Re: [R] Problem with svyvar in survey package
Thank you Sir.Now it is giving results correctly. --Amitava -- View this message in context: http://r.789695.n4.nabble.com/Problem-with-svyvar-in-survey-package-tp3932818p3984390.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Help in ranef Function
Hi I'm getting the intercepts of the Random effects as 0. Please help me to understand why this is coming Zero This is my R code Data- read.csv(C:/FE and RE.csv) Formula=Y~X2+X3+X4 + (1|State) + (0+X5|State) fit=lmer(formula=Formula,data=Data) ranef(fit). My sample Data State YearY X2 X3 X4 X5 X6 S2 196027.8397.5 42.250.778.365.8 S1 196029.9413.3 38.152 79.266.9 S2 196129.8439.2 40.354 79.267.8 S1 196130.8459.7 39.555.379.269.6 S2 196231.2492.9 37.354.777.468.7 S1 196233.3528.6 38.163.780.273.6 S2 196335.6560.3 39.369.880.476.3 S1 196336.4624.6 37.865.983.977.2 S2 196436.7666.4 38.464.585.578.1 S1 196438.4717.8 40.170 93.784.7 S2 196540.4768.2 38.673.2106.1 93.3 S1 196540.3843.3 39.867.8104.8 89.7 S2 196641.8911.6 39.779.1114 100.7 S1 196640.4931.1 52.195.4124.1 113.5 S2 196740.71021.5 48.994.2127.6 115.3 S1 196740.11165.9 58.3123.5 142.9 136.7 S2 196842.71349.6 57.9129.9 143.6 139.2 S1 196844.11449.4 56.5117.6 139.2 132 S2 196946.71575.5 63.7130.9 165.5 132.1 S1 196950.61759.1 61.6129.8 203.3 154.4 S2 197050.11994.2 58.9128 219.6 174.9 S1 197051.72258.1 66.4141 221.6 180.8 S2 197152.92478.7 70.4168.2 232.6 189.4 -- View this message in context: http://r.789695.n4.nabble.com/Help-in-ranef-Function-tp3984436p3984436.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] problem with merging two matrices
Dear all, I hope you can forgive me my stupid questions, but I am a very new R user (; So, this is my question: I have two matrices, those are: matrix1 - matrix(cbind(vector1, vector2), 1,2, dimnames = list(c(values), c(T value, p value))) matrix2 - matrix(dcbind,2,6,dimnames = list(c(x, y), c(Min, 1st qu., Median, Mean, 3rd qu., Max))) Now, I would like to merge them, but I want to receive the following result: Min 1st qu. Median Mean 3rd qu. Max x 3 34 44 4 y 3 3 3 3 3 3 t value p value value 3 3 so both vectors should stand above each other... when I use merge() I dont get this result, also not with cbind or rbind. I neither can make a a data frame of the two matrices. I think that I should use the function array with dim(6,2,2), but I dont know how that is exactly working (I couldn make it working) I would be very glad if you could let me know how to solve this problem. Cheers, maria -- View this message in context: http://r.789695.n4.nabble.com/problem-with-merging-two-matrices-tp3983136p3983136.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Creating barplot using time as X
Hello all, my data looks like this: phaseno / activity / beg / end / phasetime 1 / L / 2010-06-03 19:15:24 / 2010-06-03 21:18:14 / 7370 2 / D / 2010-06-03 21:18:15 / 2010-06-03 21:19:55 / 100 3 / W / 2010-06-03 21:19:56 / 2010-06-03 21:22:47 / 171 4 / D / 2010-06-03 21:22:48 / 2010-06-03 21:23:47 / 59 5 / W / 2010-06-03 21:23:48 / 2010-06-03 21:23:53 / 5 6 / D / 2010-06-03 21:23:54 / 2010-06-03 21:26:18 / 144 7 / W / 2010-06-03 21:26:19 / 2010-06-03 21:32:10 / 351 8 / D / 2010-06-03 21:32:11 / 2010-06-03 21:32:11 / 0 9 / W / 2010-06-03 21:32:12 / 2010-06-03 21:32:12 / 0 10 / D / 2010-06-03 21:32:13 / 2010-06-03 21:32:29 / 16 Please note that phasetime is in seconds and is only a difftime() of beg and end. I want to create a stacked bar chart that gives me the percentage of time spent doing every activity (L,D or W) for every 24h period. Example: Day 1: 20% =L; 40% = W ; 40% = D. In a graph obviously. What I was thinking was dividing my dataframe into several smaller dataframes spanning 24h. I tried with by() but I doubt this is the correct function. If possible, I would also like to separate a row the phasetime of a row if it overlaps two 24 periods. I joined part of my data where *header=TRUE and sep=\t* http://r.789695.n4.nabble.com/file/n3981961/example.txt example.txt Thank you very much for your time PEL -- View this message in context: http://r.789695.n4.nabble.com/Creating-barplot-using-time-as-X-tp3981961p3981961.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] MD5 checksum, mirror Zuerich
Hi there, I just downloaded the newest version of R for Mac from the mirror in Zuerich… checksums do not match. bye, stefan __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Extract Data from Yahoo Finance
Hi R –users, I am using R-2.14.0 on Windows XP. May I request you to assist me for the following please. I like to extract all the fields (example: a : Ask, b : Bid, ……, w : 52-week Range, x: Stock Exchange) for certain period of time, say, 1 October 2011 to 31 October 2011. Is there any R-Package(s) any R- script please? Once again, thank you very much for the time you have given. Regards, Deb __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] How much data can R process?
Would like to know how much data can R process - number of rows and columns? [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Question about Calculation of Cross Product
The function of crossprod in R puzzled me. I would like to calculate the cross product of two vectors. According to my text book, it defines like this: a = (ax, ay, az) b = (bx, by, bz) then, the cross product of a and b is: a X b = (ay*bz-az*by, az*bx-ax*bz, ax*by-ay*bz) It can also write in a determinant format. But the crossprod or tcrossprod function in R appeared not calculate the cross product. Suppose I enter this: a = c(1, 2, 3) b = c(2, 3, 4) ab = crossprod(a, b) ab [,1] [1,] 20 ab = tcrossprod(a, b) ab [,1] [,2] [,3] [1,]234 [2,]468 [3,]69 12 Does any know how to calculate cross product in R. Or I have to write the function myself? [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] text orientation in persp
I am doing a 3D plot using persp {graphics} but cannot figure out how to change
the text direction on the z-axis. The default is for it to read down. Also, I
cannot figure out how to change the position of the text itself.
Many thanks
Andrew
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[R] Problem with R CMD check and the inconsolata font business.
I have just installed R version 2.14.0 and tried to re-build and
re-check some
of the packages that I maintain.
I'm getting a warning (in the process of running R CMD check on my deldir
package):
* checking PDF version of manual ... WARNING
LaTeX errors when creating PDF version.
This typically indicates Rd problems.
LaTeX errors found:
! Font T1/fi4/m/n/10=ec-inconsolata at 10.0pt not loadable: Metric
(TFM) file n
ot found.
to be read again
relax
l.19 ...lf Turner }\email{r.tur...@auckland.ac.nz}
! \textfont 0 is undefined (character h).
\Url@FormatString ...\Url@String \UrlRight \m@th $
l.26 ...\AsIs{}\url{http://www.math.unb.ca/~rolf/}
\AsIs{}
! \textfont 0 is undefined (character t).
\Url@FormatString ...\Url@String \UrlRight \m@th $
l.26 ...\AsIs{}\url{http://www.math.unb.ca/~rolf/}
\AsIs{}
! \textfont 0 is undefined (character t).
.. etc., etc., etc., ad (almost) infinitum.
So there's some problem with a font file not being loadable.
Can anyone tell me what the expletive deleted I should ***do***
about this? I managed to install the inconsolata package from CTAN.
At least I think I managed; I downloaded the *.zip file and then unzipped
it in /usr/share/texmf/tex/latex. And ran texhash. This stopped R CMD
check from complaining that the inconsolata package could not be found,
but then led to the further complaint described above.
So how do I make the required font loadable? What files do I need?
Where do I get them? And where should I put them once I've got them?
I would be grateful for any assistance that can be rendered.
(I know it's just a warning, but I *hate* to ignore warnings!)
cheers,
Rolf Turner
P. S. I'm running Ubuntu; session info, in case it's of any relevance is:
sessionInfo()
R version 2.14.0 (2011-10-31)
Platform: i686-pc-linux-gnu (32-bit)
locale:
[1] LC_CTYPE=en_US.UTF-8 LC_NUMERIC=C
[3] LC_TIME=en_US.UTF-8LC_COLLATE=en_US.UTF-8
[5] LC_MONETARY=en_US.UTF-8LC_MESSAGES=en_US.UTF-8
[7] LC_PAPER=C LC_NAME=C
[9] LC_ADDRESS=C LC_TELEPHONE=C
[11] LC_MEASUREMENT=en_US.UTF-8 LC_IDENTIFICATION=C
attached base packages:
[1] stats graphics grDevices utils datasets methods base
other attached packages:
[1] misc_0.0-15
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Re: [R] nproc parameter in efpFunctional
Thank you. I've understood, that it should be k (number of parameters) separate Brownian bridges. Is it possible, to get such separated/disaggregated processes also in function efp()? (one can take gefp(..., family=gaussian), or construct by myself residuals(lm.model)*X, but still interesting). And on the contrary, how can I get an aggregated Brownian bridge path for all parameters together, similar to efp()$process? It is made in plot.gefp, but only for graphical visualization... Thank you in advance! Julia -- View this message in context: http://r.789695.n4.nabble.com/nproc-parameter-in-efpFunctional-tp3972419p3984605.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] MD5 checksum, mirror Zuerich
On Nov 3, 2011, at 00:56 , Stefan Bienert wrote: Hi there, I just downloaded the newest version of R for Mac from the mirror in Zuerich… checksums do not match. They do for me. Perhaps you caught it in the middle of an update. Please retry. bye, stefan __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Peter Dalgaard, Professor Center for Statistics, Copenhagen Business School Solbjerg Plads 3, 2000 Frederiksberg, Denmark Phone: (+45)38153501 Email: pd@cbs.dk Priv: pda...@gmail.com __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] non-parametric sample size calculation
Hi, I am trying to estimate the sample size needed for the comparison of two groups on a certain measurement, given some previous data at hand. I find that the data collected does not follow a normal distribution, so I would like to use a non-parametric option for sample size calculation. I found the pwr package but I don't think it has this option and on the internet found that http://www.epibiostat.ucsf.edu/biostat/sampsize.html says only PASS allows non-parametric sample size calculations (although the webpage is not updated). Any help would be greatly appreciated Thanks, Dave [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] palettes for the color-blind
Max Kuhn mxkuhn at gmail.com writes: Yes, I was aware of the different type and their respective prevalences. The dichromat package helped me find what I needed. Thanks, Max On Wed, Nov 2, 2011 at 6:38 PM, Thomas Lumley tlumley at uw.edu wrote: On Thu, Nov 3, 2011 at 11:04 AM, Carl Witthoft carl at witthoft.com wrote: Before you pick out a palette: you are aware that their are several different types of color-blindness, aren't you? Yes, but to first approximation there are only two, and they have broadly similar, though not identical impact on choice of color palettes. The dichromat package knows about them, and so does Professor Brewer. More people will be unable to read your graphs due to some kind of gross visual impairment (cataracts, uncorrected focusing problems, macular degeneration, etc) than will have tritanopia or monochromacy. -thomas Sorry to come into this late, but I was travelling. As indicated, the dichromat package will give you an excellent first order approximation as to what works or doesn't for the 3 types of congenital dichromacies, As indicated by THomas, the two most prevalent varieties, protanopia and deuteranopia, result in similar confusion axes and the third, tritanopia, is relatively rare, except in eye disease. That said, the most prevalent color deficiencies are not the dichromacies but the anomalous trichromacies. These will not necessarily lead to losses in chromatic discrimination but just shifts (i.e., one might see as orange or green what a normal trichromat sees as yellow). About 20 years ago, I was involved in an attempt to develop guidelines (or rules of thumb, at least) for display design for color deficient observers, that did not require any deep understanding of colorimetry. The distillation of this effort can be found here, for what it is worth;: http://www.lighthouse.org/accessibility/design/ accessible-print-design/effective-color-contrast The most important point, I think, was to make sure that there was a sufficient luminance contrast difference, so that in the absence of the capacity to make a chromatic discrimination, the differences would still be detectable. The principles necessary for optimizing color choices in a scatterplot will certainly be more complex, however. In this light (no pun intended), I would draw your attention to the seminal work of Berniece Rogowitz at IBM: http://www.research.ibm.com/people/l/lloydt/ color/color.HTM http://www.research.ibm.com/dx/proceedings/ pravda/truevis.htm who was (is) quite concerned with this issue, as well, as the excellent article by Zeileis, Hornik and Murrell http://statmath.wu.ac.at/~zeileis/papers/ Zeileis+Hornik+Murrell-2009.pdf HTH, Ken -- Ken Knoblauch Inserm U846 Stem-cell and Brain Research Institute Department of Integrative Neurosciences 18 avenue du Doyen Lépine 69500 Bron France tel: +33 (0)4 72 91 34 77 fax: +33 (0)4 72 91 34 61 portable: +33 (0)6 84 10 64 10 http://www.sbri.fr/members/kenneth-knoblauch.html __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] How much data can R process?
On 03/11/11 15:59, Nicholay Anne Caumeran wrote: Would like to know how much data can R process - number of rows and columns? How long is a piece of string? cheers, Rolf Turner __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] nproc parameter in efpFunctional
On Thu, 3 Nov 2011, bonda wrote: Thank you. I've understood, that it should be k (number of parameters) separate Brownian bridges. Well, if you use a process based on OLS residuals, you have always a one-dimensional process even though your model has k parameters. Hence, the two parameters are really conceptually different.. Is it possible, to get such separated/disaggregated processes also in function efp()? (one can take gefp(..., family=gaussian), or construct by myself residuals(lm.model)*X, but still interesting). Some processes that efp() computes are always 1-dimensional (namely those based on residuals) while some are k-dimensional (namely the estimates-based processes) and some are (k+1)-dimensional (the score-based processes). gefp() generalizes this concept and lets you construct the fluctuation processes fairly flexibly. And on the contrary, how can I get an aggregated Brownian bridge path for all parameters together, similar to efp()$process? It is made in plot.gefp, but only for graphical visualization... For gefp objects all aggregation is done by the efpFunctional employed. But this is really described in a fair amount of detail in the accompanying papers. Specifically, for gefp/efpFunctional in the 2006 CSDA paper. Thank you in advance! Julia -- View this message in context: http://r.789695.n4.nabble.com/nproc-parameter-in-efpFunctional-tp3972419p3984605.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] array manipulation
You seem to be looking for 'aperm'.
There is a chapter in 'S Poetry' (available
on http://www.burns-stat.com) that talks
about working with higher dimensional
arrays. I don't think any changes need
to be made for R.
On 02/11/2011 16:16, Simone Salvadei wrote:
Hello,
I'm at the very beginning of the learning process of this language.
Sorry in advance for the (possible but plausible) stupidity of my question.
I would like to find a way to permute the DIMENSIONS of an array.
Something that sounds like the function permute() in matlab.
Given an array C of dimensions c x d x T , for instance, the command
permute(C, [2 1 3])
would provide (in Matlab) an array very similar to C, but this time each
one of the T matrices c x d has changed into its transposed.
Any alternatives to the following (and primitive) 'for' cycle?
*# (previously defined) phi=array with dimensions c(c,d,T)*
*
*
*temp=array(0,dim=c(c,d,T))*
* for(i in 1:T)*
* {*
* temp[,,i]=t(phi[,,i])*
* }*
* phi=temp*
*
*
Thank you very much!
S
--
Patrick Burns
pbu...@pburns.seanet.com
twitter: @portfolioprobe
http://www.portfolioprobe.com/blog
http://www.burns-stat.com
(home of 'Some hints for the R beginner'
and 'The R Inferno')
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Re: [R] How much data can R process?
