Re: [R] How to calculate the spatial correlation of several files?
So where is the final correlation map Can we write it: to.write = file(paste(C:\\Users\\aalyaari\\desktop\\corr1.bin,sep=),wb) writeBin(as.double(results[[.f]]), to.write, size = 4) -- View this message in context: http://r.789695.n4.nabble.com/How-to-calculate-the-spatial-correlation-of-several-files-tp4651888p4652004.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] xlsx file read in R
Hi Nico, please let me know the details of sessionInfo() in R. Also, what version of Java are you running? If you could post the output of java -version from the command line that would be great. Note that XLConnect requires at least Java 1.6. Best regards, Martin -- View this message in context: http://r.789695.n4.nabble.com/xlsx-file-read-in-R-tp4651829p4652003.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] question about sum of (column) elements in R
Hi, I have the following data: 0 12 1 10 1 4 1 6 1 7 1 13 2 21 2 23 2 20 3 18 3 17 3 16 3 27 3 33 4 11 4 8 4 19 4 16 4 9 In this data file I would like to sum the numbers of second column which belong to the same number in the first column. So the output would be: 0 12 1 40 2 64 3 111 etc. Thank you. Kind regards, T. Bal [[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] question about sum of (column) elements in R
Hi, You can follow this example: test - structure(list(V1 = c(0L, 1L, 1L, 1L, 1L, 1L, 2L, 2L, 2L, 3L, 3L, 3L, 3L, 3L, 4L, 4L, 4L, 4L, 4L), V2 = c(12L, 10L, 4L, 6L, 7L, 13L, 21L, 23L, 20L, 18L, 17L, 16L, 27L, 33L, 11L, 8L, 19L, 16L, 9L)), .Names = c(V1, V2), class = data.frame, row.names = c(NA,-19L)) tapply(grid1$V2, grid1$V1, sum) 0 1 2 3 4 12 40 64 111 63 HTH Pascal Le 04/12/2012 16:59, T Bal a écrit : Hi, I have the following data: 0 12 1 10 1 4 1 6 1 7 1 13 2 21 2 23 2 20 3 18 3 17 3 16 3 27 3 33 4 11 4 8 4 19 4 16 4 9 In this data file I would like to sum the numbers of second column which belong to the same number in the first column. So the output would be: 0 12 1 40 2 64 3 111 etc. Thank you. Kind regards, T. Bal [[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] problem with factor levels
Hi I have a data.frame with 371,718 obs. of 12 variables (see below for an str). My problem is with V1, a Factor w/ 93144 levels, there should actually be 93994 levels. Each entry looks like: comp[number]_c[number]_seq[number] for example comp215489_c0_seq40 R is grouping as though the last number is a decimal for some reason, in other words comp215489_c0_seq40 and comp215489_c0_seq4 are considered to be the same. My problem is that they are not the same so when I group by this factor I am losing 800 levels. Here is an str 'data.frame': 371718 obs. of 12 variables: $ V1 : Factor w/ 93144 levels comp10_c0_seq1,..: 92271 91685 29 30 1564 1564 1623 91700 91701 91848 ... $ V2 : Factor w/ 17162 levels gi|345842331|ref|NM_001244016.1|,..: 10119 10779 13210 13210 11522 8115 13079 14493 14493 15858 ... $ V3 : num 95.5 90.2 98.7 99.2 81.4 ... $ V4 : int 335 153 237 122 258 127 306 258 120 177 ... $ V5 : int 15 15 3 1 38 19 20 23 5 9 ... $ V6 : int 0 0 0 0 4 2 0 0 0 0 ... $ V7 : int 1 45 1 43 1 129 1 54 1 70 ... $ V8 : int 335 197 237 164 254 254 306 311 120 246 ... $ V9 : int 6866 18 3172 3438 67 122 3927 42 346 195 ... $ V10: int 7200 170 3408 3559 318 247 4232 299 465 19 ... $ V11: num 7e-155 2e-46 4e-125 2e-61 3e-24 ... $ V12: num 545 184 446 234 111 69.9 448 329 198 280 .. -- View this message in context: http://r.789695.n4.nabble.com/problem-with-factor-levels-tp4652006.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] problem with factor levels
Le mardi 04 décembre 2012 à 00:34 -0800, Jeremy.Shearman a écrit : Hi I have a data.frame with 371,718 obs. of 12 variables (see below for an str). My problem is with V1, a Factor w/ 93144 levels, there should actually be 93994 levels. Each entry looks like: comp[number]_c[number]_seq[number] for example comp215489_c0_seq40 R is grouping as though the last number is a decimal for some reason, in other words comp215489_c0_seq40 and comp215489_c0_seq4 are considered to be the same. My problem is that they are not the same so when I group by this factor I am losing 800 levels. What format is your original data using? How do you import it? Please provide us with an excerpt of your original file showing at least two different values of V1 that are considered the same once imported in R (which sounds very unlikely to me...). Regards Here is an str 'data.frame': 371718 obs. of 12 variables: $ V1 : Factor w/ 93144 levels comp10_c0_seq1,..: 92271 91685 29 30 1564 1564 1623 91700 91701 91848 ... $ V2 : Factor w/ 17162 levels gi|345842331|ref|NM_001244016.1|,..: 10119 10779 13210 13210 11522 8115 13079 14493 14493 15858 ... $ V3 : num 95.5 90.2 98.7 99.2 81.4 ... $ V4 : int 335 153 237 122 258 127 306 258 120 177 ... $ V5 : int 15 15 3 1 38 19 20 23 5 9 ... $ V6 : int 0 0 0 0 4 2 0 0 0 0 ... $ V7 : int 1 45 1 43 1 129 1 54 1 70 ... $ V8 : int 335 197 237 164 254 254 306 311 120 246 ... $ V9 : int 6866 18 3172 3438 67 122 3927 42 346 195 ... $ V10: int 7200 170 3408 3559 318 247 4232 299 465 19 ... $ V11: num 7e-155 2e-46 4e-125 2e-61 3e-24 ... $ V12: num 545 184 446 234 111 69.9 448 329 198 280 .. -- View this message in context: http://r.789695.n4.nabble.com/problem-with-factor-levels-tp4652006.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] question about sum of (column) elements in R
Hi, T. Bal, homework? Take a look at ?tapply Regards -- Gerrit On Tue, 4 Dec 2012, T Bal wrote: Hi, I have the following data: 0 12 1 10 1 4 1 6 1 7 1 13 2 21 2 23 2 20 3 18 3 17 3 16 3 27 3 33 4 11 4 8 4 19 4 16 4 9 In this data file I would like to sum the numbers of second column which belong to the same number in the first column. So the output would be: 0 12 1 40 2 64 3 111 etc. Thank you. Kind regards, T. Bal [[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] partial analisys of a time series
Dear list members I want to analyze separately the months of a time series. In other words, I want to plot and fit models for each month separately. Taking the example of http://a-little-book-of-r-for-time-series.readthedocs.org/en/latest/src/timeseries.html births - scan(http://robjhyndman.com/tsdldata/data/nybirths.dat;) birthstimeseries - ts(births, frequency=12, start=c(1946,1)) birthstimeseries plot.ts(birthstimeseries) birthstimeseriesHW - HoltWinters(birthstimeseries) plot(birthstimeseriesHW) How to proceed the plotting and HoltWinters smoothing considereing only Januarys, Februarys, etc. separately. Thanks in advance. Antonio Olinto [[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] problem with factor levels
Hi -Original Message- From: r-help-boun...@r-project.org [mailto:r-help-bounces@r- project.org] On Behalf Of Jeremy.Shearman Sent: Tuesday, December 04, 2012 9:35 AM To: r-help@r-project.org Subject: [R] problem with factor levels Hi I have a data.frame with 371,718 obs. of 12 variables (see below for an str). My problem is with V1, a Factor w/ 93144 levels, there should actually be 93994 levels. Each entry looks like: comp[number]_c[number]_seq[number] for example comp215489_c0_seq40 R is grouping as though the last number is a decimal for some reason, in other words comp215489_c0_seq40 and comp215489_c0_seq4 are considered to be the same. My problem is that they are not the same so when I group by this factor I am losing 800 levels. Hm. How did you constructed those factors? factor(c(comp215489_c0_seq40, comp215489_c0_seq4) ) [1] comp215489_c0_seq40 comp215489_c0_seq4 Levels: comp215489_c0_seq4 comp215489_c0_seq40 gives me 2 levels as expected. I also doubt that R will do such stripping during reading from other file. Regards Petr Here is an str 'data.frame': 371718 obs. of 12 variables: $ V1 : Factor w/ 93144 levels comp10_c0_seq1,..: 92271 91685 29 30 1564 1564 1623 91700 91701 91848 ... $ V2 : Factor w/ 17162 levels gi|345842331|ref|NM_001244016.1|,..: 10119 10779 13210 13210 11522 8115 13079 14493 14493 15858 ... $ V3 : num 95.5 90.2 98.7 99.2 81.4 ... $ V4 : int 335 153 237 122 258 127 306 258 120 177 ... $ V5 : int 15 15 3 1 38 19 20 23 5 9 ... $ V6 : int 0 0 0 0 4 2 0 0 0 0 ... $ V7 : int 1 45 1 43 1 129 1 54 1 70 ... $ V8 : int 335 197 237 164 254 254 306 311 120 246 ... $ V9 : int 6866 18 3172 3438 67 122 3927 42 346 195 ... $ V10: int 7200 170 3408 3559 318 247 4232 299 465 19 ... $ V11: num 7e-155 2e-46 4e-125 2e-61 3e-24 ... $ V12: num 545 184 446 234 111 69.9 448 329 198 280 .. -- View this message in context: http://r.789695.n4.nabble.com/problem- with-factor-levels-tp4652006.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] problem with factor levels
Oh, your skepticism was spot on! I was using excel to check the output (silly, but I am still in the process of moving from excel to R) and there was a discrepancy in the number of output from R and excel. Turns out the problem was with excel and not with R at all. That's a relief. SOLVED -- View this message in context: http://r.789695.n4.nabble.com/problem-with-factor-levels-tp4652006p4652019.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] computing marginal values based on multiple columns?
Hello all,
I have what feels like a simple problem, but I can't find an simple
answer. Consider this data frame:
x - data.frame(sample1=c(35,176,182,193,124),
sample2=c(198,176,190,23,15), sample3=c(12,154,21,191,156),
class=c('a','a','c','b','c'))
x
sample1 sample2 sample3 class
1 35 198 12 a
2 176 176 154 a
3 182 190 21 c
4 193 23 191 b
5 124 15 156 c
Now I wish to know: for each sample, for values 20% of the sample mean,
what percentage of those are class a?
I want to end up with a table like:
sample1 sample2 sample3
1 1.0 0 0.5
I can calculate this for an individual sample using this rather clumsy
expression:
length(which(x$sample1 mean(x$sample1) x$class=='a')) /
length(which(x$sample1 mean(x$sample1)))
I'd normally propagate it across the data frame using apply, but I
can't because it depends on more than one column.
Any help much appreciated!
Cheers,
Simon
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Re: [R] partial analisys of a time series
Hi -Original Message- From: r-help-boun...@r-project.org [mailto:r-help-bounces@r- project.org] On Behalf Of Antonio Silva Sent: Tuesday, December 04, 2012 10:26 AM To: R-help@r-project.org Subject: [R] partial analisys of a time series Dear list members I want to analyze separately the months of a time series. In other words, I want to plot and fit models for each month separately. Taking the example of http://a-little-book-of-r-for-time- series.readthedocs.org/en/latest/src/timeseries.html births - scan(http://robjhyndman.com/tsdldata/data/nybirths.dat;) birthstimeseries - ts(births, frequency=12, start=c(1946,1)) birthstimeseries plot.ts(birthstimeseries) birthstimeseriesHW - HoltWinters(birthstimeseries) plot(birthstimeseriesHW) How to proceed the plotting and HoltWinters smoothing considereing only Januarys, Februarys, etc. separately. Split your data by months to a list, use lapply. using zoo package blist -split(birthstimeseries, months(as.Date(birthstimeseries))) l.blist - lapply(blist, HoltWinters) Regards Petr Thanks in advance. Antonio Olinto [[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] problem with factor levels
Hi That is quite usual. Excel is so widespread that almost everybody assumes it shall not contain mistakes and behaves correctly. The contrary is true. Spreadsheet often guess what user have on mind and corrects values to fit such assumption, let aside mistakes in coded functions. R expects it is used by clever and able people and performs just what you tell it to do, not more not less. So whenever result does not fit your expectations, first proof your expectations. Regards Petr -Original Message- From: r-help-boun...@r-project.org [mailto:r-help-bounces@r- project.org] On Behalf Of Jeremy.Shearman Sent: Tuesday, December 04, 2012 10:38 AM To: r-help@r-project.org Subject: Re: [R] problem with factor levels Oh, your skepticism was spot on! I was using excel to check the output (silly, but I am still in the process of moving from excel to R) and there was a discrepancy in the number of output from R and excel. Turns out the problem was with excel and not with R at all. That's a relief. SOLVED -- View this message in context: http://r.789695.n4.nabble.com/problem- with-factor-levels-tp4652006p4652019.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] question about sum of (column) elements in R
On 04-12-2012, at 08:59, T Bal wrote: Hi, I have the following data: 0 12 1 10 1 4 1 6 1 7 1 13 2 21 2 23 2 20 3 18 3 17 3 16 3 27 3 33 4 11 4 8 4 19 4 16 4 9 In this data file I would like to sum the numbers of second column which belong to the same number in the first column. So the output would be: 0 12 1 40 2 64 3 111 etc. Using Pascal's test object you can also use aggregate aggregate(test$V2,by=list(test$V1),FUN=sum) 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] question about sum of (column) elements in R
Hi, Imagine the data you have is in a data frame, c, with columns a and b. Then you can do this: res=tapply(b,c[,-2],sum) res[is.na(res)]-0 res 0 1 2 3 4 12 40 64 111 63 Hope it helps, 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 T Bal Sent: 04 December 2012 07:59 To: r-help@r-project.org Subject: [R] question about sum of (column) elements in R Hi, I have the following data: 0 12 1 10 1 4 1 6 1 7 1 13 2 21 2 23 2 20 3 18 3 17 3 16 3 27 3 33 4 11 4 8 4 19 4 16 4 9 In this data file I would like to sum the numbers of second column which belong to the same number in the first column. So the output would be: 0 12 1 40 2 64 3 111 etc. Thank you. Kind regards, T. Bal [[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.
Re: [R] partial analisys of a time series
Thanks Petr I thought there might be an equivalent for birthstimeseries[,1] if it were a dataframe, but split function sounds great. I could not reproduce the second line of your suggestion l.blist - lapply(blist, HoltWinters). I receive the message: Error in decompose(ts(x[1L:wind], start = start(x), frequency = f), seasonal) : time series has no or less than 2 periods What could be going wrong? Best regards Antonio 2012/12/4 PIKAL Petr petr.pi...@precheza.cz Hi -Original Message- From: r-help-boun...@r-project.org [mailto:r-help-bounces@r- project.org] On Behalf Of Antonio Silva Sent: Tuesday, December 04, 2012 10:26 AM To: R-help@r-project.org Subject: [R] partial analisys of a time series Dear list members I want to analyze separately the months of a time series. In other words, I want to plot and fit models for each month separately. Taking the example of http://a-little-book-of-r-for-time- series.readthedocs.org/en/latest/src/timeseries.html births - scan(http://robjhyndman.com/tsdldata/data/nybirths.dat;) birthstimeseries - ts(births, frequency=12, start=c(1946,1)) birthstimeseries plot.ts(birthstimeseries) birthstimeseriesHW - HoltWinters(birthstimeseries) plot(birthstimeseriesHW) How to proceed the plotting and HoltWinters smoothing considereing only Januarys, Februarys, etc. separately. Split your data by months to a list, use lapply. using zoo package blist -split(birthstimeseries, months(as.Date(birthstimeseries))) l.blist - lapply(blist, HoltWinters) Regards Petr Thanks in advance. Antonio Olinto [[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. -- Antônio Olinto Ávila da Silva Biólogo / Oceanógrafo Instituto de Pesca (Fisheries Institute) São Paulo, Brasil [[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] computing marginal values based on multiple columns?
Hello, Simon,
see below!
On Tue, 4 Dec 2012, Simon wrote:
Hello all,
I have what feels like a simple problem, but I can't find an simple
answer. Consider this data frame:
x - data.frame(sample1=c(35,176,182,193,124),
sample2=c(198,176,190,23,15), sample3=c(12,154,21,191,156),
class=c('a','a','c','b','c'))
x
sample1 sample2 sample3 class
1 35 198 12 a
2 176 176 154 a
3 182 190 21 c
4 193 23 191 b
5 124 15 156 c
Now I wish to know: for each sample, for values 20% of the sample mean,
what percentage of those are class a?
I want to end up with a table like:
sample1 sample2 sample3
1 1.0 0 0.5
I can't reproduce this result from your description above, but if I
understand the latter correctly, maybe the following does what you want:
x.wo.class - subset( x, select = -class)
# extract only the sample-columns
x.small.and.a - x.wo.class 0.2 * colMeans( x.wo.class) x$class == a
apply( x.small.and.a, 2, function( xx) mean( x$class[ xx] == a))
Hth -- Gerrit
I can calculate this for an individual sample using this rather clumsy
expression:
length(which(x$sample1 mean(x$sample1) x$class=='a')) /
length(which(x$sample1 mean(x$sample1)))
I'd normally propagate it across the data frame using apply, but I
can't because it depends on more than one column.
Any help much appreciated!
Cheers,
Simon
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and provide commented, minimal, self-contained, reproducible code.
