[R] different results of fisher.test function in R2.8.1 and R2.6.0
Hi; I use the function fisher.test to compute in R2.8.1 and R2.6.0,and the results are not identical.the last number is different. why? thank you ! Merry [[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] Weighting data with normal distribution
I have a vector of binary data – a string of 0’s and 1’s. I want to weight these inputs with a normal kernel centered around entry x so it is transformed into a new vector of data that takes into account the values of the entries around it (weighting them more heavily if they are near). Example: - - - - - 0 1 0 0 1 0 0 1 1 1 1 If x = 3, it’s current value is 0 but it’s new value with the Gaussian weighting around would be something like .1*0+.5*1+1*0+0.5*0+.1*1= 0.6 I want to be able to play with adjusting the variance to different values as well. I’ve found wkde in the mixtools library and think it may be useful but I have not figured out how to use it yet. Any tips would be appreciated. Thanks! -- View this message in context: http://www.nabble.com/Weighting-data-with-normal-distribution-tp22728289p22728289.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] interactive image graphic
Dear All I want to plot a kind of figures, which can interactive with user. For example, i have a matirx which can be showed by image function. i.e. we can compare the value depend on different colors. However, the change of colors depend on the range of value. Nowaday, i want to set a bar, which can be moved by user such that the user can obtain the appropriate range. Does anyone suggest me which function can be applied to solve this problem? Thanks __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Data manipulation - multiplicate cases
Hi listers, I am trying to arrange my data and I didn't find any information how to do it! I have a data with 3 variables: X Y Z 1-I would like to multiplicate de information of X according to the number I have for my Y variable... 2-Then I want to identify with a dicotomic variable by the number according my variable Z from X... I can do the first part by... z-rep(x,y) But I don't know how to set a dicotomic variable according to Z... Exemple... I have... X YZ 123 31 234 31 345 42 456 32 I want to get... X YZ 123 31 123 30 123 30 234 31 234 30 234 30 345 41 345 41 345 40 345 40 456 31 456 31 456 30 Thanks in advance... -- View this message in context: http://www.nabble.com/Data-manipulation---multiplicate-cases-tp22730453p22730453.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] how to quit this mailing list
Hi, I will end this mailbox and start a new one and i want to quit this list. I search the R official website with no answer. thanks in advance [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] loading and manipulating 10 data frames-simplified
I have to load 10 different data frames and then manipulate those 10 data frames but would like to do this in a more simplified code than what i am doing. I have tried a couple of approaches but cannot get it to work correctly. So the initial (bulky) code is: #Bin 1 #--- #Loads bin data frame from csv files with acres and TAZ data Bin1_main - read.csv(file=I:/Research/Samba/urb_transport_modeling/LUSDR/Workspace/BizLandPrice/data/Bin_lookup_values/Bin1_lookup.csv,head=FALSE); #Separates Acres data from main data and converts acres to square feet Bin1_Acres=Bin1_main[[1]]*43560 #Separates TAZ data from main data Bin1_TAZ=Bin1_main[[2]] #Separates TAZ data from main data and converts acres to square feet Bin1_TAZvacant=Bin1_main[[3]]*43560 #Sums each parcel acreage data of the bin Bin1Acres_sum=sum(Bin1_Acres) #Creates data frame of cumlative percentages of each parcel of bin Bin1_cumper=cumsum(Bin1_Acres/Bin1Acres_sum) #Calculates the probability of choosing particular parcel from bin Bin1_parprob=abs(1-Bin1_cumper) #Combines parcel acreage data and cumlative percentage data Bin1Main.data = cbind(Bin1_Acres,Bin1_parprob,Bin1_TAZ,Bin1_TAZvacant) #Bin 2 #--- #Loads bin data frame from csv files with acres and TAZ data Bin2_main - read.csv(file=I:/Research/Samba/urb_transport_modeling/LUSDR/Workspace/BizLandPrice/data/Bin_lookup_values/Bin2_lookup.csv,header=FALSE); #Separates Acres data from main data and converts acres to square feet Bin2_Acres=Bin2_main[[1]]*43560 #Separates TAZ data from main data Bin2_TAZ=Bin2_main[[2]] #Separates TAZ data from main data and converts acres to square feet Bin2_TAZvacant=Bin2_main[[3]]*43560 #Sums each parcel acreage data of the bin Bin2Acres_sum=sum(Bin2_Acres) #Creates data frame of cumlative percentages of each parcel of bin Bin2_cumper=cumsum(Bin2_Acres/Bin2Acres_sum) #Calculates the probability of choosing particular parcel from bin Bin2_parprob=abs(1-Bin2_cumper) #Combines parcel acreage data and cumlative percentage data Bin2Main.data = cbind(Bin2_Acres,Bin2_parprob,Bin2_TAZ,Bin2_TAZvacant) #Bin 3 #--- #Loads bin data frame from csv files with acres and TAZ data Bin3_main - read.csv(file=I:/Research/Samba/urb_transport_modeling/LUSDR/Workspace/BizLandPrice/data/Bin_lookup_values/Bin3_lookup.csv,header=FALSE); #Separates Acres data from main data and converts acres to square feet Bin3_Acres=Bin3_main[[1]]*43560 #Separates TAZ data from main data Bin3_TAZ=Bin3_main[[2]] #Separates TAZ data from main data and converts acres to square feet Bin3_TAZvacant=Bin3_main[[3]]*43560 #Sums each parcel acreage data of the bin Bin3Acres_sum=sum(Bin3_Acres) #Creates data frame of cumlative percentages of each parcel of bin Bin3_cumper=cumsum(Bin3_Acres/Bin3Acres_sum) #Calculates the probability of choosing particular parcel from bin Bin3_parprob=abs(1-Bin3_cumper) #Combines parcel acreage data and cumlative percentage data Bin3Main.data = cbind(Bin3_Acres,Bin3_parprob,Bin3_TAZ,Bin3_TAZvacant) #Bin 4 #--- #Loads bin data frame from csv files with acres and TAZ data Bin4_main - read.csv(file=I:/Research/Samba/urb_transport_modeling/LUSDR/Workspace/BizLandPrice/data/Bin_lookup_values/Bin4_lookup.csv,header=FALSE); #Separates Acres data from main data and converts acres to square feet Bin4_Acres=Bin4_main[[1]]*43560 #Separates TAZ data from main data Bin4_TAZ=Bin4_main[[2]] #Separates TAZ data from main data and converts acres to square feet Bin4_TAZvacant=Bin4_main[[3]]*43560 #Sums each parcel acreage data of the bin Bin4Acres_sum=sum(Bin4_Acres) #Creates data frame of cumlative percentages of each parcel of bin Bin4_cumper=cumsum(Bin4_Acres/Bin4Acres_sum) #Calculates the probability of choosing particular parcel from bin Bin4_parprob=abs(1-Bin4_cumper) #Combines parcel acreage data and cumlative percentage data Bin4Main.data = cbind(Bin4_Acres,Bin4_parprob,Bin4_TAZ,Bin4_TAZvacant) #Bin 5 #--- #Loads bin data frame from csv files with acres and TAZ data Bin5_main - read.csv(file=I:/Research/Samba/urb_transport_modeling/LUSDR/Workspace/BizLandPrice/data/Bin_lookup_values/Bin5_lookup.csv,header=FALSE); #Separates Acres data from main data and converts acres to square feet Bin5_Acres=Bin5_main[[1]]*43560 #Separates TAZ data from main data Bin5_TAZ=Bin5_main[[2]] #Separates TAZ data from main data and converts acres to square feet Bin5_TAZvacant=Bin5_main[[3]]*43560 #Sums each parcel acreage data of the bin Bin5Acres_sum=sum(Bin5_Acres) #Creates data frame of cumlative percentages of each parcel of bin Bin5_cumper=cumsum(Bin5_Acres/Bin5Acres_sum) #Calculates the probability of choosing particular parcel from bin Bin5_parprob=abs(1-Bin5_cumper) #Combines parcel acreage data and cumlative percentage data Bin5Main.data = cbind(Bin5_Acres,Bin5_parprob,Bin5_TAZ,Bin5_TAZvacant) #Bin 6 #--- #Loads bin data frame from csv files with acres and TAZ data Bin6_main -
[R] A beginner's question
I am a new R-language user. I have set up a data frame mydata,one of the colume of which is skill. Now I want to select the observations whose skill value is equal to 1,by what command can I get it? __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] how to quit this mailing list
Hi, https://stat.ethz.ch/mailman/listinfo/r-help and there You'll find the section: To unsubscribe from R-help, get a password reminder, or change your subscription options enter your subscription email address: Hope this helps, Kimmo __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] A beginner's question
Here's an example: mydata-data.frame(skill=c(1,2,3,4),x=c(1,1,1,1)) mydata[mydata$skill==1,] On Fri, Mar 27, 2009 at 16:40, minben minb...@gmail.com wrote: I am a new R-language user. I have set up a data frame mydata,one of the colume of which is skill. Now I want to select the observations whose skill value is equal to 1,by what command can I get it? __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] A beginner's question
Hi, minben wrote: I am a new R-language user. I have set up a data frame mydata,one of the colume of which is skill. Now I want to select the observations whose skill value is equal to 1,by what command can I get it? Try this: mydata1-mydatasubset(mydata, skill==1) Maybe You should also read this introduction: http://cran.r-project.org/doc/manuals/R-intro.pdf Kind regards, Kimmo __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] use of @ character in variable name
On Thu, 26 Mar 2009, Mike Miller wrote: Importing data with a header row using read.delim, one variable should be named @5HTT but it is automatically renamed to X.5HTT, presumably because the @ is either unacceptable or misunderstood. I've tried to find out what the rules are on variable names but have been unsuccessful. I'll bet someone here can tell me where to look. Maybe it's hidden away in here somewhere: http://cran.r-project.org/doc/manuals/R-data.pdf It's hidden away in: FAQ 7.14 What are valid names? -thomas Thomas Lumley Assoc. Professor, Biostatistics tlum...@u.washington.eduUniversity of Washington, Seattle __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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 of fisher.test function in R2.8.1 and R2.6.0
On Fri, 27 Mar 2009, 马传香 wrote: Hi; I use the function fisher.test to compute in R2.8.1 and R2.6.0,and the results are not identical.the last number is different. why? thank you ! Can you show us the data? The only change listed in the NEWS file since 2.6.0 was with the option simulate.p.values=TRUE - you don't say whether you used that option. -thomas Thomas Lumley Assoc. Professor, Biostatistics tlum...@u.washington.eduUniversity of Washington, Seattle __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] Physical or Statistical Explanation for the Funnel Plot?
On Thu, 26 Mar 2009, Jason Rupert wrote: The R code below produces (after running for a few minutes on a decent computer) the plot shown at the following location: http://n2.nabble.com/Is-there-a-physical-and-quantitative-explanation-for-this-plot--td2542321.html I'm just taking the mean of a given set of random variables, where the set size is increased. There appears to be a quick convergence and then a pretty steady variance out to a set size of 10,. Part of the convergence is just that the standard devation of a mean of N observations is proportional to 1/sqrt(N). In your case the distributions are all exactly Normal; the same convergence would occur with other distributions, but you would also see the change in shape from left to right as the distribution converged to Normal. There's also some plotting artifacts due to the size of the points. The apparent stabilization at large N (and the wide vertical bar at zero that Marc Schwartz commented on) are due partly to the slow convergence of 1/sqrt(N) but largely because the width can't be smaller than the width of a point. When I draw funnel plots like this for whole-genome association data I use the 'hexbin' package, which doesn't have these artifacts and is much faster and produces smaller graphics files. -thomas Thomas Lumley Assoc. Professor, Biostatistics tlum...@u.washington.eduUniversity of Washington, Seattle __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] R 2.8.1 and 2.9 alpha crash when running survest of Design package
The Design package is incompatible with updates to the survival package (version 2.35 and higher) that were made for version 2.9.0. It calls some internal fitting functions (coxreg.fit, agreg.fit) whose arguments have changed. According to the CRAN checks, about a dozen other packages were also affected by other changes in the update, but most just give an error message rather than crash. Maintainers of all the packages have been notified and given my best guess at the reason for their specific incompatibility (and an offer of further assistance if necessary) You may need to downgrade to version 2.34 of the survival package until Design is updated. -thomas On Fri, 27 Mar 2009, Nguyen Dinh Nguyen wrote: Dear Prof Harrell and everyone, My PC: Window XP service pack 3 and service pack 2 R version 2.8.1 and 2.9 alpha For the last 3 days, after updating R, my two computers have been facing problems when running existing and runable R commands that involves with Design package I attempt to use 'survest', but I failed all the times with R (both 2.8.1 and 2.9 alpha) being shut down immediately with following error report messages. AppName: rgui.exe AppVer: 2.90.48212.0 ModName: survival.dll ModVer: 0.0.0.0 Offset: 7749 However, if I run these commands on other computers which have not been updated for 2 week, they run OK Could you please consider the matter and give me advice I am looking forward to hearing from you soon Regards Nguyen D Nguyen Garvan Institute of Medical Research Sydney, Australia [[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. Thomas Lumley Assoc. Professor, Biostatistics tlum...@u.washington.eduUniversity of Washington, Seattle __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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 the plm package
Dear R help, I use the package plm the function plm() to analyse a panel data and estimate a fixeffect model. I use the code as follow : fe - plm(y~x+z, data, model = within) But my result have no intercept.why? I consult the paper plm.pdf , but I can't find any answer. Can I estimate a model with intercept? Thanks, Helen Chen -- View this message in context: http://www.nabble.com/problem-with-the-plm-package-tp22737251p22737251.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] Remove error data and clustering analysis
Hi, all, I’d like to do the clustering analysis in my dataset. The example data are as follows: Dataset 1: 500, 490, 486, 490, 491, 493, 480, 461, 504, 476, 434, 500, 470, 495, 3116, 3142, 12836, 3062, 3091, 3141, 3177, 3150, 3114, 3149; Dataset 2: 506, 473, 495, 494, 434, 459, 445, 475, 476, 128367, 470, 513, 466, 476,482, 1201, 469, 502; I had so many datasets like that. Basically, every dataset can classify one or two clusters (no more than 2), meanwhile, there have error data points, for example, 12836 is error data point in Dataset 1; and 128367, 1201 is error data points in dataset2. The clustered data is following the normal distribution, the standard deviation was known. That’s mean the one cluster is following the normal distribution when the dataset classified one cluster like dataset2; the two clusters are following the normal distribution respectively when the dataset classified two clusters like dataset1. Error data are far away of the mean. I am wondering is there any mathematic pipeline/function can do the analysis that removing error data, and clustering the dataset in 1 or 2 clusters? Thank you for your reply. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] Plot with nested x-labels
Antje wrote: Hi there, I was wondering wether it is possible to creeate plots with nested lables like in excel? If yes, could anyone provide me the information how to do it? I've attached an image of an Excel plot to show you, what I'd like to plot with R :-) Hi Antje, The very question asked by Ofir Levy that led to the new hierobarp function in the plotrix package. Jim __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] A beginner's question
You can do like this: 1. mydata[mydata$skill==1,] 2. mydata[mydata[,skill]==1,] /Forin On Thu, 26 Mar 2009 23:40:32 -0700 (PDT) minben minb...@gmail.com wrote: I am a new R-language user. I have set up a data frame mydata,one of the colume of which is skill. Now I want to select the observations whose skill value is equal to 1,by what command can I get it? __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Florin G. Maican == Ph.D. candidate, Department of Economics, School of Business, Economics and Law, Gothenburg University, Sweden --- P.O. Box 640 SE-405 30, Gothenburg, Sweden Mobil: +46 76 235 3039 Phone: +46 31 786 4866 Fax:+46 31 786 4154 Home Page: http://maicanfg.googlepages.com/index.html E-mail: florin.mai...@handels.gu.se Not everything that counts can be counted, and not everything that can be counted counts. --- Einstein --- __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] Sort by timestamp
Good morning Johannes,
This might help. Try:
df - data.frame(V1=as.factor(c('2008-10-14 09:10:00','2008-10-14
9:20:20','2008-10-14 08:45:00')),V2=runif(3))
df # is a dataframe, just as yours
class(df$V1) # is a factor, just as yours. See ?factor
# This will probably not be ordered
# in a way you like.
df$V1 - as.POSIXct(df$V1, tz='CET') # makes it a time. See ?POSIXct
class(df$V1) # is a POSIX time now
df2 - df[do.call(order, df), ] # see ?order
df2 # sorted in a way you like
Cheers, Arien
On Thu, March 26, 2009 08:54, j.k wrote:
#Good morning alltogheter. I'm using R for a short time to analyse
TimeSeries
and I have the following Problem:
#I have a bunch of Time Series:
#First of all I import them from a txt File
data.input01 -read.csv(./LD/20081030.txt, header = TRUE, sep = ;,
quote=\, dec=,, fill = TRUE, comment.char=)
data.input02 -read.csv(./LD/20090305.txt, header = TRUE, sep = ;,
quote=\, dec=,, fill = TRUE, comment.char=)
data.input03 -read.csv(./LD/20081114.txt, header = TRUE, sep = ;,
quote=\, dec=,, fill = TRUE, comment.char=)
data.input04 -read.csv(./LD/20081201.txt, header = TRUE, sep = ;,
quote=\, dec=,, fill = TRUE, comment.char=)
data.input05 -read.csv(./LD/20081219.txt, header = TRUE, sep = ;,
quote=\, dec=,, fill = TRUE, comment.char=)
data.input06 -read.csv(./LD/20090107.txt, header = TRUE, sep = ;,
quote=\, dec=,, fill = TRUE, comment.char=)
#After the import they look like that:
V1 V2
1 2008-10-14 08:45:00 92130.68
2 2008-10-14 08:50:00 94051.70
3 2008-10-14 08:55:00 97050.85
4 2008-10-14 09:00:00 81133.81
5 2008-10-14 09:05:00 70705.40
6 2008-10-14 09:10:00 75213.92
7 2008-10-14 09:15:00 90876.14
8 2008-10-14 09:20:00 85995.17
#Next steps are to combine them with rbind and sort duplicates out
data.troughput01 -
rbind(data.input03,data.input01,data.input04,data.input02,data.input05,data.input06)
data.troughput02 - unique(data.troughput01)
#The Problem is that the dates are mixed and I want to sort/order them by
the date and time.
#The class of the Date/time is as followed:
class(data.input01$V1)
[1] factor
# I've already tried sort and order but it didn't work
#Are there any suggestions, how I can solve this issue??
Thanks in advance
Johannes
--
View this message in context:
http://www.nabble.com/Sort-by-timestamp-tp22717322p22717322.html
Sent from the R help mailing list archive at Nabble.com.
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide
http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
--
drs. H.A. (Arien) Lam (Ph.D. student)
Department of Physical Geography
Faculty of Geosciences
Utrecht University, The Netherlands
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] Plot with nested x-labels
Hi Jim, is this plotrix package version higher than 2.2-7 ? Antje Jim Lemon schrieb: Antje wrote: Hi there, I was wondering wether it is possible to creeate plots with nested lables like in excel? If yes, could anyone provide me the information how to do it? I've attached an image of an Excel plot to show you, what I'd like to plot with R :-) Hi Antje, The very question asked by Ofir Levy that led to the new hierobarp function in the plotrix package. Jim __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Sort by timestamp
Works perfect!!
Thanks a lot...
Cheers
Johannes
Arien Lam wrote:
Good morning Johannes,
This might help. Try:
df - data.frame(V1=as.factor(c('2008-10-14 09:10:00','2008-10-14
9:20:20','2008-10-14 08:45:00')),V2=runif(3))
df # is a dataframe, just as yours
class(df$V1) # is a factor, just as yours. See ?factor
# This will probably not be ordered
# in a way you like.
df$V1 - as.POSIXct(df$V1, tz='CET') # makes it a time. See ?POSIXct
class(df$V1) # is a POSIX time now
df2 - df[do.call(order, df), ] # see ?order
df2 # sorted in a way you like
Cheers, Arien
On Thu, March 26, 2009 08:54, j.k wrote:
#Good morning alltogheter. I'm using R for a short time to analyse
TimeSeries
and I have the following Problem:
#I have a bunch of Time Series:
#First of all I import them from a txt File
data.input01 -read.csv(./LD/20081030.txt, header = TRUE, sep = ;,
quote=\, dec=,, fill = TRUE, comment.char=)
data.input02 -read.csv(./LD/20090305.txt, header = TRUE, sep = ;,
quote=\, dec=,, fill = TRUE, comment.char=)
data.input03 -read.csv(./LD/20081114.txt, header = TRUE, sep = ;,
quote=\, dec=,, fill = TRUE, comment.char=)
data.input04 -read.csv(./LD/20081201.txt, header = TRUE, sep = ;,
quote=\, dec=,, fill = TRUE, comment.char=)
data.input05 -read.csv(./LD/20081219.txt, header = TRUE, sep = ;,
quote=\, dec=,, fill = TRUE, comment.char=)
data.input06 -read.csv(./LD/20090107.txt, header = TRUE, sep = ;,
quote=\, dec=,, fill = TRUE, comment.char=)
#After the import they look like that:
V1 V2
1 2008-10-14 08:45:00 92130.68
2 2008-10-14 08:50:00 94051.70
3 2008-10-14 08:55:00 97050.85
4 2008-10-14 09:00:00 81133.81
5 2008-10-14 09:05:00 70705.40
6 2008-10-14 09:10:00 75213.92
7 2008-10-14 09:15:00 90876.14
8 2008-10-14 09:20:00 85995.17
#Next steps are to combine them with rbind and sort duplicates out
data.troughput01 -
rbind(data.input03,data.input01,data.input04,data.input02,data.input05,data.input06)
data.troughput02 - unique(data.troughput01)
#The Problem is that the dates are mixed and I want to sort/order them by
the date and time.
#The class of the Date/time is as followed:
class(data.input01$V1)
[1] factor
# I've already tried sort and order but it didn't work
#Are there any suggestions, how I can solve this issue??
Thanks in advance
Johannes
--
View this message in context:
http://www.nabble.com/Sort-by-timestamp-tp22717322p22717322.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.