On 11/03/2011 08:20 PM, Rolf Turner wrote: On 03/11/11 15:59, Nicholay Anne Caumeran wrote: Would like to know how much data can R process - number of rows and columns? How long is a piece of string? Around 10e-35 meters. Jim __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] RGoogleTrends error in getGTrends
Hi all, I've just installed RGoogleTrends Version:0.2-1 (after compiling it for windows). And when running the most basic command I get the following error: ans = getGTrends(coupon) Error in curlPerform(url = url, curl = curl, .opts = .opts) : embedded nul in string: 'fffeY' In addition: Warning message: RS-DBI driver warning: (closing pending result sets before closing this connection) Am I missing something, or is this due to some change in Rgoogletrends website? p.s: any chance there is a Rgoogleinsights alternative? (for http://www.google.com/insights/search/) I wasn't able to find any. Thanks in advance, Tal Contact Details:--- Contact me: tal.gal...@gmail.com | 972-52-7275845 Read me: www.talgalili.com (Hebrew) | www.biostatistics.co.il (Hebrew) | www.r-statistics.com (English) -- [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] mysterious warning message regarding bytecode...
On 02.11.2011 23:51, Justin Haynes wrote: While running a long script which source()s other scripts I get the following warning: Warning message: In t(object$S[[1]]) : bytecode version mismatch; using eval Are you using byte compiled code from some package or is the source code byte compiled at runtime? If the former, I guess the packages was installed under another version of R than the version of R you are using. Anyway, nothing to reproduce here nor a suficient description what you really did. Best, Uwe Ligges I cannot replicate it if I run the sourced files line by line though... What is that error? And do I care about it? It doesn't seem to affect my output as far as I can tell. Thanks! Justin sessionInfo() R version 2.13.2 (2011-09-30) Platform: x86_64-pc-linux-gnu (64-bit) locale: [1] LC_CTYPE=en_US.UTF-8 LC_NUMERIC=C LC_TIME=en_US.UTF-8LC_COLLATE=en_US.UTF-8 [5] LC_MONETARY=C LC_MESSAGES=en_US.UTF-8 LC_PAPER=en_US.UTF-8 LC_NAME=C [9] LC_ADDRESS=C LC_TELEPHONE=C LC_MEASUREMENT=en_US.UTF-8 LC_IDENTIFICATION=C attached base packages: [1] grid stats graphics grDevices utils datasets methods base other attached packages: [1] mgcv_1.7-9stringr_0.5 RPostgreSQL_0.2-0 biglm_0.8 DBI_0.2-5 doMC_1.2.3multicore_0.1-7 [8] foreach_1.3.2 codetools_0.2-8 iterators_1.0.5 cairoDevice_2.19 pixmap_0.4-11 gridExtra_0.8.5 splancs_2.01-29 [15] sp_0.9-91 ellipse_0.3-5 ggplot2_0.8.9 proto_0.3-9.2 reshape_0.8.4 plyr_1.6 MASS_7.3-14 loaded via a namespace (and not attached): [1] compiler_2.13.2 digest_0.5.1lattice_0.19-33 Matrix_1.0-1nlme_3.1-102 __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] problem with merging two matrices
Hi, On Wednesday, November 2, 2011, flokke flo...@live.de wrote: Dear all, I hope you can forgive me my stupid questions, but I am a very new R user (; So, this is my question: I have two matrices, those are: matrix1 - matrix(cbind(vector1, vector2), 1,2, dimnames = list(c(values), c(T value, p value))) matrix2 - matrix(dcbind,2,6,dimnames = list(c(x, y), c(Min, 1st qu., Median, Mean, 3rd qu., Max))) Now, I would like to merge them, but I want to receive the following result: Min 1st qu. Median Mean 3rd qu. Max x 3 34 44 4 y 3 3 3 3 3 3 t value p value value 3 3 so both vectors should stand above each other... when I use merge() I dont get this result, also not with cbind or rbind. I neither can make a a data frame of the two matrices. I think that I should use the function array with dim(6,2,2), but I dont know how that is exactly working (I couldn make it working) What are you trying to do, really? You have a matrix with one row and two columns, and a matrix with two rows and six columns. Those just don't fit together. If you're. trying to work wit them in a different form, you'll. need to give more detail. If you're just trying to display them, then print them sequentially. Otherwise, I'm perplexed. Sarah I would be very glad if you could let me know how to solve this problem. Cheers, maria View this message in context: http://r.789695.n4.nabble.com/problem-with-merging-two-matrices-tp3983136p3983136.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Sarah Goslee http://www.stringpage.com http://www.sarahgoslee.com http://www.functionaldiversity.org [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Error: serialization is too large to store in a raw vector
On 11/02/2011 03:37 PM, Alaios wrote: Dear all, I have quite large code (with lapply and mclapply) and I am getting the following error. Error: serialization is too large to store in a raw vector Is it possible to ask from R to extend the Error messages with more details? I would like to see where this problem exists. This is likely from the return value of mclapply's FUN: parallel:::sendMaster tries to serialize it, and fails. serialize(integer(.Machine$integer.max / 4), NULL, TRUE) do further data reduction before trying to return the results (probably a parallel 'best practices' anyway). Neither traceback() nor options(error=recover) deal gracefully with mclapply errors like this. Hope that helps, Martin B.R Alex [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Computational Biology Fred Hutchinson Cancer Research Center 1100 Fairview Ave. N. PO Box 19024 Seattle, WA 98109 Location: M1-B861 Telephone: 206 667-2793 __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] grep fixed (?) in 2.14
#This is probably due to my incomplete understanding of grep, but the below code has been working for some time to #search for .R with anything in front of it and return a list of scripts to source. Likely, the syntax for the #grep statement has been wrong all along. scripts2source - (c(/home/ssefick/R_scripts/Convert_package.R, /home/ssefick/R_scripts/Convert_R_CODE, /home/ssefick/R_scripts/CV.R, /home/ssefick/R_scripts/cvs.out.R, /home/ssefick/R_scripts/database_connect, /home/ssefick/R_scripts/database_connect_package.R, /home/ssefick/R_scripts/exit_db.R, /home/ssefick/R_scripts/exit.R, /home/ssefick/R_scripts/hourly_zoo.R, /home/ssefick/R_scripts/model_diag.R, /home/ssefick/R_scripts/not_numeric.R, /home/ssefick/R_scripts/num_ecol_package.R, /home/ssefick/R_scripts/NumEcolR_scripts, /home/ssefick/R_scripts/old_scripts_DELETE_AFTER_DECEMBER, /home/ssefick/R_scripts/only_numeric_dataframe.R, /home/ssefick/R_scripts/only_numeric.R, /home/ssefick/R_scripts/PCA.ve.R, /home/ssefick/R_scripts/poster_ggplot2_theme.R, /home/ssefick/R_scripts/pressure_transducer_package.R, /home/ssefick/R_scripts/Pressure_Transducer_R_CODE, /home/ssefick/R_scripts/publication_ggplot2_theme.R, /home/ssefick/R_scripts/r2test.R, /home/ssefick/R_scripts/recession_constant_package.R, /home/ssefick/R_scripts/recession_constant_R_CODE, /home/ssefick/R_scripts/serdp_name_split.R, /home/ssefick/R_scripts/setup_R.R, /home/ssefick/R_scripts/USGS.R)) scripts2source[grep(*.R, scripts2source)] #here is my problem; I would like these to be removed. scripts2source[c(2,5,13,14,20,24)] #Thanks for all of your help in advance #kindest regards, #Stephen Sefick __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] How much data can R process?
At 02:59 03/11/2011, Nicholay Anne Caumeran wrote: Would like to know how much data can R process - number of rows and columns? Nicholay, You are going to have to give us a lot more information before anyone can even attempt to answer this. Assuming you have tried to analyse your data and failed for some reason you might want to look at the High Performance Computing task view on CRAN and examine the section entitled large memory and out of memory data. [[alternative HTML version deleted]] Michael Dewey i...@aghmed.fsnet.co.uk http://www.aghmed.fsnet.co.uk/home.html __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] grep fixed (?) in 2.14
Hi, . and * both mean something different in regular expressions than in command-line wildcards. You need: grep(.*\\.R$, scripts2source) which parses to .- any character *- any number of times \\. - an actual . escaped as R requires R - the R denoting a script $ - at the end of the string grep(.*\\.R$, scripts2source) [1] 1 3 4 6 7 8 9 10 11 12 15 16 17 18 19 21 22 23 25 26 27 Sarah On Thu, Nov 3, 2011 at 8:54 AM, Stephen Sefick sas0...@auburn.edu wrote: #This is probably due to my incomplete understanding of grep, but the below code has been working for some time to #search for .R with anything in front of it and return a list of scripts to source. Likely, the syntax for the #grep statement has been wrong all along. scripts2source - (c(/home/ssefick/R_scripts/Convert_package.R, /home/ssefick/R_scripts/Convert_R_CODE, /home/ssefick/R_scripts/CV.R, /home/ssefick/R_scripts/cvs.out.R, /home/ssefick/R_scripts/database_connect, /home/ssefick/R_scripts/database_connect_package.R, /home/ssefick/R_scripts/exit_db.R, /home/ssefick/R_scripts/exit.R, /home/ssefick/R_scripts/hourly_zoo.R, /home/ssefick/R_scripts/model_diag.R, /home/ssefick/R_scripts/not_numeric.R, /home/ssefick/R_scripts/num_ecol_package.R, /home/ssefick/R_scripts/NumEcolR_scripts, /home/ssefick/R_scripts/old_scripts_DELETE_AFTER_DECEMBER, /home/ssefick/R_scripts/only_numeric_dataframe.R, /home/ssefick/R_scripts/only_numeric.R, /home/ssefick/R_scripts/PCA.ve.R, /home/ssefick/R_scripts/poster_ggplot2_theme.R, /home/ssefick/R_scripts/pressure_transducer_package.R, /home/ssefick/R_scripts/Pressure_Transducer_R_CODE, /home/ssefick/R_scripts/publication_ggplot2_theme.R, /home/ssefick/R_scripts/r2test.R, /home/ssefick/R_scripts/recession_constant_package.R, /home/ssefick/R_scripts/recession_constant_R_CODE, /home/ssefick/R_scripts/serdp_name_split.R, /home/ssefick/R_scripts/setup_R.R, /home/ssefick/R_scripts/USGS.R)) scripts2source[grep(*.R, scripts2source)] #here is my problem; I would like these to be removed. scripts2source[c(2,5,13,14,20,24)] #Thanks for all of your help in advance #kindest regards, #Stephen Sefick -- Sarah Goslee http://www.functionaldiversity.org __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] grep fixed (?) in 2.14
your syntax is wrong, you need: scripts2source[grep(*\\.R$, scripts2source)] Notice the '\\.' to escape the special meaning of '.', and the $ to anchor to the end of the line. On Thu, Nov 3, 2011 at 8:54 AM, Stephen Sefick sas0...@auburn.edu wrote: #This is probably due to my incomplete understanding of grep, but the below code has been working for some time to #search for .R with anything in front of it and return a list of scripts to source. Likely, the syntax for the #grep statement has been wrong all along. scripts2source - (c(/home/ssefick/R_scripts/Convert_package.R, /home/ssefick/R_scripts/Convert_R_CODE, /home/ssefick/R_scripts/CV.R, /home/ssefick/R_scripts/cvs.out.R, /home/ssefick/R_scripts/database_connect, /home/ssefick/R_scripts/database_connect_package.R, /home/ssefick/R_scripts/exit_db.R, /home/ssefick/R_scripts/exit.R, /home/ssefick/R_scripts/hourly_zoo.R, /home/ssefick/R_scripts/model_diag.R, /home/ssefick/R_scripts/not_numeric.R, /home/ssefick/R_scripts/num_ecol_package.R, /home/ssefick/R_scripts/NumEcolR_scripts, /home/ssefick/R_scripts/old_scripts_DELETE_AFTER_DECEMBER, /home/ssefick/R_scripts/only_numeric_dataframe.R, /home/ssefick/R_scripts/only_numeric.R, /home/ssefick/R_scripts/PCA.ve.R, /home/ssefick/R_scripts/poster_ggplot2_theme.R, /home/ssefick/R_scripts/pressure_transducer_package.R, /home/ssefick/R_scripts/Pressure_Transducer_R_CODE, /home/ssefick/R_scripts/publication_ggplot2_theme.R, /home/ssefick/R_scripts/r2test.R, /home/ssefick/R_scripts/recession_constant_package.R, /home/ssefick/R_scripts/recession_constant_R_CODE, /home/ssefick/R_scripts/serdp_name_split.R, /home/ssefick/R_scripts/setup_R.R, /home/ssefick/R_scripts/USGS.R)) scripts2source[grep(*.R, scripts2source)] #here is my problem; I would like these to be removed. scripts2source[c(2,5,13,14,20,24)] #Thanks for all of your help in advance #kindest regards, #Stephen Sefick __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Jim Holtman Data Munger Guru What is the problem that you are trying to solve? Tell me what you want to do, not how you want to do it. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Question about Calculation of Cross Product
Try library(pracma) ? cross Michael Weylandt On Thu, Nov 3, 2011 at 1:17 AM, 阮铮 rz1...@foxmail.com wrote: The function of crossprod in R puzzled me. I would like to calculate the cross product of two vectors. According to my text book, it defines like this: a = (ax, ay, az) b = (bx, by, bz) then, the cross product of a and b is: a X b = (ay*bz-az*by, az*bx-ax*bz, ax*by-ay*bz) It can also write in a determinant format. But the crossprod or tcrossprod function in R appeared not calculate the cross product. Suppose I enter this: a = c(1, 2, 3) b = c(2, 3, 4) ab = crossprod(a, b) ab [,1] [1,] 20 ab = tcrossprod(a, b) ab [,1] [,2] [,3] [1,] 2 3 4 [2,] 4 6 8 [3,] 6 9 12 Does any know how to calculate cross product in R. Or I have to write the function myself? [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Extract Data from Yahoo Finance
The quantmod package can probably do what you are asking, but it's a little hard to be certain since you provide neither a list of all the fields you are actually talking about nor a link to the page with the fields in question. Michael On Thu, Nov 3, 2011 at 12:02 AM, Deb Midya debmi...@yahoo.com wrote: Hi R –users, I am using R-2.14.0 on Windows XP. May I request you to assist me for the following please. I like to extract all the fields (example: a : Ask, b : Bid, ……, w : 52-week Range, x: Stock Exchange) for certain period of time, say, 1 October 2011 to 31 October 2011. Is there any R-Package(s) any R- script please? Once again, thank you very much for the time you have given. Regards, Deb __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Lattice plots and missing x-axis labels on second page
I believe that the explanation is: Axes, including tick labels, appear only on the edges of the trellis array, which in your case would be at the bottom of the 4th row. However, on the second page there is no 4th row, and lattice is not smart enough to put them instead at the bottom of the second. See the scales argument of xyplot, especially the alternating component, for how you can gain finer control to deal with your situation. Note: I would appreciate correction if I'm wrong about this. -- Bert On Wed, Nov 2, 2011 at 5:23 PM, Evans, David G (DFG) david.ev...@alaska.gov wrote: using -- Bert Gunter Genentech Nonclinical Biostatistics Internal Contact Info: Phone: 467-7374 Website: http://pharmadevelopment.roche.com/index/pdb/pdb-functional-groups/pdb-biostatistics/pdb-ncb-home.htm [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Select columns of a data.frame by name OR index in a function
Dear all,
Sometimes I have the situation where a function takes a data.frame and
an additional argument describing come columns. For greater flexibility
I want to allow for either column names or column indices. What I
usually do then is something like the following:
-8-
f - function(datf, cols) {
nc - seq_along(datf)
cn - colnames(datf)
colOK - (cols %in% nc) | (cols %in% cn)
if (!all(colOK)) {
badc - paste(sQuote(cols[!colOK]), collapse = , )
msg - sprintf(ngettext(sum(!colOK),
%s is not a valid column selector,
%s are not valid column selectors),
badc)
stop(msg)
}
which((nc %in% cols) | (cn %in% cols)) # with this set of indices I
would work in the rest of the code
}
dd - data.frame(a=1, b=1, c=1)
f(dd, 2:3) # [1] 2 3
f(dd, 1:4) # Error in f(dd, 1:4) : '4' is not a valid column selector
f(dd, a) # [1] 1
f(dd, c(a, d, e)) # Error in f(dd, c(a, d, e)) : 'd', 'e'
are not valid column selectors
-8-
So my question is, whether there are smarter/better/easier/more R-like
ways of doing that?