Re: [R] Resampling Help Needed
You can use, 'sample' function for sampling and may consider using partition clustering for selecting your regions, see Cluster task view: http://cran.r-project.org/web/views/Cluster.html On 4 December 2012 00:53, KoopaTrooper ncoop...@tulane.edu wrote: I am using package ks() to build 3D representations of bird territories and calculate territory volume from spatial data (simply x, y, and z coordinates). What I want to do is determine at what sample size (# locations collected) does the territory volume stop increasing. This should give me an idea of the number of points needed for future seasons. So I have a couple of birds each with 200 spatial locations (x,y,z). I want to run the following code (see below), but have R calculate territory size 100 times with 10 random points (no replacement), 100 times with 20 random points, 100 times with 30 random points, etc. I can figure out how to do this manually (i.e. create 100 individual files with 10 random points, 20 random points, etc.) but I figure there must be a way to make my life easier. Any help would be appreciated. Even pointing me in the correct direction would be a big help. Thanks! Nathan #read data files (.csv's with 200 rows of x,y,z coordinates) a-read.csv(A_PW_ASY_M_LII_2011.csv) #calls the plug-in bandwidth estimator Ha - Hpi(a) #sets min/max grid size for each dimension minX-min(a$X)-25 minY-min(a$Y)-25 minZ-0 maxX-max(a$X)+25 maxY-max(a$Y)+25 maxZ-max(a$Z)+5 #creates kernel utilization distribution fhata - kde(x=a, H=Ha, binned=FALSE, xmin=c(minX,minY,minZ), xmax=c(maxX,maxY,maxZ)) #calculates territory volume at 95% isopleth Vol95-contourSizes(fhata, cont=95) -- View this message in context: http://r.789695.n4.nabble.com/Resampling-Help-Needed-tp4651973.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] Réponse automatique
Bonjour, Je serais en congés jusqu'au Jeudi 6 Décembre. Pour des raisons d'urgence, vous pourrez me contacter par téléphone au 06 46 34 81 03. Cordialement, -- Jérôme Boutet Conservatoire d'espaces naturels de Picardie 1, place Ginkgo - village Oasis 80 044 AMIENS cedex tél : 03 22 89 84 24 From: r-help-requ...@r-project.org Subject: R-help Digest, Vol 118, Issue 4 Date: Tue, 04 Dec 2012 12:00:10 +0100 __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] Forest plot
At 20:39 03/12/2012, Min Dong wrote: Hi, I am a novice in R. It will be greatly appreciated if someone can advise me with the following questions. There are at least three packages available from CRAN (meta, metafor, rmeta) which draw forest plots so it would help us if you had told us which one you are using. 1) How to highlight reference range in forest plot? For example, if 1.5-2 is the reference range, I would like to have all the area between 1.5-2 to be highlighted (such as in grey color). Are you really using a forest plot for a meta-analysis or for some other purpose? 2) If I have handunds of objects, how to set the output plot to multiple columns? For example, I have 500 objects included, and I want to have objects 1-100 to the left (column #1), 101-200 in column#2 (next to column#1)...etc, ect, how to do it? It would be unusual to have a meta-analysis with so many studies included so I assume you are doing something else. Thank you very much! Mindy [[alternative HTML version deleted]] Michael Dewey i...@aghmed.fsnet.co.uk http://www.aghmed.fsnet.co.uk/home.html __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] partial analisys of a time series
Try period.apply() from the xts package. MW On Tue, Dec 4, 2012 at 9:26 AM, Antonio Silva aolinto@gmail.com wrote: Dear list members I want to analyze separately the months of a time series. In other words, I want to plot and fit models for each month separately. Taking the example of http://a-little-book-of-r-for-time-series.readthedocs.org/en/latest/src/timeseries.html births - scan(http://robjhyndman.com/tsdldata/data/nybirths.dat;) birthstimeseries - ts(births, frequency=12, start=c(1946,1)) birthstimeseries plot.ts(birthstimeseries) birthstimeseriesHW - HoltWinters(birthstimeseries) plot(birthstimeseriesHW) How to proceed the plotting and HoltWinters smoothing considereing only Januarys, Februarys, etc. separately. Thanks in advance. Antonio Olinto [[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] partial analisys of a time series
Hi I am not an expert in time series. The problem is that resulting list is not time series any more. So you need to convert it again to time series and you need to give it frequency greater than 1. something like l.blist - lapply(blist, function (x) HoltWinters(ts(x, frequency=2))) But this depends on how your data are structured after splitting them to months. Regards Petr From: Antonio Silva [mailto:aolinto@gmail.com] Sent: Tuesday, December 04, 2012 11:50 AM To: PIKAL Petr Cc: R-help@r-project.org Subject: Re: [R] partial analisys of a time series Thanks Petr I thought there might be an equivalent for birthstimeseries[,1] if it were a dataframe, but split function sounds great. I could not reproduce the second line of your suggestion l.blist - lapply(blist, HoltWinters). I receive the message: Error in decompose(ts(x[1L:wind], start = start(x), frequency = f), seasonal) : time series has no or less than 2 periods What could be going wrong? Best regards Antonio 2012/12/4 PIKAL Petr petr.pi...@precheza.czmailto:petr.pi...@precheza.cz Hi -Original Message- From: r-help-boun...@r-project.orgmailto:r-help-boun...@r-project.org [mailto:r-help-bounces@r-mailto:r-help-bounces@r- project.orghttp://project.org] On Behalf Of Antonio Silva Sent: Tuesday, December 04, 2012 10:26 AM To: R-help@r-project.orgmailto:R-help@r-project.org Subject: [R] partial analisys of a time series Dear list members I want to analyze separately the months of a time series. In other words, I want to plot and fit models for each month separately. Taking the example of http://a-little-book-of-r-for-time- series.readthedocs.org/en/latest/src/timeseries.htmlhttp://series.readthedocs.org/en/latest/src/timeseries.html births - scan(http://robjhyndman.com/tsdldata/data/nybirths.dat;) birthstimeseries - ts(births, frequency=12, start=c(1946,1)) birthstimeseries plot.ts(birthstimeseries) birthstimeseriesHW - HoltWinters(birthstimeseries) plot(birthstimeseriesHW) How to proceed the plotting and HoltWinters smoothing considereing only Januarys, Februarys, etc. separately. Split your data by months to a list, use lapply. using zoo package blist -split(birthstimeseries, months(as.Date(birthstimeseries))) l.blist - lapply(blist, HoltWinters) Regards Petr Thanks in advance. Antonio Olinto [[alternative HTML version deleted]] __ R-help@r-project.orgmailto:R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting- guide.html and provide commented, minimal, self-contained, reproducible code. -- Antônio Olinto Ávila da Silva Biólogo / Oceanógrafo Instituto de Pesca (Fisheries Institute) São Paulo, Brasil [[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] partial analisys of a time series
HI, You can subset by: birthstimeseriesJan-subset(birthstimeseries,cycle(birthstimeseries)==1) A.K. - Original Message - From: Antonio Silva aolinto@gmail.com To: R-help@r-project.org Cc: Sent: Tuesday, December 4, 2012 4:26 AM Subject: [R] partial analisys of a time series Dear list members I want to analyze separately the months of a time series. In other words, I want to plot and fit models for each month separately. Taking the example of http://a-little-book-of-r-for-time-series.readthedocs.org/en/latest/src/timeseries.html births - scan(http://robjhyndman.com/tsdldata/data/nybirths.dat;) birthstimeseries - ts(births, frequency=12, start=c(1946,1)) birthstimeseries plot.ts(birthstimeseries) birthstimeseriesHW - HoltWinters(birthstimeseries) plot(birthstimeseriesHW) How to proceed the plotting and HoltWinters smoothing considereing only Januarys, Februarys, etc. separately. Thanks in advance. Antonio Olinto [[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] Using multicores in R
Thanks for the help,
Perhaps I should elaborate a bit, I am working on bioinformatics project in
which I am trying to run a forward selection algorithm for machine learning
classification of two biological conditions.
At each iteration I want to find the gene that in addition to those I have
found already does the best classification.
It looks something like this:
for (j in 1:5030)
{
tp - 0;
for (i in 1:5030)
{
if (!(i %in% idx))
{
classifier-naiveBayes(trn[,c(i,idx)], trn[,20118])
tbl -table(predict(classifier, trn[,-20118]), trn[,20118])
success - (tbl[[1]] +tbl[[4]])/(tbl[[1]] +tbl[[4]]+tbl[[2]]+tbl[[3]])
if (success tp)
{
tp - success
ind - i
gene - names(trn)[i]
}
}
}
idx - c(idx,ind)
res - rbind(res, data.frame(Iteration=j,Success=tp*100,Gene=gene))
}
I am no expert when it comes to programming so I am not sure how can I
optimize my relatively primitive code in the best way...
Thanks,
Moriah
--
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[R] do.call
Hello, I have a problem with the do.call-function. I would like to merge the values of more than 30 columns, but not in all of the rows exist values, so with this commando i get a lot of ; or NA. How get i only the merge of cells with a number? datos$NEW - do.call(paste, c(datos[,19:53], sep = ;)) $ NEW : chr 218.0NA;;;NA;;;NA;NA;NA;NA;NA I hope you can help me. Thanks! Best regards, Dominic __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] question about sum of (column) elements in R
HI, You can use either ?tapply(), ?aggregate(), ?ddply() from library(plyr) dat1-read.table(text= 0 12 1 10 1 4 1 6 1 7 1 13 2 21 2 23 2 20 3 18 3 17 3 16 3 27 3 33 4 11 4 8 4 19 4 16 4 9 ,sep=,header=FALSE) with(dat1,aggregate(dat1[,2],by=list(V1=dat1[,1]),sum)) # V1 x #1 0 12 #2 1 40 #3 2 64 #4 3 111 #5 4 63 A.K. - Original Message - From: T Bal studentt...@gmail.com To: r-help@r-project.org Cc: Sent: Tuesday, December 4, 2012 2:59 AM Subject: [R] question about sum of (column) elements in R Hi, I have the following data: 0 12 1 10 1 4 1 6 1 7 1 13 2 21 2 23 2 20 3 18 3 17 3 16 3 27 3 33 4 11 4 8 4 19 4 16 4 9 In this data file I would like to sum the numbers of second column which belong to the same number in the first column. So the output would be: 0 12 1 40 2 64 3 111 etc. Thank you. Kind regards, T. Bal [[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] Chi-squared test when observed near expected
When you typed x as a command, R runs the command print(x). That function produces a summary of the results which may include round off numbers to a few decimal places to make them more readable. When you typed x$statistic, you got the unrounded version of the result 5.6e-31 which I think you will agree is pretty close to 0. -- David L Carlson Associate Professor of Anthropology Texas AM University College Station, TX 77843-4352 -Original Message- From: r-help-boun...@r-project.org [mailto:r-help-bounces@r- project.org] On Behalf Of Troy S Sent: Monday, December 03, 2012 3:41 PM To: r-help@r-project.org Subject: [R] Chi-squared test when observed near expected Dear UseRs, I'm running a chi-squared test where the expected matrix is the same as the observed, after rounding. R reports a X-squared of zero with a p value of one. I can justify this because any other result will deviate at least as much from the expected because what we observe is the expected, after rounding. But the formula for X-squared, sum (O-E)^2/E gives a positive value. What is the reason for X-Squared being zero in this case? Troy trial-as.table(matrix(c(26,16,13,7),ncol=2)) x-chisq.test(trial) x data: trial X-squared = 0, df = 1, p-value = 1 x$expected A B A 26.41935 12.580645 B 15.58065 7.419355 x$statistic X-squared 5.596653e-31 (x$observed-x$expected)^2/x$expected A B A 0.006656426 0.013978495 B 0.011286983 0.023702665 sum((x$observed-x$expected)^2/x$expected) [1] 0.05562457 [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting- guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] julia language unfair comparisons against R
Hello List, Probably many of you aware of the Julia language (http://julialang.org/), It is a promising project. However it seems like R is very slow in their benchmarks. Very important point they omit, they did not use R's own JIT ! I had a feeling that R is mistreaded there :) Also another important point is that they all use for-loops in R instead of vectorized code! Any thought on this? Are there any other improvement one can do with compiler package other then 'cmpfun'? Best, -m __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] How do I make R randomForest model size smaller?
Try the following:
set.seed(100)
rf1 - randomForest(Species ~ ., data=iris)
set.seed(100)
rf2 - randomForest(iris[1:4], iris$Species)
object.size(rf1)
object.size(rf2)
str(rf1)
str(rf2)
You can try it on your own data. That should give you some hints about why the
formula interface should be avoided with large datasets.
Andy
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
Behalf Of John Foreman
Sent: Monday, December 03, 2012 3:43 PM
To: r-help@r-project.org
Subject: [R] How do I make R randomForest model size smaller?
I've been training randomForest models on 7 million rows of data (41
features). Here's an example call:
myModel - randomForest(RESPONSE~., data=mydata, ntree=50, maxnodes=30)
I thought surely with only 50 trees and 30 terminal nodes that the memory
footprint of myModel would be small. But it's 65 megs in a dump file. The
object seems to be holding all sorts of predicted, actual, and vote data
from the training process.
What if I just want the forest and that's it? I want a tiny dump file that
I can load later to make predictions off of quickly. I feel like the forest
by itself shouldn't be all that large...
Anyone know how to strip this sucker down to just something I can make
predictions off of going forward?
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Re: [R] julia language unfair comparisons against R
There been, that done. http://stackoverflow.com/questions/9968578/speeding-up-julias-poorly-written-r-examples/10712158#10712158 MW On Tue, Dec 4, 2012 at 2:34 PM, Suzen, Mehmet msu...@gmail.com wrote: Hello List, Probably many of you aware of the Julia language (http://julialang.org/), It is a promising project. However it seems like R is very slow in their benchmarks. Very important point they omit, they did not use R's own JIT ! I had a feeling that R is mistreaded there :) Also another important point is that they all use for-loops in R instead of vectorized code! Any thought on this? Are there any other improvement one can do with compiler package other then 'cmpfun'? Best, -m __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] Subsetting columns in data.table
DT = data.frame(x=rep(c(a,b,c),each=3), y=c(1,3,6), v=1:9, w=3:11, z=LETTERS[1:9]) If I understand you right, and you want to select all rows where v3 and W10 with(DT, DT[which(v3 w10),]) x y v w z 4 b 1 4 6 D 5 b 3 5 7 E 6 b 6 6 8 F 7 c 1 7 9 G you can use colSums, rowSums on this, but after removing the categorical columns Anto -- View this message in context: http://r.789695.n4.nabble.com/Subsetting-columns-in-data-table-tp4652048p4652049.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] Winbugs from R
Hi,
I am trying to covert a Winbugs code into R code. Here is the winbugs code
model{# models likelihoodfor (i in 1:n){time[i] ~ dnorm( mu[i], tau ) #
stochastic componenent# link and linear predictormu[i] - beta0 + beta1 *
cases[i] + beta2 * distance[i]}# prior distributionstau ~ dgamma( 0.01, 0.01
)beta0 ~ dnorm( 0.0, 1.0E-4)beta1 ~ dnorm( 0.0, 1.0E-4)beta2 ~ dnorm( 0.0,
1.0E-4)# definition of sigmas2-1/taus -sqrt(s2)# calculation of the sample
variancefor (i in 1:n){ c.time[i]-time[i]-mean(time[]) }sy2 - inprod(
c.time[], c.time[] )/(n-1)# calculation of Bayesian version R squaredR2B - 1 -
s2/sy2# Expected y for a typical delivery timetypical.y - beta0 + beta1 *
mean(cases[]) + beta2 * mean(distance[])}INITSlist( tau=1, beta0=1, beta1=0,
beta2=0 )DATA (LIST)list( n=25,time = c(16.68, 11.5, 12.03, 14.88, 13.75,
18.11, 8, 17.83,79.24, 21.5, 40.33, 21, 13.5, 19.75, 24, 29, 15.35,19, 9.5,
35.1, 17.9, 52.32, 18.75, 19.83, 10.75),distance = c(560, 220, 340, 80, 150,
330, 110, 210, 1460,605, 688, 215, 255, 462, 448, 776, 200, 132,36, 770, 140,
810, 450, 635, 150),cases = c( 7, 3, 3, 4, 6, 7, 2, 7, 30, 5, 16, 10, 4, 6,
9,10, 6, 7, 3, 17, 10, 26, 9, 8, 4) )
I want to do this in R. So, I copied the model and pasted into a txt file named
reg. Here is the R code I used
time -
c(16.68,11.5,12.03,14.88,13.75,18.11,8,17.83,79.24,21.5,40.33,21,13.5,19.75,24,29,15.35,19,9.5,35.1,17.9,52.32,18.75,19.83,10.75)cases
- c(7,3,3,4,6,7,2,7,30,5,16,10,4,6,9,10,6,7,3,17,10,26,9,8,4)distance -
c(560,220,340,80,150,330,110,210,1260,605,688,215,255,462,448,776,200,132,36,770,140,810,450,635,150)
data - list(time,cases,distance)
inits - function(){list(tau=1,beta1=0,beta2=0,beta3=0)}
sim - bugs(data, inits, model.file =
C:/Users/Gunal/Desktop/dummy/reg.txt,parameters = c(beta1,
beta2,beta3),n.chains = 3, n.iter = 1000,bugs.directory =
D:/PROGRAMLAR/WinBUGS14/,debug=TRUE)
Winbugs is producing this error page
display(log)check(C:/Users/Gunal/Desktop/dummy/reg.txt)model is syntactically
correctdata(C:/Users/Gunal/AppData/Local/Temp/RtmpCYxtDZ/data.txt)data
loadedcompile(3)variable n is not
definedinits(1,C:/Users/Gunal/AppData/Local/Temp/RtmpCYxtDZ/inits1.txt)command
#Bugs:inits cannot be executed (is greyed
out)inits(2,C:/Users/Gunal/AppData/Local/Temp/RtmpCYxtDZ/inits2.txt)command
#Bugs:inits cannot be executed (is greyed
out)inits(3,C:/Users/Gunal/AppData/Local/Temp/RtmpCYxtDZ/inits3.txt)command
#Bugs:inits cannot be executed (is greyed out)gen.inits()command
#Bugs:gen.inits cannot be executed (is greyed
out)thin.updater(1)update(500)command #Bugs:update cannot be executed (is
greyed out)set(beta1)command #Bugs:set cannot be executed (is greyed
out)set(beta2)command #Bugs:set cannot be executed (is greyed
out)set(beta3)command #Bugs:set cannot be executed (is greyed
out)set(deviance)command #Bugs:set cannot be executed (is greyed
out)dic.set()command #Bugs:dic.set cannot be executed (is greyed
out)update(500)command #Bugs:update cannot be executed (is greyed
out)coda(*,C:/Users/Gunal/AppData/Local/Temp/RtmpCYxtDZ/coda)command #Bugs:coda
cannot be executed (is greyed out)stats(*)command #Bugs:stats cannot be
executed (is greyed out)dic.stats()
DIChistory(*,C:/Users/Gunal/AppData/Local/Temp/RtmpCYxtDZ/history.odc)command
#Bugs:history cannot be executed (is greyed
out)save(C:/Users/Gunal/AppData/Local/Temp/RtmpCYxtDZ/log.odc)save(C:/Users/Gunal/AppData/Local/Temp/RtmpCYxtDZ/log.txt)
Any help would be greatly appreciated.
Cheers
Gunal
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[R] Help for a function
Hello all,
I need a help.
I am modeling a disease and a create a R function like that:
Lambda-function (x,date1,r,h,a){
ndate1 - as.Date(date1, %d/%m/%Y)
t1 - as.numeric(ndate1)
x[order(x$i),]
t -x[,t]
i -x[,i]
CONTAGIEUX -x[,CONTAGIEUX]
while ( t1 min(t) ){
for (i in 1:length(i) ){
{for (j in 1:CONTAGIEUX[length(CONTAGIEUX)]){
res1[j] -(a*h)
res2 -sum( res1[j])
}
}
lambda[i] - r*res2
}
}
x-data.frame(x,lambda)
x
}
on such data :
DATEi Symptomes t Incubation CONTAGIEUX
1 2009-04-29 Canada 13 14363 13 13
2 2009-05-01 Israel 2 14365 2 2
3 2009-05-09 argentina 1 14373 1 1
5 2009-05-09 australia 1 14373 1 1
6 2009-05-10 australia 1 14374 2 2
7 2009-04-29 Austria 1 14363 1 1
8 2009-04-30 Austria 1 14364 2 2
9 2009-05-01 Austria 1 14365 2 3
10 2009-05-02 Austria 1 14366 2 4
11 2009-05-03 Austria 1 14367 2 5
17 2009-05-09 Austria 1 14373 2 7
18 2009-05-10 Austria 1 14374 2 7
19 2009-05-08 brasil 4 14372 4 4
20 2009-05-09 brazil 6 14373 6 6
21 2009-05-10 brazil 6 14374 12 12
22 2009-05-02 canada 51 14366 51 51
23 2009-05-03 canada 85 14367 136 136
24 2009-05-04 canada 101 14368 186 237
31 2009-04-27 Canada 6 14361 6 6
32 2009-04-28 Canada 6 14362 6 6
33 2009-04-30 Canada 19 14364 25 25
34 2009-05-01 Canada 34 14365 53 59
35 2009-05-01 China,HongKong, SAR 1 14365 1 1
36 2009-05-02 China,HongKong, SAR 1 14366 2 2
37 2009-05-03 China,HongKong, SAR 1 14367 2 3
38 2009-05-04 China,HongKong, SAR 1 14368 2 4
44 2009-05-10 China,HongKong, SAR 1 14374 2 7
45 2009-05-04 colombia1 14368 1 1
46 2009-05-05 colombia1 14369 2 2
47 2009-05-06 colombia
But i do not get the results,i try by all means but i d'ont understant the
problem.
