[R] How to repeat replication
Dear all, I generate data under IRT mdel. I conducted 1000 replications. When I run, some replication was not fit with my model. So, replications were fit model were less than 1000 replication. If I want its run until 1000 replicaions. How shouldI write functin. Thank you, Kamontip [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] How to subtract the counter i in for loop?
On 30-11-2012, at 07:18, C W wrote:
thanks, Berend. Both of your code works great. Is there a function that can
do it?
Something like this:
x - matrix(NA, nrow=15, ncol=2)
for(i in 1:15){
x[i,] - sample(c(NA, 20, 77), 2, prob=c(0.2, 0.3, 0.4))
}
x
[,1] [,2]
[1,] NA 77
[2,] 77 NA
[3,] NA 77
[4,] 77 20
[5,] 77 20
[6,] 77 20
[7,] 20 NA
[8,] 77 20
[9,] 77 NA
[10,] 77 NA
[11,] 77 20
[12,] 20 77
[13,] NA 77
[14,] 77 20
[15,] 77 20
I want to have a column of 15 samples without NA's. Is there an R function
like ifelse()?
It's not clear what you exactly want.
A matrix with 15 rows and some columns?
How do you want to remove the NA's in each column?
Then why don't you leave out the NA in the sample?
Berend
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Re: [R] googleVis plot and knitr/sweave
Just a lame question: is there any chance to generate SVG maps with googleVis and import that in LaTeX? Best, Gergely On Fri, Nov 30, 2012 at 2:02 AM, Yihui Xie x...@yihui.name wrote: Then you can take a look at the ggmap (on CRAN) or snippets package (on Simon's RFroge: http://rforge.net/snippets/), which supports OpenStreet Map; googleVis is HTML/SVG based. Regards, Yihui -- Yihui Xie xieyi...@gmail.com Phone: 515-294-2465 Web: http://yihui.name Department of Statistics, Iowa State University 2215 Snedecor Hall, Ames, IA On Thu, Nov 29, 2012 at 5:00 PM, Filoche pmassico...@hotmail.com wrote: Hi and thank you for your answer. In fact, I don't really need the interactive part of googleVis map. So if there's a way to plot a static version of the map it would be perfect for me. Regards, Phil __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Little's Chi Square test for MCAR?
If you mean Little's Chi Sq test in particular, go to the BaylorEdPsych package. Regards, José José Iparraguirre Chief Economist Age UK -Original Message- From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On Behalf Of Sileshi Melesse Sent: 30 November 2012 05:57 To: 'r-h...@stat.math.ethz.ch' Subject: [R] Little's Chi Square test for MCAR? Is there any further development in R to identify the missing data Mechanism? Would you please help ? === Please find our Email Disclaimer here--: http://www.ukzn.ac.za/disclaimer === [[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. A Star for Christmas Kick start the festive season by attending one of Age UK’s Carol Concerts, A Star for Christmas. Taking place at Manchester Cathedral on Saturday 1 December and London’s St Pancras Church (opposite Euston Station) on Thursday 6 December, they will feature special musical performances, readings by your favourite celebrities and carols, followed by mince pies and wine. Tickets are priced at £20 full price/ £10 concessions. For more information, please visit http://www.ageuk.org.uk/astarforchristmas or contact the Fundraising Events Team on 020 303 31725. Age UK Improving later life www.ageuk.org.uk --- Age UK is a registered charity and company limited by guarantee, (registered charity number 1128267, registered company number 6825798). Registered office: Tavis House, 1-6 Tavistock Square, London WC1H 9NA. For the purposes of promoting Age UK Insurance, Age UK is an Appointed Representative of Age UK Enterprises Limited, Age UK is an Introducer Appointed Representative of JLT Benefit Solutions Limited and Simplyhealth Access for the purposes of introducing potential annuity and health cash plans customers respectively. Age UK Enterprises Limited, JLT Benefit Solutions Limited and Simplyhealth Access are all authorised and regulated by the Financial Services Authority. -- This email and any files transmitted with it are confidential and intended solely for the use of the individual or entity to whom they are addressed. If you receive a message in error, please advise the sender and delete immediately. Except where this email is sent in the usual course of our business, any opinions expressed in this email are those of the author and do not necessarily reflect the opinions of Age UK or its subsidiaries and associated companies. Age UK monitors all e-mail transmissions passing through its network and may block or modify mails which are deemed to be unsuitable. Age Concern England (charity number 261794) and Help the Aged (charity number 272786) and their trading and other associated companies merged on 1st April 2009. Together they have formed the Age UK Group, dedicated to improving the lives of people in later life. The three national Age Concerns in Scotland, Northern Ireland and Wales have also merged with Help the Aged in these nations to form three registered charities: Age Scotland, Age NI, Age Cymru. __ 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] kernlab 0.9-15 on CRAN
kernlab package version 0.9-15 has been posted on CRAN, it contains several improvements and fixes along with new features e.g. variance estimation for Gaussian Process regression. cheers Alexandros ___ R-packages mailing list r-packa...@r-project.org https://stat.ethz.ch/mailman/listinfo/r-packages __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] worldmap_region/country problem
anna.fechner at de.pwc.com writes: Dear R community, I'm trying to graphically illustrate my data with a worlmap. Unfortunately, my data is partly on country basis and partly on regional basis (e.g. certain African countries are aggregated to one region). I am using the package rwoldmap. The data on country basis can be mapped, but our defined regions cannot be identified in R. Therefore, all the countries in these regions are not plotted. Is there a way to define regions (for example 'Rest of East-Europe') and to map a combination of these defined regions and individual countries in one worldmap? Is it possible to not display borderlines between some countries but to show the borders between others? Yes, use gUnaryUnion() in rgeos, or the equivalent unionSpatialPolygons() in maptools, which calls gUnaryUnion(). These create new, merged polygons, and should then be associated with the data for the entities you are using. Be careful not to use gpclib union functions as they are prohibited for commercial purposes. Please follow up on the R-sig-geo list if need be. Roger Thank you very much for your help! Anna Fechner PricewaterhouseCoopers Aktiengesellschaft Wirtschaftsprüfungsgesellschaft __ 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] CreateThread failure since R 2.15.2 (32-bit)
Dear R users developers, I coming across the following issue since R 2.15.2 32-bit (running on Windows XP 32.bit; some output left out for conciseness): setInternet2(TRUE) require(rJava) .jinit() getCRANmirrors() system(ls , intern = TRUE) Error in system(ls, intern = TRUE) : CreateThread failed My details are as follows: sessionInfo() R version 2.15.2 (2012-10-26) Platform: i386-w64-mingw32/i386 (32-bit) locale: [1] LC_COLLATE=German_Switzerland.1252 LC_CTYPE=German_Switzerland.1252 LC_MONETARY=German_Switzerland.1252 [4] LC_NUMERIC=CLC_TIME=German_Switzerland.1252 attached base packages: [1] stats graphics grDevices utils datasets methods base other attached packages: [1] rJava_0.9-3 The failure appears since R 2.15.2, i.e. also with R 2.15.2patched and with R-Devel. It does, however, not appear with e.g. R 2.15.1 or R 2.14.2. Also, I don't get the issue with R 2.15.2 64-bit (running on Windows 7 Professional 64-bit). Also note that the issue seems related to rJava in that running the same without the rJava part (and therefore without initializing a JVM instance) runs fine: setInternet2(TRUE) getCRANmirrors() system(ls , intern = TRUE) Any ideas on what the issue might be? Thanks Best regards, 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 on stacking matrices up
Dear All, #I have the following code Dose-1000 Tinf -0.5 INTERVAL -8 TIME8 -matrix(c((0*INTERVAL):(1*INTERVAL))) TIME7 -matrix(c((0*INTERVAL):(2*INTERVAL))) TIME6 -matrix(c((0*INTERVAL):(3*INTERVAL))) TIME5 -matrix(c((0*INTERVAL):(4*INTERVAL))) TIME4 -matrix(c((0*INTERVAL):(5*INTERVAL))) TIME3 -matrix(c((0*INTERVAL):(6*INTERVAL))) TIME2 -matrix(c((0*INTERVAL):(7*INTERVAL))) TIME1 -matrix(c((0*INTERVAL):(8*INTERVAL))) CDURINF1 -((Dose/Tinf)*(1/(0.088*76.9)))*(1-exp(-0.088*TIME1)) CAFTINF1 -((Dose/Tinf)*(1/(0.088*76.9)))*(1-exp(-0.088*Tinf))*(exp(-0.088*(TIME1))) CONC1 -ifelse(TIME1=Tinf,CDURINF1,CAFTINF1) CDURINF2 -((Dose/Tinf)*(1/(0.088*76.9)))*(1-exp(-0.088*TIME2)) CAFTINF2 -((Dose/Tinf)*(1/(0.088*76.9)))*(1-exp(-0.088*Tinf))*(exp(-0.088*(TIME2))) CONC2 -ifelse(TIME2=Tinf,CDURINF2,CAFTINF2) CDURINF3 -((Dose/Tinf)*(1/(0.088*76.9)))*(1-exp(-0.088*TIME3)) CAFTINF3 -((Dose/Tinf)*(1/(0.088*76.9)))*(1-exp(-0.088*Tinf))*(exp(-0.088*(TIME3))) CONC3 -ifelse(TIME3=Tinf,CDURINF3,CAFTINF3) CDURINF4 -((Dose/Tinf)*(1/(0.088*76.9)))*(1-exp(-0.088*TIME4)) CAFTINF4 -((Dose/Tinf)*(1/(0.088*76.9)))*(1-exp(-0.088*Tinf))*(exp(-0.088*(TIME4))) CONC4 -ifelse(TIME4=Tinf,CDURINF4,CAFTINF4) CDURINF5 -((Dose/Tinf)*(1/(0.088*76.9)))*(1-exp(-0.088*TIME5)) CAFTINF5 -((Dose/Tinf)*(1/(0.088*76.9)))*(1-exp(-0.088*Tinf))*(exp(-0.088*(TIME5))) CONC5 -ifelse(TIME5=Tinf,CDURINF5,CAFTINF5) CDURINF6 -((Dose/Tinf)*(1/(0.088*76.9)))*(1-exp(-0.088*TIME6)) CAFTINF6 -((Dose/Tinf)*(1/(0.088*76.9)))*(1-exp(-0.088*Tinf))*(exp(-0.088*(TIME6))) CONC6 -ifelse(TIME6=Tinf,CDURINF6,CAFTINF6) CDURINF7 -((Dose/Tinf)*(1/(0.088*76.9)))*(1-exp(-0.088*TIME7)) CAFTINF7 -((Dose/Tinf)*(1/(0.088*76.9)))*(1-exp(-0.088*Tinf))*(exp(-0.088*(TIME7))) CONC7 -ifelse(TIME7=Tinf,CDURINF7,CAFTINF7) CDURINF8 -((Dose/Tinf)*(1/(0.088*76.9)))*(1-exp(-0.088*TIME7)) CAFTINF8 -((Dose/Tinf)*(1/(0.088*76.9)))*(1-exp(-0.088*Tinf))*(exp(-0.088*(TIME8))) CONC8 -ifelse(TIME8=Tinf,CDURINF8,CAFTINF8) CONCF -matrix(c(CONC1+CONC2+CONC3+CONC4+CONC5+CONC6+CONC7+CONC8)) plot(CONCF,type=l) I am trying to plot CONCF in a specific way. Let me try to explain: I would like to plot it so that the x axis goes from 0 to 8*INTERVAL (here the INTERVAL represents time in hours), and then each segments of the matrix would be added to the sum of the previous ones at a specific moment in time. Assuming a much shorter example this is what it should be like where according to the same equations above each value of CONCi was calculated at 1 h intervals: INTERVAL =3 time points here where the values are calculated is 0:(3*INTERVAL) for CONC1, 0:(2*INTERVAL) for CONC2, and 0:(1*INTERVAL) for CONC3, all calculated at 1 hour intervals and lets assume calculated values are: CONC1 -c(0,1,2,3,4,5,6,7,8,9) CONC2 -c( 11,12,13,14,15,16,17) CONC3 -(21,22,23,24) . . . so I would like to get CONCF that would have the following values: CONCF -c(0,1,2,3+11,4+12,5+13,6+14,7+15,8+16,9+17, 6+14+21,7+15+22,8+16+23,9+17+23) please note that CONC2 was started to be added to CONC1 at 1*INTERVAL value estimated for CONC1, ie CONC1 has the value of 3 at hour 3, and I need to add the value of CONC2 at hour zero to it, and so on... hope I made it clear enough and thanks for the help, Andras [[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] Confidence intervals for estimates of all independent variables in WLS regression
Hello,
You don't have to exchange 'object' by the name of your model, you call
the function with the name of your model:
x - 1:20
y - x + rnorm(20)
fit - lm(y ~ x)
ci_lm(fit)
beta lowerupper
(Intercept) 0.6741130 -0.9834827 2.331709
x 0.9575906 0.8192171 1.095964
Hope this helps,
Rui Barradas
Em 30-11-2012 01:07, Torvon escreveu:
Rui,
Thank you very much. Are there other things I have to adjust except for
exchanging object by the name of my model?
Torvon
On 29 November 2012 08:17, Rui Barradas ruipbarra...@sapo.pt wrote:
ci_lm - function(object, level = 0.95){
summfit - summary(object)
beta - summfit$coefficients[, 1]
se - summfit$coefficients[, 2]
df - summfit$df[1]
alpha - 1 - level
lower - beta + qt(alpha/2, df = df)*se
upper - beta + qt(1 - alpha/2, df = df)*se
data.frame(beta, lower, upper)
}
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[R] log-scale xy-plots
Hello R-Users, Hello R-help-team. I found a nice way to create a xy-plot with a single log-axis (e.g. the y-axis). This is often needed to show biological data. First attach the desired dataset you want to plot. Now use the following commands: plot(x,y,log=y, yaxt=n, ylim=c(0.1,10), ... ) y1-floor(log10(range(WT))) pow-seq(y1[1],y1[2]+1) ticksat -as.vector(sapply(pow, function(p) (1:10)*10^p)) axis(2, 10^pow) axis(2, ticksat, labels=NA, tcl=-0.25, lwd=0, lwd.ticks=1) I want to thank Aaron from http://stackoverflow.com/users/210673/aaron. I just added the ylim option to create a nice 0.1 to 10 log-axis which is commonly used for biological datasets. Good luck, Stefan -- Dipl. Biologe Stefan Stagge Pengsjövägen 35 91133 Vännäs mobilephone: +46 762 666401 __ 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] xts indexed with Date class
Hi I see a changed behaviour in xts indexed on class Date in the latest versions, versus 2. It seems to be related to changes to/from daylight savings time, happens those weekends. Is it not intended that class Date be used like this, or is this new behaviour incorrect? Giles Example: a-as.Date(15423:15426) x-xts(seq_along(a),a) print(x) [,1] 2012-03-241 2012-03-252 2012-03-253 2012-03-264 print(index(x)) [1] 2012-03-24 2012-03-25 2012-03-25 2012-03-26 print(as.numeric(index(x))) [1] 15423 15424 15424 15425 #for reference, zoo behaves as expected: z-zoo(seq_along(a),a) print(index(z)) [1] 2012-03-24 2012-03-25 2012-03-26 2012-03-27 print(as.numeric(index(z))) [1] 15423 15424 15425 15426 Package: xts Type: Package Title: eXtensible Time Series Version: 0.8-8 Date: 2012-10-05 Windows 7 64-bit R version 2.15.2 64-bit all packages up-to-date #--previously, R 2.13.2 and xts 0.8-2 , showing expected behaviour - a-as.Date(15423:15426) x-xts(seq_along(a),a) print(x) [,1] 2012-03-241 2012-03-252 2012-03-263 2012-03-274 print(index(x)) [1] 2012-03-24 2012-03-25 2012-03-26 2012-03-27 print(as.numeric(index(x))) [1] 15423 15424 15425 15426 z-zoo(seq_along(a),a) print(index(z)) [1] 2012-03-24 2012-03-25 2012-03-26 2012-03-27 print(as.numeric(index(z))) [1] 15423 15424 15425 15426 __ 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] missed values
Hello I have dataframe 101 2008-07 0.2898966 102 2008-08 0.3101667 103 2008-09 0.3730476 104 2008-10 0.2717037 105 2008-11 0.1344286 106 2008-12 0.1375000 107 2009-01 0.1781000 108 2009-02 0.2146667 109 2009-03 0.2808235 110 2009-04 0.4326250 111 2009-05 0.3420741 112 2009-06 0.2675238 113 2009-07 0.2478667 114 2009-08 0.3147000 115 2009-09 0.3437826 116 2009-10 0.2057391 117 2009-11 0.1824737 118 2009-12 0.1520714 119 2010-01 0.1485455 120 2010-02 0.176 121 2010-03 0.3072000 122 2010-04 0.3294445 123 2010-05 0.3278125 124 2010-06 0.2865000 125 2010-07 0.3170333 126 2010-08 0.2052143 127 2010-09 0.1837368 128 2010-10 0.1652778 129 2010-11 0.1292500 130 2010-12 0.1366250 131 2011-01 0.1909231 132 2011-02 0.1841177 133 2011-03 0.2897222 134 2011-04 0.3084211 135 2011-05 0.3349600 136 2011-06 0.3157917 137 2011-07 0.2755652 138 2011-08 0.2019355 139 2011-09 0.3325556 140 2011-10 0.1724348 141 2011-11 0.1935200 142 2011-12 0.1388000 143 2012-01 0.1484500 144 2012-02 0.204 145 2012-03 NaN 146 2012-04 NaN 147 2012-05 NaN 148 2012-06 NaN 149 2012-07 NaN 150 2012-08 NaN 151 2012-09 NaN then I use function df.ts- ts(df[,2],start=c(2004,10),freq=12) and try to use stl(log(df.ts),s.window=periodic) But I have an error related with missed values. How can omit missing values? Regards Aleksander [[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] TSCov function from RTAQ package
Hi R Users! I am having some difficulties in using the TSCov function from RTAQ package that should calculate the two time scale realized volatility (Zhang et al, 2005). Let's suppose I have tick by tick data, let's say aaa and bbb. If I write in R: /stock1=aaa$PRICE stock2=bbb$PRICE TSCov(list(stock1,stock2))/ The result is: /Error in var[i] : object of type 'closure' is not subsettable/ Instead, if I write: /TSCov(stock1,stock2)/ I obtain a single value and not a 2X2 matrix (one for each day). I would only want to obtain a 2x2 matrix per day. Where is my error? It would be sufficient an example of correct syntax! :) Thank you, Vincent -- View this message in context: http://r.789695.n4.nabble.com/TSCov-function-from-RTAQ-package-tp4651402.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] subgroup-based quantiles
Dear R users, given the patient sample with their Gender and Age GENDER Age [1,] 2 45 [2,] 1 58 [3,] 1 54 [4,] 2 71 [5,] 2 64 ... I would like to create an another column which groups the patients wrt Gender specific Age quantiles, following methodology similar to: Age_group - cut(Age, labels=c(1:10), breaks=quantile(Age, seq(0,1,.1)),include.lowest = TRUE) The function above allows me to group only wrt Age quantiles. Best, Robert -- View this message in context: http://r.789695.n4.nabble.com/subgroup-based-quantiles-tp4651415.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] Shipping densitty
-- View this message in context: http://r.789695.n4.nabble.com/Shipping-densitty-tp4651404.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] Oracle database connction from R in Linux
Hi, This is sasi. I installed R 64 bit and oracle client 64 bit in fedora 17. After R is working fine. Oracle database also working fine. but problem is, If i try to connect oracle from R its not connecting. So, pls guide me how to connect from R to oracle. how, i tried:-- odbcinst.ini file code is [ORACLE2] Description = oracle driver Driver = /ibexis/app/oracle/product/11.2.0/db_1/lib/libsqora.so.11.1 DontDLClose = 1 FileUsage = 1 UsageCount = 1 I tried in R ... library(RODBC) con - odbcConnect(data source name, uid=username, pwd=password) after this line i got error, odbcConnect(india, uid=data_quality, pwd=data_quality) [1] -1 Warning messages: 1: In odbcDriverConnect(DSN=india;UID=data_quality;PWD=data_quality) : [RODBC] ERROR: state 01000, code 0, message [unixODBC][Driver Manager]Can't open lib '/ibexis/app/oracle/product/11.2.0/db_1/lib/libsqora.so.11.1' : file not found 2: In odbcDriverConnect(DSN=india;UID=data_quality;PWD=data_quality) : ODBC connection failed Please help me . Iam eagerly waiting for reply Thanks Regards M.Sasidhar [[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] Changing the base of geom_bar in ggplot
Hi, this is likely very easy to do but I don't find the right terms to search for the solution. I want to show in a bar chart how much different categories deviate from the average percentage change. Thus, in the following example I would like to set the zero line or base or whatever it is called to the average of the values (=0.82) and the bars should show the deviation from that, i.e. have the positive length of the deviation of the value from the average. I can do that by subtracting the average from each value but I would like to have the zero line as the average to make it more clear. How do you change the 'base'-line? x - data.frame(val=c(1.5, 3, -1.4, -1, 2), lab=letters[1:5]) ggplot(data=x, aes(x=lab, y=val)) + geom_bar(stat=identity) Many thanks for your help! Best regards, Werner -- View this message in context: http://r.789695.n4.nabble.com/Changing-the-base-of-geom-bar-in-ggplot-tp4651407.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] Nightingale’s Rose chart-any suggestion?
