24) (yeah, I need to update) __ R-help@stat.math.ethz.ch 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. -- Jim Holtman Cincinnati, OH +1 513 646 9390 What is the problem you are trying to solve? __ R-help@stat.math.ethz.ch 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] R 2.4.0 can not work in Chinese windows
Dear list, R 2.4.0 can not work in Chinese windows. it's a bug? Thanks! -- Sincerely yours, Liu, jcheng __ R-help@stat.math.ethz.ch 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] Function that operates on functions: it's ok, but the display isn't
Deepayan Sarkar wrote: # f1 is a function that takes one function # as argument, and returns another function f1 - function(f) function(x) f(x+1) - f(x) # h(x) is g(x+1) - g(x) or 2x + 1 h - f1(g) h function(x) f(x+1)-f(x) environment: 0264BE84 Presumably, 'f' takes the value 'g' in this environment, and indeed environment(h)$f function(y) y^2 Ok, so it's possible to recover the meaning of h. It might then be possible to write something like simplify, that would take a function and try to rewrite it using simpler blocks. In this case, simplify h would return the function h1 equal to function(x) (x+1)^2 - x^2. Not that I know much about functions and environments, but printing is ultimately a matter of aesthetics. What do you think should happen when h - f1(function(y) y^2) and you try to print h ? Yesterday, I would expect function(y) (y+1)^2 - y^2, but today I expect the same as above. Alberto Monteiro __ R-help@stat.math.ethz.ch 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 R from Java
Hello All, I am Fuad from Indonesia. I am doing a research using R. I need to access R from Java. Here are my computer specifications : 1. Windows XP 2. Java 1.4.0 3. R 2.2.1 4. rJava_0.4.3.zip ; I download the binary from CRAN 5. Rserve_0.4 ; I download the exe from Rosuda 6. JRI_0.3.5.tar.gz from Rosuda too I have successfully install and try a little code in R. I have install packages rJava from RGUI and it was successful. But when I try to load packages using library(rJava) I have error : unable to load shared library rJava.dll. There is no Jvm.dll. So I copied jvm.dll from JAVA_HOME\bin to R_HOME\bin, and it seems done well until I try to wake up Java Virtual Machine using .jinit() I received error : cannot create JVM. Is there anyway I can call R inside Java code ? Please help me. Thank you Best regards, Fuad [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch 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] R 2.4.0 can not work in Chinese windows
On 10/10/2006 8:30 PM, liu, jcheng wrote: Dear list, R 2.4.0 can not work in Chinese windows. it's a bug? Yes, the translations were mislabelled. R-patched fixes it. Duncan Murdoch __ R-help@stat.math.ethz.ch 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] Question about error of non-numeric argument to binary operator
Hi,
I have the following data and there is no binary operator contained,
however, I still receive the error message when running unitrootTest
function, could someone give me a guidance on it??
readClipboard()
[1] 245246261.5 275.5 307284.5 289313.5
323.75 334 312.5 325305.5 322.5 317310.5
[17] 302301305287277.5 271 271.5 278.5
279271263262262271262257
[33] 251.5 258 252.5 254.5 253251.5 255
253253243238.5 234229.5 230.5 237.5
235.5
[49] 238225227.5 233236.5 236.5 231.5
225221.5 221.5 221.5 221.5 221.5 221.25 221
206
[65] 207.5 196192.5 193.5 186 197.5 193.5
201195183185.5 181.5 179177173
169.5
[81] 173178169173167 158.5 169.5
164145127.5 132131131120120.5
120.25
[97] 120 112.5 114.5 106113111113118.5
131131146.5 133.5 128132 130.5 122
[113] 122.5 123.5 126140132140143
148168162.5 152.5 148144144150.5
151
[129] 156156156152156 153.5 137
135140135138.5 139130131125
125
[145] 121.5 125.5 128128129.5 133129.5
140154.5 167.5 156179 178.5 174188
214
[161] 197.5 181.5 181.5 197.5 191.5 179189.5
184.5 183182
182.5 177191198191180.5
[177] 182183.5 183182 189.5 195208
203194176.5 173173174165.5 163
162.5
[193] 159162.5 171168.5 164158147
149149.5 144141 138.5 138136138
140
[209] 135132.5 130.75 129129.5 126127 128.5
127.5 124117119120.5 122129133
[225] 136137133133127123122
117122126126133 127.5 129130
125
[241] 122126.5 136148147150.5 143.5
138.5 134135
135.75 136.5 132129127.5 118.5
x-readClipboard()
unitrootTest(x)
Error in r[i1] - r[-length(r):-(length(r) - lag + 1)] :
non-numeric argument to binary operator
I also try to do simple code with it, and getting error message as
well, such as
for(j in 1:length(x)){w-x/1}
Error in x/1 : non-numeric argument to binary operator
Thanks for your help!