Any input appreciated.
KR,
-Thorn
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Re: [R] grep fixed (?) in 2.14
That did the trick. I have read about regular expressions often, and sometimes I get them right and sometimes I don't. Is there a good reference resource that anyone could suggest? Thanks for all of the help. Stephen Sefick On 11/03/2011 08:03 AM, jim holtman wrote: your syntax is wrong, you need: scripts2source[grep(*\\.R$, scripts2source)] Notice the '\\.' to escape the special meaning of '.', and the $ to anchor to the end of the line. On Thu, Nov 3, 2011 at 8:54 AM, Stephen Seficksas0...@auburn.edu wrote: #This is probably due to my incomplete understanding of grep, but the below code has been working for some time to #search for .R with anything in front of it and return a list of scripts to source. Likely, the syntax for the #grep statement has been wrong all along. scripts2source- (c(/home/ssefick/R_scripts/Convert_package.R, /home/ssefick/R_scripts/Convert_R_CODE, /home/ssefick/R_scripts/CV.R, /home/ssefick/R_scripts/cvs.out.R, /home/ssefick/R_scripts/database_connect, /home/ssefick/R_scripts/database_connect_package.R, /home/ssefick/R_scripts/exit_db.R, /home/ssefick/R_scripts/exit.R, /home/ssefick/R_scripts/hourly_zoo.R, /home/ssefick/R_scripts/model_diag.R, /home/ssefick/R_scripts/not_numeric.R, /home/ssefick/R_scripts/num_ecol_package.R, /home/ssefick/R_scripts/NumEcolR_scripts, /home/ssefick/R_scripts/old_scripts_DELETE_AFTER_DECEMBER, /home/ssefick/R_scripts/only_numeric_dataframe.R, /home/ssefick/R_scripts/only_numeric.R, /home/ssefick/R_scripts/PCA.ve.R, /home/ssefick/R_scripts/poster_ggplot2_theme.R, /home/ssefick/R_scripts/pressure_transducer_package.R, /home/ssefick/R_scripts/Pressure_Transducer_R_CODE, /home/ssefick/R_scripts/publication_ggplot2_theme.R, /home/ssefick/R_scripts/r2test.R, /home/ssefick/R_scripts/recession_constant_package.R, /home/ssefick/R_scripts/recession_constant_R_CODE, /home/ssefick/R_scripts/serdp_name_split.R, /home/ssefick/R_scripts/setup_R.R, /home/ssefick/R_scripts/USGS.R)) scripts2source[grep(*.R, scripts2source)] #here is my problem; I would like these to be removed. scripts2source[c(2,5,13,14,20,24)] #Thanks for all of your help in advance #kindest regards, #Stephen Sefick __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] non-parametric sample size calculation
On Nov 3, 2011, at 3:47 AM, David A. wrote: Hi, I am trying to estimate the sample size needed for the comparison of two groups on a certain measurement, given some previous data at hand. I find that the data collected does not follow a normal distribution, so I would like to use a non-parametric option for sample size calculation. I found the pwr package but I don't think it has this option and on the internet found that http://www.epibiostat.ucsf.edu/biostat/sampsize.html says only PASS allows non-parametric sample size calculations (although the webpage is not updated). Any help would be greatly appreciated Thanks, Dave The first question is how non normal are your data? If you used some formal test for normality and the p value was =0.05, I would suggest that you search the R-Help archives for a plethora of discussions on testing for normality. You will find that such tests should largely not be used in deference to the question Are the data normal enough?. If they are or can be transformed reasonably, use standard functions for calculating power and sample size, such as power.t.test(). If you need to use a non-parametric test, you might want to review this page by Jerry Dallal: http://www.jerrydallal.com/LHSP/npar.htm which has some general guidelines for calculating sample size predicated upon using standard parametric tests and then adjusting the sample size using the ARE (asymptotic relative efficiency) based upon the non-parametric intended. HTH, Marc Schwartz __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] HOW TO REMOVE MTEXT FROM PLOT, plotting changing populations with titles in loop
On 02.11.2011 20:29, Sarah Goslee wrote:
It's not perfect, but you could use:
mtext(paste(this is iteration , i, sep=), col=white)
to overwrite it, or polygon() to draw a white rectangle over the text each time.
The question is if it is not better to do the whole plot again and just
add the one text in the end. At least if you want to plot into non
screen device: You end up with ll those layers of text in the output
that makes it larger and additionally slows down the rendering of the
whole plot.
Uwe Ligges
Sarah
On Wed, Nov 2, 2011 at 3:15 PM, prinzOfNorwaytorgrim...@gmail.com wrote:
is there a way to hide/undraw mtext (or lines etc.) in a loop like
plot(runif(10))
iterCol- rainbowPalette(10)
for(i in 1:10){
mtext(paste(this is iteration , i, sep=))
points(runif(10),col=iterCol[i])
Sys.sleep(1)
## UNDRAW/HIDE the text so that it does not mess up the plot in the next
iteration
}
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and provide commented, minimal, self-contained, reproducible code.
Re: [R] Creating barplot using time as X
This should get you started. PEL - read.table(http://r.789695.n4.nabble.com/file/n3981961/example.txt;, header=TRUE, sep=\t) PELtable - with(PEL, tapply(phasetime, data.frame(activity, as.Date(beg)), sum)) PELtable as.Date.beg. activity 2010-06-03 2010-06-04 2010-06-05 2010-06-06 2010-06-07 D 3407 14405 17704 16740 7096 L 7370 43544 35898 30401 NA W 5976 26611 30294 36807 14466 PELtable[2,5] - 0 as.data.frame.table(PELtable) activity as.Date.beg. Freq 1 D 2010-06-03 3407 2 L 2010-06-03 7370 3 W 2010-06-03 5976 4 D 2010-06-04 14405 5 L 2010-06-04 43544 6 W 2010-06-04 26611 7 D 2010-06-05 17704 8 L 2010-06-05 35898 9 W 2010-06-05 30294 10D 2010-06-06 16740 11L 2010-06-06 30401 12W 2010-06-06 36807 13D 2010-06-07 7096 14L 2010-06-07 0 15W 2010-06-07 14466 barchart(Freq ~ as.Date.beg. , groups=activity, data=as.data.frame.table(PELtable), + stack=TRUE, auto.key=list(reverse=TRUE)) You might wish to be more precise in your definition of a 24-hour time period. I used calendar date here. Rich On Wed, Nov 2, 2011 at 3:24 PM, PEL pierre-etienne.lessar...@ulaval.cawrote: Hello all, my data looks like this: phaseno / activity / beg / end / phasetime 1 / L / 2010-06-03 19:15:24 / 2010-06-03 21:18:14 / 7370 2 / D / 2010-06-03 21:18:15 / 2010-06-03 21:19:55 / 100 3 / W / 2010-06-03 21:19:56 / 2010-06-03 21:22:47 / 171 4 / D / 2010-06-03 21:22:48 / 2010-06-03 21:23:47 / 59 5 / W / 2010-06-03 21:23:48 / 2010-06-03 21:23:53 / 5 6 / D / 2010-06-03 21:23:54 / 2010-06-03 21:26:18 / 144 7 / W / 2010-06-03 21:26:19 / 2010-06-03 21:32:10 / 351 8 / D / 2010-06-03 21:32:11 / 2010-06-03 21:32:11 / 0 9 / W / 2010-06-03 21:32:12 / 2010-06-03 21:32:12 / 0 10 / D / 2010-06-03 21:32:13 / 2010-06-03 21:32:29 / 16 Please note that phasetime is in seconds and is only a difftime() of beg and end. I want to create a stacked bar chart that gives me the percentage of time spent doing every activity (L,D or W) for every 24h period. Example: Day 1: 20% =L; 40% = W ; 40% = D. In a graph obviously. What I was thinking was dividing my dataframe into several smaller dataframes spanning 24h. I tried with by() but I doubt this is the correct function. If possible, I would also like to separate a row the phasetime of a row if it overlaps two 24 periods. I joined part of my data where *header=TRUE and sep=\t* http://r.789695.n4.nabble.com/file/n3981961/example.txt example.txt Thank you very much for your time PEL -- View this message in context: http://r.789695.n4.nabble.com/Creating-barplot-using-time-as-X-tp3981961p3981961.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.htmlhttp://www.r-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] variable transformation for lm
Hello, I am doing a simple regression using lm(Y~X). As my response and my predictor seemed to be skewed and I can't meet the model assumptions. Therefore I need to transform my variables. I wanted to ask what is the preferred way to find out if predictor and/or response needs to be transformed and if yes how (log-transform?). I found a procedure in A modern approach to Regressoin in R (Sheather, 2009): There they suggest an approach with the function bctrans from alr3...but it seems that it is deprecated. So what is the best way (box-cox test) find the best transformation for predictor and response simultaneously? AFAIK boxcox from MASS is used only used for transformation of the predictor? Thank you very much Johannes -- __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Take variables in data.frame and create list of matrices
Hi, I have this sample data below and would like to create a list of matricies. setseed(1254) id - c(1,1,1,1 ,2,2,2) o - as.factor(c(1:4, 1, 3, 4)) r - rep(.5, 7) v - rnorm(7) s - rnorm(7) dat -data.frame(id, o, r, v, s) dat # dat # id o r v s # 1 1 0.5 0.7024631 2.0813672 # 1 2 0.5 -0.5541955 0.1095156 # 1 3 0.5 -1.0418167 0.4164930 # 1 4 0.5 1.9973868 -2.0402058 # 2 1 0.5 1.3957243 -0.2548652 # 2 3 0.5 -0.6571325 0.6946570 # 2 4 0.5 1.3998202 0.4601017 Is there an easy way (or convenient function) to convert this into a list where each unique 'id' has its own 4x4 matrix (based on the maximum number of factor levels in 'o')in the list? Each matrix would have v in the diagonal and r*s*s in the off-diagonal (e.g., .5*2.0814*.1095 = 0.1139), where r is constant (.5) and s varies based on the wthin 'id' factor level of 'o'. The output would look something like this (except values in all cells): [[id 1]] 'o'-levels [,1][,2] [,3] [,4] [1,] .702 .1139 r*s1*s3 ... [2,] .1139 .702 r*s2*s3 ... [3,] r*s3*s1.... 702 [4,] .702 [[id 2]] 'o'-levels [,1] [,2] [,3] [,4] [1,] .702 NA r*s1*s3 ... [2,] NA .702 NA ... [3,] r*s3*s1 NA .702 [4,] ....702 any assistance is much appreciated. Thank you, AC [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Sample size calculations for one sided binomial exact test
From: https://stat.ethz.ch/pipermail/r-help/2011-November/294329.html I'm trying to compute sample size requirements for a binomial exact test. we want to show that the proportion is at least 90% assuming that it is 95%, with 80% power so any asymptotic approximations are out of the questions. I was planning on using binom.test to perform the simple test against a prespecified value, but cannot find any functions for computing sample size. do any exist? Thanks, Andrew Hi, I don't have the original e-mail, so this reply will be out of the thread in the archive. I am not aware of anything pre-existing in R for this application, but stand to be corrected on that point. There are at least two non-R related options: 1. The G*Power program which is available from: http://www.psycho.uni-duesseldorf.de/abteilungen/aap/gpower3/ 2. There is a paper by A'Hern which contains sample size tables here: Sample size tables for exact single-stage phase II designs R.P. A'Hern STATISTICS IN MEDICINE Statist. Med. 2001; 20:859–866 http://stat.ethz.ch/education/semesters/as2011/bio/ahernSampleSize.pdf The author used the BINOMDIST function in Excel to derive the tables. Notwithstanding criticisms of Excel, these tables, based upon a small test sample, agree with the G*Power program, as well as my own computations using R code below. He also uses normal approximations for sample sizes 300, given the limitations found in the BINOMDIST function. Here is my R code for deriving the critical value and sample size for a one sided exact binomial test, given an alpha, a null proportion, an alternate proportion and the desired power: # The possible sample size vector N needs to be selected in such a fashion # that it covers the possible range of values that include the true # minima. My example here does with a finite range and makes the # plot easier to visualize. N - 100:200 Alpha - 0.05 Pow - 0.8 p0 - 0.90 p1 - 0.95 # Required number of events, given a vector of sample sizes (N) # to be considered at the null proportion, for the given Alpha CritVal - qbinom(p = 1 - Alpha, size = N, prob = p0) # Get Beta (Type II error) for each N at the alternate hypothesis # proportion Beta - pbinom(CritVal, N, p1) # Get the Power Power - 1 - Beta # Find the smallest sample size yielding at least the required power SampSize - min(which(Power Pow)) # Get and print the required number of events to reject the null # given the sample size required (Res - paste(CritVal[SampSize] + 1, out of, N[SampSize])) # Plot it all plot(N, Power, type = b, las = 1) title(paste(One Sided Sample Size and Critical Value for H0 =, p0, versus HA = , p1, \n, For Power = , Pow), cex.main = 0.95) points(N[SampSize], Power[SampSize], col = red, pch = 19) text(N[SampSize], Power[SampSize], col = red, label = Res, pos = 3) abline(h = Pow, lty = dashed) One thing to note here (see the plot) is the non-monotonic function describing the power at each of the values of the sample size. This is due to the discrete nature of the binomial distribution. It also generally means that you are powering the sample size calculation for an alpha at something lower than the value indicated. The G*Power program provides both the actual alpha and power, given the input values. So there is a need to search the vector of sample sizes where the power is greater than that desired, to obtain the smallest sample size required to satisfy the power desired. The above could of course be encapsulated in a function to make use easier, but the code yields values that agree with both the G*Power application and A'Hern's tables. Hope that this is helpful. Regards, Marc Schwartz __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Take variables in data.frame and create list of matrices
