Re: [R] R Error
Rolf Turner wrote: On 21/11/2008, at 10:13 AM, Steffy, Elizabeth A. wrote: I got this error for this equation and i'm not sure what it means or how to fix it: Error in S[index] = S[index - 1] + (dSi - dSo - SC) * dt : nothing to replace with Does anyone know how to fix this? No. PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. Do you see the lines immediately above? Many people on the R-help list are very clever. ***None*** of them are telepathic. No? I think she should ensure that index = 2, and that her R version is from before September 2008. -- O__ Peter Dalgaard Øster Farimagsgade 5, Entr.B c/ /'_ --- Dept. of Biostatistics PO Box 2099, 1014 Cph. K (*) \(*) -- University of Copenhagen Denmark Ph: (+45) 35327918 ~~ - ([EMAIL PROTECTED]) FAX: (+45) 35327907 __ R-help@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] matlab style of storing arrays compared to R
Hi I am trying to figure out, if the matlab style of linear indexing of an array is the same as in R. i.e. when x - array( 1:24, dim=c(2,3,4) ) x[3] 3 and if the same is true in matlab, assuming that x[n1,n2,n3] in R returns the same as y(n1,n2,n3) when y is a matrix in matlab I found the following reference http://www.mathworks.com/access/helpdesk/help/techdoc/index.html?/access/helpdesk/help/techdoc/matlab_prog/f1-86528.html#f1-86846 And it seems to be the same, but another source says it is different. Could somebody confirm, if it is the same? Thanks, Rainer -- Rainer M. Krug, PhD (Conservation Ecology, SUN), MSc (Conservation Biology, UCT), Dipl. Phys. (Germany) Centre of Excellence for Invasion Biology Faculty of Science Natural Sciences Building Private Bag X1 University of Stellenbosch Matieland 7602 South Africa __ 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 specific listcomponents to calculate the means
Dear list, I have following list [[1]] Pnr timeCA CACen 1 62083014541 0.008 TRUE 2 62083014542 0.008 TRUE 3 62083014543 0.008 FALSE 4 62083014544 0.013 TRUE 5 62083014545 0.007 FALSE [[2]] Pnr timeCA CACen 1 64031471161 0.020 FALSE 2 64031471162 0.089 FALSE 3 64031471163 0.020 FALSE 4 64031471164 0.025 FALSE 5 64031471165 0.012 FALSE [[3]] Pnr time CACACen 1 49051274131 0.008 TRUE 2 49051274132 0.007 TRUE 3 49051274133 0.003 TRUE 4 49051274134 0.006 TRUE [[4]] Pin timeCA CACen 1 50092771371 0.008 TRUE 2 50092771372 0.009 TRUE 3 50092771373 0.008 FALSE 4 50092771374 0.009 FALSE 5 50092771375 0.008 FALSE How do I tell R to select the listelements containing both TRUE and FALSE to calculate the weighted means and those with only TRUE or FALSE to calculate the aritmetic means and then put all the means together in a dataframe. The result should look like Pnr Mean 6208301454weighted mean 6403147116arith. mean 4905127413arith. mean 5009277137weighted mean Thanks for any help, Tom - Låna pengar utan säkerhet. Sök och jämför lån hos Kelkoo. [[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] R Error
Error in S[index] = S[index - 1] + (dSi - dSo - SC) * dt : nothing to replace with Peter Dalgaard wrote: ...that her R version is from before September 2008. Just curious which item in NEWS this comment refers to. Something changed in [] ? Dieter -- View this message in context: http://www.nabble.com/R-Error-tp20613209p20618176.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] generate random number
Hi Dimitris, Appreciate for your reply with detailed information, many thanks! I realize that generating random number won't be so simple more than I expected, but got some hints from the advice. I am actually hoping to do a parametric bootstrap likelihood test, because this is the way of testing for glmm result what I understood. Following is what I would like to do: # settings n - number of samples y - c(2 2 1 0 0 2 0 2 1 2 2 1 2 2 2 2 2 2 2 2 2 2 2 0 2 0 2 2) x - c(22 22 24 21 26 18 23 21 17 15 22 24 21 17 26 15 16 13 22 15 15 23 16 23 18 37 22 30) #dependent=y #independent=x Then, I would like to generate random number on glmm = random.y - rbinom (num,n,p) to do this test. However, this only hypothesized binomial distribution, not include normal one. In this case, how would you be able to do? Apologize if I didn't understand correctly what you wrote and make confuse you. Greatly appreciated if you help me. Any advice would be wonderful! Best reagards, Odette On Thu, Nov 20, 2008 at 8:24 PM, Dimitris Rizopoulos [EMAIL PROTECTED] wrote: check the following code: # settings n - 100 # number of sample units p - 10 # number of repeated measurements N - n * p # total number of measurements t.max - 3 # parameter values betas - c(0.5, 0.4, -0.5, -0.8) # fixed effects (check also 'X' below) sigma.b - 2 # random effects variance # id, treatment time id - rep(1:n, each = p) treat - rep(0:1, each = n/2) time - seq(0, t.max, length.out = p) # simulate random effects b - rnorm(n, sd = sigma.b) # simulate longitudinal process conditionally on random effects time.rep - rep(time, n) treat.rep - rep(treat, each = p) X - cbind(1, treat.rep, time.rep, treat.rep * time.rep) # fixed effects design matrix muY - plogis(c(X %*% betas) + b[id]) # conditional probabilities y - rbinom(N, 1, muY) # simulate binary responses # put the simulated data in a data.frame simulData - data.frame( id = id, y = y, treat = treat.rep, time = time.rep ) # fit the model library(glmmML) fit - glmmML(y ~ treat * time, data = simulData, cluster = id) summary(fit) I hope it helps. Best, Dimitris Odette Gaston wrote: Hi everybody, I am currently working on glmmML() and wish to generate random number to do some tests, however, glmm was hypothesized the mixed distributions with normal and binomial in terms of having a random effect. How would you be able to generate random number in this case? Is there a function in R to generate random number of mixed distribution (normal+binomial)? Any comments would be appreciated. Many thanks, Odette [[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.htmlhttp://www.r-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Dimitris Rizopoulos Assistant Professor Department of Biostatistics Erasmus Medical Center Address: PO Box 2040, 3000 CA Rotterdam, the Netherlands Tel: +31/(0)10/7043478 Fax: +31/(0)10/7043014 [[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] ggplot2 - Problem with geom_abline()
Hello, Sorry to ask again something with ggplot2... I detect something, perhaps a bug (but more probably a syntax error, I'm learning...) : # This is working : x = wt, y = mpg, abline with intercept = 20 and slope = 1 qplot(wt, mpg, data = mtcars) + geom_abline(intercept = 20, slope = 1) # This is not working : x = mpg, y = wt, abline with intercept = 3 and slope = 1 qplot(mpg, wt, data = mtcars) + geom_abline(intercept = 3, slope = 1) There is only points, no line, on the second graph. (R 2.8.0, ggplot2 0.7, windows 2000) Many thanks. david [[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] summary statistics into table/data base, many factors to analyse
Dear list, thanks to your help I managed to find means of analysing my data. However, the whole data set contains 264 variables. Of which some are factors, others are not. The factors tend to be grouped, e.g. data$f1304 to data$f1484 and data$f3204 to data$5408. But there are other types of variables in the data set as well, e.g. data$f1504. Not every spot is taken, i.e data$f1345 to data$1399 might not exist in the data set. The solution summaryBy works for cross analysis, of which there is a handful. So I am not worried here. The solution from Jorge is fine. However, I am trying to get my head around how to efficiently reduce my data set to the dependet variable and the factors such that the solution is applicable. Having to type each variable into my.reduced.data - cbind(my.data$f1001, my.data$1002, my.data$1003... is an obvious option, but does not seem to be the most efficient one. Are there better ways to go about? Thanks, Gerit -- Sensationsangebot nur bis 30.11: GMX FreeDSL - Telefonanschluss + DSL für nur 16,37 Euro/mtl.!* http://dsl.gmx.de/?ac=OM.AD.PD003K11308T4569a __ 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] R Error
Dieter Menne wrote: Error in S[index] = S[index - 1] + (dSi - dSo - SC) * dt : nothing to replace with Peter Dalgaard wrote: ...that her R version is from before September 2008. Just curious which item in NEWS this comment refers to. Something changed in [] ? No, it was just the error message, which now says that the replacement has length zero. We don't usually make a NEWS entry for changes to docs and messages. (Actually, September is not quite right. I was looking at one of the translation files. The actual message was changed on August 28...) Dieter -- O__ Peter Dalgaard Øster Farimagsgade 5, Entr.B c/ /'_ --- Dept. of Biostatistics PO Box 2099, 1014 Cph. K (*) \(*) -- University of Copenhagen Denmark Ph: (+45) 35327918 ~~ - ([EMAIL PROTECTED]) FAX: (+45) 35327907 __ R-help@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] R course in Scotland
Hello Peter, If you want to use R for bioinformatics, you probably want a course using Bioconductor (www.bioconductor.org). To combine with a introduction to R, the following should be good: http://www3.imperial.ac.uk/stathelp/courses/statisticalmicroarrayanalysisusingr but some time to wait till the next course. The following may provide a good alternative: http://www.ebi.ac.uk/training/handson/course_090119_transcriptomics.html Otherwise there are lots of materials from previous courses on http://www.bioconductor.org/workshops which you can use for self-teaching. Hope that helps, Heather -- Dr H Turner Senior Research Fellow Dept. of Statistics The University of Warwick Coventry CV4 7AL Tel: 024 76575870 Fax: 024 76524532 Url: www.warwick.ac.uk/go/heatherturner Gustavo Carvalho wrote: Hello, Take a look at this course: http://www.r4all.group.shef.ac.uk/index.html I don't think they teach tools for working with the genome, but it might be helpful anyway. On Thu, Nov 20, 2008 at 11:16 AM, Peter Saffrey [EMAIL PROTECTED] wrote: (apologies if this is the wrong list) I'm a bioinformatician looking for a course in using R, in particular the tools for working with the genome - I've heard they're lightning fast. I'm in Glasgow, but I've tried the Robertson centre for biostatistics and they use minitab. If anybody knows of a course, I would be grateful. Glasgow or Edinburgh would be preferable, but anywhere in the UK will do if it's a good course. Thanks, 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-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] a trous wavelet transform in R ?
There may be some chance that a proper wavelet transform may help me with signal features extraction. I know R contains a number of packages implementing wavelett filters and/or transforms. Is there an implementation of a trous wavelet transform ? Thank you so much, Maura Alice Messenger ;-) chatti anche con gli amici di Windows Live Messenger e tutti i telefonini TIM! Vai su http://maileservizi.alice.it/alice_messenger/index.html?pmk=footer [[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] Simulation Result display form
after obtaining the result, the result is displayed below: individual RS_number Phy_Posi LOH_intensity 1 1718890 rs1496555 2.2241110.8121 2 1668776 rs2376495 3.0849860.786 --- 3 1723597 rs4648462 3.1551270.784 --- 4 1728870 rs10492940 3.1876070.7831 5 1669946 rs10492939 3.2927310.7801 6 1708099 rs10492938 3.2941960.78 --- 7 1716798 rs10492937 3.3291990.7791 8 1701872 rs1128474 3.5354720.7733 9 1697748 rs2154068 3.7108250.7685 10-1699138 rs2887274 3.7111780.7685 my problem is in the result of LOH_intesity. i hope that 0.786 can be showed 0.7860, 0.78 can be 0.7800. so, how can I manage my result to be those? Thanks a lot... __ 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] timezone attribute lost during operations
Hi, I was just *highly* surprised to find out that R 2.8.0 (Windows XP) changes the timezone-interpretation during operations on time data, as apparently the timezone attribute is lost and then, for the next interpretation of the timezone, the system settings are used. Here is sample code (executed under a platform with the system timezone managed by Windows set to CET, but note that as the input data is GMT I also want all the calculations to occur in GMT): # input data Time = as.POSIXct(strptime(c(2007-12-12 14:30:15, 2008-04-14 15:31:34, 2008-04-14 15:31:34), format = %Y-%m-%d %H:%M:%S, tz = GMT)) Time # OK, time zone is GMT attr(Time, tzone) # OK, time zone is GMT TApply = tapply(1:3, Time, max) names(TApply) # wrong, names are converted to time zone of system UTime = unique(Time) UTime # wrong, again time zone of system is used attr(UTime, tzone) # == NULL Now the issue is not that I wouldn't know how to solve the problem (by setting TZ to GMT prior to executing the calculations), but I wonder why is R doing this mess at all? Why is it not able to maintain the timezone-information stored in my original vector Time ? Is this behavior supposed to be a feature, or is it a plain bug ? Thanks, Thomas __ 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] Extracting diagonal matrix
Dear All, I have a correlation matrix of size 100 x 100 and would like to extract the diagonal matrix from it. I have used the for loop to store tha correlation values of the diagonal matrix. Is there a 'R way' of doing this? Thanks in advance. Kind regards, Ezhil __ 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] Extracting diagonal matrix
Try
?diag
Rory Winston
RBS Global Banking Markets
Office: +44 20 7085 4476
-Original Message-
From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] On Behalf Of A Ezhil
Sent: 21 November 2008 12:28
To: r-help@r-project.org
Subject: [R] Extracting diagonal matrix
Dear All,
I have a correlation matrix of size 100 x 100 and would like to extract the
diagonal matrix from it. I have used the for loop to store tha correlation
values of the diagonal matrix. Is there a 'R way' of doing this?
Thanks in advance.
Kind regards,
Ezhil
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Re: [R] Extracting diagonal matrix
[EMAIL PROTECTED] wrote: Try ?diag Or, if he really means the diagonal of a 100x100 correlation matrix, rep(1,100) :-) Rory Winston RBS Global Banking Markets Office: +44 20 7085 4476 -Original Message- From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] On Behalf Of A Ezhil Sent: 21 November 2008 12:28 To: r-help@r-project.org Subject: [R] Extracting diagonal matrix Dear All, I have a correlation matrix of size 100 x 100 and would like to extract the diagonal matrix from it. I have used the for loop to store tha correlation values of the diagonal matrix. Is there a 'R way' of doing this? Thanks in advance. Kind regards, Ezhil -- O__ Peter Dalgaard Øster Farimagsgade 5, Entr.B c/ /'_ --- Dept. of Biostatistics PO Box 2099, 1014 Cph. K (*) \(*) -- University of Copenhagen Denmark Ph: (+45) 35327918 ~~ - ([EMAIL PROTECTED]) FAX: (+45) 35327907 __ R-help@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] ggplot2 - version 0.8
ggplot2 ggplot2 is a plotting system for R, based on the grammar of graphics, which tries to take the good parts of base and lattice graphics and avoid bad parts. It takes care of many of the fiddly details that make plotting a hassle (like drawing legends) as well as providing a powerful model of graphics that makes it easy to produce complex multi-layered graphics. Find out more at http://had.co.nz/ggplot2, and check out the nearly 500 examples of ggplot in use. If you're interested, you can also sign up to the ggplot2 mailing list at http://groups.google.com/group/ggplot2, or track development at http://github.com/hadley/ggplot2 ggplot2 0.8 (2008-11-18) The two biggest new features in this release are the (long awaited) ability to have scales that vary between facets, and a faceting system that works like lattice (facet_wrap). From qplot, you can use facet_wrap by specifying one sided formula (~ colour, as opposed to . ~ color). To see some potential uses for these new features, see the Positioning chapter of the book or the documentation for facet_wrap and facet_grid. Implementing these changes has required a rewrite of large parts of the coordinate systems code, so if anything seems strange with non-Cartesian coordinate systems, please get in touch. I've also made another round of tweaks to make the plots more aesthetically pleasing. This includes using a bright blue colour for geoms used to add statistical summaries to plots (contour, smooth, and quantiles), and tweaking the default colour scheme for the continuous colour scale. Please let me know what you think. Remember that most of these options are controllable with the theming system - see the book chapter Polishing your plots for publication. Accompanying this new release of the package is an updated and expanded version of the book. The content of the book is now largely complete (~170 pages), and over the coming months I will be working on make it polished and easy to understand. See http://had.co.nz/ggplot2/book. I love to hear your feedback about the book, but at this point please don't bother reporting minor typos, I would much rather hear about what you want to do, but can't figure out from the book. Other new features: * geom_bin2d/stat_bin2d geom_hex/stat_binhex: for 2d square and hexagon binning, particularly useful for alleviating overplotting in scatterplots * geom_freqpoly: draws frequency polygons (= stat_bin + geom_line) * scale_position: both discrete and continuous gain a new formatter argument to control the default formatting of the axis labels. See also the handy numeric formatters: dollar, comma and percent * the xlim and ylim functions now produce discrete scales when appropriate, and generate a reverse scale if the minimum is greater than the maximum Improvements * coord_map gains experimental axis labels * facet_grid: new support for varying scales in rows and columns * facet_wrap: new faceter which wraps a 1d ribbon of panels into 2d, in a similar way to lattice * geom_bin: gains a drop argument to control whether or not 0 count bins should be removed * geom_path and geom_line gain arrows argument to match geom_segment * ggsave now checks that you are using it with a ggplot plot * ggsave now produces postscript files that are suitable for embedding in another document * ggsave now recognises the .svg extension and will produce svg files, if possible * ggsave: default dpi changed to 300, on the assumption that you are saving the plot for printing * qplot: uses facet_wrap if formula looks like ~ a + b (as opposed to a ~ b) Aesthetic tweaks * geom_bar, geom_polygon, geom_rect, ...: default fill colour is now much closer to black to match the defaults in other geoms (point, line, etc) * geom_line, geom_path, geom_segment: lines have squared ends * geom_point, geom_pointrange and geom_boxplot: now use shape = 16 instead of 19. This shape does not have a border from R 2.8 on, and so will look better when displayed transparently. * geom_contour, geom_density2d, geom_quantile and geom_smooth use a bright blue colour for lines, to make them stand out when used with black points * scale_gradient: tweaked default colours to make more aesthetically pleasing * theme: new theme setting panel.margin (a unit) controls gap between panels in facetted plots (for both grid and wrap) * theme_gray: removed black border around strips * theme_bw: tweaks to make black and white theme look a little nicer Bug fixes * coord_cartesian now correctly clips instead of dropping points outside of its limits * facet_grid: margins now grouped correctly in default case (non-aesthetic variables ignored when generating default group value) * facet_grid: fix long standing bug when combining datasets with different levels of facetting variable * geom_smooth calls stat::predict explicitly to avoid conflicts with packages that override
Re: [R] Math Expression in 3D Plots
Alan Lue wrote:
Is there anyway to label axes in 3D plots with mathematical expressions?
