[R] comparing vectors with condition
Hi I have two vector x=c(10,30,40,50) total=c(20,20,0,10) Var_exceeding_total=sum(as.numeric(x total)) This will compare the vectors x and total a gives the result. if i want to compare only if the y 0 . can we apply condition in the comparison ? - Thanks in Advance Arun -- View this message in context: http://r.789695.n4.nabble.com/comparing-vectors-with-condition-tp4382946p4382946.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] Reading in csv with footer
I believe this should work d - read.table(foo.csv, header=T, sep=,, comment=T) although its spitting back a warning... this used to work for me. Noah Silverman wrote Hi, I have a CSV file that is formatted well, except that the last line is a summary not is CSV format. Toy example: label_1, label_2, label_3 1,2,3 3,2,4 2,3,4 Total Rows: 3 When I try to import this into R with: d - read.table(foo.csv, header=T, sep=,) It fails to import properly because of the last line. Currently, I have a shell script that strips the last line from the file, then it imports to R cleanly. I don't like this extra layer of processing. Is there a way to import something like this cleanly in R. Thanks! -- Noah __ R-help@ mailing list https://stat.ethz.ch/mailman/listinfo/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://r.789695.n4.nabble.com/Reading-in-csv-with-footer-tp4382441p4382980.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] comparing vectors with condition
sum(ifelse(x*total!=0, as.numeric(x total), 0)) arunkumar wrote Hi I have two vector x=c(10,30,40,50) total=c(20,20,0,10) Var_exceeding_total=sum(as.numeric(x total)) This will compare the vectors x and total a gives the result. if i want to compare only if the y 0 . can we apply condition in the comparison ? -- View this message in context: http://r.789695.n4.nabble.com/comparing-vectors-with-condition-tp4382946p4383000.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] Assigning a function to the 'times' argument of rep()
On Sun, Feb 12, 2012 at 07:56:48PM -0800, z2.0 wrote: Question: I'm trying to use paste() with rep() to reformat a series of values as zip codes. e.g., if column 1 looks like: 52775 83111 99240 4289 112 57701 20001 I want rows 4 and 5 to read, 04289 00112 My thought was this: perry_frame$zip - ifelse(nchar(as.character(perry_frame$zip))5, paste(rep(0,times=(5-nchar(as.character(perry_frame$zip,perry_frame$zip,sep=''), as.character(perry_frame$zip)) But R throws the following: Error in rep(0, times = (5 - nchar(as.character(perry_frame$zip : invalid 'times' argument Is there a reason this doesn't work? Hi. Working solutions were suggested by others. The answer to your question is that times argument should either be a single number or a vector of the same length as the first argument. However, 0 has length 1 5 - nchar(as.character(perry_frame$zip)) [1] 0 0 0 1 2 0 0 Hope this helps. Petr Savicky. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Writing output into a file
Hi everyone, I tried writing this data into a file using the save(myList, file=test1.bin) command, but unfortunately, the numerical values seem to get garbled when I do so. The numbers in my RGui look like 0, 0.5, 0, 1 etc. etc. But when I stored it into a .bin file, and retrieved it using java code, it returns data such as, 2272919233031569408 1701436416123530 -2278152494445862686 7161955281552955800 Etc. etc. I also tried the second method (using a # Open a file connection) Unfortunately, here too the data gets extremely garbled. Has anyone faced such a situation before? Any help / comments / useful links would be much appreciated Thanks and best regards, Suranga On Mon, Feb 13, 2012 at 10:37 AM, Suranga Kasthurirathne suranga...@gmail.com wrote: Hi, Thank you very much for sharing these ideas. I really appreciate them. Let me go try them out :-) On Mon, Feb 13, 2012 at 4:37 AM, Rui Barradas rui1...@sapo.pt wrote: Hello One way is # Write the file save(myList, file=test1.bin) # Reload the data, under the same name, 'myList' load(file=test1.bin) Another way is a bit more complicated # Open a file connection and write the list to it (using comma as separator) fileCon - file(test2.txt, open=wt) lapply(myList, function(x) writeLines(paste(x, collapse=,), con=fileCon)) close(fileCon) # Load the data, maybe under another name strsplit(readLines(con=test2.txt), split=,) If you use the first method, the list is retrieved as it was. If you use the second, you lose the list's members' names. Hope this helps, Rui Barradas -- View this message in context: http://r.789695.n4.nabble.com/Writing-output-into-a-file-tp4382243p4382310.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. -- Best Regards, Suranga -- Best Regards, Suranga -- Best Regards, Suranga [[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] Retrieve by Id from an R list
Hi everyone, I'm using the poLCA module for some analysis work. Basically, Im using the command poLCA(f, data=response,nclass=2) This returns a poLCA object (a list) From this data, I need to retrieve certain indexes, such as[[5]] as seen below. [[5]] Pr(1) Pr(2) Pr(3) class 1: 01.00.0 class 2: 00.50.5 1) The structure shown in [[5]] above is a matrix, isn't it ? 2) how can I specifically retrieve this matrix (retrieve by ID, I guess) ? Any help / suggestions or helpful links would be very much welcome :-) -- Thanks and Best Regards, Suranga [[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] finding and describing missing data runs in a time series
On 13-Feb-2012 Durant, James T. (ATSDR/DTEM/PRMSB) wrote: Hi - I am trying to find and describe missing data in a time series. For instance, in the library openair, there is a data frame called mydata: library(openair) head(mydata) date ws wd nox no2 o3 pm10so2 co pm25 1 1998-01-01 00:00:00 0.60 280 285 39 1 29 4.7225 3.3725 NA 2 1998-01-01 01:00:00 2.16 230 NA NA NA 37 NA NA NA 3 1998-01-01 02:00:00 2.76 190 NA NA 3 34 6.8300 9.6025 NA 4 1998-01-01 03:00:00 2.16 170 493 52 3 35 7.6625 10.2175 NA 5 1998-01-01 04:00:00 2.40 180 468 78 2 34 8.0700 8.9125 NA 6 1998-01-01 05:00:00 3.00 190 264 42 0 16 5.5050 3.0525 NA So for example, I would like to be able to detect for pm25, I would like to be able to detect that there are NA's starting at 1998-01-01 0:00:00 and runs for 2887 hourly observations. Then I would be able to know that there is an NA at 2910 and so on. The key information I am looking for is when the NA's start and their length. The closest thing I can use that I know about is timePlot in the openair package with statistic=frequency but it only gives monthly summary data, and does not tell me if the missing data are clumped together or are dispersed. VR Jim James T. Durant, MSPH CIH Emergency Response Coordinator US Agency for Toxic Substances and Disease Registry Atlanta, GA 30341 770-378-1695 You might consider an approach based on rle(is.na(mydata$pm25)) See ?rle Example: X - c(1,2,3,NA,NA,NA,4,5,NA,6,7,8,NA,NA,NA,NA,NA) X # [1] 1 2 3 NA NA NA 4 5 NA 6 7 8 NA NA NA NA NA rle(is.na(X)) # Run Length Encoding # lengths: int [1:6] 3 3 2 1 3 5 # values : logi [1:6] FALSE TRUE FALSE TRUE FALSE TRUE Ted. - E-Mail: (Ted Harding) ted.hard...@wlandres.net Date: 13-Feb-2012 Time: 08:51:19 This message was sent by XFMail __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] Release plans for 2.14.2 and 2.15.0
It is the intention of the R Core Team to release the finalized version of the 2.14.x series at the end of February, and soon thereafter to start the run-in for 2.15.0. I.e., 2.14.2 Gift-Getting Season on Feb 29 (3rd anniversary of R-1.0.0!) 2.15.0 Easter Beagle on Mar 30 (We'll see about getting the nicknames into the actual sources this time.) Further details to follow. -- Peter Dalgaard, Professor Center for Statistics, Copenhagen Business School Solbjerg Plads 3, 2000 Frederiksberg, Denmark Phone: (+45)38153501 Email: pd@cbs.dk Priv: pda...@gmail.com ___ r-annou...@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-announce __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Is it possible to run multiple instances of Tinn-R?