If i understood correctly this may help you dlply(dat, .(id)) $`1` id o r v s 1 1 1 0.5 -1.7131858 -2.0548666 2 1 2 0.5 0.6582979 0.9688939 3 1 3 0.5 -0.6861830 2.4213072 4 1 4 0.5 1.4134171 -0.4540244$`2` id o r v s 5 2 1 0.5 0.2771839 1.0688474 6 2 3 0.5 -0.7543462 -1.0371494 7 2 4 0.5 -1.9810018 -0.6742232 Date: Thu, 3 Nov 2011 08:54:52 -0700 From: de...@wisc.edu To: r-help@r-project.org Subject: [R] Take variables in data.frame and create list of matrices Hi, I have this sample data below and would like to create a list of matricies. setseed(1254) id - c(1,1,1,1 ,2,2,2) o - as.factor(c(1:4, 1, 3, 4)) r - rep(.5, 7) v - rnorm(7) s - rnorm(7) dat -data.frame(id, o, r, v, s) dat # dat # id o r v s # 1 1 0.5 0.7024631 2.0813672 # 1 2 0.5 -0.5541955 0.1095156 # 1 3 0.5 -1.0418167 0.4164930 # 1 4 0.5 1.9973868 -2.0402058 # 2 1 0.5 1.3957243 -0.2548652 # 2 3 0.5 -0.6571325 0.6946570 # 2 4 0.5 1.3998202 0.4601017 Is there an easy way (or convenient function) to convert this into a list where each unique 'id' has its own 4x4 matrix (based on the maximum number of factor levels in 'o')in the list? Each matrix would have v in the diagonal and r*s*s in the off-diagonal (e.g., .5*2.0814*.1095 = 0.1139), where r is constant (.5) and s varies based on the wthin 'id' factor level of 'o'. The output would look something like this (except values in all cells): [[id 1]] 'o'-levels [,1][,2] [,3] [,4] [1,] .702 .1139 r*s1*s3 ... [2,] .1139 .702 r*s2*s3 ... [3,] r*s3*s1.... 702 [4,] .702 [[id 2]] 'o'-levels [,1] [,2] [,3] [,4] [1,] .702 NA r*s1*s3 ... [2,] NA .702 NA ... [3,] r*s3*s1 NA .702 [4,] ....702 any assistance is much appreciated. Thank you, AC [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Take variables in data.frame and create list of matrices
I missed; using the Hadley Wickham plyr package library (plyr) dlply(dat, .(id)) $`1` id o r v s 1 1 1 0.5 -1.7131858 -2.0548666 2 1 2 0.5 0.6582979 0.9688939 3 1 3 0.5 -0.6861830 2.4213072 4 1 4 0.5 1.4134171 -0.4540244$`2` id o r v s 5 2 1 0.5 0.2771839 1.0688474 6 2 3 0.5 -0.7543462 -1.0371494 7 2 4 0.5 -1.9810018 -0.6742232 Date: Thu, 3 Nov 2011 08:54:52 -0700 From: de...@wisc.edu To: r-help@r-project.org Subject: [R] Take variables in data.frame and create list of matrices Hi, I have this sample data below and would like to create a list of matricies. setseed(1254) id - c(1,1,1,1 ,2,2,2) o - as.factor(c(1:4, 1, 3, 4)) r - rep(.5, 7) v - rnorm(7) s - rnorm(7) dat -data.frame(id, o, r, v, s) dat # dat # id o r v s # 1 1 0.5 0.7024631 2.0813672 # 1 2 0.5 -0.5541955 0.1095156 # 1 3 0.5 -1.0418167 0.4164930 # 1 4 0.5 1.9973868 -2.0402058 # 2 1 0.5 1.3957243 -0.2548652 # 2 3 0.5 -0.6571325 0.6946570 # 2 4 0.5 1.3998202 0.4601017 Is there an easy way (or convenient function) to convert this into a list where each unique 'id' has its own 4x4 matrix (based on the maximum number of factor levels in 'o')in the list? Each matrix would have v in the diagonal and r*s*s in the off-diagonal (e.g., .5*2.0814*.1095 = 0.1139), where r is constant (.5) and s varies based on the wthin 'id' factor level of 'o'. The output would look something like this (except values in all cells): [[id 1]] 'o'-levels [,1][,2] [,3] [,4] [1,] .702 .1139 r*s1*s3 ... [2,] .1139 .702 r*s2*s3 ... [3,] r*s3*s1.... 702 [4,] .702 [[id 2]] 'o'-levels [,1] [,2] [,3] [,4] [1,] .702 NA r*s1*s3 ... [2,] NA .702 NA ... [3,] r*s3*s1 NA .702 [4,] ....702 any assistance is much appreciated. Thank you, AC [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] query about counting rows of a dataframe
Dear R users, I have got the following data frame, called my_df: gender day_birth month_birth year_birth labour 1 F 22 10 2001 1 2 M29 10 2001 2 3 M 1 11 2001 1 4 F 3 11 2001 1 5 M 3 11 2001 2 6 F 4 11 2001 1 7 F 4 11 2001 2 8 F 5 12 2001 2 9 M 22 14 2001 2 10 F 29 13 2001 2 ... I need to count data in different ways: 1. count the births for each day (having 0 when necessary) independently from the value of the labour column 2. count the births for each day (having 0 when necessary), divided by the value of labour (which can have two valuers, 1 or 2) 3. count the births for each day of all the years (i.e. the 22nd of October of all the years present in the data frame) independently from the value of labour 4. count the births for each day of all the years (i.e. the 22nd of October of all the years present in the data frame), divided by the value of labour I tried with the command table(my_df$year_birth, my_df$month_birth, my_df$day_birth) which satisfies (partially) question numer 1 (I am not able to have 0 in the not available days). Is there a smart way to do that without invoking too many loops? thank you for your help Stefano Sofia AVVISO IMPORTANTE: Questo messaggio di posta elettronica può contenere informazioni confidenziali, pertanto è destinato solo a persone autorizzate alla ricezione. I messaggi di posta elettronica per i client di Regione Marche possono contenere informazioni confidenziali e con privilegi legali. Se non si è il destinatario specificato, non leggere, copiare, inoltrare o archiviare questo messaggio. Se si è ricevuto questo messaggio per errore, inoltrarlo al mittente ed eliminarlo completamente dal sistema del proprio computer. Ai sensi dell’art. 6 della DGR n. 1394/2008 si segnala che, in caso di necessità ed urgenza, la risposta al presente messaggio di posta elettronica può essere visionata da persone estranee al destinatario. IMPORTANT NOTICE: This e-mail message is intended to be received only by persons entitled to receive the confidential information it may contain. E-mail messages to clients of Regione Marche may contain information that is confidential and legally privileged. Please do not read, copy, forward, or store this message unless you are an intended recipient of it. If you have received this message in error, please forward it to the sender and delete it completely from your computer system. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Lattice plots and missing x-axis labels on second page
A follow-up to my reply below. My explanation below is correct AFAICS, but my proposed remedy using alternating is not. Instead(again, AFAICS), one must use even finer control with the at and labels components of scales.This worked for me. -- Bert On Thu, Nov 3, 2011 at 6:40 AM, Bert Gunter bgun...@gene.com wrote: I believe that the explanation is: Axes, including tick labels, appear only on the edges of the trellis array, which in your case would be at the bottom of the 4th row. However, on the second page there is no 4th row, and lattice is not smart enough to put them instead at the bottom of the second. See the scales argument of xyplot, especially the alternating component, for how you can gain finer control to deal with your situation. Note: I would appreciate correction if I'm wrong about this. -- Bert On Wed, Nov 2, 2011 at 5:23 PM, Evans, David G (DFG) david.ev...@alaska.gov wrote: using -- Bert Gunter Genentech Nonclinical Biostatistics Internal Contact Info: Phone: 467-7374 Website: http://pharmadevelopment.roche.com/index/pdb/pdb-functional-groups/pdb-biostatistics/pdb-ncb-home.htm -- Bert Gunter Genentech Nonclinical Biostatistics Internal Contact Info: Phone: 467-7374 Website: http://pharmadevelopment.roche.com/index/pdb/pdb-functional-groups/pdb-biostatistics/pdb-ncb-home.htm [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] variable transformation for lm
See the acepack package or the transace function in Hmisc for ACE or AVAS. -- Bert On Thu, Nov 3, 2011 at 8:55 AM, Johannes Radinger jradin...@gmx.at wrote: Hello, I am doing a simple regression using lm(Y~X). As my response and my predictor seemed to be skewed and I can't meet the model assumptions. Therefore I need to transform my variables. I wanted to ask what is the preferred way to find out if predictor and/or response needs to be transformed and if yes how (log-transform?). I found a procedure in A modern approach to Regressoin in R (Sheather, 2009): There they suggest an approach with the function bctrans from alr3...but it seems that it is deprecated. So what is the best way (box-cox test) find the best transformation for predictor and response simultaneously? AFAIK boxcox from MASS is used only used for transformation of the predictor? Thank you very much Johannes -- __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Bert Gunter Genentech Nonclinical Biostatistics Internal Contact Info: Phone: 467-7374 Website: http://pharmadevelopment.roche.com/index/pdb/pdb-functional-groups/pdb-biostatistics/pdb-ncb-home.htm [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Take variables in data.frame and create list of matrices
Thats a good start but I am looking for more. Specifically (from previous post): 4x4 matrix (based on the maximum number of factor levels in 'o') for each 'id' as a list. Each matrix would have v in the diagonal and r*s*s in the off-diagonal (e.g., .5*2.0814*.1095 = 0.1139), where r is constant (.5) and s varies based on the wthin 'id' factor level of 'o'. The output would look something like this (except values in all cells): [[id 1]] 'o'-levels [,1][,2] [,3] [,4] [1,] .702 .1139 r*s1*s3 ... [2,] .1139 .702 r*s2*s3 ... [3,] r*s3*s1.... 702 [4,] .702 [[id 2]] 'o'-levels [,1] [,2] [,3] [,4] [1,] .702 NA r*s1*s3 ... [2,] NA .702 NA ... [3,] r*s3*s1 NA .702 [4,] ....702 Thanks, AC On Thu, Nov 3, 2011 at 9:25 AM, jose Bartolomei surfpr...@hotmail.comwrote: I missed; using the Hadley Wickham plyr package library (plyr) dlply(dat, .(id)) $`1` id o r v s 1 1 1 0.5 -1.7131858 -2.0548666 2 1 2 0.5 0.6582979 0.9688939 3 1 3 0.5 -0.6861830 2.4213072 4 1 4 0.5 1.4134171 -0.4540244 $`2` id o r v s 5 2 1 0.5 0.2771839 1.0688474 6 2 3 0.5 -0.7543462 -1.0371494 7 2 4 0.5 -1.9810018 -0.6742232 Date: Thu, 3 Nov 2011 08:54:52 -0700 From: de...@wisc.edu To: r-help@r-project.org Subject: [R] Take variables in data.frame and create list of matrices Hi, I have this sample data below and would like to create a list of matricies. setseed(1254) id - c(1,1,1,1 ,2,2,2) o - as.factor(c(1:4, 1, 3, 4)) r - rep(.5, 7) v - rnorm(7) s - rnorm(7) dat -data.frame(id, o, r, v, s) dat # dat # id o r v s # 1 1 0.5 0.7024631 2.0813672 # 1 2 0.5 -0.5541955 0.1095156 # 1 3 0.5 -1.0418167 0.4164930 # 1 4 0.5 1.9973868 -2.0402058 # 2 1 0.5 1.3957243 -0.2548652 # 2 3 0.5 -0.6571325 0.6946570 # 2 4 0.5 1.3998202 0.4601017 Is there an easy way (or convenient function) to convert this into a list where each unique 'id' has its own 4x4 matrix (based on the maximum number of factor levels in 'o')in the list? Each matrix would have v in the diagonal and r*s*s in the off-diagonal (e.g., .5*2.0814*.1095 = 0.1139), where r is constant (.5) and s varies based on the wthin 'id' factor level of 'o'. The output would look something like this (except values in all cells): [[id 1]] 'o'-levels [,1] [,2] [,3] [,4] [1,] .702 .1139 r*s1*s3 ... [2,] .1139 .702 r*s2*s3 ... [3,] r*s3*s1 ... . 702 [4,] .702 [[id 2]] 'o'-levels [,1] [,2] [,3] [,4] [1,] .702 NA r*s1*s3 ... [2,] NA .702 NA ... [3,] r*s3*s1 NA .702 [4,] ... .702 any assistance is much appreciated. Thank you, AC [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] variable transformation for lm
On Nov 3, 2011, at 11:55 AM, Johannes Radinger jradin...@gmx.at wrote: Hello, I am doing a simple regression using lm(Y~X). As my response and my predictor seemed to be skewed and I can't meet the model assumptions. Therefore I need to transform my variables. The presence of skewness in either or both the response or predictors does NOT imply failure to meet model assumptions. The assumptions of linear regression regarding normality only apply to the residuals after the estimation of the model. -- David. I wanted to ask what is the preferred way to find out if predictor and/or response needs to be transformed and if yes how (log-transform?). I found a procedure in A modern approach to Regressoin in R (Sheather, 2009): There they suggest an approach with the function bctrans from alr3...but it seems that it is deprecated. So what is the best way (box-cox test) find the best transformation for predictor and response simultaneously? AFAIK boxcox from MASS is used only used for transformation of the predictor? Thank you very much Johannes -- __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] optimising a loop
Dear R community,
I'm trying to remove a loop from my code but I'm stock and I can't find a good
way to do it. Hopefully one of you will have something clever to propose.
Here is a simplified example:
I have a squared matrix:
nom.plac2 - c(102, 103, 301, 303,304, 403)
poids2 - matrix(NA, 6,6, dimnames=list(nom.plac2,nom.plac2))
poids2
102 103 301 303 304 403
102 NA NA NA NA NA NA
103 NA NA NA NA NA NA
301 NA NA NA NA NA NA
303 NA NA NA NA NA NA
304 NA NA NA NA NA NA
403 NA NA NA NA NA NA
I want to replace some of the NAs following specific criterion included in 2
others matrix:
wei2 -
matrix(c(.6,.4,.5,.5,.9,.1,.8,.2,.7,.3,.6,.4),6,2,dimnames=list(nom.plac2,
c(p1,p2)),byrow=T)
wei2
p1 p2
102 0.6 0.4
103 0.5 0.5
301 0.9 0.1
303 0.8 0.2
304 0.7 0.3
403 0.6 0.4
voisin - matrix(c(103,304, 303, 102, 103
,303,403,304,303,102,103 ,303),
6,2,dimnames=list(nom.plac2, c(v1,v2)),byrow=T)
voisin
v1v2
102 103 304
103 303 102
301 103 303
303 403 304
304 303 102
403 103 303
So my final result is:
102 103 301 303 304 403
102 NA 0.6 NA NA 0.4 NA
103 0.5 NA NA 0.5 NA NA
301 NA 0.9 NA 0.1 NA NA
303 NA NA NA NA 0.2 0.8
304 0.3 NA NA 0.7 NA NA
403 NA 0.6 NA 0.4 NA NA
So, globally I want to fill for each line of poids2 data from wei2
associated with the good the good identifier found in voisin.
This can easily be done by a loop:
loop - poids2
for(i in 1:6){
+ loop[i,voisin[i,]] - wei2[i,]
+ }
But I expect it to be quite slow with my larger dataset.
Does any of you has an idea how I could remove the loop and speed up the
operation?
Best regards,
Bastien Ferland-Raymond, M.Sc. Stat., M.Sc. Biol.
Division des orientations et projets spéciaux
Direction des inventaires forestiers
Ministère des Ressources naturelles et de la Faune du Québec
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and provide commented, minimal, self-contained, reproducible code.
[R] RpgSQL vs RPostgreSQL
Hello, Could someone who has experience with or knowledge regarding both RPostgreSQL and RpgSQL packages provide some feedback? Thanks! I am most interested in hearing from people who have knowledge regarding both packages, not just one. The only real difference I can see is that RpgSQL has a Java dependency, which I am not apposed to if it provides some added benefit...otherwise I will probably use the RPostgreSQL package. Both packages look to be maintained still. I have skimmed over both of these links: http://cran.r-project.org/web/packages/RpgSQL/RpgSQL.pdf http://cran.r-project.org/web/packages/RPostgreSQL/RPostgreSQL.pdf Thanks, Ben [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] does there any function like sumif in excel?
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[R] How to interpret vglm output!!