In the code below, I want to replace delta_yrsed with what \Delta
\widehat{yrsed} represents in TeX, but the [xyz]lab parameters of title3d
appear to only accept character strings.
Unfortunately, that's right: rgl doesn't have any support for plotmath
type text.
The only way to get what you want would be to produce bitmaps of the
labels, then place those in the plots as sprites or surface textures.
Duncan Murdoch
require(rgl)
fn.delta.yrsed - function(dist, delta.dist,
beta.dist=-0.1376463, beta.dist2=0.0088698) {
delta.yrsed - (beta.dist + 2*beta.dist2*dist)*delta.dist +
beta.dist2*delta.dist^2
return(delta.yrsed)
}
plot.deeffect - function(scolor=blue) {
delta.dist - dist - seq(0, 16, .5)
delta.yrsed - outer(dist, delta.dist, fn.delta.yrsed)
rgl.open()
bbox3d(xat=seq(0, 16, 2), yat=0:5, zat=seq(0, 16, 2), color=black)
title3d(main=Effect of Change in dist on yrsed,
pos=c(NA, 8, 0), color=black)
title3d(xlab=dist, pos=c(NA, 0, -3), color=black)
title3d(ylab=delta_yrsed, pos=c(12, NA, -3), color=black)
title3d(zlab=delta_dist, pos=c(-3, 0, NA), color=black)
rgl.bg(color=rep(white, 2))
rgl.surface(dist, delta.dist, delta.yrsed,
color=scolor, front=lines, back=lines)
}
Alan
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Re: [R] summary statistics into table/data base, many factors to analyse
Hi [EMAIL PROTECTED] napsal dne 21.11.2008 11:50:52: Dear list, thanks to your help I managed to find means of analysing my data. However, the whole data set contains 264 variables. Of which some are factors, others are not. The factors tend to be grouped, e.g. data$f1304 to data$f1484 and data$f3204 to data$5408. But there are other types of variables in the data set as well, e.g. data$f1504. Not every spot is taken, i.e data$f1345 to data$1399 might not exist in the data set. The solution summaryBy works for cross analysis, of which there is a handful. So I am not worried here. The solution from Jorge is fine. However, I am trying to get my head around how to efficiently reduce my data set to the dependet variable and the factors such that the solution is applicable. Having to type each variable into my.reduced.data - cbind(my.data$f1001, my.data$1002, my.data$1003... is an obvious option, but does not seem to be the most efficient one. Maybe not so obvious. How did you get your data into R? By some read.* command? Then it shall be data frame with appropriate column type. see str(mydata) and you can choose only columns you really want by mydata[, select.some.columns] If your data is a list (see Intro manual for data types and its properties), then the transformation to data frame depends partly on how it looks like and if it has the same number of values. do.call(cbind, mydata) shall combine all vectors in mydata however it will convert them to unique type as cbind produce matrix which has to have only one type of data. If all variables have same length do.call(data.frame, mydata) will produce data frame and all variables shall be preserved in their respective type. Regards Petr Are there better ways to go about? Thanks, Gerit -- Sensationsangebot nur bis 30.11: GMX FreeDSL - Telefonanschluss + DSL für nur 16,37 Euro/mtl.!* http://dsl.gmx.de/?ac=OM.AD.PD003K11308T4569a __ 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] post - plotting lines in a divided graphic device
On Fri, 21 Nov 2008, _ wrote: Hi all, I have a graphic device divided in 2 areas for a plot. Is it possible to add lines or points in the first plot after the last one have been set, without plotting all the data again ? See ?par, look at mfg. However, you would probably find screen() easier to use, and the way you have done this you have lost the coordinate system for the first plot and would need plot(1, new=T) to reset it. Example par(mfrow(2,1)) plot(1) plot(2) lines(c(1,2)) # should be visible in the first plot That's not a valid R command. Thanks for help. Greetings __ 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, [EMAIL PROTECTED] Professor of Applied Statistics, http://www.stats.ox.ac.uk/~ripley/ University of Oxford, Tel: +44 1865 272861 (self) 1 South Parks Road, +44 1865 272866 (PA) Oxford OX1 3TG, UKFax: +44 1865 272595 __ R-help@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] R course in Scotland
Thanks to the many people on the list who provided helpful responses, including those who Emailed me directly. Several people have suggested that I just pick up R and give it a try. My reluctance to do this is that I am already very familiar with my current working method (Python + Numpy) and I worry that without a course I will work in a Python-centric way, which won't be optimal. I feel it would also be useful to spend some time with an R expert so that I can find out what R does well and what it does badly. This way I'll when I should use R and when I should use some other method (like Python). I may find out that I'm better off with Python for everything I do, but I won't be confident in this conclusion unless I talk to somebody who really knows what they're doing. Anyway, I now have a few courses to choose from, so many thanks for the suggestions. 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] write every element of a variable into a separate text-file
Hello, what I want to do, is, to write every element of a variable into a separate text-file automatically: My Variable: wull [1] Hallo Leute, wie gehts denn euch seid ihr noch alle... [2] Is their anyone how can help me with... [3] mann, mann, mann... das nervt aber.. [4] how are you littele strange tiger... [5] )()()(UJKJKJIJIJJOO9989 [6] bradortslow, eiwudoiude, kdkdkdk:::idjidji I was trying to do things like that: for (i in seq(along = wull)) + write(wull[i], (C://Users//zuber//Documents//wull(1).txt), + append = F) But what I get is just the last element [6] (bradortslow, eiwudoiude, kdkdkdk:::idjidji) of my variable in just one text-file. - I do not know, how to say to R, that it should write: [1] Hallo Leute, wie gehts denn euch seid ihr noch alle... in wull(1).txt [2] Is their anyone how can help me with... in wull(2).txt [3] mann, mann, mann... das nervt aber.. in wull(3).txt ... an so on till the last element. I hope this is possible and someone can help me. Martin __ 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] Extracting diagonal matrix
Thanks. I would like to extract all the matrix entries below or above the diagnol. diag(x) simply gives diagonal elements. Thanks. Kind regards, Ezhil --- On Fri, 11/21/08, Peter Dalgaard [EMAIL PROTECTED] wrote: From: Peter Dalgaard [EMAIL PROTECTED] Subject: Re: [R] Extracting diagonal matrix To: [EMAIL PROTECTED] Cc: [EMAIL PROTECTED], r-help@r-project.org Date: Friday, November 21, 2008, 6:23 PM [EMAIL PROTECTED] wrote: Try ?diag Or, if he really means the diagonal of a 100x100 correlation matrix, rep(1,100) :-) Rory Winston RBS Global Banking Markets Office: +44 20 7085 4476 -Original Message- From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] On Behalf Of A Ezhil Sent: 21 November 2008 12:28 To: r-help@r-project.org Subject: [R] Extracting diagonal matrix Dear All, I have a correlation matrix of size 100 x 100 and would like to extract the diagonal matrix from it. I have used the for loop to store tha correlation values of the diagonal matrix. Is there a 'R way' of doing this? Thanks in advance. Kind regards, Ezhil -- O__ Peter Dalgaard Øster Farimagsgade 5, Entr.B c/ /'_ --- Dept. of Biostatistics PO Box 2099, 1014 Cph. K (*) \(*) -- University of Copenhagen Denmark Ph: (+45) 35327918 ~~ - ([EMAIL PROTECTED]) FAX: (+45) 35327907 __ R-help@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] post - plotting lines in a divided graphic device
Hi all, I have a graphic device divided in 2 areas for a plot. Is it possible to add lines or points in the first plot after the last one have been set, without plotting all the data again ? Example par(mfrow(2,1)) plot(1) plot(2) lines(c(1,2)) # should be visible in the first plot Thanks for help. Greetings __ 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] Extracting diagonal matrix
See ?upper.tri and ?lower.tri. On Fri, Nov 21, 2008 at 1:05 PM, A Ezhil [EMAIL PROTECTED] wrote: Thanks. I would like to extract all the matrix entries below or above the diagnol. diag(x) simply gives diagonal elements. Thanks. Kind regards, Ezhil --- On Fri, 11/21/08, Peter Dalgaard [EMAIL PROTECTED] wrote: From: Peter Dalgaard [EMAIL PROTECTED] Subject: Re: [R] Extracting diagonal matrix To: [EMAIL PROTECTED] Cc: [EMAIL PROTECTED], r-help@r-project.org Date: Friday, November 21, 2008, 6:23 PM [EMAIL PROTECTED] wrote: Try ?diag Or, if he really means the diagonal of a 100x100 correlation matrix, rep(1,100) :-) Rory Winston RBS Global Banking Markets Office: +44 20 7085 4476 -Original Message- From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] On Behalf Of A Ezhil Sent: 21 November 2008 12:28 To: r-help@r-project.org Subject: [R] Extracting diagonal matrix Dear All, I have a correlation matrix of size 100 x 100 and would like to extract the diagonal matrix from it. I have used the for loop to store tha correlation values of the diagonal matrix. Is there a 'R way' of doing this? Thanks in advance. Kind regards, Ezhil -- O__ Peter Dalgaard Øster Farimagsgade 5, Entr.B c/ /'_ --- Dept. of Biostatistics PO Box 2099, 1014 Cph. K (*) \(*) -- University of Copenhagen Denmark Ph: (+45) 35327918 ~~ - ([EMAIL PROTECTED]) FAX: (+45) 35327907 __ R-help@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. -- Henrique Dallazuanna Curitiba-Paraná-Brasil 25° 25' 40 S 49° 16' 22 O [[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] write every element of a variable into a separate text-file
Well, you could read the R Data Import/Export manual. Or you could reread the help for write and figure out what the append argument does. Or you could reread the help for write and notice that it should write out your entire variable without requiring a loop. Sarah On Fri, Nov 21, 2008 at 8:39 AM, [EMAIL PROTECTED] wrote: Hello, what I want to do, is, to write every element of a variable into a separate text-file automatically: -- 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] Extracting diagonal matrix
You can also do it from first principles: outer(1:100,1:100,``) * m which generalizes nicely to , =, =, !=, etc. -s On 11/21/08, Henrique Dallazuanna [EMAIL PROTECTED] wrote: See ?upper.tri and ?lower.tri. __ 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] R course in Scotland
pzs wrote: Several people have suggested that I just pick up R and give it a try. My reluctance to do this is that I am already very familiar with my current working method (Python + Numpy) and I worry that without a course I will work in a Python-centric way, which won't be optimal. I know this isn't what you asked, but as a sidenote, if you are used to working with Python and Numpy, then have you considered Sage ( http://sagemath.org/ http://sagemath.org/ )? It's a Python environment that you can run R commands from (amongst other things). That way you can keep your current working style but let R do the hard stats for you. - Regards, Richie. Mathematical Sciences Unit HSL -- View this message in context: http://www.nabble.com/R-course-in-Scotland-tp20601268p20623414.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] write every element of a variable into a separate text-file
Your code isn't changing the filename Try this for(i in seq_along(wull)) write(wull[i], paste(C://Users//zuber//Documents//wull(,i,),.txt,sep=)) -Original Message- From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] On Behalf Of [EMAIL PROTECTED] Sent: Friday, November 21, 2008 7:39 AM To: r-help@r-project.org Subject: [R] write every element of a variable into a separate text-file Hello, what I want to do, is, to write every element of a variable into a separate text-file automatically: My Variable: wull [1] Hallo Leute, wie gehts denn euch seid ihr noch alle... [2] Is their anyone how can help me with... [3] mann, mann, mann... das nervt aber.. [4] how are you littele strange tiger... [5] )()()(UJKJKJIJIJJOO9989 [6] bradortslow, eiwudoiude, kdkdkdk:::idjidji I was trying to do things like that: for (i in seq(along = wull)) + write(wull[i], (C://Users//zuber//Documents//wull(1).txt), + append = F) But what I get is just the last element [6] (bradortslow, eiwudoiude, kdkdkdk:::idjidji) of my variable in just one text-file. - I do not know, how to say to R, that it should write: [1] Hallo Leute, wie gehts denn euch seid ihr noch alle... in wull(1).txt [2] Is their anyone how can help me with... in wull(2).txt [3] mann, mann, mann... das nervt aber.. in wull(3).txt ... an so on till the last element. I hope this is possible and someone can help me. Martin __ 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] Extracting diagonal matrix
Hi Ezhil,
Maybe this will help. There might be an easier way to do this but here is
one solution.
tril - function(A)
{
A - as.matrix(A)
cmats - matrix(rep(1,length(A)),dim(A)[1],dim(A)[2])
upper.tri(cmats,diag=T)
cmats[upper.tri(cmats)] - 0
out - A*cmats
return(out)
}
tril(correlation.matrix)
will give you the output.
HTH
Best Regards
Anup
On Fri, Nov 21, 2008 at 7:28 AM, A Ezhil [EMAIL PROTECTED] wrote:
Dear All,
I have a correlation matrix of size 100 x 100 and would like to extract the
diagonal matrix from it. I have used the for loop to store tha correlation
values of the diagonal matrix. Is there a 'R way' of doing this?
Thanks in advance.
Kind regards,
Ezhil
__
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https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide
http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
[[alternative HTML version deleted]]
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and provide commented, minimal, self-contained, reproducible code.
[R] Dataframe with single level column
Dear all,
I have a dataframe with multiple observations and the levels as the last
column, as in:
d -
data.frame(A=sample(1:100,12),B=sample(1:100,12),levels=c(rep('A',4),rep('B',4),rep('C',4)))
d
A B levels
1 77 40 A
2 14 18 A
3 56 7 A
4 46 27 A
5 63 35 B
6 80 21 B
7 3 54 B
8 93 76 B
9 5 46 C
10 16 53 C
11 40 17 C
12 25 31 C
I need to run anova analyis on the group in levels against the merge data in
the first two columns. I can manually split and join the different columns as in
d.t -
rbind(data.frame(value=d[,1],ind=d[,3]),data.frame(value=d[,2],ind=d[,3]))
but I was wondering if there would be a more elegant and easy way than that
that would prevent me from hard coding the different vectors making the data
frame.