Hello, Everyone. Does anyone know how to create a Nightingales Rose chart by using R? Hopefully, the graph could be displayed like this: http://mbostock.github.com/protovis/ex/crimea-rose.html Thanks a lot. Kind regards, Henry [[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] Generating sequences of dates
Hello, I have the starting date in UNIX_TIME LONG number (number of milliseconds since 1970 jan 1st). Is there a way to generete the sequence of dates with monthly interval in POSIXct format like 2000-01-01 2000-02-01 2000-03-01 etc Thanks [[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] missed values
HI, Use, ?na.omit() na.omit(data) A.K. - Original Message - From: Vasilchenko Aleksander vasilchenko@gmail.com To: r-help@r-project.org Cc: Sent: Friday, November 30, 2012 7:07 AM Subject: [R] missed values Hello I have dataframe 101 2008-07 0.2898966 102 2008-08 0.3101667 103 2008-09 0.3730476 104 2008-10 0.2717037 105 2008-11 0.1344286 106 2008-12 0.1375000 107 2009-01 0.1781000 108 2009-02 0.2146667 109 2009-03 0.2808235 110 2009-04 0.4326250 111 2009-05 0.3420741 112 2009-06 0.2675238 113 2009-07 0.2478667 114 2009-08 0.3147000 115 2009-09 0.3437826 116 2009-10 0.2057391 117 2009-11 0.1824737 118 2009-12 0.1520714 119 2010-01 0.1485455 120 2010-02 0.176 121 2010-03 0.3072000 122 2010-04 0.3294445 123 2010-05 0.3278125 124 2010-06 0.2865000 125 2010-07 0.3170333 126 2010-08 0.2052143 127 2010-09 0.1837368 128 2010-10 0.1652778 129 2010-11 0.1292500 130 2010-12 0.1366250 131 2011-01 0.1909231 132 2011-02 0.1841177 133 2011-03 0.2897222 134 2011-04 0.3084211 135 2011-05 0.3349600 136 2011-06 0.3157917 137 2011-07 0.2755652 138 2011-08 0.2019355 139 2011-09 0.3325556 140 2011-10 0.1724348 141 2011-11 0.1935200 142 2011-12 0.1388000 143 2012-01 0.1484500 144 2012-02 0.204 145 2012-03 NaN 146 2012-04 NaN 147 2012-05 NaN 148 2012-06 NaN 149 2012-07 NaN 150 2012-08 NaN 151 2012-09 NaN then I use function df.ts- ts(df[,2],start=c(2004,10),freq=12) and try to use stl(log(df.ts),s.window=periodic) But I have an error related with missed values. How can omit missing values? Regards Aleksander [[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] how to separate stuck row elements?
Hi, I was wondering if it's possible to separate elements in multiple rows that actually should appear in different columns. I have a file where in certain lines there are elements not separated, and they certainly should appear in different columns (an example of the file is attached). The point is that I do not want to manually add a space in the txt file, however, I did not manage to do it automatically in R... Thanks in advance for any insight. Jaime -100 -100-3456-3456-3456-100 -100 23 -3456-3456-189 34 56 78 -100 34 56 21 44 65 78 __ 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] Nightingale’s Rose chart-any suggestion?
Hello, Take a look at the graph in http://gallery.r-enthusiasts.com/graph/Rose_diagram,97 Hope this helps, Rui Barradas Em 30-11-2012 12:24, Henry Smith escreveu: Hello, Everyone. Does anyone know how to create a Nightingales Rose chart by using R? Hopefully, the graph could be displayed like this: http://mbostock.github.com/protovis/ex/crimea-rose.html Thanks a lot. Kind regards, Henry [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Generating sequences of dates
Hello, Try the following. Y - 2000:2001 M - 1:12 D - 1 x - expand.grid(Y, M, D) Dates - sort(as.Date(apply(x, 1, paste, collapse = -))) Hope this helps, Rui Barradas Em 30-11-2012 12:26, Vasilchenko Aleksander escreveu: Hello, I have the starting date in UNIX_TIME LONG number (number of milliseconds since 1970 jan 1st). Is there a way to generete the sequence of dates with monthly interval in POSIXct format like 2000-01-01 2000-02-01 2000-03-01 etc Thanks [[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] how to separate stuck row elements?
Hello,
Try the following.
x - scan(text=
-100 -100-3456-3456-3456-100 -100
23 -3456-3456-189 34 56 78
-100 34 56 21 44 65 78
, what = )
fun - function(x){
f - function(.x){
if(grepl(-[[:digit:]]+, .x)){
g - gregexpr(-[[:digit:]]+, .x)
.y - as.numeric(unlist(regmatches(.x, g)))
}else{
.y - as.numeric(.x)
}
.y
}
unlist(unname(sapply(x, f)))
}
fun(x)
Hope this helps,
Rui Barradas
Em 30-11-2012 12:55, Jaime Otero Villar escreveu:
Hi,
I was wondering if it's possible to separate elements in multiple rows
that actually should appear in different columns. I have a file where
in certain lines there are elements not separated, and they certainly
should appear in different columns (an example of the file is
attached). The point is that I do not want to manually add a space in
the txt file, however, I did not manage to do it automatically in R...
Thanks in advance for any insight.
Jaime
__
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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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https://stat.ethz.ch/mailman/listinfo/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] Deleting certain observations (and their imprint?)
Hi -Original Message- From: r-help-boun...@r-project.org [mailto:r-help-bounces@r- project.org] On Behalf Of Stodola, Kirk Sent: Thursday, November 29, 2012 5:32 PM To: r-help@r-project.org Subject: [R] Deleting certain observations (and their imprint?) I'm manipulating a large dataset and need to eliminate some observations based on specific identifiers. This isn't a problem in and of itself (using which.. or subset..) but an imprint of the deleted observations seem to remain, even though they have 0 observations. This is causing me problems later on. I'll use the dataset warpbreaks to illustrate, I apologize if this isn't in the best format ##Summary of warpbreaks suggests three tension levels (H, M, L) summary(warpbreaks) breaks wool tension Min. :10.00 A:27 L:18 1st Qu.:18.25 B:27 M:18 Median :26.00 H:18 Mean :28.15 3rd Qu.:34.00 Max. :70.00 ## Subset the dataset and keep only those observations with L wb.subset - warpbreaks[which(warpbreaks$tension==L),] wb.subset - warpbreaks[which(warpbreaks$tension==L), , drop=TRUE] or warpbreaks$tension - factor(warpbreaks$tension) or change tension from factor to character vector. Regards Petr ##Summary of the subsetted data shows: L=18, M=0, H=0, Why is M and H still included? summary(wb.subset) breaks wool tension Min. :14.00 A:9 L:18 1st Qu.:26.00 B:9 M: 0 Median :29.50 H: 0 Mean :36.39 3rd Qu.:49.25 Max. :70.00 ##The subsetted dataset does not show M or H wb.subset Is there a way that M H can be completely eliminated (i.e. they don't show up in summary)? The only way I found was to export the dataset and then reimport, which seems pretty cumbersome. Thanks in advance for any help. -Kirk __ 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] help on stacking matrices up
Hi try to make your example easier and reproducible. I get CONCF -matrix(c(CONC1+CONC2+CONC3+CONC4+CONC5+CONC6+CONC7+CONC8)) Error in CONC1 + CONC2 : non-conformable arrays so I cannot understand what shall the final plot look like. Regards Petr -Original Message- From: r-help-boun...@r-project.org [mailto:r-help-bounces@r- project.org] On Behalf Of Andras Farkas Sent: Friday, November 30, 2012 12:51 PM To: r-help@r-project.org Subject: [R] help on stacking matrices up Dear All, #I have the following code Dose-1000 Tinf -0.5 INTERVAL -8 TIME8 -matrix(c((0*INTERVAL):(1*INTERVAL))) TIME7 -matrix(c((0*INTERVAL):(2*INTERVAL))) TIME6 -matrix(c((0*INTERVAL):(3*INTERVAL))) TIME5 -matrix(c((0*INTERVAL):(4*INTERVAL))) TIME4 -matrix(c((0*INTERVAL):(5*INTERVAL))) TIME3 -matrix(c((0*INTERVAL):(6*INTERVAL))) TIME2 -matrix(c((0*INTERVAL):(7*INTERVAL))) TIME1 -matrix(c((0*INTERVAL):(8*INTERVAL))) CDURINF1 -((Dose/Tinf)*(1/(0.088*76.9)))*(1-exp(-0.088*TIME1)) CAFTINF1 -((Dose/Tinf)*(1/(0.088*76.9)))*(1-exp(-0.088*Tinf))*(exp(- 0.088*(TIME1))) CONC1 -ifelse(TIME1=Tinf,CDURINF1,CAFTINF1) CDURINF2 -((Dose/Tinf)*(1/(0.088*76.9)))*(1-exp(-0.088*TIME2)) CAFTINF2 -((Dose/Tinf)*(1/(0.088*76.9)))*(1-exp(-0.088*Tinf))*(exp(- 0.088*(TIME2))) CONC2 -ifelse(TIME2=Tinf,CDURINF2,CAFTINF2) CDURINF3 -((Dose/Tinf)*(1/(0.088*76.9)))*(1-exp(-0.088*TIME3)) CAFTINF3 -((Dose/Tinf)*(1/(0.088*76.9)))*(1-exp(-0.088*Tinf))*(exp(- 0.088*(TIME3))) CONC3 -ifelse(TIME3=Tinf,CDURINF3,CAFTINF3) CDURINF4 -((Dose/Tinf)*(1/(0.088*76.9)))*(1-exp(-0.088*TIME4)) CAFTINF4 -((Dose/Tinf)*(1/(0.088*76.9)))*(1-exp(-0.088*Tinf))*(exp(- 0.088*(TIME4))) CONC4 -ifelse(TIME4=Tinf,CDURINF4,CAFTINF4) CDURINF5 -((Dose/Tinf)*(1/(0.088*76.9)))*(1-exp(-0.088*TIME5)) CAFTINF5 -((Dose/Tinf)*(1/(0.088*76.9)))*(1-exp(-0.088*Tinf))*(exp(- 0.088*(TIME5))) CONC5 -ifelse(TIME5=Tinf,CDURINF5,CAFTINF5) CDURINF6 -((Dose/Tinf)*(1/(0.088*76.9)))*(1-exp(-0.088*TIME6)) CAFTINF6 -((Dose/Tinf)*(1/(0.088*76.9)))*(1-exp(-0.088*Tinf))*(exp(- 0.088*(TIME6))) CONC6 -ifelse(TIME6=Tinf,CDURINF6,CAFTINF6) CDURINF7 -((Dose/Tinf)*(1/(0.088*76.9)))*(1-exp(-0.088*TIME7)) CAFTINF7 -((Dose/Tinf)*(1/(0.088*76.9)))*(1-exp(-0.088*Tinf))*(exp(- 0.088*(TIME7))) CONC7 -ifelse(TIME7=Tinf,CDURINF7,CAFTINF7) CDURINF8 -((Dose/Tinf)*(1/(0.088*76.9)))*(1-exp(-0.088*TIME7)) CAFTINF8 -((Dose/Tinf)*(1/(0.088*76.9)))*(1-exp(-0.088*Tinf))*(exp(- 0.088*(TIME8))) CONC8 -ifelse(TIME8=Tinf,CDURINF8,CAFTINF8) CONCF -matrix(c(CONC1+CONC2+CONC3+CONC4+CONC5+CONC6+CONC7+CONC8)) plot(CONCF,type=l) I am trying to plot CONCF in a specific way. Let me try to explain: I would like to plot it so that the x axis goes from 0 to 8*INTERVAL (here the INTERVAL represents time in hours), and then each segments of the matrix would be added to the sum of the previous ones at a specific moment in time. Assuming a much shorter example this is what it should be like where according to the same equations above each value of CONCi was calculated at 1 h intervals: INTERVAL =3 time points here where the values are calculated is 0:(3*INTERVAL) for CONC1, 0:(2*INTERVAL) for CONC2, and 0:(1*INTERVAL) for CONC3, all calculated at 1 hour intervals and lets assume calculated values are: CONC1 -c(0,1,2,3,4,5,6,7,8,9) CONC2 -c( 11,12,13,14,15,16,17) CONC3 -(21,22,23,24) . . . so I would like to get CONCF that would have the following values: CONCF -c(0,1,2,3+11,4+12,5+13,6+14,7+15,8+16,9+17, 6+14+21,7+15+22,8+16+23,9+17+23) please note that CONC2 was started to be added to CONC1 at 1*INTERVAL value estimated for CONC1, ie CONC1 has the value of 3 at hour 3, and I need to add the value of CONC2 at hour zero to it, and so on... hope I made it clear enough and thanks for the help, Andras [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] how to separate stuck row elements?