--
drs. H.A. (Arien) Lam (Ph.D. student)
Department of Physical Geography
Faculty of Geosciences
Utrecht University, The Netherlands
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide
http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
--
View this message in context:
http://www.nabble.com/Sort-by-timestamp-tp22717322p22738808.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] A beginner's question
minben schreef: I am a new R-language user. I have set up a data frame mydata,one of the colume of which is skill. Now I want to select the observations whose skill value is equal to 1,by what command can I get it? __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. To add the number of possibilities :): subset(mydata, skill == 1) cheers, Paul -- Drs. Paul Hiemstra Department of Physical Geography Faculty of Geosciences University of Utrecht Heidelberglaan 2 P.O. Box 80.115 3508 TC Utrecht Phone: +3130 274 3113 Mon-Tue Phone: +3130 253 5773 Wed-Fri http://intamap.geo.uu.nl/~paul __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] Remove error data and clustering analysis
Hi, all, I'd like to do the clustering analysis in my dataset. The example data are as follows: Dataset 1: 500, 490, 486, 490, 491, 493, 480, 461, 504, 476, 434, 500, 470, 495, 3116, 3142, 12836, 3062, 3091, 3141, 3177, 3150, 3114, 3149; Dataset 2: 506, 473, 495, 494, 434, 459, 445, 475, 476, 128367, 470, 513, 466, 476,482, 1201, 469, 502; I had so many datasets like that. Basically, every dataset can classify one or two clusters (no more than 2), meanwhile, there have error data points, for example, 12836 is error data point in Dataset 1; and 128367, 1201 is error data points in dataset2. The clustered data is following the normal distribution, the standard deviation was known. Thats mean the one cluster is following the normal distribution when the dataset classified one cluster like dataset2; the two clusters are following the normal distribution respectively when the dataset classified two clusters like dataset1. Error data are far away of the mean. I am wondering is there any mathematic pipeline/function can do the analysis that removing error data, and clustering the dataset in 1 or 2 clusters? Thank you for your reply. 2009-03-27 wanggd1983 [[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] Plot with nested x-labels
Hi there, I was wondering wether it is possible to creeate plots with nested lables like in excel? If yes, could anyone provide me the information how to do it? I've attached an image of an Excel plot to show you, what I'd like to plot with R :-) Ciao, Antje inline: plot.png__ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] Ploting a matrix
Hi evrybody, in a matrix consisting of 49 columns, I would like to plot all columns against the first in 48 different graphs. Can you help me? Thank you in advance Sebastian -- *** Dipl. Biol. Sebastian Krug PhD - student IFM - GEOMAR Leibniz Institute of Marine Sciences Research Division 2 - Marine Biogeochemistry Düsternbrooker Weg 20 D - 24105 Kiel Germany Tel.: +49 431 600-4282 Fax.: +49 431 600-4446 email: sk...@ifm-geomar.de __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] Ploting a matrix
Something like this perhaps, a - matrix(rnorm(5*49), ncol=49) pdf(width=15, height=15) par(mfrow= c(8,6)) apply(a[,-1], 2, plot, x= a[,1]) dev.off() HTH, baptiste On 27 Mar 2009, at 11:05, skrug wrote: Hi evrybody, in a matrix consisting of 49 columns, I would like to plot all columns against the first in 48 different graphs. Can you help me? Thank you in advance Sebastian -- *** Dipl. Biol. Sebastian Krug PhD - student IFM - GEOMAR Leibniz Institute of Marine Sciences Research Division 2 - Marine Biogeochemistry Düsternbrooker Weg 20 D - 24105 Kiel Germany Tel.: +49 431 600-4282 Fax.: +49 431 600-4446 email: sk...@ifm-geomar.de __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. _ Baptiste Auguié School of Physics University of Exeter Stocker Road, Exeter, Devon, EX4 4QL, UK Phone: +44 1392 264187 http://newton.ex.ac.uk/research/emag __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] 'stretching' a binomial variable
Hi, Im carrying out some Bayesian analysis using a binomial response variable (proportion: 0 to 1), but most of my observations have a value of 0 and many have very small values (i.e. 0.001). I'm having troubles getting my MCMC algorithm to converge, so I have decided to try normalising my response variable to see if this helps. I want it to stay between 0 and 1 but to have a larger range of values, or just for them all to be slightly higher. Does anyone know the best way to acheive this? I could just add a value to each observation (say 10 to increase the proportion a bit, but ensuring it would still be between 0 and 1) - would that be ok? Or is there a better way to stretch the values up? Sorry - i know its not really an R specific question, but I have never found a forum with as many stats litterate people as this one :-) Cheers - any advice much appreciated! nicola -- View this message in context: http://www.nabble.com/%27stretching%27-a-binomial-variable-tp22740114p22740114.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] ICC question: Interrater and intrarater variability (intraclass correlation coefficients)
Hello dear R help group. I encountered this old thread (http://tinyurl.com/dklgsk) containing the a similar question to the one I have, but left without an answer. I am and hoping one of you might help. A simplified situation: I have a factorial design (with 2^3 experiment combinations), for 167 subjects, each one has answered the same question twice (out of a bunch of types of questions). Each answer could get an integer number between 0 to 3. I wish to combine the two answers, but first to be sure I could, I would have liked to run an ICC (Intraclass correlation) check on the two answers. Naturally, I would use the irr, condord, or psy packages (as John Fox suggested back then), but I can't because of the repetitions of different design question for each patient. Since the mentioned packages (irr, condord, and psy) Can take only a n*m matrix for subjects and raters. But no place is given for the repetitions as data and therefore it will be impossible to get results for the INTRArater reliability. Thanks, Tal [[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] loading and manipulating 10 data frames-simplified
Put the data.frames as elements in a list and loop / sapply() over that list. Uwe Ligges PDXRugger wrote: I have to load 10 different data frames and then manipulate those 10 data frames but would like to do this in a more simplified code than what i am doing. I have tried a couple of approaches but cannot get it to work correctly. So the initial (bulky) code is: #Bin 1 #--- #Loads bin data frame from csv files with acres and TAZ data Bin1_main - read.csv(file=I:/Research/Samba/urb_transport_modeling/LUSDR/Workspace/BizLandPrice/data/Bin_lookup_values/Bin1_lookup.csv,head=FALSE); #Separates Acres data from main data and converts acres to square feet Bin1_Acres=Bin1_main[[1]]*43560 #Separates TAZ data from main data Bin1_TAZ=Bin1_main[[2]] #Separates TAZ data from main data and converts acres to square feet Bin1_TAZvacant=Bin1_main[[3]]*43560 #Sums each parcel acreage data of the bin Bin1Acres_sum=sum(Bin1_Acres) #Creates data frame of cumlative percentages of each parcel of bin Bin1_cumper=cumsum(Bin1_Acres/Bin1Acres_sum) #Calculates the probability of choosing particular parcel from bin Bin1_parprob=abs(1-Bin1_cumper) #Combines parcel acreage data and cumlative percentage data Bin1Main.data = cbind(Bin1_Acres,Bin1_parprob,Bin1_TAZ,Bin1_TAZvacant) #Bin 2 #--- #Loads bin data frame from csv files with acres and TAZ data Bin2_main - read.csv(file=I:/Research/Samba/urb_transport_modeling/LUSDR/Workspace/BizLandPrice/data/Bin_lookup_values/Bin2_lookup.csv,header=FALSE); #Separates Acres data from main data and converts acres to square feet Bin2_Acres=Bin2_main[[1]]*43560 #Separates TAZ data from main data Bin2_TAZ=Bin2_main[[2]] #Separates TAZ data from main data and converts acres to square feet Bin2_TAZvacant=Bin2_main[[3]]*43560 #Sums each parcel acreage data of the bin Bin2Acres_sum=sum(Bin2_Acres) #Creates data frame of cumlative percentages of each parcel of bin Bin2_cumper=cumsum(Bin2_Acres/Bin2Acres_sum) #Calculates the probability of choosing particular parcel from bin Bin2_parprob=abs(1-Bin2_cumper) #Combines parcel acreage data and cumlative percentage data Bin2Main.data = cbind(Bin2_Acres,Bin2_parprob,Bin2_TAZ,Bin2_TAZvacant) #Bin 3 #--- #Loads bin data frame from csv files with acres and TAZ data Bin3_main - read.csv(file=I:/Research/Samba/urb_transport_modeling/LUSDR/Workspace/BizLandPrice/data/Bin_lookup_values/Bin3_lookup.csv,header=FALSE); #Separates Acres data from main data and converts acres to square feet Bin3_Acres=Bin3_main[[1]]*43560 #Separates TAZ data from main data Bin3_TAZ=Bin3_main[[2]] #Separates TAZ data from main data and converts acres to square feet Bin3_TAZvacant=Bin3_main[[3]]*43560 #Sums each parcel acreage data of the bin Bin3Acres_sum=sum(Bin3_Acres) #Creates data frame of cumlative percentages of each parcel of bin Bin3_cumper=cumsum(Bin3_Acres/Bin3Acres_sum) #Calculates the probability of choosing particular parcel from bin Bin3_parprob=abs(1-Bin3_cumper) #Combines parcel acreage data and cumlative percentage data Bin3Main.data = cbind(Bin3_Acres,Bin3_parprob,Bin3_TAZ,Bin3_TAZvacant) #Bin 4 #--- #Loads bin data frame from csv files with acres and TAZ data Bin4_main - read.csv(file=I:/Research/Samba/urb_transport_modeling/LUSDR/Workspace/BizLandPrice/data/Bin_lookup_values/Bin4_lookup.csv,header=FALSE); #Separates Acres data from main data and converts acres to square feet Bin4_Acres=Bin4_main[[1]]*43560 #Separates TAZ data from main data Bin4_TAZ=Bin4_main[[2]] #Separates TAZ data from main data and converts acres to square feet Bin4_TAZvacant=Bin4_main[[3]]*43560 #Sums each parcel acreage data of the bin Bin4Acres_sum=sum(Bin4_Acres) #Creates data frame of cumlative percentages of each parcel of bin Bin4_cumper=cumsum(Bin4_Acres/Bin4Acres_sum) #Calculates the probability of choosing particular parcel from bin Bin4_parprob=abs(1-Bin4_cumper) #Combines parcel acreage data and cumlative percentage data Bin4Main.data = cbind(Bin4_Acres,Bin4_parprob,Bin4_TAZ,Bin4_TAZvacant) #Bin 5 #--- #Loads bin data frame from csv files with acres and TAZ data Bin5_main - read.csv(file=I:/Research/Samba/urb_transport_modeling/LUSDR/Workspace/BizLandPrice/data/Bin_lookup_values/Bin5_lookup.csv,header=FALSE); #Separates Acres data from main data and converts acres to square feet Bin5_Acres=Bin5_main[[1]]*43560 #Separates TAZ data from main data Bin5_TAZ=Bin5_main[[2]] #Separates TAZ data from main data and converts acres to square feet Bin5_TAZvacant=Bin5_main[[3]]*43560 #Sums each parcel acreage data of the bin Bin5Acres_sum=sum(Bin5_Acres) #Creates data frame of cumlative percentages of each parcel of bin Bin5_cumper=cumsum(Bin5_Acres/Bin5Acres_sum) #Calculates the probability of choosing particular parcel from bin Bin5_parprob=abs(1-Bin5_cumper) #Combines parcel acreage data and cumlative percentage data Bin5Main.data = cbind(Bin5_Acres,Bin5_parprob,Bin5_TAZ,Bin5_TAZvacant) #Bin 6 #---
Re: [R] Data manipulation - multiplicate cases
Is this what you are looking for: x X Y Z 1 123 3 1 2 234 3 1 3 345 4 2 4 456 3 2 new.x - x[rep(seq(nrow(x)), times=x$Y),] new.x X Y Z 1 123 3 1 1.1 123 3 1 1.2 123 3 1 2 234 3 1 2.1 234 3 1 2.2 234 3 1 3 345 4 2 3.1 345 4 2 3.2 345 4 2 3.3 345 4 2 4 456 3 2 4.1 456 3 2 4.2 456 3 2 new.x$Z - ave(new.x$Z, new.x$X, FUN=function(z) c(rep(1,z[1]), rep(0, length(z) - z[1]))) new.x X Y Z 1 123 3 1 1.1 123 3 0 1.2 123 3 0 2 234 3 1 2.1 234 3 0 2.2 234 3 0 3 345 4 1 3.1 345 4 1 3.2 345 4 0 3.3 345 4 0 4 456 3 1 4.1 456 3 1 4.2 456 3 0 On Thu, Mar 26, 2009 at 4:26 PM, MarcioRibeiro mes...@pop.com.br wrote: Hi listers, I am trying to arrange my data and I didn't find any information how to do it! I have a data with 3 variables: X Y Z 1-I would like to multiplicate de information of X according to the number I have for my Y variable... 2-Then I want to identify with a dicotomic variable by the number according my variable Z from X... I can do the first part by... z-rep(x,y) But I don't know how to set a dicotomic variable according to Z... Exemple... I have... X Y Z 123 3 1 234 3 1 345 4 2 456 3 2 I want to get... X Y Z 123 3 1 123 3 0 123 3 0 234 3 1 234 3 0 234 3 0 345 4 1 345 4 1 345 4 0 345 4 0 456 3 1 456 3 1 456 3 0 Thanks in advance... -- View this message in context: http://www.nabble.com/Data-manipulation---multiplicate-cases-tp22730453p22730453.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Jim Holtman Cincinnati, OH +1 513 646 9390 What is the problem that you are trying to solve? __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Some install package fixes for Ubuntu Hardy
Thanks Dirk. I should have noted the debian sig. My question -- as Dirk recognized -- was where to let folk know. Part of the issue is that Ubuntu users who are new to R may not find this stuff. My workaround may also be a pointer for those with other Linux distros. There is now another issue that I don't find r-cran-java as an available package when I run the apt-cache search. (I do find r-cran-rgl; rgl popped up as a dependency in the install.packages, but I should have looked for the debian package.) I'll post my r-cran-java query to the r-sig-debian. JN Dirk Eddelbuettel wrote: On 26 March 2009 at 09:45, John C Nash wrote: | I encountered some failures in using install.packages() to install rgl | and rJava in some of my (multiple) Ubuntu Hardy systems. A quick search | of the 'Net did not show any debian packages for these. The | install.packages messages said header or other files were missing, | suggesting path and related woes. Email with Duncan Murdoch (thanks!) | pointed the way with rgl and led to a fix for rJava in similar fashion. | It may save others some frustration to know my resolution. See below. | | However, I do have a question which a brief rummage of r-project did not | answer. Where should information like this be put? My opinion is that it Maybe on the r-sig-debian list that is dedicated to Debian / Ubuntu and R? | should go on the wiki, but possibly there is a better solution if we can | get the right messages into the package installers, though I recognize | the load that puts on maintainers. | | Cheers, JN | | Ubuntu Hardy rgl install fix: | | The headers gl.h and glu.h are installed with the dev packages | libgl1-mesa-dev and libglu1-mesa-dev. So the fix is to run (in at | terminal as root) | | apt-get install libgl1-mesa-dev | apt-get install libglu1-mesa-dev Yes, which is why the r-cran-rgl package (available in Debian for over five years now, and hence in Ubuntu for probably 4 1/2) has the following Build-Depends (with my manual indentation here): Build-Depends: debhelper (= 5.0.0), r-base-dev (= 2.8.1), cdbs, \ libgl1-mesa-dev | libgl-dev, libglu1-mesa-dev | libglu-dev, \ libpng12-dev, libx11-dev, libxt-dev, x11proto-core-dev | then | R | . | install.packages(rgl) | | etc. Let's not forget the 'sudo apt-get install r-cran-rgl' alternative. | Ubuntu Hardy rJava install fix: | | Needed to get Sun JDK (not JRE) | | Then add new | ln -s /usr/java/jdkx/bin/java java | and | ln -s /usr/java/jdkx/bin/javac javac | | where xx is the version information on the jdk directory name -- in | my case 1.6.0_13 (see below) | | Then | | R CMD javareconf | | still fails to find the java compiler. | | Seems $JAVA_HOME may not be defined. | | Try | export JAVA_HOME=/usr/java/jdk1.6.0_13/ | | Then (as root) | | R CMD javareconf | | seems to work. | Then rJava installed OK. I was then able to install RWeka (my original | objective) and it seems to run OK. Likewise, the r-cran-rjava package has Build-Depends: debhelper (= 7.0.0), r-base-dev (= 2.8.1), cdbs, \ openjdk-6-jdk, automake and R is now configured for this Java version at the built. Again, questions on the r-sig-debian list may have been of help. Hope this helps, Dirk __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] anova with means and SEMs (no raw data)
hi, I have only the means and standard errors of this means of different groups and different conditions (and the group sizes). Is there a function which can compute me an anova out of this information? thanks! __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] pgmm (Blundell-Bond) sample needed
Dear Ivo, please find below some answers to your pgmm-related questions. ## Was: Message: 70 Date: Thu, 26 Mar 2009 21:39:19 + From: ivo...@gmail.com Subject: [R] pgmm (Blundell-Bond) sample needed To: r-help r-h...@stat.math.ethz.ch Message-ID: 0016361e8962dfdfd704660c7...@google.com Content-Type: text/plain Dear R Experts--- Sorry for all the questions yesterday and today. I am trying to use Yves Croissant's pgmm function in the plm package with Blundell-Bond moments. I have read the Blundell-Bond paper, and want to run the simplest model first, d[i,t] = a*d[i,t-1] + fixed[i] + u[i,t] . no third conditioning variables yet. the full set of moment conditions recommended for system-GMM, which is (T-1)*(T-2)/2+(T-3), in which the u's interact with all possible lagged y's and delta y's. I believe that pgmm operates by demanding that firm (i) and year (t) be the first two columns in the data set. Almost correct: this is the easiest way. Else you can supply data organized as you like but then you have to specify who the index is. See vignette(plm), § 4 library(plm) NF=20; NT=10 d= data.frame( firm= rep(1:NF, each=NT), year= rep( 1:NT, NF), x=rnorm(NF*NT) ); # the following fails, because dynformula magic is required; learned this the hard way # v=pgmm( x ~ lag(x), data=d, gmm.inst=~x, lag.gmm=c(2,99), transformation=ld ) The reason for 'dynformula magic' is that lags in panel data are only well defined in conjunction with the group and time indices; therefore in 'plm' lags (and first differences) are best supplied through a 'dynformula' interface inside a model. else you get the standard time-series lag, which is incorrect here. formula= dynformula( x ~ 1, list(1)); # this creates x ~ lag(x) v=pgmm( formula, data=d, gmm.inst=~x, lag.gmm=c(2,99), transformation=ld ) Error in solve.default(suml(Vi)) : system is computationally singular: reciprocal condition number = 8.20734e-20 obviously, I am confused. You should not, as you yourself state that the full set of moment conditions recommended for system-GMM [...] is (T-1)*(T-2)/2+(T-3). If T=10 then you have the equivalent of 9*8/2+7 = 43 regressors (instruments). That's why N=20 is way too little. The original Arellano and Bond example in UKEmpl (which is actually called 'EmplUK'!) has N=140, T=9. I already pointed this out in another r-help post, not many days ago (March 9th, 17:59). May I suggest you give a further look at Arellano's panel data book? This would probably clarify how the instrumments are constructed (by the way, that's also what I am currently reading in my spare time). See also Greene, Econometric analysis, § 18.5 and the Z matrix in particular. (Yves Croissant has put this down nicely in the package vignette as well). when I execute the same command on the included UKEmpl data set, it works. however, my inputs would seem perfectly reasonable. I would hope that the procedure could produce a lag(x) coefficient estimate of around 0, and then call it a day. would be nice; but your troubles aren't over yet :^) could someone please tell me how to instruct pgmm to just estimate this simplest of all BB models? OK, you found out by yourself. Just for the benefit of other list readers, I reproduce the lines you sent us by private email (comments are mine): lagformula= dynformula(x ~ 1, list(1)) # reproduces x~lag(x, 1) in standard OLS parlance v=pgmm(lagformula, data=d, gmm.inst=~x, lag.gmm=c(1,99), transformation=ld ) # means the GMM-system estimator # where you use both levels and differences as instruments. [My ultimate goal is to replicate what another author has run via xtabond2 d ld, gmm(L.(d), lag(1 3)) robust in Stata; if you know the magic of moving this statement into pgmm syntax, I would be even more grateful. Right now, I am so stuck on square 1 that I do not know how to move towards figuring out where I ultimately need to go.] GMM are a tricky subject I still don't master. I'll try to figure out what both Stata and plm do with the instruments and let you know. Anyway, the 'plm' equivalent of Stata's Robust option, which uses the Windmeijer correction if I'm not mistaken, is to specify a robust covariance via vcovHC(). Now to your second message: # Was: Message: 82 Date: Thu, 26 Mar 2009 21:45:49 -0400 From: ivo welch ivo...@gmail.com Subject: Re: [R] pgmm (blundell-bond) help needed To: r-help r-h...@stat.math.ethz.ch Message-ID: 50d1c22d0903261845m7d8b321fq97faab26542a...@mail.gmail.com Content-Type: text/plain; charset=ISO-8859-1 I have been playing with more examples, and I now know that with larger NF's my example code actually produces a result, instead of a singular matrix error. interestingly, stata's xtabond2 command seems ok with these sorts of data sets. either R has more stringent requirements, or stata is too casual.