Thanks
--
Marco Blanchette, Ph.D.
Assistant Investigator
Stowers Institute for Medical Research
1000 East 50th St.
Kansas City, MO 64110
Tel: 816-926-4071
Cell: 816-726-8419
Fax: 816-926-2018
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[R] Breakdown of Vector
Hi,
I have a vector of Size 7420. I wanna break down in such a way that every 20
elements of it should be as elements of an list.
Ex
EXAM1
ABC, SDF, LMN,ERF,EGC,EFG,WER,FRE,QWE,ERT,DGW,QWE,YUR,ERT,GHJ,FHH,7420
what i want is
Breakdown.list
[[1]]
ABC,SDF,.20
[[2]]
21.40
[[3]]
4150
i thought of using a for loop but i am wondering how to incerment
test.breakdown.list-list()
for( i =0;i=length(EXAM1);i+20)
{
test.breakdown.list-Exam1[c(i:i+20)]
print (paste(i))
}
Ias this how we do...please correct me if am wrong.
Regards
Ramya
--
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http://www.nabble.com/Breakdown-of-Vector-tp20623764p20623764.html
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
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Re: [R] glmer for cauchit link function
On Thu, Nov 20, 2008 at 5:38 PM, Lizz Metcalfe [EMAIL PROTECTED] wrote: I am trying to fit a generalized linear mixed effects model with a binomial link function, my response data is binary, using the lme4 R package, for the glmer model but with the cauchit link function (CDF of Cauchy distribution), under the package this has not yet been coded and was wondering if anyone knew a way in which I could incorporate this link function into the code. The current version of lme4 uses C code for the inverse link function so you would need to modify the function lme4_muEta in the file lme4/src/lmer.c If you look at the other inverse link functions defined in that C function you will see that it is not a matter of simply writing down the inverse link. You need to be very careful of the edge conditions and most of the code is devoted to those kinds of checks. __ 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] Dataframe with single level column
I need to run anova analyis on the group in levels against the merge data in the first two columns. I can manually split and join the different columns as in d.t - rbind(data.frame(value=d[,1],ind=d[,3]),data.frame(value=d[,2],ind=d[,3])) but I was wondering if there would be a more elegant and easy way than that that would prevent me from hard coding the different vectors making the data frame. ?reshape hth Stefan -- Sensationsangebot nur bis 30.11: GMX FreeDSL - Telefonanschluss + DSL für nur 16,37 Euro/mtl.!* http://dsl.gmx.de/?ac=OM.AD.PD003K11308T4569a __ 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] matlab style of storing arrays compared to R
R uses column-major ordering of multidimensional arrays, as in Fortran, unlike the row-major ordering of C multidimensional arrays. This is because the numerical linear algebra code used in R is from the Eispack, Linpack, BLAS, Lapack family of Fortran subroutine packages. Even when implementation languages other than Fortran are used, the Fortran storage conventions prevail because they affect the design of algorithms. Matlab is based on the same code (Matlab began as an interactive wrapper around Eispack and Linpack) and uses the same representation of arrays. On Fri, Nov 21, 2008 at 2:41 AM, Rainer M Krug [EMAIL PROTECTED] wrote: Hi I am trying to figure out, if the matlab style of linear indexing of an array is the same as in R. i.e. when x - array( 1:24, dim=c(2,3,4) ) x[3] 3 and if the same is true in matlab, assuming that x[n1,n2,n3] in R returns the same as y(n1,n2,n3) when y is a matrix in matlab I found the following reference http://www.mathworks.com/access/helpdesk/help/techdoc/index.html?/access/helpdesk/help/techdoc/matlab_prog/f1-86528.html#f1-86846 And it seems to be the same, but another source says it is different. Could somebody confirm, if it is the same? Thanks, Rainer -- Rainer M. Krug, PhD (Conservation Ecology, SUN), MSc (Conservation Biology, UCT), Dipl. Phys. (Germany) Centre of Excellence for Invasion Biology Faculty of Science Natural Sciences Building Private Bag X1 University of Stellenbosch Matieland 7602 South Africa __ 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] syntax and package for generalized linear mixed models
On Thu, Nov 20, 2008 at 2:54 PM, Jeff Evans [EMAIL PROTECTED] wrote: I am making the switch to R and uncertain which of the several packages for mixed models is appropriate for my analysis. I am waiting for Pinheiro and Bates' book to arrive via inter-library loan, but it will be a week or more before it arrives. I am trying to fit a generalized linear mixed model of survival data (successes/trials) as a function of several categorical fixed and nested random effects and a covariate. Additionally, the residual variance in the data is a negative function of the covariate, which I would like to model as well. Working in SAS, I was able to do this on transformed data in PROC MIXED, but ran into trouble trying to run it as a logistic regression in the original scale in GLIMMIX. Can glmer in lme4 do this? It seems that weights in lme4 refers to weighting of observations rather than modeling the variance-covariate, as it does in nlme. I tried running nlme, but I'm stuck on syntax, particularly with regards to how the fixed and random statements should be constructed separate from the model statement (not sure how this is supposed to work). Generally, I believe my model in lme4 should look like this: gm1 = glmer(cbind(#successes,#trials) ~ A*B + covariate + (1|B/C), family = binomial, link=logit, data=mydata, weights=varExp(form = ~covariate)) I'm sorry to say that this is not a valid model specification for glmer. As you have surmised, lme4 does not allow a general weights specification like this. Failure to accept a specification like this is not just an oversight or an unimplemented feature. This isn't a valid model specification because this isn't a valid model. If the conditional distribution of the response, given the value of the random effects, is Bernoulli (or Poisson) then it is completely specified by the conditional mean. You can't separately specify the mean and the variance for a Bernoulli or a Poisson distribution as you can for a Gaussian distribution. As tempting as it may be to want to have several dials and knobs on statistical models to tune their behavior we still need to be careful to specify a mathematical model that is consistent. where #trials is the number of subjects at the beginning of the experiment in each experimental unit, #successes is the number of survivors, A and B are fully crossed fixed categorical factors, C is a categorical random factor nested within B (i.e. random site within year), and covariate is a continuous numerical variable ranging from 1- +inf. Can someone please suggest (a) which package to use for this analysis and (b) perhaps share some dummy code using my mock variables above? Many thanks, Jeff Evans PhD Candidate Department of Entomology EEBB Graduate Program Michigan State University [[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] Breakdown of Vector
No need to use a for loop. Try something like this
dim(EXAM1) - c(20,7420/20)
test.breakdown.list - as.list(data.frame(EXAM1))
-Original Message-
From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED]
On Behalf Of Rajasekaramya
Sent: Friday, November 21, 2008 9:48 AM
To: r-help@r-project.org
Subject: [R] Breakdown of Vector
Hi,
I have a vector of Size 7420. I wanna break down in such a way that
every 20 elements of it should be as elements of an list.
Ex
EXAM1
ABC, SDF,
LMN,ERF,EGC,EFG,WER,FRE,QWE,ERT,DGW,QWE,YUR,ERT,GHJ,FHH,7420
what i want is
Breakdown.list
[[1]]
ABC,SDF,.20
[[2]]
21.40
[[3]]
4150
i thought of using a for loop but i am wondering how to incerment
test.breakdown.list-list()
for( i =0;i=length(EXAM1);i+20)
{
test.breakdown.list-Exam1[c(i:i+20)]
print (paste(i))
}
Ias this how we do...please correct me if am wrong.
Regards
Ramya
--
View this message in context:
http://www.nabble.com/Breakdown-of-Vector-tp20623764p20623764.html
Sent from the R help mailing list archive at Nabble.com.
__
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PLEASE do read the posting guide
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and provide commented, minimal, self-contained, reproducible code.
__
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] Breakdown of Vector
Also something along the following lines:
x - 1:100
y - split(x, (seq(along = x) - 1) %/% 5)
HTH,
Giovanni Petris
Date: Fri, 21 Nov 2008 07:48:09 -0800 (PST)
From: Rajasekaramya [EMAIL PROTECTED]
Sender: [EMAIL PROTECTED]
Precedence: list
Hi,
I have a vector of Size 7420. I wanna break down in such a way that every 20
elements of it should be as elements of an list.
Ex
EXAM1
ABC, SDF, LMN,ERF,EGC,EFG,WER,FRE,QWE,ERT,DGW,QWE,YUR,ERT,GHJ,FHH,7420
what i want is
Breakdown.list
[[1]]
ABC,SDF,.20
[[2]]
21.40
[[3]]
4150
i thought of using a for loop but i am wondering how to incerment
test.breakdown.list-list()
for( i =0;i=length(EXAM1);i+20)
{
test.breakdown.list-Exam1[c(i:i+20)]
print (paste(i))
}
Ias this how we do...please correct me if am wrong.
Regards
Ramya
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Ph: (479) 575-6324, 575-8630 (fax)
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Re: [R] Breakdown of Vector
another way:
exam.list - split(EXAM, floor((seq_along(EXAM) - 1) / 20))
On Fri, Nov 21, 2008 at 10:48 AM, Rajasekaramya [EMAIL PROTECTED] wrote:
Hi,
I have a vector of Size 7420. I wanna break down in such a way that every 20
elements of it should be as elements of an list.
Ex
EXAM1
ABC, SDF, LMN,ERF,EGC,EFG,WER,FRE,QWE,ERT,DGW,QWE,YUR,ERT,GHJ,FHH,7420
what i want is
Breakdown.list
[[1]]
ABC,SDF,.20
[[2]]
21.40
[[3]]
4150
i thought of using a for loop but i am wondering how to incerment
test.breakdown.list-list()
for( i =0;i=length(EXAM1);i+20)
{
test.breakdown.list-Exam1[c(i:i+20)]
print (paste(i))
}
Ias this how we do...please correct me if am wrong.
Regards
Ramya
--
View this message in context:
http://www.nabble.com/Breakdown-of-Vector-tp20623764p20623764.html
Sent from the R help mailing list archive at Nabble.com.
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+1 513 646 9390
What is the problem that you are trying to solve?
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[R] lsoda warning too much accuracy requested
Dear list -
Does anyone have any ideas / comments about why I am receiving the following
warning when I run lsoda:
1: lsoda-- at t (=r1), too much accuracy requested in: lsoda(start, times,
model, parms)
2: for precision of machine.. see tolsf (=r2) in: lsoda(start, times,
model, parms)
I have tried changing both rtol and atol but without success. I saw the
thread in the R-archive of 11 June 2004 but this has not helped me.
I have built the model in stages and the problem only occurs when the
exponent beta in the third DE is anything other than 0 or 1. If beta = 0 or
1 then the solver gives me perfectly justifiable results. Just changing
beta to 0.9 or similar causes the problem.
I am still new to R so I am unsure if it is my programming or my
understanding of the way lsoda works.
Any comments or input would be welcome.
Many thanks
Colleen
___
My code is:
library(odesolve)
SI - 80
model - function(t, x, parms) {
H - x[1]
BA - x[2]
N - x[3]
with(as.list(parms), {
dHdt - (b/c)*(((a**c)*((H)**(1-c))-H))
dBAdt - -(BA*b)*(c0+(c1*SI)-log(BA))/(log(1-((H/a)**c)))
dNdt - N*alpha*(((log(1-((H/a)**c)))/b)**beta) - (gamma*BA)
list(c(dHdt, dBAdt, dNdt))
})
}
times - seq(0, 40, 1)
parms - c(a=(SI*1.258621)-1.32759, b=0.1, c=0.4, c0=4.6012, c1=0.013597,
alpha=0.0005, beta=0.5, gamma=0.01)
start - c(H=0.1, BA=0.1, N=600)
out - as.data.frame(lsoda(start, times, model, parms))
[[alternative HTML version deleted]]
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[R] plots of ACF
Hello. I have one Model (M3) fitted using the lme package, and I have checked the correlation structure of within-group errors using plot(ACF (M3,maxLag=10),alpha=0.05) But now I am not sure how to interpret this plot for the empirical autocorrelation function. The problem is that I am used to see/interpret diagrams in which all the autocorrelation Lags, except lag-1, are inside the confidence envelopes, or those plots where only Lags 1 and 2 are outside those envelopes. But how should I interpret my ACF plot, where Lags 1, 7, 8 and 9, are those outside those envelopes? Is there any correlation structure? Which one might that be? AR1()? Also, I have used the plot of the empirical ACF of the normalized residuals, but it gives exactly the same results. Could you please help me? Best regards Sara Mouro Sara Maltez Mouro Centro de Ecologia Funcional Departamento de Botânica Universidade de Coimbra [EMAIL PROTECTED] www.uc.pt/ecology/saramaltezmouro Sara Maltez Mouro Centro de Ecologia Funcional Departamento de Botânica Universidade de Coimbra [EMAIL PROTECTED] www.uc.pt/ecology/saramaltezmouro [[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-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] lsoda warning too much accuracy requested
Hi Colleen,
this error was not uncommon and is usually a sign of a numerically
problematic or wrongly implemented model. Please use package deSolve,
the successor of odesolve, that is more robust and has also a bunch
alternative solvers for difficult cases.
I tested your code with deSolve (on R 2.8.0, Windows) and it runs
without problems.
Thomas Petzoldt
BTW: your system worked also with odesolve, so my question: which
versions (R, odesolve) and operating system are you using?
Colleen Carlson wrote:
Dear list -
Does anyone have any ideas / comments about why I am receiving the following
warning when I run lsoda:
1: lsoda-- at t (=r1), too much accuracy requested in: lsoda(start, times,
model, parms)
2: for precision of machine.. see tolsf (=r2) in: lsoda(start, times,
model, parms)
I have tried changing both rtol and atol but without success. I saw the
thread in the R-archive of 11 June 2004 but this has not helped me.
I have built the model in stages and the problem only occurs when the
exponent beta in the third DE is anything other than 0 or 1. If beta = 0 or
1 then the solver gives me perfectly justifiable results. Just changing
beta to 0.9 or similar causes the problem.
I am still new to R so I am unsure if it is my programming or my
understanding of the way lsoda works.
Any comments or input would be welcome.
Many thanks
Colleen
___
My code is:
library(odesolve)
SI - 80
model - function(t, x, parms) {
H - x[1]
BA - x[2]
N - x[3]
with(as.list(parms), {
dHdt - (b/c)*(((a**c)*((H)**(1-c))-H))
dBAdt - -(BA*b)*(c0+(c1*SI)-log(BA))/(log(1-((H/a)**c)))
dNdt - N*alpha*(((log(1-((H/a)**c)))/b)**beta) - (gamma*BA)
list(c(dHdt, dBAdt, dNdt))
})
}
times - seq(0, 40, 1)
parms - c(a=(SI*1.258621)-1.32759, b=0.1, c=0.4, c0=4.6012, c1=0.013597,
alpha=0.0005, beta=0.5, gamma=0.01)
start - c(H=0.1, BA=0.1, N=600)
out - as.data.frame(lsoda(start, times, model, parms))
[[alternative HTML version deleted]]
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--
Thomas Petzoldt
Technische Universitaet Dresden
Institut fuer Hydrobiologie[EMAIL PROTECTED]
01062 Dresden http://tu-dresden.de/hydrobiologie/
GERMANY
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Re: [R] Simulation Result display form
Use something like sprintf if you want trailing zeros. x [1] 0.78 sprintf(%.4f, x) [1] 0.7800 On Fri, Nov 21, 2008 at 5:31 AM, Abelian [EMAIL PROTECTED] wrote: after obtaining the result, the result is displayed below: individual RS_number Phy_Posi LOH_intensity 1 1718890 rs1496555 2.2241110.8121 2 1668776 rs2376495 3.0849860.786 --- 3 1723597 rs4648462 3.1551270.784 --- 4 1728870 rs10492940 3.1876070.7831 5 1669946 rs10492939 3.2927310.7801 6 1708099 rs10492938 3.2941960.78 --- 7 1716798 rs10492937 3.3291990.7791 8 1701872 rs1128474 3.5354720.7733 9 1697748 rs2154068 3.7108250.7685 10-1699138 rs2887274 3.7111780.7685 my problem is in the result of LOH_intesity. i hope that 0.786 can be showed 0.7860, 0.78 can be 0.7800. so, how can I manage my result to be those? Thanks a lot... __ 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 Cincinnati, OH +1 513 646 9390 What is the problem that you are trying to solve? __ 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] Extracting diagonal matrix
Hi,
Thank you very much for the help. c[lower.tri(c,diag=FALSE)] works fine for me.