Yulei
[[alternative HTML version deleted]]
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Re: [R] Question about error of non-numeric argument to binary operator
On Tue, 2006-10-10 at 22:35 -0400, Yulei Gong wrote:
Hi,
I have the following data and there is no binary operator contained,
however, I still receive the error message when running unitrootTest
function, could someone give me a guidance on it??
readClipboard()
[1] 245246261.5 275.5 307284.5 289313.5
323.75 334 312.5 325305.5 322.5 317310.5
[17] 302301305287277.5 271 271.5 278.5
279271263262262271262257
[33] 251.5 258 252.5 254.5 253251.5 255
253253243238.5 234229.5 230.5 237.5
235.5
[49] 238225227.5 233236.5 236.5 231.5
225221.5 221.5 221.5 221.5 221.5 221.25 221
206
[65] 207.5 196192.5 193.5 186 197.5 193.5
201195183185.5 181.5 179177173
169.5
[81] 173178169173167 158.5 169.5
164145127.5 132131131120120.5
120.25
[97] 120 112.5 114.5 106113111113118.5
131131146.5 133.5 128132 130.5 122
[113] 122.5 123.5 126140132140143
148168162.5 152.5 148144144150.5
151
[129] 156156156152156 153.5 137
135140135138.5 139130131125
125
[145] 121.5 125.5 128128129.5 133129.5
140154.5 167.5 156179 178.5 174188
214
[161] 197.5 181.5 181.5 197.5 191.5 179189.5
184.5 183182
182.5 177191198191180.5
[177] 182183.5 183182 189.5 195208
203194176.5 173173174165.5 163
162.5
[193] 159162.5 171168.5 164158147
149149.5 144141 138.5 138136138
140
[209] 135132.5 130.75 129129.5 126127 128.5
127.5 124117119120.5 122129133
[225] 136137133133127123122
117122126126133 127.5 129130
125
[241] 122126.5 136148147150.5 143.5
138.5 134135
135.75 136.5 132129127.5 118.5
x-readClipboard()
unitrootTest(x)
Error in r[i1] - r[-length(r):-(length(r) - lag + 1)] :
non-numeric argument to binary operator
I also try to do simple code with it, and getting error message as
well, such as
for(j in 1:length(x)){w-x/1}
Error in x/1 : non-numeric argument to binary operator
Thanks for your help!
Yulei
Your 'x' is a character vector, not numeric. The tip is that the values
are surrounded by double quotes. Thus, you are trying to use operators
intended for numerics on characters.
For example:
x - c(122, 126.5, 136, 148, 147, 150.5, 143.5)
x
[1] 122 126.5 136 148 147 150.5 143.5
x / 1
Error in x/1 : non-numeric argument to binary operator
is.numeric(x)
[1] FALSE
is.character(x)
[1] TRUE
It's not clear from your post how you read in the data originally. You
should review that process or alternatively, coerce 'x' to numeric:
x.num - as.numeric(x)
x.num
[1] 122.0 126.5 136.0 148.0 147.0 150.5 143.5
x.num / 1
[1] 122.0 126.5 136.0 148.0 147.0 150.5 143.5
HTH,
Marc Schwartz
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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] during fitting of successive datasets, stall crashes iterations
Hi all,
I am trying to do fitting of large sets of timeseries data, and error
messages derail the process when I encounter a dataset that doesn't fit. I'd
like to ignore those misfits and try another equation or move to the next
dataset. Any ideas? (piece of code below)
Thanks,--Warren
##The code looks something like this:
attach(zf)
x - hours
n-length(zf[,2])
for (i in 2:n)
{y - zf[,i]
plot(y ~ x, xlab = Time (h), ylab = Lumi, pch = 15)
# fitting algorithm looks like this:
sine- nls (y ~ baseline + trend*x + amplitude*(sin(2*pi*(x+phase)/period)),
start=list(baseline=4000, amplitude=4000, phase=0, period=24, trend=0),
trace=TRUE)
...
}
--
Warren G. Lewis
[EMAIL PROTECTED]
[[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.