Hello , I need to estimate the parameters of generalized poisson regression model. I found that I could use : vglm(formula, family,.., data), but I dont know how to interpret the output!!! min 1Q Median 3Q Max elogit (lambda)-.66 -.61-.46 -.06 35.33 log(theta) -10.2-.02 .11 .64 1.2 coefficients: value stdt.value intercept:1 .76 .019 39.32 intercept:2 1.02 .0239.88 sex .55 1.216.00 log-likelihood -990.83, I dont know what is the exactly meaning of these two intercept? Thanks in advance for your help. Akram [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Sum with condition
Problem solved, thanks everyone for your help. Céline -- View this message in context: http://r.789695.n4.nabble.com/Sum-with-condition-tp3972839p3984909.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Extract Data from Yahoo Finance
Vikram,  Thanks for this.  The field names we put like this: quote = c(Open, High, Low, Close)  But in order to download all the fields, what are the names I need to use in the object quote. In Yahho Finance these are special tags.  The special tags are: a=Ask, b= Bid, ..., v= Volume, W= 52-week Range, x= Stock EXchange  quote= c(Open, High, Low, Close, Ask, Bid, , W, x). What are the names I can use to download Ask, Bid, ..., Stock Exchange?  Once again, thank you very much for the time you have given.  Regards,  Deb  From: Vikram Bahure economics.vik...@gmail.com Sent: Thursday, 3 November 2011 7:15 PM Subject: Re: [R] Extract Data from Yahoo Finance Hi, You can install the following library: library(tseries) and use the following command: ?get.hist.quote Regards Vikram Hi R âusers,  I am using R-2.14.0 on Windows XP.  May I request you to assist me for the following please.  I like to extract all the fields (example: a : Ask, b : Bid, â¦â¦, w : 52-week Range, x: Stock Exchange)  for certain period of time, say, 1 October 2011 to 31 October 2011.  Is there any R-Package(s) any R- script please?  Once again, thank you very much for the time you have given.  Regards,  Deb __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Why can't this function be used with the 'by' command?
Why can't this function be used with the 'by' command?
x = array(runif(16), dim=c(8,2))
x = data.frame(x)
x$group = rep(c('wt', 'app'), each=4)
shapiro.p = function(x) shapiro.test(x)[[2]]
apply(x[,1:2], 2, shapiro.p)
X1X2
0.4126345 0.2208781
by(x[,1:2], x$group, shapiro.p)
Error in `[.data.frame`(x, complete.cases(x)) :
undefined columns selected
[[alternative HTML version deleted]]
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[R] EMD arguments default values
Hi, i'm using this command line with EMD package filename-emd(y,x,boundary=wave,plot.imf=F) i'd like to know if this EMD uses default values for non-specified arguments. i'm especially interested in stoprule, max.imf, tol behavior in such cases. Thanks __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] data frame to workspace
Is there a way to import a data frame into a workspace? I created a data frame and from my understanding, a data frame is a type of object, and that the workspace stores the current session's objects. Wondering why my data frame is not showing up... Any thoughts/suggestions? -- View this message in context: http://r.789695.n4.nabble.com/data-frame-to-workspace-tp3986415p3986415.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Comparing negative binomial models
Hi, I am trying to compare negative binomial models for the prediction of sports games (I know that Poisson models would be better but I'm just trying Negative Binomial at the moment). But, to compare the models I need them to have the same theta value. How can I change the explanatory variables while maintaining the theta value? Thanks for the help. David -- View this message in context: http://r.789695.n4.nabble.com/Comparing-negative-binomial-models-tp3985353p3985353.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] take me off the list
thank you. Alan Gao University of Michigan '12 Statistics, B.S. 239.682.3509 __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Fit continuous distribution to truncated empirical values
Hi all, I am trying to fit a distribution to some data about survival times. I am interested only in a specific interval, e.g., while the data lies in the interval (0,, 600), I want the best for the interval (0,..., 24). I have tried both fitdistr (MASS package) and fitdist (from the fitdistrplus package), but I could not get them working, e.g. fitdistr(left, weibull, upper=24) Error in optim(x = c(529L, 528L, 527L, 526L, 525L, 524L, 523L, 522L, 521L, : L-BFGS-B needs finite values of 'fn' In addition: Warning message: In dweibull(x, shape, scale, log) : NaNs produced Am I doing something wrong? Thanks, Michele p.s. I have seen similar posts, e.g., http://tolstoy.newcastle.edu.au/R/help/05/02/11558.html, but I am not sure whether I can apply the same approach here. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] problem with merging two matrices
Dear Sarah, THanks for your answer! Sorry that my thread is somehow not clear, that's because I am not really experienced with R and dont know yet how to put thinings in words.. I am not trying to work with them differently, I am just trying to print them as the result of a function. But as the result() function does allow only one argument I have to somehow merge these two matrcices. I know that I could use the list() function as well but I wanted to have some nicer output than that, that's why I created these two matrices. Want I want to get in the end is that can call my function(x,y) and then you get the print of the two matrices, above each other. -- View this message in context: http://r.789695.n4.nabble.com/problem-with-merging-two-matrices-tp3983136p3986638.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Histograms in R
We have a histogram of our observed response and we want to overlay the corresponding poisson distribution with respect to our poisson model. -- View this message in context: http://r.789695.n4.nabble.com/Histograms-in-R-tp3985397p3985397.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Back-transforming in lme
Hello I am analysing aboveground biomass data from revegetation testplots which I constructed in a split-plot design using the function lme. For the experiment, the three factors are amelioration (2 levels), fertilizer (2 levels) and treatment (7 levels). Each testplot (block) has a singlereplicate of each treatment (total of 8 testplots). The blocks were constructed of topsoil. Each block was divided in two subplots in which half were ameliorated the other half was not (amelioration - yes or no). Each subplot was divided again into two subplots (fertilizer - yes or no). Then for eachfertilizer subplot, the area was divided into 7 seeding treatments (treatments). I have transformed my data to meet model assumptions and simplified my model using stepwise elimination. Here is my minimal model: fm13- lme(sqrt(abovegroundbiomass.m.2)~amelioration+fertilizer+treatment,random=~1|block, method=ML,na.action = na.omit) Here is a summary output: summary(fm13)$tTable Value Std.Error DF t-value p-value (Intercept) 1.6196612 0.9058574100 1.7879869 7.680688e-02 ameliorationyes 1.0155789 0.5970789100 1.7009125 9.206718e-02 fertilizeryes 4.31538410.5970789100 7.2274943 9.985161e-11 treatmentelymgla ecovar 5.01255841.1170323100 4.4873891 1.931291e-05 treatmentelymgla local 2.76990721.1170323100 2.4797020 1.482059e-02 treatmentfestsax ecovar 0.6953266 1.1170323100 0.6224768 5.350453e-01 treatmentlupiarc local 1.13918401.11703231001.0198309 3.102696e-01 treatmentnativemix ecovar 2.10916831.11703231001.8881892 6.189917e-02 treatmentnativemix local 3.21031661.11703231002.8739694 4.951797e-03 I would like to obtain the back-transformed estimates but I cannot seem to find the a way to do this. I have a copy of Crawley (The R Book) but could not find instructions on how to back-transform in lme. I have also look for a copy of Zuur (Mixed Effects Models and Extensions in Ecology with R) but cannot seem to obtain a copy. Any guidance in how to backtranform these estimates would be greatly appreciated. Thanks. Allan CarsonGraduate StudentUniversity of Northern British Columbia [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Uploding package help
Hello All I want to upload a R package in CRAN. Kindly tell me how to upload a new package in CRAN REGARDS~ Reema Singh [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] data frame to workspace
Hi, When you saves a workspace, all the objects created are saved inside the workspace. When you restarts R, you are loading this workspace? (Se quiser que eu te responda em português, me envie um e-mali direto para saldanha.plan...@gmail.com). On Thu, Nov 3, 2011 at 1:46 PM, playballa23 cj...@mail.usp.edu wrote: Is there a way to import a data frame into a workspace? I created a data frame and from my understanding, a data frame is a type of object, and that the workspace stores the current session's objects. Wondering why my data frame is not showing up... Any thoughts/suggestions? -- View this message in context: http://r.789695.n4.nabble.com/data-frame-to-workspace-tp3986415p3986415.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Atenciosamente, Raphael Saldanha saldanha.plan...@gmail.com [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] XLConnect Error
Hi, I am having the same issue as described by the original subscriber can you offer any guidance on how to install Java ??? Do I need to do this from the R command line ??? or do I need to do this outside of the application?? Thanks Daniel -- View this message in context: http://r.789695.n4.nabble.com/XLConnect-Error-tp3628528p3986491.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] optimising a loop
Try replacing your for loop with the line
loop[cbind(as.vector(row(voisin)), match(voisin, nom.plac2))] -
as.vector(wei2)
Look help(Subscript) to see how subscripting an n-way array by
an n-column integer matrix works.
Bill Dunlap
Spotfire, TIBCO Software
wdunlap tibco.com
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
Behalf Of Bastien.Ferland-
raym...@mrnf.gouv.qc.ca
Sent: Thursday, November 03, 2011 10:25 AM
To: r-help@r-project.org
Subject: [R] optimising a loop
Dear R community,
I'm trying to remove a loop from my code but I'm stock and I can't find a
good way to do it.
Hopefully one of you will have something clever to propose.
Here is a simplified example:
I have a squared matrix:
nom.plac2 - c(102, 103, 301, 303,304, 403)
poids2 - matrix(NA, 6,6, dimnames=list(nom.plac2,nom.plac2))
poids2
102 103 301 303 304 403
102 NA NA NA NA NA NA
103 NA NA NA NA NA NA
301 NA NA NA NA NA NA
303 NA NA NA NA NA NA
304 NA NA NA NA NA NA
403 NA NA NA NA NA NA
I want to replace some of the NAs following specific criterion included in 2
others matrix:
wei2 -
matrix(c(.6,.4,.5,.5,.9,.1,.8,.2,.7,.3,.6,.4),6,2,dimnames=list(nom.plac2,
c(p1,p2)),byrow=T)
wei2
p1 p2
102 0.6 0.4
103 0.5 0.5
301 0.9 0.1
303 0.8 0.2
304 0.7 0.3
403 0.6 0.4
voisin - matrix(c(103,304, 303, 102, 103
,303,403,304,303,102,103 ,303),
6,2,dimnames=list(nom.plac2, c(v1,v2)),byrow=T)
voisin
v1v2
102 103 304
103 303 102
301 103 303
303 403 304
304 303 102
403 103 303
So my final result is:
102 103 301 303 304 403
102 NA 0.6 NA NA 0.4 NA
103 0.5 NA NA 0.5 NA NA
301 NA 0.9 NA 0.1 NA NA
303 NA NA NA NA 0.2 0.8
304 0.3 NA NA 0.7 NA NA
403 NA 0.6 NA 0.4 NA NA
So, globally I want to fill for each line of poids2 data from wei2
associated with the good the
good identifier found in voisin.
This can easily be done by a loop:
loop - poids2
for(i in 1:6){
+ loop[i,voisin[i,]] - wei2[i,]
+ }
But I expect it to be quite slow with my larger dataset.
Does any of you has an idea how I could remove the loop and speed up the
operation?
Best regards,
Bastien Ferland-Raymond, M.Sc. Stat., M.Sc. Biol.
Division des orientations et projets spéciaux
Direction des inventaires forestiers
Ministère des Ressources naturelles et de la Faune du Québec
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
[R] L1 penalization for proportional odds logistic regression
Dear community, I am currently attempting to perform a (L1) penalized ordinal logistic regression with proportional odds. For the moment I only found R packages allowing to perform forward or backward continuation ratio model with several penalizations. Does anyone have a clue of what R package I could use ? I am not even quite sure that penalized logistic regression with proportional odds has already been developed theoretically... Thanks a lot for your help ! -- View this message in context: http://r.789695.n4.nabble.com/L1-penalization-for-proportional-odds-logistic-regression-tp3986573p3986573.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Uploding package help
See 'Writing R Extensions', http://cran.r-project.org/doc/manuals/R-exts.html#Submitting-a-package-to-CRAN However, you can only submit a package: other people actually upload it to CRAN, if they accept it. On Thu, 3 Nov 2011, Reema Singh wrote: Hello All I want to upload a R package in CRAN. Kindly tell me how to upload a new package in CRAN REGARDS~ Reema Singh [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Brian D. Ripley, rip...@stats.ox.ac.uk Professor of Applied Statistics, http://www.stats.ox.ac.uk/~ripley/ University of Oxford, Tel: +44 1865 272861 (self) 1 South Parks Road, +44 1865 272866 (PA) Oxford OX1 3TG, UKFax: +44 1865 272595 __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Creating barplot using time as X
Thank you very much. This is exactly what I needed. Problem solved! Thanks again PEL -- View this message in context: http://r.789695.n4.nabble.com/Creating-barplot-using-time-as-X-tp3981961p3986882.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] take me off the list
On Thu, 3 Nov 2011, Alan Gao wrote: Alan, This is a self-service mail list; no maids to do the work for you. Go to the Web site http://www.r-project.org/, navigate to the mail list page, and unsubscribe yourself. Rich __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] optimising a loop
try this:
nom.plac2 - c(102, 103, 301, 303,304, 403)
poids2 - matrix(NA, 6,6, dimnames=list(nom.plac2,nom.plac2))
poids2
102 103 301 303 304 403
102 NA NA NA NA NA NA
103 NA NA NA NA NA NA
301 NA NA NA NA NA NA
303 NA NA NA NA NA NA
304 NA NA NA NA NA NA
403 NA NA NA NA NA NA
wei2 -
matrix(c(.6,.4,.5,.5,.9,.1,.8,.2,.7,.3,.6,.4),6,2,dimnames=list(nom.plac2,
c(p1,p2)),byrow=T)
voisin - matrix(c(103,304, 303, 102, 103
,303,403,304,303,102,103 ,303),
+ 6,2,dimnames=list(nom.plac2, c(v1,v2)),byrow=T)
# do matrix addressing by converting names to numbers
indx - rbind(
+ cbind(match(rownames(voisin), rownames(poids2))
+, match(voisin[, 1], colnames(poids2))
+)
+ , cbind(match(rownames(voisin), rownames(poids2))
+, match(voisin[, 2], colnames(poids2))
+)
+ )
indx
[,1] [,2]
[1,]12
[2,]24
[3,]32
[4,]46
[5,]54
[6,]62
[7,]15
[8,]21
[9,]34
[10,]45
[11,]51
[12,]64
# change the data
poids2[indx] - c(wei2[,1], wei2[,2])
poids2
102 103 301 303 304 403
102 NA 0.6 NA NA 0.4 NA
103 0.5 NA NA 0.5 NA NA
301 NA 0.9 NA 0.1 NA NA
303 NA NA NA NA 0.2 0.8
304 0.3 NA NA 0.7 NA NA
403 NA 0.6 NA 0.4 NA NA
On Thu, Nov 3, 2011 at 1:25 PM,
bastien.ferland-raym...@mrnf.gouv.qc.ca wrote:
Dear R community,
I'm trying to remove a loop from my code but I'm stock and I can't find a
good way to do it. Hopefully one of you will have something clever to
propose.
Here is a simplified example:
I have a squared matrix:
nom.plac2 - c(102, 103, 301, 303,304, 403)
poids2 - matrix(NA, 6,6, dimnames=list(nom.plac2,nom.plac2))
poids2
102 103 301 303 304 403
102 NA NA NA NA NA NA
103 NA NA NA NA NA NA
301 NA NA NA NA NA NA
303 NA NA NA NA NA NA
304 NA NA NA NA NA NA
403 NA NA NA NA NA NA
I want to replace some of the NAs following specific criterion included in 2
others matrix:
wei2 -
matrix(c(.6,.4,.5,.5,.9,.1,.8,.2,.7,.3,.6,.4),6,2,dimnames=list(nom.plac2,
c(p1,p2)),byrow=T)
wei2
p1 p2
102 0.6 0.4
103 0.5 0.5
301 0.9 0.1
303 0.8 0.2
304 0.7 0.3
403 0.6 0.4
voisin - matrix(c(103,304, 303, 102, 103
,303,403,304,303,102,103 ,303),
6,2,dimnames=list(nom.plac2, c(v1,v2)),byrow=T)
voisin
v1 v2
102 103 304
103 303 102
301 103 303
303 403 304
304 303 102
403 103 303
So my final result is:
102 103 301 303 304 403
102 NA 0.6 NA NA 0.4 NA
103 0.5 NA NA 0.5 NA NA
301 NA 0.9 NA 0.1 NA NA
303 NA NA NA NA 0.2 0.8
304 0.3 NA NA 0.7 NA NA
403 NA 0.6 NA 0.4 NA NA
So, globally I want to fill for each line of poids2 data from wei2
associated with the good the good identifier found in voisin.