Thanks again.
Kind regards,
Ezhil
--- On Fri, 11/21/08, Anup Menon [EMAIL PROTECTED] wrote:
From: Anup Menon [EMAIL PROTECTED]
Subject: Re: [R] Extracting diagonal matrix
To: [EMAIL PROTECTED]
Cc: r-help@r-project.org
Date: Friday, November 21, 2008, 9:16 PM
Hi Ezhil,
Maybe this will help. There might be an easier way to do
this but here is
one solution.
tril - function(A)
{
A - as.matrix(A)
cmats - matrix(rep(1,length(A)),dim(A)[1],dim(A)[2])
upper.tri(cmats,diag=T)
cmats[upper.tri(cmats)] - 0
out - A*cmats
return(out)
}
tril(correlation.matrix)
will give you the output.
HTH
Best Regards
Anup
On Fri, Nov 21, 2008 at 7:28 AM, A Ezhil
[EMAIL PROTECTED] wrote:
Dear All,
I have a correlation matrix of size 100 x 100 and
would like to extract the
diagonal matrix from it. I have used the for loop to
store tha correlation
values of the diagonal matrix. Is there a 'R
way' of doing this?
Thanks in advance.
Kind regards,
Ezhil
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Re: [R] matlab style of storing arrays compared to R
On Fri, Nov 21, 2008 at 6:12 PM, Douglas Bates [EMAIL PROTECTED] wrote: R uses column-major ordering of multidimensional arrays, as in Fortran, unlike the row-major ordering of C multidimensional arrays. This is because the numerical linear algebra code used in R is from the Eispack, Linpack, BLAS, Lapack family of Fortran subroutine packages. Even when implementation languages other than Fortran are used, the Fortran storage conventions prevail because they affect the design of algorithms. Matlab is based on the same code (Matlab began as an interactive wrapper around Eispack and Linpack) and uses the same representation of arrays. Thanks a lot for your detailed explanation. This will make my life much easier. Rainer On Fri, Nov 21, 2008 at 2:41 AM, Rainer M Krug [EMAIL PROTECTED] wrote: Hi I am trying to figure out, if the matlab style of linear indexing of an array is the same as in R. i.e. when x - array( 1:24, dim=c(2,3,4) ) x[3] 3 and if the same is true in matlab, assuming that x[n1,n2,n3] in R returns the same as y(n1,n2,n3) when y is a matrix in matlab I found the following reference http://www.mathworks.com/access/helpdesk/help/techdoc/index.html?/access/helpdesk/help/techdoc/matlab_prog/f1-86528.html#f1-86846 And it seems to be the same, but another source says it is different. Could somebody confirm, if it is the same? Thanks, Rainer -- Rainer M. Krug, PhD (Conservation Ecology, SUN), MSc (Conservation Biology, UCT), Dipl. Phys. (Germany) Centre of Excellence for Invasion Biology Faculty of Science Natural Sciences Building Private Bag X1 University of Stellenbosch Matieland 7602 South Africa __ 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. -- Rainer M. Krug, PhD (Conservation Ecology, SUN), MSc (Conservation Biology, UCT), Dipl. Phys. (Germany) Centre of Excellence for Invasion Biology Faculty of Science Natural Sciences Building Private Bag X1 University of Stellenbosch Matieland 7602 South Africa __ 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] capturing stderr/stdout
Hi, Prof. Ripley,
Thanks for the reply. Mostly I want to capture output as it is written
to the stream. For example, I quite often do the following to view the
progress of a log file from a computationally intensive script.
1. Open a console and type:
Rscript script.R script.log
which directs both stdout and stderr to script.log
2. Open another console and type:
tail -f script.log
This way I get both the script's log file and its current progress.
I guess my question is: Is there a way to accomplish the tail -f
command in R?
Thanks,
--sundar
Prof Brian Ripley said the following on 11/20/2008 11:43 PM:
I am not sure what the issue is here. Do you want to capture both stderr
and stdout (use 21 in the command with an sh-like shell), or is the
problem that you don't get immediate output?
The latter is a Perl issue: you need to circumvent output buffering.
See e.g
http://perl.plover.com/FAQs/Buffering.html
Sundar Dorai-Raj wrote:
Hi,
I have an application in perl that prints some output to either stderr
or stdout.
Here's an example:
# tmp.pl
print STDERR starting iterator\n;
for(my $i = 0; $i 100; $i++) {
print $i . \n;
}
# tmp.R
con - pipe(perl tmp.pl)
r - readLines(con, n = -1)
close(con)
However, the second line stalls until the perl for-loop finishes. What
I would like is to process each line as it comes. E.g. something like:
while(TRUE) {
r - readLines(con, n = 1) # read one line
if(r == 1) print(r)
if(length(r) == 0) break
}
Of course, this won't work since I'm not calling readLines
appropriately. Answers must work on Windows but may include cygwin
utilities if necessary. Any advice would be appreciated. Version info
at the end if it matters.
Thanks, --sundar
version
_
platform i386-pc-mingw32
arch i386
os mingw32
system i386, mingw32
status
major 2
minor 8.0
year 2008
month 10
day20
svn rev46754
language R
version.string R version 2.8.0 (2008-10-20)
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[R] Basic question on concatenating factors
Hi all, I hope it's not too trivial for the list - I'm trying to concatenate two factor arrays, and obtain the following: f1-factor(c(a,a,b)) f1 [1] a a b Levels: a b f2-factor(c(b,b,a)) f2 [1] b b a Levels: a b c(f1,f2) [1] 1 1 2 2 2 1 Instead of getting: [1] a a b b b a Levels: a b a related question is: how do I add a level which does not exists yet in a factored vector, so I'll be able to add later these values, without getting: In `[-.factor`(`*tmp*`, 2, value = c) : invalid factor level, NAs generated Thanks, EC __ 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] Discrepancy in the regression coefficients for Cox regression - PBC data set
Hi, When I run the following Cox proportional hazards model on the Mayo clinic's PBC data set (given in the survival package), the regression coefficients do not agree with the results presented in Table 4.6.3 (p. 195) of Fleming Harrington's book. library(survival) data(pbc) ans.cox - coxph(Surv(time, status) ~ log(bili) + log(alb) + age + log(protime) + edema) ans.cox ans.cox - coxph(Surv(time, status) ~ log(bili) + log(alb) + age + log(protime) + edema) ans.cox Call: coxph(formula = Surv(time, status) ~ log(bili) + log(alb) + age + log(protime) + edema) coef exp(coef) se(coef) z p log(bili) 0.8975 2.453 0.08271 10.85 0.0e+00 log(alb) -2.4524 0.086 0.65707 -3.73 1.9e-04 age 0.0382 1.039 0.00768 4.97 6.5e-07 log(protime) 2.345810.442 0.77425 3.03 2.4e-03 edema 0.6613 1.937 0.20595 3.21 1.3e-03 Likelihood ratio test=234 on 5 df, p=0 n= 418 These coefficients, however, are significantly different (i.e. the differences can't be just attributed to round-off's) from that reported in Table 4.6.3 (in the Final model column) of Fleming and Harrington (p. 195). The coefficients reported are: 0.8707, -2.533, 0.0394, 2.380, 0.8592. Note the big difference for the edema variable. It seems like the data set considered in the book and that available in survival package are the same (with n=418). I also re-ran the Cox PH model with the 2 data-errors discussed in p.188 of FH, but still I could not match the results in Table 4.6.3. Is it possible that the results could be explained due to difference in convergence during maximization of partial likelihood? Can anyone help me figure out why this diescrepancy exists? Thanks very much, Ravi. --- Ravi Varadhan, Ph.D. Assistant Professor, The Center on Aging and Health Division of Geriatric Medicine and Gerontology Johns Hopkins University Ph: (410) 502-2619 Fax: (410) 614-9625 Email: [EMAIL PROTECTED] Webpage: http://www.jhsph.edu/agingandhealth/People/Faculty/Varadhan.html [[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] plots of ACF
There are a variety of reasons that a question might go unanswered for over a day on R-help. The question may be so technical or narrow that only one or two people may be equipped to answer it. Or it may look like homework. Or it may lack sufficient detail one which to base an answer. Or it may sound like a request for statistical consultation more than a request for help with the correct use of R (which is the specified purpose of the list.) I am not a statistician or an r-expert and my experience with time series analysis is over 20 years ago, but my impression is that a mixture of some of those latter reasons may have been the reason for this identical question being unanswered yesterday. You have not specified any detail about the units of the serially correlated variable. You have not offered any output text. You have not described the research question. And there is no real R content. I can tell you that that time series analysis is well described in Modern Applied Statistics in S which some people use as their beginning R text. If time series was not in your intorductory text then maybe you should get one that handles it. As a complete guess I would speculate that seasonality might explain a series with significant correlations at lags of 1, 7, 8, and 9 but who knows? You have not given your audience much to work with. And you may be outside the boundaries of the list's purpose for existence. http://www.r-project.org/posting-guide.html http://www.catb.org/~esr/faqs/smart-questions.html -- Regards; David Winsemius, MD On Nov 21, 2008, at 11:45 AM, Sara Mouro wrote: Hello. I have one Model (M3) fitted using the lme package, and I have checked the correlation structure of within-group errors using plot(ACF (M3,maxLag=10),alpha=0.05) But now I am not sure how to interpret this plot for the empirical autocorrelation function. The problem is that I am used to see/interpret diagrams in which all the autocorrelation Lags, except lag-1, are inside the confidence envelopes, or those plots where only Lags 1 and 2 are outside those envelopes. But how should I interpret my ACF plot, where Lags 1, 7, 8 and 9, are those outside those envelopes? Is there any correlation structure? Which one might that be? AR1()? Also, I have used the plot of the empirical ACF of the normalized residuals, but it gives exactly the same results. Could you please help me? Best regards Sara Mouro Sara Maltez Mouro Centro de Ecologia Funcional Departamento de Botânica Universidade de Coimbra [EMAIL PROTECTED] www.uc.pt/ecology/saramaltezmouro Sara Maltez Mouro Centro de Ecologia Funcional Departamento de Botânica Universidade de Coimbra [EMAIL PROTECTED] www.uc.pt/ecology/saramaltezmouro [[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] Basic question on concatenating factors
Hi, I have a solution to concatenate two factors in one but I don't believe it is the best one: factor(c(as.character(f1),as.character(f2))) [1] a a b b b a Levels: a b You can always add a level by assigning a new vector at the level vector: levels(f1) - c(a,b,c) f1 [1] a a b Levels: a b c udi cohen wrote: Hi all, I hope it's not too trivial for the list - I'm trying to concatenate two factor arrays, and obtain the following: f1-factor(c(a,a,b)) f1 [1] a a b Levels: a b f2-factor(c(b,b,a)) f2 [1] b b a Levels: a b c(f1,f2) [1] 1 1 2 2 2 1 Instead of getting: [1] a a b b b a Levels: a b a related question is: how do I add a level which does not exists yet in a factored vector, so I'll be able to add later these values, without getting: In `[-.factor`(`*tmp*`, 2, value = c) : invalid factor level, NAs generated Thanks, EC __ 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. -- Alain Guillet Statistician and Computer Scientist SMCS - Institut de statistique - Université catholique de Louvain Bureau d.126 Voie du Roman Pays, 20 B-1348 Louvain-la-Neuve Belgium tel: +32 10 47 30 50 [[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] Discrepancy in the regression coefficients for Cox regression - PBC data set
There is a discussion in Appendix D.3 of Modeling Survival Data by Thereau and Grambsch regarding the differences in the datasets including the fact that there was significantly more follow-up for many patients at the time this dataset was assembled. I do not see a material difference in the estimates. -- David Winsemius, MD Heritage Labs On Nov 21, 2008, at 12:16 PM, Ravi Varadhan wrote: Hi, When I run the following Cox proportional hazards model on the Mayo clinic's PBC data set (given in the survival package), the regression coefficients do not agree with the results presented in Table 4.6.3 (p. 195) of Fleming Harrington's book. library(survival) data(pbc) ans.cox - coxph(Surv(time, status) ~ log(bili) + log(alb) + age + log(protime) + edema) ans.cox ans.cox - coxph(Surv(time, status) ~ log(bili) + log(alb) + age + log(protime) + edema) ans.cox Call: coxph(formula = Surv(time, status) ~ log(bili) + log(alb) + age + log(protime) + edema) coef exp(coef) se(coef) z p log(bili) 0.8975 2.453 0.08271 10.85 0.0e+00 log(alb) -2.4524 0.086 0.65707 -3.73 1.9e-04 age 0.0382 1.039 0.00768 4.97 6.5e-07 log(protime) 2.345810.442 0.77425 3.03 2.4e-03 edema 0.6613 1.937 0.20595 3.21 1.3e-03 Likelihood ratio test=234 on 5 df, p=0 n= 418 These coefficients, however, are significantly different (i.e. the differences can't be just attributed to round-off's) from that reported in Table 4.6.3 (in the Final model column) of Fleming and Harrington (p. 195). The coefficients reported are: 0.8707, -2.533, 0.0394, 2.380, 0.8592. Note the big difference for the edema variable. It seems like the data set considered in the book and that available in survival package are the same (with n=418). I also re-ran the Cox PH model with the 2 data-errors discussed in p.188 of FH, but still I could not match the results in Table 4.6.3. Is it possible that the results could be explained due to difference in convergence during maximization of partial likelihood? Can anyone help me figure out why this diescrepancy exists? Thanks very much, Ravi. --- Ravi Varadhan, Ph.D. Assistant Professor, The Center on Aging and Health Division of Geriatric Medicine and Gerontology Johns Hopkins University Ph: (410) 502-2619 Fax: (410) 614-9625 Email: [EMAIL PROTECTED] Webpage: http://www.jhsph.edu/agingandhealth/People/Faculty/Varadhan.html [[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] capturing stderr/stdout
Sundar Dorai-Raj wrote:
Hi, Prof. Ripley,
Thanks for the reply. Mostly I want to capture output as it is written
to the stream. For example, I quite often do the following to view the
progress of a log file from a computationally intensive script.
1. Open a console and type:
Rscript script.R script.log
which directs both stdout and stderr to script.log
2. Open another console and type:
tail -f script.log
This way I get both the script's log file and its current progress.
The usual way to so this is via 'tee' .
I guess my question is: Is there a way to accomplish the tail -f
command in R?
Yes, but why would you want to? Note that sink() has a 'split' option,
so you could in principle do all this in your script.R.
Otherwise you can write a simple R loop to do readLines(n=1);
writeLines() to emulate tail -F.
Thanks,
--sundar
Prof Brian Ripley said the following on 11/20/2008 11:43 PM:
I am not sure what the issue is here. Do you want to capture both
stderr and stdout (use 21 in the command with an sh-like shell), or
is the
problem that you don't get immediate output?
The latter is a Perl issue: you need to circumvent output buffering.
See e.g
http://perl.plover.com/FAQs/Buffering.html
Sundar Dorai-Raj wrote:
Hi,
I have an application in perl that prints some output to either
stderr or stdout.
Here's an example:
# tmp.pl
print STDERR starting iterator\n;
for(my $i = 0; $i 100; $i++) {
print $i . \n;
}
# tmp.R
con - pipe(perl tmp.pl)
r - readLines(con, n = -1)
close(con)
However, the second line stalls until the perl for-loop finishes.
What I would like is to process each line as it comes. E.g. something
like:
while(TRUE) {
r - readLines(con, n = 1) # read one line
if(r == 1) print(r)
if(length(r) == 0) break
}
Of course, this won't work since I'm not calling readLines
appropriately. Answers must work on Windows but may include cygwin
utilities if necessary. Any advice would be appreciated. Version info
at the end if it matters.