Re: [R] during fitting of successive datasets, stall crashes iterations
See ?try or ?tryCatch. The basic idiom is given here:
https://stat.ethz.ch/pipermail/r-help/2005-May/072035.html
On 10/10/06, Warren [EMAIL PROTECTED] wrote:
Hi all,
I am trying to do fitting of large sets of timeseries data, and error
messages derail the process when I encounter a dataset that doesn't fit. I'd
like to ignore those misfits and try another equation or move to the next
dataset. Any ideas? (piece of code below)
Thanks,--Warren
##The code looks something like this:
attach(zf)
x - hours
n-length(zf[,2])
for (i in 2:n)
{y - zf[,i]
plot(y ~ x, xlab = Time (h), ylab = Lumi, pch = 15)
# fitting algorithm looks like this:
sine- nls (y ~ baseline + trend*x + amplitude*(sin(2*pi*(x+phase)/period)),
start=list(baseline=4000, amplitude=4000, phase=0, period=24, trend=0),
trace=TRUE)
...
}
--
Warren G. Lewis
[EMAIL PROTECTED]
[[alternative HTML version deleted]]
__
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[R] Using integrate() with vectors as boundaries rather than scalars
Hi all
my apologies if above title is misleading, but here is my problem anyways:
I need to evaluate an integral n times. Since I can't get my head around
vectorization as of yet, I have coded it up in a loop, i.e.:
for (i in 1:n)
{
z[i] - integrate(dnorm,x[i],Inf)
}
Since n is quite large in my operation, ~4, I would rather stack all the
elements of x into a vector and then evaluate the integral for each vector
element in x up to infinity. So that the operation would look something like
this:
z - integrate(dnorm,x,Inf)
I am not sure if this works, as I am new to R, however managed the
equivalent in MatLab. Is this possible at all? If yes, your input would be
highly appreciated.
Regards
Tobias
--
View this message in context:
http://www.nabble.com/Using-integrate%28%29-with-vectors-as-boundaries-rather-than-scalars-tf2421074.html#a6749926
Sent from the R help mailing list archive at Nabble.com.
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
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Re: [R] Accessing R from Java [Broadcast]
I've had good luck with Rserve. http://stats.math.uni-augsburg.de/Rserve/
Hope this helps,
Matt
-Original Message-
From: [EMAIL PROTECTED]
[mailto:[EMAIL PROTECTED] On Behalf Of fuad
Sent: Tuesday, October 10, 2006 9:58 PM
To: R-help@stat.math.ethz.ch
Subject: [R] Accessing R from Java [Broadcast]
Hello All,
I am Fuad from Indonesia. I am doing a research using R. I need to access R
from Java.
Here are my computer specifications :
1. Windows XP
2. Java 1.4.0
3. R 2.2.1
4. rJava_0.4.3.zip ; I download the binary from CRAN
5. Rserve_0.4 ; I download the exe from Rosuda
6. JRI_0.3.5.tar.gz from Rosuda too
I have successfully install and try a little code in R.
I have install packages rJava from RGUI and it was successful.
But when I try to load packages using library(rJava) I have error : unable
to load shared library rJava.dll. There is no Jvm.dll.
So I copied jvm.dll from JAVA_HOME\bin to R_HOME\bin, and it seems done well
until I try to wake up Java Virtual Machine using .jinit() I received error
: cannot create JVM.
Is there anyway I can call R inside Java code ?
Please help me. Thank you
Best regards,
Fuad
[[alternative HTML version deleted]]
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Notice: This e-mail message, together with any attachments,...{{dropped}}
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[R] for loop not working in function
I'm trying to write a small function (below) to compute Box Cox
transformations of x for arbitrary values of lambda. I'd like to
specify a range of values for lamba (min,max,step) and am having trouble
getting the for loop to work. Suggestions?
Any pointers to resources for learning to write functions in R for
neophyte programmers? Thanks. --Dale
boxcox - function(x,min,max,step) {
lambda - seq(min,max,step)
s - length(lambda)
for (lambda in 1:s)
n - nrow(x)
if(lambda ==0) xL - log(x) else
xL - ((x^lambda) - 1)/lambda
xLbar - mean(xL)
t1 - (-n/2)* log((1/n)*sum((xL - xLbar)^2))
t2 - (lambda - 1)*sum(log(x))
l= t1 + t2
l
}
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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] Using integrate() with vectors as boundaries rather than scalars
I think I have figured it out myself, would however like to know the opinion
of more experienced coders. Is this a good way of approaching this:
cumdnorm1 - function(x) {integrate(dnorm,x,Inf)}
evalvec - function(x) {lapply(x,cumdnorm1)}
It does return the desired result.
Tobias wrote:
Hi all
my apologies if above title is misleading, but here is my problem anyways:
I need to evaluate an integral n times. Since I can't get my head around
vectorization as of yet, I have coded it up in a loop, i.e.:
for (i in 1:n)
{
z[i] - integrate(dnorm,x[i],Inf)
}
Since n is quite large in my operation, ~4, I would rather stack all
the elements of x into a vector and then evaluate the integral for each
vector element in x up to infinity. So that the operation would look
something like this:
z - integrate(dnorm,x,Inf)
I am not sure if this works, as I am new to R, however managed the
equivalent in MatLab. Is this possible at all? If yes, your input would be
highly appreciated.