Dear All, is there anyone using Tinn R, who knows if it's possible to run more than one instance of Tinn R on one machine? I'd like to run more than one R process (R Term) at a time from Tinn R. Any help will be appreciated. Thanks. -- View this message in context: http://r.789695.n4.nabble.com/Is-it-possible-to-run-multiple-instances-of-Tinn-R-tp4383244p4383244.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] dotplots with error bars
On 02/13/2012 09:51 AM, Colin Wahl wrote: Does anyone have any recommendations for producing dotplots with error bars? Are there packages available for this? I searched far and wide and cannot find a suitable option. I am trying to produce publication-quality figures for my thesis results. Dotplots (Cleveland dotplots) are a much better form of summarizing barchart-type data. It does not appear that any of the main plotting packages in r support dotplots with error bars. Considering the benefit of these plots, I find it difficult to believe that they have not been fully integrated into R. I did find a function dotplots.errors available here: http://agrobiol.sggw.waw.pl/~cbcs/articles/CBCS_5_2_2.pdf. However, I have found this function absurdly difficult to use when customizing figures (ordering displays properly, or just simple getting the function to work.) I've been struggling for the last few hours to figure out the error: error using packet 1 sum not meaningful for factors. Unlike other packages, this function doesnt have a ?dotplots.errors to help guide troubleshooting. I presume this is a technicality due to the a numeric variable being identified as a factor. However, I've double checked that all the numeric columns in the data frame are not factors, and the error persists. I'd really prefer not just calling it quits and resorting to old-school sloppy bar charts, but if thats what I need to do to finish this in a timely manner, then so be it. Hi Colin, I am grateful that Marcin Kozak gave plotrix a plug in the paper, and to show my gratitude, I'll explain how to use centipede.plot to get the illustration in the paper. Assume that you have the data frame shown on p70 of the paper: plant_height-read.csv(plant_height.csv) Now, to echo Marcin, let us produce the plot: library(plotrix) centipede.plot(t(plant_height[,c(3,2,4)]), left.labels=plant_height$group,bg=black, right.labels=rep(,13),xlab=Mean plant height (cm) +- SE) If you want the mean value line: abline(v=mean(plant_height$est),col=lightgray) The grid lines are a bit more difficult. You could insert a line into the function just after the call to box() to draw grid lines under each dot: abline(h=1:dim(x)[2],col=lightgray,lty=2) However, this looks like such a good idea that I will add two arguments to the function to do the vertical line(s) and horizontal grid automatically, and this option will appear in the next version of plotrix. 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] dotplots with error bars
Colin Wahl biowahl at gmail.com writes: Does anyone have any recommendations for producing dotplots with error bars? Are there packages available for this? I searched far and wide and cannot find a suitable option. Dear Colin, have a look at this page from the R wiki: http://rwiki.sciviews.org/doku.php?id=tips:graphics-base:errbars Also, if you want more details on how to do it with ggplot2, a very nice graphic package, you can have a look here: http://wiki.stdout.org/rcookbook/Graphs/Plotting%20means%20and%20error%20bars%20(ggplot 2)/ HTH Matthieu __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] see NA
Dear All I want to chose just spacial columns in R. (read table) data1- read.table(/home/Documents/data.txt,header=F,sep = \t, as.is =F) data.2-data1[sub.data[,1],] The dimension of data.2 is correct but my data are transformed NA head(data.2) V1 V2 V3 V4 V5 V6 V7 V8 V9 V10 V11 V12 V13 V14 V15 V16 V17 V18 V19 V20 V21 V22 V23 V24 NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA.1 NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA.2 NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA.3 NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA.4 NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA.5 NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA when I used as.is =T Error in read.table(/home/Documents/data.txt, : invalid numeric 'as.is' expression I will appreciate if you help me. Soheila [[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] Retrieve by Id from an R list
On Mon, Feb 13, 2012 at 02:13:41PM +0530, Suranga Kasthurirathne wrote: Hi everyone, I'm using the poLCA module for some analysis work. Basically, Im using the command poLCA(f, data=response,nclass=2) This returns a poLCA object (a list) From this data, I need to retrieve certain indexes, such as[[5]] as seen below. [[5]] Pr(1) Pr(2) Pr(3) class 1: 01.00.0 class 2: 00.50.5 1) The structure shown in [[5]] above is a matrix, isn't it ? 2) how can I specifically retrieve this matrix (retrieve by ID, I guess) ? Hi. If the list is in the variable poLCA, then try a - poLCA[[5]] a Whether this is a matrix may be checked using class(a) Hope this helps. Petr Savicky. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] see NA
GHard to tell given the file is not available nor do we see the first lines of it. Uwe Ligges On 13.02.2012 10:51, Soheila Khodakarim wrote: Dear All I want to chose just spacial columns in R. (read table) data1- read.table(/home/Documents/data.txt,header=F,sep = \t, as.is =F) data.2-data1[sub.data[,1],] The dimension of data.2 is correct but my data are transformed NA head(data.2) V1 V2 V3 V4 V5 V6 V7 V8 V9 V10 V11 V12 V13 V14 V15 V16 V17 V18 V19 V20 V21 V22 V23 V24 NANA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA.1NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA.2NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA.3NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA.4NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA.5NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA when I used as.is =T Error in read.table(/home/Documents/data.txt, : invalid numeric 'as.is' expression I will appreciate if you help me. Soheila [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] logical operator for different dimensions
I have some dataset
sci.pre - 0.300.380.500.650.801.031.331.72
2.22
2.873.815.066.759.00 11.97 14.15 16.34 19.04
22.27 25.49 29.72 34.67 40.47 47.29 55.29 64.67 75.6
88.50 103.50 121.10 141.70 165.80 194.00 227.00 265.00 308.00
356.50 411.00 441.60 472.20 506.35 540.50 578.55 616.60 637.75
658.90 680.05 701.20 724.65 748.10 771.55 795.00 820.95 846.90
872.85 898.80 927.35 955.90 984.45 1013.00 ( length is 60)
sci.avgkernal - 0.300.380.500.650.801.031.33
1.722.22
2.873.815.066.759.00 11.97 14.15 16.34 19.04
22.27 25.49 29.72 34.67 40.47 47.29 55.29 64.67 75.65
88.50 103.50 121.10 141.70 165.80 194.00 227.00 265.00 308.00
356.50 411.00 441.60 472.20 506.35 540.50 578.55 616.60 637.75
658.90 680.05 701.20 724.65 748.10 771.55 795.00 820.95 846.90
872.85 898.80 927.35 955.90 984.45 1013.00 ( length is 60)
pres.interptimes - [,1] [,2] [,3] [,4] [,5]
[,6] [,7] [,8]
[1,] 1016.1267 1005.9741 989.9127 970.0237 945.6067 880.5082 790.4647
675.8315
[2,] 875.6320 866.8767 853.0258 835.8741 814.8176 758.6784 681.0275
582.1712
[3,] 996.0351 986.0758 970.3201 950.8098 926.8576 862.9984 774.6692
662.2184
[4,] 996.0353 986.0760 970.3203 950.8100 926.8578 862.9987 774.6694
662.2187
[5,] 1008.0222 997.9431 981.9978 962.2527 938.0123 873.3847 783.9926
670.1888
[6,] 999.8343 989.8371 974.0214 954.4367 930.3932 866.2906 777.6247
664.7453
[,9][,10][,11][,12][,13][,14][,15][,16]
[1,] 544.3248 410.3611 289.4130 237.2794 191.5622 152.3020 119.1842 91.68080
[2,] 468.7636 353.2371 248.9350 203.9765 164.5513 130.6945 102.1346 78.41645
[3,] 533.2153 401.8020 283.1566 232.0154 187.1686 148.6559 116.1686 89.18875
[4,] 533.2155 401.8022 283.1568 232.0157 187.1688 148.6561 116.1688 89.18898
[5,] 539.6334 406.6388 286.5657 234.8092 189.4227 150.4466 117.5683 90.26387
[6,] 535.2505 403.3363 284.2386 232.9026 187.8848 149.2253 116.6141 89.53148
[,17][,18][,19] [,20]
[1,] 69.12170 50.81654 24.30808 0.8657024
[2,] 58.96213 43.17629 20.31615 0.1001228
[3,] 67.05913 49.10247 23.09867 0.1025881
[4,] 67.05936 49.10270 23.09890 0.1028196
[5,] 67.86797 49.69524 23.37854 0.1057444
[6,] 67.31751 49.29240 23.18949 0.1057560 ( dim is 6 20)
sci.prediff - diff(sci.pre)
sci.prediff - c(sci.pre[1],sci.prediff)
sum(sci.avgkernal*sci.prediff )/sum(sci.prediff )
pres.interptimes - pres.interptime[,-20]#skip last level
tm3.avgkernal -array(NA,c(length(1:nobs),19))
for (k in 1:nobs){
for (h in 1:19){
sel- sci.pre = pres.interptimes[k,h] sci.pre
pres.interptimes[k,h+1]
tm3.avgkernal[k,h] - sum((sci.avgkernal * sci.prediff)[sel]) /
sum(sci.prediff[sel])
}
}
after running code I get error
Error: subscript out of bounds
How to fix this error ?