Thanks, --sundar
version
_
platform i386-pc-mingw32
arch i386
os mingw32
system i386, mingw32
status
major 2
minor 8.0
year 2008
month 10
day20
svn rev46754
language R
version.string R version 2.8.0 (2008-10-20)
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--
Brian D. Ripley, [EMAIL PROTECTED]
Professor of Applied Statistics, http://www.stats.ox.ac.uk/~ripley/
University of Oxford, Tel: +44 1865 272861 (self)
1 South Parks Road, +44 1865 272866 (PA)
Oxford OX1 3TG, UKFax: +44 1865 272595
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Re: [R] Discrepancy in the regression coefficients for Cox regression - PBC data set
Hi David, I did look at Appendix D.3 of TG, but am not sure if the data set analyzed in FH and that attached with survival are different. They both have n=418 (312 from RCT and 106 observational). There is a major difference in the coefficient for edema 0.66 vs 0.86. In any case, the point is not whether the differences in coefficient affect interpretation of the model, but to understand why there are differences in the results. Best, Ravi. --- Ravi Varadhan, Ph.D. Assistant Professor, The Center on Aging and Health Division of Geriatric Medicine and Gerontology Johns Hopkins University Ph: (410) 502-2619 Fax: (410) 614-9625 Email: [EMAIL PROTECTED] Webpage: http://www.jhsph.edu/agingandhealth/People/Faculty/Varadhan.html -Original Message- From: David Winsemius [mailto:[EMAIL PROTECTED] Sent: Friday, November 21, 2008 12:34 PM To: Ravi Varadhan Cc: r-help@r-project.org Subject: Re: [R] Discrepancy in the regression coefficients for Cox regression - PBC data set There is a discussion in Appendix D.3 of Modeling Survival Data by Thereau and Grambsch regarding the differences in the datasets including the fact that there was significantly more follow-up for many patients at the time this dataset was assembled. I do not see a material difference in the estimates. -- David Winsemius, MD Heritage Labs On Nov 21, 2008, at 12:16 PM, Ravi Varadhan wrote: Hi, When I run the following Cox proportional hazards model on the Mayo clinic's PBC data set (given in the survival package), the regression coefficients do not agree with the results presented in Table 4.6.3 (p. 195) of Fleming Harrington's book. library(survival) data(pbc) ans.cox - coxph(Surv(time, status) ~ log(bili) + log(alb) + age + log(protime) + edema) ans.cox ans.cox - coxph(Surv(time, status) ~ log(bili) + log(alb) + age + log(protime) + edema) ans.cox Call: coxph(formula = Surv(time, status) ~ log(bili) + log(alb) + age + log(protime) + edema) coef exp(coef) se(coef) z p log(bili) 0.8975 2.453 0.08271 10.85 0.0e+00 log(alb) -2.4524 0.086 0.65707 -3.73 1.9e-04 age 0.0382 1.039 0.00768 4.97 6.5e-07 log(protime) 2.345810.442 0.77425 3.03 2.4e-03 edema 0.6613 1.937 0.20595 3.21 1.3e-03 Likelihood ratio test=234 on 5 df, p=0 n= 418 These coefficients, however, are significantly different (i.e. the differences can't be just attributed to round-off's) from that reported in Table 4.6.3 (in the Final model column) of Fleming and Harrington (p. 195). The coefficients reported are: 0.8707, -2.533, 0.0394, 2.380, 0.8592. Note the big difference for the edema variable. It seems like the data set considered in the book and that available in survival package are the same (with n=418). I also re-ran the Cox PH model with the 2 data-errors discussed in p.188 of FH, but still I could not match the results in Table 4.6.3. Is it possible that the results could be explained due to difference in convergence during maximization of partial likelihood? Can anyone help me figure out why this diescrepancy exists? Thanks very much, Ravi. -- -- --- Ravi Varadhan, Ph.D. Assistant Professor, The Center on Aging and Health Division of Geriatric Medicine and Gerontology Johns Hopkins University Ph: (410) 502-2619 Fax: (410) 614-9625 Email: [EMAIL PROTECTED] Webpage: http://www.jhsph.edu/agingandhealth/People/Faculty/Varadhan.html -- -- [[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.
[R] question about shapiro.test()
Hi all! I tried to perform Shapiro-Wilk test for my sample of 243 values. Us [1] -10.4 -13.1 -12.2 38.1 -18.8 -13.3 -11.7 29.3 49.7 6.8 12.7 16.3 [13] 5.8 -0.7 -29.4 4.1 38.8 -1.4 8.8 15.6 32.9 -5.3 19.1 35.8 [25] 4.0 -1.5 0.6 -4.2 -10.0 -4.0 1.1 48.9 -21.0 -5.3 5.8 -10.8 [37] 21.9 8.2 -3.2 -3.9 -2.3 12.6 -4.7 -8.0 11.8 27.4 -9.5 -20.8 [49] -8.1 -5.4 0.7 -13.2 -6.1 -5.7 -9.6 0.3 -2.1 -15.2 1.7 -15.2 [61] 7.4 -16.0 13.1 8.7 -8.9 -21.2 29.8 -22.6 10.5 7.5 9.9 -13.0 [73] -8.3 -22.9 -9.7 -3.7 8.2 -9.0 -21.8 28.5 -10.2 8.8 -10.1 15.3 [85] 5.5 -35.0 -14.5 -61.2 -8.3 -17.5 -16.3 -8.5 2.0 -17.5 -12.2 -10.8 [97] -16.7 6.8 -12.4 -1.9 -13.3 -8.2 -22.6 20.7 -12.5 -21.9 -6.0 9.2 [109] -1.1 -17.5 -13.5 6.1 -18.5 18.1 6.3 -13.1 -16.2 -30.5 -23.4 4.3 [121] -11.2 -18.5 -17.7 26.7 2.2 -9.4 1.8 -7.7 -5.1 6.3 7.1 7.4 [133] -9.2 24.8 53.5 23.9 0.0 -25.7 -33.3 6.9 -9.7 -22.6 -1.5 18.0 [145] -20.5 -12.9 -14.3 3.9 5.5 -6.3 2.1 -11.2 -4.9 6.1 3.3 7.0 [157] 4.1 0.3 -0.7 -6.4 -5.1 -17.6 -3.3 6.7 1.5 -8.8 31.2 -30.5 [169] -16.3 -23.4 -30.7 -3.2 8.3 18.1 -17.0 12.1 0.6 -1.6 -1.4 6.3 [181] 3.6 5.3 1.6 1.8 23.7 -14.1 1.1 12.4 18.9 -11.5 -1.8 -21.1 [193] 2.4 23.4 -11.4 -5.9 -5.2 16.9 -19.8 19.5 -4.4 0.5 -1.4 -8.1 [205] 5.4 -2.7 -2.3 -2.9 4.4 8.0 -7.8 32.0 5.4 2.7 3.6 14.9 [217] -7.0 3.6 -12.1 3.8 47.8 -5.8 3.0 12.6 -1.1 2.0 1.5 -0.1 [229] 11.2 -1.4 -3.1 22.0 -14.4 -10.5 -4.1 -37.5 -17.6 -3.3 3.0 -0.6 [241] 33.8 1.6 -3.9 Output is: shapiro.test(Us) Shapiro-Wilk normality test data: Us W = 0.9686, p-value = 3.417e-05 So, if I understand correctly, output inform me that H0 hypotesis of normality of my sample shoud be rejected because p-value 0.05 (at 5% level). Is'n it? But actually my data distributed very close to the normal distribution. Thanks in advance. Eugene -- --- e-mail: [EMAIL PROTECTED] Special Astrophysical Observatory RAS, WWW: http://tiger.sao.ru/ Nizhnij Arkhyz phone: +7 87878 46 5 77 Karachai-Chercassian Republic, fax: +7 87878 46 5 27 Russia, 369167 __ 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] Equivalent to apply(x[,2:5],1,sum) for dataframe
What's the most correct way of doing the equivalent to apply(x[,2:5],1,sum) if x is dataframe in which the only numeric fields are in columns 2:5 ? (using apply returns a character vector) Thanks Agus -- Dr. Agustin Lobo Institut de Ciencies de la Terra Jaume Almera (CSIC) LLuis Sole Sabaris s/n 08028 Barcelona Spain Tel. 34 934095410 Fax. 34 934110012 email: [EMAIL PROTECTED] http://www.ija.csic.es/gt/obster __ 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 data.frame()
Given str(x) 'data.frame': 5284 obs. of 5 variables: $ COD : chr 0800101001 0800101002 0800101003 0800101004 ... $ 0-4 : num 79 215 84 58 127 134 15 122 101 99 ... $ 5-9 : num 76 180 32 56 81 106 10 112 128 96 ... $ 10-14: num 68 145 39 46 78 81 8 92 142 107 ... $ 15-19: num 73 170 49 52 103 77 10 116 129 129 ... I get str(data.matrix(x)) num [1:5284, 1:5] 8e+08 8e+08 8e+08 8e+08 8e+08 ... - attr(*, dimnames)=List of 2 ..$ : chr [1:5284] 3 6 9 12 ... ..$ : chr [1:5] COD 0-4 5-9 10-14 ... Should not data.matrix() return a numeric matrix? And I'm even more confused by this: x2 - censocatT[,2:5] str(x2) 'data.frame': 5284 obs. of 4 variables: $ 0-4 : num 79 215 84 58 127 134 15 122 101 99 ... $ 5-9 : num 76 180 32 56 81 106 10 112 128 96 ... $ 10-14: num 68 145 39 46 78 81 8 92 142 107 ... $ 15-19: num 73 170 49 52 103 77 10 116 129 129 ... str(data.matrix(x2)) num [1:5284, 1:4] 79 215 84 58 127 134 15 122 101 99 ... - attr(*, dimnames)=List of 2 ..$ : chr [1:5284] 3 6 9 12 ... ..$ : chr [1:4] 0-4 5-9 10-14 15-19 Thanks for any clarification Agus -- Dr. Agustin Lobo Institut de Ciencies de la Terra Jaume Almera (CSIC) LLuis Sole Sabaris s/n 08028 Barcelona Spain Tel. 34 934095410 Fax. 34 934110012 email: [EMAIL PROTECTED] http://www.ija.csic.es/gt/obster __ 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] list creation interpolation
Hello all, I apologize if this is simple or has already been answered somewhere, but I'm not sure what to search for although I have tried and didn't come up with anything so.. Here's my question. How can I interpolate list names or do I have to do it post list creation. Since that's not very clear here is some sample code of what I wanted to do: resultlist-list() a-c(Book, video, radio, mp3) foo-rnorm(10) resultlist$a[1]-foo Warning message: In resultlist$a[1] - foo : number of items to replace is not a multiple of replacement length What I wanted here was a named list so that I would have 4 values the list would be resultlist$Book resultlist$video resultlist$radio resultlist$mp3 Any idea on how to do this other than doing it post creation ie resultlist[[1]]-foo resultlist[[2]]-rnorm(20) etc.. names(resultlist)-a Cheers, Paul [[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] Ryan's Q Post-Hoc for ANOVA
I realize this is a little late, but I recently ended up rolling my own Ryan's Q in R. If anyone is interested, or wishes to make improvements, you can find it here: http://homes.msi.ucsb.edu/~byrnes/r_files/ryans_q.r On Oct 25, 2005, at 10:15 AM, Jarrett Byrnes wrote: I'm using lm to run an ANOVA, and would like to use Ryan's Q as my post-hoc (as recommended by Day and Quinn, 1989, Ecological Monographs). I can't seem to find any methods in the base stats package that implement this post-hoc. Is there a good package of post-hoc methods out there, or has someone written a method for Ryan's Q previously? Thanks! -Jarrett __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.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] Fitting a sine wave using solver
Ben Zuckerberg bz73 at cornell.edu writes: I have several sets of oscillation data and would like to estimate the parameters of a sine function to each set (and hopefully automate this). There is an example using lme (yes, LINEAR) fit on page 239 of Pinheiro/Bates Mixed Effects Book (ovarian cycle). If you do not have the book at hand, have a look at library/nlme/scripts/ch05.r. Dieter __ 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 data.frame()
Agustin Lobo-4 wrote: Given str(x) 'data.frame': 5284 obs. of 5 variables: $ COD : chr 0800101001 0800101002 0800101003 0800101004 ... $ 0-4 : num 79 215 84 58 127 134 15 122 101 99 ... $ 5-9 : num 76 180 32 56 81 106 10 112 128 96 ... $ 10-14: num 68 145 39 46 78 81 8 92 142 107 ... $ 15-19: num 73 170 49 52 103 77 10 116 129 129 ... I get str(data.matrix(x)) num [1:5284, 1:5] 8e+08 8e+08 8e+08 8e+08 8e+08 ... - attr(*, dimnames)=List of 2 ..$ : chr [1:5284] 3 6 9 12 ... ..$ : chr [1:5] COD 0-4 5-9 10-14 ... Should not data.matrix() return a numeric matrix? It does, only the dimnames are non-numeric, and you first column is a little ugly in floating point because the numbers are so big. Do you really want to convert these to doubles? Try dm= data.matrix(x) dm[,2] which probably looks quite good. Dieter -- View this message in context: http://www.nabble.com/Problem-with-data.frame%28%29-tp20626551p20627068.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] Equivalent to apply(x[,2:5],1,sum) for dataframe
Agustin Lobo-4 wrote: What's the most correct way of doing the equivalent to apply(x[,2:5],1,sum) if x is dataframe in which the only numeric fields are in columns 2:5 ? (using apply returns a character vector) Could it be that you meant apply(x[,2:5],2,sum)? Not very easy to remember, when to use 2 or 1. Also check rowSums and colSums (note the plural forms!) that can be MUCH faster. Dieter -- View this message in context: http://www.nabble.com/Equivalent-to-apply%28x-%2C2%3A5-%2C1%2Csum%29-for-dataframe-tp20626483p20627203.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] Help with avoiding For Loop
Dear Jorge,
Thanks a lot.
Regards
Anup
On Thu, Nov 20, 2008 at 10:19 PM, Jorge Ivan Velez [EMAIL PROTECTED]
wrote:
Dear Anup,
Try this:
# Data
A - c(1,2,3,4,5)
B - rnorm(100)
# Results
t(apply(sapply(A,function(x) pnorm(x*B)),2,function(x)
c(Mean=mean(x),Var=var(x
HTH,
Jorge
On Thu, Nov 20, 2008 at 10:09 PM, Anup Menon [EMAIL PROTECTED]wrote:
Dear Friends,
I'm trying to see if there is some possibility that I can do the following
computations without a loop. I have attached a toy example below.
A - c(1,2,3,4,5)
B - rnorm(100)
store - matrix(0,5,2)
for (i in 1:5)
{
store[i,1] - mean(pnorm(A[i]*B))
store[i,2] - var(pnorm(A[i]*B))
}
store
Thanks in advance for your help.
Kind Regards
Anup
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and provide commented, minimal, self-contained, reproducible code.