Re: [R] 'stretching' a binomial variable
At 06:49 AM 3/27/2009, imicola wrote: Hi, Im carrying out some Bayesian analysis using a binomial response variable (proportion: 0 to 1), but most of my observations have a value of 0 and many have very small values (i.e. 0.001). I'm having troubles getting my MCMC algorithm to converge, so I have decided to try normalising my response variable to see if this helps. I want it to stay between 0 and 1 but to have a larger range of values, or just for them all to be slightly higher. Does anyone know the best way to acheive this? I could just add a value to each observation (say 10 to increase the proportion a bit, but ensuring it would still be between 0 and 1) - would that be ok? Or is there a better way to stretch the values up? Sorry - i know its not really an R specific question, but I have never found a forum with as many stats litterate people as this one :-) Cheers - any advice much appreciated! nicola Work with events instead of proportions, and use a Poisson model. Robert A. LaBudde, PhD, PAS, Dpl. ACAFS e-mail: r...@lcfltd.com Least Cost Formulations, Ltd.URL: http://lcfltd.com/ 824 Timberlake Drive Tel: 757-467-0954 Virginia Beach, VA 23464-3239Fax: 757-467-2947 Vere scire est per causas scire __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] loading and manipulating 10 data frames-simplified
Look at using a 'list' as obtained from 'lapply'
fileNames - 'your files to be read'
Bin_main - lapply(fileNames, function(.file){
input - read.csv(fileNames, )
# all your calculations; e.g.,
acre - ...
cbind(acres, parprob, ...)
})
Now look at the structure ('str(Bin_main)') and it should have 10 (or
how ever many files you have) elements with the data you want.
On Thu, Mar 26, 2009 at 5:25 PM, PDXRugger j_r...@hotmail.com wrote:
I have to load 10 different data frames and then manipulate those 10 data
frames but would like to do this in a more simplified code than what i am
doing. I have tried a couple of approaches but cannot get it to work
correctly.
So the initial (bulky) code is:
#Bin 1
#---
#Loads bin data frame from csv files with acres and TAZ data
Bin1_main -
read.csv(file=I:/Research/Samba/urb_transport_modeling/LUSDR/Workspace/BizLandPrice/data/Bin_lookup_values/Bin1_lookup.csv,head=FALSE);
#Separates Acres data from main data and converts acres to square feet
Bin1_Acres=Bin1_main[[1]]*43560
#Separates TAZ data from main data
Bin1_TAZ=Bin1_main[[2]]
#Separates TAZ data from main data and converts acres to square feet
Bin1_TAZvacant=Bin1_main[[3]]*43560
#Sums each parcel acreage data of the bin
Bin1Acres_sum=sum(Bin1_Acres)
#Creates data frame of cumlative percentages of each parcel of bin
Bin1_cumper=cumsum(Bin1_Acres/Bin1Acres_sum)
#Calculates the probability of choosing particular parcel from bin
Bin1_parprob=abs(1-Bin1_cumper)
#Combines parcel acreage data and cumlative percentage data
Bin1Main.data = cbind(Bin1_Acres,Bin1_parprob,Bin1_TAZ,Bin1_TAZvacant)
#Bin 2
#---
#Loads bin data frame from csv files with acres and TAZ data
Bin2_main -
read.csv(file=I:/Research/Samba/urb_transport_modeling/LUSDR/Workspace/BizLandPrice/data/Bin_lookup_values/Bin2_lookup.csv,header=FALSE);
#Separates Acres data from main data and converts acres to square feet
Bin2_Acres=Bin2_main[[1]]*43560
#Separates TAZ data from main data
Bin2_TAZ=Bin2_main[[2]]
#Separates TAZ data from main data and converts acres to square feet
Bin2_TAZvacant=Bin2_main[[3]]*43560
#Sums each parcel acreage data of the bin
Bin2Acres_sum=sum(Bin2_Acres)
#Creates data frame of cumlative percentages of each parcel of bin
Bin2_cumper=cumsum(Bin2_Acres/Bin2Acres_sum)
#Calculates the probability of choosing particular parcel from bin
Bin2_parprob=abs(1-Bin2_cumper)
#Combines parcel acreage data and cumlative percentage data
Bin2Main.data = cbind(Bin2_Acres,Bin2_parprob,Bin2_TAZ,Bin2_TAZvacant)
#Bin 3
#---
#Loads bin data frame from csv files with acres and TAZ data
Bin3_main -
read.csv(file=I:/Research/Samba/urb_transport_modeling/LUSDR/Workspace/BizLandPrice/data/Bin_lookup_values/Bin3_lookup.csv,header=FALSE);
#Separates Acres data from main data and converts acres to square feet
Bin3_Acres=Bin3_main[[1]]*43560
#Separates TAZ data from main data
Bin3_TAZ=Bin3_main[[2]]
#Separates TAZ data from main data and converts acres to square feet
Bin3_TAZvacant=Bin3_main[[3]]*43560
#Sums each parcel acreage data of the bin
Bin3Acres_sum=sum(Bin3_Acres)
#Creates data frame of cumlative percentages of each parcel of bin
Bin3_cumper=cumsum(Bin3_Acres/Bin3Acres_sum)
#Calculates the probability of choosing particular parcel from bin
Bin3_parprob=abs(1-Bin3_cumper)
#Combines parcel acreage data and cumlative percentage data
Bin3Main.data = cbind(Bin3_Acres,Bin3_parprob,Bin3_TAZ,Bin3_TAZvacant)
#Bin 4
#---
#Loads bin data frame from csv files with acres and TAZ data
Bin4_main -
read.csv(file=I:/Research/Samba/urb_transport_modeling/LUSDR/Workspace/BizLandPrice/data/Bin_lookup_values/Bin4_lookup.csv,header=FALSE);
#Separates Acres data from main data and converts acres to square feet
Bin4_Acres=Bin4_main[[1]]*43560
#Separates TAZ data from main data
Bin4_TAZ=Bin4_main[[2]]
#Separates TAZ data from main data and converts acres to square feet
Bin4_TAZvacant=Bin4_main[[3]]*43560
#Sums each parcel acreage data of the bin
Bin4Acres_sum=sum(Bin4_Acres)
#Creates data frame of cumlative percentages of each parcel of bin
Bin4_cumper=cumsum(Bin4_Acres/Bin4Acres_sum)
#Calculates the probability of choosing particular parcel from bin
Bin4_parprob=abs(1-Bin4_cumper)
#Combines parcel acreage data and cumlative percentage data
Bin4Main.data = cbind(Bin4_Acres,Bin4_parprob,Bin4_TAZ,Bin4_TAZvacant)
#Bin 5
#---
#Loads bin data frame from csv files with acres and TAZ data
Bin5_main -
read.csv(file=I:/Research/Samba/urb_transport_modeling/LUSDR/Workspace/BizLandPrice/data/Bin_lookup_values/Bin5_lookup.csv,header=FALSE);
#Separates Acres data from main data and converts acres to square feet
Bin5_Acres=Bin5_main[[1]]*43560
#Separates TAZ data from main data
Bin5_TAZ=Bin5_main[[2]]
#Separates TAZ data from main data and converts acres to square feet
Bin5_TAZvacant=Bin5_main[[3]]*43560
#Sums
Re: [R] Snow Parallel R: makeCluster with more nodes than available
Ubuntu Diego wrote: Hi all, I would like to know what would happen if using snow I create a cluster of size 50, for example using makeCluster(50,type='SOCK') on a machine with 2 Cores and run a function. Does snow run 25 and 25 functions on each of my 2 real processors or it just run 50 functions in one processor ? It will run the 50 in parallel and is not advisable to do so on a machine with 3 cores - a slow down due to administrative overhead is expected. Uwe Ligges Thanks. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] color vectors other than gray()
I'm trying to create a graph where different cells of a grid (a shapefile) will be painted with a color share scale, where the most easy way is to use gray(). Can I somehow get a vector (gradient) of colors, a vector of colors with other methods but gray()? I'm doing this until now quad_N_sp - merge(sp_dist[sp_dist$sp==splist[i],],grelha_ID,by.x=quad,by.y=quadricula ,all.y=T,) quad_N_sp$x[is.na(quad_N_sp$x)] - 0 quad_N_sp - quad_N_sp[order(quad_N_sp$id),] paleta - gray(1-(quad_N_sp$x)/max(quad_N_sp$x)) #! Tons de cinzento win.graph(4,5) plot(grelha,ol=grey80, #! Gráfico com grelha de amostragem e gradiente de abundância fg=paleta, cex.lab=0.7, cex.axis=0.7, cex.main=0.7, xlab=Coord X, ylab=Coord Y, main=paste(Espécie: ,splist[i]), xlim=c(21,24) ) col_lab - c(max(quad_N_sp$x),min(quad_N_sp$x)) #! Vector com os limites min e max do N de indivíduos observados color.legend(248000,12,25,128000,col_lab,sort(unique(paleta)),gradie nt=y,cex=0.6)#! Legenda text(245300,130500,Nº Indivíduos,cex=0.6) plot(blocos,ol=grey40,fg=NA,add=T) I'd like to replace the grey shade by other colors. Thanks in advance Paulo E. Cardoso [[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] Some install package fixes for Ubuntu Hardy
On 27 March 2009 at 08:59, John C Nash wrote: | Thanks Dirk. | | I should have noted the debian sig. My question -- as Dirk recognized -- | was where to let folk know. Part of the issue is that Ubuntu users who are | new to R may not find this stuff. No disrespect --- but we feel that it is easier for new (and even experienced) users to install the r-cran-* binaries, not compile from source. That is why we have been providing binary packages for all these years. | workaround may also be a pointer for those with other Linux distros. | | There is now another issue that I don't find r-cran-java as an available | package when I run the apt-cache search. (I do find r-cran-rgl; rgl popped up as a | dependency in the install.packages, but I should have looked for the debian | package.) It is called r-cran-rjava, but as it was added to Debian only this winter it has not yet made it into Ubuntu 8.10. I expect it to be in 9.04 which will be released next month. Dirk | I'll post my r-cran-java query to the r-sig-debian. | | JN | | Dirk Eddelbuettel wrote: | On 26 March 2009 at 09:45, John C Nash wrote: | | I encountered some failures in using install.packages() to install rgl | | and rJava in some of my (multiple) Ubuntu Hardy systems. A quick search | | of the 'Net did not show any debian packages for these. The | | install.packages messages said header or other files were missing, | | suggesting path and related woes. Email with Duncan Murdoch (thanks!) | | pointed the way with rgl and led to a fix for rJava in similar fashion. | | It may save others some frustration to know my resolution. See below. | | | | However, I do have a question which a brief rummage of r-project did not | | answer. Where should information like this be put? My opinion is that it | | Maybe on the r-sig-debian list that is dedicated to Debian / Ubuntu and R? | | | should go on the wiki, but possibly there is a better solution if we can | | get the right messages into the package installers, though I recognize | | the load that puts on maintainers. | | | | Cheers, JN | | | | Ubuntu Hardy rgl install fix: | | | | The headers gl.h and glu.h are installed with the dev packages | | libgl1-mesa-dev and libglu1-mesa-dev. So the fix is to run (in at | | terminal as root) | | | | apt-get install libgl1-mesa-dev | | apt-get install libglu1-mesa-dev | | Yes, which is why the r-cran-rgl package (available in Debian for over five | years now, and hence in Ubuntu for probably 4 1/2) has the following | Build-Depends (with my manual indentation here): | |Build-Depends: debhelper (= 5.0.0), r-base-dev (= 2.8.1), cdbs, \ |libgl1-mesa-dev | libgl-dev, libglu1-mesa-dev | libglu-dev, \ |libpng12-dev, libx11-dev, libxt-dev, x11proto-core-dev | | | then | | R | | . | | install.packages(rgl) | | | | etc. | | Let's not forget the 'sudo apt-get install r-cran-rgl' alternative. | | | Ubuntu Hardy rJava install fix: | | | | Needed to get Sun JDK (not JRE) | | | | Then add new | | ln -s /usr/java/jdkx/bin/java java | | and | | ln -s /usr/java/jdkx/bin/javac javac | | | | where xx is the version information on the jdk directory name -- in | | my case 1.6.0_13 (see below) | | | | Then | | | | R CMD javareconf | | | | still fails to find the java compiler. | | | | Seems $JAVA_HOME may not be defined. | | | | Try | | export JAVA_HOME=/usr/java/jdk1.6.0_13/ | | | | Then (as root) | | | | R CMD javareconf | | | | seems to work. | | Then rJava installed OK. I was then able to install RWeka (my original | | objective) and it seems to run OK. | | Likewise, the r-cran-rjava package has | | Build-Depends: debhelper (= 7.0.0), r-base-dev (= 2.8.1), cdbs, \ |openjdk-6-jdk, automake | | and R is now configured for this Java version at the built. | | | Again, questions on the r-sig-debian list may have been of help. | | Hope this helps, Dirk | | | | __ | R-help@r-project.org mailing list | https://stat.ethz.ch/mailman/listinfo/r-help | PLEASE do read the posting guide http://www.R-project.org/posting-guide.html | and provide commented, minimal, self-contained, reproducible code. -- Three out of two people have difficulties with fractions. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] Snow Parallel R: makeCluster with more nodes than available
On 27 March 2009 at 13:19, Uwe Ligges wrote: | | Ubuntu Diego wrote: | Hi all, | I would like to know what would happen if using snow I create a cluster | of size 50, for example using makeCluster(50,type='SOCK') on a machine | with 2 Cores and run a function. Does snow run 25 and 25 functions on | each of my 2 real processors or it just run 50 functions in one | processor ? | | It will run the 50 in parallel and is not advisable to do so on a | machine with 3 cores - a slow down due to administrative overhead is | expected. Morevoer those 50 session have to share the existing memory allocation -- which is hardly likely to be large enough. Dirk | Uwe Ligges | | | | Thanks. | | __ | R-help@r-project.org mailing list | https://stat.ethz.ch/mailman/listinfo/r-help | PLEASE do read the posting guide http://www.R-project.org/posting-guide.html | and provide commented, minimal, self-contained, reproducible code. | | __ | R-help@r-project.org mailing list | https://stat.ethz.ch/mailman/listinfo/r-help | PLEASE do read the posting guide http://www.R-project.org/posting-guide.html | and provide commented, minimal, self-contained, reproducible code. -- Three out of two people have difficulties with fractions. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] adding matrices with common column names
folks, if i have three matrices, a, b, cc with some colnames in common, and i want to create a matrix which consists of the common columns added up, and the other columns tacked on, what's a good way to do it? i've got the following roundabout code for two matrices, but if the number of matrices increases, then i'm a bit stymied. a - matrix(1:20,ncol=4); colnames(a) - c(a,b,c,d) b - matrix(1:20,ncol=4); colnames(b) - c(b,c,d, e) cbind(a[,!(colnames(a) %in% colnames(b)), drop = FALSE], a[,intersect(colnames(a),colnames(b))] + b[,intersect(colnames(a),colnames(b)), drop = FALSE], b[,!(colnames(b) %in% colnames(a)), drop = FALSE]) a b c d e [1,] 1 7 17 27 16 [2,] 2 9 19 29 17 [3,] 3 11 21 31 18 [4,] 4 13 23 33 19 [5,] 5 15 25 35 20 now, what if i had a matrix cc? i want to perform the above operation on all three matrices a, b, cc. cc - matrix(1:10,ncol=2); colnames(cc) - c(e,f) i need to end up with: a b c d e f [1,] 1 7 17 27 17 6 [2,] 2 9 19 29 19 7 [3,] 3 11 21 31 21 8 [4,] 4 13 23 33 23 9 [5,] 5 15 25 35 25 10 and, in general, with multiple matrices with intersecting colnames? thanks, murali __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] color vectors other than gray()
?colorRamp Hope this helps, baptiste On 27 Mar 2009, at 13:16, Paulo E. Cardoso wrote: I'm trying to create a graph where different cells of a grid (a shapefile) will be painted with a color share scale, where the most easy way is to use gray(). Can I somehow get a vector (gradient) of colors, a vector of colors with other methods but gray()? I'm doing this until now quad_N_sp - merge(sp_dist[sp_dist $sp==splist[i],],grelha_ID,by.x=quad,by.y=quadricula ,all.y=T,) quad_N_sp$x[is.na(quad_N_sp$x)] - 0 quad_N_sp - quad_N_sp[order(quad_N_sp$id),] paleta - gray(1-(quad_N_sp$x)/max(quad_N_sp$x)) #! Tons de cinzento win.graph(4,5) plot(grelha,ol=grey80, #! Gráfico com grelha de amostragem e gradiente de abundância fg=paleta, cex.lab=0.7, cex.axis=0.7, cex.main=0.7, xlab=Coord X, ylab=Coord Y, main=paste(Espécie: ,splist[i]), xlim=c(21,24) ) col_lab - c(max(quad_N_sp$x),min(quad_N_sp$x)) #! Vector com os limites min e max do N de indivíduos observados color .legend (248000,12,25,128000,col_lab,sort(unique(paleta)),gradie nt=y,cex=0.6)#! Legenda text(245300,130500,Nº Indivíduos,cex=0.6) plot(blocos,ol=grey40,fg=NA,add=T) I'd like to replace the grey shade by other colors. Thanks in advance Paulo E. Cardoso [[alternative HTML version deleted]] ATT1.txt _ Baptiste Auguié School of Physics University of Exeter Stocker Road, Exeter, Devon, EX4 4QL, UK Phone: +44 1392 264187 http://newton.ex.ac.uk/research/emag __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] anova with means and SEMs (no raw data)
Martin Batholdy wrote: hi, I have only the means and standard errors of this means of different groups and different conditions (and the group sizes). Is there a function which can compute me an anova out of this information? There's an example in my book (Section 12.4), or look at fake.trypsin.R in the rawdata subdirectory of the ISwR package. -- O__ Peter Dalgaard Øster Farimagsgade 5, Entr.B c/ /'_ --- Dept. of Biostatistics PO Box 2099, 1014 Cph. K (*) \(*) -- University of Copenhagen Denmark Ph: (+45) 35327918 ~~ - (p.dalga...@biostat.ku.dk) FAX: (+45) 35327907 __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] color vectors other than gray()
See ?colorRampPalette and ?colorRamp colorRampPalette( c(blue, white, red) )(100) colorRamp( c(blue, white, red) )( 0:10 / 10 ) Romain Paulo E. Cardoso wrote: I'm trying to create a graph where different cells of a grid (a shapefile) will be painted with a color share scale, where the most easy way is to use gray(). Can I somehow get a vector (gradient) of colors, a vector of colors with other methods but gray()? I'm doing this until now quad_N_sp - merge(sp_dist[sp_dist$sp==splist[i],],grelha_ID,by.x=quad,by.y=quadricula ,all.y=T,) quad_N_sp$x[is.na(quad_N_sp$x)] - 0 quad_N_sp - quad_N_sp[order(quad_N_sp$id),] paleta - gray(1-(quad_N_sp$x)/max(quad_N_sp$x)) #! Tons de cinzento win.graph(4,5) plot(grelha,ol=grey80, #! Gráfico com grelha de amostragem e gradiente de abundância fg=paleta, cex.lab=0.7, cex.axis=0.7, cex.main=0.7, xlab=Coord X, ylab=Coord Y, main=paste(Espécie: ,splist[i]), xlim=c(21,24) ) col_lab - c(max(quad_N_sp$x),min(quad_N_sp$x)) #! Vector com os limites min e max do N de indivíduos observados color.legend(248000,12,25,128000,col_lab,sort(unique(paleta)),gradie nt=y,cex=0.6)#! Legenda text(245300,130500,Nº Indivíduos,cex=0.6) plot(blocos,ol=grey40,fg=NA,add=T) I'd like to replace the grey shade by other colors. Thanks in advance Paulo E. Cardoso -- Romain Francois Independent R Consultant +33(0) 6 28 91 30 30 http://romainfrancois.blog.free.fr __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] color vectors other than gray()
Hi, Have a look at: ?rainbow ?rgb ?heatmap In my opinion this would've likely popped up with just a little effort of searching. In fact, the help of grey() (?grey) already gives pointers to the other color functions. Please show that you at least have tried to find answers before posting questions on a user list. Tsjerk On Fri, Mar 27, 2009 at 2:16 PM, Paulo E. Cardoso pecard...@netcabo.pt wrote: I'm trying to create a graph where different cells of a grid (a shapefile) will be painted with a color share scale, where the most easy way is to use gray(). Can I somehow get a vector (gradient) of colors, a vector of colors with other methods but gray()? I'm doing this until now quad_N_sp - merge(sp_dist[sp_dist$sp==splist[i],],grelha_ID,by.x=quad,by.y=quadricula ,all.y=T,) quad_N_sp$x[is.na(quad_N_sp$x)] - 0 quad_N_sp - quad_N_sp[order(quad_N_sp$id),] paleta - gray(1-(quad_N_sp$x)/max(quad_N_sp$x)) #! Tons de cinzento win.graph(4,5) plot(grelha,ol=grey80, #! Gráfico com grelha de amostragem e gradiente de abundância fg=paleta, cex.lab=0.7, cex.axis=0.7, cex.main=0.7, xlab=Coord X, ylab=Coord Y, main=paste(Espécie: ,splist[i]), xlim=c(21,24) ) col_lab - c(max(quad_N_sp$x),min(quad_N_sp$x)) #! Vector com os limites min e max do N de indivíduos observados color.legend(248000,12,25,128000,col_lab,sort(unique(paleta)),gradie nt=y,cex=0.6)#! Legenda text(245300,130500,Nº Indivíduos,cex=0.6) plot(blocos,ol=grey40,fg=NA,add=T) I'd like to replace the grey shade by other colors. Thanks in advance Paulo E. Cardoso [[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. -- Tsjerk A. Wassenaar, Ph.D. Junior UD (post-doc) Biomolecular NMR, Bijvoet Center Utrecht University Padualaan 8 3584 CH Utrecht The Netherlands P: +31-30-2539931 F: +31-30-2537623 __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] adding matrices with common column names
one approach is:
a - matrix(1:20,ncol=4); colnames(a) - c(a,b,c,d)
b - matrix(1:20,ncol=4); colnames(b) - c(b,c,d, e)
cc - matrix(1:10,ncol=2); colnames(cc) - c(e,f)
f - function (...) {
mat.lis - list(...)
unq.cnams - unique(unlist(lapply(mat.lis, colnames)))
out - matrix(0, nrow(mat.lis[[1]]), length(unq.cnams),
dimnames = list(NULL, unq.cnams))
for (i in seq_along(mat.lis)) {
mm - mat.lis[[i]]
out[, colnames(mm)] - out[, colnames(mm)] + mm
}
out
}
f(a, b)
f(a, cc)
f(a, b, cc)
I hope it helps.