On 30-11-2012, at 13:55, Jaime Otero Villar wrote: Hi, I was wondering if it's possible to separate elements in multiple rows that actually should appear in different columns. I have a file where in certain lines there are elements not separated, and they certainly should appear in different columns (an example of the file is attached). The point is that I do not want to manually add a space in the txt file, however, I did not manage to do it automatically in R... Taking into account your description simulate file. Use readLines to read the file into a vector of lines. Use gsub() to replace each - with a single space. Finally use read.table to get a dataframe. # use the example consisting of 3 lines data.text - -100 -100-3456-3456-3456-100 -100 23 -3456-3456-189 34 56 78 -100 34 56 21 44 65 78 x.lines - readLines(textConnection(data.text)) x.lines # replace - with single space x.1 - gsub(-, ,x.lines) x.1 read.table(text=x.1, header=FALSE) Berend __ 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] bootstrapped cox regression (rms package)
Thanks Eric. It would be good to show your entire script next time as stated in the posting guidance. Regarding matching with SPSS please describe the bootstrapping algorithm used there. In rms I do the unconditional bootstrap, i.e., I sample with replacement from the rows of the raw data. And also make sure that SPSS ran a large number of bootstrap replications. Frank Eric Claus wrote Hi Frank, Below is the actual output from the anova(out) command. I had copied in the p-values and from the previous output from anova(out) and the confidence intervals from print(quantile(out$boot.Coef[,i], c(.025, .975))) to illustrate that the confidence intervals were similar to SPSS while the p-values were not. Actual output from anova.rms(out): Wald Statistics Response: Surv(months, recidivate) Factor Chi-Square d.f. P fac1 0.27 10.6055 fac2 0.20 10.6514 fac3 0.01 10.9338 fac4 0.05 10.8311 fac5 1.06 10.3036 fac6 0.33 10.5647 fac7 0.81 10.3670 fac8 0.30 10.5832 TOTAL 1.48 80.9930 Regarding your second question, it looks like SPSS is using the original estimate of Cox beta coefficients in the test (i.e. a new point estimate is not generated for the statistical test) Thanks again, Eric - Frank Harrell Department of Biostatistics, Vanderbilt University -- View this message in context: http://r.789695.n4.nabble.com/bootstrapped-cox-regression-rms-package-tp4651306p4651438.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] Changing the base of geom_bar in ggplot
I think your request is internally inconsistent: you want units of percent but you want to show the average(I assume you intend mean)? You can subtract the mean from each value and then divide by the mean, then subtract 1, then multiply by 100. You inherently lose visibility to what the mean was, but that is the nature of percentages. --- Jeff NewmillerThe . . Go Live... DCN:jdnew...@dcn.davis.ca.usBasics: ##.#. ##.#. Live Go... Live: OO#.. Dead: OO#.. Playing Research Engineer (Solar/BatteriesO.O#. #.O#. with /Software/Embedded Controllers) .OO#. .OO#. rocks...1k --- Sent from my phone. Please excuse my brevity. Werner pensterfuz...@yahoo.de wrote: Hi, this is likely very easy to do but I don't find the right terms to search for the solution. I want to show in a bar chart how much different categories deviate from the average percentage change. Thus, in the following example I would like to set the zero line or base or whatever it is called to the average of the values (=0.82) and the bars should show the deviation from that, i.e. have the positive length of the deviation of the value from the average. I can do that by subtracting the average from each value but I would like to have the zero line as the average to make it more clear. How do you change the 'base'-line? x - data.frame(val=c(1.5, 3, -1.4, -1, 2), lab=letters[1:5]) ggplot(data=x, aes(x=lab, y=val)) + geom_bar(stat=identity) Many thanks for your help! Best regards, Werner -- View this message in context: http://r.789695.n4.nabble.com/Changing-the-base-of-geom-bar-in-ggplot-tp4651407.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] subset data frame by variable with missing value
Hello, I have a variable in a data frame that contains NA values. I just want to subset so that I get the obs where that variable is missing. In SAS I would do: data missing; set test; if myvalue=' '; run; How can I perform this simple task in R? Thanks in advance for your help. -- View this message in context: http://r.789695.n4.nabble.com/subset-data-frame-by-variable-with-missing-value-tp4651439.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] subset data frame by variable with missing value
On Nov 30, 2012, at 9:27 AM, ramoss ramine.mossad...@finra.org wrote: Hello, I have a variable in a data frame that contains NA values. I just want to subset so that I get the obs where that variable is missing. In SAS I would do: data missing; set test; if myvalue=' '; run; How can I perform this simple task in R? Thanks in advance for your help. The easiest is probably: NewDF - subset(DF, is.na(myvalue)) See ?is.na Regards, Marc Schwartz __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] For loop
Hello user, I have large data containing subject id, time and response where subjects are measured repeatedly. However some time are duplicates. I only want data with unique time points per id. I mean if time is repeated, then take only one. Here is a sample data. id timeres 1 2 0.64 1 3 0.78 1 3 6.5 1 3 4.5 1 4 4 1 5 3.4 2 10 5.7 2 11 5.8 2 11 9.3 2 11 3.4 2 12 3.4 2 13 6.7 3 3 5.6 3 3 3.4 3 4 2.3 3 5 5.6 3 12 9.8 3 10 7 3 24 6 3 16 4 for 1st subject I want this, id timeres 1 2 0.64 1 3 0.78 1 4 4 1 5 3.4 Any suggestions are much appreciated! Thanks, Bikek __ 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 on stacking matrices up
Hi Andras Now it works but I suppose you shall simplify or explain it a bit more. CONC7-c(rep(0,3),CONC7) CONC8-c(rep(0,6),CONC8) CONC6+CONC7+CONC8 [,1] [1,] 0.00 [2,] 2.330056 [3,] 2.133774 [4,] 1.954027 [5,] 4.119477 [6,] 3.772456 [7,] 3.454667 [8,] 5.493705 [9,] 5.030920 [10,] 4.607120 gives you desired result easier. So for summing CONC objects just add to beginning of each object so many zeroes as there are missing values. You can check dimension by dim or nrow and use it to estimate how many zeroes you need to add. Besides, you probably do not need matrices, vectors can do it too. This CONC6 -ifelse(TIME6=Tinf,CDURINF6,CAFTINF6) in your case just change only the first value, but I presume in reality you can have Tinf such that it results in changing more values. If this is not you are after, please be more specific. Regards Petr From: Andras Farkas [mailto:motyoc...@yahoo.com] Sent: Friday, November 30, 2012 3:52 PM To: PIKAL Petr; r-help@r-project.org Subject: Re: [R] help on stacking matrices up Petr, sorry, here is a better example that should work: Dose-200 Tinf -0.5 INTERVAL -3 TIME8 -matrix(c((0*INTERVAL):(1*INTERVAL))) TIME7 -matrix(c((0*INTERVAL):(2*INTERVAL))) TIME6 -matrix(c((0*INTERVAL):(3*INTERVAL))) CDURINF6 -((Dose/Tinf)*(1/(0.088*76.9)))*(1-exp(-0.088*TIME6)) CAFTINF6 -((Dose/Tinf)*(1/(0.088*76.9)))*(1-exp(-0.088*Tinf))*(exp(-0.088*(TIME6))) CONC6 -ifelse(TIME6=Tinf,CDURINF6,CAFTINF6) CDURINF7 -((Dose/Tinf)*(1/(0.088*76.9)))*(1-exp(-0.088*TIME7)) CAFTINF7 -((Dose/Tinf)*(1/(0.088*76.9)))*(1-exp(-0.088*Tinf))*(exp(-0.088*(TIME7))) CONC7 -ifelse(TIME7=Tinf,CDURINF7,CAFTINF7) CDURINF8 -((Dose/Tinf)*(1/(0.088*76.9)))*(1-exp(-0.088*TIME8)) CAFTINF8 -((Dose/Tinf)*(1/(0.088*76.9)))*(1-exp(-0.088*Tinf))*(exp(-0.088*(TIME8))) CONC8 -ifelse(TIME8=Tinf,CDURINF8,CAFTINF8) CONCF -matrix(c(CONC6[1,],CONC6[2,],CONC6[3,],CONC6[4,]+CONC7[1,],CONC6[5,]+CONC7[2,],CONC6[6,]+CONC7[3,],CONC6[7,]+CONC7[4,]+CONC8[1,],CONC6[8,]+CONC7[5,]+CONC8[2,],CONC6[9,]+CONC7[6,]+CONC8[3,],CONC6[10,]+CONC7[7,]+CONC8[4,])) plot(TIME6,CONCF,type=b) the plot is what it should look like, keeping in mind that the INTERVAL will change, and that is the part i amtrying to figure out, ie: how can i make that dynamic. Is this better? thanks, Andras From: PIKAL Petr petr.pi...@precheza.czmailto:petr.pi...@precheza.cz To: Andras Farkas motyoc...@yahoo.commailto:motyoc...@yahoo.com; r-help@r-project.orgmailto:r-help@r-project.org r-help@r-project.orgmailto:r-help@r-project.org Sent: Friday, November 30, 2012 8:51 AM Subject: RE: [R] help on stacking matrices up Hi try to make your example easier and reproducible. I get CONCF -matrix(c(CONC1+CONC2+CONC3+CONC4+CONC5+CONC6+CONC7+CONC8)) Error in CONC1 + CONC2 : non-conformable arrays so I cannot understand what shall the final plot look like. Regards Petr -Original Message- From: r-help-boun...@r-project.orgmailto:r-help-boun...@r-project.org [mailto:r-help-bounces@r- project.orgmailto:r-help-bounces@r-%0b%3e%20project.org] On Behalf Of Andras Farkas Sent: Friday, November 30, 2012 12:51 PM To: r-help@r-project.orgmailto:r-help@r-project.org Subject: [R] help on stacking matrices up Dear All, #I have the following code Dose-1000 Tinf -0.5 INTERVAL -8 TIME8 -matrix(c((0*INTERVAL):(1*INTERVAL))) TIME7 -matrix(c((0*INTERVAL):(2*INTERVAL))) TIME6 -matrix(c((0*INTERVAL):(3*INTERVAL))) TIME5 -matrix(c((0*INTERVAL):(4*INTERVAL))) TIME4 -matrix(c((0*INTERVAL):(5*INTERVAL))) TIME3 -matrix(c((0*INTERVAL):(6*INTERVAL))) TIME2 -matrix(c((0*INTERVAL):(7*INTERVAL))) TIME1 -matrix(c((0*INTERVAL):(8*INTERVAL))) CDURINF1 -((Dose/Tinf)*(1/(0.088*76.9)))*(1-exp(-0.088*TIME1)) CAFTINF1 -((Dose/Tinf)*(1/(0.088*76.9)))*(1-exp(-0.088*Tinf))*(exp(- 0.088*(TIME1))) CONC1 -ifelse(TIME1=Tinf,CDURINF1,CAFTINF1) CDURINF2 -((Dose/Tinf)*(1/(0.088*76.9)))*(1-exp(-0.088*TIME2)) CAFTINF2 -((Dose/Tinf)*(1/(0.088*76.9)))*(1-exp(-0.088*Tinf))*(exp(- 0.088*(TIME2))) CONC2 -ifelse(TIME2=Tinf,CDURINF2,CAFTINF2) CDURINF3 -((Dose/Tinf)*(1/(0.088*76.9)))*(1-exp(-0.088*TIME3)) CAFTINF3 -((Dose/Tinf)*(1/(0.088*76.9)))*(1-exp(-0.088*Tinf))*(exp(- 0.088*(TIME3))) CONC3 -ifelse(TIME3=Tinf,CDURINF3,CAFTINF3) CDURINF4 -((Dose/Tinf)*(1/(0.088*76.9)))*(1-exp(-0.088*TIME4)) CAFTINF4 -((Dose/Tinf)*(1/(0.088*76.9)))*(1-exp(-0.088*Tinf))*(exp(- 0.088*(TIME4))) CONC4 -ifelse(TIME4=Tinf,CDURINF4,CAFTINF4) CDURINF5 -((Dose/Tinf)*(1/(0.088*76.9)))*(1-exp(-0.088*TIME5)) CAFTINF5 -((Dose/Tinf)*(1/(0.088*76.9)))*(1-exp(-0.088*Tinf))*(exp(- 0.088*(TIME5))) CONC5 -ifelse(TIME5=Tinf,CDURINF5,CAFTINF5) CDURINF6 -((Dose/Tinf)*(1/(0.088*76.9)))*(1-exp(-0.088*TIME6)) CAFTINF6 -((Dose/Tinf)*(1/(0.088*76.9)))*(1-exp(-0.088*Tinf))*(exp(- 0.088*(TIME6))) CONC6 -ifelse(TIME6=Tinf,CDURINF6,CAFTINF6) CDURINF7 -((Dose/Tinf)*(1/(0.088*76.9)))*(1-exp(-0.088*TIME7)) CAFTINF7 -((Dose/Tinf)*(1/(0.088*76.9)))*(1-exp(-0.088*Tinf))*(exp(- 0.088*(TIME7))) CONC7
Re: [R] subset data frame by variable with missing value
Hi see ?is.na x -sample(c(1:3, NA), 20, replace=T) x [1] 2 NA 2 3 3 3 3 2 3 NA 3 2 1 2 NA 3 3 3 2 2 y-rnorm(20) y[is.na(x)] [1] 0.1600417 1.3264063 -0.6175832 Regards Petr -Original Message- From: r-help-boun...@r-project.org [mailto:r-help-bounces@r- project.org] On Behalf Of ramoss Sent: Friday, November 30, 2012 4:27 PM To: r-help@r-project.org Subject: [R] subset data frame by variable with missing value Hello, I have a variable in a data frame that contains NA values. I just want to subset so that I get the obs where that variable is missing. In SAS I would do: data missing; set test; if myvalue=' '; run; How can I perform this simple task in R? Thanks in advance for your help. -- View this message in context: http://r.789695.n4.nabble.com/subset- data-frame-by-variable-with-missing-value-tp4651439.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] subset data frame by variable with missing value
Le vendredi 30 novembre 2012 à 07:27 -0800, ramoss a écrit : Hello, I have a variable in a data frame that contains NA values. I just want to subset so that I get the obs where that variable is missing. In SAS I would do: data missing; set test; if myvalue=' '; run; How can I perform this simple task in R? missing - test[is.na(test$myvalue),] or missing - subset(test, is.na(myvalue)) Regards Thanks in advance for your help. -- View this message in context: http://r.789695.n4.nabble.com/subset-data-frame-by-variable-with-missing-value-tp4651439.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] Generating sequences of dates
?seq.Date seq(as.Date(2000-01-01), len=5, by=1 mon) [1] 2000-01-01 2000-02-01 2000-03-01 2000-04-01 2000-05-01 -Original Message- From: r-help-boun...@r-project.org [mailto:r-help-bounces@r- project.org] On Behalf Of Vasilchenko Aleksander Sent: Friday, November 30, 2012 6:27 AM To: r-help@r-project.org Subject: [R] Generating sequences of dates Hello, I have the starting date in UNIX_TIME LONG number (number of milliseconds since 1970 jan 1st). Is there a way to generete the sequence of dates with monthly interval in POSIXct format like 2000-01-01 2000-02-01 2000-03-01 etc Thanks [[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] For loop
Hello Bikek, please use dput() next time to provide the data, its easier to use that. also: looking at the data provided, how would you want to decide which value of the non-unique times to retain? Just take the first one? They aren't all the same. On 30.11.2012, at 16:59, bibek sharma wrote: Hello user, I have large data containing subject id, time and response where subjects are measured repeatedly. However some time are duplicates. I only want data with unique time points per id. I mean if time is repeated, then take only one. Here is a sample data. idtimeres 1 2 0.64 1 3 0.78 1 3 6.5 1 3 4.5 1 4 4 1 5 3.4 2 10 5.7 2 11 5.8 2 11 9.3 2 11 3.4 2 12 3.4 2 13 6.7 3 3 5.6 3 3 3.4 3 4 2.3 3 5 5.6 3 12 9.8 3 10 7 3 24 6 3 16 4 for 1st subject I want this, idtimeres 1 2 0.64 1 3 0.78 1 4 4 1 5 3.4 Any suggestions are much appreciated! Thanks, Bikek __ 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 separate stuck row elements?
Hi, Try this: Lines--100 -100-3456-3456-3456-100 -100 23 -3456-3456-189 34 56 78 -100 34 56 21 44 65 78 res-unlist(strsplit(gsub(\\-, -,Lines),\n)) res1-do.call(rbind,lapply(lapply(split(res,seq_along(res)),function(x) unlist(strsplit(x, ))),function(x) as.numeric(x[x!=]))) res1 ## [,1] [,2] [,3] [,4] [,5] [,6] [,7] #1 -100 -100 -3456 -3456 -3456 -100 -100 #2 23 -3456 -3456 -189 34 56 78 #3 -100 34 56 21 44 65 78 res2-data.frame(res1) A.K. - Original Message - From: Jaime Otero Villar j.o.vil...@bio.uio.no To: r-help@r-project.org Cc: Sent: Friday, November 30, 2012 7:55 AM Subject: [R] how to separate stuck row elements? Hi, I was wondering if it's possible to separate elements in multiple rows that actually should appear in different columns. I have a file where in certain lines there are elements not separated, and they certainly should appear in different columns (an example of the file is attached). The point is that I do not want to manually add a space in the txt file, however, I did not manage to do it automatically in R... Thanks in advance for any insight. Jaime __ 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] For loop
Hi, Try this: dat1-read.table(text= id time res 1 2 0.64 1 3 0.78 1 3 6.5 1 3 4.5 1 4 4 1 5 3.4 2 10 5.7 2 11 5.8 2 11 9.3 2 11 3.4 2 12 3.4 2 13 6.7 3 3 5.6 3 3 3.4 3 4 2.3 3 5 5.6 3 12 9.8 3 10 7 3 24 6 3 16 4 ,sep=,header=TRUE,stringsAsFactors=TRUE) res-dat1[!duplicated(dat1[,1:2]),] res # id time res #1 1 2 0.64 #2 1 3 0.78 #5 1 4 4.00 #6 1 5 3.40 #7 2 10 5.70 #8 2 11 5.80 #11 2 12 3.40 #12 2 13 6.70 #13 3 3 5.60 #15 3 4 2.30 #16 3 5 5.60 #17 3 12 9.80 #18 3 10 7.00 #19 3 24 6.00 #20 3 16 4.00 row.names(res)-1:nrow(res) A.K. - Original Message - From: bibek sharma mbhpat...@gmail.com To: R-help@r-project.org Cc: Sent: Friday, November 30, 2012 10:59 AM Subject: [R] For loop Hello user, I have large data containing subject id, time and response where subjects are measured repeatedly. However some time are duplicates. I only want data with unique time points per id. I mean if time is repeated, then take only one. Here is a sample data. id time res 1 2 0.64 1 3 0.78 1 3 6.5 1 3 4.5 1 4 4 1 5 3.4 2 10 5.7 2 11 5.8 2 11 9.3 2 11 3.4 2 12 3.4 2 13 6.7 3 3 5.6 3 3 3.4 3 4 2.3 3 5 5.6 3 12 9.8 3 10 7 3 24 6 3 16 4 for 1st subject I want this, id time res 1 2 0.64 1 3 0.78 1 4 4 1 5 3.4 Any suggestions are much appreciated! Thanks, Bikek __ 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] For loop
Hi -Original Message- From: r-help-boun...@r-project.org [mailto:r-help-bounces@r- project.org] On Behalf Of bibek sharma Sent: Friday, November 30, 2012 5:00 PM To: R-help@r-project.org Subject: [R] For loop Hello user, I have large data containing subject id, time and response where subjects are measured repeatedly. However some time are duplicates. I only want data with unique time points per id. I mean if time is repeated, then take only one. Which one? The first one? aggregate(test$res, list(time=test$time, id=test$id), function(x) x[1]) Regards Petr Here is a sample data. idtimeres 1 2 0.64 1 3 0.78 1 3 6.5 1 3 4.5 1 4 4 1 5 3.4 2 10 5.7 2 11 5.8 2 11 9.3 2 11 3.4 2 12 3.4 2 13 6.7 3 3 5.6 3 3 3.4 3 4 2.3 3 5 5.6 3 12 9.8 3 10 7 3 24 6 3 16 4 for 1st subject I want this, idtimeres 1 2 0.64 1 3 0.78 1 4 4 1 5 3.4 Any suggestions are much appreciated! Thanks, Bikek __ 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] subset data frame by variable with missing value
I found the answer; Its mymissing - subset(mydata,is.na(myvar)) -- View this message in context: http://r.789695.n4.nabble.com/subset-data-frame-by-variable-with-missing-value-tp4651439p4651440.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] Standard errors for predictions of zero-inflated models
Dear all, I am using the zeroinfl() function from the pscl package to develop a zero-inflated Poisson GLM. I would like to calculate the standard errors of predicted values. I've tried code posted in a previous discussion on this topic (https://stat.ethz.ch/pipermail/r-help/2008-December/182806.html), and I don't understand the results. Before I apply this code, I get the predicted value for the data 'newdat' using the following: - predict(bestfit,type=response,newdata=newdat) 109 198.5146 The predicted value is approximately 199. When I apply the previously mentioned code (predict.zeroinfl), I can now calculate the standard error: - predict(bestfit,type=response,se=TRUE,newdata=newdat) MC iterate 1 of 1000 MC iterate 2 of 1000 ... MC iterate 1000 of 1000 [[1]] 109 0.00016793 [[2]] [[2]]$lower [1] 9.151924e-05 [[2]]$upper [1] 0.0002945504 [[2]]$se [1] 5.296472e-05 However, the predicted value is now 1.7x10^-4. The standard error value makes sense for this predicted value, but I'm not sure why the predicted value has changed. I would appreciate any assistance. Cheers, Laura [[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] googleVis plot and knitr/sweave
On 11/29/2012 2:53 PM, Filoche wrote: Dear R users. I'm currently making a report with knitr (RStudio) where I would like to plot a googleVis map. However, the map generated is an HTML file which I don't know how to integrate it in my report. So my question is how to include a map generated with googleVis in a PDF created with knitr/sweave. Since others have told you that you can't do this directly, one recommended kludge is to use a static graph in the report, e.g., a screen capture from the HTML page, perhaps with a link to the 'live' version if you are putting this online. You can still show the googleVis code if you want using ... eval=FALSE -- Michael Friendly Email: friendly AT yorku DOT ca Professor, Psychology Dept. Chair, Quantitative Methods York University Voice: 416 736-2100 x66249 Fax: 416 736-5814 4700 Keele StreetWeb: http://www.datavis.ca Toronto, ONT M3J 1P3 CANADA __ 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] For loop
You could use the duplicated function maybe: mtest id time value [1,] 13 1 [2,] 13 2 [3,] 12 3 [4,] 11 4 [5,] 21 5 [6,] 23 6 [7,] 23 7 [8,] 23 8 duplicated(mtest[,1:2]) [1] FALSE TRUE FALSE FALSE FALSE FALSE TRUE TRUE mtest[!duplicated(mtest[,1:2]),] id time value [1,] 13 1 [2,] 12 3 [3,] 11 4 [4,] 21 5 [5,] 23 6 On 30.11.2012, at 17:15, Jessica Streicher wrote: Hello Bikek, please use dput() next time to provide the data, its easier to use that. also: looking at the data provided, how would you want to decide which value of the non-unique times to retain? Just take the first one? They aren't all the same. On 30.11.2012, at 16:59, bibek sharma wrote: Hello user, I have large data containing subject id, time and response where subjects are measured repeatedly. However some time are duplicates. I only want data with unique time points per id. I mean if time is repeated, then take only one. Here is a sample data. id timeres 12 0.64 13 0.78 13 6.5 13 4.5 14 4 15 3.4 210 5.7 211 5.8 211 9.3 211 3.4 212 3.4 213 6.7 33 5.6 33 3.4 34 2.3 35 5.6 312 9.8 310 7 324 6 316 4 for 1st subject I want this, id timeres 12 0.64 13 0.78 14 4 15 3.4 Any suggestions are much appreciated! Thanks, Bikek __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] how to separate stuck row elements?