Re: [R] Vector lty argrument for lines or plot
Thank you for the suggestion, baptiste. segments() does do exactly what I was wanting and matplot()/matlines() is probably a better solution to what I was trying to do. However, I am still concerned about the discrepancy between the documentation in ?par and the behavior of lines(). Should lines() be changed to cycle over a vector of lty (so that it agrees with the documentation in ?par)? Should the documentation of par be changed to use a different example of a function that cycles over a vector of lty (segments() being a good candidate)? Or are both lines() and ?par correct and there is a situation which lines() does cycle over a vector of lty that I (and at least baptiste as well) do not understand? The middle option is certainly the easiest, and I think the correct one, but I wanted to rule out the last one before filing a bug report. -- Brian Diggs, Ph.D. Senior Research Associate, Department of Surgery, Oregon Health Science University baptiste auguie wrote: Hi, If you wish to connect each point to the next with a different linetype, I think your best bet is to use segments() x - stats::runif(12); y - stats::rnorm(12) i - order(x,y); x - x[i]; y - y[i] plot(x, y) s - seq(length(x)-1) segments(x[s], y[s], x[s+1], y[s+1], lty=1:10) If, however, you wish to plot several groups of lines with different linetypes, then matlines() should do the job. Both of these make actual use of lty as a vector, while polygon(), abline(), plot(), lines() will only use the first value (as far as i can see). Hope this helps, baptiste On 20 Nov 2008, at 20:24, Brian Diggs wrote: I am confused by the behavior of the lines function when the lty argument is a vector. ?lines indicates that lty is a valid parameter, but says nothing else about it. ?plot.xy (which I think is what gets called) refers back to ?lines. ?plot.default says to see ?par. In ?par, about lty it says Some functions such as lines accept a vector of values which are recycled. Other uses will take just the first value if a vector of length greater than one is supplied. However, I cannot get lines to use more than one type of line. Some example code: pt - runif(10) plot(pt) lines(pt, type=c, lty=1:10) I expected each subsequent line segment to be in a different style. Only the first seems to be used. The same is true for plot: plot(pt, type=b, lty=1:10) uses only one style of line segment (although no documentation says explicitly that the others would be used). It doesn't matter the order or manner of specification, only the first is used. plot(pt) lines(pt, type=c, lty=c(dashed,solid)) plot(pt) lines(pt, type=c, lty=c(FF, 11)) I have used a vector of lty before (in legend) and it cycled through all the values. Am I misunderstanding what a vector lty to lines means, or is this a bug? I'm running on Windows XP Pro, if that might matter. sessionInfo() R version 2.8.0 (2008-10-20) i386-pc-mingw32 locale: LC_COLLATE=English_United States.1252;LC_CTYPE=English_United States.1252;LC_MONETARY=English_United States.1252;LC_NUMERIC=C;LC_TIME=English_United States.1252 attached base packages: [1] stats graphics grDevices utils datasets methods base -- Brian Diggs, Ph.D. Senior Research Associate, Department of Surgery, Oregon Health Science University __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. _ Baptiste Auguié School of Physics University of Exeter Stocker Road, Exeter, Devon, EX4 4QL, UK Phone: +44 1392 264187 http://newton.ex.ac.uk/research/emag __ 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] Conditionally Removing data from a vector
I am having trouble removing entries from a vector based on the value of the
vector and another object value. It works in my pseudo test run:
DEV=400
#Y
CANDS=c(100:105)
#Z
DEVS=c(120,220,320,420,520)
if(DEVDEVS)
(CANDS=CANDS[which(DEVDEVS)])
CANDS DEVS
[1,] 100 120
[2,] 101 220
[3,] 102 320
[4,] 103 420
[5,] 104 520
[6,] 105 620
So the result CANDS is 103, 104 and 105 b/c the corresponding DEVS are
larger than the DEV(400 in this case).
Now i am trying to implement this into my working code but it doenst seem to
work.
#Loads TAZ and corresponding vacant acres data
TAZ_VAC_ACRES= read.csv(file=I:/Research.csv,header=TRUE);
#Converts vacant acres by TAZ to vacant square footage by TAZ
TAZ_VAC_FEET=cbind(TAZ_VAC_ACRES[1],TAZ_VAC_ACRES[2]*43560)
#Creates test Candidates Vector
Candidates=c(100,101,102,103,104,105)
#Creates object equaling the number of candidate TAZs from the main script
Location Choice Model
NumCands1=length(Candidates)
Dev..At=999
for(i in 1:NumCands1){
#Renames each candidate TAZ to function variable
Loc_Mod_TAZ=Candidates[i]
#Converts Development size from main script to Development density
function
format
Dev_Size=Dev..At
#Determines vacant square feet by TAZ
TAZDetermine_FEET=TAZ_VAC_FEET[TAZ_VAC_FEET$TAZ==Loc_Mod_TAZ,2]
#Determines if the Candidate TAZs have sufficient
vacancy and creates
vector with TAZs that can accomadate development
if(Dev_SizeTAZDetermine_FEET)
(Candidates=Candidates[which(Dev_SizeTAZDetermine_FEET)])
}
So basically i am starting out with a list of CAndidate TAZs and i want to
remove the TAZS that are not big enough to accomadate the
Development(Dev..At). My test code works but again i cant get it to run in
my workiong code. Anything stupid im missing or maybe a better approach
than the which function. Remove only seem useful for entire objects so
not sure what else would work. Thanks guys and gals
Cheers,
JR
--
View this message in context:
http://www.nabble.com/Conditionally-Removing-data-from-a-vector-tp20627244p20627244.html
Sent from the R help mailing list archive at Nabble.com.
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[R] Saving color2D.matplot plots to a file
Hi, I am trying to figure out how to save plots produced with color2D.matplot to a file, any format. I have tried: jpeg() / png() color2D.matplot(x,c(1,0),c(1,0),c(1,0)) dev.off() as well as dev.new() dev.print(device=png,file=myfreakinplot2.png) #(and jpeg, ps, pdf, eps, together with dev.copy() as well...) color2D.matplot(x,c(1,0),c(1,0),c(1,0),show.legend=TRUE) dev.off() to no avail. Does anyone have any suggestions? Thanks! Bernardo __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Suppression anova message
Dear all, I am running anova(lm()) on a series of different data frame and I am getting the following message Using dataFrame$levels as id variables 1. Why am I getting that message 2. How do I suppress it (or correct it). Thanks Marco -- Marco Blanchette, Ph.D. Assistant Investigator Stowers Institute for Medical Research 1000 East 50th St. Kansas City, MO 64110 Tel: 816-926-4071 Cell: 816-726-8419 Fax: 816-926-2018 __ 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] Suppression anova message
Dear all, I am running anova(lm()) on a series of different data frame and I am getting the following message Using dataFrame$levels as id variables 1. Why am I getting that message 2. How do I suppress it (or correct it). Thanks Marco -- Marco Blanchette, Ph.D. Assistant Investigator Stowers Institute for Medical Research 1000 East 50th St. Kansas City, MO 64110 Tel: 816-926-4071 Cell: 816-726-8419 Fax: 816-926-2018 __ 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] log likelihood
Hi all, I ran a Weibull model, and I am wondering if there is any way to extract the log likelihood. I tried loglik(model) but it does not seem to work any help would be greatly appreciated joe [[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] log likelihood
Hi all, I ran a Weibull model, and I am wondering if there is any way to extract the log likelihood. I tried loglik(model) but it does not seem to work any help would be greatly appreciated joe [[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] Discrepancy in the regression coefficients for Cox regression - PBC data set
Ravi Varadhan wrote: Hi David, I did look at Appendix D.3 of TG, but am not sure if the data set analyzed in FH and that attached with survival are different. They both have n=418 (312 from RCT and 106 observational). Well, as David implies, if the observation times are longer and a few more people died, that could easily explain the differences. Someone borrowed our copy of FH so I can't check, but presumably you have one (and it is your problem anyway...). There is a major difference in the coefficient for edema 0.66 vs 0.86. In any case, the point is not whether the differences in coefficient affect interpretation of the model, but to understand why there are differences in the results. Best, Ravi. --- Ravi Varadhan, Ph.D. Assistant Professor, The Center on Aging and Health Division of Geriatric Medicine and Gerontology Johns Hopkins University Ph: (410) 502-2619 Fax: (410) 614-9625 Email: [EMAIL PROTECTED] Webpage: http://www.jhsph.edu/agingandhealth/People/Faculty/Varadhan.html -Original Message- From: David Winsemius [mailto:[EMAIL PROTECTED] Sent: Friday, November 21, 2008 12:34 PM To: Ravi Varadhan Cc: r-help@r-project.org Subject: Re: [R] Discrepancy in the regression coefficients for Cox regression - PBC data set There is a discussion in Appendix D.3 of Modeling Survival Data by Thereau and Grambsch regarding the differences in the datasets including the fact that there was significantly more follow-up for many patients at the time this dataset was assembled. I do not see a material difference in the estimates. -- David Winsemius, MD Heritage Labs On Nov 21, 2008, at 12:16 PM, Ravi Varadhan wrote: Hi, When I run the following Cox proportional hazards model on the Mayo clinic's PBC data set (given in the survival package), the regression coefficients do not agree with the results presented in Table 4.6.3 (p. 195) of Fleming Harrington's book. library(survival) data(pbc) ans.cox - coxph(Surv(time, status) ~ log(bili) + log(alb) + age + log(protime) + edema) ans.cox ans.cox - coxph(Surv(time, status) ~ log(bili) + log(alb) + age + log(protime) + edema) ans.cox Call: coxph(formula = Surv(time, status) ~ log(bili) + log(alb) + age + log(protime) + edema) coef exp(coef) se(coef) z p log(bili) 0.8975 2.453 0.08271 10.85 0.0e+00 log(alb) -2.4524 0.086 0.65707 -3.73 1.9e-04 age 0.0382 1.039 0.00768 4.97 6.5e-07 log(protime) 2.345810.442 0.77425 3.03 2.4e-03 edema 0.6613 1.937 0.20595 3.21 1.3e-03 Likelihood ratio test=234 on 5 df, p=0 n= 418 These coefficients, however, are significantly different (i.e. the differences can't be just attributed to round-off's) from that reported in Table 4.6.3 (in the Final model column) of Fleming and Harrington (p. 195). The coefficients reported are: 0.8707, -2.533, 0.0394, 2.380, 0.8592. Note the big difference for the edema variable. It seems like the data set considered in the book and that available in survival package are the same (with n=418). I also re-ran the Cox PH model with the 2 data-errors discussed in p.188 of FH, but still I could not match the results in Table 4.6.3. Is it possible that the results could be explained due to difference in convergence during maximization of partial likelihood? Can anyone help me figure out why this diescrepancy exists? Thanks very much, Ravi. -- -- --- Ravi Varadhan, Ph.D. Assistant Professor, The Center on Aging and Health Division of Geriatric Medicine and Gerontology Johns Hopkins University Ph: (410) 502-2619 Fax: (410) 614-9625 Email: [EMAIL PROTECTED] Webpage: http://www.jhsph.edu/agingandhealth/People/Faculty/Varadhan.html -- -- [[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. -- O__ Peter Dalgaard Øster Farimagsgade 5, Entr.B c/ /'_ --- Dept. of Biostatistics PO Box 2099, 1014 Cph. K (*) \(*) -- University of Copenhagen Denmark Ph: (+45)
Re: [R] Vector lty argrument for lines or plot
Brian Diggs wrote: Thank you for the suggestion, baptiste. segments() does do exactly what I was wanting and matplot()/matlines() is probably a better solution to what I was trying to do. However, I am still concerned about the discrepancy between the documentation in ?par and the behavior of lines(). Should lines() be changed to cycle over a vector of lty (so that it agrees with the documentation in ?par)? Should the documentation of par be changed to use a different example of a function that cycles over a vector of lty (segments() being a good candidate)? Or are both lines() and ?par correct and there is a situation which lines() does cycle over a vector of lty that I (and at least baptiste as well) do not understand? The middle option is certainly the easiest, and I think the correct one, but I wanted to rule out the last one before filing a bug report. The documentation for lines has 'lwd' can be a vector: its first element will apply to lines but the whole vector to symbols (recycled as necessary). which really is true, and you might expect something similar for lty, but lty does not apply to symbols. So the text in ?par is most likely a copy-paste blunder. In any case, this usage is somewhat far-fetched and a reference matplot() or segments() would, in both cases, express more clearly what ?par is trying to say. -- O__ Peter Dalgaard Øster Farimagsgade 5, Entr.B c/ /'_ --- Dept. of Biostatistics PO Box 2099, 1014 Cph. K (*) \(*) -- University of Copenhagen Denmark Ph: (+45) 35327918 ~~ - ([EMAIL PROTECTED]) FAX: (+45) 35327907 __ R-help@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] Dendrogram labels
Is there any way to change the orientation of the labels on the end of the
dendrograms to horizontal rather than vertical? If so, how can I do that.
_
Patrick Richardson
Biostatistician - Program of Translational Medicine
Van Andel Research Institute - Webb Lab
333 Bostwick Avenue NE
Grand Rapids, MI 49503
ph. 616.234.5787
This email message, including any attachments, is for th...{{dropped:9}}
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
[R] Dendrogram labels
Is there any way to change the orientation of the labels on the end of the
dendrograms to horizontal rather than vertical? If so, how can I do that.
Below is my code, but I'm not sure which argument(s) I can use to change the
label(s) (if it is possible to do).
a - hclust(z, method = complete)
# Plots hclust dendrogram #
plot(a, frame.plot=TRUE, lwd=2)
Many thanks,
_
Patrick Richardson
Biostatistician - Program of Translational Medicine
Van Andel Research Institute - Webb Lab
333 Bostwick Avenue NE
Grand Rapids, MI 49503
ph. 616.234.5787
This email message, including any attachments, is for th...{{dropped:9}}
__
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] Conditionally Removing data from a vector
I am having trouble removing entries from a vector based on the value of the vector and another object value. It works in my pseudo test run: DEV=400 #Y CANDS=c(100:105) #Z DEVS=c(120,220,320,420,520) if(DEVDEVS) (CANDS=CANDS[which(DEVDEVS)]) CANDS DEVS [1,] 100 120 [2,] 101 220 [3,] 102 320 [4,] 103 420 [5,] 104 520 [6,] 105 620 The above code does not produce the output you show when I paste it into R. It would be highly desirable to have a short, documented, working example... So the result CANDS is 103, 104 and 105 b/c the corresponding DEVS are larger than the DEV(400 in this case). How do they correspond? Do you mean they are both columns of a data.frame as your example output suggests? But CANDS and DEVS are of differnt length, so this is probably not what you meant. Finally, your code seems to indicate that you want all elements of CANDS for which the corresponding (?) value of DEVS is less than threshold DEV, but only if DEVDEVS. The latter produces a warning, as you are comparing a single value to a vector of 5, which is most likely not what you want. At this point my confusion is perfect and I give up. I think you need to give us a better explanantion of what data you have and what exactly you want to accomplish. cu Philipp -- Dr. Philipp Pagel Lehrstuhl für Genomorientierte Bioinformatik Technische Universität München Wissenschaftszentrum Weihenstephan 85350 Freising, Germany http://mips.gsf.de/staff/pagel __ 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] rgl lighting question
Hi, I'm using rgl to generate a 3D surface plot and I'm struggling to get the lighting correct. Currently the surface gets plotted, but is very 'shiny'. On rotating the view, I get to see parts of the surface - but overall I don't see much detail because of the spotlight like lighting. I've played around with the specular, ambient and diffuse but I can't bring out the details of the surface. Could anybody point me to some examples of how to make a plain matte surface, which isn't obscured by specular reflections? Thanks, --- Rajarshi Guha [EMAIL PROTECTED] GPG Fingerprint: D070 5427 CC5B 7938 929C DD13 66A1 922C 51E7 9E84 --- If you don't get a good night kiss, you get Kafka dreams. -Hobbes __ 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] Discrepancy in the regression coefficients for Cox regression - PBC data set
Peter, I did check the data in the Appendix of FH with the data in survival package. I couldn't find any differences in the time and status variables. May be Terry Therneau knows the answer?! Ravi. --- Ravi Varadhan, Ph.D. Assistant Professor, The Center on Aging and Health Division of Geriatric Medicine and Gerontology Johns Hopkins University Ph: (410) 502-2619 Fax: (410) 614-9625 Email: [EMAIL PROTECTED] Webpage: http://www.jhsph.edu/agingandhealth/People/Faculty/Varadhan.html -Original Message- From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] On Behalf Of Peter Dalgaard Sent: Friday, November 21, 2008 1:58 PM To: Ravi Varadhan Cc: r-help@r-project.org Subject: Re: [R] Discrepancy in the regression coefficients for Cox regression - PBC data set Ravi Varadhan wrote: Hi David, I did look at Appendix D.3 of TG, but am not sure if the data set analyzed in FH and that attached with survival are different. They both have n=418 (312 from RCT and 106 observational). Well, as David implies, if the observation times are longer and a few more people died, that could easily explain the differences. Someone borrowed our copy of FH so I can't check, but presumably you have one (and it is your problem anyway...). There is a major difference in the coefficient for edema 0.66 vs 0.86. In any case, the point is not whether the differences in coefficient affect interpretation of the model, but to understand why there are differences in the results. Best, Ravi. -- -- --- Ravi Varadhan, Ph.D. Assistant Professor, The Center on Aging and Health Division of Geriatric Medicine and Gerontology Johns Hopkins University Ph: (410) 502-2619 Fax: (410) 614-9625 Email: [EMAIL PROTECTED] Webpage: http://www.jhsph.edu/agingandhealth/People/Faculty/Varadhan.html -- -- -Original Message- From: David Winsemius [mailto:[EMAIL PROTECTED] Sent: Friday, November 21, 2008 12:34 PM To: Ravi Varadhan Cc: r-help@r-project.org Subject: Re: [R] Discrepancy in the regression coefficients for Cox regression - PBC data set There is a discussion in Appendix D.3 of Modeling Survival Data by Thereau and Grambsch regarding the differences in the datasets including the fact that there was significantly more follow-up for many patients at the time this dataset was assembled. I do not see a material difference in the estimates. -- David Winsemius, MD Heritage Labs On Nov 21, 2008, at 12:16 PM, Ravi Varadhan wrote: Hi, When I run the following Cox proportional hazards model on the Mayo clinic's PBC data set (given in the survival package), the regression coefficients do not agree with the results presented in Table 4.6.3 (p. 195) of Fleming Harrington's book. library(survival) data(pbc) ans.cox - coxph(Surv(time, status) ~ log(bili) + log(alb) + age + log(protime) + edema) ans.cox ans.cox - coxph(Surv(time, status) ~ log(bili) + log(alb) + age + log(protime) + edema) ans.cox Call: coxph(formula = Surv(time, status) ~ log(bili) + log(alb) + age + log(protime) + edema) coef exp(coef) se(coef) z p log(bili) 0.8975 2.453 0.08271 10.85 0.0e+00 log(alb) -2.4524 0.086 0.65707 -3.73 1.9e-04 age 0.0382 1.039 0.00768 4.97 6.5e-07 log(protime) 2.345810.442 0.77425 3.03 2.4e-03 edema 0.6613 1.937 0.20595 3.21 1.3e-03 Likelihood ratio test=234 on 5 df, p=0 n= 418 These coefficients, however, are significantly different (i.e. the differences can't be just attributed to round-off's) from that reported in Table 4.6.3 (in the Final model column) of Fleming and Harrington (p. 195). The coefficients reported are: 0.8707, -2.533, 0.0394, 2.380, 0.8592. Note the big difference for the edema variable. It seems like the data set considered in the book and that available in survival package are the same (with n=418). I also re-ran the Cox PH model with the 2 data-errors discussed in p.188 of FH, but still I could not match the results in Table 4.6.3. Is it possible that the results could be explained due to difference in convergence during maximization of partial likelihood? Can anyone help me figure out why this diescrepancy exists? Thanks very much, Ravi. - - -- --- Ravi Varadhan, Ph.D. Assistant Professor, The Center on Aging and Health Division of Geriatric Medicine and Gerontology Johns Hopkins University Ph: (410) 502-2619 Fax: (410) 614-9625 Email: [EMAIL PROTECTED] Webpage:
Re: [R] Dendrogram labels
This does work, but also orients the entire dendrogram as horizontal. If I
could, I'd like the dendrogram to stay vertical and just orient the labels as
horizontal.