This can easily be done by a loop:
loop - poids2
for(i in 1:6){
+ loop[i,voisin[i,]] - wei2[i,]
+ }
But I expect it to be quite slow with my larger dataset.
Does any of you has an idea how I could remove the loop and speed up the
operation?
Best regards,
Bastien Ferland-Raymond, M.Sc. Stat., M.Sc. Biol.
Division des orientations et projets spéciaux
Direction des inventaires forestiers
Ministère des Ressources naturelles et de la Faune du Québec
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What is the problem that you are trying to solve?
Tell me what you want to do, not how you want to do it.
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Re: [R] EMD arguments default values
The help shows: emd(xt, tt=NULL, tol=sd(xt)*0.1^2, max.sift=20, stoprule=type1, boundary=periodic, smlevels=c(1), sm=none, spar=NA, weight=20, check=FALSE, max.imf=10, plot.imf=TRUE, interm=NULL) You can use ?emd to show the help and see the defaults. On Thu, Nov 3, 2011 at 9:13 AM, Sara sarah_...@yahoo.it wrote: Hi, i'm using this command line with EMD package filename-emd(y,x,boundary=wave,plot.imf=F) i'd like to know if this EMD uses default values for non-specified arguments. i'm especially interested in stoprule, max.imf, tol behavior in such cases. Thanks __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Atenciosamente, Raphael Saldanha saldanha.plan...@gmail.com [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] any updates w.r.t. lapply, sapply, apply retaining classes
Hi All, I don't have a I need help question, so much as a query into any update whether 'R' has made any progress with some of the core functions retaining classes. As an example, because it's one of the cases that most egregiously impacts me my work and keeps pushing me away from 'R' and into other numerical languages (such as NumPy in python), I will use sapply / lapply to demonstrate, but this behavior is ubiquitous throughout 'R'. Let's say I have a class which is theoretically supported, but not one of the core numeric or character classes (and, to some degree, factor classes). Many of the basic functions will convert my desired class into either numeric or character, so that my returned answer is gibberish. E.g.: test= as.difftime(c(1, 1, 8, 0.25, 8, 1.25), units= days) ## create a small array of time differences class(test) ## this will return the proper class, difftime class(test[1] ) ## this will also return the proper class, difftime sapply(test, class) ## this will return *numerics* for all of the classes. Ack!! In the example I give above, the impact might seem small, but the implications are *huge*. This means that I am, in effect, not allowed to use *any* of the vectoring functions in 'R', which avoid performing loops thereby speeding up process time extraordinarily. Many can sympathize that 'R' is ridiculously slow with for loops, compared to other languages. But that's theoretically OK, a good statistician or data analyst should be able to work comfortably with matrices and vectors. However, *'R' cannot work comfortably* with matrices or vectors, *unless* they are using the numeric or character classes. Many of the classes suffer the problem I just described, although I only used difftime in the example. Factors seem a bit more comfortable, and can be handled most of the time, but not as well as numerics, and at times functions working on factors can return the numerical representation of the factor instead of the original factor. Is there any progress in guaranteeing that all core functions either (a) ideally return exactly the classes, and hierarchy of classes, that they received (e.g., a list of data frames with difftimes dates characters would return a list of data frames with difftimes dates characters), or (b) barring that, the function should at least error out with a clear error explaining that sapply, for example, cannot vectorize on the class being used? Returning incorrect answers is far worse than returning an error, from a perspective of stability. This is, by far, the largest Achilles' heel to 'R'. Personally, as my career advances and I work on more technical things, I am finding that I have to leave 'R' by the wayside and use other languages for robust numerical calculations and programming. This saddens me, because there are so many wonderful packages developed by the community. The example above came up because I am using the forecast library to great effect in predicting how long our product cycle time will be. However, I spend much of my time fighting all these class typing bugs in 'R' (and we have to start recognizing that they are bugs, otherwise they may never get resolved), such that many of the improvements in my productivity due to all the wonderful computational packages are entirely offset by the time I spend fighting this issue of poor classes. Thanks Regards! Mike --- XKCD http://www.xkcd.com [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Create design matrix
Greetings useRs, What is the easiest way to create a design matrix of several factor variables? Function gendata in Design seems to do that for a fitted model, but how to do that only on several factor vectors?? The result should be a df with one row for each distinct combination of levels of factors eg for (M,F) (Y,O) We get M Y M O F Y F O In reality I will have more than 1000 rows so doing by hand not good. Maybe there is a way with outer, but I couldn't see it. All the best to everybody. Stephen __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] problem with merging two matrices
Hi, On Thu, Nov 3, 2011 at 12:31 PM, flokke flo...@live.de wrote: Dear Sarah, THanks for your answer! Sorry that my thread is somehow not clear, that's because I am not really experienced with R and dont know yet how to put thinings in words.. I am not trying to work with them differently, I am just trying to print them as the result of a function. But as the result() function does allow only one argument I have to somehow merge these two matrcices. I know that I could use the list() function as well but I wanted to have some nicer output than that, that's why I created these two matrices. Want I want to get in the end is that can call my function(x,y) and then you get the print of the two matrices, above each other. See, that would have been useful to know. If you wish to return two such matrices, you need to use a list. If you are determined that they appear in a particular way, you could also print them from within the function, so that particular things appear onscreen regardless of the way in which the results are returned. But, in that case, you may be better off reconsidering what your actual objectives are, and whether separating form and function might not be a more effective course. Sarah -- Sarah Goslee http://www.functionaldiversity.org __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Create design matrix
?expand.grid expand.grid(c(M,F),c(Y,O)) Var1 Var2 1MY 2FY 3MO 4FO Justin On Thu, Nov 3, 2011 at 10:56 AM, Bond, Stephen stephen.b...@cibc.com wrote: Greetings useRs, What is the easiest way to create a design matrix of several factor variables? Function gendata in Design seems to do that for a fitted model, but how to do that only on several factor vectors?? The result should be a df with one row for each distinct combination of levels of factors eg for (M,F) (Y,O) We get M Y M O F Y F O In reality I will have more than 1000 rows so doing by hand not good. Maybe there is a way with outer, but I couldn't see it. All the best to everybody. Stephen __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Create design matrix
?expand.grid is what you're looking for, I think. (which, I agree, is not something one could easily guess) -- Bert On Thu, Nov 3, 2011 at 10:56 AM, Bond, Stephen stephen.b...@cibc.comwrote: Greetings useRs, What is the easiest way to create a design matrix of several factor variables? Function gendata in Design seems to do that for a fitted model, but how to do that only on several factor vectors?? The result should be a df with one row for each distinct combination of levels of factors eg for (M,F) (Y,O) We get M Y M O F Y F O In reality I will have more than 1000 rows so doing by hand not good. Maybe there is a way with outer, but I couldn't see it. All the best to everybody. Stephen __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Bert Gunter Genentech Nonclinical Biostatistics Internal Contact Info: Phone: 467-7374 Website: http://pharmadevelopment.roche.com/index/pdb/pdb-functional-groups/pdb-biostatistics/pdb-ncb-home.htm [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] For loop to cycle through datasets of differing lengths
I have encountered this problem on several occasions and am not sure how to
handle it. I use for-loops to cycle through datasets. When each dataset is
of equal length, it works fine as I can combine the datasets and have each
loop pick up a different column, but when the datasets are differing
lengths, I am struggling. Here is an example:
A-1:10
B-1:15
C-1:18
Set1-data.frame(A,runif(10))
Set2-data.frame(B,runif(15))
Set3-data.frame(C,runif(18))
for (i in 1:3){
if (i==1) Data-Set1 else if (i==2) Data-Set2 else Data-Set3
dev.new()
plot(Data[,1],Data[,2])
}
I don't always want to plot them and instead do other things, such as fit a
non-linear equation to the dataset, etc. I end up using that if statement
to cycle through the datasets and was hoping there is an easier method.
Maybe one would be to add extra zeros until they are the same length and
then take out the extra zeros in the first step. Any help would be
appreciated.
-
In theory, practice and theory are the same. In practice, they are not - Albert
Einstein
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Re: [R] For loop to cycle through datasets of differing lengths
On Thu, Nov 3, 2011 at 11:41 AM, Schatzi adele_thomp...@cargill.com wrote:
I have encountered this problem on several occasions and am not sure how to
handle it. I use for-loops to cycle through datasets. When each dataset is
of equal length, it works fine as I can combine the datasets and have each
loop pick up a different column, but when the datasets are differing
lengths, I am struggling. Here is an example:
A-1:10
B-1:15
C-1:18
Set1-data.frame(A,runif(10))
Set2-data.frame(B,runif(15))
Set3-data.frame(C,runif(18))
for (i in 1:3){
if (i==1) Data-Set1 else if (i==2) Data-Set2 else Data-Set3
dev.new()
plot(Data[,1],Data[,2])
}
I don't always want to plot them and instead do other things, such as fit a
non-linear equation to the dataset, etc. I end up using that if statement
to cycle through the datasets and was hoping there is an easier method.
Maybe one would be to add extra zeros until they are the same length and
then take out the extra zeros in the first step. Any help would be
appreciated.
Well, you can start by putting your data sets in a list:
allSets = list(Set1, Set2, Set3)
then in your loop say
Data = allSets[[i]]
and you're done.
If you have an apriori unknown number of data sets, you can use get.
Before the loop you define the names,
setNames = c(Set1, Set2, Set3)
Then in the loop, you can use
Data = get(setNames[i])
But I'm not sure why this would be more useful than the list example
unless perhaps if you `load()' previously `save()'d data sets (the
load() function can give you the names of loaded variables). In any
case, if you repeat the same analysis on a number of data sets with
different dimensions, list() is IMHO the way to go.
HTH,
Peter
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[R] Reclassify string values
Hi All, Is there a simple way to convert a string such as c(A, B ,C, D) to a string of c(Group1, Group1, Group2, Group2). Naturally I could use the factor function as below but I don't like seeing that warning message (and I don't want to turn off warning messages). Perhaps a function called reclassify or recategorize? Zev x-LETTERS[1:4] x2-as.character(factor(x, levels=LETTERS[1:4], labels=rep(c(Group1, Group2), each=2))) Warning message: In `levels-`(`*tmp*`, value = c(Group1, Group1, Group2, Group2 : duplicated levels will not be allowed in factors anymore -- Zev Ross ZevRoss Spatial Analysis 120 N Aurora, Suite 3A Ithaca, NY 14850 607-277-0004 (phone) 866-877-3690 (fax, toll-free) zev at zevross.com __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Plotting skewed normal distribution with a bar plot
Hi, I need to create a plot (type = h) and then overlay a skewed-normal curve on this distribution, but I'm not finding a procedure to accomplish this. I want to use the plot function here in order to control the bin distributions. I have explored the sn library and found the dsn function. dsn uses known location, scaling and shape parameters associated with a given input vector of probabilities. However, how can I calculate the skewed-normal curve if I don't know these parameters in advance? Is there another function to calculate the skew-normal, perhaps in a different package? I'm working with R 2.13.2 on a windows based machine. Steve Friedman Ph. D. Ecologist / Spatial Statistical Analyst Everglades and Dry Tortugas National Park 950 N Krome Ave (3rd Floor) Homestead, Florida 33034 steve_fried...@nps.gov Office (305) 224 - 4282 Fax (305) 224 - 4147 __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] data frame to workspace
What did you do to create/save/load your data frame? There are n - infty ways to do all three of those steps and it's hard to give meaningful help without knowing what you tried. M On Nov 3, 2011, at 11:46 AM, playballa23 cj...@mail.usp.edu wrote: Is there a way to import a data frame into a workspace? I created a data frame and from my understanding, a data frame is a type of object, and that the workspace stores the current session's objects. Wondering why my data frame is not showing up... Any thoughts/suggestions? -- View this message in context: http://r.789695.n4.nabble.com/data-frame-to-workspace-tp3986415p3986415.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Is it possible to vectorize/accelerate this?
Dear Members,
I work on a simulaton experiment but it has an bottleneck. It's quite fast
because of R and vectorizing, but it has a very slow for loop. The adjacent
element of a vector (in terms of index number) depends conditionally on the
former value of itself. Like a simple cumulating function (eg. cumsum) but with
condition. Let's show me an example:
a_vec = rnorm(100)
b_vec = rep(0, 100)
b_vec[1]=a_vec[1]
for (i in 2:100){b_vec[i]=ifelse(abs(b_vec[i-1]+a_vec[i])1, a_vec[i],
b_vec[i-1]+a_vec[i])}
print(b_vec)
(The behaviour is like cumsum's, but when the value would excess 1.0 then it
has another value from a_vec.)
Is it possible to make this faster? I experienced that my way is even slower
than in Excel! Programming in C would my last try...
Any suggestions?
Than you,
Peter
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Re: [R] Reclassify string values
On Thu, Nov 3, 2011 at 11:59 AM, Zev Ross z...@zevross.com wrote: Hi All, Is there a simple way to convert a string such as c(A, B ,C, D) to a string of c(Group1, Group1, Group2, Group2). Naturally I could use the factor function as below but I don't like seeing that warning message (and I don't want to turn off warning messages). Perhaps a function called reclassify or recategorize? Zev x-LETTERS[1:4] x2-as.character(factor(x, levels=LETTERS[1:4], labels=rep(c(Group1, Group2), each=2))) Warning message: In `levels-`(`*tmp*`, value = c(Group1, Group1, Group2, Group2 : duplicated levels will not be allowed in factors anymore If you want to translate, why not first build a translation table tt = cbind(LETTERS[1:4], c(group1, group1, group2, group2)) then apply it on an example: xx = sample(LETTERS[1:4], 20, replace = TRUE) translation = tt[ match(xx, tt[, 1]), 2] translation [1] group2 group2 group2 group2 group2 group1 group2 group1 [9] group2 group1 group1 group2 group2 group2 group1 group2 [17] group2 group1 group1 group2 Or did I misunderstand your intent? Peter __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] [R-pkgs] new version of FRAILTYPACK: general frailty models
Dear R users, We are pleased to tell you that FRAILTYPACK has been updated. FRAILTYPACK stands now for general frailty models estimated with a semi-parametrical penalized likelihood, but also with a parametrical approach. In case of comments/corrections/remarks/suggestions -- which are very welcome --please contact the maintainer directly. Kind regards, The FRAILTYPACK team. ___ R-packages mailing list r-packa...@r-project.org https://stat.ethz.ch/mailman/listinfo/r-packages __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Searching elements in list
Hi all, I have a list ov vectors of enequal lenght and need to check is the given vector in list. v1-c(1,2) v2-c(1,2,3) v3-c(1,3) allv-list(v1,v2,v3) somev-c(1,2) somev%in%allv [1] FALSE FALSE Hence, %in% checks that elements of vector somev are in list. How it is possible to check that somev is element of list? -- View this message in context: http://r.789695.n4.nabble.com/Searching-elements-in-list-tp3987066p3987066.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Searching elements in list
I forget to mention that the vectors are ordered. I think that one of the possible solutions is to use lapply, i.e., T %in% lapply(allv, function(x,y) all.equal(x,y),y=somev) [1] TRUE -- View this message in context: http://r.789695.n4.nabble.com/Searching-elements-in-list-tp3987066p3987487.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Is it possible to vectorize/accelerate this?
I don't immediately see a good trick for vectorization so this seems to me to
be a good candidate for work in a lower-level language. Staying within R, I'd
suggest you use if and else rather than ifelse() since your computation isn't
vectorized: this will eliminate a small amount over overhead. Since you also
always add a_vec, you could also define b_vec as a copy of a to avoid all those
calls to subset a, but I don't think the effects will be large and the code
might not be as clear.
You indicated that you may be comfortable with writing C, but I'd suggest you
look into the Rcpp/Inline package pair which make the whole process much easier
than it would otherwise be.