Thanks for your help.
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Re: [R] r function definition
Are you using windows? If you are you may want to try to run your R code from a batch file: REM on Microsoft Windows (adjust the path to R.exe as needed) C:\Program Files\R\R-2.13.2\bin\x64\R.exe CMD BATCH C:\Users\Frank\Documents\R\Projects\Heinrich\Heinrich.txt C:\Users\Frank\Documents\R\Projects\Heinrich\Heinrich.out PAUSE Where ...Heinrich.txt is the following R code in a text file: pdf(Heinrich.pdf) ## load Pearson package library(PearsonDS) ## calculate probability density function Fig. 1 x - seq(0.0, +20.0, length=100) hx- dpearsonIV(x,m=2.25,nu=5,location=17,scale=2) k-0.0206631 plot(x, hx , type=l, lwd=2, tck=1, xlab=x value, ylab=Density, main=J. Heinrich Fig. 1) ## calculate probability density function Fig 2 hx- dpearsonIV(x,m=0.75,nu=1.0,location=15,scale=0.5) k-0.218991 plot(x, hx , type=l, lwd=2, tck=1, xlab=x value, ylab=Density, main=J. Heinrich Fig. 2) quit() and Heinrich.out will contain the output of the results of the commands. Look at Heinrich.out for results and errors. Then correct the text file and rerun. GL Frank Chicago Date: Mon, 3 Dec 2012 12:30:31 -0800 From: q...@hotmail.com To: r-help@r-project.org Subject: [R] r function definition I am a very new R user. I am trying to write functons and debug functions. One problem for me is that I need to alwasy copy the whole function body and resubmit to R console every time I changed even one line of the function. Because I have long algorithm function, copying and pasting is very tedious for me. I assume if I save the function files, R should be able to just use the new function body since it is a scripting language. Can somebody let me know his best practice of using R function? -- View this message in context: http://r.789695.n4.nabble.com/r-function-definition-tp4651943.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] help with if statement
DT-data.frame(time=c(0,1,5,24,36,48,72),DV=seq(0,60,10)) time DV 10 0 21 10 35 20 4 48 30 5 84 40 6 96 50 7 120 60 You want to add 24 to values that are =24 in 'time' DT[DT$time=24,'time']-DT[DT$time=24,'time']+24 time DV 10 0 21 10 35 20 4 48 30 5 60 40 6 72 50 7 96 60 Is this what you were looking for? A. -- View this message in context: http://r.789695.n4.nabble.com/help-with-if-statement-tp4652015p4652055.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] partial analisys of a time series
HI, I am getting an error message: l.blist-lapply(blist,HoltWinters) #Error in decompose(ts(x[1L:wind], start = start(x), frequency = f), seasonal) : # time series has no or less than 2 periods A.K. - Original Message - From: PIKAL Petr petr.pi...@precheza.cz To: Antonio Silva aolinto@gmail.com; R-help@r-project.org R-help@r-project.org Cc: Sent: Tuesday, December 4, 2012 5:07 AM Subject: Re: [R] partial analisys of a time series Hi -Original Message- From: r-help-boun...@r-project.org [mailto:r-help-bounces@r- project.org] On Behalf Of Antonio Silva Sent: Tuesday, December 04, 2012 10:26 AM To: R-help@r-project.org Subject: [R] partial analisys of a time series Dear list members I want to analyze separately the months of a time series. In other words, I want to plot and fit models for each month separately. Taking the example of http://a-little-book-of-r-for-time- series.readthedocs.org/en/latest/src/timeseries.html births - scan(http://robjhyndman.com/tsdldata/data/nybirths.dat;) birthstimeseries - ts(births, frequency=12, start=c(1946,1)) birthstimeseries plot.ts(birthstimeseries) birthstimeseriesHW - HoltWinters(birthstimeseries) plot(birthstimeseriesHW) How to proceed the plotting and HoltWinters smoothing considereing only Januarys, Februarys, etc. separately. Split your data by months to a list, use lapply. using zoo package blist -split(birthstimeseries, months(as.Date(birthstimeseries))) l.blist - lapply(blist, HoltWinters) Regards Petr Thanks in advance. Antonio Olinto [[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-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] do.call
Hi, Try this: set.seed(15) datos-as.data.frame(matrix(sample(c(1:20,NA),30,replace=TRUE),ncol=6)) do.call(paste,c(na.omit(datos),sep=;)) #[1] 5;18;14;10;17;3 14;15;15;3;2;20 8;18;19;17;12;11 A.K. - Original Message - From: Dominic Roye dominic.r...@gmail.com To: r-help@r-project.org Cc: Sent: Tuesday, December 4, 2012 7:38 AM Subject: [R] do.call Hello, I have a problem with the do.call-function. I would like to merge the values of more than 30 columns, but not in all of the rows exist values, so with this commando i get a lot of ; or NA. How get i only the merge of cells with a number? datos$NEW - do.call(paste, c(datos[,19:53], sep = ;)) $ NEW : chr 218.0NA;;;NA;;;NA;NA;NA;NA;NA I hope you can help me. Thanks! Best regards, Dominic __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] partial analisys of a time series
Hi, If the frequency is 1, the error message will be gone. For e.g. birthstimeseriesJanFeb-subset(birthstimeseries,cycle(birthstimeseries)==c(1,2)) birthstimeseriesJanFeb1-ts(birthstimeseriesJanFeb,frequency=2,start=c(1946,1)) plot.ts(birthstimeseriesJanFeb1) birthstimeseriesJanFebHW-HoltWinters(birthstimeseriesJanFeb1) plot(birthstimeseriesJanFebHW) A.K. - Original Message - From: Antonio Silva aolinto@gmail.com To: PIKAL Petr petr.pi...@precheza.cz Cc: R-help@r-project.org R-help@r-project.org Sent: Tuesday, December 4, 2012 5:50 AM Subject: Re: [R] partial analisys of a time series Thanks Petr I thought there might be an equivalent for birthstimeseries[,1] if it were a dataframe, but split function sounds great. I could not reproduce the second line of your suggestion l.blist - lapply(blist, HoltWinters). I receive the message: Error in decompose(ts(x[1L:wind], start = start(x), frequency = f), seasonal) : time series has no or less than 2 periods What could be going wrong? Best regards Antonio 2012/12/4 PIKAL Petr petr.pi...@precheza.cz Hi -Original Message- From: r-help-boun...@r-project.org [mailto:r-help-bounces@r- project.org] On Behalf Of Antonio Silva Sent: Tuesday, December 04, 2012 10:26 AM To: R-help@r-project.org Subject: [R] partial analisys of a time series Dear list members I want to analyze separately the months of a time series. In other words, I want to plot and fit models for each month separately. Taking the example of http://a-little-book-of-r-for-time- series.readthedocs.org/en/latest/src/timeseries.html births - scan(http://robjhyndman.com/tsdldata/data/nybirths.dat;) birthstimeseries - ts(births, frequency=12, start=c(1946,1)) birthstimeseries plot.ts(birthstimeseries) birthstimeseriesHW - HoltWinters(birthstimeseries) plot(birthstimeseriesHW) How to proceed the plotting and HoltWinters smoothing considereing only Januarys, Februarys, etc. separately. Split your data by months to a list, use lapply. using zoo package blist -split(birthstimeseries, months(as.Date(birthstimeseries))) l.blist - lapply(blist, HoltWinters) Regards Petr Thanks in advance. Antonio Olinto [[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. -- Antônio Olinto Ávila da Silva Biólogo / Oceanógrafo Instituto de Pesca (Fisheries Institute) São Paulo, Brasil [[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] Different results from random.Forest with test option and using predict function
Without data to reproduce what you saw, we can only guess.
One possibility is due to tie-breaking. There are several places where ties
can occur and are broken at random, including at the prediction step. One
difference between the two ways of doing prediction is that when it's all done
within randomForest(), the test set prediction is performed as each tree is
grown. If there is any tie that needs to be broken at any prediction step, it
will affect the RNG stream used by the subsequent tree growing step.
You can also inspect/compare the forest components of the randomForest
objects to see if they are the same. At least the first tree in both should be
identical.
Andy
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
Behalf Of tdbuskirk
Sent: Monday, December 03, 2012 6:31 PM
To: r-help@r-project.org
Subject: [R] Different results from random.Forest with test option and using
predict function
Hello R Gurus,
I am perplexed by the different results I obtained when I ran code like
this:
set.seed(100)
test1-randomForest(BinaryY~., data=Xvars, trees=51, mtry=5, seed=200)
predict(test1, newdata=cbind(NewBinaryY, NewXs), type=response)
and this code:
set.seed(100)
test2-randomForest(BinaryY~., data=Xvars, trees=51, mtry=5, seed=200,
xtest=NewXs, ytest=NewBinarY)
The confusion matrices for the two forests I thought would be the same by
virtue of the same seed settings, but they differ as do the predicted values
as well as the votes. At first I thought it was just the way ties were
broken, so I changed the number of trees to an odd number so there are no
ties anymore.
Can anyone shed light on what I am hoping is a simple oversight? I just
can't figure out why the results of the predictions from these two forests
applied to the NewBinaryYs and NewX data sets would not be the same.
Thanks for any hints and help.
Sincerely,
Trent Buskirk
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Re: [R] Histogram plot help
Thanks, Rui and David! -- View this message in context: http://r.789695.n4.nabble.com/Histogram-plot-help-tp4651958p4652065.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] Spatial AND Temporal Dependency with lme, or other function?
Hi, I've got a new dataset which I don't know how to analyze with R. My knowledge about R is limited for this kind of problem. I've tried to find a solution with some spatio temporel packages and lme/lmer functions, but didn't find any similar example. I've got 10 locations on a coast on which the number of walrus was noted. I've got a distance measure between each location. We also have an indicator of the presence of tourists, and the number of seals. These information on each location were noted 10 times ( 2 summers * 5 weeks). We want to know how the number of walrus is affected by the tourists, the number of seals, the week of the summer and the location. For example, we think that they are moving east at the end of the summer, when there are more tourist in the west, and less seals in the east. My problem is how to take into account the spatial and temporal dependency. The locations are beside one another, so the number of walrus in one location is closely related to the closest location. I would like to take into account the distance between each site for the dependency. And we took 10 measures in time at each location, so I should take into account this correlation too. I've looked at lme function to model the dependency. It's correct with only spatial or temporal, but I don't know how to take into account both dependencies at the same time. If you know how to do it with lme, or with any other function/package, let me know! Thanks in advance, and have a nice day! Sophie [[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] model selection with spg and AIC (or, convert list to fitted model object)
Adam,
Getting the variance of MLE estimator when the true parameter is on the
boundary is a very difficult problem. It is known that the standard bootstrap
does not work. There are some sub-sampling approaches (Springer book:
Politis, Romano, Wolff), but I am not an expert on this.
However, an important question is how do we know that the truth is supposedly
on the boundary. If you can assume that the truth is in the interior, then you
can use standard approaches: observed hessian or bootstrap to get standard
errors.
Best,
Ravi
From: Adam Zeilinger [mailto:ad...@ucr.edu]
Sent: Sunday, December 02, 2012 5:04 PM
To: Ravi Varadhan
Cc: Adam Zeilinger (zeil0...@umn.edu); r-help@r-project.org
Subject: Re: [R] model selection with spg and AIC (or, convert list to fitted
model object)
Dear Ravi,
Thank you so much for the help. I switched to using the optimx function but I
continue to use the spg method (for the most part) because I found that only
spg consistently converges give different datasets. I also decided to use AIC
rather that a likelihood ratio test.
I have a new question. I would like to construct 95% confidence intervals for
the parameter estimates from the best model. From a previous R Help thread,
you said that it was a bad idea to use the Hessian matrix computed from
optim/optimx or hessian() [numDeriv package] when the optimization is
constrained and parameter estimates are on the boundary because the MLE is
likely at a local minimum:
http://tolstoy.newcastle.edu.au/R/e15/help/11/09/6673.html
In the same thread, you suggest using the Hessian matrix from augmented
Lagrangian optimization with auglag() [alabama package] (with some caveats). I
would like to construct 95% CI with auglag, but I don't understand how to write
the inequality constraints (hin) function. Could you please help me write the
hin function?
Below is R code for my MLE problem, using data that results in parameter
estimates on the boundary, and my unsuccessful attempt at auglag()
optimization. Note: I have gradient functions for NLL1 and NLL2 but they're
very large and don't seem to improve optimization, so they are not included
here. I can supply them if needed for the auglag() function. Also, I am
running R 2.15.2 on a Windows 7 64-bit machine.
##
library(optimx)
library(alabama)
# define multinomial distribution
dmnom2 - function(x,prob,log=FALSE) {
r - lgamma(sum(x) + 1) + sum(x * log(prob) - lgamma(x + 1))
if (log) r else exp(r)
}
# data frame y
y - structure(list(t = c(0.167, 0.5, 1, 12, 18, 24, 36), n1 = c(1L,
1L, 1L, 8L, 9L, 10L, 12L), n2 = c(0L, 1L, 2L, 6L, 5L, 3L, 2L),
n3 = c(13L, 12L, 11L, 0L, 0L, 1L, 0L)), .Names = c(t, n1,
n2, n3), class = data.frame, row.names = 36:42)
# Negative log-likelihood functions
NLL1 - function(par, y) {
p1 - par[1]
p2 - p1
mu1 - par[2]
mu2 - mu1
t - y$t
n1 - y$n1
n2 - y$n2
n3 - y$n3
P1 - (p1*((-1 + exp(sqrt((mu1 + mu2 + p1 + p2)^2 -
4*(mu2*p1 + mu1*(mu2 + p2)))*t))*((-mu2)*(mu2 - p1 + p2) +
mu1*(mu2 + 2*p2)) - mu2*sqrt((mu1 + mu2 + p1 + p2)^2 -
4*(mu2*p1 + mu1*(mu2 + p2))) -
exp(sqrt((mu1 + mu2 + p1 + p2)^2 - 4*(mu2*p1 + mu1*(mu2 + p2)))*t)*
mu2*sqrt((mu1 + mu2 + p1 + p2)^2 - 4*(mu2*p1 + mu1*(mu2 + p2))) +
2*exp((1/2)*(mu1 + mu2 + p1 + p2 + sqrt((mu1 + mu2 + p1 + p2)^2 -
4*(mu2*p1 + mu1*(mu2 + p2*t)*mu2*
sqrt((mu1 + mu2 + p1 + p2)^2 - 4*(mu2*p1 + mu1*(mu2 + p2)/
exp((1/2)*(mu1 + mu2 + p1 + p2 + sqrt((mu1 + mu2 + p1 + p2)^2 -
4*(mu2*p1 + mu1*(mu2 + p2*t)/(2*(mu2*p1 + mu1*(mu2 + p2))*
sqrt((mu1 + mu2 + p1 + p2)^2 - 4*(mu2*p1 + mu1*(mu2 + p2
P2 - (p2*((-1 + exp(sqrt((mu1 + mu2 + p1 + p2)^2 -
4*(mu2*p1 + mu1*(mu2 + p2)))*t))*(-mu1^2 + 2*mu2*p1 +
mu1*(mu2 - p1 + p2)) - mu1*sqrt((mu1 + mu2 + p1 + p2)^2 -
4*(mu2*p1 + mu1*(mu2 + p2))) -
exp(sqrt((mu1 + mu2 + p1 + p2)^2 - 4*(mu2*p1 + mu1*(mu2 + p2)))*t)*
mu1*sqrt((mu1 + mu2 + p1 + p2)^2 - 4*(mu2*p1 + mu1*(mu2 + p2))) +
2*exp((1/2)*(mu1 + mu2 + p1 + p2 + sqrt((mu1 + mu2 + p1 + p2)^2 -
4*(mu2*p1 + mu1*(mu2 + p2*t)*mu1*
sqrt((mu1 + mu2 + p1 + p2)^2 - 4*(mu2*p1 + mu1*(mu2 + p2)/
exp((1/2)*(mu1 + mu2 + p1 + p2 + sqrt((mu1 + mu2 + p1 + p2)^2 -
4*(mu2*p1 + mu1*(mu2 + p2*t)/(2*(mu2*p1 + mu1*(mu2 + p2))*
sqrt((mu1 + mu2 + p1 + p2)^2 - 4*(mu2*p1 + mu1*(mu2 + p2
P3 - 1 - P1 - P2
p.all - c(P1, P2, P3)
if(all(p.all 0 p.all 1)) -sum(dmnom2(c(n1, n2, n3), prob = p.all, log =
TRUE)) else 1e07
}
NLL2 - function(par, y) {
p1 - par[1]
p2 - par[2]
mu1 - par[3]
mu2 - par[4]
t - y$t
n1 - y$n1
n2 - y$n2
n3 - y$n3
P1 - (p1*((-1 + exp(sqrt((mu1 + mu2 + p1 + p2)^2 -
4*(mu2*p1 + mu1*(mu2 + p2)))*t))*((-mu2)*(mu2 - p1 + p2) +
mu1*(mu2 + 2*p2)) - mu2*sqrt((mu1 + mu2 + p1 +
[R] lme: subject-specific slopes.
I am running a random intercept random slope regression: fitRIRT - lme(echogen~time,random=~ 1+time|subject,data=repeatdata,na.action=na.omit) summary(fitRIRT) I would like to get the subject-specific slopes, i.e. the slope that the model computes for each subject. If I have 10-subjects I should have 10-slopes. I don't see the slope when I look at the items listed in names(summary(fitRIRT) nor when I look at the items listed in names(fitRIRT) Thanks John John David Sorkin M.D., Ph.D. Chief, Biostatistics and Informatics University of Maryland School of Medicine Division of Gerontology Baltimore VA Medical Center 10 North Greene Street GRECC (BT/18/GR) Baltimore, MD 21201-1524 (Phone) 410-605-7119 (Fax) 410-605-7913 (Please call phone number above prior to faxing) Confidentiality Statement: This email message, including any attachments, is for the sole use of the intended recipient(s) and may contain confidential and privileged information. Any unauthorized use, disclosure or distribution is prohibited. If you are not the intended recipient, please contact the sender by reply email and destroy all copies of the original message. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] lme: subject-specific slopes.