--
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and provide commented, minimal, self-contained, reproducible code.
[R] Error from GNLS (undefined columns selected)
Dear R-helpers, I'm a new R-user and I was trying to gain some experience with the GNLS function of the NLME package. This is an extract from my dataset (it's a 432x6 data.frame) called input, in the first column I have the values that I need to fit, while the remaining columns are input variables for the theoretical model, the function mymodel (which returns a 432x1 vector of fitted values): V0V1V2V3 V4 V5 0.56374863838.6875400.1041095890.0572495640 1.47392252639.125400.2767123290.0553929370 0.51704068441.0625450.1808219180.0532145910 1.38181318737.5625400.3534246580.0505213210 and this is my call of gnls: fm1 - gnls(V0 ~ mymodel(V1,V2,V3,V4,V5,par1,par2,par3,par4,par5,par6,par7,par8), data=input,start=list(par1=0.2,par2=0.4,par3=0.8,par4=0.2,par5=10,par6=0.8,par7=0.9,par8=-0.5)) and I get the following error: Error in `[.data.frame`(eval(model, data.frame(data, getParsGnls(plist, : undefined columns selected My guess is that R is not able to create that data.frame since getParsGnls, an internal function of gnls, does not provide its output (obviously the error is mine, but I can't clearly identify it). Indeed, in the previous lines of the gnls code, I don't have any problems when R creates a data.frame using the dataframe data and pars, as follows: res - eval(model, data.frame(data, pars)) I would be grateful to receive any help or hint. Best regards, Louis Salomon [[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] Discrete Event Simulation problem
I have made some chances and I believe now the only problem is making the
system reorder. Please any help would be great
test - function(seed = 123456789, maxStock= 100, minStock =
20,t.max=1100,inventory =50)
{
LAST = t.max
START = 0
t.demand = START
t.supply = START
t.current = START
GetDemand-function()
{
t.demand - t.demand + runif(1,min=0,max=5)
return(t.demand)
}
GetSupply -function(){
if (inventory minStock)
{
t.supply - t.supply + 1.0
}
else
t.supply - Inf
return(t.supply)
}
main - function(seed)
{
if(seed 0)
set.seed(seed)
index = 0
t.current = START Starting
Conditions
t.demand = START
t.supply = START
minStock = 20
maxStock = 100
inventory = 50
order_costs = 0
storage_costs = 0
sum = list(inventory = 50,order_costs = 0,storage_costs = 0)
while(index LAST){
index = index + 1
t.demand = GetDemand() ### expected time to next sale
t.supply = GetSupply() ### expected time to arrival of order,
Infinity as
order has not been placed
t.next =min(t.demand,t.supply)###next event either sale or supply is
the one with imminent arrival
k = maxStock - inventory
t.current = t.next -min(t.demand,t.max)
if(inventory 0) {
storage_costs = (t.next-t.current)*0.10*inventory
}
if (inventory minStock)
{###Need to
Order
k = maxStock - inventory
order_costs = 50 + 0.02*k
sum$order_costs = sum$order_costs + order_costs
t.supply = GetSupply()
}
if(t.next ==t.demand)
{
inventory - inventory - 1
Sale made
sum$inventory = sum$inventory - 1.0
t.demand = GetDemand()
}
if(t.next == t.supply)
{
Order Arrives
sum$inventory = sum$inventory + k
k = 0
t.supply = GetSupply()
}
if(inventory maxStock)
{
k = maxStock - inventory
sum$storage_costs = sum$storage_costs + storage_costs
sum$order_costs = sum$order_costs + order_costs
}
}
options(digits = 5)
sis = list(Time = index,StorageCosts =sum$storage_costs,OrderCosts=
sum$order_costs,AverageCosts =((sum$order_costs +
sum$storage_costs)/index),Inventory = sum$inventory)
return(sis)
}
return(main(seed))
}
--
View this message in context:
http://r.789695.n4.nabble.com/Discrete-Event-Simulation-problem-tp4377276p4383464.html
Sent from the R help mailing list archive at Nabble.com.
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Re: [R] R-help Digest, Vol 108, Issue 13
dear: i want to know how to get a survival curve of the Cox proportional risk regression, thank you. [[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] logical operator for different dimensions
On Mon, Feb 13, 2012 at 02:08:52AM -0800, uday wrote:
I have some dataset
sci.pre - 0.300.380.500.650.801.031.331.72
2.22
2.873.815.066.759.00 11.97 14.15 16.34 19.04
22.27 25.49 29.72 34.67 40.47 47.29 55.29 64.67 75.6
88.50 103.50 121.10 141.70 165.80 194.00 227.00 265.00 308.00
356.50 411.00 441.60 472.20 506.35 540.50 578.55 616.60 637.75
658.90 680.05 701.20 724.65 748.10 771.55 795.00 820.95 846.90
872.85 898.80 927.35 955.90 984.45 1013.00 ( length is 60)
sci.avgkernal - 0.300.380.500.650.801.031.33
1.722.22
2.873.815.066.759.00 11.97 14.15 16.34 19.04
22.27 25.49 29.72 34.67 40.47 47.29 55.29 64.67 75.65
88.50 103.50 121.10 141.70 165.80 194.00 227.00 265.00 308.00
356.50 411.00 441.60 472.20 506.35 540.50 578.55 616.60 637.75
658.90 680.05 701.20 724.65 748.10 771.55 795.00 820.95 846.90
872.85 898.80 927.35 955.90 984.45 1013.00 ( length is 60)
pres.interptimes - [,1] [,2] [,3] [,4] [,5]
[,6] [,7] [,8]
[1,] 1016.1267 1005.9741 989.9127 970.0237 945.6067 880.5082 790.4647
675.8315
[2,] 875.6320 866.8767 853.0258 835.8741 814.8176 758.6784 681.0275
582.1712
[3,] 996.0351 986.0758 970.3201 950.8098 926.8576 862.9984 774.6692
662.2184
[4,] 996.0353 986.0760 970.3203 950.8100 926.8578 862.9987 774.6694
662.2187
[5,] 1008.0222 997.9431 981.9978 962.2527 938.0123 873.3847 783.9926
670.1888
[6,] 999.8343 989.8371 974.0214 954.4367 930.3932 866.2906 777.6247
664.7453
[,9][,10][,11][,12][,13][,14][,15][,16]
[1,] 544.3248 410.3611 289.4130 237.2794 191.5622 152.3020 119.1842 91.68080
[2,] 468.7636 353.2371 248.9350 203.9765 164.5513 130.6945 102.1346 78.41645
[3,] 533.2153 401.8020 283.1566 232.0154 187.1686 148.6559 116.1686 89.18875
[4,] 533.2155 401.8022 283.1568 232.0157 187.1688 148.6561 116.1688 89.18898
[5,] 539.6334 406.6388 286.5657 234.8092 189.4227 150.4466 117.5683 90.26387
[6,] 535.2505 403.3363 284.2386 232.9026 187.8848 149.2253 116.6141 89.53148
[,17][,18][,19] [,20]
[1,] 69.12170 50.81654 24.30808 0.8657024
[2,] 58.96213 43.17629 20.31615 0.1001228
[3,] 67.05913 49.10247 23.09867 0.1025881
[4,] 67.05936 49.10270 23.09890 0.1028196
[5,] 67.86797 49.69524 23.37854 0.1057444
[6,] 67.31751 49.29240 23.18949 0.1057560 ( dim is 6 20)
sci.prediff - diff(sci.pre)
sci.prediff - c(sci.pre[1],sci.prediff)
sum(sci.avgkernal*sci.prediff )/sum(sci.prediff )
pres.interptimes - pres.interptime[,-20]#skip last level
tm3.avgkernal -array(NA,c(length(1:nobs),19))
for (k in 1:nobs){
for (h in 1:19){
sel- sci.pre = pres.interptimes[k,h] sci.pre
pres.interptimes[k,h+1]
tm3.avgkernal[k,h] - sum((sci.avgkernal * sci.prediff)[sel]) /
sum(sci.prediff[sel])
}
}
after running code I get error
Error: subscript out of bounds
How to fix this error ?