Many thanks,
_
Patrick Richardson
Biostatistician - Program of Translational Medicine
Van Andel Research Institute - Webb Lab
333 Bostwick Avenue NE
Grand Rapids, MI 49503
ph. 616.234.5787
-Original Message-
From: Fan Yang [mailto:[EMAIL PROTECTED]
Sent: Friday, November 21, 2008 2:28 PM
To: Richardson, Patrick
Cc: 'r-help@r-project.org'
Subject: Re: [R] Dendrogram labels
Try:
b-as.dendrogram(a); plot(b, horiz=TRUE).
that should help.
fan
On Nov 21, 2008, at 2:14 PM, Richardson, Patrick wrote:
Is there any way to change the orientation of the labels on the end
of the dendrograms to horizontal rather than vertical? If so, how
can I do that. Below is my code, but I'm not sure which argument(s)
I can use to change the label(s) (if it is possible to do).
a - hclust(z, method = complete)
# Plots hclust dendrogram #
plot(a, frame.plot=TRUE, lwd=2)
Many thanks,
_
Patrick Richardson
Biostatistician - Program of Translational Medicine
Van Andel Research Institute - Webb Lab
333 Bostwick Avenue NE
Grand Rapids, MI 49503
ph. 616.234.5787
This email message, including any attachments, is for th...{{dropped:
9}}
__
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.
This email message, including any attachments, is for th...{{dropped:6}}
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and provide commented, minimal, self-contained, reproducible code.
Re: [R] question about shapiro.test()
Why are you doing the test for normality? The 2 most common reasons are: 1. want to test if data comes from an exact normal distribution. 2. want to know if tests based on normal assumptions are reasonable to use with this data. Technically, the Shapiro-Wilk test does not do either of the above, but further suggestions depend on what you are trying to accomplish, or what question you want answered. -- Gregory (Greg) L. Snow Ph.D. Statistical Data Center Intermountain Healthcare [EMAIL PROTECTED] 801.408.8111 -Original Message- From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] project.org] On Behalf Of Eugene A. Semenko Sent: Friday, November 21, 2008 8:19 AM To: [EMAIL PROTECTED] Subject: [R] question about shapiro.test() Hi all! I tried to perform Shapiro-Wilk test for my sample of 243 values. Us [1] -10.4 -13.1 -12.2 38.1 -18.8 -13.3 -11.7 29.3 49.7 6.8 12.7 16.3 [13] 5.8 -0.7 -29.4 4.1 38.8 -1.4 8.8 15.6 32.9 -5.3 19.1 35.8 [25] 4.0 -1.5 0.6 -4.2 -10.0 -4.0 1.1 48.9 -21.0 -5.3 5.8 -10.8 [37] 21.9 8.2 -3.2 -3.9 -2.3 12.6 -4.7 -8.0 11.8 27.4 -9.5 -20.8 [49] -8.1 -5.4 0.7 -13.2 -6.1 -5.7 -9.6 0.3 -2.1 -15.2 1.7 -15.2 [61] 7.4 -16.0 13.1 8.7 -8.9 -21.2 29.8 -22.6 10.5 7.5 9.9 -13.0 [73] -8.3 -22.9 -9.7 -3.7 8.2 -9.0 -21.8 28.5 -10.2 8.8 -10.1 15.3 [85] 5.5 -35.0 -14.5 -61.2 -8.3 -17.5 -16.3 -8.5 2.0 -17.5 -12.2 -10.8 [97] -16.7 6.8 -12.4 -1.9 -13.3 -8.2 -22.6 20.7 -12.5 -21.9 -6.0 9.2 [109] -1.1 -17.5 -13.5 6.1 -18.5 18.1 6.3 -13.1 -16.2 -30.5 -23.4 4.3 [121] -11.2 -18.5 -17.7 26.7 2.2 -9.4 1.8 -7.7 -5.1 6.3 7.1 7.4 [133] -9.2 24.8 53.5 23.9 0.0 -25.7 -33.3 6.9 -9.7 -22.6 -1.5 18.0 [145] -20.5 -12.9 -14.3 3.9 5.5 -6.3 2.1 -11.2 -4.9 6.1 3.3 7.0 [157] 4.1 0.3 -0.7 -6.4 -5.1 -17.6 -3.3 6.7 1.5 -8.8 31.2 -30.5 [169] -16.3 -23.4 -30.7 -3.2 8.3 18.1 -17.0 12.1 0.6 -1.6 -1.4 6.3 [181] 3.6 5.3 1.6 1.8 23.7 -14.1 1.1 12.4 18.9 -11.5 -1.8 -21.1 [193] 2.4 23.4 -11.4 -5.9 -5.2 16.9 -19.8 19.5 -4.4 0.5 -1.4 -8.1 [205] 5.4 -2.7 -2.3 -2.9 4.4 8.0 -7.8 32.0 5.4 2.7 3.6 14.9 [217] -7.0 3.6 -12.1 3.8 47.8 -5.8 3.0 12.6 -1.1 2.0 1.5 -0.1 [229] 11.2 -1.4 -3.1 22.0 -14.4 -10.5 -4.1 -37.5 -17.6 -3.3 3.0 -0.6 [241] 33.8 1.6 -3.9 Output is: shapiro.test(Us) Shapiro-Wilk normality test data: Us W = 0.9686, p-value = 3.417e-05 So, if I understand correctly, output inform me that H0 hypotesis of normality of my sample shoud be rejected because p-value 0.05 (at 5% level). Is'n it? But actually my data distributed very close to the normal distribution. Thanks in advance. Eugene -- --- e-mail: [EMAIL PROTECTED] Special Astrophysical Observatory RAS, WWW: http://tiger.sao.ru/ Nizhnij Arkhyz phone: +7 87878 46 5 77 Karachai-Chercassian Republic, fax: +7 87878 46 5 27 Russia, 369167 __ 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] list creation interpolation
The $ syntax for working with elements of a list is a magical shortcut for [[]]. It is a great shortcut when used as intended, but trying to force magical shortcuts to do things that they were not intended for usually results in the programming equivalent of turning yourself into a toad. The correct approach is to not use the shortcut, but the thing that it is a shortcut for. Did you try: resultlist[[ a[1] ] - foo resultlist[[ a[2] ] - rnorm(20) Hope this helps, -- Gregory (Greg) L. Snow Ph.D. Statistical Data Center Intermountain Healthcare [EMAIL PROTECTED] 801.408.8111 -Original Message- From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] project.org] On Behalf Of H. Paul Benton Sent: Friday, November 21, 2008 10:58 AM To: r-help@r-project.org Subject: [R] list creation interpolation Hello all, I apologize if this is simple or has already been answered somewhere, but I'm not sure what to search for although I have tried and didn't come up with anything so.. Here's my question. How can I interpolate list names or do I have to do it post list creation. Since that's not very clear here is some sample code of what I wanted to do: resultlist-list() a-c(Book, video, radio, mp3) foo-rnorm(10) resultlist$a[1]-foo Warning message: In resultlist$a[1] - foo : number of items to replace is not a multiple of replacement length What I wanted here was a named list so that I would have 4 values the list would be resultlist$Book resultlist$video resultlist$radio resultlist$mp3 Any idea on how to do this other than doing it post creation ie resultlist[[1]]-foo resultlist[[2]]-rnorm(20) etc.. names(resultlist)-a Cheers, Paul [[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.
[R] Bug in Kendall for n4?
library(Kendall) Kendall(1:3,1:3) WARNING: Error exit, tauk2. IFAULT = 12 tau = 1, 2-sided pvalue =1 I believe Kendall tau is well-defined for this case and the reported value is correct; isn't it a bug to give a warning? (And if, e.g., the pvalue is not well-defined in this case, wouldn't it be better to return NA or NaN or something?) Also, shouldn't the error code be given in plain English -- or at least the meaning of IFAULT = 12 documented on the help page? A somewhat less clear case is Kendall(1:2,1:2), which gives the same error. Though the usual formula for Kendall tau has a zero in the denominator in this case, I'd think the correct generalization is 1 if the two elements are in the same order, and -1 if they are not (the only possibilities). But perhaps I don't fully understand the interpretation of this statistic. -s __ 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] Dendrogram labels
hah, I see what you mean. try:
plot(a, leaflab=c(textlike)).
fan
On Nov 21, 2008, at 2:30 PM, Richardson, Patrick wrote:
This does work, but also orients the entire dendrogram as
horizontal. If I could, I'd like the dendrogram to stay vertical
and just orient the labels as horizontal.
Many thanks,
_
Patrick Richardson
Biostatistician - Program of Translational Medicine
Van Andel Research Institute - Webb Lab
333 Bostwick Avenue NE
Grand Rapids, MI 49503
ph. 616.234.5787
-Original Message-
From: Fan Yang [mailto:[EMAIL PROTECTED]
Sent: Friday, November 21, 2008 2:28 PM
To: Richardson, Patrick
Cc: 'r-help@r-project.org'
Subject: Re: [R] Dendrogram labels
Try:
b-as.dendrogram(a); plot(b, horiz=TRUE).
that should help.
fan
On Nov 21, 2008, at 2:14 PM, Richardson, Patrick wrote:
Is there any way to change the orientation of the labels on the end
of the dendrograms to horizontal rather than vertical? If so, how
can I do that. Below is my code, but I'm not sure which argument(s)
I can use to change the label(s) (if it is possible to do).
a - hclust(z, method = complete)
# Plots hclust dendrogram #
plot(a, frame.plot=TRUE, lwd=2)
Many thanks,
_
Patrick Richardson
Biostatistician - Program of Translational Medicine
Van Andel Research Institute - Webb Lab
333 Bostwick Avenue NE
Grand Rapids, MI 49503
ph. 616.234.5787
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Re: [R] rgl lighting question
2008/11/21 Rajarshi Guha [EMAIL PROTECTED]: Hi, I'm using rgl to generate a 3D surface plot and I'm struggling to get the lighting correct. Currently the surface gets plotted, but is very 'shiny'. On rotating the view, I get to see parts of the surface - but overall I don't see much detail because of the spotlight like lighting. I've played around with the specular, ambient and diffuse but I can't bring out the details of the surface. Could anybody point me to some examples of how to make a plain matte surface, which isn't obscured by specular reflections? Have you tried lit=FALSE? This sphere is so unspecular it appears flat: spheres3d(0,0,0,color=red,lit=FALSE) Barry __ 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] list of list objects
hi there, I have a list of list objects i need to remove the top layer [[1]] [1].0 ABC DEFLMN [1].1 WER ERT TRY [[2]] [2].0 ASD,werqwe [2].1 wdvghjggj I wanna avoid the top layer...that is [[1]] [[2]] shouldnt be there just a simple list is wat i need. [1].0 ABC DEFLMN [1].1 WER ERT TRY [2].0 ASD,werqwe [2].1 wdvghjggj kindly let me know hoe to go abt it regards ramya -- View this message in context: http://www.nabble.com/list-of-list-objects-tp20628759p20628759.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] equation of a line or curve
Hi R users I used the function line(x,y) and line(lowess(x,y)) to see the correlation between 2 variables (x,y). Here is my question: is there a way to ask R to tell me the equation of -this line : line(x,y) -this curve: line(lowess(x,y)) Best regards -- View this message in context: http://www.nabble.com/equation-of-a-line-or-curve-tp20628845p20628845.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] list of list objects
Will one of these two solutions work for you: x - list(list(1,2), list(3,4), list(5,6)) x [[1]] [[1]][[1]] [1] 1 [[1]][[2]] [1] 2 [[2]] [[2]][[1]] [1] 3 [[2]][[2]] [1] 4 [[3]] [[3]][[1]] [1] 5 [[3]][[2]] [1] 6 lapply(x, [[, 1) [[1]] [1] 1 [[2]] [1] 3 [[3]] [1] 5 lapply(x, '[', 1) [[1]] [[1]][[1]] [1] 1 [[2]] [[2]][[1]] [1] 3 [[3]] [[3]][[1]] [1] 5 On Fri, Nov 21, 2008 at 3:14 PM, Rajasekaramya [EMAIL PROTECTED] wrote: hi there, I have a list of list objects i need to remove the top layer [[1]] [1].0 ABC DEFLMN [1].1 WER ERT TRY [[2]] [2].0 ASD,werqwe [2].1 wdvghjggj I wanna avoid the top layer...that is [[1]] [[2]] shouldnt be there just a simple list is wat i need. [1].0 ABC DEFLMN [1].1 WER ERT TRY [2].0 ASD,werqwe [2].1 wdvghjggj kindly let me know hoe to go abt it regards ramya -- View this message in context: http://www.nabble.com/list-of-list-objects-tp20628759p20628759.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. -- Jim Holtman Cincinnati, OH +1 513 646 9390 What is the problem that you are trying to solve? __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] list of list objects
?unlist -Original Message- From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] On Behalf Of Rajasekaramya Sent: Friday, November 21, 2008 2:14 PM To: r-help@r-project.org Subject: [R] list of list objects hi there, I have a list of list objects i need to remove the top layer [[1]] [1].0 ABC DEFLMN [1].1 WER ERT TRY [[2]] [2].0 ASD,werqwe [2].1 wdvghjggj I wanna avoid the top layer...that is [[1]] [[2]] shouldnt be there just a simple list is wat i need. [1].0 ABC DEFLMN [1].1 WER ERT TRY [2].0 ASD,werqwe [2].1 wdvghjggj kindly let me know hoe to go abt it regards ramya -- View this message in context: http://www.nabble.com/list-of-list-objects-tp20628759p20628759.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] list of list objects
?unlist -Original Message- From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] On Behalf Of Rajasekaramya Sent: Friday, November 21, 2008 2:14 PM To: r-help@r-project.org Subject: [R] list of list objects hi there, I have a list of list objects i need to remove the top layer [[1]] [1].0 ABC DEFLMN [1].1 WER ERT TRY [[2]] [2].0 ASD,werqwe [2].1 wdvghjggj I wanna avoid the top layer...that is [[1]] [[2]] shouldnt be there just a simple list is wat i need. [1].0 ABC DEFLMN [1].1 WER ERT TRY [2].0 ASD,werqwe [2].1 wdvghjggj kindly let me know hoe to go abt it regards ramya -- View this message in context: http://www.nabble.com/list-of-list-objects-tp20628759p20628759.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] How to convert S4 class slots into data.frame or how to assign variables of type 'Date'
Hi,
I created a class (S4) with some slots like value, date, description
(it's actually a financial transaction class). Now I need a method
to convert this class forth and back into a single row data.frame,
where every slots represents a column. This method looks at the
moment like this:
setMethod(as.data.frame, Transaction,
function(x, row.names = NULL, optional = FALSE, ...){
slotnames - slotNames(x)
slotlist -
data.frame(rbind(1:length(slotnames)))
names(slotlist) - slotnames
for(i in slotnames) {
slotlist[1, i] - slot(x, i)
}
return(slotlist)
}
)
This method doesn't require predetermined slotnames or types, which
is important to me. The method works quite good but the problem is
that I have slots of type 'Date' and this method doesn't preserve
the type but converts it to numeric.