Ken, Thank you for your help. ranef(fitRIRT) does not give me what I expect. The subject-specific slopes, and subject-specific intercepts are not anywhere close to what I would expect them to be; the mean of the subject-specfic values should be close to those reported by summary(fitRIRT) and they are not. As you will see by examining the material below, the subject-specific slopes are off by many order of magnitude. The intercepts are also far from the value reported in summary(fitRIRT). Do you have any thoughts? Thanks, John fitRIRT - lme(echogen~time,random=~ 1+time|subject,data=repeatdata,na.action=na.omit) summary(fitRIRT)Linear mixed-effects model fit by REML Data: repeatdata AIC BIClogLik 495.097 507.2491 -241.5485 Random effects: Formula: ~1 + time | subject Structure: General positive-definite, Log-Cholesky parametrization StdDev Corr (Intercept) 1.917511e+01 (Intr) time2.032276e-04 0 Residual1.044601e+01 Fixed effects: echogen ~ time Value Std.Error DF t-value p-value (Intercept) 64.54864 4.258235 32 15.158543 0. time 0.35795 0.227080 32 1.576307 0.1248 Correlation: (Intr) time -0.242 Standardized Within-Group Residuals: Min Q1 Med Q3 Max -1.61362755 -0.52710871 0.02948008 0.41793322 1.77340082 Number of Observations: 58 Number of Groups: 25 ranef(fitRIRT) (Intercept) time 1-3.278112 2.221016e-09 2 -35.400618 4.314995e-08 311.493110 -6.797543e-09 4 -16.209586 -7.070834e-08 5 3.585227 -2.389705e-08 6 1.614320 -1.967700e-09 7 8.346905 5.827094e-08 830.917812 -3.768584e-08 9-0.394101 -9.158251e-09 104.437509 -4.057971e-08 11 31.956597 -2.126275e-08 12 41.567402 -4.853942e-08 13 -10.723993 1.307152e-08 14 -4.554837 5.551908e-09 15 -4.554501 4.815086e-08 16 13.296985 -3.743967e-08 17 -8.255439 1.733238e-08 18 -21.317239 2.203885e-08 19 -13.480159 2.194016e-08 20 -13.044766 2.269168e-08 21 11.639198 -1.418706e-08 22 -27.457388 -1.154099e-08 232.194001 -5.509119e-09 24 -3.992646 7.682188e-08 251.614320 -1.967700e-09 John David Sorkin M.D., Ph.D. Chief, Biostatistics and Informatics University of Maryland School of Medicine Division of Gerontology Baltimore VA Medical Center 10 North Greene Street GRECC (BT/18/GR) Baltimore, MD 21201-1524 (Phone) 410-605-7119 (Fax) 410-605-7913 (Please call phone number above prior to faxing) Kenneth Frost kfr...@wisc.edu 12/4/2012 10:44 AM I think this might be close to what you want. ranef(fitRIRT) On 12/04/12, John Sorkin wrote: I am running a random intercept random slope regression: fitRIRT - lme(echogen~time,random=~ 1+time|subject,data=repeatdata,na.action=na.omit) summary(fitRIRT) I would like to get the subject-specific slopes, i.e. the slope that the model computes for each subject. If I have 10-subjects I should have 10-slopes. I don't see the slope when I look at the items listed in names(summary(fitRIRT) nor when I look at the items listed in names(fitRIRT) Thanks John John David Sorkin M.D., Ph.D. Chief, Biostatistics and Informatics University of Maryland School of Medicine Division of Gerontology Baltimore VA Medical Center 10 North Greene Street GRECC (BT/18/GR) Baltimore, MD 21201-1524 (Phone) 410-605-7119 (Fax) 410-605-7913 (Please call phone number above prior to faxing) Confidentiality Statement: This email message, including any attachments, is for the sole use of the intended recipient(s) and may contain confidential and privileged information. Any unauthorized use, disclosure or distribution is prohibited. If you are not the intended recipient, please contact the sender by reply email and destroy all copies of the original message. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Kenneth Frost Graduate Research Assistant - Dept. of Plant Pathology University of Wisconsin - Madison Lab: (608) 262-9914 Mobile: (608) 556-9637 kfr...@wisc.edu Confidentiality Statement: This email message, including any attachments, is for the sole use of the intended recipient(s) and may contain confidential and privileged information. Any unauthorized use, disclosure or distribution is prohibited. If you are not the intended recipient, please contact the sender by reply email and destroy all copies of the original message. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] lme: subject-specific slopes.
I think the random effects represent the subject adjustments to the population averages. You may have to do the addition yourself to get the subject specific slopes and intercepts. Someone will hopefully correct me if I'm wrong. On 12/04/12, John Sorkin wrote: Ken, Thank you for your help. ranef(fitRIRT) does not give me what I expect. The subject-specific slopes, and subject-specific intercepts are not anywhere close to what I would expect them to be; the mean of the subject-specfic values should be close to those reported by summary(fitRIRT) and they are not. As you will see by examining the material below, the subject-specific slopes are off by many order of magnitude. The intercepts are also far from the value reported in summary(fitRIRT). Do you have any thoughts? Thanks, John fitRIRT - lme(echogen~time,random=~ 1+time|subject,data=repeatdata,na.action=na.omit) summary(fitRIRT) Linear mixed-effects model fit by REML Data: repeatdata AIC BIC logLik 495.097 507.2491 -241.5485 Random effects: Formula: ~1 + time | subject Structure: General positive-definite, Log-Cholesky parametrization StdDev Corr (Intercept) 1.917511e+01 (Intr) time 2.032276e-04 0 Residual 1.044601e+01 Fixed effects: echogen ~ time Value Std.Error DF t-value p-value (Intercept) 64.54864 4.258235 32 15.158543 0. time 0.35795 0.227080 32 1.576307 0.1248 Correlation: (Intr) time -0.242 Standardized Within-Group Residuals: Min Q1 Med Q3 Max -1.61362755 -0.52710871 0.02948008 0.41793322 1.77340082 Number of Observations: 58 Number of Groups: 25 ranef(fitRIRT) (Intercept) time 1 -3.278112 2.221016e-09 2 -35.400618 4.314995e-08 3 11.493110 -6.797543e-09 4 -16.209586 -7.070834e-08 5 3.585227 -2.389705e-08 6 1.614320 -1.967700e-09 7 8.346905 5.827094e-08 8 30.917812 -3.768584e-08 9 -0.394101 -9.158251e-09 10 4.437509 -4.057971e-08 11 31.956597 -2.126275e-08 12 41.567402 -4.853942e-08 13 -10.723993 1.307152e-08 14 -4.554837 5.551908e-09 15 -4.554501 4.815086e-08 16 13.296985 -3.743967e-08 17 -8.255439 1.733238e-08 18 -21.317239 2.203885e-08 19 -13.480159 2.194016e-08 20 -13.044766 2.269168e-08 21 11.639198 -1.418706e-08 22 -27.457388 -1.154099e-08 23 2.194001 -5.509119e-09 24 -3.992646 7.682188e-08 25 1.614320 -1.967700e-09 John David Sorkin M.D., Ph.D. Chief, Biostatistics and Informatics University of Maryland School of Medicine Division of Gerontology Baltimore VA Medical Center 10 North Greene Street GRECC (BT/18/GR) Baltimore, MD 21201-1524 (Phone) 410-605-7119 (Fax) 410-605-7913 (Please call phone number above prior to faxing) Kenneth Frost kfr...@wisc.edu 12/4/2012 10:44 AM I think this might be close to what you want. ranef(fitRIRT) On 12/04/12, John Sorkin wrote: I am running a random intercept random slope regression: fitRIRT - lme(echogen~time,random=~ 1+time|subject,data=repeatdata,na.action=na.omit) summary(fitRIRT) I would like to get the subject-specific slopes, i.e. the slope that the model computes for each subject. If I have 10-subjects I should have 10-slopes. I don't see the slope when I look at the items listed in names(summary(fitRIRT) nor when I look at the items listed in names(fitRIRT) Thanks John John David Sorkin M.D., Ph.D. Chief, Biostatistics and Informatics University of Maryland School of Medicine Division of Gerontology Baltimore VA Medical Center 10 North Greene Street GRECC (BT/18/GR) Baltimore, MD 21201-1524 (Phone) 410-605-7119 (Fax) 410-605-7913 (Please call phone number above prior to faxing) Confidentiality Statement: This email message, including any attachments, is for the sole use of the intended recipient(s) and may contain confidential and privileged information. Any unauthorized use, disclosure or distribution is prohibited. If you are not the intended recipient, please contact the sender by reply email and destroy all copies of the original message. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Kenneth Frost Graduate Research Assistant - Dept. of Plant Pathology University of Wisconsin - Madison Lab: (608) 262-9914 Mobile: (608) 556-9637 kfr...@wisc.edu Confidentiality Statement: This email message, including any attachments, is for the sole use of the intended recipient(s) and may contain confidential and privileged information. Any unauthorized use, disclosure or distribution is prohibited. If you are not the intended recipient, please contact the sender by reply email and destroy all copies of the original message. -- Kenneth Frost Graduate Research Assistant - Dept. of Plant Pathology University of Wisconsin
Re: [R] lme: subject-specific slopes.
Yes, you are correct. Thanks, John John David Sorkin M.D., Ph.D. Chief, Biostatistics and Informatics University of Maryland School of Medicine Division of Gerontology Baltimore VA Medical Center 10 North Greene Street GRECC (BT/18/GR) Baltimore, MD 21201-1524 (Phone) 410-605-7119 (Fax) 410-605-7913 (Please call phone number above prior to faxing) Kenneth Frost kfr...@wisc.edu 12/4/2012 11:07 AM I think the random effects represent the subject adjustments to the population averages. You may have to do the addition yourself to get the subject specific slopes and intercepts. Someone will hopefully correct me if I'm wrong. On 12/04/12, John Sorkin wrote: Ken, Thank you for your help. ranef(fitRIRT) does not give me what I expect. The subject-specific slopes, and subject-specific intercepts are not anywhere close to what I would expect them to be; the mean of the subject-specfic values should be close to those reported by summary(fitRIRT) and they are not. As you will see by examining the material below, the subject-specific slopes are off by many order of magnitude. The intercepts are also far from the value reported in summary(fitRIRT). Do you have any thoughts? Thanks, John fitRIRT - lme(echogen~time,random=~ 1+time|subject,data=repeatdata,na.action=na.omit) summary(fitRIRT) Linear mixed-effects model fit by REML Data: repeatdata AIC BIC logLik 495.097 507.2491 -241.5485 Random effects: Formula: ~1 + time | subject Structure: General positive-definite, Log-Cholesky parametrization StdDev Corr (Intercept) 1.917511e+01 (Intr) time 2.032276e-04 0 Residual 1.044601e+01 Fixed effects: echogen ~ time Value Std.Error DF t-value p-value (Intercept) 64.54864 4.258235 32 15.158543 0. time 0.35795 0.227080 32 1.576307 0.1248 Correlation: (Intr) time -0.242 Standardized Within-Group Residuals: Min Q1 Med Q3 Max -1.61362755 -0.52710871 0.02948008 0.41793322 1.77340082 Number of Observations: 58 Number of Groups: 25 ranef(fitRIRT) (Intercept) time 1 -3.278112 2.221016e-09 2 -35.400618 4.314995e-08 3 11.493110 -6.797543e-09 4 -16.209586 -7.070834e-08 5 3.585227 -2.389705e-08 6 1.614320 -1.967700e-09 7 8.346905 5.827094e-08 8 30.917812 -3.768584e-08 9 -0.394101 -9.158251e-09 10 4.437509 -4.057971e-08 11 31.956597 -2.126275e-08 12 41.567402 -4.853942e-08 13 -10.723993 1.307152e-08 14 -4.554837 5.551908e-09 15 -4.554501 4.815086e-08 16 13.296985 -3.743967e-08 17 -8.255439 1.733238e-08 18 -21.317239 2.203885e-08 19 -13.480159 2.194016e-08 20 -13.044766 2.269168e-08 21 11.639198 -1.418706e-08 22 -27.457388 -1.154099e-08 23 2.194001 -5.509119e-09 24 -3.992646 7.682188e-08 25 1.614320 -1.967700e-09 John David Sorkin M.D., Ph.D. Chief, Biostatistics and Informatics University of Maryland School of Medicine Division of Gerontology Baltimore VA Medical Center 10 North Greene Street GRECC (BT/18/GR) Baltimore, MD 21201-1524 (Phone) 410-605-7119 (Fax) 410-605-7913 (Please call phone number above prior to faxing) Kenneth Frost kfr...@wisc.edu 12/4/2012 10:44 AM I think this might be close to what you want. ranef(fitRIRT) On 12/04/12, John Sorkin wrote: I am running a random intercept random slope regression: fitRIRT - lme(echogen~time,random=~ 1+time|subject,data=repeatdata,na.action=na.omit) summary(fitRIRT) I would like to get the subject-specific slopes, i.e. the slope that the model computes for each subject. If I have 10-subjects I should have 10-slopes. I don't see the slope when I look at the items listed in names(summary(fitRIRT) nor when I look at the items listed in names(fitRIRT) Thanks John John David Sorkin M.D., Ph.D. Chief, Biostatistics and Informatics University of Maryland School of Medicine Division of Gerontology Baltimore VA Medical Center 10 North Greene Street GRECC (BT/18/GR) Baltimore, MD 21201-1524 (Phone) 410-605-7119 (Fax) 410-605-7913 (Please call phone number above prior to faxing) Confidentiality Statement: This email message, including any attachments, is for the sole use of the intended recipient(s) and may contain confidential and privileged information. Any unauthorized use, disclosure or distribution is prohibited. If you are not the intended recipient, please contact the sender by reply email and destroy all copies of the original message. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Kenneth Frost Graduate Research Assistant - Dept. of Plant Pathology University of Wisconsin - Madison Lab: (608) 262-9914 Mobile: (608) 556-9637 kfr...@wisc.edu Confidentiality Statement: This email message, including any
[R] How to reset R settings to original state other than remove .Rdata and .Rhistory
Dear all,
Do you know how to return all the R settings to original state? Other than
.Rdata and .Rhistory
Some weid thing happened to my machine. I was trying to get shaded confidence
band ploted using survplot from rms liberary.
It worked on one machine, but not on the other. I tried unstall R and reinstall
R and remove. Rdata and .Rhisotory. Nothing helped so far.
Thanks for your input.
Best regards,
Hong
Codes are
library(rms)
data generation
n - 1000
set.seed(731)
age - 50 + 12*rnorm(n)
label(age) - Age
sex - factor(sample(c('male','female'), n, TRUE))
cens - 15*runif(n)
h - .02*exp(.04*(age-50)+.8*(sex=='female'))
dt - -log(runif(n))/h
label(dt) - 'Follow-up Time'
e - ifelse(dt = cens,1,0)
dt - pmin(dt, cens)
units(dt) - 'Year'
dd - datadist(age, sex)
options(datadist='dd')
S - Surv(dt,e)
tte-data.frame(dt,e,age,sex)
male group KM curve with bands#
f - survfit(Surv(dt,e)~1, data=tte,subset=sex=='male')
survplot(f,conf=bands)
female group KM curve with bands#
g - survfit(Surv(dt,e)~1, data=tte,subset=sex!='male')
survplot(g,conf=bands, col=3, add=T)
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and provide commented, minimal, self-contained, reproducible code.
[R] Solve system of equations (nleqslv) only returns origin
I'm solving 4 complex equations simultaneously. Code is below. The code
returns only zero's for the solution though there should also be a non-zero
result. I'm pretty confident that the equations are correct because they
are straight from a published paper and I checked them pretty thoroughly.
The parameter values I used are from the published paper as well. Any
suggestions for how I can find the maximum non-zero solution instead of
going straight to the minimum?
Thanks!
Alicia
install.packages(nleqslv,
lib=Users/Alicia/Documents/R.Codes/R.Packages/)
require(nleqslv)
## Global Parameters
beeta=0.8
pq=1
L=12600
theta=0.6
psale=0.6
mu=psale*(1-theta)
alphah=0.15
Cg=6240
Cs=2820
A= 100
D=0.0001
greekp=0.43
K=10
# Species Parameters ##
b1=0.38
p1=16654
v1 = 0.28
N1=6000
g1=1
delta1=1
b2=0.4
p2=2797
v2 = 0.31
N2=1
g2=1
delta2=1
### Define functions with vector x = c(Lg, Ls, gamma1, gamma2, lamda)
firstordercond - function (x) {
y=numeric(4)
y[1]=(alphah/x[3])-(x[5]*((p1-(((theta+mu)*(((N1/A)*g1^greekp*x[1]^b1)+K))+((theta+mu)*(((1-exp(-2*D*v1*N1))*x[2])+K*(((N1/A)*g1^(greekp))*x[1]^b1+(2*v1*N1*D)*x[2])
+
delta1*theta*(((N1/A)*g1^(greekp))*x[1]^b1+(2*v1*N1*D)*x[2])))
y[2]=(alphah/x[4])-(x[5]*((p2-(((theta+mu)*(((N2/A)*g2^greekp*x[1]^b2)+K))+((theta+mu)*(((1-exp(-2*D*v2*N2))*x[2])+K*(((N2/A)*g2^(greekp))*x[1]^b2+(2*v2*N2*D)*x[2])
+
delta2*theta*(((N2/A)*g2^(greekp))*x[1]^b2+(2*v2*N2*D)*x[2])))
y[3]=((alphah*((N1/A)*g1^(greekp))*b1*x[1]^(b1-1))/(((N1/A)*g1^(greekp))*x[1]^b1+(2*v1*N1*D)*x[2]))
+
((alphah*((N2/A)*g2^(greekp))*b2*x[1]^(b2-1))/(((N2/A)*g2^(greekp))*x[1]^b2+(2*v2*N2*D)*x[2]))
+
x[5]*(((-1*beeta*pq*(L-x[1]-x[2])^(beeta-1)) +
(b1*(1-x[3])*(p1-(((theta+mu)*(((N1/A)*g1^greekp*x[1]^b1)+K))+((theta+mu)*(((1-exp(-2*D*v1*N1))*x[2])+K*((N1/A)*g1^(greekp))*x[1]^(b1-1))
+
(b2*(1-x[4])*(p2-(((theta+mu)*(((N2/A)*g2^greekp*x[1]^b2)+K))+((theta+mu)*(((1-exp(-2*D*v2*N2))*x[2])+K*((N2/A)*g2^(greekp))*x[1]^(b2-1))
-
(delta1*theta*x[3]*((N1/A)*g1^(greekp))*b1*x[1]^(b1-1)) -
(delta2*theta*x[4]*((N2/A)*g2^(greekp))*b2*x[1]^(b2-1)))-Cg*(((N1/A)*g1^(greekp))*b1*x[1]^(b1-1)+((N2/A)*g2^(greekp))*b2*x[1]^(b2-1)))
y[4]=((alphah*(2*v1*N1*D))/(((N1/A)*g1^(greekp))*x[1]^b1+(2*v1*N1*D)*x[2]))
+ ((alphah*(2*v2*N2*D))/(((N2/A)*g2^(greekp))*x[1]^b2+(2*v2*N2*D)*x[2])) +
x[5]*(((-1*beeta*pq*(L-x[1]-x[2])^(beeta-1)) +
((1-x[3])*(p1-(((theta+mu)*(((N1/A)*g1^greekp*x[1]^b1)+K))+((theta+mu)*(((1-exp(-2*D*v1*N1))*x[2])+K*(2*v1*N1*D))
+
((1-x[4])*(p2-(((theta+mu)*(((N2/A)*g2^greekp*x[1]^b2)+K))+((theta+mu)*(((1-exp(-2*D*v2*N2))*x[2])+K*(2*v2*N2*D))
- (delta1*theta*x[3]*(2*v1*N1*D)) -
(delta2*theta*x[4]*(2*v2*N2*D)))-Cs*((2*v1*N1*D)+(2*v2*N2*D)))
result=(x)
}
Xstart=c(1, 200, 0.5, 0.5, 12)
fstart= firstordercond(Xstart)
nleqslv(Xstart, firstordercond)
--
Alicia Ellis
Postdoc
Gund Institute for Ecological Economics
University of Vermont
617 Main Street
Burlington, VT 05405
(802) 656-1046
http://www.uvm.edu/~aellis5/
http://entomology.ucdavis.edu/faculty/scott/aellis/
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and provide commented, minimal, self-contained, reproducible code.
Re: [R] lme: subject-specific slopes.