Hi, You could also try this: Lines--100 -100-3456-3456-3456-100 -100 23 -3456-3456-189 34 56 78 -100 34 56 21 44 65 78 Lines1-readLines(textConnection(Lines)) res1-unlist(strsplit(gsub([-], -,Lines2), )) matrix(as.numeric(res1[res1!=]),nrow=length(Lines1),byrow=TRUE) [,1] [,2] [,3] [,4] [,5] [,6] [,7] #[1,] -100 -100 -3456 -3456 -3456 -100 -100 #[2,] 23 -3456 -3456 -189 34 56 78 #[3,] -100 34 56 21 44 65 78 A.K. - Original Message - From: Jaime Otero Villar j.o.vil...@bio.uio.no To: r-help@r-project.org Cc: Sent: Friday, November 30, 2012 7:55 AM Subject: [R] how to separate stuck row elements? Hi, I was wondering if it's possible to separate elements in multiple rows that actually should appear in different columns. I have a file where in certain lines there are elements not separated, and they certainly should appear in different columns (an example of the file is attached). The point is that I do not want to manually add a space in the txt file, however, I did not manage to do it automatically in R... Thanks in advance for any insight. Jaime __ 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] help on stacking matrices up
Hi ifelse shall work just fine. x-1:10 a-rnorm(10) b-rnorm(10)+100 tinf-5 ifelse(xtinf, a,b) [1] -0.32157732 -0.11065857 0.07127818 0.35928359 100.06200507 [6] 98.77752570 100.92743376 99.73918386 99.79837539 101.32640626 It is only your value of Tinf which is 0.5 compared to TIME which is 0:10 and therefore your logical construction selects only values lower than Tinf see TIME6=Tinf result If you encounter strange result of complicated code skin it to bones and separate to pieces line by line and even to line parts if possible. This will give you insight how your code really works. Just like disassembling old clock (those which have cogwheels) to see and understand how it works. Regards Petr From: Andras Farkas [mailto:motyoc...@yahoo.com] Sent: Friday, November 30, 2012 5:37 PM To: PIKAL Petr; r-help@r-project.org Subject: Re: [R] help on stacking matrices up Petr, thanks for the solution, putting the zeros in there makes it work just how I want it. Also, thanks for picking up on the error in my code below... I did not notice that it only changes the 1st value, and yes, your assumption is right: there is a chance where more values should be changed based on Tinf... I have tryed to work with if ... else..., but that does not give me correct answer either, to my suprise... Would you be willing to provide some insigths on how I could go about solving that? thanks, Andras From: PIKAL Petr petr.pi...@precheza.czmailto:petr.pi...@precheza.cz To: Andras Farkas motyoc...@yahoo.commailto:motyoc...@yahoo.com; r-help@r-project.orgmailto:r-help@r-project.org r-help@r-project.orgmailto:r-help@r-project.org Sent: Friday, November 30, 2012 10:58 AM Subject: RE: [R] help on stacking matrices up Hi Andras Now it works but I suppose you shall simplify or explain it a bit more. CONC7-c(rep(0,3),CONC7) CONC8-c(rep(0,6),CONC8) CONC6+CONC7+CONC8 [,1] [1,] 0.00 [2,] 2.330056 [3,] 2.133774 [4,] 1.954027 [5,] 4.119477 [6,] 3.772456 [7,] 3.454667 [8,] 5.493705 [9,] 5.030920 [10,] 4.607120 gives you desired result easier. So for summing CONC objects just add to beginning of each object so many zeroes as there are missing values. You can check dimension by dim or nrow and use it to estimate how many zeroes you need to add. Besides, you probably do not need matrices, vectors can do it too. This CONC6 -ifelse(TIME6=Tinf,CDURINF6,CAFTINF6) in your case just change only the first value, but I presume in reality you can have Tinf such that it results in changing more values. If this is not you are after, please be more specific. Regards Petr From: Andras Farkas [mailto:motyoc...@yahoo.com] Sent: Friday, November 30, 2012 3:52 PM To: PIKAL Petr; r-help@r-project.orgmailto:r-help@r-project.org Subject: Re: [R] help on stacking matrices up Petr, sorry, here is a better example that should work: Dose-200 Tinf -0.5 INTERVAL -3 TIME8 -matrix(c((0*INTERVAL):(1*INTERVAL))) TIME7 -matrix(c((0*INTERVAL):(2*INTERVAL))) TIME6 -matrix(c((0*INTERVAL):(3*INTERVAL))) CDURINF6 -((Dose/Tinf)*(1/(0.088*76.9)))*(1-exp(-0.088*TIME6)) CAFTINF6 -((Dose/Tinf)*(1/(0.088*76.9)))*(1-exp(-0.088*Tinf))*(exp(-0.088*(TIME6))) CONC6 -ifelse(TIME6=Tinf,CDURINF6,CAFTINF6) CDURINF7 -((Dose/Tinf)*(1/(0.088*76.9)))*(1-exp(-0.088*TIME7)) CAFTINF7 -((Dose/Tinf)*(1/(0.088*76.9)))*(1-exp(-0.088*Tinf))*(exp(-0.088*(TIME7))) CONC7 -ifelse(TIME7=Tinf,CDURINF7,CAFTINF7) CDURINF8 -((Dose/Tinf)*(1/(0.088*76.9)))*(1-exp(-0.088*TIME8)) CAFTINF8 -((Dose/Tinf)*(1/(0.088*76.9)))*(1-exp(-0.088*Tinf))*(exp(-0.088*(TIME8))) CONC8 -ifelse(TIME8=Tinf,CDURINF8,CAFTINF8) CONCF -matrix(c(CONC6[1,],CONC6[2,],CONC6[3,],CONC6[4,]+CONC7[1,],CONC6[5,]+CONC7[2,],CONC6[6,]+CONC7[3,],CONC6[7,]+CONC7[4,]+CONC8[1,],CONC6[8,]+CONC7[5,]+CONC8[2,],CONC6[9,]+CONC7[6,]+CONC8[3,],CONC6[10,]+CONC7[7,]+CONC8[4,])) plot(TIME6,CONCF,type=b) the plot is what it should look like, keeping in mind that the INTERVAL will change, and that is the part i amtrying to figure out, ie: how can i make that dynamic. Is this better? thanks, Andras From: PIKAL Petr petr.pi...@precheza.czmailto:petr.pi...@precheza.cz To: Andras Farkas motyoc...@yahoo.commailto:motyoc...@yahoo.com; r-help@r-project.orgmailto:r-help@r-project.org r-help@r-project.orgmailto:r-help@r-project.org Sent: Friday, November 30, 2012 8:51 AM Subject: RE: [R] help on stacking matrices up Hi try to make your example easier and reproducible. I get CONCF -matrix(c(CONC1+CONC2+CONC3+CONC4+CONC5+CONC6+CONC7+CONC8)) Error in CONC1 + CONC2 : non-conformable arrays so I cannot understand what shall the final plot look like. Regards Petr -Original Message- From: r-help-boun...@r-project.orgmailto:r-help-boun...@r-project.org [mailto:r-help-bounces@r- project.orgmailto:r-help-bounces@r-%0b%3e%20project.org] On Behalf Of Andras Farkas Sent: Friday, November 30, 2012 12:51 PM To: r-help@r-project.orgmailto:r-help@r-project.org Subject: [R] help on stacking matrices up
Re: [R] missed values
Hi,
May be this helps:
dat1-read.table(text=
---data---
,header=TRUE,stringsAsFactors=FALSE)
library(zoo)
dat1$date-as.yearmon(dat1$date,format=%Y-%m)
lm1-lm(value~date,dat1)
dat2-data.frame(date=dat1[,1])
dat1$fit-predict(lm1,newdata=dat2)
dat1-within(dat1,{newvalue-ifelse(is.na(value)==T,fit,value)})
dat1new-dat1[,c(1:2,4)]
dat1new[80:86,]
# date value newvalue
#80 Oct 2006 0.1577647 0.1577647
#81 Nov 2006 NaN 0.2782320
#82 Dec 2006 NaN 0.2773986
#83 Jan 2007 NaN 0.2765651
#84 Feb 2007 NaN 0.2757317
#85 Mar 2007 0.2956429 0.2956429
#86 Apr 2007 0.3767500 0.3767500
A.K.
- Original Message -
From: Vasilchenko Aleksander vasilchenko@gmail.com
To: arun smartpink...@yahoo.com
Cc:
Sent: Friday, November 30, 2012 8:22 AM
Subject: Re: [R] missed values
date value
1 2000-03 0.3425263
2 2000-04 0.461
3 2000-05 0.459
4 2000-06 0.3952500
5 2000-07 0.3497931
6 2000-08 0.458
7 2000-09 0.3281667
8 2000-10 0.3025263
9 2000-11 0.1706000
10 2000-12 0.1474118
11 2001-01 0.124
12 2001-02 0.2417333
13 2001-03 0.2615882
14 2001-04 0.3127778
15 2001-05 0.4201250
16 2001-06 0.3875000
17 2001-07 0.4382400
18 2001-08 0.4810345
19 2001-09 0.2790476
20 2001-10 0.2814483
21 2001-11 0.1588125
22 2001-12 0.1346429
23 2002-01 0.2103750
24 2002-02 0.2967000
25 2002-03 0.4348334
26 2002-04 0.475
27 2002-05 0.3669000
28 2002-06 0.4609600
29 2002-07 0.4170909
30 2002-08 0.4220435
31 2002-09 0.3641053
32 2002-10 0.1797308
33 2002-11 0.2112500
34 2002-12 0.1486250
35 2003-01 0.1800667
36 2003-02 0.2997857
37 2003-03 0.3697500
38 2003-04 0.495
39 2003-05 0.5344348
40 2003-06 0.5187334
41 2003-07 0.413
42 2003-08 0.5350715
43 2003-09 0.2706316
44 2003-10 0.2185333
45 2003-11 0.2260625
46 2003-12 0.1523750
47 2004-01 0.183
48 2004-02 0.2055455
49 2004-03 0.3884118
50 2004-04 0.367
51 2004-05 0.3104211
52 2004-06 0.3226818
53 2004-07 0.3570741
54 2004-08 0.3387097
55 2004-09 0.3168750
56 2004-10 0.2618000
57 2004-11 0.1487500
58 2004-12 0.138
59 2005-01 0.1768125
60 2005-02 0.2206000
61 2005-03 0.3592353
62 2005-04 0.3589500
63 2005-05 0.3481250
64 2005-06 0.3983077
65 2005-07 0.3612857
66 2005-08 0.3426539
67 2005-09 0.3435000
68 2005-10 0.2008636
69 2005-11 0.1967333
70 2005-12 0.111
71 2006-01 0.2308125
72 2006-02 0.1883125
73 2006-03 0.3261000
74 2006-04 0.2914118
75 2006-05 0.3331852
76 2006-06 0.4564348
77 2006-07 0.3920968
78 2006-08 0.3059259
79 2006-09 0.2567917
80 2006-10 0.1577647
81 2006-11 NaN
82 2006-12 NaN
83 2007-01 NaN
84 2007-02 NaN
85 2007-03 0.2956429
86 2007-04 0.3767500
87 2007-05 0.3727391
88 2007-06 0.4370800
89 2007-07 0.3504194
90 2007-08 0.3266400
91 2007-09 0.2328400
92 2007-10 0.209
93 2007-11 0.1631667
94 2007-12 0.1302143
95 2008-01 0.1131539
96 2008-02 0.1982727
97 2008-03 0.2568000
98 2008-04 0.2892000
99 2008-05 0.3523158
100 2008-06 0.358
101 2008-07 0.2898966
102 2008-08 0.3101667
103 2008-09 0.3730476
104 2008-10 0.2717037
105 2008-11 0.1344286
106 2008-12 0.1375000
107 2009-01 0.1781000
108 2009-02 0.2146667
109 2009-03 0.2808235
110 2009-04 0.4326250
111 2009-05 0.3420741
112 2009-06 0.2675238
113 2009-07 0.2478667
114 2009-08 0.3147000
115 2009-09 0.3437826
116 2009-10 0.2057391
117 2009-11 0.1824737
118 2009-12 0.1520714
119 2010-01 0.1485455
120 2010-02 0.176
121 2010-03 0.3072000
122 2010-04 0.3294445
123 2010-05 0.3278125
124 2010-06 0.2865000
125 2010-07 0.3170333
126 2010-08 0.2052143
127 2010-09 0.1837368
128 2010-10 0.1652778
129 2010-11 0.1292500
130 2010-12 0.1366250
131 2011-01 0.1909231
132 2011-02 0.1841177
133 2011-03 0.2897222
134 2011-04 0.3084211
135 2011-05 0.3349600
136 2011-06 0.3157917
137 2011-07 0.2755652
138 2011-08 0.2019355
139 2011-09 0.3325556
140 2011-10 0.1724348
141 2011-11 0.1935200
142 2011-12 0.1388000
143 2012-01 0.1484500
144 2012-02 0.204
I want to obtain these values by linear interpolation between 80
2006-10 0.1577647 and 85 2007-03 0.2956429
81 2006-11 NaN
82 2006-12 NaN
83 2007-01 NaN
84 2007-02 NaN
This time series has only one such block, but there can be several
such blocks in general
thanks
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[R] Reading .gsheet within R
Hello R-experts, I would like to know if there is a solution to read files with extension .gsheet directly into R - see http://www.fileinfo.com/extension/gsheet for more info on this file format. Thank you, Luca Mr. Luca Meyer www.lucameyer.com R 2.15.1 Mac OS X 10.8.2 [[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] repeating matrices in a list
Suppose I have the following square, non-negative matrices g=matrix(c(0,2,4,0.25,0,0,0,0.6,0),3,3,byrow=T); I want to create a list where this matrix is repeated multiple times. if I do this brute force (manually), using env - list(g,g,g) works fine. Yields [[1]] [,1] [,2] [,3] [1,] 0.00 2.04 [2,] 0.25 0.00 [3,] 0.00 0.60 [[2]] [,1] [,2] [,3] [1,] 0.00 2.04 [2,] 0.25 0.00 [3,] 0.00 0.60 [[3]] [,1] [,2] [,3] [1,] 0.00 2.04 [2,] 0.25 0.00 [3,] 0.00 0.60 But - for a variety of purposes, I need to 'automate' building said list. I tried using rep env - list(rep(g,each=3)) but this yields [1] 0.00 0.00 0.00 0.25 0.25 0.25 0.00 0.00 0.00 2.00 2.00 2.00 0.00 0.00 0.00 [16] 0.60 0.60 0.60 4.00 4.00 4.00 0.00 0.00 0.00 0.00 0.00 0.00 Any suggestions/pointers to the obvious? Thanks in advance... [[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] repeating matrices in a list
You are so close: rep(list(g), 3) [[1]] [,1] [,2] [,3] [1,] 0.00 2.04 [2,] 0.25 0.00 [3,] 0.00 0.60 [[2]] [,1] [,2] [,3] [1,] 0.00 2.04 [2,] 0.25 0.00 [3,] 0.00 0.60 [[3]] [,1] [,2] [,3] [1,] 0.00 2.04 [2,] 0.25 0.00 [3,] 0.00 0.60 Sarah On Fri, Nov 30, 2012 at 12:50 PM, Anser Chen anser.c...@gmail.com wrote: Suppose I have the following square, non-negative matrices g=matrix(c(0,2,4,0.25,0,0,0,0.6,0),3,3,byrow=T); I want to create a list where this matrix is repeated multiple times. if I do this brute force (manually), using env - list(g,g,g) works fine. Yields [[1]] [,1] [,2] [,3] [1,] 0.00 2.04 [2,] 0.25 0.00 [3,] 0.00 0.60 [[2]] [,1] [,2] [,3] [1,] 0.00 2.04 [2,] 0.25 0.00 [3,] 0.00 0.60 [[3]] [,1] [,2] [,3] [1,] 0.00 2.04 [2,] 0.25 0.00 [3,] 0.00 0.60 But - for a variety of purposes, I need to 'automate' building said list. I tried using rep env - list(rep(g,each=3)) but this yields [1] 0.00 0.00 0.00 0.25 0.25 0.25 0.00 0.00 0.00 2.00 2.00 2.00 0.00 0.00 0.00 [16] 0.60 0.60 0.60 4.00 4.00 4.00 0.00 0.00 0.00 0.00 0.00 0.00 Any suggestions/pointers to the obvious? Thanks in advance... -- 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] repeating matrices in a list
Hi, Try this: lapply(1:3,function(x) g) A.K. - Original Message - From: Anser Chen anser.c...@gmail.com To: r-help@r-project.org Cc: Sent: Friday, November 30, 2012 12:50 PM Subject: [R] repeating matrices in a list Suppose I have the following square, non-negative matrices g=matrix(c(0,2,4,0.25,0,0,0,0.6,0),3,3,byrow=T); I want to create a list where this matrix is repeated multiple times. if I do this brute force (manually), using env - list(g,g,g) works fine. Yields [[1]] [,1] [,2] [,3] [1,] 0.00 2.0 4 [2,] 0.25 0.0 0 [3,] 0.00 0.6 0 [[2]] [,1] [,2] [,3] [1,] 0.00 2.0 4 [2,] 0.25 0.0 0 [3,] 0.00 0.6 0 [[3]] [,1] [,2] [,3] [1,] 0.00 2.0 4 [2,] 0.25 0.0 0 [3,] 0.00 0.6 0 But - for a variety of purposes, I need to 'automate' building said list. I tried using rep env - list(rep(g,each=3)) but this yields [1] 0.00 0.00 0.00 0.25 0.25 0.25 0.00 0.00 0.00 2.00 2.00 2.00 0.00 0.00 0.00 [16] 0.60 0.60 0.60 4.00 4.00 4.00 0.00 0.00 0.00 0.00 0.00 0.00 Any suggestions/pointers to the obvious? Thanks in advance... [[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] what is the cost in cv.glm?