I suspect that the error is generated at
pres.interptimes[k,h+1]
since h+1 goes up to 20 and pres.interptimes was restricted to 19
columns in
pres.interptimes - pres.interptime[,-20]#skip last level
(assuming that pres.interptime is a typo and is pres.interptimes
in fact).
Hope this helps.
Petr Savicky.
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Re: [R] function arrows.circular not working
Indeed. Thanks for the solution :) -Original Message- From: Sarah Goslee [mailto:sarah.gos...@gmail.com] Sent: Friday, February 10, 2012 1:01 PM To: Chosid, David (FWE); r-help Subject: Re: [R] function arrows.circular not working It's a good idea to acknowledge that you've found a solution on the R-help list, rather than just to me. That way the answer appears in the list archives, and other people will know you no longer need help with this problem. Sarah On Fri, Feb 10, 2012 at 12:57 PM, Chosid, David (MISC) david.cho...@state.ma.us wrote: Ah. Thanks. I didn't realize that I needed an updated R... Just the library. Works now. -Original Message- From: Sarah Goslee [mailto:sarah.gos...@gmail.com] Sent: Friday, February 10, 2012 12:32 PM To: Chosid, David (FWE); r-help Subject: Re: [R] function arrows.circular not working Hi, On Fri, Feb 10, 2012 at 12:15 PM, Chosid, David (MISC) david.cho...@state.ma.us wrote: Yes, sorry for not being more clear. Here is the sessionInfo(): sessionInfo() R version 2.9.1 (2009-06-26) i386-pc-mingw32 That is definitely the place to start. Your version of R is 2.5 years old, and circular is up to version 0.4-3 on CRAN. I just checked, and circular 0.3-8 doesn't *have* arrows.circular. So you must be using some documentation from the internet for a newer version, rather than using the docs that go with what you have installed on your computer. You need to update R and your packages. Sarah locale: LC_COLLATE=English_United States.1252;LC_CTYPE=English_United States.1252;LC_MONETARY=English_United States.1252;LC_NUMERIC=C;LC_TIME=English_United States.1252 attached base packages: [1] grid tcltk stats graphics grDevices utils datasets [8] methods base other attached packages: [1] relimp_1.0-1 Rcmdr_1.4-10 car_1.2-14 circular_0.3-8 [5] boot_1.2-37 lattice_0.17-25 RODBC_1.3-0 -Original Message- From: Sarah Goslee [mailto:sarah.gos...@gmail.com] Sent: Friday, February 10, 2012 12:02 PM To: Chosid, David (FWE) Cc: r-help@r-project.org Subject: Re: [R] function arrows.circular not working Your sessionInfo() would be helpful. Also, just to check: you did do library(circular) right? Sarah On Fri, Feb 10, 2012 at 11:55 AM, Chosid, David (MISC) david.cho...@state.ma.us wrote: I have started using the circular package but it is not recognizing the function arrows.circular. I attempted to use the example provided in the circular manual. Here is the example code using the circular package: plot(rvonmises(10, circular(0), kappa=1)) arrows.circular(rvonmises(10, circular(0), kappa=1)) arrows.circular(rvonmises(10, circular(0), kappa=1), y=runif(10), col=2) arrows.circular(rvonmises(10, circular(0), kappa=1), y=runif(10), x0=runif(10, -1, 1), y0=runif(10, -1, 1), col=3) My error is: Error: could not find function arrows.circular Any help would be greatly appreciated. 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] Any package for best subset selection on random effects model?
zbleach zt020200 at gmail.com writes: Hi Pros, I know leaps() computes the best subset selection for linear model, and the bestglm() computes the best subset selection for generalized linear model. Is there any package for best subset selection on random effects model, or mixed effects model? glmmLasso on CRAN ; possibly try the MuMIn package (specifically the dredge command). You might have better luck with this question on the r-sig-mixed-mod...@r-project.org list. Ben Bolker __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] best option for big 3D arrays?
I've been investigating and I partially respond myself. I tried the
packages 'bigmemory' and 'ff' and for me the latter did the work I need
pretty straightforward. I create the array in filebacked form with the
function ff, and it seems that the usual R indexing works well. I have yet
to see the limitations, but I hope it helps.
a foo example:
myArr - ff(NA, dim=rep(904,3), filename=arr.ffd, vmode=double)
myMat - matrix(1:904^2, ncol=904)
for ( i in 1:904 ) {
myArr[,,i] - myMat
}
Thanks all,
2012/2/11 Duncan Murdoch murdoch.dun...@gmail.com
On 12-02-10 9:12 AM, Djordje Bajic wrote:
Hi all,
I am trying to fill a 904x904x904 array, but at some point of the loop R
states that the 5.5Gb sized vector is too big to allocate. I have looked
at
packages such as bigmemory, but I need help to decide which is the best
way to store such an object. It would be perfect to store it in this
cube
form (for indexing and computation purpouses). If not possible, maybe the
best is to store the 904 matrices separately and read them individually
when needed?
Never dealed with such a big dataset, so any help will be appreciated
(R+ESS, Debian 64bit, 4Gb RAM, 4core)
I'd really recommend getting more RAM, so you can have the whole thing
loaded in memory. 16 Gb would be nice, but even 8Gb should make a
substantial difference. It's going to be too big to store as an array
since arrays have a limit of 2^31-1 entries, but you could store it as a
list of matrices, e.g.
x - vector(list, 904)
for (i in 1:904)
x[[i]] - matrix(0, 904,904)
and then refer to entry i,j,k as x[[i]][j,k].
Duncan Murdoch
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[R] kernlab - rvm error message: Error in if (length(data) != vl)
Hi,
I am trying to perform relevance vector machines with the rvm-function from
kernlab.
On one dataset I get this message:
Setting default kernel parameters
Error in if (length(data) != vl) { :
RMate stopped at line 0 of selection
missing value where TRUE/FALSE needed
Calls: rvm ... .local - backsolve - as.matrix - chol - diag - array
can someone explain this error message?
It works for other data-sets with the same feature / example-space.
But for one particular trainingset I get this error message...
thanks!