John: I think you want the output from coef(fitRIRT). The ranef(fitRIRT) will give you the subject specific random effect deviations from the fixed effects means. The coef(fitRIRT) will give you the combination of the fixed effect means with the subject specific random effect deviations and, thus, in the scale you are expecting. 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 From: John Sorkin jsor...@grecc.umaryland.edu To: r-help@r-project.org, Kenneth Frost kfr...@wisc.edu Date: 12/04/2012 09:10 AM Subject: Re: [R] lme: subject-specific slopes. Sent by: r-help-boun...@r-project.org Ken, Thank you for your help. ranef(fitRIRT) does not give me what I expect. The subject-specific slopes, and subject-specific intercepts are not anywhere close to what I would expect them to be; the mean of the subject-specfic values should be close to those reported by summary(fitRIRT) and they are not. As you will see by examining the material below, the subject-specific slopes are off by many order of magnitude. The intercepts are also far from the value reported in summary(fitRIRT). Do you have any thoughts? Thanks, John fitRIRT - lme(echogen~time,random=~ 1+time|subject,data=repeatdata,na.action=na.omit) summary(fitRIRT)Linear mixed-effects model fit by REML Data: repeatdata AIC BIClogLik 495.097 507.2491 -241.5485 Random effects: Formula: ~1 + time | subject Structure: General positive-definite, Log-Cholesky parametrization StdDev Corr (Intercept) 1.917511e+01 (Intr) time2.032276e-04 0 Residual1.044601e+01 Fixed effects: echogen ~ time Value Std.Error DF t-value p-value (Intercept) 64.54864 4.258235 32 15.158543 0. time 0.35795 0.227080 32 1.576307 0.1248 Correlation: (Intr) time -0.242 Standardized Within-Group Residuals: Min Q1 Med Q3 Max -1.61362755 -0.52710871 0.02948008 0.41793322 1.77340082 Number of Observations: 58 Number of Groups: 25 ranef(fitRIRT) (Intercept) time 1-3.278112 2.221016e-09 2 -35.400618 4.314995e-08 311.493110 -6.797543e-09 4 -16.209586 -7.070834e-08 5 3.585227 -2.389705e-08 6 1.614320 -1.967700e-09 7 8.346905 5.827094e-08 830.917812 -3.768584e-08 9-0.394101 -9.158251e-09 104.437509 -4.057971e-08 11 31.956597 -2.126275e-08 12 41.567402 -4.853942e-08 13 -10.723993 1.307152e-08 14 -4.554837 5.551908e-09 15 -4.554501 4.815086e-08 16 13.296985 -3.743967e-08 17 -8.255439 1.733238e-08 18 -21.317239 2.203885e-08 19 -13.480159 2.194016e-08 20 -13.044766 2.269168e-08 21 11.639198 -1.418706e-08 22 -27.457388 -1.154099e-08 232.194001 -5.509119e-09 24 -3.992646 7.682188e-08 251.614320 -1.967700e-09 John David Sorkin M.D., Ph.D. Chief, Biostatistics and Informatics University of Maryland School of Medicine Division of Gerontology Baltimore VA Medical Center 10 North Greene Street GRECC (BT/18/GR) Baltimore, MD 21201-1524 (Phone) 410-605-7119 (Fax) 410-605-7913 (Please call phone number above prior to faxing) Kenneth Frost kfr...@wisc.edu 12/4/2012 10:44 AM I think this might be close to what you want. ranef(fitRIRT) On 12/04/12, John Sorkin wrote: I am running a random intercept random slope regression: fitRIRT - lme(echogen~time,random=~ 1+time|subject,data=repeatdata,na.action=na.omit) summary(fitRIRT) I would like to get the subject-specific slopes, i.e. the slope that the model computes for each subject. If I have 10-subjects I should have 10-slopes. I don't see the slope when I look at the items listed in names(summary(fitRIRT) nor when I look at the items listed in names(fitRIRT) Thanks John John David Sorkin M.D., Ph.D. Chief, Biostatistics and Informatics University of Maryland School of Medicine Division of Gerontology Baltimore VA Medical Center 10 North Greene Street GRECC (BT/18/GR) Baltimore, MD 21201-1524 (Phone) 410-605-7119 (Fax) 410-605-7913 (Please call phone number above prior to faxing) Confidentiality Statement: This email message, including any attachments, is for the sole use of the intended recipient(s) and may contain confidential and privileged information. Any unauthorized use, disclosure or distribution is prohibited. If you are not the intended recipient, please contact the sender by reply email and destroy all copies of the original message. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Kenneth Frost Graduate Research Assistant - Dept. of Plant Pathology University of Wisconsin - Madison Lab:
Re: [R] computing marginal values based on multiple columns?
HI,
I am not sure the output you wanted is correct:
sample1 sample2 sample3
1 1.0 0 0.5
because
0.2*colMeans(x[,-4])
sample1 sample2 sample3
# 28.40 24.08 21.36
This might help you:
apply(x[-4],2,function(y) length(y[y 0.2*mean(y)
x$class==a])/length(x[x$class==a]))
#sample1 sample2 sample3
# 0.0 0.0 0.5
A.K.
- Original Message -
From: Simon simonzm...@gmail.com
To: r-help@r-project.org
Cc:
Sent: Tuesday, December 4, 2012 4:49 AM
Subject: [R] computing marginal values based on multiple columns?
Hello all,
I have what feels like a simple problem, but I can't find an simple
answer. Consider this data frame:
x - data.frame(sample1=c(35,176,182,193,124),
sample2=c(198,176,190,23,15), sample3=c(12,154,21,191,156),
class=c('a','a','c','b','c'))
x
sample1 sample2 sample3 class
1 35 198 12 a
2 176 176 154 a
3 182 190 21 c
4 193 23 191 b
5 124 15 156 c
Now I wish to know: for each sample, for values 20% of the sample mean,
what percentage of those are class a?
I want to end up with a table like:
sample1 sample2 sample3
1 1.0 0 0.5
I can calculate this for an individual sample using this rather clumsy
expression:
length(which(x$sample1 mean(x$sample1) x$class=='a')) /
length(which(x$sample1 mean(x$sample1)))
I'd normally propagate it across the data frame using apply, but I
can't because it depends on more than one column.
Any help much appreciated!
Cheers,
Simon
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Re: [R] Help for a function
What are you expecting?
What do you get?
What is the problem?
J
On 4 December 2012 06:01, anoumou teko_maur...@yahoo.fr wrote:
Hello all,
I need a help.
I am modeling a disease and a create a R function like that:
Lambda-function (x,date1,r,h,a){
ndate1 - as.Date(date1, %d/%m/%Y)
t1 - as.numeric(ndate1)
x[order(x$i),]
t -x[,t]
i -x[,i]
CONTAGIEUX -x[,CONTAGIEUX]
while ( t1 min(t) ){
for (i in 1:length(i) ){
{for (j in 1:CONTAGIEUX[length(CONTAGIEUX)]){
res1[j] -(a*h)
res2 -sum( res1[j])
}
}
lambda[i] - r*res2
}
}
x-data.frame(x,lambda)
x
}
on such data :
DATEi Symptomes t Incubation CONTAGIEUX
1 2009-04-29 Canada 13 14363 13 13
2 2009-05-01 Israel 2 14365 2 2
3 2009-05-09 argentina 1 14373 1 1
5 2009-05-09 australia 1 14373 1 1
6 2009-05-10 australia 1 14374 2 2
7 2009-04-29 Austria 1 14363 1 1
8 2009-04-30 Austria 1 14364 2 2
9 2009-05-01 Austria 1 14365 2 3
10 2009-05-02 Austria 1 14366 2 4
11 2009-05-03 Austria 1 14367 2 5
17 2009-05-09 Austria 1 14373 2 7
18 2009-05-10 Austria 1 14374 2 7
19 2009-05-08 brasil 4 14372 4 4
20 2009-05-09 brazil 6 14373 6 6
21 2009-05-10 brazil 6 14374 12 12
22 2009-05-02 canada 51 14366 51 51
23 2009-05-03 canada 85 14367 136 136
24 2009-05-04 canada 101 14368 186 237
31 2009-04-27 Canada 6 14361 6 6
32 2009-04-28 Canada 6 14362 6 6
33 2009-04-30 Canada 19 14364 25 25
34 2009-05-01 Canada 34 14365 53 59
35 2009-05-01 China,HongKong, SAR 1 14365 1 1
36 2009-05-02 China,HongKong, SAR 1 14366 2 2
37 2009-05-03 China,HongKong, SAR 1 14367 2 3
38 2009-05-04 China,HongKong, SAR 1 14368 2 4
44 2009-05-10 China,HongKong, SAR 1 14374 2 7
45 2009-05-04 colombia1 14368 1 1
46 2009-05-05 colombia1 14369 2 2
47 2009-05-06 colombia
But i do not get the results,i try by all means but i d'ont understant the
problem.
Thanks for your help.
--
View this message in context:
http://r.789695.n4.nabble.com/Help-for-a-function-tp4652054.html
Sent from the R help mailing list archive at Nabble.com.
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[R] Labelling x axis in plot function
Hi, In the plot function I want to label x axis as the numbers between 1 and 12 (so 1, 2, 3, 4, 5, ..., 12). How should I do it? The range of x values are different than this range. Thanks! Kind regards, T. Bal [[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] Labelling x axis in plot function
Hello, You should provide sample data and code. plot(3:10, xlim = c(0, 12), xaxt = n) axis(1, at = 1:12) See the help page ?par for a description of the graphical parameters 'xlim' and 'xaxt', and of ?axis. Hope this helps, Rui Barradas Em 04-12-2012 17:24, T Bal escreveu: Hi, In the plot function I want to label x axis as the numbers between 1 and 12 (so 1, 2, 3, 4, 5, ..., 12). How should I do it? The range of x values are different than this range. Thanks! Kind regards, T. Bal [[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] Solve system of equations (nleqslv) only returns origin
On 04-12-2012, at 17:06, Alicia Ellis wrote:
I'm solving 4 complex equations simultaneously. Code is below. The code
returns only zero's for the solution though there should also be a non-zero
result. I'm pretty confident that the equations are correct because they
are straight from a published paper and I checked them pretty thoroughly.
The parameter values I used are from the published paper as well. Any
suggestions for how I can find the maximum non-zero solution instead of
going straight to the minimum?
Are you trying to minimize a function by solving the first order condition?
If yes then you should try functions such optim, nlminb and there are many more.
See below for more comments.
Thanks!
Alicia
install.packages(nleqslv,
lib=Users/Alicia/Documents/R.Codes/R.Packages/)
require(nleqslv)
## Global Parameters
beeta=0.8
pq=1
L=12600
theta=0.6
psale=0.6
mu=psale*(1-theta)
alphah=0.15
Cg=6240
Cs=2820
A= 100
D=0.0001
greekp=0.43
K=10
# Species Parameters ##
b1=0.38
p1=16654
v1 = 0.28
N1=6000
g1=1
delta1=1
b2=0.4
p2=2797
v2 = 0.31
N2=1
g2=1
delta2=1
### Define functions with vector x = c(Lg, Ls, gamma1, gamma2, lamda)
firstordercond - function (x) {
y=numeric(4)
y[1]=(alphah/x[3])-(x[5]*((p1-(((theta+mu)*(((N1/A)*g1^greekp*x[1]^b1)+K))+((theta+mu)*(((1-exp(-2*D*v1*N1))*x[2])+K*(((N1/A)*g1^(greekp))*x[1]^b1+(2*v1*N1*D)*x[2])
+
delta1*theta*(((N1/A)*g1^(greekp))*x[1]^b1+(2*v1*N1*D)*x[2])))
y[2]=(alphah/x[4])-(x[5]*((p2-(((theta+mu)*(((N2/A)*g2^greekp*x[1]^b2)+K))+((theta+mu)*(((1-exp(-2*D*v2*N2))*x[2])+K*(((N2/A)*g2^(greekp))*x[1]^b2+(2*v2*N2*D)*x[2])
+
delta2*theta*(((N2/A)*g2^(greekp))*x[1]^b2+(2*v2*N2*D)*x[2])))
y[3]=((alphah*((N1/A)*g1^(greekp))*b1*x[1]^(b1-1))/(((N1/A)*g1^(greekp))*x[1]^b1+(2*v1*N1*D)*x[2]))
+
((alphah*((N2/A)*g2^(greekp))*b2*x[1]^(b2-1))/(((N2/A)*g2^(greekp))*x[1]^b2+(2*v2*N2*D)*x[2]))
+
x[5]*(((-1*beeta*pq*(L-x[1]-x[2])^(beeta-1)) +
(b1*(1-x[3])*(p1-(((theta+mu)*(((N1/A)*g1^greekp*x[1]^b1)+K))+((theta+mu)*(((1-exp(-2*D*v1*N1))*x[2])+K*((N1/A)*g1^(greekp))*x[1]^(b1-1))
+
(b2*(1-x[4])*(p2-(((theta+mu)*(((N2/A)*g2^greekp*x[1]^b2)+K))+((theta+mu)*(((1-exp(-2*D*v2*N2))*x[2])+K*((N2/A)*g2^(greekp))*x[1]^(b2-1))
-
(delta1*theta*x[3]*((N1/A)*g1^(greekp))*b1*x[1]^(b1-1)) -
(delta2*theta*x[4]*((N2/A)*g2^(greekp))*b2*x[1]^(b2-1)))-Cg*(((N1/A)*g1^(greekp))*b1*x[1]^(b1-1)+((N2/A)*g2^(greekp))*b2*x[1]^(b2-1)))
y[4]=((alphah*(2*v1*N1*D))/(((N1/A)*g1^(greekp))*x[1]^b1+(2*v1*N1*D)*x[2]))
+ ((alphah*(2*v2*N2*D))/(((N2/A)*g2^(greekp))*x[1]^b2+(2*v2*N2*D)*x[2])) +
x[5]*(((-1*beeta*pq*(L-x[1]-x[2])^(beeta-1)) +
((1-x[3])*(p1-(((theta+mu)*(((N1/A)*g1^greekp*x[1]^b1)+K))+((theta+mu)*(((1-exp(-2*D*v1*N1))*x[2])+K*(2*v1*N1*D))
+
((1-x[4])*(p2-(((theta+mu)*(((N2/A)*g2^greekp*x[1]^b2)+K))+((theta+mu)*(((1-exp(-2*D*v2*N2))*x[2])+K*(2*v2*N2*D))
- (delta1*theta*x[3]*(2*v1*N1*D)) -
(delta2*theta*x[4]*(2*v2*N2*D)))-Cs*((2*v1*N1*D)+(2*v2*N2*D)))
result=(x)
what is this statement supposed to do?
You are actually returning the input.
And why like that?
Just x or return(x) is quite sufficient.
You should return the vector y i.e. function values.
But y has length 4 and x has length 4.
So where is the fifth value for y?
}
Xstart=c(1, 200, 0.5, 0.5, 12)
fstart= firstordercond(Xstart)
If you print fstart you will see that it is identical to Xstart.
You need to rethink you firstordercond() function.
It's not correct.
Berend
nleqslv(Xstart, firstordercond)
--
Alicia Ellis
Postdoc
Gund Institute for Ecological Economics
University of Vermont
617 Main Street
Burlington, VT 05405
(802) 656-1046
http://www.uvm.edu/~aellis5/
http://entomology.ucdavis.edu/faculty/scott/aellis/
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[R] odd behavior of browser()
Hi everyone, I normally include a call to browser() as I'm working out the kinks in my scripts, and I am always able to step through each line by hitting Return, but for some reason, in the scripts I'm working on now, hitting Return seems to cause execution of *all* the lines in my script. I've restarted R several times in case it was stuck in a bad state for some reason, but I'm consistently getting this behavior anyway. Has anyone run into this problem before? Maybe I inadvertently reset preferences? An example which produces this behavior is the following: file bugcheck.r: browser() a - 1 b - 2 source(bugcheck.r) Called from: eval(expr, envir, enclos) Browse[1] Return ls() [1] a b a [1] 1 b [1] 2 I'd be grateful for any help in resolving this! Thanks, David Romano [[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] Solve system of equations (nleqslv) only returns origin
On 04-12-2012, at 18:50, Berend Hasselman wrote: You should return the vector y i.e. function values. But y has length 4 and x has length 4. x has length 5 of course. Berend So where is the fifth value for y? } Xstart=c(1, 200, 0.5, 0.5, 12) fstart= firstordercond(Xstart) If you print fstart you will see that it is identical to Xstart. You need to rethink you firstordercond() function. It's not correct. Berend nleqslv(Xstart, firstordercond) -- Alicia Ellis Postdoc Gund Institute for Ecological Economics University of Vermont 617 Main Street Burlington, VT 05405 (802) 656-1046 http://www.uvm.edu/~aellis5/ http://entomology.ucdavis.edu/faculty/scott/aellis/ [[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-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] error reading xlsm file with read.xls
Dear all, I cannot reading a .xlsm file using read.xls. I executed: read.xls(resultados.xlsm, colNames = TRUE, sheet = 1, type = data.frame, from = 1, rowNames = NA, colClasses = character, checkNames = TRUE, dateTime = numeric, naStrings = NA, stringsAsFactors = F) Error: Call(ReadXls, file, colNames, sheet, type, from, rowNames, : Incorrect number of arguments (11), expecting 10 for 'ReadXls' If I just write read.xls(resultados.xlsm) It give me the same error. Regards, Sebastián. [[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] odd behavior of browser()
Untested, I think it's the blank line in your script which exits the debugger and then you're seeing regular execution. MW On Tue, Dec 4, 2012 at 5:54 PM, David Romano drom...@stanford.edu wrote: Hi everyone, I normally include a call to browser() as I'm working out the kinks in my scripts, and I am always able to step through each line by hitting Return, but for some reason, in the scripts I'm working on now, hitting Return seems to cause execution of *all* the lines in my script. I've restarted R several times in case it was stuck in a bad state for some reason, but I'm consistently getting this behavior anyway. Has anyone run into this problem before? Maybe I inadvertently reset preferences? An example which produces this behavior is the following: file bugcheck.r: browser() a - 1 b - 2 source(bugcheck.r) Called from: eval(expr, envir, enclos) Browse[1] Return ls() [1] a b a [1] 1 b [1] 2 I'd be grateful for any help in resolving this! Thanks, David Romano [[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] Read.csv
Hi list, I am using read.csv to read data from csf files, but noticed that the numeric data (those larger than 10 power 9) are rounded to the nearest million (10 power 6). Any solution? Thanks Arvin -- Sent from my mobile device __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Read.csv
On 04/12/2012 1:02 PM, Torus Insurance wrote: Hi list, I am using read.csv to read data from csf files, but noticed that the numeric data (those larger than 10 power 9) are rounded to the nearest million (10 power 6). Any solution? What makes you think that is happening? R rounds values for printing, but generally maintains about 15 digits internally. 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] Read.csv
Hi Arvin, How are you preparing the data to be read: a spreadsheet? How are you reading the data? How have you verified that the CSV file is correct? How have you verified that the data frame is incorrect? Can you provide a reproducible example using a small portion of your dataset? Sarah On Tue, Dec 4, 2012 at 1:02 PM, Torus Insurance torusinsuran...@gmail.com wrote: Hi list, I am using read.csv to read data from csf files, but noticed that the numeric data (those larger than 10 power 9) are rounded to the nearest million (10 power 6). Any solution? Thanks Arvin -- Sent from my mobile device -- 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.