Hi, I have a question regarding the cv.glm function in the package boot. What is exactly the cost? Is it the threshold value for an estimated value to be classified as either 0 or 1? I have troubles understanding the explanation in R. Lets say I want all estimations 0.65 to be classified as 1s and 0.35 as 0s, how do I write that? And if the cost is something else, how do I set the threshold value? Thanks in advance! Anna -- View this message in context: http://r.789695.n4.nabble.com/what-is-the-cost-in-cv-glm-tp4651435.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 on stacking matrices up
Petr, sorry, here is a better example that should work: Dose-200 Tinf -0.5 INTERVAL -3 TIME8 -matrix(c((0*INTERVAL):(1*INTERVAL))) TIME7 -matrix(c((0*INTERVAL):(2*INTERVAL))) TIME6 -matrix(c((0*INTERVAL):(3*INTERVAL))) CDURINF6 -((Dose/Tinf)*(1/(0.088*76.9)))*(1-exp(-0.088*TIME6)) CAFTINF6 -((Dose/Tinf)*(1/(0.088*76.9)))*(1-exp(-0.088*Tinf))*(exp(-0.088*(TIME6))) CONC6 -ifelse(TIME6=Tinf,CDURINF6,CAFTINF6) CDURINF7 -((Dose/Tinf)*(1/(0.088*76.9)))*(1-exp(-0.088*TIME7)) CAFTINF7 -((Dose/Tinf)*(1/(0.088*76.9)))*(1-exp(-0.088*Tinf))*(exp(-0.088*(TIME7))) CONC7 -ifelse(TIME7=Tinf,CDURINF7,CAFTINF7) CDURINF8 -((Dose/Tinf)*(1/(0.088*76.9)))*(1-exp(-0.088*TIME8)) CAFTINF8 -((Dose/Tinf)*(1/(0.088*76.9)))*(1-exp(-0.088*Tinf))*(exp(-0.088*(TIME8))) CONC8 -ifelse(TIME8=Tinf,CDURINF8,CAFTINF8) CONCF -matrix(c(CONC6[1,],CONC6[2,],CONC6[3,],CONC6[4,]+CONC7[1,],CONC6[5,]+CONC7[2,],CONC6[6,]+CONC7[3,],CONC6[7,]+CONC7[4,]+CONC8[1,],CONC6[8,]+CONC7[5,]+CONC8[2,],CONC6[9,]+CONC7[6,]+CONC8[3,],CONC6[10,]+CONC7[7,]+CONC8[4,])) plot(TIME6,CONCF,type=b) the plot is what it should look like, keeping in mind that the INTERVAL will change, and that is the part i amtrying to figure out, ie: how can i make that dynamic. Is this better? thanks, Andras From: PIKAL Petr petr.pi...@precheza.cz roject.org Sent: Friday, November 30, 2012 8:51 AM Subject: RE: [R] help on stacking matrices up Hi try to make your example easier and reproducible. I get CONCF -matrix(c(CONC1+CONC2+CONC3+CONC4+CONC5+CONC6+CONC7+CONC8)) Error in CONC1 + CONC2 : non-conformable arrays so I cannot understand what shall the final plot look like. Regards Petr -Original Message- From: r-help-boun...@r-project.org [mailto:r-help-bounces@r- project.org] On Behalf Of Andras Farkas Sent: Friday, November 30, 2012 12:51 PM To: r-help@r-project.org Subject: [R] help on stacking matrices up Dear All, #I have the following code Dose-1000 Tinf -0.5 INTERVAL -8 TIME8 -matrix(c((0*INTERVAL):(1*INTERVAL))) TIME7 -matrix(c((0*INTERVAL):(2*INTERVAL))) TIME6 -matrix(c((0*INTERVAL):(3*INTERVAL))) TIME5 -matrix(c((0*INTERVAL):(4*INTERVAL))) TIME4 -matrix(c((0*INTERVAL):(5*INTERVAL))) TIME3 -matrix(c((0*INTERVAL):(6*INTERVAL))) TIME2 -matrix(c((0*INTERVAL):(7*INTERVAL))) TIME1 -matrix(c((0*INTERVAL):(8*INTERVAL))) CDURINF1 -((Dose/Tinf)*(1/(0.088*76.9)))*(1-exp(-0.088*TIME1)) CAFTINF1 -((Dose/Tinf)*(1/(0.088*76.9)))*(1-exp(-0.088*Tinf))*(exp(- 0.088*(TIME1))) CONC1 -ifelse(TIME1=Tinf,CDURINF1,CAFTINF1) CDURINF2 -((Dose/Tinf)*(1/(0.088*76.9)))*(1-exp(-0.088*TIME2)) CAFTINF2 -((Dose/Tinf)*(1/(0.088*76.9)))*(1-exp(-0.088*Tinf))*(exp(- 0.088*(TIME2))) CONC2 -ifelse(TIME2=Tinf,CDURINF2,CAFTINF2) CDURINF3 -((Dose/Tinf)*(1/(0.088*76.9)))*(1-exp(-0.088*TIME3)) CAFTINF3 -((Dose/Tinf)*(1/(0.088*76.9)))*(1-exp(-0.088*Tinf))*(exp(- 0.088*(TIME3))) CONC3 -ifelse(TIME3=Tinf,CDURINF3,CAFTINF3) CDURINF4 -((Dose/Tinf)*(1/(0.088*76.9)))*(1-exp(-0.088*TIME4)) CAFTINF4 -((Dose/Tinf)*(1/(0.088*76.9)))*(1-exp(-0.088*Tinf))*(exp(- 0.088*(TIME4))) CONC4 -ifelse(TIME4=Tinf,CDURINF4,CAFTINF4) CDURINF5 -((Dose/Tinf)*(1/(0.088*76.9)))*(1-exp(-0.088*TIME5)) CAFTINF5 -((Dose/Tinf)*(1/(0.088*76.9)))*(1-exp(-0.088*Tinf))*(exp(- 0.088*(TIME5))) CONC5 -ifelse(TIME5=Tinf,CDURINF5,CAFTINF5) CDURINF6 -((Dose/Tinf)*(1/(0.088*76.9)))*(1-exp(-0.088*TIME6)) CAFTINF6 -((Dose/Tinf)*(1/(0.088*76.9)))*(1-exp(-0.088*Tinf))*(exp(- 0.088*(TIME6))) CONC6 -ifelse(TIME6=Tinf,CDURINF6,CAFTINF6) CDURINF7 -((Dose/Tinf)*(1/(0.088*76.9)))*(1-exp(-0.088*TIME7)) CAFTINF7 -((Dose/Tinf)*(1/(0.088*76.9)))*(1-exp(-0.088*Tinf))*(exp(- 0.088*(TIME7))) CONC7 -ifelse(TIME7=Tinf,CDURINF7,CAFTINF7) CDURINF8 -((Dose/Tinf)*(1/(0.088*76.9)))*(1-exp(-0.088*TIME7)) CAFTINF8 -((Dose/Tinf)*(1/(0.088*76.9)))*(1-exp(-0.088*Tinf))*(exp(- 0.088*(TIME8))) CONC8 -ifelse(TIME8=Tinf,CDURINF8,CAFTINF8) CONCF -matrix(c(CONC1+CONC2+CONC3+CONC4+CONC5+CONC6+CONC7+CONC8)) plot(CONCF,type=l) I am trying to plot CONCF in a specific way. Let me try to explain: I would like to plot it so that the x axis goes from 0 to 8*INTERVAL (here the INTERVAL represents time in hours), and then each segments of the matrix would be added to the sum of the previous ones at a specific moment in time. Assuming a much shorter example this is what it should be like where according to the same equations above each value of CONCi was calculated at 1 h intervals: INTERVAL =3 time points here where the values are calculated is 0:(3*INTERVAL) for CONC1, 0:(2*INTERVAL) for CONC2, and 0:(1*INTERVAL) for CONC3, all calculated at 1 hour intervals and lets assume calculated values are: CONC1 -c(0,1,2,3,4,5,6,7,8,9) CONC2 -c( 11,12,13,14,15,16,17) CONC3 -(21,22,23,24) . . . so I would like to get CONCF that would have the following values: CONCF -c(0,1,2,3+11,4+12,5+13,6+14,7+15,8+16,9+17, 6+14+21,7+15+22,8+16+23,9+17+23) please note that CONC2 was
[R] can't integrate in loop
Hi guys!
I have to compute something and i don't know what i'm doing wrong. my code
is a bit complex, but imagine that is something like this:
a = c(1 2 3 4)
ia = length(a)
x = seq(1,100,length=0.1)
ib = length(x)
for(j in 1:ia) {
H = function(x) {sen(x) + a[j]}
for(i in 1:ib) {
int = function(x) { integrate(H, lower = 0, upper = x[i])}
int1[i] = int(1)
}
end
int1 = unlist(int1)
ss[j] = sum(int1)
}
end
if i try this code without the for loop it's ok, but when i put the cycle
on, i get all sort of errors...
Thank you for your help :)
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[R] loop function and integrate?
Hi guys!
I have to compute something and i don't know what i'm doing wrong. my code
is a bit complex, but imagine that is something like this:
a = c(1,2,3,4)
ia = length(a)
x = seq(1,100,length=0.1)
ib = length(x)
int1 = numeric(ib)
b = numeric(ib)
for(j in 1:ia) {
H = function(x) {sin(x + a[j])}
for(i in 1:ib) {
int = integrate(H, lower = 0, upper = x[i])
int1[i] = int[1]
b[i] = 1 + a[i]
}
end
int1 = unlist(int1)
int2 = int1*b
ss[j] = sum(int2)
}
end
if i try this code without the for loop it's ok, but when i put the for on,
i get all sort of errors...
Thank you for your help :)
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[R] Help regarding calendar heatmap
Hi, I would like to create calendarheat map with Hours of day on Y axis and Date on X axis. Is there any function to do so? Can someone help me with reference. Thanks Sharan [[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] subset data frame by variable with missing value
Hi, YOu could also use: set.seed(5) dat1-data.frame(col1=sample(c(1:4,NA),10,replace=TRUE),col2=runif(10,0,1)) dat1[!complete.cases(dat1),] # col1 col2 #3 NA 0.3184040 #8 NA 0.8878698 #9 NA 0.5549226 A.K. - Original Message - From: ramoss ramine.mossad...@finra.org To: r-help@r-project.org Cc: Sent: Friday, November 30, 2012 10:27 AM Subject: [R] subset data frame by variable with missing value Hello, I have a variable in a data frame that contains NA values. I just want to subset so that I get the obs where that variable is missing. In SAS I would do: data missing; set test; if myvalue=' '; run; How can I perform this simple task in R? Thanks in advance for your help. -- View this message in context: http://r.789695.n4.nabble.com/subset-data-frame-by-variable-with-missing-value-tp4651439.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] Nightingale’s Rose chart-any suggestion?
Hi, thanks a lot. The package from your link seems could plot only one type of variable. If I have following three variables:If I have following three variables: a b c 1 3 7 8 2 5 4 1 3 5 1 7 4 3 1 5 5 1 7 3 6 7 1 1 7 1 1 1 8 3 3 2 Do you have any idea how to plot it by using variable color and radii in one rose chart like here? http://gallery.r-enthusiasts.com/graph/Rose_diagram,97 On Fri, Nov 30, 2012 at 1:55 PM, Rui Barradas ruipbarra...@sapo.pt wrote: Hello, Take a look at the graph in http://gallery.r-enthusiasts.com/graph/Rose_diagram,97 Hope this helps, Rui Barradas Em 30-11-2012 12:24, Henry Smith escreveu: Hello, Everyone. Does anyone know how to create a Nightingales Rose chart by using R? Hopefully, the graph could be displayed like this:http://mbostock.github.com/protovis/ex/crimea-rose.html Thanks a lot. Kind regards, Henry [[alternative HTML version deleted]] __r-h...@r-project.org mailing listhttps://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] help on stacking matrices up
Petr, thanks for the solution, putting the zeros in there makes it work just how I want it. Also, thanks for picking up on the error in my code below... I did not notice that it only changes the 1st value, and yes, your assumption is right: there is a chance where more values should be changed based on Tinf... I have tryed to work with if ... else..., but that does not give me correct answer either, to my suprise... Would you be willing to provide some insigths on how I could go about solving that? thanks, Andras From: PIKAL Petr petr.pi...@precheza.cz roject.org Sent: Friday, November 30, 2012 10:58 AM Subject: RE: [R] help on stacking matrices up Hi Andras Now it works but I suppose you shall simplify or explain it a bit more. CONC7-c(rep(0,3),CONC7) CONC8-c(rep(0,6),CONC8) CONC6+CONC7+CONC8 [,1] [1,] 0.00 [2,] 2.330056 [3,] 2.133774 [4,] 1.954027 [5,] 4.119477 [6,] 3.772456 [7,] 3.454667 [8,] 5.493705 [9,] 5.030920 [10,] 4.607120 gives you desired result easier. So for summing CONC objects just add to beginning of each object so many zeroes as there are missing values. You can check dimension by dim or nrow and use it to estimate how many zeroes you need to add. Besides, you probably do not need matrices, vectors can do it too. This CONC6 -ifelse(TIME6=Tinf,CDURINF6,CAFTINF6) in your case just change only the first value, but I presume in reality you can have Tinf such that it results in changing more values. If this is not you are after, please be more specific. Regards Petr Sent: Friday, November 30, 2012 3:52 PM To: PIKAL Petr; r-help@r-project.org Subject: Re: [R] help on stacking matrices up Petr, sorry, here is a better example that should work: Dose-200 Tinf -0.5 INTERVAL -3 TIME8 -matrix(c((0*INTERVAL):(1*INTERVAL))) TIME7 -matrix(c((0*INTERVAL):(2*INTERVAL))) TIME6 -matrix(c((0*INTERVAL):(3*INTERVAL))) CDURINF6 -((Dose/Tinf)*(1/(0.088*76.9)))*(1-exp(-0.088*TIME6)) CAFTINF6 -((Dose/Tinf)*(1/(0.088*76.9)))*(1-exp(-0.088*Tinf))*(exp(-0.088*(TIME6))) CONC6 -ifelse(TIME6=Tinf,CDURINF6,CAFTINF6) CDURINF7 -((Dose/Tinf)*(1/(0.088*76.9)))*(1-exp(-0.088*TIME7)) CAFTINF7 -((Dose/Tinf)*(1/(0.088*76.9)))*(1-exp(-0.088*Tinf))*(exp(-0.088*(TIME7))) CONC7 -ifelse(TIME7=Tinf,CDURINF7,CAFTINF7) CDURINF8 -((Dose/Tinf)*(1/(0.088*76.9)))*(1-exp(-0.088*TIME8)) CAFTINF8 -((Dose/Tinf)*(1/(0.088*76.9)))*(1-exp(-0.088*Tinf))*(exp(-0.088*(TIME8))) CONC8 -ifelse(TIME8=Tinf,CDURINF8,CAFTINF8) CONCF -matrix(c(CONC6[1,],CONC6[2,],CONC6[3,],CONC6[4,]+CONC7[1,],CONC6[5,]+CONC7[2,],CONC6[6,]+CONC7[3,],CONC6[7,]+CONC7[4,]+CONC8[1,],CONC6[8,]+CONC7[5,]+CONC8[2,],CONC6[9,]+CONC7[6,]+CONC8[3,],CONC6[10,]+CONC7[7,]+CONC8[4,])) plot(TIME6,CONCF,type=b) the plot is what it should look like, keeping in mind that the INTERVAL will change, and that is the part i amtrying to figure out, ie: how can i make that dynamic. Is this better? thanks, Andras From:PIKAL Petr petr.pi...@precheza.cz roject.org Sent: Friday, November 30, 2012 8:51 AM Subject: RE: [R] help on stacking matrices up Hi try to make your example easier and reproducible. I get CONCF -matrix(c(CONC1+CONC2+CONC3+CONC4+CONC5+CONC6+CONC7+CONC8)) Error in CONC1 + CONC2 : non-conformable arrays so I cannot understand what shall the final plot look like. Regards Petr -Original Message- From: r-help-boun...@r-project.org [mailto:r-help-bounces@r- project.org] On Behalf Of Andras Farkas Sent: Friday, November 30, 2012 12:51 PM To: r-help@r-project.org Subject: [R] help on stacking matrices up Dear All, #I have the following code Dose-1000 Tinf -0.5 INTERVAL -8 TIME8 -matrix(c((0*INTERVAL):(1*INTERVAL))) TIME7 -matrix(c((0*INTERVAL):(2*INTERVAL))) TIME6 -matrix(c((0*INTERVAL):(3*INTERVAL))) TIME5 -matrix(c((0*INTERVAL):(4*INTERVAL))) TIME4 -matrix(c((0*INTERVAL):(5*INTERVAL))) TIME3 -matrix(c((0*INTERVAL):(6*INTERVAL))) TIME2 -matrix(c((0*INTERVAL):(7*INTERVAL))) TIME1 -matrix(c((0*INTERVAL):(8*INTERVAL))) CDURINF1 -((Dose/Tinf)*(1/(0.088*76.9)))*(1-exp(-0.088*TIME1)) CAFTINF1 -((Dose/Tinf)*(1/(0.088*76.9)))*(1-exp(-0.088*Tinf))*(exp(- 0.088*(TIME1))) CONC1 -ifelse(TIME1=Tinf,CDURINF1,CAFTINF1) CDURINF2 -((Dose/Tinf)*(1/(0.088*76.9)))*(1-exp(-0.088*TIME2)) CAFTINF2 -((Dose/Tinf)*(1/(0.088*76.9)))*(1-exp(-0.088*Tinf))*(exp(- 0.088*(TIME2))) CONC2 -ifelse(TIME2=Tinf,CDURINF2,CAFTINF2) CDURINF3 -((Dose/Tinf)*(1/(0.088*76.9)))*(1-exp(-0.088*TIME3)) CAFTINF3 -((Dose/Tinf)*(1/(0.088*76.9)))*(1-exp(-0.088*Tinf))*(exp(- 0.088*(TIME3))) CONC3 -ifelse(TIME3=Tinf,CDURINF3,CAFTINF3) CDURINF4 -((Dose/Tinf)*(1/(0.088*76.9)))*(1-exp(-0.088*TIME4)) CAFTINF4 -((Dose/Tinf)*(1/(0.088*76.9)))*(1-exp(-0.088*Tinf))*(exp(- 0.088*(TIME4))) CONC4 -ifelse(TIME4=Tinf,CDURINF4,CAFTINF4) CDURINF5 -((Dose/Tinf)*(1/(0.088*76.9)))*(1-exp(-0.088*TIME5)) CAFTINF5 -((Dose/Tinf)*(1/(0.088*76.9)))*(1-exp(-0.088*Tinf))*(exp(- 0.088*(TIME5))) CONC5
[R] flip heatmap (pheatmap)
When plotting a heatmap with heatmap.2 the shortest euclidean distance cluster is plotted in the top left corner of the heatmap http://r.789695.n4.nabble.com/file/n4651430/heatmap.2.jpeg when plotting the same data with pheatmap the very same cluster is in the bottom right corner http://r.789695.n4.nabble.com/file/n4651430/pheatmap.jpeg I am just wondering what I would need to change in my pheatmap code to have the same orientation as with heamap.2 (eg the ? Any comments much appreciated. Thanks, DKV -- View this message in context: http://r.789695.n4.nabble.com/flip-heatmap-pheatmap-tp4651430.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] segfault debugging
Hello everybody, I have written a script with two inline cfunctions. The script crashes from time to time with: *** caught segfault *** address 0x10, cause 'memory not mapped' The crashs happen within R code after the cfunctions were executed. Nevertheless I think that the pointers in my cfunctions are not used correctly. I tried to find some examples for debugging tools. I found something like R -d gdb. But I could not find an example how to use it with #!/usr/bin/Rscript. And I cannot source the script within a running R session because I must run it with some args. Can anybody tell me as a non nerd step by step what I can do to locate the problem? Can you recommend tools? If so how are they used with Rscript? Many many thanks in advance Donatella __ 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] Nightingale’s Rose chart-any suggestion?