Best,
Dimitris
murali.me...@fortisinvestments.com wrote:
folks,
if i have three matrices, a, b, cc with some colnames in common, and i
want to create a matrix which consists of the common columns added up,
and the other columns tacked on, what's a good way to do it? i've got
the following roundabout code for two matrices, but if the number of
matrices increases, then i'm a bit stymied.
a - matrix(1:20,ncol=4); colnames(a) - c(a,b,c,d) b -
matrix(1:20,ncol=4); colnames(b) - c(b,c,d, e)
cbind(a[,!(colnames(a) %in% colnames(b)), drop = FALSE],
a[,intersect(colnames(a),colnames(b))] +
b[,intersect(colnames(a),colnames(b)), drop = FALSE],
b[,!(colnames(b) %in% colnames(a)), drop = FALSE])
a b c d e
[1,] 1 7 17 27 16
[2,] 2 9 19 29 17
[3,] 3 11 21 31 18
[4,] 4 13 23 33 19
[5,] 5 15 25 35 20
now, what if i had a matrix cc? i want to perform the above operation on
all three matrices a, b, cc.
cc - matrix(1:10,ncol=2); colnames(cc) - c(e,f)
i need to end up with:
a b c d e f
[1,] 1 7 17 27 17 6
[2,] 2 9 19 29 19 7
[3,] 3 11 21 31 21 8
[4,] 4 13 23 33 23 9
[5,] 5 15 25 35 25 10
and, in general, with multiple matrices with intersecting colnames?
thanks,
murali
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
--
Dimitris Rizopoulos
Assistant Professor
Department of Biostatistics
Erasmus University Medical Center
Address: PO Box 2040, 3000 CA Rotterdam, the Netherlands
Tel: +31/(0)10/7043478
Fax: +31/(0)10/7043014
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/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] 'stretching' a binomial variable
On 3/27/2009 7:49 AM, imicola wrote: Hi, Im carrying out some Bayesian analysis using a binomial response variable (proportion: 0 to 1), but most of my observations have a value of 0 and many have very small values (i.e. 0.001). I'm having troubles getting my MCMC algorithm to converge, so I have decided to try normalising my response variable to see if this helps. It seems to me that the problem in a situation like this is with the algorithm, not with the data. Can't you modify it to get better convergence? For example, set your target to be the square root of your posterior (or some other power between 0 and 1); this is more diffuse, so it's easier to sample from. Then use importance sampling to reweight the sample. Duncan Murdoch I want it to stay between 0 and 1 but to have a larger range of values, or just for them all to be slightly higher. Does anyone know the best way to acheive this? I could just add a value to each observation (say 10 to increase the proportion a bit, but ensuring it would still be between 0 and 1) - would that be ok? Or is there a better way to stretch the values up? Sorry - i know its not really an R specific question, but I have never found a forum with as many stats litterate people as this one :-) Cheers - any advice much appreciated! nicola __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] adding matrices with common column names
Shucks, Dimitris beat me to it. And his code is a bit more elegant than
mine. But since I did the work I may as well post it, right?
This version incorporates a couple of error checks to make sure all your
arguments are matrices with the same number of rows.
add.by.name - function(...){
args - list(...)
mat.test - sapply(args,is.matrix)
if(FALSE %in% mat.test) stop(All arguments must be matrices)
mat.row - unique(sapply(args,nrow))
if(length(mat.row)1) stop(All matrices must have the same number of
rows)
all.names - unique(as.vector(sapply(args,colnames)))
sum.mat - matrix(0,nrow=mat.row,ncol=length(all.names))
colnames(sum.mat) - all.names
for(i in 1:length(args)){
tmp - args[[i]]
sum.mat[,colnames(tmp)] - sum.mat[,colnames(tmp)] + tmp
}
return(sum.mat)
}
m1 - matrix(1:20,ncol=4); colnames(m1) - c(a,b,c,d)
m2 - matrix(1:20,ncol=4); colnames(m2) - c(b,c,d,e)
m3 - matrix(1:20,ncol=4); colnames(m3) - c(a,b,d,e)
add.by.name(m1,m2,m3)
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org]
On Behalf Of murali.me...@fortisinvestments.com
Sent: Friday, March 27, 2009 9:25 AM
To: r-help@r-project.org
Subject: [R] adding matrices with common column names
folks,
if i have three matrices, a, b, cc with some colnames in common, and i
want to create a matrix which consists of the common columns added up,
and the other columns tacked on, what's a good way to do it? i've got
the following roundabout code for two matrices, but if the number of
matrices increases, then i'm a bit stymied.
a - matrix(1:20,ncol=4); colnames(a) - c(a,b,c,d) b -
matrix(1:20,ncol=4); colnames(b) - c(b,c,d, e)
cbind(a[,!(colnames(a) %in% colnames(b)), drop = FALSE],
a[,intersect(colnames(a),colnames(b))] +
b[,intersect(colnames(a),colnames(b)), drop = FALSE],
b[,!(colnames(b) %in% colnames(a)), drop = FALSE])
a b c d e
[1,] 1 7 17 27 16
[2,] 2 9 19 29 17
[3,] 3 11 21 31 18
[4,] 4 13 23 33 19
[5,] 5 15 25 35 20
now, what if i had a matrix cc? i want to perform the above operation on
all three matrices a, b, cc.
cc - matrix(1:10,ncol=2); colnames(cc) - c(e,f)
i need to end up with:
a b c d e f
[1,] 1 7 17 27 17 6
[2,] 2 9 19 29 19 7
[3,] 3 11 21 31 21 8
[4,] 4 13 23 33 23 9
[5,] 5 15 25 35 25 10
and, in general, with multiple matrices with intersecting colnames?
thanks,
murali
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide
http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
===
P Please consider the environment before printing this e-mail
Cleveland Clinic is ranked one of the top hospitals
in America by U.S. News World Report (2008).
Visit us online at http://www.clevelandclinic.org for
a complete listing of our services, staff and
locations.
Confidentiality Note: This message is intended for use\...{{dropped:13}}
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/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] 3D PLOT
Hello,
I would like to create a 3D plot with the following data formats:
a - 1:100
b - 1:100
c - matrix(, 100, 100)
i.e.
c(i,j) = f ( a(i) , b(j) )
each of the 1 elements i,j in matrix c is a function of a(i) and
b(j). I would like to have a,b on the x and z axis and c on the y-axis.
Does anybody have an idea how to accomplish that? Thanks in advance.
Regards
BO
#adBox3 {display:none;}
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/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] Competing risks Kalbfleisch Prentice method
Ravi's last note finished with I am wondering why Terry Therneau's survival package doesn't have this option. The short answer is that there are only so many hours in a day. I've recently moved the code base from an internal Mayo repository to R-forge, one long term goal with this is to broaden the developer base to n2 (me and Thomas Lumley). A longer statistical answer: I'm not sure if the this of Ravi's question is a. smoothed hazards, b. the KP cumulative incidence or c. the Fine Gray model. b. I like the CI model and am using it more. We also have local code. The latest version of survival (on rforge, likely in the next default R release) has added simple CI curves to the survfit function. Adding code for survfit on Cox models is on the todo list. But -- this release also fixes up survfit.coxph to handle weighted Cox models and that was on my list for approx 10 years, i.e., don't hold your breath. I don't release something until it also has a set of worked out test cases to add to the 'tests' directory. a. smoothed hazards. For the case at hand I don't see any particular advantage of this. On the other hand, I often would like to display hazard functions instead of CI functions for Cox models; with time dependent covariates I don't think a survival curve makes sense. But I haven't had the time to think through exactly which methods should be added. c. Fine Gray model, i.e., where covariates have a direct influence on the competing risk. I find the model completely untenable from a biologic point of view, so have no interest in adding it. (Due to finite time, everything in the survival package is code that I needed for an analysis; medical research is what pays my salary.) Assume that I have competing processes/risks, say progression of a tumor and heart disease; I expect that the tumor process pays no attention whatsoever to what is going on in the heart. But this is necessary if type=squamous is modeled as an absolute beta=__ increase in the CI for cancer. The squamous cells need to step up the pace of invasion if heart failure threatens, like jockeys in a horse race. Terry T. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Reading in files with variable parts to names
Dear all, Thanks for the help in the previous posts. I've considered each one and have nearly managed to get it working. The structure of the filelist being produced is correct, except for a single space which I can't seem to eradicate! This is my amended code, followed by the first twelve rows of the output (it really goes up to 120 rows). filelist - paste(C:\\Documents and Settings\\Data\\comp_runoff_hd_,do.call(paste, expand.grid(year = sprintf(%04d, seq(1986,1995)), month = sprintf(%02d,1:12))),.asc, sep=) filelist [1] C:\\Documents and Settings\\Data\\comp1986 01.asc [2] C:\\Documents and Settings\\Data\\comp1987 01.asc [3] C:\\Documents and Settings\\Data\\comp1988 01.asc [4] C:\\Documents and Settings\\Data\\comp1989 01.asc [5] C:\\Documents and Settings\\Data\\comp1990 01.asc [6] C:\\Documents and Settings\\Data\\comp1991 01.asc [7] C:\\Documents and Settings\\Data\\comp1992 01.asc [8] C:\\Documents and Settings\\Data\\comp1993 01.asc [9] C:\\Documents and Settings\\Data\\comp1994 01.asc [10] C:\\Documents and Settings\\Data\\comp1995 01.asc [11] C:\\Documents and Settings\\Data\\comp1986 02.asc [12] C:\\Documents and Settings\\Data\\comp1987 02.asc I've tried inserting sep= after the 'month=sprintf(%02d,1:12)' but this doesn't appear to solve the problem - in fact it doesn't change the output at all...! Any help would be much appreciated, Steve __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] snow Error.
Hello, I have a program that used to run well in October, it uses library snow. Since then, one change has ocurred (snow library has been updated) and another could have ocurred (I've unadvertently modified something). Anyway, now when I make the call: parallel.model.results - clusterApply(cl,processors.struct,MCexe) exactly as I used to do, where MCexe is my function and processors.struct is a list containing everything required by MCexe, I obtain the following error: Error in checkForRemoteErrors(val) : 2 nodes produced errors; first error: incorrect number of dimensions Please, do you have any clue about what could be the error? Best regards, Javier García-Pintado __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] 3D PLOT
On 3/27/2009 9:57 AM, bastian250...@freenet.de wrote: Hello, I would like to create a 3D plot with the following data formats: a - 1:100 b - 1:100 c - matrix(, 100, 100) i.e. c(i,j) = f ( a(i) , b(j) ) each of the 1 elements i,j in matrix c is a function of a(i) and b(j). I would like to have a,b on the x and z axis and c on the y-axis. Does anybody have an idea how to accomplish that? Thanks in advance. persp, contour, image, wireframe, contourplot, etc. (in graphics and lattice); persp3d (in rgl). 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.
[R] Sweave-output causes error-message in pdflatex
Dear list, Latex/Sweave has trouble processing Sveave-output coming from the summary-command of a linear Model. summary(lmRub) The output line causing the trouble looks in R like this Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1 In my Sweaved Tex-file that line looks like this Signif. codes: 0 *** 0.001 ** 0.01 * 0.05 . 0.1 1 (actually in the editor the quotation signs are replaced by bars, but they got lost through copy paste. I don't know if that says anything about my problem.) In the error message produced through pdflatex, the quotation signs reappear. Latex error-message: ! Package inputenc Error: Keyboard character used is undefined (inputenc) in inputencoding `Latin1'. See the inputenc package documentation for explanation. Type H return for immediate help. ... l.465 ...*’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1 You need to provide a definition with \DeclareInputText or \DeclareInputMath before using this key. I hope anybody knows how I can prevent that error message. Thanks in advance. Gerrit __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] R 2.8.1 and 2.9 alpha crash when running survest of Design package
A couple additions to Thomas's message. The 'survest' function in design directly called C routines in the survival package. The argument list to the routines changed due to the addition of weights; calling a C routine with the wrong arguments is one of the more reliable ways to crash a program. The simplest (short term) solution is to use survfit for your curves rather than survest. Frank Harrell has been aware of the issue for several weeks and is working hard on solving it. The simple fix is a few minutes, but he's thinking about how to avoid any future problems. The C routines in survival change arguments VERY rarely, but direcly calling the routines of another package is considered dangerous in general. Most breakage was less severe. For instance there were a couple of errors in the PBC data set. I fixed these, and also replaced all the 999 codes with NA to make it easier to use. Some other packages use this data. (My name is on most of the PBC papers and I have the master PBC data with all labs, patient id, etc, but I was not the source of the first data set). We'll be keeping an eye on the R list as the package rolls out; sending a message directly to Thomas and/or I would also be appreciated for issues like this. Terry Therneau __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] color vectors other than gray()
On Fri, 27 Mar 2009, Tsjerk Wassenaar wrote: Hi, Have a look at: ?rainbow ?rgb ?heatmap Furthermore, the packages colorspace, ggplot, plotrix, and RColorBrewer have useful tools for this. For the ides behind the palettes in colorspace, see Achim Zeileis, Kurt Hornik, and Paul Murrell (2009). Escaping RGBland: Selecting Colors for Statistical Graphics. Computational Statistics Data Analysis, Forthcoming. doi:10.1016/j.csda.2008.11.033 Preprint: http://statmath.wu-wien.ac.at/~zeileis/papers/Zeileis+Hornik+Murrell-2008.pdf hth, Z In my opinion this would've likely popped up with just a little effort of searching. In fact, the help of grey() (?grey) already gives pointers to the other color functions. Please show that you at least have tried to find answers before posting questions on a user list. Tsjerk On Fri, Mar 27, 2009 at 2:16 PM, Paulo E. Cardoso pecard...@netcabo.pt wrote: I'm trying to create a graph where different cells of a grid (a shapefile) will be painted with a color share scale, where the most easy way is to use gray(). Can I somehow get a vector (gradient) of colors, a vector of colors with other methods but gray()? I'm doing this until now quad_N_sp - merge(sp_dist[sp_dist$sp==splist[i],],grelha_ID,by.x=quad,by.y=quadricula ,all.y=T,) quad_N_sp$x[is.na(quad_N_sp$x)] - 0 quad_N_sp - quad_N_sp[order(quad_N_sp$id),] paleta - gray(1-(quad_N_sp$x)/max(quad_N_sp$x)) #! Tons de cinzento win.graph(4,5) plot(grelha,ol=grey80, #! Gráfico com grelha de amostragem e gradiente de abundância fg=paleta, cex.lab=0.7, cex.axis=0.7, cex.main=0.7, xlab=Coord X, ylab=Coord Y, main=paste(Espécie: ,splist[i]), xlim=c(21,24) ) col_lab - c(max(quad_N_sp$x),min(quad_N_sp$x)) #! Vector com os limites min e max do N de indivíduos observados color.legend(248000,12,25,128000,col_lab,sort(unique(paleta)),gradie nt=y,cex=0.6)#! Legenda text(245300,130500,Nº Indivíduos,cex=0.6) plot(blocos,ol=grey40,fg=NA,add=T) I'd like to replace the grey shade by other colors. Thanks in advance Paulo E. Cardoso [[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. -- Tsjerk A. Wassenaar, Ph.D. Junior UD (post-doc) Biomolecular NMR, Bijvoet Center Utrecht University Padualaan 8 3584 CH Utrecht The Netherlands P: +31-30-2539931 F: +31-30-2537623 __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Reading in files with variable parts to names
Does this give you what you want (just did it in two steps); x - expand.grid(year = sprintf(%04d, seq(1986, 1995)), month = sprintf(%02d, 1:12)) filelist - paste(C:\\Documents and Settings\\Data\\comp_runoff_hd_, paste(x$year, x$month, sep=''), '.asc', sep='') filelist [1] C:\\Documents and Settings\\Data\\comp_runoff_hd_198601.asc C:\\Documents and Settings\\Data\\comp_runoff_hd_198701.asc C:\\Documents and Settings\\Data\\comp_runoff_hd_198801.asc [4] C:\\Documents and Settings\\Data\\comp_runoff_hd_198901.asc C:\\Documents and Settings\\Data\\comp_runoff_hd_199001.asc C:\\Documents and Settings\\Data\\comp_runoff_hd_199101.asc [7] C:\\Documents and Settings\\Data\\comp_runoff_hd_199201.asc C:\\Documents and Settings\\Data\\comp_runoff_hd_199301.asc C:\\Documents and Settings\\Data\\comp_runoff_hd_199401.asc [10] C:\\Documents and Settings\\Data\\comp_runoff_hd_199501.asc C:\\Documents and Settings\\Data\\comp_runoff_hd_198602.asc C:\\Documents and Settings\\Data\\comp_runoff_hd_198702.asc [13] C:\\Documents and Settings\\Data\\comp_runoff_hd_198802.asc C:\\Documents and Settings\\Data\\comp_runoff_hd_198902.asc C:\\Documents and Settings\\Data\\comp_runoff_hd_199002.asc [16] C:\\Documents and Settings\\Data\\comp_runoff_hd_199102.asc C:\\Documents and Settings\\Data\\comp_runoff_hd_199202.asc C:\\Documents and Settings\\Data\\comp_runoff_hd_199302.asc [19] C:\\Documents and Settings\\Data\\comp_runoff_hd_199402.asc C:\\Documents and Settings\\Data\\comp_runoff_hd_199502.asc C:\\Documents and Settings\\Data\\comp_runoff_hd_198603.asc [22] C:\\Documents and Settings\\Data\\comp_runoff_hd_198703.asc C:\\Documents and Settings\\Data\\comp_runoff_hd_198803.asc C:\\Documents and Settings\\Data\\comp_runoff_hd_198903.asc [25] C:\\Documents and Settings\\Data\\comp_runoff_hd_199003.asc C:\\Documents and Settings\\Data\\comp_runoff_hd_199103.asc C:\\Documents and Settings\\Data\\comp_runoff_hd_199203.asc [28] C:\\Documents and Settings\\Data\\comp_runoff_hd_199303.asc C:\\Documents and Settings\\Data\\comp_runoff_hd_199403.asc C:\\Documents and Settings\\Data\\comp_runoff_hd_199503.asc [31] C:\\Documents and Settings\\Data\\comp_runoff_hd_198604.asc C:\\Documents and Settings\\Data\\comp_runoff_hd_198704.asc C:\\Documents and Settings\\Data\\comp_runoff_hd_198804.asc [34] C:\\Documents and Settings\\Data\\comp_runoff_hd_198904.asc C:\\Documents and Settings\\Data\\comp_runoff_hd_199004.asc C:\\Documents and Settings\\Data\\comp_runoff_hd_199104.asc [37] C:\\Documents and Settings\\Data\\comp_runoff_hd_199204.asc C:\\Documents and Settings\\Data\\comp_runoff_hd_199304.asc C:\\Documents and Settings\\Data\\comp_runoff_hd_199404.asc [40] C:\\Documents and Settings\\Data\\comp_runoff_hd_199504.asc C:\\Documents and Settings\\Data\\comp_runoff_hd_198605.asc C:\\Documents and Settings\\Data\\comp_runoff_hd_198705.asc [43] C:\\Documents and Settings\\Data\\comp_runoff_hd_198805.asc C:\\Documents and Settings\\Data\\comp_runoff_hd_198905.asc C:\\Documents and Settings\\Data\\comp_runoff_hd_199005.asc On Fri, Mar 27, 2009 at 9:56 AM, Steve Murray smurray...@hotmail.com wrote: Dear all, Thanks for the help in the previous posts. I've considered each one and have nearly managed to get it working. The structure of the filelist being produced is correct, except for a single space which I can't seem to eradicate! This is my amended code, followed by the first twelve rows of the output (it really goes up to 120 rows). filelist - paste(C:\\Documents and Settings\\Data\\comp_runoff_hd_,do.call(paste, expand.grid(year = sprintf(%04d, seq(1986,1995)), month = sprintf(%02d,1:12))),.asc, sep=) filelist [1] C:\\Documents and Settings\\Data\\comp1986 01.asc [2] C:\\Documents and Settings\\Data\\comp1987 01.asc [3] C:\\Documents and Settings\\Data\\comp1988 01.asc [4] C:\\Documents and Settings\\Data\\comp1989 01.asc [5] C:\\Documents and Settings\\Data\\comp1990 01.asc [6] C:\\Documents and Settings\\Data\\comp1991 01.asc [7] C:\\Documents and Settings\\Data\\comp1992 01.asc [8] C:\\Documents and Settings\\Data\\comp1993 01.asc [9] C:\\Documents and Settings\\Data\\comp1994 01.asc [10] C:\\Documents and Settings\\Data\\comp1995 01.asc [11] C:\\Documents and Settings\\Data\\comp1986 02.asc [12] C:\\Documents and Settings\\Data\\comp1987 02.asc I've tried inserting sep= after the 'month=sprintf(%02d,1:12)' but this doesn't appear to solve the problem - in fact it doesn't change the output at all...! Any help would be much appreciated, Steve __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Jim Holtman Cincinnati, OH +1 513 646 9390 What is the problem that you are trying to solve?