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[R] kernlab - error message: array(0, c(n, p)) : 'dim' specifies too large an array
Hi, For another trainingset I get this error message, which again is rather cryptic to me: Setting default kernel parameters Error in array(0, c(n, p)) : 'dim' specifies too large an array RMate stopped at line 0 of selection Calls: rvm ... .local - backsolve - as.matrix - chol - diag - array thanks for any suggestions! __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] see NA
On Feb 13, 2012, at 4:51 AM, Soheila Khodakarim wrote: Dear All I want to chose just spacial columns in R. (read table) data1- read.table(/home/Documents/data.txt,header=F,sep = \t, as.is =F) data.2-data1[sub.data[,1],] What is sub.data? are you migrating to R from a language where sub. would be some kind of operation on data? -- David. The dimension of data.2 is correct but my data are transformed NA head(data.2) V1 V2 V3 V4 V5 V6 V7 V8 V9 V10 V11 V12 V13 V14 V15 V16 V17 V18 V19 V20 V21 V22 V23 V24 NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA.1 NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA.2 NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA.3 NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA.4 NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA.5 NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA when I used as.is =T Error in read.table(/home/Documents/data.txt, : invalid numeric 'as.is' expression I will appreciate if you help me. Soheila [[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. David Winsemius, MD West Hartford, CT __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] kernlab - error message: array(0, c(n, p)) : 'dim' specifies too large an array
I am using a linear kernel (vanilladot). By switching the kernel, I actually get rid of the error message, but I would like to stick to the linear one ... On 13.02.2012, at 16:23, Martin Batholdy wrote: Hi, For another trainingset I get this error message, which again is rather cryptic to me: Setting default kernel parameters Error in array(0, c(n, p)) : 'dim' specifies too large an array RMate stopped at line 0 of selection Calls: rvm ... .local - backsolve - as.matrix - chol - diag - array thanks for any suggestions! __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] kernlab - error message: array(0, c(n, p)) : 'dim' specifies too large an array
On Feb 13, 2012, at 10:23 AM, Martin Batholdy wrote: Hi, For another trainingset I get this error message, which again is rather cryptic to me: Just imagine how it seems to us! Setting default kernel parameters Error in array(0, c(n, p)) : 'dim' specifies too large an array RMate stopped at line 0 of selection Calls: rvm ... .local - backsolve - as.matrix - chol - diag - array You are on you way to the prize for the greatest number of cryptic (your word) postings in a short interval. (And this one doesn't even have the context of your posting of 8 minutes ago.) thanks for any suggestions! More details about data and code. -- David Winsemius, MD West Hartford, CT __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] R-help Digest, Vol 108, Issue 13
Le lundi 13 février 2012 à 19:48 +0800, 丁飞 a écrit : dear: i want to know how to get a survival curve of the Cox proportional risk regression, thank you. See the relevant part of the Survival Task View here: http://cran.r-project.org/web/views/Survival.html And more specifically, see the coxph() function in the survival package. A tutorial is available from: http://cran.r-project.org/doc/contrib/Fox-Companion/appendix-cox-regression.pdf (But RSiteSearch(coxph) or a simple Google search will give you many more results.) Cheers __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] multi-regression with more than 50 independent variables
Hi R Users, I am going to run a multiple linear regression with around 57 independent variables. Each time I run the model with just 11 variables, the results are reasonable. With increasing the number of independent variables more than 11, the coefficients will get “NA” in the output. Is there any limitation for the number of independent variables in multiple linear regressions in R? I attached my dataset as well as R codes below: mlr.data- read.table(./multiple.txt,header=T) mlr.output- lm(formula = CaV ~ SHG + TrD+ CrH+ SPAD+ FlN+ FrN+ YT+ LA+ LDMP+B+Cu+ Zn+ Mn + Fe+ K + P+ N +Clay30 +Silt30 +Sand30 +Clay60 +Silt60 +Sand60 +ESP30 +NaEx30+ CEC30+Cl30+ SAR30 +KSol30+ NaSol30 +CaMgSol3 +ZnAv30 +FeAv30 +OC30 +PAv30 +KAv30 +TNV30+ pH30+ EC30 +SP30 +ESP60 +NaEx60 +CEC60 +Cl60 +SAR60 +KSol60 +NaSol60 +CaMgSol6 +ZnAv60+FeAv60 +OC60 +PAv60 +KAv60 +TNV60 +pH60 + EC60 +SP60, data=mlr.data) summary (mlr.output) Regards, Reza CaV SHG TrD CrH SPADFlN FrN YT LA LDMP B Cu Zn Mn Fe K P N Clay30 Silt30 Sand30 Clay60 Silt60 Sand60 ESP30 NaEx30 CEC30 Cl30SAR30 KSol30 NaSol30 CaMgSol3ZnAv30 FeAv30 OC30PAv30 KAv30 TNV30 pH30 EC30SP30ESP60 NaEx60 CEC60 Cl60SAR60 KSol60 NaSol60 CaMgSol6ZnAv60 FeAv60 OC60PAv60 KAv60 TNV60 pH60EC60 SP60 49.83 15.822.32 45.882.7126 5 55.834.757.7 17.98.4 14.750.5144.7 0.9 0.1 1.7 14 50 36 10 26 64 3.2 0.3 11.987.75.6 2.5 25.769.82.9 3.1 0.6664.4360 14.89.940.6 39.64.8 0.2 12.251.46.9 1.4 33.469.91.8 3.4 0.2 3.5 290 11.57.193.2140.1 37.85 18.920.89 53.275.9169 4 67.440.158.6 15.87.9 12.153.2141.6 0.7 0.1 1.9 13 51 34 11 28 64.01.9 0.2 12.545.77.8 3.8 22.977.93.1 2.5 0.6945.6390 14.87.2 1.1 40.15.4 0.1 13.359.97.9 1.7 56.658.91.5 5.5 0.6 1.9 350 10.27.1 3.7 42.1 64.12 20.719.04 43.974.3133 5 55.938.760.0 12.16.3 12.647.4159.5 1.0 0.1 1.6 12 48 33 13 26 62.06.4 1.1 12.633.210.11.8 43.899.31.5 2.8 0.5720.5470 15.17.1 0.7 39.84.9 0.3 10.965.57.1 2.3 42.166.41.1 3.9 0.7 2.9 288.9 12.27.2 4.1 35.6 90.28 14.619.52 56.961.9145 7 66.533.259.1 13.74.7 10.052.1241.1 0.8 0.1 1.9 13 47 32 10 30 57.05.5 0.3 11.731.23.5 2.4 50.465.71.9 2.5 0.8841.9398 14.37 2.4 38.73.1 0.5 14.171.98.4 1.9 36.759.10.9 2.6 0.5 2.7 290.7 13.67 2.6 43.7 111.18 13.216.53 61.378 127 6 49.941.751.6 14.74.7 10.055.8148.9 0.7 0.1 1.9 14 46 37 11 28 60.03.9 0.7 11 21.74.3 0.9 33.773.62.3 2.7 0.7833.4349 15.27.4 0.9 39.12.8 0.1 12.576.68.9 3.1 32.171.40.5 6.9 0.5 2.5 256.9 15.17.5 1.7 36.7 59.11 12.921.34 45.378.9178 6 63.339.852.0 19.54.2 8.9 54.7229.5 0.7 0.1 1.7 13 46 36 12 30 62.02.7 0.7 12.919.52.6 2.8 61.286.92.2 3.7 0.8627.8400.5 17.17.1 2.1 39.93.9 0.3 11.957.46.7 1.6 30.189.91.8 5.8 0.3 3.7 224.8 12.97.3 5.5 34.9 80.89 17.915.86 40.366.8154 7 45.636.847.8 21.63.7 12.654.7162.1 0.7 0.1 1.9 11 50 35 13 31 61.02.9 0.4 10.937.97.1 1.9 19.555.82.8 2.9 0.6645.1459 15.67.2 0.8 36.15.1 0.4 13.185.55.7 2.1 29.192.11.9 7.8 0.7 2.8 278.9 11.87.2 6.1 32.7 122.74 16.617.29 43 77.8140 6 32.937.755.6 20.04.2 12.147.4152.6 0.9 0.1 1.7 14 49 36 11 25 58.03.2 1.5 11.524.53.9 3.7 20.856.91.9 2.6 0.7240.7
[R] Wavelet and inverse wavelet to filter data
Dear Friends of R, I would like to filter high frequent turbulence data with wavelet analysis. Therefore i want to calculate the wavelet coefficients of the raw data. And then the inverse for a special frequency band. How can i do this in the best and easiest way? Many thanks for your answers and help, Best regards [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Error in apply(x2, 1, diff) : dim(X) must have a positive length
Thank you for your help, David. I was trying to run bootrapping on the
dataset shoes from MASS package. But I still have some problem here.
as.matrix(data.frame(shoes),nrows=10,ncols=2,byrow=T)
A B
[1,] 13.2 14.0
[2,] 8.2 8.8
[3,] 10.9 11.2
[4,] 14.3 14.2
[5,] 10.7 11.8
[6,] 6.6 6.4
[7,] 9.5 9.8
[8,] 10.8 11.3
[9,] 8.8 9.3
[10,] 13.3 13.6
y=as.matrix(data.frame(shoes),nrows=10,ncols=2,byrow=T)
class(y)
[1] matrix
apply(y,1,diff)
[1] 0.8 0.6 0.3 -0.1 1.1 -0.2 0.3 0.5 0.5 0.3
dif.mns - function(x2,tr1=.1,tr2=.2){
+ diffs=apply(x2,1,diff)
+ mn1=mean(diffs)
+ mn2=mean(diffs,tr=.1)
+ mn3=mean(diffs,tr=.2)
+ mn4=median(diffs)
+ mns=c(mn1,mn2,mn3,mn4)
+ list(mnds=round(mns,3))}
dif.mns(y,tr1=.1,tr2=.2)
$mnds
[1] 0.410 0.400 0.417 0.400
bootstrap(y,nboot=1000,theta=dif.mns)
Error in apply(x2, 1, diff) : dim(X) must have a positive length
I am not sure what I am doing wrong. When I run apply(y,1,diff) I am able to
get output, and when I run dif.mns with just the original dataset, I am also
able to get output.