Dear Dennis, Many thanks! Yes, ggplot2 could be used to illustrate a simple rose chart (category + one type variable) If I have a data frame like this: DF - data.frame(month = factor(month.abb, levels = month.abb), freq1 = rpois(12, 80),freq2=rpois(12, 80),freq3=rpois(12, 80)) Do you have any idea how to plot the other 2 variables freq2 and freq3 by R? Nightingales Rose chart actually represent 3 variables here: blue is disease, red is wounds, and black is uncategorized. I checked plotrix; however, it seems it could not fulfill my purpose too. Actually, what I am looking for is how to render a heatmap (not a bar chart) in a polar coordinate system. Thanks again. Best, Henry On Fri, Nov 30, 2012 at 2:37 PM, Dennis Murphy djmu...@gmail.com wrote: A simple version of what can be done in ggplot2 is illustrated in the following toy example: DF - data.frame(month = factor(month.abb, levels = month.abb), freq = rpois(12, 80)) ggplot(DF, aes(x = month, y = freq)) + theme_bw() + geom_bar(stat = identity, fill = pink2) + coord_polar() Another option might be to use the plotrix package and to look into the radial.plot function, but I don't know offhand if it will produce rose diagrams without some effort. In ggplot2, a rose diagram is a bar chart rendered in a polar coordinate system, consistent with Wilkinson's grammar of graphics (whence the gg in ggplot2). Dennis On Fri, Nov 30, 2012 at 4:24 AM, Henry Smith henry.helsi...@gmail.com wrote: Hello, Everyone. Does anyone know how to create a Nightingales Rose chart by using R? Hopefully, the graph could be displayed like this: http://mbostock.github.com/protovis/ex/crimea-rose.html Thanks a lot. Kind regards, Henry [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] googleVis plot and knitr/sweave
Thank you Michael. I'll try that. Maybe I can find a solution to convert the HTML file into a let's say SVG file. Regards, Phil -- View this message in context: http://r.789695.n4.nabble.com/googleVis-plot-and-knitr-sweave-tp4651341p4651463.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] googleVis plot and knitr/sweave
Thank everyone for the answers. I looked into ggmap but could not find how to plot a map that would like this: http://www.google.ca/trends/explore#q=flu I looked into get_map but it can only be of this type: maptype = c(terrain, satellite, roadmap, hybrid, toner, watercolor) Any suggestions would be appreciated. Regards, Phil -- View this message in context: http://r.789695.n4.nabble.com/googleVis-plot-and-knitr-sweave-tp4651341p4651434.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] loop function and integrate?
Now i've managed to do this:
funcs - list()
funcs[]
# loop through to define functions
for(i in 1:ib-1){
# Make function name
funcName - paste( 'func', i, sep = '' )
# make function
func = paste('function(x){sin(x + a[', i,'])))}',sep = '')
funcs[[funcName]] = eval(parse(text=func))
}
end
but still cant apply the integrate in a loop for all the different
functions. :/
help me out guys pretty please :)
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and provide commented, minimal, self-contained, reproducible code.
[R] qbinom
a=c(0.1, 0.2, 0.3, 0.4, 0.5, 0.6, 0.7, 0.8, 0.9) b=c(0.9, 0.8, 0.7, 0.6, 0.5, 0.4, 0.3, 0.2, 0.1) cor(a,b)= -1 a'=qbinom(a, 1, 0.5) b'=qbinom(b, 1, 0.5) why cor(a',b') becomes -0.5 ? -- View this message in context: http://r.789695.n4.nabble.com/qbinom-tp4651460.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] subgroup-based quantiles
Hello,
If you want Age quantiles by gender, you have to split the data by
gender, apply the same code then recombine the result.
fun - function(x){
Age_group - cut(x[, Age], labels=c(1:10),
breaks=quantile(x[, Age], seq(0,1,.1)),
include.lowest = TRUE)
cbind(x, Age_group)
}
result - do.call(rbind, lapply(split(dat, dat[, GENDER]), fun))
rownames(result) - seq_len(nrow(result))
result
Hope this helps,
Rui Barradas
Em 30-11-2012 12:18, R Kozarski escreveu:
Dear R users,
given the patient sample with their Gender and Age
GENDER Age
[1,] 2 45
[2,] 1 58
[3,] 1 54
[4,] 2 71
[5,] 2 64
...
I would like to create an another column which groups the patients wrt
Gender specific Age quantiles, following methodology similar to:
Age_group - cut(Age, labels=c(1:10), breaks=quantile(Age,
seq(0,1,.1)),include.lowest = TRUE)
The function above allows me to group only wrt Age quantiles.
Best, Robert
--
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http://r.789695.n4.nabble.com/subgroup-based-quantiles-tp4651415.html
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Re: [R] repeating matrices in a list
HI, You could also use: res1-sapply(1:3,function(x) g,simplify=FALSE) #or res2-replicate(3,g,simplify=FALSE) identical(res1,res2) #[1] TRUE A.K. - Original Message - From: Anser Chen anser.c...@gmail.com To: r-help@r-project.org Cc: Sent: Friday, November 30, 2012 12:50 PM Subject: [R] repeating matrices in a list Suppose I have the following square, non-negative matrices g=matrix(c(0,2,4,0.25,0,0,0,0.6,0),3,3,byrow=T); I want to create a list where this matrix is repeated multiple times. if I do this brute force (manually), using env - list(g,g,g) works fine. Yields [[1]] [,1] [,2] [,3] [1,] 0.00 2.0 4 [2,] 0.25 0.0 0 [3,] 0.00 0.6 0 [[2]] [,1] [,2] [,3] [1,] 0.00 2.0 4 [2,] 0.25 0.0 0 [3,] 0.00 0.6 0 [[3]] [,1] [,2] [,3] [1,] 0.00 2.0 4 [2,] 0.25 0.0 0 [3,] 0.00 0.6 0 But - for a variety of purposes, I need to 'automate' building said list. I tried using rep env - list(rep(g,each=3)) but this yields [1] 0.00 0.00 0.00 0.25 0.25 0.25 0.00 0.00 0.00 2.00 2.00 2.00 0.00 0.00 0.00 [16] 0.60 0.60 0.60 4.00 4.00 4.00 0.00 0.00 0.00 0.00 0.00 0.00 Any suggestions/pointers to the obvious? Thanks in advance... [[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] loop function and integrate?
Hello,
Your code doesn't run without initializing 'ss' to something. And I've
made some changes, but I don't understand what you are trying to do. See
comments inline.
a = c(1,2,3,4)
ia = length(a)
x = seq(1, 100, by=0.1) # It was 'length = 0.1' (!)
ib = length(x)
ss - numeric(ia) # New, 'ss' must exist.
int1 = numeric(ib)
b = numeric(ib)
for(j in 1:ia) {
H = function(x) {sin(x + a[j])}
for(i in 1:ib) {
int = integrate(H, lower = 0, upper = x[i])
int1[i] = int[1]
b[i] = 1 + a[j] # It was 'a[i]', didn't make any sense.
}
int1 = unlist(int1)
int2 = int1*b
ss[j] = sum(int2)
}
ss
And for loops (or any other type of loops) do _not_ end with 'end'.
Hope this helps,
Rui Barradas
Em 30-11-2012 15:08, faeriewhisper escreveu:
Hi guys!
I have to compute something and i don't know what i'm doing wrong. my code
is a bit complex, but imagine that is something like this:
a = c(1,2,3,4)
ia = length(a)
x = seq(1,100,length=0.1)
ib = length(x)
int1 = numeric(ib)
b = numeric(ib)
for(j in 1:ia) {
H = function(x) {sin(x + a[j])}
for(i in 1:ib) {
int = integrate(H, lower = 0, upper = x[i])
int1[i] = int[1]
b[i] = 1 + a[i]
}
end
int1 = unlist(int1)
int2 = int1*b
ss[j] = sum(int2)
}
end
if i try this code without the for loop it's ok, but when i put the for on,
i get all sort of errors...
Thank you for your help :)
--
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Re: [R] loop function and integrate?
On 30-11-2012, at 16:08, faeriewhisper wrote:
Hi guys!
I have to compute something and i don't know what i'm doing wrong. my code
is a bit complex, but imagine that is something like this:
a = c(1,2,3,4)
ia = length(a)
x = seq(1,100,length=0.1)
ib = length(x)
int1 = numeric(ib)
b = numeric(ib)
for(j in 1:ia) {
H = function(x) {sin(x + a[j])}
for(i in 1:ib) {
int = integrate(H, lower = 0, upper = x[i])
int1[i] = int[1]
b[i] = 1 + a[i]
}
end
int1 = unlist(int1)
int2 = int1*b
ss[j] = sum(int2)
}
end
What are you doing?
What's the end doing in your code in two places.
end is a function to extract and encode the last observation of a time series
object. See ?end
It makes absolutely no sense to put them in your code.
Remove them immediately.
You haven't declared ss to be a vector.
So before the start of the j loop insert ss - numeric(ia)
And simplify your code:
ss - numeric(ia)
for(j in 1:ia) {
H = function(x) {sin(x + a[j])}
for(i in 1:ib) {
int = integrate(H, lower = 0, upper = x[i])
int1[i] = int$value
b[i] = 1 + a[i]
}
ss[j] = sum(int1*b)
}
ss
And more simplification is possible by eliminating b, which I leave to you.
Berend
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Re: [R] can't integrate in loop
Thats not a very precise question. I'll try anyway..
- if you use c, you need to separate the values by commas
- i think you mean seq(1,100,0.1), otherwise x only has one value
- function sen is not defined
- If you call int(1), upper will be 1, not x[i]
- why are you making a function and calling it instead of just : int1[i] -
integrate(H, lower = 0, upper = x[i]) ?
- looking at ?integral, you may rather want integral(..)$value
On 30.11.2012, at 13:32, faeriewhisper wrote:
Hi guys!
I have to compute something and i don't know what i'm doing wrong. my code
is a bit complex, but imagine that is something like this:
a = c(1 2 3 4)
ia = length(a)
x = seq(1,100,length=0.1)
ib = length(x)
for(j in 1:ia) {
H = function(x) {sen(x) + a[j]}
for(i in 1:ib) {
int = function(x) { integrate(H, lower = 0, upper = x[i])}
int1[i] = int(1)
}
end
int1 = unlist(int1)
ss[j] = sum(int1)
}
end
if i try this code without the for loop it's ok, but when i put the cycle
on, i get all sort of errors...
Thank you for your help :)
--
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http://r.789695.n4.nabble.com/can-t-integrate-in-loop-tp4651416.html
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Re: [R] loop function and integrate?
On 30-11-2012, at 19:34, Berend Hasselman wrote:
On 30-11-2012, at 16:08, faeriewhisper wrote:
Hi guys!
I have to compute something and i don't know what i'm doing wrong. my code
is a bit complex, but imagine that is something like this:
a = c(1,2,3,4)
ia = length(a)
x = seq(1,100,length=0.1)
ib = length(x)
int1 = numeric(ib)
b = numeric(ib)
for(j in 1:ia) {
H = function(x) {sin(x + a[j])}
for(i in 1:ib) {
int = integrate(H, lower = 0, upper = x[i])
int1[i] = int[1]
b[i] = 1 + a[i]
}
end
int1 = unlist(int1)
int2 = int1*b
ss[j] = sum(int2)
}
end
What are you doing?
What's the end doing in your code in two places.
end is a function to extract and encode the last observation of a time series
object. See ?end
It makes absolutely no sense to put them in your code.
Remove them immediately.
You haven't declared ss to be a vector.
So before the start of the j loop insert ss - numeric(ia)
And simplify your code:
ss - numeric(ia)
for(j in 1:ia) {
H = function(x) {sin(x + a[j])}
for(i in 1:ib) {
int = integrate(H, lower = 0, upper = x[i])
int1[i] = int$value
b[i] = 1 + a[i]
}
ss[j] = sum(int1*b)
}
ss
And more simplification is possible by eliminating b, which I leave to you.
There is no need to define the function H in the j loop.
You could do this:
H = function(x,A) {sin(x + A)}
for(j in 1:ia) {
aj - a[j]
for(i in 1:ib) {
int = integrate(H, lower = 0, upper = x[i], A=aj)
int1[i] = (1 + a[i])* int$value
}
ss[j] = sum(int1)
}
And if Rui is right a[i] can be replaced by aj.
Berend
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Re: [R] segfault debugging
On 30/11/2012 12:02 PM, Donatella Quagli wrote: Hello everybody, I have written a script with two inline cfunctions. The script crashes from time to time with: *** caught segfault *** address 0x10, cause 'memory not mapped' The crashs happen within R code after the cfunctions were executed. Nevertheless I think that the pointers in my cfunctions are not used correctly. I tried to find some examples for debugging tools. I found something like R -d gdb. But I could not find an example how to use it with #!/usr/bin/Rscript. And I cannot source the script within a running R session because I must run it with some args. Can anybody tell me as a non nerd step by step what I can do to locate the problem? Can you recommend tools? If so how are they used with Rscript? This isn't easy, but what I would do is add something to your script to get it to pause (or to wait for some file to be created, or some other signal), then figure out the process number, and tell gdb to connect to it after it has already started. The ps command will list all the running processes, for some definition of all. Then gdb --pid=PID will connect to the process with id PID. I believe it will interrupt the process at that point, but you can restart it, send the signal to it to get out of its loop), and hopefully see the segfault in action. Duncan Murdoch __ 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] qbinom
On Fri, Nov 30, 2012 at 9:47 AM, jaybell stephe...@yahoo.com.tw wrote: a=c(0.1, 0.2, 0.3, 0.4, 0.5, 0.6, 0.7, 0.8, 0.9) b=c(0.9, 0.8, 0.7, 0.6, 0.5, 0.4, 0.3, 0.2, 0.1) cor(a,b)= -1 a'=qbinom(a, 1, 0.5) b'=qbinom(b, 1, 0.5) why cor(a',b') becomes -0.5 ? On my computer the correlation is -0.8. It is not 1 because you did a non-linear transformation of a and b. Plot a vs. b, then plot a' vs b' and you will see why the correlation is not -1: a vs b is a straight line, a' vs b' is not a straight line. 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.
Re: [R] xts indexed with Date class
Hi, You could try this: Sys.setenv(TZ=GMT) a-as.Date(15423:15426) x-xts(seq_along(a),a) print(index(x)) #[1] 2012-03-24 2012-03-25 2012-03-26 2012-03-27 print(as.numeric(index(x))) #[1] 15423 15424 15425 15426 A.K. - Original Message - From: Giles giles.heyw...@cantab.net To: r-help@r-project.org Cc: Sent: Friday, November 30, 2012 4:32 AM Subject: [R] xts indexed with Date class Hi I see a changed behaviour in xts indexed on class Date in the latest versions, versus 2. It seems to be related to changes to/from daylight savings time, happens those weekends. Is it not intended that class Date be used like this, or is this new behaviour incorrect? Giles Example: a-as.Date(15423:15426) x-xts(seq_along(a),a) print(x) [,1] 2012-03-24 1 2012-03-25 2 2012-03-25 3 2012-03-26 4 print(index(x)) [1] 2012-03-24 2012-03-25 2012-03-25 2012-03-26 print(as.numeric(index(x))) [1] 15423 15424 15424 15425 #for reference, zoo behaves as expected: z-zoo(seq_along(a),a) print(index(z)) [1] 2012-03-24 2012-03-25 2012-03-26 2012-03-27 print(as.numeric(index(z))) [1] 15423 15424 15425 15426 Package: xts Type: Package Title: eXtensible Time Series Version: 0.8-8 Date: 2012-10-05 Windows 7 64-bit R version 2.15.2 64-bit all packages up-to-date #--previously, R 2.13.2 and xts 0.8-2 , showing expected behaviour - a-as.Date(15423:15426) x-xts(seq_along(a),a) print(x) [,1] 2012-03-24 1 2012-03-25 2 2012-03-26 3 2012-03-27 4 print(index(x)) [1] 2012-03-24 2012-03-25 2012-03-26 2012-03-27 print(as.numeric(index(x))) [1] 15423 15424 15425 15426 z-zoo(seq_along(a),a) print(index(z)) [1] 2012-03-24 2012-03-25 2012-03-26 2012-03-27 print(as.numeric(index(z))) [1] 15423 15424 15425 15426 __ 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] subgroup-based quantiles
Hello,
Arun's code is much better.