[R] monte carlo simulation on R
Hello, How can I make a monte carlo simulation on R? Regards Adel -- PhD candidate in Computer Science Address 3 avenue lamine, cité ezzahra, Sousse 4000 Tunisia tel: +216 97 246 706 (+33640302046 jusqu'au 15/6) fax: +216 71 391 166 [[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] odd behavior of browser()
On 04/12/2012 12:54 PM, David Romano wrote: Hi everyone, I normally include a call to browser() as I'm working out the kinks in my scripts, and I am always able to step through each line by hitting Return, but for some reason, in the scripts I'm working on now, hitting Return seems to cause execution of *all* the lines in my script. I've restarted R several times in case it was stuck in a bad state for some reason, but I'm consistently getting this behavior anyway. Has anyone run into this problem before? Maybe I inadvertently reset preferences? I wouldn't have expected that to work. Calling browser() from within a function will let you step through the function, but calling it from within a script doesn't. Do you really have some scripts where this worked? Duncan Murdoch An example which produces this behavior is the following: file bugcheck.r: browser() a - 1 b - 2 source(bugcheck.r) Called from: eval(expr, envir, enclos) Browse[1] Return ls() [1] a b a [1] 1 b [1] 2 I'd be grateful for any help in resolving this! Thanks, David Romano [[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] monte carlo simulation on R
replicate(1000, sum(rnorm(50)^2-rchisq(50, 3))) Or you know, many other things... Michael On Tuesday, December 4, 2012, Adel ESSAFI wrote: Hello, How can I make a monte carlo simulation on R? Regards Adel -- PhD candidate in Computer Science Address 3 avenue lamine, cité ezzahra, Sousse 4000 Tunisia tel: +216 97 246 706 (+33640302046 jusqu'au 15/6) fax: +216 71 391 166 [[alternative HTML version deleted]] [[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] Winbugs from R
This is a BUGS problem, not an R problem, so would be better asked on the
BUGS list (a practical hint: if it's not working through R, run the code
directly in BUGS: the error reporting is better). Anyway, here BUGS manages
to tell R to tell you what's wrong:
variable n is not defined
And looking at your data, it's right. n is not defined
On 4 December 2012 14:17, Günal Bilek gunal...@hotmail.com wrote:
Hi,
I am trying to covert a Winbugs code into R code. Here is the winbugs code
model{# models likelihoodfor (i in 1:n){time[i] ~ dnorm( mu[i], tau ) #
stochastic componenent# link and linear predictormu[i] - beta0 + beta1 *
cases[i] + beta2 * distance[i]}# prior distributionstau ~ dgamma( 0.01,
0.01 )beta0 ~ dnorm( 0.0, 1.0E-4)beta1 ~ dnorm( 0.0, 1.0E-4)beta2 ~ dnorm(
0.0, 1.0E-4)# definition of sigmas2-1/taus -sqrt(s2)# calculation of the
sample variancefor (i in 1:n){ c.time[i]-time[i]-mean(time[]) }sy2 -
inprod( c.time[], c.time[] )/(n-1)# calculation of Bayesian version R
squaredR2B - 1 - s2/sy2# Expected y for a typical delivery timetypical.y
- beta0 + beta1 * mean(cases[]) + beta2 * mean(distance[])}INITSlist(
tau=1, beta0=1, beta1=0, beta2=0 )DATA (LIST)list( n=25,time = c(16.68,
11.5, 12.03, 14.88, 13.75, 18.11, 8, 17.83,79.24, 21.5, 40.33, 21, 13.5,
19.75, 24, 29, 15.35,19, 9.5, 35.1, 17.9, 52.32, 18.75, 19.83,
10.75),distance = c(560, 220, 340, 80, 150, 330, 110, 210, 1460,605, 688,
215, 255, 462, 448, 776, 200, 132,36, 770, 140, 810, 450, 635, 150),cases =
c( 7, 3, 3, 4, 6, 7, 2, 7, 30, 5, 16, 10, 4, 6, 9,10, 6, 7, 3, 17, 10, 26,
9, 8, 4) )
I want to do this in R. So, I copied the model and pasted into a txt file
named reg. Here is the R code I used
time -
c(16.68,11.5,12.03,14.88,13.75,18.11,8,17.83,79.24,21.5,40.33,21,13.5,19.75,24,29,15.35,19,9.5,35.1,17.9,52.32,18.75,19.83,10.75)cases
- c(7,3,3,4,6,7,2,7,30,5,16,10,4,6,9,10,6,7,3,17,10,26,9,8,4)distance -
c(560,220,340,80,150,330,110,210,1260,605,688,215,255,462,448,776,200,132,36,770,140,810,450,635,150)
data - list(time,cases,distance)
inits - function(){list(tau=1,beta1=0,beta2=0,beta3=0)}
sim - bugs(data, inits, model.file =
C:/Users/Gunal/Desktop/dummy/reg.txt,parameters = c(beta1,
beta2,beta3),n.chains = 3, n.iter = 1000,bugs.directory =
D:/PROGRAMLAR/WinBUGS14/,debug=TRUE)
Winbugs is producing this error page
display(log)check(C:/Users/Gunal/Desktop/dummy/reg.txt)model is
syntactically
correctdata(C:/Users/Gunal/AppData/Local/Temp/RtmpCYxtDZ/data.txt)data
loadedcompile(3)variable n is not
definedinits(1,C:/Users/Gunal/AppData/Local/Temp/RtmpCYxtDZ/inits1.txt)command
#Bugs:inits cannot be executed (is greyed
out)inits(2,C:/Users/Gunal/AppData/Local/Temp/RtmpCYxtDZ/inits2.txt)command
#Bugs:inits cannot be executed (is greyed
out)inits(3,C:/Users/Gunal/AppData/Local/Temp/RtmpCYxtDZ/inits3.txt)command
#Bugs:inits cannot be executed (is greyed out)gen.inits()command
#Bugs:gen.inits cannot be executed (is greyed
out)thin.updater(1)update(500)command #Bugs:update cannot be executed (is
greyed out)set(beta1)command #Bugs:set cannot be executed (is greyed
out)set(beta2)command #Bugs:set cannot be executed (is greyed
out)set(beta3)command #Bugs:set cannot be executed (is greyed
out)set(deviance)command #Bugs:set cannot be executed (is greyed
out)dic.set()command #Bugs:dic.set cannot be executed (is greyed
out)update(500)command #Bugs:update cannot be executed (is greyed
out)coda(*,C:/Users/Gunal/AppData/Local/Temp/RtmpCYxtDZ/coda)command
#Bugs:coda cannot be executed (is greyed out)stats(*)command #Bugs:stats
cannot be executed (is greyed out)dic.stats()
DIChistory(*,C:/Users/Gunal/AppData/Local/Temp/RtmpCYxtDZ/history.odc)command
#Bugs:history cannot be executed (is greyed
out)save(C:/Users/Gunal/AppData/Local/Temp/RtmpCYxtDZ/log.odc)save(C:/Users/Gunal/AppData/Local/Temp/RtmpCYxtDZ/log.txt)
Any help would be greatly appreciated.
Cheers
Gunal
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--
Bob O'Hara
Biodiversity and Climate Research Centre
Senckenberganlage 25
D-60325 Frankfurt am Main,
Germany
Tel: +49 69 798 40226
Mobile: +49 1515 888 5440
WWW: http://www.bik-f.de/root/index.php?page_id=219
Blog: http://occamstypewriter.org/boboh/
Journal of Negative Results - EEB: www.jnr-eeb.org
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and provide commented, minimal, self-contained, reproducible code.
[R] Large loops in R
Dear all, I need to access data from a large matrix (48000 x 48000) and to do it I am trying to run two loops using for command. Surely it is been a very slow job. I heard that for is not the best option to perform large loops in R, but I don't really know what would be the best (fast) option. sapply? vapply? Could anyone help me with this issue, please? Thank you very much for your attention and for any help! Best regards, Charles -- Um axé! :) -- Charles Novaes de Santana http://www.imedea.uib-csic.es/~charles PhD student - Global Change Laboratorio Internacional de Cambio Global Department of Global Change Research Instituto Mediterráneo de Estudios Avanzados(CSIC/UIB) Calle Miquel Marques 21, 07190 Esporles - Islas Baleares - España Office phone - +34 971 610 896 Cell phone - +34 660 207 940 __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] Large loops in R
On Dec 4, 2012, at 6:47 PM, Charles Novaes de Santana charles.sant...@gmail.com wrote: Dear all, I need to access data from a large matrix (48000 x 48000) and to do it I am trying to run two loops using for command. Surely it is been a very slow job. I heard that for is not the best option to perform large loops in R, but I don't really know what would be the best (fast) option. sapply? vapply? Could anyone help me with this issue, please? What exactly are you trying to do? It's likely doable with a few vectorized operations. Michael Thank you very much for your attention and for any help! Best regards, Charles -- Um axé! :) -- Charles Novaes de Santana http://www.imedea.uib-csic.es/~charles PhD student - Global Change Laboratorio Internacional de Cambio Global Department of Global Change Research Instituto Mediterráneo de Estudios Avanzados(CSIC/UIB) Calle Miquel Marques 21, 07190 Esporles - Islas Baleares - España Office phone - +34 971 610 896 Cell phone - +34 660 207 940 __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] control point size of superscript when labeling axes with title()
Hi-
A journal has asked me to make all of my text annotations on a figure at
10-point size. For the most part this is easy, e.g. by creating figures with:
pdf(..., family='Times', pointsize=10)
But where I have superscripts (or subscripts) in axis labels, the default seems
to be to shrink the superscripted text slightly. For example this code:
title(ylab=expression(paste('Respiration (mg ',O[2],' ',L^-1,'
',d^-1,')',sep=' ')),outer=T,line=0.3)
produces superscripted numbers at approximately 7 point.
I have been poking around for a solution but not having much luck. The
textstyle() function keeps superscripted text at original size if you use it
within a text() call, but I can't figure out an equivalent solution within a
title() call.
Thanks for any suggestions
Chris
--
Dr. Chris Solomon
Assistant Professor
Dept. of Natural Resource Sciences
McGill University
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Re: [R] Large loops in R
Dear Michael,
Thank you for your answer.
I have 2 matrices. Each position of the matrices is a weight. And I
need to calculate the following sum of differences:
Considering:
mat1 and mat2 - two matrices (each of them 48000 x 48000).
d1 and d2 - two constant values.
sum-0;
for(i in 1:nrows1){
for(j in 1:nrows2){
sum-sum+ ( ( (mat1(i,j)/d1) -
(mat2(i,j)/d2) )^2 )
}
}
}
I was wondering if there is a better way to do this sum.
Thank you for your attention!
Best,
Charles
On Tue, Dec 4, 2012 at 7:54 PM, R. Michael Weylandt
michael.weyla...@gmail.com michael.weyla...@gmail.com wrote:
On Dec 4, 2012, at 6:47 PM, Charles Novaes de Santana
charles.sant...@gmail.com wrote:
Dear all,
I need to access data from a large matrix (48000 x 48000) and to do it
I am trying to run two loops using for command. Surely it is been a
very slow job.
I heard that for is not the best option to perform large loops in R,
but I don't really know what would be the best (fast) option. sapply?
vapply? Could anyone help me with this issue, please?
What exactly are you trying to do? It's likely doable with a few vectorized
operations.
Michael
Thank you very much for your attention and for any help!
Best regards,
Charles
--
Um axé! :)
--
Charles Novaes de Santana
http://www.imedea.uib-csic.es/~charles
PhD student - Global Change
Laboratorio Internacional de Cambio Global
Department of Global Change Research
Instituto Mediterráneo de Estudios Avanzados(CSIC/UIB)
Calle Miquel Marques 21, 07190
Esporles - Islas Baleares - España
Office phone - +34 971 610 896
Cell phone - +34 660 207 940
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
--
Um axé! :)
--
Charles Novaes de Santana
http://www.imedea.uib-csic.es/~charles
PhD student - Global Change
Laboratorio Internacional de Cambio Global
Department of Global Change Research
Instituto Mediterráneo de Estudios Avanzados(CSIC/UIB)
Calle Miquel Marques 21, 07190
Esporles - Islas Baleares - España
Office phone - +34 971 610 896
Cell phone - +34 660 207 940
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/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] Large loops in R
Without a reproducible example it's hard to tell for certain, but what
about simply (assuming nrows2 is actually columns):
sum((mat1/d1 - mat2/d2)^2)
R is smart enough to understand elementwise manipulation of a matrix:
you shouldn't need a loop at all.
Sarah
On Tue, Dec 4, 2012 at 2:27 PM, Charles Novaes de Santana
charles.sant...@gmail.com wrote:
Dear Michael,
Thank you for your answer.
I have 2 matrices. Each position of the matrices is a weight. And I
need to calculate the following sum of differences:
Considering:
mat1 and mat2 - two matrices (each of them 48000 x 48000).
d1 and d2 - two constant values.
sum-0;
for(i in 1:nrows1){
for(j in 1:nrows2){
sum-sum+ ( ( (mat1(i,j)/d1) -
(mat2(i,j)/d2) )^2 )
}
}
}
I was wondering if there is a better way to do this sum.
Thank you for your attention!
Best,
Charles
On Tue, Dec 4, 2012 at 7:54 PM, R. Michael Weylandt
michael.weyla...@gmail.com michael.weyla...@gmail.com wrote:
On Dec 4, 2012, at 6:47 PM, Charles Novaes de Santana
charles.sant...@gmail.com wrote:
Dear all,
I need to access data from a large matrix (48000 x 48000) and to do it
I am trying to run two loops using for command. Surely it is been a
very slow job.
I heard that for is not the best option to perform large loops in R,
but I don't really know what would be the best (fast) option. sapply?
vapply? Could anyone help me with this issue, please?
What exactly are you trying to do? It's likely doable with a few vectorized
operations.
Michael
--
Sarah Goslee
http://www.functionaldiversity.org
__
R-help@r-project.org mailing list
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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] Large loops in R
On Tue, Dec 4, 2012 at 11:27 AM, Charles Novaes de Santana
charles.sant...@gmail.com wrote:
Dear Michael,
Thank you for your answer.
I have 2 matrices. Each position of the matrices is a weight. And I
need to calculate the following sum of differences:
Considering:
mat1 and mat2 - two matrices (each of them 48000 x 48000).
d1 and d2 - two constant values.
sum-0;
for(i in 1:nrows1){
for(j in 1:nrows2){
sum-sum+ ( ( (mat1(i,j)/d1) -
(mat2(i,j)/d2) )^2 )
}
}
}
I was wondering if there is a better way to do this sum.
sum( (mat1/d1-mat2/d2)^2)
Correct me if I'm wrong though - aren't matrices of 48x times 48k
larger than what R can handle at present?
HTH
Peter
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[R] list to matrix?
How do I convert a list to a matrix? --8---cut here---start-8--- list(c(5, 101), c(1e+05, 46), c(15, 31), c(2e+05, 17), c(25, 19), c(3e+05, 11), c(35, 12), c(4e+05, 25), c(45, 19), c(5e+05, 16)) as.matrix(a) [,1] [1,] Numeric,2 [2,] Numeric,2 [3,] Numeric,2 [4,] Numeric,2 [5,] Numeric,2 [6,] Numeric,2 [7,] Numeric,2 [8,] Numeric,2 [9,] Numeric,2 --8---cut here---end---8--- thanks! -- Sam Steingold (http://sds.podval.org/) on Ubuntu 12.04 (precise) X 11.0.11103000 http://www.childpsy.net/ http://palestinefacts.org http://dhimmi.com http://jihadwatch.org http://www.PetitionOnline.com/tap12009/ http://memri.org Rhinoceros has poor vision, but, due to his size, it's not his problem. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] monte carlo simulation on R
There is a book on MC with R: Introducing Monte Carlo Methods with R by Robert/Casella: http://www.springer.com/statistics/computational+statistics/book/978-1-4419-1575-7 On 4 December 2012 19:21, Adel ESSAFI adeless...@gmail.com wrote: Hello, How can I make a monte carlo simulation on R? Regards Adel -- PhD candidate in Computer Science Address 3 avenue lamine, cité ezzahra, Sousse 4000 Tunisia tel: +216 97 246 706 (+33640302046 jusqu'au 15/6) fax: +216 71 391 166 [[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] Large loops in R
On Tue, Dec 4, 2012 at 8:43 PM, Peter Langfelder
peter.langfel...@gmail.com wrote:
On Tue, Dec 4, 2012 at 11:27 AM, Charles Novaes de Santana
charles.sant...@gmail.com wrote:
Dear Michael,
Thank you for your answer.
I have 2 matrices. Each position of the matrices is a weight. And I
need to calculate the following sum of differences:
Considering:
mat1 and mat2 - two matrices (each of them 48000 x 48000).
d1 and d2 - two constant values.
sum-0;
for(i in 1:nrows1){
for(j in 1:nrows2){
sum-sum+ ( ( (mat1(i,j)/d1) -
(mat2(i,j)/d2) )^2 )
}
}
}
I was wondering if there is a better way to do this sum.
sum( (mat1/d1-mat2/d2)^2)
Correct me if I'm wrong though - aren't matrices of 48x times 48k
larger than what R can handle at present?
HTH
Peter
hmmm I didn't know that the limitation of R was below this value. I
found this error message:
Error in matrix(0, 48000, 48000) : too many elements specified
but I thought it was a machine limitation (and I was asking for access
to a better machine in my labs...). Thanks for clarifying it.
Well, when Sarah gave me the answer for my problem, I got a new one :)
Thank you, Sarah and Peter.
So, is there any other way to trick R and allocate such large matrices?
Best,
Charles
--
Um axé! :)
--
Charles Novaes de Santana
http://www.imedea.uib-csic.es/~charles
PhD student - Global Change
Laboratorio Internacional de Cambio Global
Department of Global Change Research
Instituto Mediterráneo de Estudios Avanzados(CSIC/UIB)
Calle Miquel Marques 21, 07190
Esporles - Islas Baleares - España
Office phone - +34 971 610 896
Cell phone - +34 660 207 940
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] list to matrix?
Try: matrix(unlist(a), ncol=2, byrow=T) --Mark Lamias From: Sam Steingold s...@gnu.org To: r-help@r-project.org Sent: Tuesday, December 4, 2012 3:09 PM Subject: [R] list to matrix? How do I convert a list to a matrix? --8---cut here---start-8--- list(c(5, 101), c(1e+05, 46), c(15, 31), c(2e+05, 17), c(25, 19), c(3e+05, 11), c(35, 12), c(4e+05, 25), c(45, 19), c(5e+05, 16)) as.matrix(a) [,1] [1,] Numeric,2 [2,] Numeric,2 [3,] Numeric,2 [4,] Numeric,2 [5,] Numeric,2 [6,] Numeric,2 [7,] Numeric,2 [8,] Numeric,2 [9,] Numeric,2 --8---cut here---end---8--- thanks! -- Sam Steingold (http://sds.podval.org/) on Ubuntu 12.04 (precise) X 11.0.11103000 http://www.childpsy.net/ http://palestinefacts.org http://dhimmi.com http://jihadwatch.org http://www.PetitionOnline.com/tap12009/ http://memri.org Rhinoceros has poor vision, but, due to his size, it's not his problem. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] list to matrix?
On Tue, Dec 4, 2012 at 8:09 PM, Sam Steingold s...@gnu.org wrote: How do I convert a list to a matrix? --8---cut here---start-8--- list(c(5, 101), c(1e+05, 46), c(15, 31), c(2e+05, 17), c(25, 19), c(3e+05, 11), c(35, 12), c(4e+05, 25), c(45, 19), c(5e+05, 16)) do.call(rbind, LIST) or do.call(cbind, LIST) depending on your desired direction. The do.call syntax takes a function name and uses a list as arguments to that function, returning the result. Super useful for these situations where you collect things and something like cbind(x[[1]],x[[2]], x[[3]]...) isn't feasible or possible. MW __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] Large loops in R
Thank you, Sarah! It is a wonderful new!!! :)
Now I need to solve the other question hehe How to allocate such large matrix :)
best,
Charles
On Tue, Dec 4, 2012 at 8:39 PM, Sarah Goslee sarah.gos...@gmail.com wrote:
Without a reproducible example it's hard to tell for certain, but what
about simply (assuming nrows2 is actually columns):
sum((mat1/d1 - mat2/d2)^2)
R is smart enough to understand elementwise manipulation of a matrix:
you shouldn't need a loop at all.
Sarah
On Tue, Dec 4, 2012 at 2:27 PM, Charles Novaes de Santana
charles.sant...@gmail.com wrote:
Dear Michael,
Thank you for your answer.