Re: [R] Reading in files with variable parts to names
That's because do.call wants a list: what about this one: do.call( sprintf, append( list(C:\\Documents and Settings\\Data\\comp_runoff_hd_%04d%02d.asc), expand.grid( seq(1986,1995), 1:12) ) ) Romain Steve Murray wrote: Dear all, Thanks for the help in the previous posts. I've considered each one and have nearly managed to get it working. The structure of the filelist being produced is correct, except for a single space which I can't seem to eradicate! This is my amended code, followed by the first twelve rows of the output (it really goes up to 120 rows). filelist - paste(C:\\Documents and Settings\\Data\\comp_runoff_hd_,do.call(paste, expand.grid(year = sprintf(%04d, seq(1986,1995)), month = sprintf(%02d,1:12))),.asc, sep=) filelist [1] C:\\Documents and Settings\\Data\\comp1986 01.asc [2] C:\\Documents and Settings\\Data\\comp1987 01.asc [3] C:\\Documents and Settings\\Data\\comp1988 01.asc [4] C:\\Documents and Settings\\Data\\comp1989 01.asc [5] C:\\Documents and Settings\\Data\\comp1990 01.asc [6] C:\\Documents and Settings\\Data\\comp1991 01.asc [7] C:\\Documents and Settings\\Data\\comp1992 01.asc [8] C:\\Documents and Settings\\Data\\comp1993 01.asc [9] C:\\Documents and Settings\\Data\\comp1994 01.asc [10] C:\\Documents and Settings\\Data\\comp1995 01.asc [11] C:\\Documents and Settings\\Data\\comp1986 02.asc [12] C:\\Documents and Settings\\Data\\comp1987 02.asc I've tried inserting sep= after the 'month=sprintf(%02d,1:12)' but this doesn't appear to solve the problem - in fact it doesn't change the output at all...! Any help would be much appreciated, Steve __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Romain Francois Independent R Consultant +33(0) 6 28 91 30 30 http://romainfrancois.blog.free.fr __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] pca vs. pfa: dimension reduction
At 18:22 25/03/2009, Jonathan Baron wrote: On 03/25/09 19:06, soeren.vo...@eawag.ch wrote: Can't make sense of calculated results and hope I'll find help here. I've collected answers from about 600 persons concerning three variables. I hypothesise those three variables to be components (or indicators) of one latent factor. In order to reduce data (vars), I had the following idea: Calculate the factor underlying these three vars. Use the loadings and the original var values to construct an new (artificial) var: (B1 * X1) + (B2 * X2) + (B3 * X3) = ArtVar (brackets for readability). Use ArtVar for further analysis of the data, that is, as predictor etc. In my (I realise, elementary) psychological statistics readings I was taught to use pca for these problems. Referring to Venables Ripley (2002, chapter 11), I applied princomp to my vars. But the outcome shows 4 components -- which is obviously not what I want. Reading further I found factanal, which produces loadings on the one specified factor very fine. But since this is a contradiction to theoretical introductions in so many texts I'm completely confused whether I'm right with these calculations. Perhaps I am missing something here but how do you get four components with three variables? (1) Is there an easy example, which explains the differences between pca and pfa? (2) Which R procedure should I use to get what I want? Possibly what you want is the first principal component, which the weighted sum that accounts for the most variance of the three variables. It does essentially what you say in your first paragraph. So you want something like p1 - princomp(cbind(X1,X2,X3),scores=TRUE) p1$scores[,1] The trouble with factanal is that it does a rotation, and the default is varimax. The first factor will usually not be the same as the first principal component (I think). Perhaps there is another rotation option that will give you this, but why bother even to look? (I didn't, obviously.) Jon -- Jonathan Baron, Professor of Psychology, University of Pennsylvania Home page: http://www.sas.upenn.edu/~baron Michael Dewey http://www.aghmed.fsnet.co.uk __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] cannot reproduce matlab wavelet results with R
Dear R users, I want to get the D1 details reconstructed to the level of my time series. My original time series is NH$annual[,] and it has 118 elements. This is the code I use and the results: library(wavelets) NHj - extend.series(X=NH$annual[,], method=reflection, length=powerof2, j=7); detach(package:wavelets) attributes(mra(X=NHj, filter=d4, n.levels=7, boundary=reflection, fast=TRUE, method=dwt))$D[[1]][1:20] [1] -0.0363166651 0.0793856487 0.0229855716 -0.0863862067 0.0586129763 [6] -0.0552697096 0.0049741291 0.0327406169 0.0006990289 -0.0150729128 [11] -0.0203610433 0.0424490289 -0.0103379856 -0.0072504717 -0.0310480084 [16] 0.0543716287 0.0491613932 -0.1187032803 0.0317373320 0.0132924682 detach(package:wavelets) library(waveslim) mra(x=NHj, wf = d4, J = 7, method = dwt, boundary = reflection)$D1[1:20] [1] -0.0363166651 0.0793856487 0.0229855716 -0.0863862067 0.0586129763 [6] -0.0552697096 0.0049741291 0.0327406169 0.0006990289 -0.0150729128 [11] -0.0203610433 0.0424490289 -0.0103379856 -0.0072504717 -0.0310480084 [16] 0.0543716287 0.0491613932 -0.1187032803 0.0317373320 0.0132924682 detach(package:waveslim) library(wmtsa) wavMRDSum(x=NHj, wavelet=d4,levels=1, xform=dwt, reflect=FALSE, keep.smooth=FALSE, keep.details=TRUE)[1:20] t=0 t=1 t=2 t=3 t=4 -0.1449234169 0.0166815113 0.0229855716 -0.0863862067 0.0586129763 t=5 t=6 t=7 t=8 t=9 -0.0552697096 0.0049741291 0.0327406169 0.0006990289 -0.0150729128 t=10 t=11 t=12 t=13 t=14 -0.0203610433 0.0424490289 -0.0103379856 -0.0072504717 -0.0310480084 t=15 t=16 t=17 t=18 t=19 0.0543716287 0.0491613932 -0.1187032803 0.0317373320 0.0132924682 detach(package:wmtsa) library(wavethresh) NHwd.obj - wd(data=NHj, filter.number=4, family=DaubExPhase, type=wavelet, bc=symmetric, verbose=TRUE); NHwd.objA0 - putC(wd=NHwd.obj, level=6, v=rep(0,2^6), boundary=FALSE, index=FALSE); D1 - accessC(wd=wr(wd=NHwd.objA0, start.level = 6, return.object = TRUE, verbose = TRUE),level=7,boundary=FALSE); D1[1:20] [1] -0.25283845 0.06657357 0.03389600 -0.04797488 0.05665413 -0.09317851 [7] 0.06466827 0.06839502 -0.07792329 -0.06458924 0.07678030 0.04101479 [13] -0.08070069 0.06491276 -0.02459910 -0.05140745 0.07088627 -0.03537575 [19] 0.01366095 -0.01599816 As you can see, with wavethresh the results are quite different. Have I messed something up? Is this the correct way of getting the D1 details? I am a former matlab user, and the results I get with it are only reproducible in R with the other 3 packages and the d2 or haar wavelets. With any other wavelet, e.g. d4 I get different results.I would be very thankful to you if you give me some clue. I really apologize for taking some of your precious time. I wish you fruitful work. Regards, Martin 27.03.2009 __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] color vectors other than gray()
I'm certainly missing something. In fact the ramp I need must be scaled according to a vector of values (in this case species abundance in each grid cell), as in the example vector below: length(quad_N_sp$x) # where x is the abundance value [1] 433 quad_N_sp$x [1] 0 0 0 0 0 0 5 0 0 0 0 0 0 0 0 0 0 0 2 0 0 0 0 3 0 0 0 0 0 0 0 0 0 0 0 3 2 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 [101] 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 [201] 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 [301] 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 2 0 0 0 0 0 0 0 1 0 2 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 [401] 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 I need to discriminate shading level accordingly to the abundance value (level). I don't know how to proceed. Paulo E. Cardoso -Mensagem original- De: baptiste auguie [mailto:ba...@exeter.ac.uk] Enviada: sexta-feira, 27 de Março de 2009 13:30 Para: Paulo E. Cardoso Cc: r-h...@stat.math.ethz.ch; r-help@r-project.org Assunto: Re: [R] color vectors other than gray() ?colorRamp Hope this helps, baptiste On 27 Mar 2009, at 13:16, Paulo E. Cardoso wrote: I'm trying to create a graph where different cells of a grid (a shapefile) will be painted with a color share scale, where the most easy way is to use gray(). Can I somehow get a vector (gradient) of colors, a vector of colors with other methods but gray()? I'm doing this until now quad_N_sp - merge(sp_dist[sp_dist $sp==splist[i],],grelha_ID,by.x=quad,by.y=quadricula ,all.y=T,) quad_N_sp$x[is.na(quad_N_sp$x)] - 0 quad_N_sp - quad_N_sp[order(quad_N_sp$id),] paleta - gray(1-(quad_N_sp$x)/max(quad_N_sp$x)) #! Tons de cinzento win.graph(4,5) plot(grelha,ol=grey80, #! Gráfico com grelha de amostragem e gradiente de abundância fg=paleta, cex.lab=0.7, cex.axis=0.7, cex.main=0.7, xlab=Coord X, ylab=Coord Y, main=paste(Espécie: ,splist[i]), xlim=c(21,24) ) col_lab - c(max(quad_N_sp$x),min(quad_N_sp$x)) #! Vector com os limites min e max do N de indivíduos observados color .legend (248000,12,25,128000,col_lab,sort(unique(paleta)),gradie nt=y,cex=0.6)#! Legenda text(245300,130500,Nº Indivíduos,cex=0.6) plot(blocos,ol=grey40,fg=NA,add=T) I'd like to replace the grey shade by other colors. Thanks in advance Paulo E. Cardoso [[alternative HTML version deleted]] ATT1.txt _ Baptiste Auguié School of Physics University of Exeter Stocker Road, Exeter, Devon, EX4 4QL, UK Phone: +44 1392 264187 http://newton.ex.ac.uk/research/emag __ No virus found in this incoming message. Checked by AVG - www.avg.com 03/27/09 07:13:00 __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] Competing risks Kalbfleisch Prentice method
Dear Prof. Therneau, Thank you for your views on this subject. I think all R users who play with survival analysis are most grateful for the functions you have already supplied us with. I'm guessing Ravi is wondering why you have not implemented the smoothing of the baseline hazard from the Cox model. I actually tried to do this originally, inspired from this thread (i.e use sm.spline to smooth the hazard): https://stat.ethz.ch/pipermail/r-help/2004-July/053843.html but it overestimated the CI (perhaps I implemented it wrong). I was then advised to treat CI as a step function, rather than continuous, which means that F(t+1, cause k)-F(t, cause k) will be 0 unless an event of cause k has occurred in that interval (see also Competing Risks, by Melanie Pintilie, page 62). This is obviously problematic if one wants to estimate the CI at times that are not close to observed events for either cause (perhaps a parametric model could be used in this case). But then again, this was not an issue wtih my data. Eleni Rapsomaniki Research Associate Strangeways Research Laboratory Department of Public Health and Primary Care University of Cambridge -Original Message- From: Terry Therneau [mailto:thern...@mayo.edu] Sent: 27 March 2009 13:53 To: Eleni Rapsomaniki; tuech...@gmx.at; Ravi Varadhan Cc: r-help@r-project.org Subject: RE: Competing risks Kalbfleisch Prentice method Ravi's last note finished with I am wondering why Terry Therneau's survival package doesn't have this option. The short answer is that there are only so many hours in a day. I've recently moved the code base from an internal Mayo repository to R-forge, one long term goal with this is to broaden the developer base to n2 (me and Thomas Lumley). A longer statistical answer: I'm not sure if the this of Ravi's question is a. smoothed hazards, b. the KP cumulative incidence or c. the Fine Gray model. b. I like the CI model and am using it more. We also have local code. The latest version of survival (on rforge, likely in the next default R release) has added simple CI curves to the survfit function. Adding code for survfit on Cox models is on the todo list. But -- this release also fixes up survfit.coxph to handle weighted Cox models and that was on my list for approx 10 years, i.e., don't hold your breath. I don't release something until it also has a set of worked out test cases to add to the 'tests' directory. a. smoothed hazards. For the case at hand I don't see any particular advantage of this. On the other hand, I often would like to display hazard functions instead of CI functions for Cox models; with time dependent covariates I don't think a survival curve makes sense. But I haven't had the time to think through exactly which methods should be added. c. Fine Gray model, i.e., where covariates have a direct influence on the competing risk. I find the model completely untenable from a biologic point of view, so have no interest in adding it. (Due to finite time, everything in the survival package is code that I needed for an analysis; medical research is what pays my salary.) Assume that I have competing processes/risks, say progression of a tumor and heart disease; I expect that the tumor process pays no attention whatsoever to what is going on in the heart. But this is necessary if type=squamous is modeled as an absolute beta=__ increase in the CI for cancer. The squamous cells need to step up the pace of invasion if heart failure threatens, like jockeys in a horse race. Terry T. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] Competing risks Kalbfleisch Prentice method
Hi Terry, My this was your (a), i.e. the smoothed hazard rate function. I apologize if I came across as being rude. I was only curious to see if you had any scientific/statistical rationale for not including the smoothed hazard option in your survival package, which is, by far, the most widely used tool for time-to-event analysis in R. Therefore, I just felt that having this, fairly useful, capability in survival would be nice. I have a couple of questions related to your two other points: point (b): How would you estimate the effect of a treatment on the cumulative incidence of primary outcome, adjusted for covariates, using the KP approach (both point and interval estimation)? point (c): I don't quite understand why you find the FG model completely biologically untenable. I view it as mathematical trickery to obtain a compact summary of the impact of a covariate on the cumulative incidence. The FG model is especially useful in estimating covariate adjusted treatment effect, provided the proportionality assumption on the sub-distribution hazard is reasonable. The KP approach does not provide such compactness as you have to model all the cause-specific hazards. Best, Ravi. Ravi Varadhan, Ph.D. Assistant Professor, Division of Geriatric Medicine and Gerontology School of Medicine Johns Hopkins University Ph. (410) 502-2619 email: rvarad...@jhmi.edu - Original Message - From: Terry Therneau thern...@mayo.edu Date: Friday, March 27, 2009 9:52 am Subject: RE: Competing risks Kalbfleisch Prentice method To: er...@medschl.cam.ac.uk, tuech...@gmx.at, Ravi Varadhan rvarad...@jhmi.edu Cc: r-help@r-project.org Ravi's last note finished with I am wondering why Terry Therneau's survival package doesn't have this option. The short answer is that there are only so many hours in a day. I've recently moved the code base from an internal Mayo repository to R-forge, one long term goal with this is to broaden the developer base to n2 (me and Thomas Lumley). A longer statistical answer: I'm not sure if the this of Ravi's question is a. smoothed hazards, b. the KP cumulative incidence or c. the Fine Gray model. b. I like the CI model and am using it more. We also have local code. The latest version of survival (on rforge, likely in the next default R release) has added simple CI curves to the survfit function. Adding code for survfit on Cox models is on the todo list. But -- this release also fixes up survfit.coxph to handle weighted Cox models and that was on my list for approx 10 years, i.e., don't hold your breath. I don't release something until it also has a set of worked out test cases to add to the 'tests' directory. a. smoothed hazards. For the case at hand I don't see any particular advantage of this. On the other hand, I often would like to display hazard functions instead of CI functions for Cox models; with time dependent covariates I don't think a survival curve makes sense. But I haven't had the time to think through exactly which methods should be added. c. Fine Gray model, i.e., where covariates have a direct influence on the competing risk. I find the model completely untenable from a biologic point of view, so have no interest in adding it. (Due to finite time, everything in the survival package is code that I needed for an analysis; medical research is what pays my salary.) Assume that I have competing processes/risks, say progression of a tumor and heart disease; I expect that the tumor process pays no attention whatsoever to what is going on in the heart. But this is necessary if type=squamous is modeled as an absolute beta=__ increase in the CI for cancer. The squamous cells need to step up the pace of invasion if heart failure threatens, like jockeys in a horse race. Terry T. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] color vectors other than gray()
Can you provide a minimal example that we can run directly after copy and paste (using a standard data set or dummy data)? It's always helpful to try and nail down the core of your question (often you'll find the answer while formulating your question in minimal terms). baptiste On 27 Mar 2009, at 14:36, Paulo E. Cardoso wrote: I'm certainly missing something. In fact the ramp I need must be scaled according to a vector of values (in this case species abundance in each grid cell), as in the example vector below: length(quad_N_sp$x) # where x is the abundance value [1] 433 quad_N_sp$x [1] 0 0 0 0 0 0 5 0 0 0 0 0 0 0 0 0 0 0 2 0 0 0 0 3 0 0 0 0 0 0 0 0 0 0 0 3 2 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 [101] 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 [201] 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 [301] 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 2 0 0 0 0 0 0 0 1 0 2 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 [401] 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 I need to discriminate shading level accordingly to the abundance value (level). I don't know how to proceed. Paulo E. Cardoso -Mensagem original- De: baptiste auguie [mailto:ba...@exeter.ac.uk] Enviada: sexta-feira, 27 de Março de 2009 13:30 Para: Paulo E. Cardoso Cc: r-h...@stat.math.ethz.ch; r-help@r-project.org Assunto: Re: [R] color vectors other than gray() ?colorRamp Hope this helps, baptiste On 27 Mar 2009, at 13:16, Paulo E. Cardoso wrote: I'm trying to create a graph where different cells of a grid (a shapefile) will be painted with a color share scale, where the most easy way is to use gray(). Can I somehow get a vector (gradient) of colors, a vector of colors with other methods but gray()? I'm doing this until now quad_N_sp - merge(sp_dist[sp_dist $sp==splist[i],],grelha_ID,by.x=quad,by.y=quadricula ,all.y=T,) quad_N_sp$x[is.na(quad_N_sp$x)] - 0 quad_N_sp - quad_N_sp[order(quad_N_sp$id),] paleta - gray(1-(quad_N_sp$x)/max(quad_N_sp$x)) #! Tons de cinzento win.graph(4,5) plot(grelha,ol=grey80, #! Gráfico com grelha de amostragem e gradiente de abundância fg=paleta, cex.lab=0.7, cex.axis=0.7, cex.main=0.7, xlab=Coord X, ylab=Coord Y, main=paste(Espécie: ,splist[i]), xlim=c(21,24) ) col_lab - c(max(quad_N_sp$x),min(quad_N_sp$x)) #! Vector com os limites min e max do N de indivíduos observados color .legend (248000,12,25,128000,col_lab,sort(unique(paleta)),gradie nt=y,cex=0.6)#! Legenda text(245300,130500,Nº Indivíduos,cex=0.6) plot(blocos,ol=grey40,fg=NA,add=T) I'd like to replace the grey shade by other colors. Thanks in advance Paulo E. Cardoso [[alternative HTML version deleted]] ATT1.txt _ Baptiste Auguié School of Physics University of Exeter Stocker Road, Exeter, Devon, EX4 4QL, UK Phone: +44 1392 264187 http://newton.ex.ac.uk/research/emag __ No virus found in this incoming message. Checked by AVG - www.avg.com Version: 8.0.238 / Virus Database: 270.11.30/2026 - Release Date: 03/27/09 07:13:00 _ Baptiste Auguié School of Physics University of Exeter Stocker Road, Exeter, Devon, EX4 4QL, UK Phone: +44 1392 264187 http://newton.ex.ac.uk/research/emag __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Ploting a matrix
Unfortunately, I could not solve the problem of plotting all columns of a matrix against the first column I used: b=read.table(d:\\programme\\R\\übungen\\Block 1b.txt, header=T) b is a table with the first column using Dates and the following columns with vectors. apply(b[,-1], 2, plot, x= b[,1]) Also all columns have the same length, [R] states that the length are different. Can you help me? baptiste auguie schrieb: Something like this perhaps, a - matrix(rnorm(5*49), ncol=49) pdf(width=15, height=15) par(mfrow= c(8,6)) apply(a[,-1], 2, plot, x= a[,1]) dev.off() HTH, baptiste On 27 Mar 2009, at 11:05, skrug wrote: Hi evrybody, in a matrix consisting of 49 columns, I would like to plot all columns against the first in 48 different graphs. Can you help me? Thank you in advance Sebastian -- *** Dipl. Biol. Sebastian Krug PhD - student IFM - GEOMAR Leibniz Institute of Marine Sciences Research Division 2 - Marine Biogeochemistry Düsternbrooker Weg 20 D - 24105 Kiel Germany Tel.: +49 431 600-4282 Fax.: +49 431 600-4446 email: sk...@ifm-geomar.de __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. _ Baptiste Auguié School of Physics University of Exeter Stocker Road, Exeter, Devon, EX4 4QL, UK Phone: +44 1392 264187 http://newton.ex.ac.uk/research/emag __ -- *** Dipl. Biol. Sebastian Krug PhD - student IFM - GEOMAR Leibniz Institute of Marine Sciences Research Division 2 - Marine Biogeochemistry Düsternbrooker Weg 20 D - 24105 Kiel Germany Tel.: +49 431 600-4282 Fax.: +49 431 600-4446 email: sk...@ifm-geomar.de __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] R: plm and pgmm
dear giovanni---
thanks for answering on r-help to me as well as privately. I very much
appreciate your responding. I read the plm vignette. I don't have the book,
so I can't consult it. :-(. I am going to post this message now (rather
than just email it privately), because other amateurs may have similar
questions in the future, and find this message and your answers via google.
Real Statisticians---please don't waste your time.
so here is my amateur interpretation of GMM in general and Arellano-Bond
and Blundell-Bond specifically. I will do an example with T=4. The model is
x(i,t) = a*x(i,t-1) + u(i,t)
ie
x(i,2) = a*x(i,1) + u(i,2)
x(i,3) = a*x(i,2) + u(i,3)
x(i,4) = a*x(i,4) + u(i,4)
I view u(i,t) as a function of a: u(i,t)[a] = x(i,t)-a*x(i,t-1) . the
Arellano-Bond method then claims that u(i,3) should be uncorrelated with
x(i,1); u(i,4) should be uncorrelated with x(i,1) and also with x(i,2).
Blundell Bond adds the further condition that u(i,4) should be uncorrelated
with x(i,2)-x(i,1). so, I think of having four sums, each over all firms
i's. Let me call cross-sectional summing as sumi. the penalty function to
minimize is
sumi u(i,3)[a]*x(i,1) + sumi u(i,4)[a]*x(i,1) + sumi u(i,4)[a]*x(i,2) +
sumi u(i,4)*(x(i,2)-x(i,1))
I am missing the correct H weights on the terms in this sum, which is some
GMM magic that I do not understand (though I can copy it from their
article). for this post, the exact moment weights are not conceptually
important. now, for this sum to be well-defined, I should not need very
many observations at all. even with, say, N=7 firms, there should be no
problem in finding an a that minimizes the sum. (To me, it seems that the
more moment conditions I have, the merrier.) I was a little more encouraged
to make such daring statements, because stata seemed able of running this
and producing output.