Rui Barradas
Em 30-11-2012 18:50, arun escreveu:
Hi,
You could also try ?ave()
dat$Age_group-ave(dat$Age,dat$GENDER,FUN=function(x){cut(x,labels=1:10,breaks=quantile(x,seq(0,1,.1)),include.lowest=TRUE)})
dat
# GENDER Age Age_group
#1 2 45 1
#2 1 5810
#3 1 54 1
#4 2 7110
#5 2 64 5
A.K.
- Original Message -
From: Rui Barradas ruipbarra...@sapo.pt
To: R Kozarski r.kozar...@gmail.com
Cc: r-help@r-project.org
Sent: Friday, November 30, 2012 1:07 PM
Subject: Re: [R] subgroup-based quantiles
Hello,
If you want Age quantiles by gender, you have to split the data by
gender, apply the same code then recombine the result.
fun - function(x){
Age_group - cut(x[, Age], labels=c(1:10),
breaks=quantile(x[, Age], seq(0,1,.1)),
include.lowest = TRUE)
cbind(x, Age_group)
}
result - do.call(rbind, lapply(split(dat, dat[, GENDER]), fun))
rownames(result) - seq_len(nrow(result))
result
Hope this helps,
Rui Barradas
Em 30-11-2012 12:18, R Kozarski escreveu:
Dear R users,
given the patient sample with their Gender and Age
GENDER Age
[1,] 2 45
[2,] 1 58
[3,] 1 54
[4,] 2 71
[5,] 2 64
...
I would like to create an another column which groups the patients wrt
Gender specific Age quantiles, following methodology similar to:
Age_group - cut(Age, labels=c(1:10), breaks=quantile(Age,
seq(0,1,.1)),include.lowest = TRUE)
The function above allows me to group only wrt Age quantiles.
Best, Robert
--
View this message in context:
http://r.789695.n4.nabble.com/subgroup-based-quantiles-tp4651415.html
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__
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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.
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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] Nightingale’s Rose chart-any suggestion?
I think this does it. require(reshape2) df1 - melt(DF, id=month) ggplot(df1, aes(x = month, y = value, fill = variable)) + theme_bw() + geom_bar(stat = identity) + coord_polar() John Kane Kingston ON Canada -Original Message- From: henry.helsi...@gmail.com Sent: Fri, 30 Nov 2012 14:56:51 +0100 To: djmu...@gmail.com, r-help@r-project.org Subject: Re: [R] Nightingale’s Rose chart-any suggestion? Dear Dennis, Many thanks! Yes, ggplot2 could be used to illustrate a simple rose chart (category + one type variable) If I have a data frame like this: DF - data.frame(month = factor(month.abb, levels = month.abb), freq1 = rpois(12, 80),freq2=rpois(12, 80),freq3=rpois(12, 80)) Do you have any idea how to plot the other 2 variables freq2 and freq3 by R? Nightingale?s Rose chart actually represent 3 variables here: blue is disease, red is wounds, and black is uncategorized. I checked plotrix; however, it seems it could not fulfill my purpose too. Actually, what I am looking for is how to render a heatmap (not a bar chart) in a polar coordinate system. Thanks again. Best, Henry On Fri, Nov 30, 2012 at 2:37 PM, Dennis Murphy djmu...@gmail.com wrote: A simple version of what can be done in ggplot2 is illustrated in the following toy example: DF - data.frame(month = factor(month.abb, levels = month.abb), freq = rpois(12, 80)) ggplot(DF, aes(x = month, y = freq)) + theme_bw() + geom_bar(stat = identity, fill = pink2) + coord_polar() Another option might be to use the plotrix package and to look into the radial.plot function, but I don't know offhand if it will produce rose diagrams without some effort. In ggplot2, a rose diagram is a bar chart rendered in a polar coordinate system, consistent with Wilkinson's grammar of graphics (whence the gg in ggplot2). Dennis On Fri, Nov 30, 2012 at 4:24 AM, Henry Smith henry.helsi...@gmail.com wrote: Hello, Everyone. Does anyone know how to create a Nightingale?s Rose chart by using R? Hopefully, the graph could be displayed like this: http://mbostock.github.com/protovis/ex/crimea-rose.html Thanks a lot. Kind regards, Henry [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. FREE 3D MARINE AQUARIUM SCREENSAVER - Watch dolphins, sharks orcas on your desktop! __ 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] subgroup-based quantiles
Hi,
You could also try ?ave()
dat$Age_group-ave(dat$Age,dat$GENDER,FUN=function(x){cut(x,labels=1:10,breaks=quantile(x,seq(0,1,.1)),include.lowest=TRUE)})
dat
# GENDER Age Age_group
#1 2 45 1
#2 1 58 10
#3 1 54 1
#4 2 71 10
#5 2 64 5
A.K.
- Original Message -
From: Rui Barradas ruipbarra...@sapo.pt
To: R Kozarski r.kozar...@gmail.com
Cc: r-help@r-project.org
Sent: Friday, November 30, 2012 1:07 PM
Subject: Re: [R] subgroup-based quantiles
Hello,
If you want Age quantiles by gender, you have to split the data by
gender, apply the same code then recombine the result.
fun - function(x){
Age_group - cut(x[, Age], labels=c(1:10),
breaks=quantile(x[, Age], seq(0,1,.1)),
include.lowest = TRUE)
cbind(x, Age_group)
}
result - do.call(rbind, lapply(split(dat, dat[, GENDER]), fun))
rownames(result) - seq_len(nrow(result))
result
Hope this helps,
Rui Barradas
Em 30-11-2012 12:18, R Kozarski escreveu:
Dear R users,
given the patient sample with their Gender and Age
GENDER Age
[1,] 2 45
[2,] 1 58
[3,] 1 54
[4,] 2 71
[5,] 2 64
...
I would like to create an another column which groups the patients wrt
Gender specific Age quantiles, following methodology similar to:
Age_group - cut(Age, labels=c(1:10), breaks=quantile(Age,
seq(0,1,.1)),include.lowest = TRUE)
The function above allows me to group only wrt Age quantiles.
Best, Robert
--
View this message in context:
http://r.789695.n4.nabble.com/subgroup-based-quantiles-tp4651415.html
Sent from the R help mailing list archive at Nabble.com.
__
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Re: [R] bootstrapped cox regression (rms package)
Hi Frank,
My apologies for not posting the entire script - I have repasted it below.
library(rms)
library(foreign)
temp=read.spss('coxdata.sav', to.data.frame=T)
formula=Surv(months, recidivate) ~ fac1 + fac2 + fac3 + fac4 + fac5 + fac6 +
fac7 + fac8
fit=cph(formula, data=temp, x=T, y=T)
val.out=validate(fit, method=boot, B=, bw=F, type=residual,
sls=0.05, aics=0,force=NULL, estimates=TRUE, pr=FALSE)
out=bootcov(fit, B=, pr=F, coef.reps=T, loglik=F)
anova(out)
Factor Chi-Square d.f. P
fac1 0.27 10.6055
fac2 0.20 10.6514
fac3 0.01 10.9338
fac4 0.05 10.8311
fac5 1.06 10.3036
fac6 0.33 10.5647
fac7 0.81 10.3670
fac8 0.30 10.5832
TOTAL 1.48 80.9930
for (i in 1:8) {
print(quantile(out$boot.Coef[,i], c(.025, .975)))
}
2.5% 97.5%
-9.236751 20.772061
2.5% 97.5%
-8.841030 3.094755
2.5% 97.5%
-1.834436 2.161983
2.5% 97.5%
-0.1800666 0.0871867
2.5% 97.5%
-3.2129636 0.4783566
2.5% 97.5%
-0.04157389 0.07130994
2.5% 97.5%
-0.6415962 0.1001843
2.5% 97.5%
-0.01529467 0.21055259
Again, the SPSS output I am trying to match is here:
variable low CI high CI p-value
fac1 -8.474 20.020 .456
fac2 -8.206 3.093 .524
fac3 -1.829 2.087 .900
fac4 -.173 .083 .749
fac5 -2.945 .450 .143
fac6 -.035 .070 .306
fac7 -.626 .092 .189
fac8 -.017 .203 .247
In looking through the SPSS syntax, my colleague is using SIMPLE resampling,
which is doing sampling with replacement from the original data set.
bootstrap replications are being used, the same as what I have used in the
bootcov command. The piece of the SPSS output that is not clear is the
generation of p-values from the distribution of parameter estimates; spss
appears to be testing the parameter estimate from the original cox
regression, but the method of testing that parameter is not clear.
Eric
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Re: [R] loop function and integrate?
hello guys!
thank u for the help, but u didnt understood what i need.
1st, it is a[i] cuz i want to sum 1 + x[i], for all i's not j.
but i've solved it! :)
like i said, my code is more complex, but, if you need to integrate several
functions in a loop, thats what you should do:
w2 = seq(-1,-1/3,length=100)
ib = length(w2)
bin = w2[2] - w2[1]
w3 - numeric(ib-1)
for (h in 1:ib-1)
w3[h] = (w2[h] + w2[h+1])/2
end
probt - numeric(ib)
di2 - numeric(ia)
a2 - numeric(ia)
ic = ib-1
arealog - numeric(ic)
log_probt - numeric(ic)
funcs - list()
funcs[]
# loop through to define functions
for(i in 1:ic){
funcName - paste( 'func', i, sep = '' )
func = paste('function(z){c/(Ho*sqrt(dens*(1+z)^3 +
(1-dens)*(1+z)^(3*(1+w3[', i,']}',sep = '')
funcs[[funcName]] = eval(parse(text=func))
}
end
for(j in 1:ic){
for(i in 1:ia){
d2 = integrate(funcs[[j]], lower = 0, upper = z[i])
di2[i] = d2[1]
a2[i] = 1 + z[i]
}
end
di2 = unlist(di2)
dist2 = di2*a2
mag2 = 5*log(dist2) + 25
prob2 = (1/disp*sqrt(2*pi))*exp( - ((mag2 - probv)^2)/2*disp^2)
log_prob2 = log(prob2)
log_proba2 = sum(log_prob2)
log_probt[j] = log_proba2
arealog[j] = bin*(log_proba2)
}
end
log_probt2 = sum(arealog)/(w2[ib] - w2[1])
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[R] Choleski decomposition
m - matrix(nrow=5, ncol=5) m - ifelse(row(m)==col(m), 1, 0.2) c - chol(m)# Choleski decomposition u - matrix(rnorm(2000*5), ncol=5) uc - u %*% c cr - pnorm(uc) cr - qbinom(cr,1,0.5) cor(cr) I expected that the cor(cr) to be 0.2 as i set in m, but the result is around 0.1. Why is that? Thanks -- View this message in context: http://r.789695.n4.nabble.com/Choleski-decomposition-tp4651475.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] Choleski decomposition
On Nov 30, 2012, at 10:34 AM, jaybell wrote: m - matrix(nrow=5, ncol=5) m - ifelse(row(m)==col(m), 1, 0.2) c - chol(m)# Choleski decomposition u - matrix(rnorm(2000*5), ncol=5) uc - u %*% c cr - pnorm(uc) cr - qbinom(cr,1,0.5) cor(cr) I expected that the cor(cr) to be 0.2 as i set in m, but the result is around 0.1. Why is that? Thanks Cross-posting to rhelp and Stackoverflow is deprecated (at least on Rhelp). If you didn't get effective advice after a reasonable interval (measured in days) it would make sense to look elsewhere. But this is essentially a statistics questions rather than a coding question and you were advised on SO to post the question on a website that is designed for statistics questions. Rhelp is not that site. -- David Winsemius, MD Alameda, CA, USA __ 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] bootstrapped cox regression (rms package)
It will be crucial to know the details of the test statistic and P-value
calculations from SPSS. It's also running anova on both the bootcov and the
original fits to see if SPSS is ignoring the bootstrap when computing the
covariance matrix.
Frank
Eric Claus wrote
Hi Frank,
My apologies for not posting the entire script - I have repasted it below.
library(rms)
library(foreign)
temp=read.spss('coxdata.sav', to.data.frame=T)
formula=Surv(months, recidivate) ~ fac1 + fac2 + fac3 + fac4 + fac5 + fac6
+ fac7 + fac8
fit=cph(formula, data=temp, x=T, y=T)
val.out=validate(fit, method=boot, B=, bw=F, type=residual,
sls=0.05, aics=0,force=NULL, estimates=TRUE, pr=FALSE)
out=bootcov(fit, B=, pr=F, coef.reps=T, loglik=F)
anova(out)
Factor Chi-Square d.f. P
fac1 0.27 10.6055
fac2 0.20 10.6514
fac3 0.01 10.9338
fac4 0.05 10.8311
fac5 1.06 10.3036
fac6 0.33 10.5647
fac7 0.81 10.3670
fac8 0.30 10.5832
TOTAL 1.48 80.9930
for (i in 1:8) {
print(quantile(out$boot.Coef[,i], c(.025, .975)))
}
2.5% 97.5%
-9.236751 20.772061
2.5% 97.5%
-8.841030 3.094755
2.5% 97.5%
-1.834436 2.161983
2.5% 97.5%
-0.1800666 0.0871867
2.5% 97.5%
-3.2129636 0.4783566
2.5% 97.5%
-0.04157389 0.07130994
2.5% 97.5%
-0.6415962 0.1001843
2.5% 97.5%
-0.01529467 0.21055259
Again, the SPSS output I am trying to match is here:
variable low CI high CI p-value
fac1 -8.474 20.020 .456
fac2 -8.206 3.093 .524
fac3 -1.829 2.087 .900
fac4 -.173 .083 .749
fac5 -2.945 .450 .143
fac6 -.035 .070 .306
fac7 -.626 .092 .189
fac8 -.017 .203 .247
In looking through the SPSS syntax, my colleague is using SIMPLE
resampling, which is doing sampling with replacement from the original
data set. bootstrap replications are being used, the same as what I
have used in the bootcov command. The piece of the SPSS output that is
not clear is the generation of p-values from the distribution of parameter
estimates; spss appears to be testing the parameter estimate from the
original cox regression, but the method of testing that parameter is not
clear.
Eric
-
Frank Harrell
Department of Biostatistics, Vanderbilt University
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[R] Line numbers with errors and warnings?
Is it possible to get a line number with an error report? I have a long script and an error: Error in `[.xts`(x, xsubset) : subscript out of bounds It would be very helpful, and save a lot of time, if there was some indication in the error message which line the error was. I can find it using binary search but that is a painful process. cheers Worik [[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] Line numbers with errors and warnings?
One thing that I do when I have a long script is to put progress report messages. These have some comments so I can chart the progress and also print out the current CPU and memory usage so I can also isolate where potential problems might be. This will help narrow down the section of code where the error occurred. Are you also running with options(error=utils::recover) so that when the error occurs, you get the stack and current environment so that debugging is easier? On Fri, Nov 30, 2012 at 4:22 PM, Worik R wor...@gmail.com wrote: Is it possible to get a line number with an error report? I have a long script and an error: Error in `[.xts`(x, xsubset) : subscript out of bounds It would be very helpful, and save a lot of time, if there was some indication in the error message which line the error was. I can find it using binary search but that is a painful process. cheers Worik [[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. -- Jim Holtman Data Munger Guru What is the problem that you are trying to solve? Tell me what you want to do, not how you want to do it. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Line numbers with errors and warnings?
On Nov 30, 2012, at 1:22 PM, Worik R wrote: Is it possible to get a line number with an error report? I have a long script and an error: Error in `[.xts`(x, xsubset) : subscript out of bounds It would be very helpful, and save a lot of time, if there was some indication in the error message which line the error was. I can find it using binary search but that is a painful process. Generally if one is in an interactive mode, one is advised to use the error recovery process provided by: options(error=recover) You should then be dropped into the browser in the environment of the error. Execute ls() to see what variables are accessible in the local environment. So you also need to consult: ?browser -- David Winsemius, MD Alameda, CA, USA __ 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] (no subject)
Hello R usuer, The code given below superimposes a pie diagram on another plot containing some points. However, I would like to center the pie diagram on the xy location on the plot, but not on the center. is there any way to re-center pic diagram. Any suggestion or better alternative are highly appreciated. Thank you in advance for your help. Regards, Bibke library(visualFields) library(car) a-saplocmap$p24d2 ordinate-data.frame(a$xod,a$yod) plot( a$xod,a$yod, xlim=c(-30,30), ylim=c(-30,30),xlab=,ylab=,type=p, data=ordinate) abline(v=0,h=0) par(new=TRUE) # Define cars vector with 8 values cars - c(1,1,1,1,1,1,1,1) # Define colors in each sector color1 - c(grey80,grey80,grey80,grey80,gray80,grey30,grey20,gray10 ) ## gray10=0.0001, gray200.001, gray300.01, gray40=0.05, gray 80=0.05 # Putting sector levels car_labels - c(1,2,3,4,5,6,7,8) car_labels - paste(car_labels,sep=) # Create a pie chart with defined heading and custom colors # and labels pie(cars, main=Sectors, col=color1, labels= , radius=0.05,border=NA, cex=0.8) __ 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] Reading .gsheet within R
Hello,
Package sos does a good job at finding things available for R.
library(sos)
findFn('gsheet')
found 0 matches
x has zero rows; nothing to display.
Warning message:
In findFn(gsheet) : HIT not found in HTML; processing one page only.
So I guess there's nothing yet.
Hope this helps,
Rui Barradas
Em 30-11-2012 17:43, Luca Meyer escreveu:
Hello R-experts,
I would like to know if there is a solution to read files with extension
.gsheet directly into R - see http://www.fileinfo.com/extension/gsheet for more
info on this file format.