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View this message in context:
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and provide commented, minimal, self-contained, reproducible code.
Re: [R] Writing output into a file
Hello, I tried writing this data into a file using the save(myList, file=test1.bin) command, but unfortunately, the numerical values seem to get garbled when I do so. The numbers in my RGui look like 0, 0.5, 0, 1 etc. etc. But when I stored it into a .bin file, and retrieved it using java code, it returns data such as, The problem should be in the use of java, 'save' uses a R format , RDA. You can use 'ascii=TRUE'and see it with a text editor. Also see ?save I also tried the second method (using a # Open a file connection) Unfortunately, here too the data gets extremely garbled. Don't understand why, check the output file with a text editor and let us know what is wrong. The problem I've seen is that the use of 'strsplit' coerses the numeric data to character, but this is easy to solve. Does your list have sub-lists? Rui Barradas -- View this message in context: http://r.789695.n4.nabble.com/Writing-output-into-a-file-tp4382243p4383741.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] logical operator for different dimensions
Hi Petr, You were correct , thats was the mistake . I am sorry for last reply. now its working. Cheers Uday -- View this message in context: http://r.789695.n4.nabble.com/logical-operator-for-different-dimensions-tp4383316p4384143.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] How to import time-series data
Gabor Grothendieck wrote Try this xyplot.zoo lattice graph. Time series are represented in columns so we transpose the data and convert it to zoo. The screen= argument available in xyplot.zoo groups series into panels: Lines - plant,aphid,1,2,3,4 pumpkin,1-2,0.065566,0.057844,0.08,0.086879 pumpkin,1-3,0.107612,0.097272,0.11663,0.160499 squash,1-4,0.126939,0.115003,0.140275,0.188829 library(zoo) library(lattice) DF - read.csv(text = Lines) z - zoo(t(DF[3:6])) colnames(z) - DF$aphid xyplot(z, screen = DF$plant) Thank you! That gets the data in exactly the shape I was expecting. I was hoping to get the graph showing all lines on one plot, but coloured according to plant. I tried to change screen=DF$plant for groups=DF$plant, but it doesn't work, and I can't figure out from the documentation why it doesn't work (I think I need to more thoroughly understand data types first). Could you point me in the right direction? Thanks so much for your help so far. -- View this message in context: http://r.789695.n4.nabble.com/How-to-import-time-series-data-tp4381372p4383740.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] non linear quantile regression - Median not plotting where it should
Hi, I'm attempting to calculate the 0.25 and 0.97 quantiles for tree height (0-50 meters) against tree age (0-300 years) and I am running into some difficulty with the plotted grafic. I've run the examples in the quantreg help and can get those to work properly and by plugging in my data I can also get the lines plotted on my dataset. Unfortunately I'm running into a problem with the median and other regression lines with tree age younger than 50 years, basically the median is in this range overestimated and even comes out of the rage of oberservations. here is the code I'm using. # then fit the median using nlrq spruce.nlrq - nlrq(height ~ SSlogis(age, Asym, mid, scal),data=spruce, tau=0.5, trace=TRUE) lines(1:200, predict(spruce.nlrq, newdata=list(age=1:200)), col=2) I believe this has something to do with the SSlogis, as this gives the parameters for an S curve. My data set does not have the typical S curve shape, instead you could image it as starting at the inflection point of an S curve. This is what I expect the .025 quantile to be similar to: x - seq(1,100,1) plot(log(x)) the 0,975 should also have a logarithmic shape but no such a steep incline. Ive tried using different self starting models as found under: apropos(^ss) however I have not gotten them to work, or at least to fix the problem. Curves similar to mine look like these site index curves: http://www.extension.umn.edu/distribution/naturalresources/components/3473-13.html In the example from the nlrq help the lines of the median and the various quantiles all start from the same location, tha is basically (x=0, y=0) in the coodinate plane. With my problem, the lines start to be drawn from various different positions ( the lines always start at age(x)=0, but the height(y) can range between 0 and 15). Additionally, the data set is quite large. with about 50,000 oberservations on age and height. Does anyone have a suggestion on how to fix this problem? Thanks in advance Dan [[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] Two surfaces in one plot with visibility
Hi, I would like plot two surfaces which are each given by vectors x and y, and a matrix m(x,y) representing the z coordinate. With persp() I can plot both, using par(new=TRUE) I can put it in one plot. However, I would like to have the visibility of the surfaces taken into account as if they are solid thin surfaces, so that for example the order of the plot commands does not matter. Any idea how to do that? Thanks a lot, Sebastian signature.asc Description: OpenPGP digital signature __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] MCMCglmm with cross-classified random effects
Dear R-users, I would like to fit a glmm with cross-classified random effects with the function MCMCglmm. Something along the lines: model1-MCMCglmm(response~pred1, random=~re1+re2, data=data) where re1 and re2 should be crossed random effects. I was wondering whether you could tell me specifying cross-classified random effects in MCMCglmm requires a particular syntax? Are there any examples somewhere? I have had a look at the manual and the package vignette, but I have not been able to find any examples relevant to what I want to do. Thanks, Agostino __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] Deleting rows and columns containing NA's and only
Hello, I use read.xls from the gdata package to read in xlsx files. Sometimes these data.frames contain NA columns and rows only. I know how to get rid of those ones but here is the R output of a test data set read in with read.xls t1 A B X D X.1 X.2 1 test 1 NANA 2 NA asdNA asdasdNA 3 NA asdasdNA 4 NANA NA t1[1,2], t1[4,5] and t1[4,6] are NA in text form in the excel sheet. I don't understand why in the first column it is NA while in the last two is not. I basically want to get rid of column 5 and 6 and row 4 as they do not contain any relevant information. If i do a is.na.data.frame(t1): is.na.data.frame(t1) A BX D X.1 X.2 [1,] FALSE FALSE TRUE FALSE TRUE FALSE [2,] TRUE FALSE TRUE FALSE TRUE FALSE [3,] FALSE FALSE TRUE FALSE TRUE FALSE [4,] FALSE FALSE TRUE FALSE TRUE FALSE does not give me the result I hoped to get. It seems that NA and NA are treated as NA but in t1[4,6] it is treated as FALSE because if I do as.character(t1[4,6]) [1] NA one can see that there is a whitespace after NA which is, however, not in the excel sheet for sure. I do not know how to deal with that... Cheers -- View this message in context: http://r.789695.n4.nabble.com/Deleting-rows-and-columns-containing-NA-s-and-only-tp4384173p4384173.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] kernlab - error message: array(0, c(n, p)) : 'dim' specifies too large an array
Ok, I am sorry, My trainingset consists of a 60 x 204 matrix (independent_training – 204 features). I have 60 continuous labels (dependent_training, ranging from 2.25 to 135). this is all the code I use: library(kernlab) rvm(as.matrix(independent_training), dependent_training, type=regression, kernel = vanilladot) On 13.02.2012, at 16:40, David Winsemius wrote: On Feb 13, 2012, at 10:23 AM, Martin Batholdy wrote: Hi, For another trainingset I get this error message, which again is rather cryptic to me: Just imagine how it seems to us! Setting default kernel parameters Error in array(0, c(n, p)) : 'dim' specifies too large an array RMate stopped at line 0 of selection Calls: rvm ... .local - backsolve - as.matrix - chol - diag - array You are on you way to the prize for the greatest number of cryptic (your word) postings in a short interval. (And this one doesn't even have the context of your posting of 8 minutes ago.) thanks for any suggestions! More details about data and code. -- David Winsemius, MD West Hartford, CT __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Writing output into a file
Hi, many thanks for the reply. I really appreciate it. Since I'm still very new to R, I think I should take some time to research what you suggested. (I don't want to keep posting basic questions to the list all the time) But still, thank you so much for being helpful... On Mon, Feb 13, 2012 at 7:23 PM, Rui Barradas rui1...@sapo.pt wrote: Hello, I tried writing this data into a file using the save(myList, file=test1.bin) command, but unfortunately, the numerical values seem to get garbled when I do so. The numbers in my RGui look like 0, 0.5, 0, 1 etc. etc. But when I stored it into a .bin file, and retrieved it using java code, it returns data such as, The problem should be in the use of java, 'save' uses a R format , RDA. You can use 'ascii=TRUE'and see it with a text editor. Also see ?save I also tried the second method (using a # Open a file connection) Unfortunately, here too the data gets extremely garbled. Don't understand why, check the output file with a text editor and let us know what is wrong. The problem I've seen is that the use of 'strsplit' coerses the numeric data to character, but this is easy to solve. Does your list have sub-lists? Rui Barradas -- View this message in context: http://r.789695.n4.nabble.com/Writing-output-into-a-file-tp4382243p4383741.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. -- Best Regards, Suranga [[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] Two surfaces in one plot with visibility
On 13/02/2012 9:24 AM, Sebastian Schubert wrote: Hi, I would like plot two surfaces which are each given by vectors x and y, and a matrix m(x,y) representing the z coordinate. With persp() I can plot both, using par(new=TRUE) I can put it in one plot. However, I would like to have the visibility of the surfaces taken into account as if they are solid thin surfaces, so that for example the order of the plot commands does not matter. Any idea how to do that? That's really hard in persp(). You'd have to plot the facets of the surfaces from back to front, and there's no easy way to do that. I'd recommend using rgl::persp3d, where your graphics hardware will do the computations of which surface is in front. 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] how to plot a nice legend?