On the other hand, the exact NF number at which pgmm() dies does suggest
that you are right.
function( NF=7, NT=4 ) {
d= data.frame( firm= rep(1:NF, each=NT), year= rep( 1:NT, NF),
x=rnorm(NF*NT) )
lagformula= dynformula( x ~ 1, list(1) )
v=pgmm( lagformula, data=d, gmm.inst=~x, model=onestep, effect=NULL,
lag.gmm=c(1,99), transformation=ld )
}
with NF=8, it works; and with NF=7, it dies. With NF=7, I have 28 data
points in levels and 21 data points in differences, which are used to
estimate only one auto-coefficient via 4 moment conditions. (Is this
correct?)
my best guess now is that even though one can get the GMM estimates with 7
firms, one cannot use the two-step method to learn how to best weight the
different moment conditions. the only method that may work is the one-step
matrix. of course, all of this is about conceptual tryouts, not about real
data. these methods work only well when NF is very large.
now, for the plm package: the non-descriptive error messages are also what
creates confusion when amateurs like myself want to create simple examples
[not real data] to understand how to provide proper inputs. if one needs a
minimum number of N, then may I strongly suggest that you trap this with a
descriptive error message at the outset? similarly, I would add an error
message if the formula provided to pgmm is not a dynformula, but a plain
formula. just die with please use dynformula instead.
there is also a small bug in the documentation. the vignette says that NULL
is a possible input to effect, while the standard docs mention
only individual or twoways.
I also emailed Yves that it would be great if you could provide a wrapper
for your more general function that does the simple estimation that 99% of
all end users would ever want. this would have the following inputs:
[a] method = arellano-bond or blundell-bond
[b] fixed effects or not
[c] a set of totally exogenous variables
[d] the number of lags of the dependent variable, defaults to 1
the version omits the GMM instrument vs. non-GMM instrument lingo, (though
after reading the vignette I have more of an inkling that all I need is to
not tell the function about exogenous variables and leave them in the
model), and knows that the dependent variable is dynamic by assumption, so
no more gmm.inst specification is required. yes, it is great to have the
implementation built on more heavy artillery that the statisticians can use
for more flexible estimations; but for end-users, having this simplified
function would really be terrific. (it would presumably default to using
the two-step method, which has more intelligent standard errors.) with such
a function wrapper, using these dynamic panel methods would become really
easy. just a suggestion...
May I end with stating that writing such a general plm seems like a
Herculean tasks, and that I want to express my thanks on behalf of many R
users that will benefit from it.
regards,
/ivo
[[alternative HTML version deleted]]
[R] Re sults sometimes in seconds with difftime unit=mins
Hello, I'm trying to calculate an integration and x-axis is a time (format : %Y-%m-%d %H:%M:%S). I use diff(date, units=mins) in a loop for but sometimes the results stay in seconds (95% is ok). Examples for 2 sets of data are given below (first result stays in seconds whereas the second in minutes as expected). Have you already seen this behaviour ? Any idea to solve this problem ? Thanks in advance. Have a good week-end, Ptit Bleu. strptime(datajour$Date, format=%Y-%m-%d %H:%M:%S) [1] 2009-03-26 11:21:31 2009-03-26 11:22:17 2009-03-26 11:27:18 2009-03-26 11:36:59 2009-03-26 11:41:59 2009-03-26 11:46:59 [7] 2009-03-26 11:51:59 2009-03-26 11:57:00 2009-03-26 12:02:00 2009-03-26 12:07:00 2009-03-26 12:12:00 2009-03-26 12:17:00 [13] 2009-03-26 12:22:00 2009-03-26 12:27:01 2009-03-26 12:32:01 2009-03-26 12:37:01 2009-03-26 12:42:01 2009-03-26 12:47:01 [19] 2009-03-26 12:52:01 2009-03-26 12:57:01 2009-03-26 13:02:02 2009-03-26 13:07:02 2009-03-26 13:12:03 2009-03-26 13:17:03 [25] 2009-03-26 13:22:03 2009-03-26 13:27:03 2009-03-26 13:32:03 2009-03-26 13:37:03 2009-03-26 13:42:03 2009-03-26 13:47:03 [31] 2009-03-26 13:52:03 2009-03-26 13:57:04 2009-03-26 14:01:02 2009-03-26 14:06:05 2009-03-26 14:11:05 2009-03-26 14:16:06 [37] 2009-03-26 14:21:06 2009-03-26 14:26:08 2009-03-26 14:31:09 2009-03-26 14:36:10 2009-03-26 14:41:10 2009-03-26 14:46:15 [43] 2009-03-26 14:51:15 2009-03-26 14:56:15 2009-03-26 15:01:15 2009-03-26 15:06:17 2009-03-26 15:11:17 2009-03-26 15:16:19 [49] 2009-03-26 15:21:19 2009-03-26 15:26:19 2009-03-26 15:31:22 2009-03-26 15:36:23 2009-03-26 15:41:24 2009-03-26 15:46:24 [55] 2009-03-26 15:51:25 2009-03-26 15:56:25 2009-03-26 16:01:25 2009-03-26 16:06:26 2009-03-26 16:11:26 2009-03-26 16:16:26 [61] 2009-03-26 16:21:27 2009-03-26 16:26:27 2009-03-26 16:31:28 2009-03-26 16:36:28 2009-03-26 16:41:29 2009-03-26 16:46:30 [67] 2009-03-26 16:51:31 2009-03-26 16:56:31 2009-03-26 17:01:32 2009-03-26 17:06:32 2009-03-26 17:11:33 2009-03-26 17:16:33 [73] 2009-03-26 17:21:33 2009-03-26 17:26:35 2009-03-26 17:31:36 2009-03-26 17:36:36 2009-03-26 17:41:36 2009-03-26 17:46:36 [79] 2009-03-26 17:51:39 2009-03-26 17:56:40 2009-03-26 18:01:40 2009-03-26 18:06:40 2009-03-26 18:11:40 2009-03-26 18:16:40 [85] 2009-03-26 18:21:41 2009-03-26 18:26:41 2009-03-26 18:31:41 2009-03-26 18:36:41 2009-03-26 18:41:41 2009-03-26 18:46:41 [91] 2009-03-26 18:51:42 2009-03-26 18:56:42 2009-03-26 19:06:42 as.numeric(diff(strptime(datajour$Date, format=%Y-%m-%d %H:%M:%S), units=mins)) [1] 46 301 581 300 300 300 301 300 300 300 300 300 301 300 300 300 300 300 300 301 300 301 300 300 300 300 300 300 300 300 301 238 303 300 301 300 302 301 [39] 301 300 305 300 300 300 302 300 302 300 300 303 301 301 300 301 300 300 301 300 300 301 300 301 300 301 301 301 300 301 300 301 300 300 302 301 300 300 [77] 300 303 301 300 300 300 300 301 300 300 300 300 300 301 300 600 strptime(datajour$Date, format=%Y-%m-%d %H:%M:%S) [1] 2009-03-26 11:22:24 2009-03-26 11:27:25 2009-03-26 11:37:04 2009-03-26 11:42:04 2009-03-26 11:47:04 2009-03-26 11:52:04 [7] 2009-03-26 11:57:04 2009-03-26 12:02:05 2009-03-26 12:07:06 2009-03-26 12:12:06 2009-03-26 12:17:06 2009-03-26 12:22:06 [13] 2009-03-26 12:27:07 2009-03-26 12:32:07 2009-03-26 12:37:07 2009-03-26 12:42:07 2009-03-26 12:47:07 2009-03-26 12:52:08 [19] 2009-03-26 12:57:08 2009-03-26 13:02:08 2009-03-26 13:07:09 2009-03-26 13:12:09 2009-03-26 13:17:09 2009-03-26 13:22:09 [25] 2009-03-26 13:27:09 2009-03-26 13:32:09 2009-03-26 13:37:09 2009-03-26 13:42:09 2009-03-26 13:47:09 2009-03-26 13:52:09 [31] 2009-03-26 13:57:10 2009-03-26 14:01:08 2009-03-26 14:06:11 2009-03-26 14:11:11 2009-03-26 14:16:12 2009-03-26 14:21:12 [37] 2009-03-26 14:26:15 2009-03-26 14:31:18 2009-03-26 14:36:18 2009-03-26 14:41:19 2009-03-26 14:46:22 2009-03-26 14:51:22 [43] 2009-03-26 14:56:23 2009-03-26 15:01:24 2009-03-26 15:06:24 2009-03-26 15:11:24 2009-03-26 15:16:24 2009-03-26 15:21:24 [49] 2009-03-26 15:26:24 2009-03-26 15:31:28 2009-03-26 15:36:29 2009-03-26 15:41:29 2009-03-26 15:46:30 2009-03-26 15:51:33 [55] 2009-03-26 15:56:33 2009-03-26 16:01:33 2009-03-26 16:06:33 2009-03-26 16:11:34 2009-03-26 16:16:34 2009-03-26 16:21:35 [61] 2009-03-26 16:26:35 2009-03-26 16:31:35 2009-03-26 16:36:35 2009-03-26 16:41:35 2009-03-26 16:46:36 2009-03-26 16:51:37 [67] 2009-03-26 16:56:37 2009-03-26 17:01:37 2009-03-26 17:06:38 2009-03-26 17:11:38 2009-03-26 17:16:38 2009-03-26 17:21:38 [73] 2009-03-26 17:26:40 2009-03-26 17:31:43 2009-03-26 17:36:43 2009-03-26 17:41:43 2009-03-26 17:46:44 2009-03-26 17:51:45 [79] 2009-03-26 17:56:46 2009-03-26 18:01:46 2009-03-26 18:06:46 2009-03-26 18:11:47 2009-03-26 18:16:47 2009-03-26 18:21:48 [85] 2009-03-26 18:26:48 2009-03-26 18:31:48 2009-03-26 18:36:48 2009-03-26 18:41:48 2009-03-26 18:46:48 2009-03-26 18:51:48 [91] 2009-03-26
Re: [R] color vectors other than gray()
Hi r-help-boun...@r-project.org napsal dne 27.03.2009 15:36:23: I'm certainly missing something. In fact the ramp I need must be scaled according to a vector of values (in this case species abundance in each grid cell), as in the example vector below: length(quad_N_sp$x) # where x is the abundance value [1] 433 quad_N_sp$x [1] 0 0 0 0 0 0 5 0 0 0 0 0 0 0 0 0 0 0 2 0 0 0 0 3 0 0 0 0 0 0 0 0 0 0 0 3 2 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 [101] 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 [201] 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 [301] 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 2 0 0 0 0 0 0 0 1 0 2 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 [401] 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 I need to discriminate shading level accordingly to the abundance value (level). If I understand correctly pal-grey(0:max(quad_N_sp$x)/max(quad_N_sp$x)) shall give you vector of equally spaced grey values pal[quad_N_sp$x+1] shall give you shadings for each quad_N_sp$x value Regards Petr I don't know how to proceed. Paulo E. Cardoso -Mensagem original- De: baptiste auguie [mailto:ba...@exeter.ac.uk] Enviada: sexta-feira, 27 de Março de 2009 13:30 Para: Paulo E. Cardoso Cc: r-h...@stat.math.ethz.ch; r-help@r-project.org Assunto: Re: [R] color vectors other than gray() ?colorRamp Hope this helps, baptiste On 27 Mar 2009, at 13:16, Paulo E. Cardoso wrote: I'm trying to create a graph where different cells of a grid (a shapefile) will be painted with a color share scale, where the most easy way is to use gray(). Can I somehow get a vector (gradient) of colors, a vector of colors with other methods but gray()? I'm doing this until now quad_N_sp - merge(sp_dist[sp_dist $sp==splist[i],],grelha_ID,by.x=quad,by.y=quadricula ,all.y=T,) quad_N_sp$x[is.na(quad_N_sp$x)] - 0 quad_N_sp - quad_N_sp[order(quad_N_sp$id),] paleta - gray(1-(quad_N_sp$x)/max(quad_N_sp$x)) #! Tons de cinzento win.graph(4,5) plot(grelha,ol=grey80, #! Gráfico com grelha de amostragem e gradiente de abundância fg=paleta, cex.lab=0.7, cex.axis=0.7, cex.main=0.7, xlab=Coord X, ylab=Coord Y, main=paste(Espécie: ,splist[i]), xlim=c(21,24) ) col_lab - c(max(quad_N_sp$x),min(quad_N_sp$x)) #! Vector com os limites min e max do N de indivíduos observados color .legend (248000,12,25,128000,col_lab,sort(unique(paleta)),gradie nt=y,cex=0.6)#! Legenda text(245300,130500,Nº Indivíduos,cex=0.6) plot(blocos,ol=grey40,fg=NA,add=T) I'd like to replace the grey shade by other colors. Thanks in advance Paulo E. Cardoso [[alternative HTML version deleted]] ATT1.txt _ Baptiste Auguié School of Physics University of Exeter Stocker Road, Exeter, Devon, EX4 4QL, UK Phone: +44 1392 264187 http://newton.ex.ac.uk/research/emag __ No virus found in this incoming message. Checked by AVG - www.avg.com 03/27/09 07:13:00 __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] Ploting a matrix
the result of read.table is a data.frame, not a matrix as you first suggested. Can you copy the result of str(b) so we know what your data is made of? I'm guessing the most elegant solution will be to use the reshape package, followed by ggplot2 or lattice. baptiste On 27 Mar 2009, at 14:54, skrug wrote: Unfortunately, I could not solve the problem of plotting all columns of a matrix against the first column I used: b=read.table(d:\\programme\\R\\übungen\\Block 1b.txt, header=T) b is a table with the first column using Dates and the following columns with vectors. apply(b[,-1], 2, plot, x= b[,1]) Also all columns have the same length, [R] states that the length are different. Can you help me? baptiste auguie schrieb: Something like this perhaps, a - matrix(rnorm(5*49), ncol=49) pdf(width=15, height=15) par(mfrow= c(8,6)) apply(a[,-1], 2, plot, x= a[,1]) dev.off() HTH, baptiste On 27 Mar 2009, at 11:05, skrug wrote: Hi evrybody, in a matrix consisting of 49 columns, I would like to plot all columns against the first in 48 different graphs. Can you help me? Thank you in advance Sebastian -- *** Dipl. Biol. Sebastian Krug PhD - student IFM - GEOMAR Leibniz Institute of Marine Sciences Research Division 2 - Marine Biogeochemistry Düsternbrooker Weg 20 D - 24105 Kiel Germany Tel.: +49 431 600-4282 Fax.: +49 431 600-4446 email: sk...@ifm-geomar.de __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. _ Baptiste Auguié School of Physics University of Exeter Stocker Road, Exeter, Devon, EX4 4QL, UK Phone: +44 1392 264187 http://newton.ex.ac.uk/research/emag __ -- *** Dipl. Biol. Sebastian Krug PhD - student IFM - GEOMAR Leibniz Institute of Marine Sciences Research Division 2 - Marine Biogeochemistry Düsternbrooker Weg 20 D - 24105 Kiel Germany Tel.: +49 431 600-4282 Fax.: +49 431 600-4446 email: sk...@ifm-geomar.de _ Baptiste Auguié School of Physics University of Exeter Stocker Road, Exeter, Devon, EX4 4QL, UK Phone: +44 1392 264187 http://newton.ex.ac.uk/research/emag __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] functions flagged for debugging
Hi, Is there a way to find which functions are flagged for debugging in a given session? Thank you. -Christos [[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] Ploting a matrix
Sorry for the mistake. As you probably already guesed, I am just starting using R. I could not name the difference between a matrix and a data.frame. str(b) 'data.frame': 9 obs. of 7 variables: $ Datum: Factor w/ 9 levels 06.03.,07.03.,..: 1 2 3 4 5 6 7 8 9 $ X1 : int 408 335 2123 4685 7669 17060 31330 70730 109667 $ X2 : int 230 241 1509 2226 7839 13997 24797 53133 93061 $ X3 : int 25 16 38 61 114 299 140 172 196 $ X4 : int 248 588 2083 2071 4563 9798 17611 38554 82354 $ X5 : int 407 201 1339 3699 8375 19200 36563 83993 123167 $ X6 : int 248 730 3056 2327 4092 8905 15931 37895 84565 Thanks baptiste auguie schrieb: the result of read.table is a data.frame, not a matrix as you first suggested. Can you copy the result of str(b) so we know what your data is made of? I'm guessing the most elegant solution will be to use the reshape package, followed by ggplot2 or lattice. baptiste On 27 Mar 2009, at 14:54, skrug wrote: Unfortunately, I could not solve the problem of plotting all columns of a matrix against the first column I used: b=read.table(d:\\programme\\R\\übungen\\Block 1b.txt, header=T) b is a table with the first column using Dates and the following columns with vectors. apply(b[,-1], 2, plot, x= b[,1]) Also all columns have the same length, [R] states that the length are different. Can you help me? baptiste auguie schrieb: Something like this perhaps, a - matrix(rnorm(5*49), ncol=49) pdf(width=15, height=15) par(mfrow= c(8,6)) apply(a[,-1], 2, plot, x= a[,1]) dev.off() HTH, baptiste On 27 Mar 2009, at 11:05, skrug wrote: Hi evrybody, in a matrix consisting of 49 columns, I would like to plot all columns against the first in 48 different graphs. Can you help me? Thank you in advance Sebastian -- *** Dipl. Biol. Sebastian Krug PhD - student IFM - GEOMAR Leibniz Institute of Marine Sciences Research Division 2 - Marine Biogeochemistry Düsternbrooker Weg 20 D - 24105 Kiel Germany Tel.: +49 431 600-4282 Fax.: +49 431 600-4446 email: sk...@ifm-geomar.de __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. _ Baptiste Auguié School of Physics University of Exeter Stocker Road, Exeter, Devon, EX4 4QL, UK Phone: +44 1392 264187 http://newton.ex.ac.uk/research/emag __ -- *** Dipl. Biol. Sebastian Krug PhD - student IFM - GEOMAR Leibniz Institute of Marine Sciences Research Division 2 - Marine Biogeochemistry Düsternbrooker Weg 20 D - 24105 Kiel Germany Tel.: +49 431 600-4282 Fax.: +49 431 600-4446 email: sk...@ifm-geomar.de _ Baptiste Auguié School of Physics University of Exeter Stocker Road, Exeter, Devon, EX4 4QL, UK Phone: +44 1392 264187 http://newton.ex.ac.uk/research/emag __ -- *** Dipl. Biol. Sebastian Krug PhD - student IFM - GEOMAR Leibniz Institute of Marine Sciences Research Division 2 - Marine Biogeochemistry Düsternbrooker Weg 20 D - 24105 Kiel Germany Tel.: +49 431 600-4282 Fax.: +49 431 600-4446 email: sk...@ifm-geomar.de __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] Weighting data with normal distribution
Alice Lin alice.ly at gmail.com writes: I have a vector of binary data – a string of 0’s and 1’s. I want to weight these inputs with a normal kernel centered around entry x so it is transformed into a new vector of data that takes into account the values of the entries around it (weighting them more heavily if they are near). Example: - - - - - 0 1 0 0 1 0 0 1 1 1 1 If x = 3, it’s current value is 0 but it’s new value with the Gaussian weighting around would be something like .1*0+.5*1+1*0+0.5*0+.1*1= 0.6 I want to be able to play with adjusting the variance to different values as well. I’ve found wkde in the mixtools library and think it may be useful but I have not figured out how to use it yet. Any tips would be appreciated. Thanks! I don't know anything about wkde. But the filter function in stats package should do what you want. x - c(0, 1, 0, 0, 1, 0, 0, 1, 1, 1, 1) filter(x, c(.1, .5, 1, .5, .1)) Time Series: Start = 1 End = 11 Frequency = 1 [1] NA NA 0.6 0.6 1.0 0.6 0.7 1.6 2.1 NA NA In the signal package, there is also a variety of windows, including the gausswin function. However, the filter function in the signal package masks the filter function from the stats package stats::filter(x, gausswin(5, 2.68)) Mark Lyman Statistician, ATK Launch Systems __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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 when setting up Rcmd BATCH on new computer
Hello, I got a new computer, and am trying to reinstall R and have run into a bit of a problem when running the BATCH command. For reference, the OS is Windows Vista, 64 bit. I installed R 2.8.1 and have the 4-3 files from the following link extracted with the containing folder in my system PATH variable. However, when I try to run the following command from the dos prompt: Rcmd BATCH TestBatch.R testoutput.txt Note: TestBatch.R is simply a file containing the statement: print(hello world) I get the error: \Common was unexpected at this time. If anyone can provide any insight into this problem, I would really appreciate it as I thought I remembered all the steps from when I set this all up on my old computer... Thanks, Brigid __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] color vectors other than gray()