A couple of tests showed that this is actually a problem of
assigning values to data.frame column. Something like this:
slotlist$DayOfTransaction - slot(Transaction, DateOfTransaction)
would preserve the type of DateOfTransaction as 'Date'.
But I don't see a way to use this assigning scheme in my method
without using the actual slotnames and giving up a lot of flexibility.
Do you have any suggestions? Is there maybe even a simple way to
convert S4 slots into data.frames?
Thanks in advance
Ronny
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Does nobody has an idea?
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[R] Growth rate determination using ANCOVA
I'm a programmer in a biology lab who is starting to use R to automate some of our statistical analysis of growth rate determination. But I'm running into some problems as I re-code. 1) Hypotheses concerning Slope similarity/difference: I'm using R's anova(lm()) methods to analyse a model which looks like this: growth.metric ~ time * test.tube I understand that testing the the interaction between time and tube (time:test.tube) will tell us if the growth rates (for the last three test tubes) are significantly different from one another (Ho=slopes are the same). The purpose of doing this test is so that we can be certain our cultures have fully acclimated to the treatment and aren't going to change much if we stop measuring. This is an important cost saving practice too as measurements can go on for years. Yet I'm worried that our null and alternative hypotheses should be swapped so that our test is more conservative (Ho=slopes are different ... ie still acclimating.) Is there a way to specify my model that flips these hypotheses? Should I be using a different method? Is this even appropriate? 2) Growth Rate is confounded with Variance of Growth Rate I'm also worried about the fact that rates for cultures with faster growth are calculated using fewer data points (assuming similar sampling times between treatments) . The result is that growth ~ var (growth). Not only does this put a wrinkle in my analysis between treatments, but it also biases the growth acclimation determining ANCOVA test above. Faster growing cultures will usually pass the no significant difference between slopes test more easily because there are fewer points from which to be certain about rejecting Ho. Is there a way to control for this? Perhaps I could include the number of points in my model? 3) Statistical validity of using subsets of growth.metric measurements within a test tube There are some lab members who insist that we can throw out the beginning and end of our log transformed growth.metric measurements because they are outliers in determining maximum growth.I've proposed looping through all possible combinations of 3 or more points within the growth curve and using the highest or best fitting (best R- squared) slope. But this idea has been rejected by our PI as not be a valid thing to do. Ideas here? Thank you. Dave __ 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] Conditionally Removing data from a vector
I greatly apoligize, clear in my own mind, just didnt explain well enough. DEVS should equal 6 values the last being 620, like the data frame: CANDS DEVS [1,] 100 120 [2,] 101 220 [3,] 102 320 [4,] 103 420 [5,] 104 520 [6,] 105 620 And yes, my first question is do i need a if statement at all as the which command seems to do the same thing. IN my working code i start with a vector of Candidates, a process occurs that looks up a value associated with each of those Candidates. If the corresponding value is equal to or larger than the starting value, in the test case the DEV value, then that Candidate is kept and the next Candidate is analyzed. The final product is a new vector of Candidates, those that are equal to or larger than the DEV value. Hope i didnt further the perfection of your confucion. Thanks for your time, let me know if i need to ask a better question Cheers, JR Philipp Pagel-5 wrote: I am having trouble removing entries from a vector based on the value of the vector and another object value. It works in my pseudo test run: DEV=400 #Y CANDS=c(100:105) #Z DEVS=c(120,220,320,420,520) if(DEVDEVS) (CANDS=CANDS[which(DEVDEVS)]) CANDS DEVS [1,] 100 120 [2,] 101 220 [3,] 102 320 [4,] 103 420 [5,] 104 520 [6,] 105 620 The above code does not produce the output you show when I paste it into R. It would be highly desirable to have a short, documented, working example... So the result CANDS is 103, 104 and 105 b/c the corresponding DEVS are larger than the DEV(400 in this case). How do they correspond? Do you mean they are both columns of a data.frame as your example output suggests? But CANDS and DEVS are of differnt length, so this is probably not what you meant. Finally, your code seems to indicate that you want all elements of CANDS for which the corresponding (?) value of DEVS is less than threshold DEV, but only if DEVDEVS. The latter produces a warning, as you are comparing a single value to a vector of 5, which is most likely not what you want. At this point my confusion is perfect and I give up. I think you need to give us a better explanantion of what data you have and what exactly you want to accomplish. cu Philipp -- Dr. Philipp Pagel Lehrstuhl für Genomorientierte Bioinformatik Technische Universität München Wissenschaftszentrum Weihenstephan 85350 Freising, Germany http://mips.gsf.de/staff/pagel __ 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. -- View this message in context: http://www.nabble.com/Conditionally-Removing-data-from-a-vector-tp20627244p20628349.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] How can I access the list argument within a for function call
Greetings,
I have been playing around with the R/Parallel package, which can farm out the
computation of a for-loop among multiple worker processes. Each worker gets a
chunk of the for-loop iterations; for example, if you have two workers and
for(x in 1:1000){...}, one worker would typically get iterations 1:500, while
the second does 501:1000.
I would like to be able to preallocate matrices to capture the results within
the for loop, on the first iteration. Is there a way to access the list
argument of the for function programmatically from within the loop? I am
referring to the for arguments shown on page 51 of the R Language Definition
manual:
for(var,list,expr)
I think I have a work-around, but was curious whether this was possible...
Thanks,
Mike
[[alternative HTML version deleted]]
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[R] Generalized Outer
Outer's arguments are restricted to atomic vectors or arrays built on atomic vectors (though the documentation is not explicit on this point). What is the equivalent for lists or arrays built on lists? My particular application was testing the Kendall tau function. I tried this outer( permn(3), permn(3), function(a,b)as.double(Kendall(a,b)$tau) ) I'd have hoped for something like matrix( c( Kendall(permn(3)[[1]], permn(3)[[1]] )$tau, Kendall(permn(3)[[1]], permn(3)[[2]] ) $tau, ...), length(permn(3)), length(permn(3)) ) but of course permn(3) is a list of vectors, not an atomic list. If I want to define a version of outer for lists myself, I suppose the clean way is to overload outer for the list class. But I'm not sure how to do that since outer is not generic. Does this mean I need to do something like outer.vector - outer; outer - define generic outer etc.? Should I be redefining a built-in this way? Can someone point me to documentation on this? Thanks, -s PS Should Kendall really be returning CSingles? The documentation seems to say that they are to be used only in interfaces to outside 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] How to convert S4 class slots into data.frame or how to assign variables of type 'Date'
Hi,
I created a class (S4) with some slots like value, date, description
(it's actually a financial transaction class). Now I need a method
to convert this class forth and back into a single row data.frame,
where every slots represents a column. This method looks at the
moment like this:
setMethod(as.data.frame, Transaction,
function(x, row.names = NULL, optional = FALSE, ...){
slotnames - slotNames(x)
slotlist -
data.frame(rbind(1:length(slotnames)))
names(slotlist) - slotnames
for(i in slotnames) {
slotlist[1, i] - slot(x, i)
}
return(slotlist)
}
)
This method doesn't require predetermined slotnames or types, which
is important to me. The method works quite good but the problem is
that I have slots of type 'Date' and this method doesn't preserve
the type but converts it to numeric.
You would probably have gotten a quicker response if you had made a
reproducible example as requested in the posting guide. However, the following
example should give you a solution.
tmp - new(numWithId, 1, id = Sys.Date())
slotnames - slotNames(tmp)
slotlist - vector(list, length(slotnames))
names(slotlist) - slotnames
for(i in slotnames) slotlist[[i]] - slot(tmp, i)
as.data.frame(slotlist)
Take a look at the coercion section of ?[.data.frame. I belive your Date is
being converted to numeric to match the class of what it is replacing.
Mark Lyman
Statistician, ATK
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[R] OT statistics question on TukeyHSD
Hi, I have a dataset that is stratified by 3 levels, with an unequal number of observations per strata: AB C 182 18291 The data are approximately normally distrubuted. Is there any reason why I can NOT use TukeyHSD to do pair-wise comparison of means? Any thoughts or references would be greatly appreciated. Thanks, Dylan -- Dylan Beaudette Soil Resource Laboratory http://casoilresource.lawr.ucdavis.edu/ University of California at Davis 530.754.7341 __ 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 with CCLUS
Hi, I am using the following syntax to enter data and perform a cluster analysis: x - read.table (clstrdbt.csv, header=TRUE, sep = ,,fill = TRUE) cl-cclust(x,4,20,verbose=TRUE,method=kmeans) This is the result I receive: Error in cclust(x, 4, 20, verbose = TRUE, method = kmeans) : (list) object cannot be coerced to type 'double' Please help. Regards, Erik [[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] Conditionally Removing data from a vector
DEVS should equal 6 values the last being 620, like the data frame: CANDS DEVS [1,] 100 120 [2,] 101 220 [3,] 102 320 [4,] 103 420 [5,] 104 520 [6,] 105 620 And yes, my first question is do i need a if statement at all as the which command seems to do the same thing. You don't - and you don't need which() either. R is pretty good at indexing things in very conveniant ways. I'll assume we have the data you gave as an example in a data.frame - I'll call 'df': # create example data df - data.frame(cands=100:105, devs=c(120, 220, 320, 420, 520, 620)) threshold - 400 # get all rows with devs above threshold df[df$devsthreshold, ] # alternative way giving the same result subset(df, devsthreshold) The above methods will work if devs and cands are separate vectors, too: # get two vectors of equal length cands - df$cands devs - df$devs cands[devsthreshold] # alternative subset(cands, devsthreshold) For details and more examples have a look at the Introduction to R - especially section 2.7 Index vectors; selecting and modifying subsets of a data set. cu Philipp -- Dr. Philipp Pagel Lehrstuhl für Genomorientierte Bioinformatik Technische Universität München Wissenschaftszentrum Weihenstephan 85350 Freising, Germany http://mips.gsf.de/staff/pagel __ 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] HELP
Cannot reproduce. Van den Berge Joke wrote: __ 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] [R-pkgs] Update of X2R (with FishGraph) at CRAN, 20 Nov 2008
X2R is a bundle of three software libraries for passing complicated data structures from Fortran, C/C++, or AD Model Builder to R. An update has been sent to CRAN and should be available from mirrors shortly. From the menu at the left of the CRAN home page, look under Software / Other. We also have updated FishGraph, a compatible set of R functions for examining output from fish population models. * * * Changes in this Update The update adds minor bug fixes only. In For2R, nested lists are now handled correctly. In FishGraph, correct axis units are now used in the CLD.plots (catch, landings, discards) routine. * * * More detail, for those interested: X2R is three independent but related software libraries: C2R, ADMB2R, and For2R (together, X2R). Each contains output routines to simplify transfer of complicated data structures from models written in a compiled language to R (note 1). Through calls to X2R routines, the user's data is written as a structured ASCII file. That file can be read by R with a single dget() function call to create an R data object of type list. The list may contain components such as vectors, data frames, matrices, and other lists. These are NOT R packages; rather they are subroutine libraries to be used with programmers' own modeling codes. Limited testing indicates compatibility with S-PLUS, as well (note 2). Languages supported are Fortran 95 (with For2R), C and C++ (with C2R) and AD Model Builder (with ADMB2R) (note 3). Source code and detailed users' manuals are supplied. ADMB2R has been tested with ADMB versions 6.03 and 7.71 and recent versions of the gcc and Borland C++ compilers. The compatible software FishGraph is a set of R functions providing exploratory and presentation graphics of fishery catch-at-age or catch-at-length models. By taking its data from an R list assumed to have a certain structure (diagram provided), FishGraph determines which graphs should be generated. The required data structure may be generated with X2R or within R itself and may contain any amount of additional data. Most FishGraph routines have options to control titles, colors, and reference lines. The combination of X2R and FishGraph allows automating graphics from routine fish stock assessments. The FishGraph functions can be modified or supplemented to reflect the needs of the analysis at hand. X2R is supplied as files X2R.zip and X2R.tar.gz, which are equivalent. Version and release date are found in file VersionInfo.txt in the root of each archive. The new version is dated November 19, 2008. FishGraph (same CRAN directory) is supplied as a Windows installer. We will gladly collaborate with anyone interested in adapting FishGraph to other operating systems. This work has been tested and is regularly used by the authors. However, any software may contain bugs, and these works are classified by NOAA as Experimental Products. THIS SOFTWARE IS SUPPLIED WITH NO WARRANTY OF ANY KIND. The authors will endeavor to fix bugs promptly and to add requested features. Send bug reports, suggestions, and extensions to Michael H. Prager - [EMAIL PROTECTED] Southeast Fisheries Science Center National Marine Fisheries Service, NOAA 101 Pivers Island Road Beaufort, North Carolina 28516 USA * Note 1. Use of product names (commercial or otherwise) does not imply endorsement or recommendation by any U.S. government agency, nor by the authors in their government capacities. * Note 2. S-PLUS is a commercial product that requires licensing by the user. * Note 3. AD Model Builder, formerly a commercial product, is now an open-source free-software project. See http://admb-project.org/. ___ R-packages mailing list [EMAIL PROTECTED] https://stat.ethz.ch/mailman/listinfo/r-packages __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Dendrogram labels
Take a look at the las() function o ?par. It permits you to choice
labels directions.
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Rodrigo Aluizio
Em 21/11/2008, às 17:14, Richardson, Patrick [EMAIL PROTECTED]
g escreveu:
!#x000a
Is there any way to change the orientation of the labels on the end
of the dendrograms to horizontal rather than vertical? If so, how
can I do that. Below is my code, but I'm not sure which argument(s)
I can use to change the label(s) (if it is possible to do).
a - hclust(z, method = complete)
# Plots hclust dendrogram #
plot(a, frame.plot=TRUE, lwd=2)
Many thanks,
_
Patrick Richardson
Biostatistician - Program of Translational Medicine
Van Andel Research Institute - Webb Lab
333 Bostwick Avenue NE
Grand Rapids, MI 49503
ph. 616.234.5787
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[R] Overlay two plots on the same device with axes having the same scale.
R 2.6.0
Windows XP
I have crated a file containing data with x and y values and have plotted the
data using the output of gam. I would like to overlay an x,y plot of the data
on top of the line returned by gam. I have succeeded in doing this using
par(new=TRUE), unfortunately the y axis of the overlayed plots are of slightly
different scale. How can I set the axes of the overlay plots so the scales are
exactly the same?
y-matrix(nrow=200,ncol=2)
# Create data
y[1:200,2]-sin((0:99)/(0.5*10*pi))+rnorm(100,0,.2)
y[,1]-1:200
# Plot of raw data.
plot(y[,1],y[,2])
oldpar-par(new=TRUE)
fit1-gam(y[,2]~s(y[,1]))
#Plot of fit - overlays plot of raw data.
#Note scale of overaly is not the same as the original raw data plot.
summary(fit1)
plot(fit1)
Thanks,
John
John David Sorkin M.D., Ph.D.
Chief, Biostatistics and Informatics
University of Maryland School of Medicine Division of Gerontology
Baltimore VA Medical Center
10 North Greene Street
GRECC (BT/18/GR)
Baltimore, MD 21201-1524
(Phone) 410-605-7119
(Fax) 410-605-7913 (Please call phone number above prior to faxing)
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