I have 2 matrices. Each position of the matrices is a weight. And I
need to calculate the following sum of differences:
Considering:
mat1 and mat2 - two matrices (each of them 48000 x 48000).
d1 and d2 - two constant values.
sum-0;
for(i in 1:nrows1){
for(j in 1:nrows2){
sum-sum+ ( ( (mat1(i,j)/d1) -
(mat2(i,j)/d2) )^2 )
}
}
}
I was wondering if there is a better way to do this sum.
Thank you for your attention!
Best,
Charles
On Tue, Dec 4, 2012 at 7:54 PM, R. Michael Weylandt
michael.weyla...@gmail.com michael.weyla...@gmail.com wrote:
On Dec 4, 2012, at 6:47 PM, Charles Novaes de Santana
charles.sant...@gmail.com wrote:
Dear all,
I need to access data from a large matrix (48000 x 48000) and to do it
I am trying to run two loops using for command. Surely it is been a
very slow job.
I heard that for is not the best option to perform large loops in R,
but I don't really know what would be the best (fast) option. sapply?
vapply? Could anyone help me with this issue, please?
What exactly are you trying to do? It's likely doable with a few vectorized
operations.
Michael
--
Sarah Goslee
http://www.functionaldiversity.org
--
Um axé! :)
--
Charles Novaes de Santana
http://www.imedea.uib-csic.es/~charles
PhD student - Global Change
Laboratorio Internacional de Cambio Global
Department of Global Change Research
Instituto Mediterráneo de Estudios Avanzados(CSIC/UIB)
Calle Miquel Marques 21, 07190
Esporles - Islas Baleares - España
Office phone - +34 971 610 896
Cell phone - +34 660 207 940
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/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] Large loops in R
On Tue, Dec 4, 2012 at 8:14 PM, Charles Novaes de Santana charles.sant...@gmail.com wrote: Error in matrix(0, 48000, 48000) : too many elements specified but I thought it was a machine limitation (and I was asking for access to a better machine in my labs...). Thanks for clarifying it. Well, when Sarah gave me the answer for my problem, I got a new one :) Thank you, Sarah and Peter. So, is there any other way to trick R and allocate such large matrices? Either 1) Use the development version of R which has large-vector support (matrices are vectors under the hood) or 2) A large matrix is usually sparse in structure: use one of the many sparse representation packages (e.g., Matrix) available. MW __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] Large loops in R
I don't think there's any reason for the calculation you're doing that
you must have the whole matrix in memory, is there?
Unless there's something more than what you've shown us, you're just
taking the sum of elementwise operations. You can read the matrix in
in manageable chunks, take the sum of that chunk and save the single
value. Repeat, then add them all up at the end.
Sarah
On Tue, Dec 4, 2012 at 3:14 PM, Charles Novaes de Santana
charles.sant...@gmail.com wrote:
On Tue, Dec 4, 2012 at 8:43 PM, Peter Langfelder
peter.langfel...@gmail.com wrote:
On Tue, Dec 4, 2012 at 11:27 AM, Charles Novaes de Santana
charles.sant...@gmail.com wrote:
Dear Michael,
Thank you for your answer.
I have 2 matrices. Each position of the matrices is a weight. And I
need to calculate the following sum of differences:
Considering:
mat1 and mat2 - two matrices (each of them 48000 x 48000).
d1 and d2 - two constant values.
sum-0;
for(i in 1:nrows1){
for(j in 1:nrows2){
sum-sum+ ( ( (mat1(i,j)/d1) -
(mat2(i,j)/d2) )^2 )
}
}
}
I was wondering if there is a better way to do this sum.
sum( (mat1/d1-mat2/d2)^2)
Correct me if I'm wrong though - aren't matrices of 48x times 48k
larger than what R can handle at present?
HTH
Peter
hmmm I didn't know that the limitation of R was below this value. I
found this error message:
Error in matrix(0, 48000, 48000) : too many elements specified
but I thought it was a machine limitation (and I was asking for access
to a better machine in my labs...). Thanks for clarifying it.
Well, when Sarah gave me the answer for my problem, I got a new one :)
Thank you, Sarah and Peter.
So, is there any other way to trick R and allocate such large matrices?
Best,
Charles
--
Um axé! :)
--
Sarah Goslee
http://www.functionaldiversity.org
__
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[R] How to find matching columns in a matrix of lists?
Dear R users, I have a matrix composed of lists: m - matrix( list(), nrow=1, ncol=3 ) m[[ 1, 1 ]] - list(A, B) m[[ 1, 2 ]] - list(A, C) m[[ 1, 3 ]] - list(A, B) and want to get the sub-matrix where cells contain B. But m[ , B %in% m[ 1, ], drop=F ] as well as m[ , B %in% m[ 1, ][], drop=F ] return empty matrices. Any ideas, hints and help will be very much appreciated! Kind regards! Josef [[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] reformatting some data
Hello, I am trying to reformat some data so that it is organized by group in the columns. The data currently looks like this: group X3.Hydroxybutyrate X3.Hydroxyisovalerate ADP 347 4 4e-04 3e-04 5e-04 353 3 5e-04 3e-04 6e-04 359 4 4e-04 3e-04 6e-04 365 4 6e-04 3e-04 5e-04 371 4 5e-04 3e-04 7e-04 377 2 7e-04 4e-04 7e-04 I would like to reformat it so it is like this: 2 3 4 var1 var2 var3 I realize that there unequal numbers in each group but I would like to none-the-less if possible. Here is a subset of the data: structure(list(group = c(4L, 3L, 4L, 4L, 4L, 2L), X3.Hydroxybutyrate = c(4e-04, 5e-04, 4e-04, 6e-04, 5e-04, 7e-04), X3.Hydroxyisovalerate = c(3e-04, 3e-04, 3e-04, 3e-04, 3e-04, 4e-04), ADP = c(5e-04, 6e-04, 6e-04, 5e-04, 7e-04, 7e-04)), .Names = c(group, X3.Hydroxybutyrate, X3.Hydroxyisovalerate, ADP), row.names = c(347L, 353L, 359L, 365L, 371L, 377L), class = data.frame) Any insight is truly appreciated, Regards, Charles [[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] reformatting some data
It's not clear to me what it is you are attempting to do, as you switch from a very specific example to some general example with the vague terms var1 var2, and var3. It sounds like you might be trying to do something similar to what would be available in the shape package using the melt function. Try ?melt.data.frame. --Mark Lamias From: Charles Determan Jr deter...@umn.edu To: r-help@r-project.org Sent: Tuesday, December 4, 2012 4:17 PM Subject: [R] reformatting some data Hello, I am trying to reformat some data so that it is organized by group in the columns. The data currently looks like this: group X3.Hydroxybutyrate X3.Hydroxyisovalerate ADP 347 4 4e-04 3e-04 5e-04 353 3 5e-04 3e-04 6e-04 359 4 4e-04 3e-04 6e-04 365 4 6e-04 3e-04 5e-04 371 4 5e-04 3e-04 7e-04 377 2 7e-04 4e-04 7e-04 I would like to reformat it so it is like this: 2 3 4 var1 var2 var3 I realize that there unequal numbers in each group but I would like to none-the-less if possible. Here is a subset of the data: structure(list(group = c(4L, 3L, 4L, 4L, 4L, 2L), X3.Hydroxybutyrate = c(4e-04, 5e-04, 4e-04, 6e-04, 5e-04, 7e-04), X3.Hydroxyisovalerate = c(3e-04, 3e-04, 3e-04, 3e-04, 3e-04, 4e-04), ADP = c(5e-04, 6e-04, 6e-04, 5e-04, 7e-04, 7e-04)), .Names = c(group, X3.Hydroxybutyrate, X3.Hydroxyisovalerate, ADP), row.names = c(347L, 353L, 359L, 365L, 371L, 377L), class = data.frame) Any insight is truly appreciated, Regards, Charles [[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] reformatting some data
That should read the reshape package -- not the shape package. My apologies. To: Charles Determan Jr deter...@umn.edu; r-help@r-project.org r-help@r-project.org Sent: Tuesday, December 4, 2012 4:36 PM Subject: Re: [R] reformatting some data It's not clear to me what it is you are attempting to do, as you switch from a very specific example to some general example with the vague terms var1 var2, and var3. It sounds like you might be trying to do something similar to what would be available in the shape package using the melt function. Try ?melt.data.frame. --Mark Lamias From: Charles Determan Jr deter...@umn.edu To: r-help@r-project.org Sent: Tuesday, December 4, 2012 4:17 PM Subject: [R] reformatting some data Hello, I am trying to reformat some data so that it is organized by group in the columns. The data currently looks like this: group X3.Hydroxybutyrate X3.Hydroxyisovalerate ADP 347 4 4e-04 3e-04 5e-04 353 3 5e-04 3e-04 6e-04 359 4 4e-04 3e-04 6e-04 365 4 6e-04 3e-04 5e-04 371 4 5e-04 3e-04 7e-04 377 2 7e-04 4e-04 7e-04 I would like to reformat it so it is like this: 2 3 4 var1 var2 var3 I realize that there unequal numbers in each group but I would like to none-the-less if possible. Here is a subset of the data: structure(list(group = c(4L, 3L, 4L, 4L, 4L, 2L), X3.Hydroxybutyrate = c(4e-04, 5e-04, 4e-04, 6e-04, 5e-04, 7e-04), X3.Hydroxyisovalerate = c(3e-04, 3e-04, 3e-04, 3e-04, 3e-04, 4e-04), ADP = c(5e-04, 6e-04, 6e-04, 5e-04, 7e-04, 7e-04)), .Names = c(group, X3.Hydroxybutyrate, X3.Hydroxyisovalerate, ADP), row.names = c(347L, 353L, 359L, 365L, 371L, 377L), class = data.frame) Any insight is truly appreciated, Regards, Charles [[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] monte carlo simulation on R
There is a book on MC with R: Introducing Monte Carlo Methods with R by Robert/Casella: http://www.springer.com/statistics/computational+statistics/book/978-1-4419-1575-7 On 4 December 2012 19:38, R. Michael Weylandt michael.weyla...@gmail.com wrote: replicate(1000, sum(rnorm(50)^2-rchisq(50, 3))) Or you know, many other things... Michael On Tuesday, December 4, 2012, Adel ESSAFI wrote: Hello, How can I make a monte carlo simulation on R? Regards Adel -- PhD candidate in Computer Science Address 3 avenue lamine, cité ezzahra, Sousse 4000 Tunisia tel: +216 97 246 706 (+33640302046 jusqu'au 15/6) fax: +216 71 391 166 [[alternative HTML version deleted]] [[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] Large loops in R
HI,
I just wonder whether your code worked or not.
set.seed(8)
mat1-matrix(sample(1:80,40,replace=TRUE),ncol=8)
set.seed(25)
mat2-matrix(sample(1:160,40,replace=TRUE),ncol=8)
#Since the dimensions are the same,
m-1:5
n-1:8
sum1-0
for(i in 1:length(m)){
for(j in 1:length(n)){
sum1-sum1+(((mat1[i,j]/d1)-(mat2[i,j]/d2))^2)
}
}
sum1
#[1] 15192.89
#Sara's code:
sum((mat1/d1 - mat2/d2)^2)
#[1] 15192.89
A.K.
- Original Message -
From: Charles Novaes de Santana charles.sant...@gmail.com
To: r-help@r-project.org r-help@r-project.org
Cc:
Sent: Tuesday, December 4, 2012 2:27 PM
Subject: Re: [R] Large loops in R
Dear Michael,
Thank you for your answer.
I have 2 matrices. Each position of the matrices is a weight. And I
need to calculate the following sum of differences:
Considering:
mat1 and mat2 - two matrices (each of them 48000 x 48000).
d1 and d2 - two constant values.
sum-0;
for(i in 1:nrows1){
for(j in 1:nrows2){
sum-sum+ ( ( (mat1(i,j)/d1) -
(mat2(i,j)/d2) )^2 )
}
}
}
I was wondering if there is a better way to do this sum.
Thank you for your attention!
Best,
Charles
On Tue, Dec 4, 2012 at 7:54 PM, R. Michael Weylandt
michael.weyla...@gmail.com michael.weyla...@gmail.com wrote:
On Dec 4, 2012, at 6:47 PM, Charles Novaes de Santana
charles.sant...@gmail.com wrote:
Dear all,
I need to access data from a large matrix (48000 x 48000) and to do it
I am trying to run two loops using for command. Surely it is been a
very slow job.
I heard that for is not the best option to perform large loops in R,
but I don't really know what would be the best (fast) option. sapply?
vapply? Could anyone help me with this issue, please?
What exactly are you trying to do? It's likely doable with a few vectorized
operations.
Michael
Thank you very much for your attention and for any help!
Best regards,
Charles
--
Um axé! :)
--
Charles Novaes de Santana
http://www.imedea.uib-csic.es/~charles
PhD student - Global Change
Laboratorio Internacional de Cambio Global
Department of Global Change Research
Instituto Mediterráneo de Estudios Avanzados(CSIC/UIB)
Calle Miquel Marques 21, 07190
Esporles - Islas Baleares - España
Office phone - +34 971 610 896
Cell phone - +34 660 207 940
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
--
Um axé! :)
--
Charles Novaes de Santana
http://www.imedea.uib-csic.es/~charles
PhD student - Global Change
Laboratorio Internacional de Cambio Global
Department of Global Change Research
Instituto Mediterráneo de Estudios Avanzados(CSIC/UIB)
Calle Miquel Marques 21, 07190
Esporles - Islas Baleares - España
Office phone - +34 971 610 896
Cell phone - +34 660 207 940
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/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] computing marginal values based on multiple columns?
Thanks to both of you! I learned something new from both of your posts :-)
And you were both right, my example numbers were wrong! I accidentally
computed them based on 0.25 * mean instead of 0.2 * mean.
Thanks again!
Simon
On Wed, Dec 5, 2012 at 4:07 AM, arun smartpink...@yahoo.com wrote:
HI,
I am not sure the output you wanted is correct:
sample1 sample2 sample3
1 1.0 0 0.5
because
0.2*colMeans(x[,-4])
sample1 sample2 sample3
# 28.40 24.08 21.36
This might help you:
apply(x[-4],2,function(y) length(y[y 0.2*mean(y)
x$class==a])/length(x[x$class==a]))
#sample1 sample2 sample3
# 0.0 0.0 0.5
A.K.
- Original Message -
From: Simon simonzm...@gmail.com
To: r-help@r-project.org
Cc:
Sent: Tuesday, December 4, 2012 4:49 AM
Subject: [R] computing marginal values based on multiple columns?
Hello all,
I have what feels like a simple problem, but I can't find an simple
answer. Consider this data frame:
x - data.frame(sample1=c(35,176,182,193,124),
sample2=c(198,176,190,23,15), sample3=c(12,154,21,191,156),
class=c('a','a','c','b','c'))
x
sample1 sample2 sample3 class
1 35 198 12 a
2 176 176 154 a
3 182 190 21 c
4 193 23 191 b
5 124 15 156 c
Now I wish to know: for each sample, for values 20% of the sample mean,
what percentage of those are class a?
I want to end up with a table like:
sample1 sample2 sample3
1 1.0 0 0.5
I can calculate this for an individual sample using this rather clumsy
expression:
length(which(x$sample1 mean(x$sample1) x$class=='a')) /
length(which(x$sample1 mean(x$sample1)))
I'd normally propagate it across the data frame using apply, but I
can't because it depends on more than one column.
Any help much appreciated!
Cheers,
Simon
[[alternative HTML version deleted]]
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[[alternative HTML version deleted]]
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Re: [R] list to matrix?
Hi, Try this: list1-list(c(5, 101), c(1e+05, 46), c(15, 31), c(2e+05, 17), c(25, 19), c(3e+05, 11), c(35, 12), c(4e+05, 25), c(45, 19), c(5e+05, 16)) res-t(sapply(list1,function(x) x)) res # [,1] [,2] #[1,] 5 101 #[2,] 10 46 #[3,] 15 31 #[4,] 20 17 #[5,] 25 19 #[6,] 30 11 #[7,] 35 12 #[8,] 40 25 #[9,] 45 19 #[10,] 50 16 is.matrix(res) #[1] TRUE #or res1-sapply(list1,function(x) x) is.matrix(res1) #[1] TRUE A.K. - Original Message - From: Sam Steingold s...@gnu.org To: r-help@r-project.org Cc: Sent: Tuesday, December 4, 2012 3:09 PM Subject: [R] list to matrix? How do I convert a list to a matrix? --8---cut here---start-8--- list(c(5, 101), c(1e+05, 46), c(15, 31), c(2e+05, 17), c(25, 19), c(3e+05, 11), c(35, 12), c(4e+05, 25), c(45, 19), c(5e+05, 16)) as.matrix(a) [,1] [1,] Numeric,2 [2,] Numeric,2 [3,] Numeric,2 [4,] Numeric,2 [5,] Numeric,2 [6,] Numeric,2 [7,] Numeric,2 [8,] Numeric,2 [9,] Numeric,2 --8---cut here---end---8--- thanks! -- Sam Steingold (http://sds.podval.org/) on Ubuntu 12.04 (precise) X 11.0.11103000 http://www.childpsy.net/ http://palestinefacts.org http://dhimmi.com http://jihadwatch.org http://www.PetitionOnline.com/tap12009/ http://memri.org Rhinoceros has poor vision, but, due to his size, it's not his problem. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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 do I get internal nodes of dendograms produced by R?
I am using R for hierarchical clustering of a number of documents. I have a distance matrix on which I have applied hclust method. When I plot the result of hclust method, I can see the dendogram plotted. What I need now is the dendogram stored as a tree in a data structure. My goal is to automatically label all internal nodes. For that, I need to know, which leaf nodes make a first level cluster, and which first level clusters make a second level cluster and so on. Is there a way in R to get this information? -- Sagnik Ray Choudhury sag...@psu.edu [[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 find matching columns in a matrix of lists?