OK I got it working partially. The plot results in a ramp varying from white - red - black, doing this: plot(grelha,ol=grey80, #! Gráfico com grelha de amostragem e gradiente de abundância fg=color.scale(((1-(quad_N_sp$x)/max(quad_N_sp$x))),c(0,1,1),c(0,1),c(0,1)), cex.lab=0.7, cex.axis=0.7, cex.main=0.7, xlab=Coord X, ylab=Coord Y, main=paste(Espécie: ,splist[i]), xlim=c(21,24) ) Where the ramp results from the red, green, blue ranges used in: color.scale(((1-(quad_N_sp$x)/max(quad_N_sp$x))),c(0,1,1),c(0,1),c(0,1)) I don't know how to control the ramp. In this particular case it results reasonably well but I ave no idea how to control the RGB channels to produce a ramp-of-interest. Attached workspace contains a data sample to create the plot. Use the code below: library(maptools) library(plotrix) pallete - color.scale(vec.ab.01,c(0,1,1),c(0,1),c(0,1)) win.graph(4,5) plot(grelha,ol=grey80, #! Gráfico com grelha de amostragem e gradiente de abundância fg=pallete, cex.lab=0.7, cex.axis=0.7, cex.main=0.7, xlab=Coord X, ylab=Coord Y, main=paste(Espécie: ,Carduelis carduelis), xlim=c(21,24) ) col_lab - c(max(vec.ab),min(vec.ab)) #! Vector com os limites min e max do N de indivíduos observados color.legend(248000,12,25,128000,col_lab,sort(unique(pallete)),gradi ent=y,cex=0.7)#! Legenda text(245300,130500,Nº Indivíduos,cex=0.6) plot(blocos,ol=grey40,fg=NA,add=T) Paulo E. Cardoso -Mensagem original- De: baptiste auguie [mailto:ba...@exeter.ac.uk] Enviada: sexta-feira, 27 de Março de 2009 14:50 Para: Paulo E. Cardoso Cc: r-h...@stat.math.ethz.ch; r-help@r-project.org Assunto: Re: [R] color vectors other than gray() Can you provide a minimal example that we can run directly after copy and paste (using a standard data set or dummy data)? It's always helpful to try and nail down the core of your question (often you'll find the answer while formulating your question in minimal terms). baptiste On 27 Mar 2009, at 14:36, Paulo E. Cardoso wrote: I'm certainly missing something. In fact the ramp I need must be scaled according to a vector of values (in this case species abundance in each grid cell), as in the example vector below: length(quad_N_sp$x) # where x is the abundance value [1] 433 quad_N_sp$x [1] 0 0 0 0 0 0 5 0 0 0 0 0 0 0 0 0 0 0 2 0 0 0 0 3 0 0 0 0 0 0 0 0 0 0 0 3 2 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 [101] 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 [201] 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 [301] 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 2 0 0 0 0 0 0 0 1 0 2 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 [401] 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 I need to discriminate shading level accordingly to the abundance value (level). I don't know how to proceed. Paulo E. Cardoso -Mensagem original- De: baptiste auguie [mailto:ba...@exeter.ac.uk] Enviada: sexta-feira, 27 de Março de 2009 13:30 Para: Paulo E. Cardoso Cc: r-h...@stat.math.ethz.ch; r-help@r-project.org Assunto: Re: [R] color vectors other than gray() ?colorRamp Hope this helps, baptiste On 27 Mar 2009, at 13:16, Paulo E. Cardoso wrote: I'm trying to create a graph where different cells of a grid (a shapefile) will be painted with a color share scale, where the most easy way is to use gray(). Can I somehow get a vector (gradient) of colors, a vector of colors with other methods but gray()? I'm doing this until now quad_N_sp - merge(sp_dist[sp_dist $sp==splist[i],],grelha_ID,by.x=quad,by.y=quadricula ,all.y=T,) quad_N_sp$x[is.na(quad_N_sp$x)] - 0 quad_N_sp - quad_N_sp[order(quad_N_sp$id),] paleta - gray(1-(quad_N_sp$x)/max(quad_N_sp$x)) #! Tons de cinzento win.graph(4,5) plot(grelha,ol=grey80, #! Gráfico com grelha de amostragem e gradiente de abundância fg=paleta, cex.lab=0.7, cex.axis=0.7, cex.main=0.7, xlab=Coord X, ylab=Coord Y, main=paste(Espécie: ,splist[i]), xlim=c(21,24) ) col_lab - c(max(quad_N_sp$x),min(quad_N_sp$x)) #! Vector com os limites min e max do N de indivíduos observados color .legend (248000,12,25,128000,col_lab,sort(unique(paleta)),gradie nt=y,cex=0.6)#! Legenda text(245300,130500,Nº Indivíduos,cex=0.6) plot(blocos,ol=grey40,fg=NA,add=T) I'd like to replace the grey shade by other colors. Thanks in advance
Re: [R] Ploting a matrix
Here's my suggestion using the ggplot2 package (but you may prefer to stick with base functions), date = factor(letters[1:9]) d - data.frame(x1=seq(1, 9), x2=seq(2, 10), date=date) head(d) # dummy data that resembles yours str(d) library(reshape) md - melt(d, id=date) # creates a data.frame in the long format head(md) library(ggplot2) qplot(date, value, data=md, geom=point) + facet_wrap(~variable) # the layout is done automatically for you # see Hadley's book for customisations # http://had.co.nz/ggplot2/facet_wrap.html HTH, baptiste On 27 Mar 2009, at 15:19, skrug wrote: Sorry for the mistake. As you probably already guesed, I am just starting using R. I could not name the difference between a matrix and a data.frame. str(b) 'data.frame': 9 obs. of 7 variables: $ Datum: Factor w/ 9 levels 06.03.,07.03.,..: 1 2 3 4 5 6 7 8 9 $ X1 : int 408 335 2123 4685 7669 17060 31330 70730 109667 $ X2 : int 230 241 1509 2226 7839 13997 24797 53133 93061 $ X3 : int 25 16 38 61 114 299 140 172 196 $ X4 : int 248 588 2083 2071 4563 9798 17611 38554 82354 $ X5 : int 407 201 1339 3699 8375 19200 36563 83993 123167 $ X6 : int 248 730 3056 2327 4092 8905 15931 37895 84565 Thanks baptiste auguie schrieb: the result of read.table is a data.frame, not a matrix as you first suggested. Can you copy the result of str(b) so we know what your data is made of? I'm guessing the most elegant solution will be to use the reshape package, followed by ggplot2 or lattice. baptiste On 27 Mar 2009, at 14:54, skrug wrote: Unfortunately, I could not solve the problem of plotting all columns of a matrix against the first column I used: b=read.table(d:\\programme\\R\\übungen\\Block 1b.txt, header=T) b is a table with the first column using Dates and the following columns with vectors. apply(b[,-1], 2, plot, x= b[,1]) Also all columns have the same length, [R] states that the length are different. Can you help me? baptiste auguie schrieb: Something like this perhaps, a - matrix(rnorm(5*49), ncol=49) pdf(width=15, height=15) par(mfrow= c(8,6)) apply(a[,-1], 2, plot, x= a[,1]) dev.off() HTH, baptiste On 27 Mar 2009, at 11:05, skrug wrote: Hi evrybody, in a matrix consisting of 49 columns, I would like to plot all columns against the first in 48 different graphs. Can you help me? Thank you in advance Sebastian -- *** Dipl. Biol. Sebastian Krug PhD - student IFM - GEOMAR Leibniz Institute of Marine Sciences Research Division 2 - Marine Biogeochemistry Düsternbrooker Weg 20 D - 24105 Kiel Germany Tel.: +49 431 600-4282 Fax.: +49 431 600-4446 email: sk...@ifm-geomar.de __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. _ Baptiste Auguié School of Physics University of Exeter Stocker Road, Exeter, Devon, EX4 4QL, UK Phone: +44 1392 264187 http://newton.ex.ac.uk/research/emag __ -- *** Dipl. Biol. Sebastian Krug PhD - student IFM - GEOMAR Leibniz Institute of Marine Sciences Research Division 2 - Marine Biogeochemistry Düsternbrooker Weg 20 D - 24105 Kiel Germany Tel.: +49 431 600-4282 Fax.: +49 431 600-4446 email: sk...@ifm-geomar.de _ Baptiste Auguié School of Physics University of Exeter Stocker Road, Exeter, Devon, EX4 4QL, UK Phone: +44 1392 264187 http://newton.ex.ac.uk/research/emag __ -- *** Dipl. Biol. Sebastian Krug PhD - student IFM - GEOMAR Leibniz Institute of Marine Sciences Research Division 2 - Marine Biogeochemistry Düsternbrooker Weg 20 D - 24105 Kiel Germany Tel.: +49 431 600-4282 Fax.: +49 431 600-4446 email: sk...@ifm-geomar.de _ Baptiste Auguié School of Physics University of Exeter Stocker Road, Exeter, Devon, EX4 4QL, UK Phone: +44 1392 264187 http://newton.ex.ac.uk/research/emag __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Passing parameters to R script in Rgui
How do I pass parameters to R script in Rgui ? Currently, I am using source(foo.R). Thanks __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Reading in files with variable parts to names
Thanks, that's great - just what I was looking for. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] color vectors other than gray()
Petr, I'd like to be able to change the ramp to other than grey shades. Please see my previous message with some data. Paulo E. Cardoso -Mensagem original- De: Petr PIKAL [mailto:petr.pi...@precheza.cz] Enviada: sexta-feira, 27 de Março de 2009 15:12 Para: Paulo E. Cardoso Cc: r-h...@stat.math.ethz.ch Assunto: Re: [R] color vectors other than gray() Hi r-help-boun...@r-project.org napsal dne 27.03.2009 15:36:23: I'm certainly missing something. In fact the ramp I need must be scaled according to a vector of values (in this case species abundance in each grid cell), as in the example vector below: length(quad_N_sp$x) # where x is the abundance value [1] 433 quad_N_sp$x [1] 0 0 0 0 0 0 5 0 0 0 0 0 0 0 0 0 0 0 2 0 0 0 0 3 0 0 0 0 0 0 0 0 0 0 0 3 2 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 [101] 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 [201] 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 [301] 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 2 0 0 0 0 0 0 0 1 0 2 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 [401] 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 I need to discriminate shading level accordingly to the abundance value (level). If I understand correctly pal-grey(0:max(quad_N_sp$x)/max(quad_N_sp$x)) shall give you vector of equally spaced grey values pal[quad_N_sp$x+1] shall give you shadings for each quad_N_sp$x value Regards Petr I don't know how to proceed. Paulo E. Cardoso -Mensagem original- De: baptiste auguie [mailto:ba...@exeter.ac.uk] Enviada: sexta-feira, 27 de Março de 2009 13:30 Para: Paulo E. Cardoso Cc: r-h...@stat.math.ethz.ch; r-help@r-project.org Assunto: Re: [R] color vectors other than gray() ?colorRamp Hope this helps, baptiste On 27 Mar 2009, at 13:16, Paulo E. Cardoso wrote: I'm trying to create a graph where different cells of a grid (a shapefile) will be painted with a color share scale, where the most easy way is to use gray(). Can I somehow get a vector (gradient) of colors, a vector of colors with other methods but gray()? I'm doing this until now quad_N_sp - merge(sp_dist[sp_dist $sp==splist[i],],grelha_ID,by.x=quad,by.y=quadricula ,all.y=T,) quad_N_sp$x[is.na(quad_N_sp$x)] - 0 quad_N_sp - quad_N_sp[order(quad_N_sp$id),] paleta - gray(1-(quad_N_sp$x)/max(quad_N_sp$x)) #! Tons de cinzento win.graph(4,5) plot(grelha,ol=grey80, #! Gráfico com grelha de amostragem e gradiente de abundância fg=paleta, cex.lab=0.7, cex.axis=0.7, cex.main=0.7, xlab=Coord X, ylab=Coord Y, main=paste(Espécie: ,splist[i]), xlim=c(21,24) ) col_lab - c(max(quad_N_sp$x),min(quad_N_sp$x)) #! Vector com os limites min e max do N de indivíduos observados color .legend (248000,12,25,128000,col_lab,sort(unique(paleta)),gradie nt=y,cex=0.6)#! Legenda text(245300,130500,Nº Indivíduos,cex=0.6) plot(blocos,ol=grey40,fg=NA,add=T) I'd like to replace the grey shade by other colors. Thanks in advance Paulo E. Cardoso [[alternative HTML version deleted]] ATT1.txt _ Baptiste Auguié School of Physics University of Exeter Stocker Road, Exeter, Devon, EX4 4QL, UK Phone: +44 1392 264187 http://newton.ex.ac.uk/research/emag __ No virus found in this incoming message. Checked by AVG - www.avg.com 03/27/09 07:13:00 __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. No virus found in this incoming message. Checked by AVG - www.avg.com 03/27/09 07:13:00 __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] interactive image graphic
Here is some code that may get you started (if I am understanding your question
correctly):
library(TeachingDemos)
myfunc - function(zmin=90, zmax=195, ncol=100, pal='heat') {
cols - switch(pal,
heat=heat.colors(ncol),
terrain=terrain.colors(ncol),
topo=topo.colors(ncol),
cm=cm.colors(ncol)
)
image(volcano, col=cols, zlim=c(zmin,zmax))
}
mylist - list(
zmin=list('slider',from=0, to=90, init=90, resolution=5),
zmax=list('slider',from=195, to=250, init=195, resolution=5),
ncol=list('spinbox', init=100, from=2, to=150, increment=5),
pal=list('radiobuttons', values=c('heat','terrain','topo','cm'),
init='heat'))
tkexamp(myfunc,mylist)
hope this helps,
--
Gregory (Greg) L. Snow Ph.D.
Statistical Data Center
Intermountain Healthcare
greg.s...@imail.org
801.408.8111
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-
project.org] On Behalf Of Abelian
Sent: Thursday, March 26, 2009 9:43 PM
To: r-help@r-project.org
Subject: [R] interactive image graphic
Dear All
I want to plot a kind of figures, which can interactive with user.
For example, i have a matirx which can be showed by image function.
i.e. we can compare the value depend on different colors.
However, the change of colors depend on the range of value.
Nowaday, i want to set a bar, which can be moved by user such that the
user can obtain the appropriate range.
Does anyone suggest me which function can be applied to solve this
problem?
Thanks
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-
guide.html
and provide commented, minimal, self-contained, reproducible code.
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/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] Passing parameters to R script in Rgui
You are misusing source. Write a function to do what you want. An Introduction to R documents how. Have you read it? -- Bert Bert Gunter Genentech Nonclinical Biostatistics 650-467-7374 -Original Message- From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On Behalf Of Daren Tan Sent: Friday, March 27, 2009 8:45 AM To: r-help@r-project.org Subject: [R] Passing parameters to R script in Rgui How do I pass parameters to R script in Rgui ? Currently, I am using source(foo.R). Thanks __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] about the Choleski factorization
Hi there, Given a positive definite symmetric matrix, I can use chol(x) to obtain U where U is upper triangular and x=U'U. For example, x=matrix(c(5,1,2,1,3,1,2,1,4),3,3) U=chol(x) U # [,1] [,2] [,3] #[1,] 2.236068 0.4472136 0.8944272 #[2,] 0.00 1.6733201 0.3585686 #[3,] 0.00 0.000 1.7525492 t(U)%*%U # this is exactly x Does anyone know how to obtain L such that L is lower triangular and x=L'L? Thank you. Alex __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] print table (data.frame) to pdf
How can I print a data.frame to a PDF with pdf()...dev.off() Paulo E. Cardoso [[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] Re sults sometimes in seconds with difftime unit=mins
I think the problem is is that 'diff' does not have a 'units' parameter; 'difftime' does. Here is a way of doing it: x [1] 2009-03-27 13:00:00 EDT 2009-03-27 13:00:35 EDT 2009-03-27 13:01:10 EDT 2009-03-27 13:01:45 EDT 2009-03-27 13:02:20 EDT 2009-03-27 13:02:55 EDT 2009-03-27 13:03:30 EDT [8] 2009-03-27 13:04:05 EDT 2009-03-27 13:04:40 EDT 2009-03-27 13:05:15 EDT difftime(tail(x, -1), head(x, -1), units='mins') Time differences in mins [1] 0.583 0.583 0.583 0.583 0.583 0.583 0.583 0.583 0.583 attr(,tzone) [1] diff(x, units='mins') # 'units' ignored Time differences in secs [1] 35 35 35 35 35 35 35 35 35 attr(,tzone) [1] On Fri, Mar 27, 2009 at 11:04 AM, Ptit_Bleu ptit_b...@yahoo.fr wrote: Hello, I'm trying to calculate an integration and x-axis is a time (format : %Y-%m-%d %H:%M:%S). I use diff(date, units=mins) in a loop for but sometimes the results stay in seconds (95% is ok). Examples for 2 sets of data are given below (first result stays in seconds whereas the second in minutes as expected). Have you already seen this behaviour ? Any idea to solve this problem ? Thanks in advance. Have a good week-end, Ptit Bleu. strptime(datajour$Date, format=%Y-%m-%d %H:%M:%S) [1] 2009-03-26 11:21:31 2009-03-26 11:22:17 2009-03-26 11:27:18 2009-03-26 11:36:59 2009-03-26 11:41:59 2009-03-26 11:46:59 [7] 2009-03-26 11:51:59 2009-03-26 11:57:00 2009-03-26 12:02:00 2009-03-26 12:07:00 2009-03-26 12:12:00 2009-03-26 12:17:00 [13] 2009-03-26 12:22:00 2009-03-26 12:27:01 2009-03-26 12:32:01 2009-03-26 12:37:01 2009-03-26 12:42:01 2009-03-26 12:47:01 [19] 2009-03-26 12:52:01 2009-03-26 12:57:01 2009-03-26 13:02:02 2009-03-26 13:07:02 2009-03-26 13:12:03 2009-03-26 13:17:03 [25] 2009-03-26 13:22:03 2009-03-26 13:27:03 2009-03-26 13:32:03 2009-03-26 13:37:03 2009-03-26 13:42:03 2009-03-26 13:47:03 [31] 2009-03-26 13:52:03 2009-03-26 13:57:04 2009-03-26 14:01:02 2009-03-26 14:06:05 2009-03-26 14:11:05 2009-03-26 14:16:06 [37] 2009-03-26 14:21:06 2009-03-26 14:26:08 2009-03-26 14:31:09 2009-03-26 14:36:10 2009-03-26 14:41:10 2009-03-26 14:46:15 [43] 2009-03-26 14:51:15 2009-03-26 14:56:15 2009-03-26 15:01:15 2009-03-26 15:06:17 2009-03-26 15:11:17 2009-03-26 15:16:19 [49] 2009-03-26 15:21:19 2009-03-26 15:26:19 2009-03-26 15:31:22 2009-03-26 15:36:23 2009-03-26 15:41:24 2009-03-26 15:46:24 [55] 2009-03-26 15:51:25 2009-03-26 15:56:25 2009-03-26 16:01:25 2009-03-26 16:06:26 2009-03-26 16:11:26 2009-03-26 16:16:26 [61] 2009-03-26 16:21:27 2009-03-26 16:26:27 2009-03-26 16:31:28 2009-03-26 16:36:28 2009-03-26 16:41:29 2009-03-26 16:46:30 [67] 2009-03-26 16:51:31 2009-03-26 16:56:31 2009-03-26 17:01:32 2009-03-26 17:06:32 2009-03-26 17:11:33 2009-03-26 17:16:33 [73] 2009-03-26 17:21:33 2009-03-26 17:26:35 2009-03-26 17:31:36 2009-03-26 17:36:36 2009-03-26 17:41:36 2009-03-26 17:46:36 [79] 2009-03-26 17:51:39 2009-03-26 17:56:40 2009-03-26 18:01:40 2009-03-26 18:06:40 2009-03-26 18:11:40 2009-03-26 18:16:40 [85] 2009-03-26 18:21:41 2009-03-26 18:26:41 2009-03-26 18:31:41 2009-03-26 18:36:41 2009-03-26 18:41:41 2009-03-26 18:46:41 [91] 2009-03-26 18:51:42 2009-03-26 18:56:42 2009-03-26 19:06:42 as.numeric(diff(strptime(datajour$Date, format=%Y-%m-%d %H:%M:%S), units=mins)) [1] 46 301 581 300 300 300 301 300 300 300 300 300 301 300 300 300 300 300 300 301 300 301 300 300 300 300 300 300 300 300 301 238 303 300 301 300 302 301 [39] 301 300 305 300 300 300 302 300 302 300 300 303 301 301 300 301 300 300 301 300 300 301 300 301 300 301 301 301 300 301 300 301 300 300 302 301 300 300 [77] 300 303 301 300 300 300 300 301 300 300 300 300 300 301 300 600 strptime(datajour$Date, format=%Y-%m-%d %H:%M:%S) [1] 2009-03-26 11:22:24 2009-03-26 11:27:25 2009-03-26 11:37:04 2009-03-26 11:42:04 2009-03-26 11:47:04 2009-03-26 11:52:04 [7] 2009-03-26 11:57:04 2009-03-26 12:02:05 2009-03-26 12:07:06 2009-03-26 12:12:06 2009-03-26 12:17:06 2009-03-26 12:22:06 [13] 2009-03-26 12:27:07 2009-03-26 12:32:07 2009-03-26 12:37:07 2009-03-26 12:42:07 2009-03-26 12:47:07 2009-03-26 12:52:08 [19] 2009-03-26 12:57:08 2009-03-26 13:02:08 2009-03-26 13:07:09 2009-03-26 13:12:09 2009-03-26 13:17:09 2009-03-26 13:22:09 [25] 2009-03-26 13:27:09 2009-03-26 13:32:09 2009-03-26 13:37:09 2009-03-26 13:42:09 2009-03-26 13:47:09 2009-03-26 13:52:09 [31] 2009-03-26 13:57:10 2009-03-26 14:01:08 2009-03-26 14:06:11 2009-03-26 14:11:11 2009-03-26 14:16:12 2009-03-26 14:21:12 [37] 2009-03-26 14:26:15 2009-03-26 14:31:18 2009-03-26 14:36:18 2009-03-26 14:41:19 2009-03-26 14:46:22 2009-03-26 14:51:22 [43] 2009-03-26 14:56:23 2009-03-26 15:01:24 2009-03-26 15:06:24 2009-03-26 15:11:24 2009-03-26 15:16:24 2009-03-26 15:21:24 [49] 2009-03-26 15:26:24 2009-03-26 15:31:28 2009-03-26 15:36:29 2009-03-26
Re: [R] Ploting a matrix
Dear Sebastian,
Consider matplot() for this. Here is an example (taken from Baptiste
Auguie's post):
date - factor(letters[1:9])
d - data.frame(x1=seq(1, 9), x2=seq(2, 10), date=date)
matplot(d[,-3],pch=16,xaxt='n',las=1,ylab='Some label here',xlab='Date')
axis(1,d[,3],d[,3])
legend('topleft',c('x1','x2'),pch=16,col=1:2)
See ?matplot, ?axis and ?legend for more information.
HTH,
Jorge
On Fri, Mar 27, 2009 at 7:05 AM, skrug sk...@ifm-geomar.de wrote:
Hi evrybody,
in a matrix consisting of 49 columns, I would like to plot all columns
against the first in 48 different graphs.
Can you help me?