Jonas,
A relatively simple way to get the legend to plot beside a graph is to use
the layout function to create two plot areas, create the graph in the
first area and the legend in the second area. You can fool around with the
location of the legend by changing the parameters to layout:
layout(matrix(c(1,2,1,3),nrow=2,ncol=2,byrow=TRUE),widths=c(8,4),heights=c(5,10));
plot(1:10,1:10,lty=1); #main graph
plot(c(0,1),c(0,1),type='n',bty='n',ann=FALSE,xaxt='n',yaxt='n');
legend('topleft',c('txt 1','txt 2'),pch=c(21,22));
Note that you can use 'layout.show(n=3)' (no quotes, of course) after
calling layout above to let you see the 3 plot areas (you only use the
1st two in the example above). You can spcify the plot widths/heights in
absolute units in layout using lcm().
Hope this helps.
Buck Stockhausen
***
* Dr. William T. Stockhausen *
***
* Resource Ecology and Fisheries Management *
* Alaska Fisheries Science Center *
* National Marine Fisheries Service *
* National Oceanic and Atmospheric Administration *
* 7600 Sand Point Way N.E.*
* Seattle, Washington 98115-6349 *
***
* email: william.stockhau...@noaa.gov *
* voice: 206-526-4241 fax: 206-526-6723 *
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***
All models are wrong, some are useful.--G.E.P. Box
Beware of geeks bearing equations.--W. Buffett
***
Disclaimer: The opinions expressed above are personal
and do not necessarily reflect official NOAA policy.
On Sun, Feb 12, 2012 at 4:07 PM, Duncan Murdoch murdoch.dun...@gmail.comwrote:
On 12-02-11 9:07 PM, Jonas Stein wrote:
Wrong? Nothing. You told R to put the legend at c(1,3)
so it did. If you want it elsewhere you need to specify that.
legend(-1,3, legend=c(one, two), inset=-1, xpd=NA)
maybe, or some other location?
ok that works fine. Now i understand how to use it.
If i create several plots it would be nice if all legends would
have the same distance to plots with different scaling.
Can the legend be placed vertically centered, 1cm right to the plot aera?
I'm not sure what position you mean, but legend(top, ...) puts it on the
top edge of the plot, and the inset argument gives detailed positioning.
The units aren't cm, but the grDevices package (or is it grid?) has
functions to convert between units.
Duncan Murdoch
Can i include a legend like this in a standard plot like
plot(1:10) too?
http://www.r-bloggers.com/wp-**content/uploads/2011/03/**heatmap.pnghttp://www.r-bloggers.com/wp-content/uploads/2011/03/heatmap.png
Yes.
What part of that do you want to duplicate?
The coloured squares.
for the reader who got to this article and had the same question:
I have just found another nice solution for a colour legend a minute ago
http://www.r-bloggers.com/**rethinking-loess-for-binomial-**
response-pitch-fx-strike-zone-**maps/http://www.r-bloggers.com/rethinking-loess-for-binomial-response-pitch-fx-strike-zone-maps/
You can specify colors, symbols, labels, etc. in legend().
can i even invent my own symbols?
Also, please link to the original blog post, not just the figure, so that
the author gets some credit and we can see the code used.
sure
http://www.r-bloggers.com/**ggheat-a-ggplot2-style-**heatmap-function/http://www.r-bloggers.com/ggheat-a-ggplot2-style-heatmap-function/
kind regards,
__**
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Re: [R] multi-regression with more than 50 independent variables
You need to spend some time reading about multiple regression. In statistics there is always what is possible and what is advisable. I'm not going to address whether a regression of 57 independent variables is advisable, only possible. For your data, it is not possible. The attached data contain only 13 observations so the maximum number of independent variables you can use is 13. Consider the following example: example - data.frame(y=rnorm(3), x1=rnorm(3), x2=rnorm(3), x3=rnorm(3)) lm(y~x1 + x2, example) lm(y~x1 + x2 + x3, example) We create four variables using random normal numbers for 3 cases (rows). The first regression (2 independent variables works (i.e. there are no NA's). The second produces an NA for the third independent variable. In my example, the three random variables are not correlated with one another. In your data there must be correlations among the 57 variables so that you are only getting slope values for 11. -- David L Carlson Associate Professor of Anthropology Texas AM University College Station, TX 77843-4352 -Original Message- From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On Behalf Of R DF Sent: Monday, February 13, 2012 9:19 AM To: r-help@r-project.org Subject: [R] multi-regression with more than 50 independent variables Hi R Users, I am going to run a multiple linear regression with around 57 independent variables. Each time I run the model with just 11 variables, the results are reasonable. With increasing the number of independent variables more than 11, the coefficients will get NA in the output. Is there any limitation for the number of independent variables in multiple linear regressions in R? I attached my dataset as well as R codes below: mlr.data- read.table(./multiple.txt,header=T) mlr.output- lm(formula = CaV ~ SHG + TrD+ CrH+ SPAD+ FlN+ FrN+ YT+ LA+ LDMP+B+Cu+ Zn+ Mn + Fe+ K + P+ N +Clay30 +Silt30 +Sand30 +Clay60 +Silt60 +Sand60 +ESP30 +NaEx30+ CEC30+Cl30+ SAR30 +KSol30+ NaSol30 +CaMgSol3 +ZnAv30 +FeAv30 +OC30 +PAv30 +KAv30 +TNV30+ pH30+ EC30 +SP30 +ESP60 +NaEx60 +CEC60 +Cl60 +SAR60 +KSol60 +NaSol60 +CaMgSol6 +ZnAv60+FeAv60 +OC60 +PAv60 +KAv60 +TNV60 +pH60 + EC60 +SP60, data=mlr.data) summary (mlr.output) Regards, Reza __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] non linear quantile regression - Median not plotting where it should
Dan, It is hard to say without being able to reproduce your example. If you send me the data I could try to advise something. Roger url:www.econ.uiuc.edu/~rogerRoger Koenker emailrkoen...@uiuc.eduDepartment of Economics vox: 217-333-4558University of Illinois fax: 217-244-6678Urbana, IL 61801 On Feb 13, 2012, at 7:50 AM, Dan Morovitz wrote: Hi, I'm attempting to calculate the 0.25 and 0.97 quantiles for tree height (0-50 meters) against tree age (0-300 years) and I am running into some difficulty with the plotted grafic. I've run the examples in the quantreg help and can get those to work properly and by plugging in my data I can also get the lines plotted on my dataset. Unfortunately I'm running into a problem with the median and other regression lines with tree age younger than 50 years, basically the median is in this range overestimated and even comes out of the rage of oberservations. here is the code I'm using. # then fit the median using nlrq spruce.nlrq - nlrq(height ~ SSlogis(age, Asym, mid, scal),data=spruce, tau=0.5, trace=TRUE) lines(1:200, predict(spruce.nlrq, newdata=list(age=1:200)), col=2) I believe this has something to do with the SSlogis, as this gives the parameters for an S curve. My data set does not have the typical S curve shape, instead you could image it as starting at the inflection point of an S curve. This is what I expect the .025 quantile to be similar to: x - seq(1,100,1) plot(log(x)) the 0,975 should also have a logarithmic shape but no such a steep incline. Ive tried using different self starting models as found under: apropos(^ss) however I have not gotten them to work, or at least to fix the problem. Curves similar to mine look like these site index curves: http://www.extension.umn.edu/distribution/naturalresources/components/3473-13.html In the example from the nlrq help the lines of the median and the various quantiles all start from the same location, tha is basically (x=0, y=0) in the coodinate plane. With my problem, the lines start to be drawn from various different positions ( the lines always start at age(x)=0, but the height(y) can range between 0 and 15). Additionally, the data set is quite large. with about 50,000 oberservations on age and height. Does anyone have a suggestion on how to fix this problem? Thanks in advance Dan [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Suggests/Enhances or .onLoad for S3-methods for optional package