Regards
Tobias
--
View this message in context:
http://www.nabble.com/Using-integrate%28%29-with-vectors-as-boundaries-rather-than-scalars-tf2421074.html#a6750170
Sent from the R help mailing list archive at Nabble.com.
__
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and provide commented, minimal, self-contained, reproducible code.
[R] How to solve Different Correlation Results
Howdy Gurus ! I have a different correlation result from the same data. The corridor1 string variable is expressed as a number like the corridor2 number variable. -- levels(corridor1) [1] A B C D E F levels(as.factor(corridor2)) [1] 0 1 2 3 4 -- I have the correlation results followings using cor() function. -- cor(jh1_1, as.factor(corridor1)) [1] 0.01528538 cor(jh1_1, as.factor(corridor2)) [1] -0.4972571 -- I donot know why the above correlation coefficients used the same data are different. They are 0.015 from as.factor(corridor1), -0.497 from as,factor(corridor2). The string variable corridor1 is the same catergory data with the variable corridor2. The difference is that A is replaced with 0, B with 1, C with 2, . Could you tell me why they are different, and which correlation coefficient is correct? Thank in advance, -- Kum-Hoe Hwang, Ph.D.Phone : 82-31-250-3516Email : [EMAIL PROTECTED] [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch 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] Using integrate() with vectors as boundaries rather than scalars
Or just:
lapply(x, integrate, f = dnorm, upper = Inf)
On 10/11/06, Tobias [EMAIL PROTECTED] wrote:
I think I have figured it out myself, would however like to know the opinion
of more experienced coders. Is this a good way of approaching this:
cumdnorm1 - function(x) {integrate(dnorm,x,Inf)}
evalvec - function(x) {lapply(x,cumdnorm1)}
It does return the desired result.
Tobias wrote:
Hi all
my apologies if above title is misleading, but here is my problem anyways:
I need to evaluate an integral n times. Since I can't get my head around
vectorization as of yet, I have coded it up in a loop, i.e.:
for (i in 1:n)
{
z[i] - integrate(dnorm,x[i],Inf)
}
Since n is quite large in my operation, ~4, I would rather stack all
the elements of x into a vector and then evaluate the integral for each
vector element in x up to infinity. So that the operation would look
something like this:
z - integrate(dnorm,x,Inf)
I am not sure if this works, as I am new to R, however managed the
equivalent in MatLab. Is this possible at all? If yes, your input would be
highly appreciated.
Regards
Tobias
--
View this message in context:
http://www.nabble.com/Using-integrate%28%29-with-vectors-as-boundaries-rather-than-scalars-tf2421074.html#a6750170
Sent from the R help mailing list archive at Nabble.com.
__
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and provide commented, minimal, self-contained, reproducible code.
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Re: [R] rarefy a matrix of counts
Hi I am not experienced in Matlab and from your explanation I do not understand what exactly do you want. It seems that you want randomly choose a sample of 100 rows from your martix, what can be achived by sample. DF-data.frame(rnorm(100), 1:100, 101:200, 201:300) DF[sample(1:100, 10),] If you want to do this several times, you need to save your result and than it depends on what you want to do next. One suitable form is list of matrices the other is array and you can use for loop for completing it. HTH Petr On 10 Oct 2006 at 17:40, Brian Frappier wrote: Date sent: Tue, 10 Oct 2006 17:40:47 -0400 From: Brian Frappier [EMAIL PROTECTED] To: r-help@stat.math.ethz.ch Subject:[R] rarefy a matrix of counts Hi all, I have a matrix of counts for objects (rows) by samples (columns). I aimed for about 500 counts in each sample (I have about 80 samples) and would now like to rarefy these down to 100 counts in each sample using simple random sampling without replacement. I plan on rarefying several times for each sample. I could do the tedious looping task of making a list of all objects (with its associated identifier) in each sample and then use the wonderful sampling package to select a sub-sample of 100 for each sample and thereby get a logical vector of inclusions. I would then regroup the resulting logical vector into a vector of counts by object, rinse and repeat several times for each sample. Alternately, using the same list, I could create a random index of integers between 1 and the number of objects for a sample (without repeats) and then select those objects from the list. Again, rinse and repeat several time for each sample. Is there a way to directly rarefy a matrix of counts without having to create a list of objects first? I am trying to switch to R from Matlab and am trying to pick up good programming habits from the start. Much appreciation! [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch 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. Petr Pikal [EMAIL PROTECTED] __ R-help@stat.math.ethz.ch 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.