Hello, m[ , sapply(1:ncol(m), function(j) sapply(B, `%in%`, m[[1 , j]])), drop=F ] It indeed does. Thank you very much! [[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] reformatting some data
Hi, Not sure whether this helps: library(reshape2) dat1-structure(list(group = c(4L, 3L, 4L, 4L, 4L, 2L), X3.Hydroxybutyrate = c(4e-04, 5e-04, 4e-04, 6e-04, 5e-04, 7e-04), X3.Hydroxyisovalerate = c(3e-04, 3e-04, 3e-04, 3e-04, 3e-04, 4e-04), ADP = c(5e-04, 6e-04, 6e-04, 5e-04, 7e-04, 7e-04)), .Names = c(group, X3.Hydroxybutyrate, X3.Hydroxyisovalerate, ADP), row.names = c(347L, 353L, 359L, 365L, 371L, 377L), class = data.frame) datM-melt(dat1,id.var=group) dcast(datM,variable~group,length) #or dcast(datM,variable~group,mean) # variable 2 3 4 #1 X3.Hydroxybutyrate 7e-04 5e-04 0.000475 #2 X3.Hydroxyisovalerate 4e-04 3e-04 0.000300 #3 ADP 7e-04 6e-04 0.000575 A.K - Original Message - From: Charles Determan Jr deter...@umn.edu To: r-help@r-project.org Cc: Sent: Tuesday, December 4, 2012 4:17 PM Subject: [R] reformatting some data Hello, I am trying to reformat some data so that it is organized by group in the columns. The data currently looks like this: group X3.Hydroxybutyrate X3.Hydroxyisovalerate ADP 347 4 4e-04 3e-04 5e-04 353 3 5e-04 3e-04 6e-04 359 4 4e-04 3e-04 6e-04 365 4 6e-04 3e-04 5e-04 371 4 5e-04 3e-04 7e-04 377 2 7e-04 4e-04 7e-04 I would like to reformat it so it is like this: 2 3 4 var1 var2 var3 I realize that there unequal numbers in each group but I would like to none-the-less if possible. Here is a subset of the data: structure(list(group = c(4L, 3L, 4L, 4L, 4L, 2L), X3.Hydroxybutyrate = c(4e-04, 5e-04, 4e-04, 6e-04, 5e-04, 7e-04), X3.Hydroxyisovalerate = c(3e-04, 3e-04, 3e-04, 3e-04, 3e-04, 4e-04), ADP = c(5e-04, 6e-04, 6e-04, 5e-04, 7e-04, 7e-04)), .Names = c(group, X3.Hydroxybutyrate, X3.Hydroxyisovalerate, ADP), row.names = c(347L, 353L, 359L, 365L, 371L, 377L), class = data.frame) Any insight is truly appreciated, Regards, Charles [[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] reformatting some data
Hi, You can also do this: dat1-structure(list(group = c(4L, 3L, 4L, 4L, 4L, 2L), X3.Hydroxybutyrate = c(4e-04, 5e-04, 4e-04, 6e-04, 5e-04, 7e-04), X3.Hydroxyisovalerate = c(3e-04, 3e-04, 3e-04, 3e-04, 3e-04, 4e-04), ADP = c(5e-04, 6e-04, 6e-04, 5e-04, 7e-04, 7e-04)), .Names = c(group, X3.Hydroxybutyrate, X3.Hydroxyisovalerate, ADP), row.names = c(347L, 353L, 359L, 365L, 371L, 377L), class = data.frame) datM-melt(dat1,id.var=group) xtabs(value~variable+group,data=datM) # group #variable 2 3 4 # X3.Hydroxybutyrate 0.0007 0.0005 0.0019 # X3.Hydroxyisovalerate 0.0004 0.0003 0.0012 # ADP 0.0007 0.0006 0.0023 A.K. - Original Message - From: Charles Determan Jr deter...@umn.edu To: r-help@r-project.org Cc: Sent: Tuesday, December 4, 2012 4:17 PM Subject: [R] reformatting some data Hello, I am trying to reformat some data so that it is organized by group in the columns. The data currently looks like this: group X3.Hydroxybutyrate X3.Hydroxyisovalerate ADP 347 4 4e-04 3e-04 5e-04 353 3 5e-04 3e-04 6e-04 359 4 4e-04 3e-04 6e-04 365 4 6e-04 3e-04 5e-04 371 4 5e-04 3e-04 7e-04 377 2 7e-04 4e-04 7e-04 I would like to reformat it so it is like this: 2 3 4 var1 var2 var3 I realize that there unequal numbers in each group but I would like to none-the-less if possible. Here is a subset of the data: structure(list(group = c(4L, 3L, 4L, 4L, 4L, 2L), X3.Hydroxybutyrate = c(4e-04, 5e-04, 4e-04, 6e-04, 5e-04, 7e-04), X3.Hydroxyisovalerate = c(3e-04, 3e-04, 3e-04, 3e-04, 3e-04, 4e-04), ADP = c(5e-04, 6e-04, 6e-04, 5e-04, 7e-04, 7e-04)), .Names = c(group, X3.Hydroxybutyrate, X3.Hydroxyisovalerate, ADP), row.names = c(347L, 353L, 359L, 365L, 371L, 377L), class = data.frame) Any insight is truly appreciated, Regards, Charles [[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 find matching columns in a matrix of lists?
HI, Does this work for you? mapply(function(x) x==B,m) [,1] [,2] [,3] #[1,] FALSE FALSE FALSE #[2,] TRUE FALSE TRUE A.K. - Original Message - From: Asis Hallab asis.hal...@gmail.com To: r-help@r-project.org Cc: Sent: Tuesday, December 4, 2012 4:16 PM Subject: [R] How to find matching columns in a matrix of lists? Dear R users, I have a matrix composed of lists: m - matrix( list(), nrow=1, ncol=3 ) m[[ 1, 1 ]] - list(A, B) m[[ 1, 2 ]] - list(A, C) m[[ 1, 3 ]] - list(A, B) and want to get the sub-matrix where cells contain B. But m[ , B %in% m[ 1, ], drop=F ] as well as m[ , B %in% m[ 1, ][], drop=F ] return empty matrices. Any ideas, hints and help will be very much appreciated! Kind regards! Josef [[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 find matching columns in a matrix of lists?
Hello, Try the following. m[ , sapply(1:ncol(m), function(j) sapply(B, `%in%`, m[[1 , j]])), drop=F ] Hope this helps, Rui Barradas Em 04-12-2012 21:16, Asis Hallab escreveu: Dear R users, I have a matrix composed of lists: m - matrix( list(), nrow=1, ncol=3 ) m[[ 1, 1 ]] - list(A, B) m[[ 1, 2 ]] - list(A, C) m[[ 1, 3 ]] - list(A, B) and want to get the sub-matrix where cells contain B. But m[ , B %in% m[ 1, ], drop=F ] as well as m[ , B %in% m[ 1, ][], drop=F ] return empty matrices. Any ideas, hints and help will be very much appreciated! Kind regards! Josef [[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] list to matrix?
On Tue, Dec 4, 2012 at 8:17 PM, arun smartpink...@yahoo.com wrote: Hi, Try this: list1-list(c(5, 101), c(1e+05, 46), c(15, 31), c(2e+05, 17), c(25, 19), c(3e+05, 11), c(35, 12), c(4e+05, 25), c(45, 19), c(5e+05, 16)) res-t(sapply(list1,function(x) x)) Bah Humbug! (In Christmas cheer) No need for all this (see solutions including mine already given) -- but even without those, this is silly. An identity map is a real waste if you just want the simplification bit of sapply() -- you'd be much better just using simplify2array() Michael __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] latticeExtra tileplot question - tiles are not all the same size, need help.
Hello,
I have been creating many tileplots to try and illustrate the relative
abundance of fish through space and time. My issue is that the tiles that
border the plot are smaller than those in the center of the plot. In the
example I've provided the effect is pretty minor (I'm hoping this will be an
adequate example as I had the code already created/data uploaded). However,
I have other plots that have fewer space-time strata (fewer tiles) and the
effect is very noticeable.
Code:
##You will need to load the following packages: latticeExtra, gridExtra and
RCurl
#Upload data
cwtb_csv-getURL(https://docs.google.com/spreadsheet/pub?key=0AjzYZNH9Dw9qd
Fc5c1FXZ19uYXE3U1QwU1MxVkR4dGcsingle=truegid=0output=csv,ssl.verifypeer
= FALSE)
cwtb-read.csv(textConnection(cwtb_csv),header=T)
gsi_csv-getURL(https://docs.google.com/spreadsheet/pub?key=0AjzYZNH9Dw9qdD
NTN2djUEl3UVdhZ0t4ZXZrdHpPc2coutput=csv,ssl.verifypeer=FALSE)
gsi_s-read.csv(textConnection(gsi_csv),header=T)
##Custom color ramp...I didn't like the ones that are provided in the
package, this is similar to jet colors in matlab
jet.colors -
colorRampPalette(c(#7F, blue, #007FFF, cyan,
#7FFF7F, yellow, #FF7F00, red, #7F))
#Tileplots comparing how CWT recoveries from brood year 1974 -1977 and GSI
data
#from 2010 - 2011 represent the relative abundance of Rogue river chinook
salmon through space and time
R_g_s-tileplot(Rogue~month_num*-area_num,gsi_s,col.regions=jet.colors(256),
scales=list(x=list(at=5:9, labels=c('May'
,'June','July','August','September')),
y=list(at=(-1):-8,labels=c('T','N','C','B','KC','FB','SF','MO'))),
main = list(label=Relative abundance of Rogue River stock (WCGSI
2010-2011),cex=0.75) ,
xlab= Month,
ylab= Management Area,
border = black,
panel = function(...){
panel.fill(black)
panel.voronoi(...)
})
R_c_b-tileplot(Rogue~month_num*-area_num,cwtb,col.regions=jet.colors(256),
scales=list(x=list(at=5:9, labels=c('May'
,'June','July','August','September')),
y=list(at=(-1):-8,labels=c('T','N','C','B','KC','FB','SF','MO'))),
main = list(label=Relative abundance of Rogue River stock (CWT 1977
- 1983),cex=0.75) ,
xlab= Month,
ylab= Management Area,
border = black,
panel = function(...){
panel.fill(black)
panel.voronoi(...)
})
#plot the two tileplots side by side
grid.arrange(R_c_b,R_g_s,ncol=2)
Thank you in advance for any advice regarding how to remedy this issue.
Regards,
Ryan
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Re: [R] list to matrix?
No need for all this (see solutions including mine already given) -- but even without those, this is silly. An identity map is a real waste if you just want the simplification bit of sapply() -- you'd be much better just using simplify2array() You are right that simplify2array(p) does everything that sapply(p,function(x)x) does and does it more quickly and that matrix(unlist(p),nrow=2,byrow=TRUE) is much faster than either: p - lapply(1:1e6, function(i)c(i, log2(i))) system.time(z1 - t(sapply(p,function(x)x))) user system elapsed 1.2 0.0 1.2 system.time(z2 - t(simplify2array(p))) user system elapsed 0.910.000.90 system.time(z3 - matrix(unlist(p), ncol=2, byrow=TRUE)) user system elapsed 0.040.000.04 You can also use vapply instead of sapply - it requires that you supply the expect shape and type of FUN's output so it is doesn't have to figure this out from looking at all the outputs of FUN: system.time(z4 - t(vapply(p,FUN=function(x)x,FUN.VALUE=numeric(2 user system elapsed 0.560.000.56 An advantage of vapply is that it stops in its tracks if FUN returns an unexpected value. sapply() and simplify2array() will silently give you an unexpected result (a single column matrix of mode list instead of a vector of numbers) and matrix(unlist()...) gives you a warning if you are lucky. pBad - p ; pBad[[length(pBad)/2]] - 666 system.time(zBad1 - t(sapply(pBad,function(x)x))) user system elapsed 1.750.001.75 zBad1[,49:51] # not what we wanted [[1]] [1] 49.0 18.93157 [[2]] [1] 666 [[3]] [1] 51.0 18.93157 system.time(zBad2 - t(simplify2array(pBad))) user system elapsed 0.5 0.0 0.5 system.time(zBad3 - matrix(unlist(pBad), ncol=2, byrow=TRUE)) user system elapsed 0.030.000.03 Warning message: In matrix(unlist(pBad), ncol = 2, byrow = TRUE) : data length [199] is not a sub-multiple or multiple of the number of rows [100] # no warning if length(unlist(p)) were even system.time(zBad4 - t(vapply(pBad,function(x)x,numeric(2 Error in vapply(pBad, function(x) x, numeric(2)) : values must be length 2, but FUN(X[[50]]) result is length 1 Timing stopped at: 0.29 0 0.28 Which of the latter two methods you choose depends on how likely errors in the data are. Bill Dunlap Spotfire, TIBCO Software wdunlap tibco.com -Original Message- From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On Behalf Of R. Michael Weylandt Sent: Tuesday, December 04, 2012 2:59 PM To: arun Cc: R help; s...@gnu.org Subject: Re: [R] list to matrix? On Tue, Dec 4, 2012 at 8:17 PM, arun smartpink...@yahoo.com wrote: Hi, Try this: list1-list(c(5, 101), c(1e+05, 46), c(15, 31), c(2e+05, 17), c(25, 19), c(3e+05, 11), c(35, 12), c(4e+05, 25), c(45, 19), c(5e+05, 16)) res-t(sapply(list1,function(x) x)) Bah Humbug! (In Christmas cheer) No need for all this (see solutions including mine already given) -- but even without those, this is silly. An identity map is a real waste if you just want the simplification bit of sapply() -- you'd be much better just using simplify2array() Michael __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] partial analisys of a time series
Hi all Thanks for the attention and answers. I learned a lot I now I can go on my work. I also tryed to you de command window(). I thought it would be possible to select one column of an ts object, like we can do with a data.frame (plot(data[,2],data[,3]), to work. But as I saw we need to extract the values and create another ts. Thanks very much. All the best. Antonio 2012/12/4 arun smartpink...@yahoo.com Hi, If the frequency is 1, the error message will be gone. For e.g. birthstimeseriesJanFeb-subset(birthstimeseries,cycle(birthstimeseries)==c(1,2)) birthstimeseriesJanFeb1-ts(birthstimeseriesJanFeb,frequency=2,start=c(1946,1)) plot.ts(birthstimeseriesJanFeb1) birthstimeseriesJanFebHW-HoltWinters(birthstimeseriesJanFeb1) plot(birthstimeseriesJanFebHW) A.K. - Original Message - From: Antonio Silva aolinto@gmail.com To: PIKAL Petr petr.pi...@precheza.cz Cc: R-help@r-project.org R-help@r-project.org Sent: Tuesday, December 4, 2012 5:50 AM Subject: Re: [R] partial analisys of a time series Thanks Petr I thought there might be an equivalent for birthstimeseries[,1] if it were a dataframe, but split function sounds great. I could not reproduce the second line of your suggestion l.blist - lapply(blist, HoltWinters). I receive the message: Error in decompose(ts(x[1L:wind], start = start(x), frequency = f), seasonal) : time series has no or less than 2 periods What could be going wrong? Best regards Antonio 2012/12/4 PIKAL Petr petr.pi...@precheza.cz Hi -Original Message- From: r-help-boun...@r-project.org [mailto:r-help-bounces@r- project.org] On Behalf Of Antonio Silva Sent: Tuesday, December 04, 2012 10:26 AM To: R-help@r-project.org Subject: [R] partial analisys of a time series Dear list members I want to analyze separately the months of a time series. In other words, I want to plot and fit models for each month separately. Taking the example of http://a-little-book-of-r-for-time- series.readthedocs.org/en/latest/src/timeseries.html births - scan(http://robjhyndman.com/tsdldata/data/nybirths.dat;) birthstimeseries - ts(births, frequency=12, start=c(1946,1)) birthstimeseries plot.ts(birthstimeseries) birthstimeseriesHW - HoltWinters(birthstimeseries) plot(birthstimeseriesHW) How to proceed the plotting and HoltWinters smoothing considereing only Januarys, Februarys, etc. separately. Split your data by months to a list, use lapply. using zoo package blist -split(birthstimeseries, months(as.Date(birthstimeseries))) l.blist - lapply(blist, HoltWinters) Regards Petr Thanks in advance. Antonio Olinto [[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. -- Antônio Olinto Ávila da Silva Biólogo / Oceanógrafo Instituto de Pesca (Fisheries Institute) São Paulo, Brasil [[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. -- Antônio Olinto Ávila da Silva Biólogo / Oceanógrafo Instituto de Pesca (Fisheries Institute) São Paulo, Brasil [[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] control point size of superscript when labeling axes with title()
On Dec 4, 2012, at 11:05 AM, Chris Solomon wrote:
Hi-
A journal has asked me to make all of my text annotations on a figure at
10-point size. For the most part this is easy, e.g. by creating figures with:
pdf(..., family='Times', pointsize=10)
But where I have superscripts (or subscripts) in axis labels, the default
seems to be to shrink the superscripted text slightly. For example this code:
title(ylab=expression(paste('Respiration (mg ',O[2],' ',L^-1,'
',d^-1,')',sep=' ')),outer=T,line=0.3)
produces superscripted numbers at approximately 7 point.
I have been poking around for a solution but not having much luck. The
textstyle() function keeps superscripted text at original size if you use it
within a text() call, but I can't figure out an equivalent solution within a
title() call.
It is not the fact that you are using title, but rather (I think) the fact that
numbers are handled differently than text in plotmath. You are also
unnecessarily using 'paste' and incorrectly including 'sep' argument in
plotmath-paste, which does not honor that argument.
Regular R-paste != plotmath-paste
Quoting the exponents allows textstyle to do its job properly.
plot(1,1, ylab=)
title(ylab =
expression(Respiration~(*mg~~O[2]~~L^textstyle(-1)~d^textstyle(-1)*')')
,line= 1.5)
[[alternative HTML version deleted]]
You might want to read what the Posting Guide says about HTML posting.
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
David Winsemius, MD
Alameda, CA, USA
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Re: [R] How to find matching columns in a matrix of lists?
Hi R users following this thread!
I evaluated both solutions given by Rui and Arun.
Both work very well.
Arun's is a little faster.
I did the following time measurements on a large matrix with 2 rows,
labelled InterPro and GO, and 88664 columns, labelled with protein IDs.
I was interested in selecting those columns (proteins) that share a certain
GO annotation for molecular function.
So here is the time evaluation:
* Rui's method:
system.time(annos[ , sapply( annos[ GO, ], function(x) { GO:0005634
%in% x } ), drop=F ] )
user system elapsed
1.068 0.012 1.079
or
system.time(annos[ , unlist( lapply( annos[ GO, ], function(x) {
GO:0005634 %in% x } ) ), drop=F ])
user system elapsed
0.876 0.000 0.880
* Arun's method:
system.time( annos[ , mapply( function(x){any(x==GO:0005634)}, annos[
GO, ] ), drop=F ] )
user system elapsed
0.808 0.012 0.826
Thank you Rui and Arun very much for your help and everyone else for your
attention!
Kind regards!
2012/12/5 arun smartpink...@yahoo.com
HI,
Just tweaking my code also gives the same result:
m[,mapply(function(x) any(x==B),m),drop=F]
# [,1] [,2]
#[1,] List,2 List,2
A.K.
- Original Message -
From: Asis Hallab asis.hal...@gmail.com
To: Rui Barradas ruipbarra...@sapo.pt; r-help@r-project.org
Cc:
Sent: Tuesday, December 4, 2012 5:26 PM
Subject: Re: [R] How to find matching columns in a matrix of lists?
Hello,
m[ , sapply(1:ncol(m), function(j) sapply(B, `%in%`, m[[1 , j]])), drop=F
]
It indeed does.
Thank you very much!
[[alternative HTML version deleted]]
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--
Asis Hallab
Rothehausstr. 6 - 12
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Skype: asis.hallab.cgn
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Re: [R] control point size of superscript when labeling axes with title()
Fantastic - that does the trick, David, thanks for setting me straight!
Chris
-Original Message-
From: David Winsemius [mailto:dwinsem...@comcast.net]
Sent: Tuesday, December 04, 2012 8:19 PM
To: Chris Solomon
Cc: r-help@r-project.org
Subject: Re: [R] control point size of superscript when labeling axes with
title()
On Dec 4, 2012, at 11:05 AM, Chris Solomon wrote:
Hi-
A journal has asked me to make all of my text annotations on a figure at
10-point size. For the most part this is easy, e.g. by creating figures with:
pdf(..., family='Times', pointsize=10)
But where I have superscripts (or subscripts) in axis labels, the default
seems to be to shrink the superscripted text slightly. For example this code:
title(ylab=expression(paste('Respiration (mg ',O[2],' ',L^-1,'
',d^-1,')',sep=' ')),outer=T,line=0.3) produces superscripted numbers at
approximately 7 point.
I have been poking around for a solution but not having much luck. The
textstyle() function keeps superscripted text at original size if you use it
within a text() call, but I can't figure out an equivalent solution within a
title() call.
It is not the fact that you are using title, but rather (I think) the fact that
numbers are handled differently than text in plotmath. You are also
unnecessarily using 'paste' and incorrectly including 'sep' argument in
plotmath-paste, which does not honor that argument.
Regular R-paste != plotmath-paste
Quoting the exponents allows textstyle to do its job properly.
plot(1,1, ylab=)
title(ylab =
expression(Respiration~(*mg~~O[2]~~L^textstyle(-1)~d^textstyle(-1
)*')') ,line= 1.5)
[[alternative HTML version deleted]]
You might want to read what the Posting Guide says about HTML posting.
PLEASE do read the posting guide
http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
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.