Thank you in advance
Sebastian
--
***
Dipl. Biol. Sebastian Krug
PhD - student
IFM - GEOMAR
Leibniz Institute of Marine Sciences
Research Division 2 - Marine Biogeochemistry
Düsternbrooker Weg 20
D - 24105 Kiel
Germany
Tel.: +49 431 600-4282
Fax.: +49 431 600-4446
email: sk...@ifm-geomar.de
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/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] R 2.8.1 and 2.9 alpha crash when running survest of Design package
We will try to quickly get out a new version of Design that checks for the version of survival that is installed and uses a different .C call accordingly. This will involve ignoring (for now) the new weights option Terry has implemented. Frank Terry Therneau wrote: A couple additions to Thomas's message. The 'survest' function in design directly called C routines in the survival package. The argument list to the routines changed due to the addition of weights; calling a C routine with the wrong arguments is one of the more reliable ways to crash a program. The simplest (short term) solution is to use survfit for your curves rather than survest. Frank Harrell has been aware of the issue for several weeks and is working hard on solving it. The simple fix is a few minutes, but he's thinking about how to avoid any future problems. The C routines in survival change arguments VERY rarely, but direcly calling the routines of another package is considered dangerous in general. Most breakage was less severe. For instance there were a couple of errors in the PBC data set. I fixed these, and also replaced all the 999 codes with NA to make it easier to use. Some other packages use this data. (My name is on most of the PBC papers and I have the master PBC data with all labs, patient id, etc, but I was not the source of the first data set). We'll be keeping an eye on the R list as the package rolls out; sending a message directly to Thomas and/or I would also be appreciated for issues like this. Terry Therneau __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Frank E Harrell Jr Professor and Chair School of Medicine Department of Biostatistics Vanderbilt University __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] print table (data.frame) to pdf
See either textplot in the gplots package or addtable2plot in the plotrix package, or for even more flexibility learn Sweave or its variants. -- Gregory (Greg) L. Snow Ph.D. Statistical Data Center Intermountain Healthcare greg.s...@imail.org 801.408.8111 -Original Message- From: r-help-boun...@r-project.org [mailto:r-help-boun...@r- project.org] On Behalf Of Paulo E. Cardoso Sent: Friday, March 27, 2009 9:53 AM To: r-help@r-project.org; r-h...@stat.math.ethz.ch Subject: [R] print table (data.frame) to pdf How can I print a data.frame to a PDF with pdf()...dev.off() Paulo E. Cardoso [[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] Sweave-output causes error-message in pdflatex
On 3/27/2009 10:12 AM, Gerrit Voigt wrote: Dear list, Latex/Sweave has trouble processing Sveave-output coming from the summary-command of a linear Model. summary(lmRub) The output line causing the trouble looks in R like this Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1 In my Sweaved Tex-file that line looks like this Signif. codes: 0 *** 0.001 ** 0.01 * 0.05 . 0.1 1 (actually in the editor the quotation signs are replaced by bars, but they got lost through copy paste. I don't know if that says anything about my problem.) In the error message produced through pdflatex, the quotation signs reappear. Latex error-message: ! Package inputenc Error: Keyboard character used is undefined (inputenc) in inputencoding `Latin1'. See the inputenc package documentation for explanation. Type H return for immediate help. ... l.465 ...*’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1 You need to provide a definition with \DeclareInputText or \DeclareInputMath before using this key. I hope anybody knows how I can prevent that error message. Thanks in advance. You are running the code on a platform where the character used for one of the quote characters is unrecognized by LaTeX. The simplest solution is to tell R not to use those characters, via executing options(useFancyQuotes = FALSE) early in your document. See ?sQuote for more details. 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] functions flagged for debugging
On 3/27/2009 11:18 AM, Christos Hatzis wrote: Hi, Is there a way to find which functions are flagged for debugging in a given session? The isdebugged() function (which is new in 2.9.0) will tell you this. 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.
[R] creating a matrix as input to lowess
I have a very large file with many rows and columns. I want to create a plot with lowess. If I try the following it works fine: data(PrecipGL) plot(PrecipGL) lines(lowess(time(PrecipGL),PrecipGL),lwd=3, col=2) In my file, 2 columns are nox and sdate, and are both typeof() = double. If I issue command Plot(nox~sdate) I can get a nice plot. However if I try lines(lowess(time(nox~sdate),nox~sdate),lwd=3, col=2) it returns an error that it is not a matrix if I try to extract these 2 columns into a matrix mdat - matrix(date,nox, byrow=TRUE) doesn't work, and search for help did not work, so am posting. Obviously I am a newbee here. Thanks for any help!! Gregory A. Graves Lead Scientist REstoration COoordination and VERification (RECOVER) Division Everglades Restoration Resource Area South Florida Water Management District Phones: DESK: 561 / 681 - 2563 x3730 CELL: 561 / 719 - 8157 [[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] Out of memory crash on 64-bit Fedora
Greetings all, First of all, thanks to all of you for creating such a useful, powerful program. I regularly work with very large datasets (several GB) in R in 64-bit Fedora 8 (details below). I'm lucky to have 16GB RAM available. However if I am not careful and load too much into R's memory, I can crash the whole system. There does not seem to be a check in place that will stop R from trying to allocate all available memory (including swap space). I have system status plots in my task bar, which I can watch to see when all the ram is taken and R then reserves all the swap space. If I don't kill the R process before the swap hits 100%, it will freeze the machine. I don't know if this is an R problem or a Fedora problem (I suppose the kernal should be killing R before it crashes, but shouldn't R stop before it takes all the memory?). To replicate this behavior, I can crash the system by allocating more and more memory in R: v1=matrix(nrow=1e5,ncol=1e4) v2=matrix(nrow=1e5,ncol=1e4) v3=matrix(nrow=1e5,ncol=1e4) v4=matrix(nrow=1e5,ncol=1e4) etc. until R claims all RAM and swap space, and crashes the machine. If I try this on a windows machine eventually the allocation fails with an error in R, Error: cannot allocate vector of size XX MB. This is much preferable to crashing the whole system. Why doesn't this happen in Linux? Is there some setting that will prevent this? I've looked though the archives and not found a similar problem. Thanks for any help. Adam The facts: sessionInfo() R version 2.8.0 (2008-10-20) x86_64-redhat-linux-gnu locale: LC_CTYPE=en_US.UTF-8;LC_NUMERIC=C;LC_TIME=en_US.UTF-8;LC_COLLATE=en_US.UTF-8;LC_MONETARY=C;LC_MESSAGES=en_US.UTF-8;LC_PAPER=en_US.UTF-8;LC_NAME=C;LC_ADDRESS=C;LC_TELEPHONE=C;LC_MEASUREMENT=en_US.UTF-8;LC_IDENTIFICATION=C attached base packages: [1] stats graphics grDevices utils datasets methods base version _ platform x86_64-redhat-linux-gnu arch x86_64 os linux-gnu system x86_64, linux-gnu status major 2 minor 8.0 year 2008 month 10 day20 svn rev46754 language R version.string R version 2.8.0 (2008-10-20) [[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] Random Forest Variable Importance
Hello, I have an object of Random Forest : iris.rf (importance = TRUE). What is the difference between iris.rf$importance and importance(iris.rf)? Thank you in advance, Best, Li GUO [[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] about the Choleski factorization
On 3/27/2009 11:46 AM, 93354504 wrote: Hi there, Given a positive definite symmetric matrix, I can use chol(x) to obtain U where U is upper triangular and x=U'U. For example, x=matrix(c(5,1,2,1,3,1,2,1,4),3,3) U=chol(x) U # [,1] [,2] [,3] #[1,] 2.236068 0.4472136 0.8944272 #[2,] 0.00 1.6733201 0.3585686 #[3,] 0.00 0.000 1.7525492 t(U)%*%U # this is exactly x Does anyone know how to obtain L such that L is lower triangular and x=L'L? Thank you. Alex __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. rev - matrix(c(0,0,1,0,1,0,1,0,0),3,3) rev [,1] [,2] [,3] [1,]001 [2,]010 [3,]100 (the matrix that reverses the row and column order when you pre and post multiply it). Then L - rev %*% chol(rev %*% x %*% rev) %*% rev is what you want, i.e. you reverse the row and column order of the Choleski square root of the reversed x: x [,1] [,2] [,3] [1,]512 [2,]131 [3,]214 L - rev %*% chol(rev %*% x %*% rev) %*% rev L [,1] [,2] [,3] [1,] 1.9771421 0.000 [2,] 0.3015113 1.6583120 [3,] 1.000 0.502 t(L) %*% L [,1] [,2] [,3] [1,]512 [2,]131 [3,]214 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.
[R] first time poster
hi, so, please bear with me as I am new to the wonderful world of computers... i am trying to answer the following question, and having no luck: Focus your analysis on a comparison between respondents labeled “Low” (coded 1) on attend4 and respondents labeled “High” (coded 4). Then, examine the variance of distributions. That is, run a command var.test. I feel like I need to recode somehow and create 2 new variables, one for the low responses, one for the high responses. I do not know how to 'get into' the variable to deal with just the answers... I hope this makes enough sense for someone out there to help me -- View this message in context: http://www.nabble.com/first-time-poster-tp22745190p22745190.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] General help for a function I'm attempting to write
Hello,
I have written a small function ('JostD' based upon a recent molecular
ecology paper) to calculate genetic distance between populations (columns in
my data set). As I have it now I have to tell it which 2 columns to use (X,
Y). I would like it to automatically calculate 'JostD' for all combinations
of columns, perhaps returning a matrix of distances. Thanks for any help or
suggestions.
Cheers
Colin
Function:
JostD - function(DF, X, Y) {
Ni1 - DF[,X]
Ni2 - DF[,Y]
N1 - sum(Ni1)
N2 - sum(Ni2)
pi1 -Ni1/N1
pi2 -Ni2/N2
pisqr - ((pi1+pi2)/2)^2
H1 - 1 - sum(pi1^2)
H2 - 1 - sum(pi2^2)
Ha - 0.5*(H1+H2)
Da - 1/(1-Ha)
Dg - 1/sum(pisqr)
Db - Dg/Da
D - -2*((1/Db) - 1)
D
}
Sample data:
e-c(0,0,0,4,27)
r-c(0,1,0,7,16)
t-c(1,0,0,16,44)
y-c(0,0,0,2,39)
df-cbind(e,r,t,y)
rownames(df)-q
colnames(df)-w
df
P01 P02 P03 P04
L01.1 0 0 1 0
L01.2 0 1 0 0
L01.3 0 0 0 0
L01.4 4 7 16 2
L01.5 27 16 44 39
JostD(df, 1, 2)
[1] 0.0535215
JostD(df, 1, 3)
[1] 0.02962404
--
Colin Garroway (PhD candidate)
Wildlife Research and Development Section
Ontario Ministry of Natural Resources
Trent University, DNA Building
2140 East Bank Drive
Peterborough, ON, K9J 7B8
Canada
[[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] how to get all iterations if I meet NaN?
hi, everybody, please help me with this question:
If I want to do iteration for 1000 times, however, for the 500th iteration,
there is NaN appears. Then the iteration will stop. If I don't want the stop
and want the all the 1000 iterations be done. What shall I do?
suppose I have x[1:1000] and z[1:1000],I want to do some calculation for all
x[1] to x[1000].
z=rep(0,1000)
for (i in 1:1000){
z[i]=sin(1/x[i])
}
if x[900] is 0, in the above code it will not stop when NaN appears. Suppose
when sin(1/x[900]) is NaN appears and the iteration will now fulfill the
rest 100 iterations. How can I write a code to let all the 1000 iterations
be done?
Thanks!
[[alternative HTML version deleted]]
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] about the Choleski factorization
You want a factorizzation of the form: A = L' L. Am I right (we may name this a Lochesky factorization)? By convention, Cholesky factorization is of the form A = L L', where L is a lower triangular matrix, and L', its transpose, is upper traingular. So, all numerical routines compute L according to this definition. R gives you U = L', which is obviously upper triangular. If you want to use a different definition: A = L' L, that is fine mathematically. Although there is no easy way to transform the result of existing routines to get what you want, you can actually derive an algorithm to convert the standard factorization to the form you want. Rather than go to this trouble, you might as well just code it up from scratch. The big question of course is why do you want the Lochesky factorization? It doesn't do anything special that the traditional Cholesky factorization can do for a symmetric, positive-definite matrix (mainly, solve a system of equations). Ravi. Ravi Varadhan, Ph.D. Assistant Professor, Division of Geriatric Medicine and Gerontology School of Medicine Johns Hopkins University Ph. (410) 502-2619 email: rvarad...@jhmi.edu - Original Message - From: 93354504 93354...@nccu.edu.tw Date: Friday, March 27, 2009 11:58 am Subject: [R] about the Choleski factorization To: r-help r-help@r-project.org Hi there, Given a positive definite symmetric matrix, I can use chol(x) to obtain U where U is upper triangular and x=U'U. For example, x=matrix(c(5,1,2,1,3,1,2,1,4),3,3) U=chol(x) U # [,1] [,2] [,3] #[1,] 2.236068 0.4472136 0.8944272 #[2,] 0.00 1.6733201 0.3585686 #[3,] 0.00 0.000 1.7525492 t(U)%*%U # this is exactly x Does anyone know how to obtain L such that L is lower triangular and x=L'L? Thank you. Alex __ R-help@r-project.org mailing list PLEASE do read the posting guide 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] about the Choleski factorization
Very nice, Duncan.
Here is a little function called loch() that implements your idea for the
Lochesky factorization:
loch - function(mat) {
n - ncol(mat)
rev - diag(1, n)[, n: 1]
rev %*% chol(rev %*% mat %*% rev) %*% rev
}
x=matrix(c(5,1,2,1,3,1,2,1,4),3,3)
L - loch(x)
all.equal(x, t(L) %*% L)
A - matrix(rnorm(36), 6, 6)
A - A %*% t(A)
L - loch(x)
all.equal(x, t(L) %*% L)
Ravi.
Ravi Varadhan, Ph.D.
Assistant Professor,
Division of Geriatric Medicine and Gerontology
School of Medicine
Johns Hopkins University
Ph. (410) 502-2619
email: rvarad...@jhmi.edu
- Original Message -
From: 93354504 93354...@nccu.edu.tw
Date: Friday, March 27, 2009 11:58 am
Subject: [R] about the Choleski factorization
To: r-help r-help@r-project.org
Hi there,
Given a positive definite symmetric matrix, I can use chol(x) to
obtain U where U is upper triangular
and x=U'U. For example,
x=matrix(c(5,1,2,1,3,1,2,1,4),3,3)
U=chol(x)
U
# [,1] [,2] [,3]
#[1,] 2.236068 0.4472136 0.8944272
#[2,] 0.00 1.6733201 0.3585686
#[3,] 0.00 0.000 1.7525492
t(U)%*%U # this is exactly x
Does anyone know how to obtain L such that L is lower triangular and
x=L'L? Thank you.
Alex
__
R-help@r-project.org mailing list
PLEASE do read the posting guide
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] about the Choleski factorization
Very nice, Duncan.
Here is a little function called loch() that implements your idea for the
Lochesky factorization:
loch - function(mat) {
n - ncol(mat)
rev - diag(1, n)[, n: 1]
rev %*% chol(rev %*% mat %*% rev) %*% rev
}
x=matrix(c(5,1,2,1,3,1,2,1,4),3,3)
L - loch(x)
all.equal(x, t(L) %*% L)
A - matrix(rnorm(36), 6, 6)
A - A %*% t(A)
L - loch(x)
all.equal(x, t(L) %*% L)
Ravi.
Ravi Varadhan, Ph.D.
Assistant Professor,
Division of Geriatric Medicine and Gerontology
School of Medicine
Johns Hopkins University
Ph. (410) 502-2619
email: rvarad...@jhmi.edu
- Original Message -
From: Duncan Murdoch murd...@stats.uwo.ca
Date: Friday, March 27, 2009 1:29 pm
Subject: Re: [R] about the Choleski factorization
To: 93354...@nccu.edu.tw
Cc: r-help r-help@r-project.org
On 3/27/2009 11:46 AM, 93354504 wrote:
Hi there,
Given a positive definite symmetric matrix, I can use chol(x) to
obtain U where U is upper triangular
and x=U'U. For example,
x=matrix(c(5,1,2,1,3,1,2,1,4),3,3)
U=chol(x)
U
# [,1] [,2] [,3]
#[1,] 2.236068 0.4472136 0.8944272
#[2,] 0.00 1.6733201 0.3585686
#[3,] 0.00 0.000 1.7525492
t(U)%*%U # this is exactly x
Does anyone know how to obtain L such that L is lower triangular
and x=L'L? Thank you.
Alex
__
R-help@r-project.org mailing list
PLEASE do read the posting guide
and provide commented, minimal, self-contained, reproducible code.
rev - matrix(c(0,0,1,0,1,0,1,0,0),3,3)
rev
[,1] [,2] [,3]
[1,]001
[2,]010
[3,]100
(the matrix that reverses the row and column order when you pre and
post
multiply it).
Then
L - rev %*% chol(rev %*% x %*% rev) %*% rev
is what you want, i.e. you reverse the row and column order of the
Choleski square root of the reversed x:
x
[,1] [,2] [,3]
[1,]512
[2,]131
[3,]214
L - rev %*% chol(rev %*% x %*% rev) %*% rev
L
[,1] [,2] [,3]
[1,] 1.9771421 0.000
[2,] 0.3015113 1.6583120
[3,] 1.000 0.502
t(L) %*% L
[,1] [,2] [,3]
[1,]512
[2,]131
[3,]214
Duncan Murdoch
__
R-help@r-project.org mailing list
PLEASE do read the posting guide
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] [R-sig-hpc] snow Error.
Recent versons of snow signal an error if the value returned from a worker indicates an error. The error handling facilities in snow are still evolving; for now id you don't want an error on a worker to become an error an the master you need to catch the error in the worker yourself and produce an appropriate result, e.g. by replacing MCexe by something like function(...) tryCatch(MCexe(...), error = function(e) NULL) if NULL is OK as a result when the MCexe computation produces an error. luke On Fri, 27 Mar 2009, jgar...@ija.csic.es wrote: Hello, I have a program that used to run well in October, it uses library snow. Since then, one change has ocurred (snow library has been updated) and another could have ocurred (I've unadvertently modified something). Anyway, now when I make the call: parallel.model.results - clusterApply(cl,processors.struct,MCexe) exactly as I used to do, where MCexe is my function and processors.struct is a list containing everything required by MCexe, I obtain the following error: Error in checkForRemoteErrors(val) : 2 nodes produced errors; first error: incorrect number of dimensions Please, do you have any clue about what could be the error? Best regards, Javier García-Pintado ___ R-sig-hpc mailing list r-sig-...@r-project.org https://stat.ethz.ch/mailman/listinfo/r-sig-hpc -- Luke Tierney Chair, Statistics and Actuarial Science Ralph E. Wareham Professor of Mathematical Sciences University of Iowa Phone: 319-335-3386 Department of Statistics andFax: 319-335-3017 Actuarial Science 241 Schaeffer Hall email: l...@stat.uiowa.edu Iowa City, IA 52242 WWW: http://www.stat.uiowa.edu__ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] constraint optimization: solving large scale general nonlinear problems
Hi I need advice regarding constraint optimization with large number of variables. I need to solve the following problem max f(x1,...,xn) x1,..xn x1=g1(x1,...,xn) . . xn=gn(x1,...,xn) I am using Rdonlp2 package which works well until 40 variables in my case. I need to solve this problem with over 300 variables. In this case Rdonlp2 is very very slowly. I know that in Matlab exists Knitro (http://www.ziena.com/knitro.htm.) for large optimization problems. It will be great if you can suggest me some alternatives solutions. Thanks in advance, Florin -- Florin G. Maican == Ph.D. candidate, Department of Economics, School of Business, Economics and Law, Gothenburg University, Sweden --- P.O. Box 640 SE-405 30, Gothenburg, Sweden Mobil: +46 76 235 3039 Phone: +46 31 786 4866 Fax:+46 31 786 4154 Home Page: http://maicanfg.googlepages.com/index.html E-mail: florin.mai...@handels.gu.se Not everything that counts can be counted, and not everything that can be counted counts. --- Einstein --- __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] Efficiency: speeding up unlist that is currently running by row
Hello everyone!
I have a piece of code that works and does what I need but...:
# I have 3 slots:
nr.of.slots-3
# My data frame is new.a:
new.a-data.frame(x=c(john,
mary),y=c(pete,john),z=c(mary,pete),stringsAsFactors=FALSE)
print(new.a)
# Creating all possible combinations of the rows of new.a with all
possible combinations of p1 and p2 in 3 locations (3 new columns):
big.a-cbind(new.a[rep(1:nrow(new.a),each=8),],expand.grid(paste(p,1:2,sep=),paste(p,1:2,sep=),paste(p,1:2,sep=))[rep(1:8,nrow(new.a)),])
print(big.a)
# Making sure the last 3 columns are characters, not factors:
for(i in 1:nr.of.slots) { big.a[[(i+3)]]-as.character(big.a[[(i+3)]]) }
str(big.a)
# Creating a final dataframe with as many columns as slots (i.e., 3);
each cell contains a name of a person and p1 or p2:
output-data.frame(matrix(nrow = nrow(big.a), ncol = nr.of.slots))
for(i in 1:nr.of.slots) {
names(output)[i]-paste(slot,i,sep=.)
}
# THIS IS THE SECTION OF THE CODE I HAVE A QUESTION ABOUT:
for(i in 1:nr.of.slots) {
output[[i]]-lapply(1:nrow(big.a),function(x){
out-unlist(c(big.a[x,i],big.a[x,i+nr.of.slots]))
return(out)
})
}
print(output)
# This is exactly the output I am looking for: Each cell of output
contains just 2 words:
print(output[1,1])
str(output[1,1])
MY QUESTION:
The section of the code above, in which I am running an unlist is
looping through rows. My problem is that in my real data frame I'll
have over a million of rows and more than 3 columns in output. It's
very slow. Is it at all possible to speed it up somehow? Somehow merge
(pairwise) the whole columns of the dataframe and not row by row?
Thank you very much for any adivce!
--
Dimitri Liakhovitski
MarketTools, Inc.
dimitri.liakhovit...@markettools.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.