I'm not at a computer write now or I'd write a fuller example, but the
documentation for those packages is uncommonly good an you should be able to
easily get it down into C++. If you aren't able to get it by tomorrow, let me
know and I can help troubleshoot. The only things I foresee that you'll need to
change are zero-basing, C's loop syntax, and (I think) the call to abs(). (I
always forget where abs() lives in c++ )
The only possible hold up is that you need to be at a computer with a C compiler
Hope this helps,
Michael
On Nov 3, 2011, at 3:10 PM, hihi v.p.m...@freemail.hu wrote:
Dear Members,
I work on a simulaton experiment but it has an bottleneck. It's quite fast
because of R and vectorizing, but it has a very slow for loop. The adjacent
element of a vector (in terms of index number) depends conditionally on the
former value of itself. Like a simple cumulating function (eg. cumsum) but
with condition. Let's show me an example:
a_vec = rnorm(100)
b_vec = rep(0, 100)
b_vec[1]=a_vec[1]
for (i in 2:100){b_vec[i]=ifelse(abs(b_vec[i-1]+a_vec[i])1, a_vec[i],
b_vec[i-1]+a_vec[i])}
print(b_vec)
(The behaviour is like cumsum's, but when the value would excess 1.0 then it
has another value from a_vec.)
Is it possible to make this faster? I experienced that my way is even slower
than in Excel! Programming in C would my last try...
Any suggestions?
Than you,
Peter
__
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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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and provide commented, minimal, self-contained, reproducible code.
Re: [R] Histograms in R
Try something like this Lam - 3 X - rpois(500, Lam) hist(X, freq = F) x - seq(min(X), max(X), length = 500) lines(x, dpois(x, Lam), col=2) Adapt as necessary Michael On Nov 3, 2011, at 8:03 AM, kerry1912 kerry1...@hotmail.com wrote: We have a histogram of our observed response and we want to overlay the corresponding poisson distribution with respect to our poisson model. -- View this message in context: http://r.789695.n4.nabble.com/Histograms-in-R-tp3985397p3985397.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Comparing negative binomial models
Davg davidgrimsey at hotmail.com writes: I am trying to compare negative binomial models for the prediction of sports games (I know that Poisson models would be better but I'm just trying Negative Binomial at the moment). But, to compare the models I need them to have the same theta value. How can I change the explanatory variables while maintaining the theta value? library(MASS) modelfit - glm(..., family=negative.binomial(theta=theta_value)) I believe that drop1() applied to a glm.nb() fit does this correctly (i.e. holds the theta parameter fixed at the estimate from the most complex model). __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Is it possible to vectorize/accelerate this?
Thank you, I will try as soon as possible...
Regards,
Peter
2011/11/3 Mark Leeds marklee...@gmail.com
hi hihi: you're not using the ifelse construct correctly because it's
already vectorized
so there's no need to use a loop.
check if below works AND if it's fast enough because I didn't check either
one. Also,
i bet someone else can send something better so I would wait anyway. good
luck.
set.seed(1)
avec = rnorm(100)
bvec = rep(0, 100)
bvec[1]=avec[1]
bveclagged - c(999,head(bvec,-1))
bvec - ifelse((abs(bveclagged + avec) 1), avec, bveclagged + avec)
On Thu, Nov 3, 2011 at 7:10 PM, hihi v.p.m...@freemail.hu wrote:
Dear Members,
I work on a simulaton experiment but it has an bottleneck. It's quite
fast because of R and vectorizing, but it has a very slow for loop. The
adjacent element of a vector (in terms of index number) depends
conditionally on the former value of itself. Like a simple cumulating
function (eg. cumsum) but with condition. Let's show me an example:
a_vec = rnorm(100)
b_vec = rep(0, 100)
b_vec[1]=a_vec[1]
for (i in 2:100){b_vec[i]=ifelse(abs(b_vec[i-1]+a_vec[i])1, a_vec[i],
b_vec[i-1]+a_vec[i])}
print(b_vec)
(The behaviour is like cumsum's, but when the value would excess 1.0 then
it has another value from a_vec.)
Is it possible to make this faster? I experienced that my way is even
slower than in Excel! Programming in C would my last try...
Any suggestions?
Than you,
Peter
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Re: [R] Is it possible to vectorize/accelerate this?
Hi:
You're doing the right thing in R by pre-allocating memory for the
result, but ifelse() is a vectorized function and your loop is
operating elementwise, so if-else is more appropriate. Try
for (i in 2:100){
b_vec[i] - if(abs(b_vec[i-1] + a_vec[i]) 1) a_vec[i] else
b_vec[i-1] + a_vec[i]
}
If speed is an issue, then I echo Michael's suggestion to write a
C(++) function and call it within R. The inline package is good for
this kind of thing.
HTH,
Dennis
On Thu, Nov 3, 2011 at 12:10 PM, hihi v.p.m...@freemail.hu wrote:
Dear Members,
I work on a simulaton experiment but it has an bottleneck. It's quite fast
because of R and vectorizing, but it has a very slow for loop. The adjacent
element of a vector (in terms of index number) depends conditionally on the
former value of itself. Like a simple cumulating function (eg. cumsum) but
with condition. Let's show me an example:
a_vec = rnorm(100)
b_vec = rep(0, 100)
b_vec[1]=a_vec[1]
for (i in 2:100){b_vec[i]=ifelse(abs(b_vec[i-1]+a_vec[i])1, a_vec[i],
b_vec[i-1]+a_vec[i])}
print(b_vec)
(The behaviour is like cumsum's, but when the value would excess 1.0 then it
has another value from a_vec.)
Is it possible to make this faster? I experienced that my way is even slower
than in Excel! Programming in C would my last try...
Any suggestions?
Than you,
Peter
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Re: [R] Reclassify string values
Hi Peter, Thanks for the response. What you've suggested works fine but I'm looking for something that is simpler than my solution and avoids the pesky warning message. Your response avoids the warning message but just as complex (if not more). I just assumed there would be a function along the lines of: mydata - c(A, C, A, D, B, B) reclassify(mydata, inCategories=c(A, B ,C, D), outCategories=c(Group1, Group1, Group2, Group2)) [1] Group1 Group2 Group1 Group2 Group1 Group1 Zev On 11/3/2011 3:13 PM, Peter Langfelder wrote: On Thu, Nov 3, 2011 at 11:59 AM, Zev Rossz...@zevross.com wrote: Hi All, Is there a simple way to convert a string such as c(A, B ,C, D) to a string of c(Group1, Group1, Group2, Group2). Naturally I could use the factor function as below but I don't like seeing that warning message (and I don't want to turn off warning messages). Perhaps a function called reclassify or recategorize? Zev x-LETTERS[1:4] x2-as.character(factor(x, levels=LETTERS[1:4], labels=rep(c(Group1, Group2), each=2))) Warning message: In `levels-`(`*tmp*`, value = c(Group1, Group1, Group2, Group2 : duplicated levels will not be allowed in factors anymore If you want to translate, why not first build a translation table tt = cbind(LETTERS[1:4], c(group1, group1, group2, group2)) then apply it on an example: xx = sample(LETTERS[1:4], 20, replace = TRUE) translation = tt[ match(xx, tt[, 1]), 2] translation [1] group2 group2 group2 group2 group2 group1 group2 group1 [9] group2 group1 group1 group2 group2 group2 group1 group2 [17] group2 group1 group1 group2 Or did I misunderstand your intent? Peter -- Zev Ross ZevRoss Spatial Analysis 120 N Aurora, Suite 3A Ithaca, NY 14850 607-277-0004 (phone) 866-877-3690 (fax, toll-free) z...@zevross.com __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Problem with R CMD check and the inconsolata font business.
On 11/3/2011 12:37 AM, Rolf Turner wrote:
I have just installed R version 2.14.0 and tried to re-build and
re-check some
of the packages that I maintain.
I'm getting a warning (in the process of running R CMD check on my deldir
package):
* checking PDF version of manual ... WARNING
LaTeX errors when creating PDF version.
This typically indicates Rd problems.
LaTeX errors found:
! Font T1/fi4/m/n/10=ec-inconsolata at 10.0pt not loadable: Metric
(TFM) file n
ot found.
to be read again
relax
l.19 ...lf Turner }\email{r.tur...@auckland.ac.nz}
! \textfont 0 is undefined (character h).
\Url@FormatString ...\Url@String \UrlRight \m@th $
l.26 ...\AsIs{}\url{http://www.math.unb.ca/~rolf/}
\AsIs{}
! \textfont 0 is undefined (character t).
\Url@FormatString ...\Url@String \UrlRight \m@th $
l.26 ...\AsIs{}\url{http://www.math.unb.ca/~rolf/}
\AsIs{}
! \textfont 0 is undefined (character t).
.. etc., etc., etc., ad (almost) infinitum.
So there's some problem with a font file not being loadable.
Can anyone tell me what the expletive deleted I should ***do***
about this? I managed to install the inconsolata package from CTAN.
At least I think I managed; I downloaded the *.zip file and then unzipped
it in /usr/share/texmf/tex/latex. And ran texhash. This stopped R CMD
check from complaining that the inconsolata package could not be found,
but then led to the further complaint described above.
So how do I make the required font loadable? What files do I need?
Where do I get them? And where should I put them once I've got them?
I would be grateful for any assistance that can be rendered.
(I know it's just a warning, but I *hate* to ignore warnings!)
cheers,
Rolf Turner
P. S. I'm running Ubuntu; session info, in case it's of any relevance is:
sessionInfo()
R version 2.14.0 (2011-10-31)
Platform: i686-pc-linux-gnu (32-bit)
locale:
[1] LC_CTYPE=en_US.UTF-8 LC_NUMERIC=C
[3] LC_TIME=en_US.UTF-8 LC_COLLATE=en_US.UTF-8
[5] LC_MONETARY=en_US.UTF-8 LC_MESSAGES=en_US.UTF-8
[7] LC_PAPER=C LC_NAME=C
[9] LC_ADDRESS=C LC_TELEPHONE=C
[11] LC_MEASUREMENT=en_US.UTF-8 LC_IDENTIFICATION=C
attached base packages:
[1] stats graphics grDevices utils datasets methods base
other attached packages:
[1] misc_0.0-15
I'm having a similar, though different, problem with inconsolata. It
may be the same root problem. The error on R CMD check I get is:
* checking PDF version of manual ... WARNING
LaTeX errors when creating PDF version.
This typically indicates Rd problems.
LaTeX errors found:
!pdfTeX error: pdflatex.EXE (file ec-inconsolata): Font ec-inconsolata
at 540 n
ot found
== Fatal error occurred, no output PDF file produced!
Now, oddly, there _is_ a -manual.pdf file in the .Rcheck directory
(despite the message saying no PDf was produced). If I just try to run
the pdflatex on the -manual.tex file in that directory, I get the same
error and no PDF (?!)
I am on Windows 7 64 bit using MiKTeX 2.9; I get the same error whether
I use the 32 or 64 bit version of R for R CMD check.
sessionInfo()
R version 2.14.0 (2011-10-31)
Platform: x86_64-pc-mingw32/x64 (64-bit)
locale:
[1] LC_COLLATE=English_United States.1252
[2] LC_CTYPE=English_United States.1252
[3] LC_MONETARY=English_United States.1252
[4] LC_NUMERIC=C
[5] LC_TIME=English_United States.1252
attached base packages:
[1] stats graphics grDevices utils datasets methods base
Some info from my tex installation (showing I have inconsolata installed
and that I am pulling the correct Rd.sty):
C:\kpsewhich Rd.sty
C:/Program Files/R/R-2.14.0/share/texmf/tex/latex/Rd.sty
C:\kpsewhich inconsolata.sty
C:/Program Files (x86)/MiKTeX 2.9/tex/latex/inconsolata/inconsolata.sty
C:\kpsewhich ec-inconsolata.tfm
C:/Program Files (x86)/MiKTeX
2.9/fonts/tfm/public/inconsolata/ec-inconsolata.tfm
--
Brian S. Diggs, PhD
Senior Research Associate, Department of Surgery
Oregon Health Science University
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Re: [R] Reclassify string values
On Thu, Nov 3, 2011 at 1:31 PM, Zev Ross z...@zevross.com wrote:
Hi Peter,
Thanks for the response. What you've suggested works fine but I'm looking
for something that is simpler than my solution and avoids the pesky warning
message. Your response avoids the warning message but just as complex (if
not more). I just assumed there would be a function along the lines of:
mydata - c(A, C, A, D, B, B)
reclassify(mydata, inCategories=c(A, B ,C, D),
outCategories=c(Group1, Group1, Group2, Group2))
[1] Group1 Group2 Group1 Group2 Group1 Group1
But of course, except sometimes you have to write the function yourself.
reclassify = function(data, inCategories, outCategories)
{
outCategories[ match(data, inCategories)]
}
Sorry I can't make it any simpler than a 1-line solution :)
Feel free to add some checking of input validity, if you need that.
Peter
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Re: [R] Searching elements in list
On Nov 3, 2011, at 3:21 PM, mstepano wrote: I forget to mention that the vectors are ordered. I think that one of the possible solutions is to use lapply, i.e., T %in% lapply(allv, function(x,y) all.equal(x,y),y=somev) [1] TRUE all.equal worked fine when the answer was 'true' but not so well in my hands when it was 'false'. Also not a good idea to replace the base definition of %in%. any( sapply( allv, identical, v1) ) [1] TRUE any( sapply( allv, identical, c(3,3,4)) ) [1] FALSE `%l.in%` - function(vec,lis) any( sapply( lis, identical, vec) ) Use: v1 %l.in% allv [1] TRUE -- David Winsemius, MD Heritage Laboratories West Hartford, CT __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Reclassify string values
On Nov 3, 2011, at 4:40 PM, Peter Langfelder wrote:
On Thu, Nov 3, 2011 at 1:31 PM, Zev Ross z...@zevross.com wrote:
Hi Peter,
Thanks for the response. What you've suggested works fine but I'm
looking
for something that is simpler than my solution and avoids the pesky
warning
message. Your response avoids the warning message but just as
complex (if
not more). I just assumed there would be a function along the lines
of:
mydata - c(A, C, A, D, B, B)
reclassify(mydata, inCategories=c(A, B ,C, D),
outCategories=c(Group1, Group1, Group2, Group2))
[1] Group1 Group2 Group1 Group2 Group1 Group1
But of course, except sometimes you have to write the function
yourself.
reclassify = function(data, inCategories, outCategories)
{
outCategories[ match(data, inCategories)]
}
Sorry I can't make it any simpler than a 1-line solution :)
It will be difficult to beat a oneliner like that. If Zev is still
holding out for a canned solution he might look in the 'car'' package
where there is at least one function that does releveling and
grouping. I foget its name at the moment but it wouldn't hurt a new
learneR to scroll through the entire 'car' suite of functions.
--
David Winsemius, MD
Heritage Laboratories
West Hartford, CT
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Re: [R] Kaplan Meier - not for dates
Thanks for the reply.
The treatment is effectively for a chronic condition - so you stay on the
treatment till it stops working. We know from trials how long that should be
and we know the theoretical cost of that treatment but that's based on the text
book dose (patients dose reduce and delay treatment and its based on weight so
variable). We've been asked to provide our national planning team with an
average cost based on our early experiences. So we have suggested to them we
might be able to get a median cost. Some patients will stay on treatment
several years so it will be impossible to get an average for years.
So the censored patients will be those still on treatment (the event being
stopping treatment)
I'll give what you've suggested a go.
Thanks
Calum Polwart BSc(Hons) MSc MRPharmS SPres IPres
Network Pharmacist - NECN and Pharmacy Clinical Team Manager (Cancer Aseptic
Services) - CDDFT
Our website has now been unlocked and updated. Should you require contacts,
meeting details, publications etc, please visit us on www.cancernorth.nhs.uk
From: Lancaster, Robert (Orbitz) [robert.lancas...@orbitz.com]
Sent: 03 November 2011 19:55
To: Polwart Calum (COUNTY DURHAM AND DARLINGTON NHS FOUNDATION TRUST);
r-help@r-project.org
Subject: RE: Kaplan Meier - not for dates
I think it really depends on what your event of interest is. If your event is
that the patient got better and left treatment then I think this could work.