Thank you,
Luca
Mr. Luca Meyer
www.lucameyer.com
R 2.15.1
Mac OS X 10.8.2
[[alternative HTML version deleted]]
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Re: [R] (no subject)
Try subplot() from the TeachingDemos package. E.g., plot(c(1,2,3), c(1,3,2), xlim=c(0,5), ylim=c(0,10)) subplot(pie(1:5), x=4, y=6) abline(h=6, v=4) subplot(pie(1:5), x=1, y=8, size=c(0.5, 0.5)) Bill Dunlap Spotfire, TIBCO Software wdunlap tibco.com -Original Message- From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On Behalf Of bibek sharma Sent: Friday, November 30, 2012 2:07 PM To: R help Subject: [R] (no subject) Hello R usuer, The code given below superimposes a pie diagram on another plot containing some points. However, I would like to center the pie diagram on the xy location on the plot, but not on the center. is there any way to re-center pic diagram. Any suggestion or better alternative are highly appreciated. Thank you in advance for your help. Regards, Bibke library(visualFields) library(car) a-saplocmap$p24d2 ordinate-data.frame(a$xod,a$yod) plot( a$xod,a$yod, xlim=c(-30,30), ylim=c(-30,30),xlab=,ylab=,type=p, data=ordinate) abline(v=0,h=0) par(new=TRUE) # Define cars vector with 8 values cars - c(1,1,1,1,1,1,1,1) # Define colors in each sector color1 - c(grey80,grey80,grey80,grey80,gray80,grey30,grey20,gray10 ) ## gray10=0.0001, gray200.001, gray300.01, gray40=0.05, gray 80=0.05 # Putting sector levels car_labels - c(1,2,3,4,5,6,7,8) car_labels - paste(car_labels,sep=) # Create a pie chart with defined heading and custom colors # and labels pie(cars, main=Sectors, col=color1, labels= , radius=0.05,border=NA, cex=0.8) __ 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] Reading .gsheet within R
Hi,
I don't know if the information is still current, but there were some
discussions a while back pertaining to reading Google Docs documents in R.
There is a blog post here from David Smith that might be helpful:
http://blog.revolutionanalytics.com/2009/09/how-to-use-a-google-spreadsheet-as-data-in-r.html
I would be sure to read the comments as well.
Regards,
Marc Schwartz
On Nov 30, 2012, at 4:12 PM, Rui Barradas ruipbarra...@sapo.pt wrote:
Hello,
Package sos does a good job at finding things available for R.
library(sos)
findFn('gsheet')
found 0 matches
x has zero rows; nothing to display.
Warning message:
In findFn(gsheet) : HIT not found in HTML; processing one page only.
So I guess there's nothing yet.
Hope this helps,
Rui Barradas
Em 30-11-2012 17:43, Luca Meyer escreveu:
Hello R-experts,
I would like to know if there is a solution to read files with extension
.gsheet directly into R - see http://www.fileinfo.com/extension/gsheet for
more info on this file format.
Thank you,
Luca
Mr. Luca Meyer
www.lucameyer.com
R 2.15.1
Mac OS X 10.8.2
__
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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] Workarounds to Rd file name conflict when extending a S4 method of some other package
Dear list, // IN SHORT // What are possible workarounds to consolidate documentation for S4 methods that are scattered across different packages (generic and some custom methods in one package, additional custom methods in another package) in a *single* Rd help file while using package 'roxygen2' to generate the actual Rd files? // ADDITIONAL INFORMATION // First of, here are the three facts that cause my problem: 1) I'd like the software, i.e. the packages, that I write to be as extendable as possible 2) I pretty much ended up using S4 functionality in everything I do 3) I'm a big fan of in-file documentation and package 'roxygen2' does a wonderful job in helping me out keeping my Rd help files synced That being said, it is a quite common scenario that some package (say 'pkga') contains the *generic* method/function 'foo()' as well as *some* custom methods (for different data types of the available signature arguments). Now, let's suppose that someone using 'pkga' and building a new package (say 'pkgb') would like to build upon the generic method 'pkga::foo()' and provide some more custom methods for it. When he sticks to the suggested workflow (especially with respect to the way the roxygen2 code is written), then R CMD check will rightly(!) complain that an Rd file with the respective name (generated by 'roxygenize()') already exists (because it is already part of 'pkga'). My question is hence twofold: 1) What would be possible workarounds that allow me to a) keep using 'roxygen2' and link documentation of pkga::foo() with that of pkgb::foo() (as they do belong together conecptionally) 2) Is there a need to address this problem on a higher level in the future? My feeling is that more people are starting to use S4 which, IMHO, is a good thing as it allows to systematically build upon code of other programmers. But then I guess we would need some sort of an inter-package check and/or help-file consolidation to present the user a single source of documentation for some S4 method. I tried to illustrate the problem with actual code in this post at Stackoverflow: http://stackoverflow.com/questions/13137912/rd-file-name-conflict-when-s4-generic-and-methods-are-not-in-the-same-package Best regards, Janko __ 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] qbinom
sorry, I repost the question again a=c(0.1, 0.2, 0.3, 0.4, 0.5, 0.6, 0.7, 0.8, 0.9) b=c(0.9, 0.8, 0.7, 0.6, 0.5, 0.4, 0.3, 0.2, 0.1) cor(a,b)= -1 pa=qbinom(a, 1, 0.5) pb=qbinom(b, 1, 0.5) cor(pa,pb)=-0.8 but when pa=qbinom(a,10,0.5) pb=qbinom(b,10,0.5) cor(pa,pb) becomes -1 again -- View this message in context: http://r.789695.n4.nabble.com/qbinom-tp4651460p4651496.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] protentially serious R error
Hi guy,
I have recently encountered a problem while I was just trying to generate
some random numbers with the function rnorm, the problem is shown below:
case 1
rnorm(20*0.2)
[1] -1.2765922 -0.5732654 -1.2246126 -0.4734006
case 2###
* rnorm(20*(1-0.8))
[1] -0.62036668 0.04211587 -0.91092165*
#case 3
a-0.2
rnorm(20*a)
[1] 0.1580288 -0.6545846 1.7672873 0.7167075
#case 4#
* b-1-0.8
rnorm(20*b)
[1] 0.9101742 0.3841854 1.6821761*
I was expecting the 4 cases should do the same job--generate 4 random
numbers. But in case 2 and 4 I only get 3. Has anyone else seen this problem
before? Thanks. (I have tried with other functions i.e rchisq,rexp ...)
###
One of my colleague also have a problem that we think it might be related
with the problem I addressed above:
test1 - runif(10,0,1)
test1
[1] 0.3868379 0.1587814 0.8140483 0.7796691 0.5357628 0.2431110 0.1782747
0.3906829 0.5262615 0.7440143
test2 - NULL
for(i in seq(0.01,1,length=100)){
+ test2[i*100] - sum(test1i)
+ }
test2
[1] 0 0 0 0 0 0 *NA* 0 0 0 0 0 0 0 0 1 1 2 2 2 2 2 2
2 3 3 3 3 3 3 3 3 3 3 3 3 3 3 4 5 5 5 5 5
[45] 5 5 5 5 5 5 5 5 6 7 7 7 7 7 7 7 7 7 7 7 7 7 7
7 7 7 7 7 7 7 8 8 8 9 9 9 9 10 10 10 10 10 10 10
[89] 10 10 10 10 10 10 10 10 10 10 10 10
Every time he re-runs the code there always always a NA(highlighted). Does
any one know why? Your help is greatly appreciated.
Xiaofeng
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and provide commented, minimal, self-contained, reproducible code.
[R] quantreg installation and conflicts with R 2.15.2
I recently lost the partitions on my hard drive (second time in 6 months) so I had to have our IT folks image all my files over to a new drive. I completely reinstalled R (now 2.15.2) and all my libraries to my computer (Dell Latitude running Windows 7). A few of my previous workspaces (created with R 2.14.1) can't be restored, reporting an error similar to the one I get when I try to load quantreg package which requires SparseM (see below). So, not only will quantreg not load but some of my workspaces can't be restored when being loaded (see below). Not sure about what this is about. I asked Roger Koenker, the package maintainer, but he is on travel and won't have chance to seriously investigate this for awhile. So I thought I would put it out to the R community and see if anyone has any suggestions about why this conflict might be occurring. library(quantreg) Loading required package: SparseM Error : object ?kronecker? is not exported by 'namespace:methods' Error: package ?SparseM? could not be loaded or Error: object ?kronecker? is not exported by 'namespace:methods' During startup - Warning message: unable to restore saved data in C:\CADESTUFF\DATA\BarryNoon\.RData Brian Brian S. Cade, PhD U. S. Geological Survey Fort Collins Science Center 2150 Centre Ave., Bldg. C Fort Collins, CO 80526-8818 email: brian_c...@usgs.gov tel: 970 226-9326 [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] How to add widgets of gWidgets to widgets of rgtk2 ???
For this it would be best to insert an intermediate group container, as its add method allows you to add the RGtk2 widgets. Something like: library(gWidgets) options(guiToolkit=RGtk2) library(RGtk2) w - gwindow() lyt - glayout(cont=w) lyt[1,1] - (g - ggroup(cont=lyt)) widget - gtkButton(click me) add(g, widget) Otherwise you can get the underlying GtkTable widget with: getToolkitWidget(lyt) -- View this message in context: http://r.789695.n4.nabble.com/How-to-add-widgets-of-gWidgets-to-widgets-of-rgtk2-tp4651396p4651491.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] subgroup-based quantiles
Dear Arun and Rui, thank you for replying, your commands are very helpful.
Before you replied, Ive solved the issue with the following approach (for
quartiles):
myFun - function(x, GENDER)
{
x.male - cut(x, labels=c(1:4), breaks=quantile(split(x,GENDER) $ '1',
seq(0,1,.25), na.rm=TRUE),include.lowest = TRUE)
x.female - cut(x, labels=c(1:4), breaks=quantile(split(x,GENDER) $ '2',
seq(0,1,.25), na.rm=TRUE),include.lowest = TRUE)
ifelse(GENDER==1, x.male, x.female)
}
Robert
--
View this message in context:
http://r.789695.n4.nabble.com/subgroup-based-quantiles-tp4651415p4651494.html
Sent from the R help mailing list archive at Nabble.com.
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[R] Fw: quantreg installation and conflicts with R 2.15.2
Just noticed that I get a similar error about object 'kronecker' in Matrix package when trying to load lme4. So this is a more pervasive problem. Brian Brian S. Cade, PhD U. S. Geological Survey Fort Collins Science Center 2150 Centre Ave., Bldg. C Fort Collins, CO 80526-8818 email: brian_c...@usgs.gov tel: 970 226-9326 - Forwarded by Brian S Cade/BRD/USGS/DOI on 11/30/2012 02:07 PM - From: Brian S Cade/BRD/USGS/DOI To: R-help@r-project.org Date: 11/30/2012 01:16 PM Subject: quantreg installation and conflicts with R 2.15.2 I recently lost the partitions on my hard drive (second time in 6 months) so I had to have our IT folks image all my files over to a new drive. I completely reinstalled R (now 2.15.2) and all my libraries to my computer (Dell Latitude running Windows 7). A few of my previous workspaces (created with R 2.14.1) can't be restored, reporting an error similar to the one I get when I try to load quantreg package which requires SparseM (see below). So, not only will quantreg not load but some of my workspaces can't be restored when being loaded (see below). Not sure about what this is about. I asked Roger Koenker, the package maintainer, but he is on travel and won't have chance to seriously investigate this for awhile. So I thought I would put it out to the R community and see if anyone has any suggestions about why this conflict might be occurring. library(quantreg) Loading required package: SparseM Error : object ?kronecker? is not exported by 'namespace:methods' Error: package ?SparseM? could not be loaded or Error: object ?kronecker? is not exported by 'namespace:methods' During startup - Warning message: unable to restore saved data in C:\CADESTUFF\DATA\BarryNoon\.RData Brian Brian S. Cade, PhD U. S. Geological Survey Fort Collins Science Center 2150 Centre Ave., Bldg. C Fort Collins, CO 80526-8818 email: brian_c...@usgs.gov tel: 970 226-9326 [[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] Training, Validation and Test Sets and LDA with MASS package
Hi all, please, can someone give me an example of LDA with MASS package with training, validation and test sets? I never used it, so I need and example to avoid to make errors. Thank you very much. Best, Roberto -- View this message in context: http://r.789695.n4.nabble.com/Training-Validation-and-Test-Sets-and-LDA-with-MASS-package-tp4651497.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] protentially serious R error
Potentially serious? Only to you: this looks like R FAQ 7.31 striking again.
Here's a hint:
identical(20*(1-0.8), 4)
and another
rnorm(round(20*(1-0.8)))
Sarah
On Fri, Nov 30, 2012 at 4:05 PM, liuxf li...@math.mcmaster.ca wrote:
Hi guy,
I have recently encountered a problem while I was just trying to generate
some random numbers with the function rnorm, the problem is shown below:
case 1
rnorm(20*0.2)
[1] -1.2765922 -0.5732654 -1.2246126 -0.4734006
case 2###
* rnorm(20*(1-0.8))
[1] -0.62036668 0.04211587 -0.91092165*
#case 3
a-0.2
rnorm(20*a)
[1] 0.1580288 -0.6545846 1.7672873 0.7167075
#case 4#
* b-1-0.8
rnorm(20*b)
[1] 0.9101742 0.3841854 1.6821761*
I was expecting the 4 cases should do the same job--generate 4 random
numbers. But in case 2 and 4 I only get 3. Has anyone else seen this problem
before? Thanks. (I have tried with other functions i.e rchisq,rexp ...)
###
One of my colleague also have a problem that we think it might be related
with the problem I addressed above:
test1 - runif(10,0,1)
test1
[1] 0.3868379 0.1587814 0.8140483 0.7796691 0.5357628 0.2431110 0.1782747
0.3906829 0.5262615 0.7440143
test2 - NULL
for(i in seq(0.01,1,length=100)){
+ test2[i*100] - sum(test1i)
+ }
test2
[1] 0 0 0 0 0 0 *NA* 0 0 0 0 0 0 0 0 1 1 2 2 2 2 2 2
2 3 3 3 3 3 3 3 3 3 3 3 3 3 3 4 5 5 5 5 5
[45] 5 5 5 5 5 5 5 5 6 7 7 7 7 7 7 7 7 7 7 7 7 7 7
7 7 7 7 7 7 7 8 8 8 9 9 9 9 10 10 10 10 10 10 10
[89] 10 10 10 10 10 10 10 10 10 10 10 10
Every time he re-runs the code there always always a NA(highlighted). Does
any one know why? Your help is greatly appreciated.
Xiaofeng
--
Sarah Goslee
http://www.functionaldiversity.org
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Re: [R] protentially serious R error
[See in-line below]
On 30-Nov-2012 21:05:23 liuxf wrote:
Hi guy,
I have recently encountered a problem while I was just trying to generate
some random numbers with the function rnorm, the problem is shown below:
case 1
rnorm(20*0.2)
[1] -1.2765922 -0.5732654 -1.2246126 -0.4734006
case 2###
* rnorm(20*(1-0.8))
[1] -0.62036668 0.04211587 -0.91092165*
The key to this case can be seen in:
20*0.2 - 4
# [1] 0
20*(1-0.8)-4
# [1] -8.881784e-16
0.2 - (1-0.8)
# [1] 5.551115e-17
So you have fallen victim to the fact that, in general, floating-point
arithmetic is not exact (and this is not a feature only of R, but
of other packages that use standard floating-point CPU arithmetic.
#case 3
a-0.2
rnorm(20*a)
[1] 0.1580288 -0.6545846 1.7672873 0.7167075
#case 4#
* b-1-0.8
rnorm(20*b)
[1] 0.9101742 0.3841854 1.6821761*
I was expecting the 4 cases should do the same job--generate 4 random
numbers. But in case 2 and 4 I only get 3. Has anyone else seen this problem
before? Thanks. (I have tried with other functions i.e rchisq,rexp ...)
###
One of my colleague also have a problem that we think it might be related
with the problem I addressed above:
test1 - runif(10,0,1)
test1
[1] 0.3868379 0.1587814 0.8140483 0.7796691 0.5357628 0.2431110 0.1782747
0.3906829 0.5262615 0.7440143
test2 - NULL
for(i in seq(0.01,1,length=100)){
+ test2[i*100] - sum(test1i)
+ }
test2
[1] 0 0 0 0 0 0 *NA* 0 0 0 0 0 0 0 0 1 1 2 2 2 2 2 2
2 3 3 3 3 3 3 3 3 3 3 3 3 3 3 4 5 5 5 5 5
[45] 5 5 5 5 5 5 5 5 6 7 7 7 7 7 7 7 7 7 7 7 7 7 7
7 7 7 7 7 7 7 8 8 8 9 9 9 9 10 10 10 10 10 10 10
[89] 10 10 10 10 10 10 10 10 10 10 10 10
Every time he re-runs the code there always always a NA(highlighted). Does
any one know why? Your help is greatly appreciated.
Xiaofeng
This is effectovely the same issue:
test1 - runif(10,0,1)
test1
# [1] 0.3868379 0.1587814 0.8140483 0.7796691 0.5357628 0.2431110
# 0.1782747 0.3906829 0.5262615 0.7440143
test2 - NULL
for(i in seq(0.01,1,length=100)){
test2[i*100] - sum(test1i)
}
test2
# [1] 0 0 0 0 0 0 *NA* 0 0 0 0 0 0 0 0 1 1 2 2
# 2 2 2 2 2 3 3 3 3 3 3 3 3 3 3 3 3 3 3 4 5 5
# 5 5 5 5 5 5 5 5 5 5 5 6 7 7 7 7 7 7 7 7 7 7
# 7 7 7 7 7 7 7 7 7 7 7 8 8 8 9 9 9 9 10 10 10 10
# 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10
S - seq(0.01,1,length=100))
100*S[7] - 7
# [1] -8.881784e-16
which(100*S (1:100))
# [1] 7
Why not simply use expressions which evaluate to exact integers?
There is no obvious reason in the examples given to do otherwise.
But, if you must adopt your approach, then consider:
round(100*S)
100*S - (1:100)
# [1] 0.00e+00 0.00e+00 0.00e+00 0.00e+00
# [5] 0.00e+00 8.881784e-16 -8.881784e-16 0.00e+00
# [9] 0.00e+00 0.00e+00 0.00e+00 0.00e+00
# [13] 0.00e+00 1.776357e-15 1.776357e-15 0.00e+00
# [17] 0.00e+00 3.552714e-15 0.00e+00 0.00e+00
# [etc ]
round(100*S) - (1:100)
# [1] 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
# [26] 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
# [51] 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
# [76] 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
Hoping this helps,
Ted.
-
E-Mail: (Ted Harding) ted.hard...@wlandres.net
Date: 30-Nov-2012 Time: 23:26:41
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