You would have to mark as censored any patient still in treatment or any
patient that stopped treatment w/o getting better (e.g. in the case of death).
You would then be predicting the cost required to make the patient well enough
to leave treatment. It is a little non-standard to use $ instead of time, but
time is money after all.
You could set up your data frame with two columns: 1) cost 2) event/censored.
Then create your survival object:
mySurv = Surv(my_data$cost,my_data$event)
And then use survfit to create your KM curves:
myFit = survfit(mySurv~NULL)
If you have other explanatory variables that you think may influence the cost,
you can of course add them to your data frame and change the formula you use in
survfit. For instance, you could have some severity measure, e.g. High,
Medium, Low. You could then do:
myFit = survfit(mySurv~my_data$severity)
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
Behalf Of Polwart Calum (COUNTY DURHAM AND DARLINGTON NHS FOUNDATION TRUST)
Sent: Monday, October 31, 2011 1:29 PM
To: r-help@r-project.org
Subject: [R] Kaplan Meier - not for dates
I have some data which is censored and I want to determine the median. Its
actually cost data for a cohort of patients, many of whom are still on
treatment and so are censored.
I can do the same sort of analysis for a survival curve and get the median
survival... ...but can I just use the survival curve functions to plot an X
axis that is $ rather than date? If not is there some other way to achieve this?
Thanks
Calum
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Re: [R] Problem with R CMD check and the inconsolata font business.
On 03/11/11 20:37, Rolf Turner wrote: I have just installed R version 2.14.0 and tried to re-build and re-check some of the packages that I maintain. I'm getting a warning (in the process of running R CMD check on my deldir package): * checking PDF version of manual ... WARNING LaTeX errors when creating PDF version. This typically indicates Rd problems. LaTeX errors found: ! Font T1/fi4/m/n/10=ec-inconsolata at 10.0pt not loadable: Metric (TFM) file n ot found. to be read again SNIP I received two replies off-line and one that was sent to the list. The gist of the off-line replies was ``you need to install some fonts'', and (unusually, and very kindly! :-) ) the responders told me *how* to do so. The first message that I read was from John Nash who wrote: Have you tried sudo apt-get install texlive-fonts-extra to get the font in? I had to do this just yesterday myself. If that works, maybe put a note on R-help. I've not sent to R-help in case it doesn't work, and we get a bunch of unnecessary noise. Well it *did* work; thank you very much John! A similar message from G. Jay Kerns said: It looks like you are missing some font files. Did you do: apt-get install texlive-fonts-recommended ? The third message (which *was* posted to the list) was from Brian Diggs, who said that he was having a related problem: The error on R CMD check I get is: * checking PDF version of manual ... WARNING LaTeX errors when creating PDF version. This typically indicates Rd problems. LaTeX errors found: !pdfTeX error: pdflatex.EXE (file ec-inconsolata): Font ec-inconsolata at 540 n ot found == Fatal error occurred, no output PDF file produced! I think this is the same as/similar to the problem I had before I installed the inconsolata (LaTeX) package from CTAN. Since Brian does Windoze I cannot advise as to how to install the package on his system, but I presume it is not too difficult --- once you know how! :-) I'm posting this so that John Nash's and Jay Kerns' solutions to my problem are there in the archives ``for the record''. Thanks to all who replied. cheers, Rolf Turner __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Histograms in R
The lines() command doesn't work and histogram combines categories unless you specify the number. How about a barplot Lam - 3 X - table(rpois(500, Lam)) Max - length(X)-1 barplot(rbind(X, 500*dpois(0:Max, Lam)), beside=TRUE, legend.text=c(Observed, Expected)) or a rootogram library(vcd) rootogram(X, dpois(0:Max, Lam)*500) -Original Message- From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On Behalf Of R. Michael Weylandt michael.weyla...@gmail.com Sent: Thursday, November 03, 2011 2:55 PM To: kerry1912 Cc: r-help@r-project.org Subject: Re: [R] Histograms in R Try something like this Lam - 3 X - rpois(500, Lam) hist(X, freq = F) x - seq(min(X), max(X), length = 500) lines(x, dpois(x, Lam), col=2) Adapt as necessary Michael On Nov 3, 2011, at 8:03 AM, kerry1912 kerry1...@hotmail.com wrote: We have a histogram of our observed response and we want to overlay the corresponding poisson distribution with respect to our poisson model. -- View this message in context: http://r.789695.n4.nabble.com/Histograms-in-R-tp3985397p3985397.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Is it possible to vectorize/accelerate this?
Yes -- if else is much faster than ifelse() because if is a
primitive while ifelse() is a whole function call (in fact, you can
see the code by typing ifelse into the prompt and see that it has two
if calls within it.
Michael
On Thu, Nov 3, 2011 at 4:38 PM, hihi v.p.m...@freemail.hu wrote:
Hi,
thank you for your very immediate response. :-) Is if than and else faster
than ifelse? I'm wondering (or not knowing something)
Best regards,
Peter
2011/11/3 R. Michael Weylandt michael.weyla...@gmail.com
michael.weyla...@gmail.com
I don't immediately see a good trick for vectorization so this seems to me
to be a good candidate for work in a lower-level language. Staying within R,
I'd suggest you use if and else rather than ifelse() since your computation
isn't vectorized: this will eliminate a small amount over overhead. Since
you also always add a_vec, you could also define b_vec as a copy of a to
avoid all those calls to subset a, but I don't think the effects will be
large and the code might not be as clear.
You indicated that you may be comfortable with writing C, but I'd suggest
you look into the Rcpp/Inline package pair which make the whole process much
easier than it would otherwise be.
I'm not at a computer write now or I'd write a fuller example, but the
documentation for those packages is uncommonly good an you should be able to
easily get it down into C++. If you aren't able to get it by tomorrow, let
me know and I can help troubleshoot. The only things I foresee that you'll
need to change are zero-basing, C's loop syntax, and (I think) the call to
abs(). (I always forget where abs() lives in c++ )
The only possible hold up is that you need to be at a computer with a C
compiler
Hope this helps,
Michael
On Nov 3, 2011, at 3:10 PM, hihi v.p.m...@freemail.hu wrote:
Dear Members,
I work on a simulaton experiment but it has an bottleneck. It's quite
fast because of R and vectorizing, but it has a very slow for loop. The
adjacent element of a vector (in terms of index number) depends
conditionally on the former value of itself. Like a simple cumulating
function (eg. cumsum) but with condition. Let's show me an example:
a_vec = rnorm(100)
b_vec = rep(0, 100)
b_vec[1]=a_vec[1]
for (i in 2:100){b_vec[i]=ifelse(abs(b_vec[i-1]+a_vec[i])1, a_vec[i],
b_vec[i-1]+a_vec[i])}
print(b_vec)
(The behaviour is like cumsum's, but when the value would excess 1.0
then it has another value from a_vec.)
Is it possible to make this faster? I experienced that my way is even
slower than in Excel! Programming in C would my last try...
Any suggestions?
Than you,
Peter
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Re: [R] any updates w.r.t. lapply, sapply, apply retaining classes
Hi Mike,
This isn't really an answer to your question, but perhaps will serve
to continue discussion. I think that there are some fundamental
issues when working special classes. As a thought example, suppose I
wrote a class, posreal, which inherits from the numeric class. It
is only valid for positive, real numbers. I use it in a package, but
do not develop methods for it. A user comes along and creates a
vector, x that is a posreal. Then tries: mean(x * -3). Since I never
bothered to write a special method for mean for my class, R falls back
to the inherited numeric, but gives a value that is clearly not valid
for posreal. What should happen? S3 methods do not really have
validation, so in principle, one could write a function like:
f - function(x) {
vclass - class(x)
res - mean(x)
class(res) - vclass
return(res)
}
which retains the appropriate class, but in name only. R core
cannot possibly know or imagine all classes that may be written that
inherit from more basic types but with possible special aspects and
requirements. I think the inherited is considered to be more generic
and that is returned. It is usually up to the user to ensure that the
function (whose methods were not specific to that special class but
the inherited) is valid for that class and can manually convert it
back:
res - as.posreal(res)
What about lapply and sapply? Neither are generic or have methods for
difftime, and so do some unexpected/desirable things. Again, without
methods defined for a particular class, they cannot know what is
special or appropriate way to handle it, they use defaults which
sometimes work but may give unexpected or undesirable results, but
what else can be done? (okay, they could just throw an error) If a
function is naive about a class, it does not seem right to operate on
it using unknown methods and then pretend to be returning the same
type of data. As it stands, they convert to a data type they know and
return that.
Now, you mention that for loops are slow in R, and this is true to a
degree. However, the *apply functions are basically just internal
loops, so they do not really save you (they are certainly not
vectorized!), though they are more elegant than explicit loops IMO.
One way to use them while retaining class would be like:
sapply(seq_along(test), function(i) class(test[i]))
this is less efficient then sapply(test, class), but the overhead
drops considerably as the function does nontrivial calculations.
Finally, I find the (relatively) new compiler package really shines at
making functions that are just wrappers for for loops more efficient.
Take a look at the examples from:
require(compiler)
?cmpfun
I am not familiar with numPy so I do not know how it handles new
classes, but with some tweaks to my workflow, I do not find myself
running into problems with how R handles them. I definitely
appreciate your position because I have been there...as I became more
familiar with R, classes, and methods, I find I work in a way that
avoids passing objects to functions that do not know how to handle
them properly.
Cheers,
Josh
On Thu, Nov 3, 2011 at 11:08 AM, Mike Williamson this.is@gmail.com wrote:
Hi All,
I don't have a I need help question, so much as a query into any
update whether 'R' has made any progress with some of the core functions
retaining classes. As an example, because it's one of the cases that most
egregiously impacts me my work and keeps pushing me away from 'R' and
into other numerical languages (such as NumPy in python), I will use sapply
/ lapply to demonstrate, but this behavior is ubiquitous throughout 'R'.
Let's say I have a class which is theoretically supported, but not one
of the core numeric or character classes (and, to some degree, factor
classes). Many of the basic functions will convert my desired class into
either numeric or character, so that my returned answer is gibberish.
E.g.:
test= as.difftime(c(1, 1, 8, 0.25, 8, 1.25), units= days) ## create a
small array of time differences
class(test) ## this will return the proper class, difftime
class(test[1] ) ## this will also return the proper class, difftime
sapply(test, class) ## this will return *numerics* for all of the classes.
Ack!!
In the example I give above, the impact might seem small, but the
implications are *huge*. This means that I am, in effect, not allowed to
use *any* of the vectoring functions in 'R', which avoid performing loops
thereby speeding up process time extraordinarily. Many can sympathize that
'R' is ridiculously slow with for loops, compared to other languages.
But that's theoretically OK, a good statistician or data analyst should be
able to work comfortably with matrices and vectors. However, *'R' cannot
work comfortably* with matrices or vectors, *unless* they are using the
numeric or character classes. Many of the classes suffer the problem I
just described, although I only used difftime in the
Re: [R] Is it possible to vectorize/accelerate this?
You should get familiar with some basic timing tools
and techniques so you can investigate things like this
yourself.
system.time is the most basic timing tool. E.g.,
system.time(for(i in 1:1000)f0(a))
user system elapsed
22.920 0.000 22.932
means it took c. 23 seconds of real time to run f0(a)
1000 times.
When comparing timing, it makes things easier to define
a series of functions that implement the various algorithms
but have the same inputs and outputs. E.g., for your problem
f0 - function(a_vec) {
b_vec - a_vec
for (i in 2:length(b_vec)){
b_vec[i] - ifelse(abs(b_vec[i-1] + a_vec[i]) 1, a_vec[i], b_vec[i-1]
+ a_vec[i])
}
b_vec
}
f1 - function(a_vec) {
b_vec - a_vec
for (i in 2:length(b_vec)){
b_vec[i] - if(abs(b_vec[i-1] + a_vec[i]) 1) a_vec[i] else b_vec[i-1]
+ a_vec[i]
}
b_vec
}
f2 - function(a_vec) {
b_vec - a_vec
for (i in 2:length(b_vec)){
if(abs(s - b_vec[i-1] + a_vec[i]) = 1) b_vec[i] - s
}
b_vec
}
Then run them with the same dataset:
a - runif(1000, 0, .3)
system.time(for(i in 1:1000)f0(a))
user system elapsed
22.920 0.000 22.932
system.time(for(i in 1:1000)f1(a))
user system elapsed
5.510 0.000 5.514
system.time(for(i in 1:1000)f2(a))
user system elapsed
4.210 0.000 4.217
(The rbenchmark package's benchmark function encapsulates this idiom.)
It pays to use a dataset similar to the one you will ultimately be using,
where similar depends on the context. E.g., the algorithm in f2 is relatively
faster when the cumsum exceeds 1 most of the time
a - runif(1000, 0, 10)
system.time(for(i in 1:1000)f0(a))
user system elapsed
21.900 0.000 21.912
system.time(for(i in 1:1000)f1(a))
user system elapsed
4.610 0.000 4.609
system.time(for(i in 1:1000)f2(a))
user system elapsed
2.490 0.000 2.494
If you will be working with large datasets, you should look at how the
time grows as the size of the dataset grows. If the time looks quadratic
between,
say, length 100 and length 200, don't waste your time testing it for length
100.
For algorithms that work on data.frames (or matrices), the relative speed ofen
depends on the ratio of the number of rows and the number of columns of data.
Check that out. For these sorts of tests it is worthwhile to make a function to
generate typical looking data of any desired size.
It doesn't take too long to do this once you have the right mindset. Once you
do you don't have to rely on folklore like never use loops and instead do
evidence-based computing.
Bill Dunlap
Spotfire, TIBCO Software
wdunlap tibco.com
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
Behalf Of R. Michael
Weylandt
Sent: Thursday, November 03, 2011 2:51 PM
To: hihi; r-help
Subject: Re: [R] Is it possible to vectorize/accelerate this?
Yes -- if else is much faster than ifelse() because if is a
primitive while ifelse() is a whole function call (in fact, you can
see the code by typing ifelse into the prompt and see that it has two
if calls within it.
Michael
On Thu, Nov 3, 2011 at 4:38 PM, hihi v.p.m...@freemail.hu wrote:
Hi,
thank you for your very immediate response. :-) Is if than and else faster
than ifelse? I'm wondering (or not knowing something)
Best regards,
Peter
2011/11/3 R. Michael Weylandt michael.weyla...@gmail.com
michael.weyla...@gmail.com
I don't immediately see a good trick for vectorization so this seems to me
to be a good candidate for work in a lower-level language. Staying within
R,
I'd suggest you use if and else rather than ifelse() since your computation
isn't vectorized: this will eliminate a small amount over overhead. Since
you also always add a_vec, you could also define b_vec as a copy of a to
avoid all those calls to subset a, but I don't think the effects will be
large and the code might not be as clear.
You indicated that you may be comfortable with writing C, but I'd suggest
you look into the Rcpp/Inline package pair which make the whole process
much
easier than it would otherwise be.
I'm not at a computer write now or I'd write a fuller example, but the
documentation for those packages is uncommonly good an you should be able
to
easily get it down into C++. If you aren't able to get it by tomorrow, let
me know and I can help troubleshoot. The only things I foresee that you'll
need to change are zero-basing, C's loop syntax, and (I think) the call to
abs(). (I always forget where abs() lives in c++ )
The only possible hold up is that you need to be at a computer with a C
compiler
Hope this helps,
Michael
On Nov 3, 2011, at 3:10 PM, hihi v.p.m...@freemail.hu wrote:
Dear Members,
I work on a simulaton experiment but it has an bottleneck. It's quite
fast because of R and vectorizing, but it has a very slow for loop. The
adjacent