Hi, I am developing a package B that, among other things, also offers some extra S3-methods for functions in package A if the user has installed A. I do not want to list A under Depends of B, as the dependency list of A is rather long, and most potential users of B will not be interested in package A. Unfortunately I struggle with doing this right. So far I have listed A under Suggests, and had a .onLoad function in B with if (require(A)) registerS3methods(newMethodsMatrix, package = A, env = environment(B)) But starting with R 2.13 or R 2.14, R CMD check complain about this require call in .onLoad, referring to the good practice-section of .onAttach. I guess this could also be a problem when uploading the package to CRAN, so I am trying to find another solution. So far I have tried: List A under Suggest of B, with a conditional import in NAMESPACE. If I build a Windows-binary from this when A is installed, this package can be installed but not loaded on computers where A is not installed. List A under Enhances of B. This seems to be the right thing, as the R extensions manual says: the ‘Enhances’ field lists packages “enhanced” by the package at hand, e.g., by providing methods for classes from these packages. However, although it seems I can install and load package B when I conditionally import package A in the NAMESPACE, R CMD check stops with the error: Namespace dependency not required: A If I remove the import, R CMD check is happier, but I cannot load the package after installing. I have read about the use of Suggest, Enhances etc in Writing R Extensions, but could not figure out the right way to do this. I am sure there is something I am missing here. Thanks, Jon -- Jon Olav Skøien Joint Research Centre - European Commission Institute for Environment and Sustainability (IES) Land Resource Management Unit Via Fermi 2749, TP 440, I-21027 Ispra (VA), ITALY jon.sko...@jrc.ec.europa.eu Tel: +39 0332 789206 Disclaimer: Views expressed in this email are those of the individual and do not necessarily represent official views of the European Commission. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] kernlab - error message: array(0, c(n, p)) : 'dim' specifies too large an array
Hi, On Mon, Feb 13, 2012 at 10:53 AM, Martin Batholdy batho...@googlemail.com wrote: Ok, I am sorry, My trainingset consists of a 60 x 204 matrix (independent_training – 204 features). I have 60 continuous labels (dependent_training, ranging from 2.25 to 135). this is all the code I use: library(kernlab) rvm(as.matrix(independent_training), dependent_training, type=regression, kernel = vanilladot) Can you call `traceback()` after you get the error to see if you can follow the code path that results in the explosion? Downloading the kernlab src package will be helpful while your smoking out the error so you can look at the entire source code, too. In my .Rprofile, I actually have something like so: options(error=utils:::dum.frames) Which allows me to call `debugger()` after an error is thrown and drops me into the location that threw the error (most of the time (I think)), allowing me to poke around and see who's who, and what's what. HTH, -steve -- Steve Lianoglou Graduate Student: Computational Systems Biology | Memorial Sloan-Kettering Cancer Center | Weill Medical College of Cornell University Contact Info: http://cbio.mskcc.org/~lianos/contact __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] kernlab - error message: array(0, c(n, p)) : 'dim' specifies too large an array
Sorry, this: options(error=utils:::dum.frames) Should be: options(error=utils:::dump.frames) -steve -- Steve Lianoglou Graduate Student: Computational Systems Biology | Memorial Sloan-Kettering Cancer Center | Weill Medical College of Cornell University Contact Info: http://cbio.mskcc.org/~lianos/contact __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Two surfaces in one plot with visibility
Hi! On 13/02/12 17:15, Duncan Murdoch wrote: I would like plot two surfaces which are each given by vectors x and y, and a matrix m(x,y) representing the z coordinate. [..] I would like to have the visibility of the surfaces taken into account as if they are solid thin surfaces, so that for example the order of the plot commands does not matter. That's really hard in persp(). You'd have to plot the facets of the surfaces from back to front, and there's no easy way to do that. I'd recommend using rgl::persp3d, where your graphics hardware will do the computations of which surface is in front. Thanks a lot! Works really well and created a wow effect (especially when comparing to the manual tuning of the theta and phi angles in persp() ). Great job! Sebastian signature.asc Description: OpenPGP digital signature __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help 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 import time-series data
On Mon, Feb 13, 2012 at 8:23 AM, RichardSmith richardsmith...@gmail.com wrote: Gabor Grothendieck wrote Try this xyplot.zoo lattice graph. Time series are represented in columns so we transpose the data and convert it to zoo. The screen= argument available in xyplot.zoo groups series into panels: Lines - plant,aphid,1,2,3,4 pumpkin,1-2,0.065566,0.057844,0.08,0.086879 pumpkin,1-3,0.107612,0.097272,0.11663,0.160499 squash,1-4,0.126939,0.115003,0.140275,0.188829 library(zoo) library(lattice) DF - read.csv(text = Lines) z - zoo(t(DF[3:6])) colnames(z) - DF$aphid xyplot(z, screen = DF$plant) Thank you! That gets the data in exactly the shape I was expecting. I was hoping to get the graph showing all lines on one plot, but coloured according to plant. I tried to change screen=DF$plant for groups=DF$plant, but it doesn't work, and I can't figure out from the documentation why it doesn't work (I think I need to more thoroughly understand data types first). Could you point me in the right direction? 1. Try this which uses lattice zoo graphics: xyplot(z, screen = 1, col = DF$plant) or with a legend: key - list(space = top, text = levels(DF$plant), points = FALSE, lines = TRUE, col = 1:nlevels(DF$plant)) xyplot(z, screen = 1, col = DF$plant, auto.key = key) 2. or using classic zoo graphics plot(z, screen = 1, col = DF$plant) To add a legend: legend(topleft, legend = levels(DF$plant), col = 1:nlevels(DF$plant), lty = 1) -- Statistics Software Consulting GKX Group, GKX Associates Inc. tel: 1-877-GKX-GROUP email: ggrothendieck at gmail.com __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Deleting rows and columns containing NA's and only
On Mon, Feb 13, 2012 at 07:48:11AM -0800, syrvn wrote: Hello, I use read.xls from the gdata package to read in xlsx files. Sometimes these data.frames contain NA columns and rows only. I know how to get rid of those ones but here is the R output of a test data set read in with read.xls t1 A B X D X.1 X.2 1 test 1 NANA 2 NA asdNA asdasdNA 3 NA asdasdNA 4 NANA NA t1[1,2], t1[4,5] and t1[4,6] are NA in text form in the excel sheet. I don't understand why in the first column it is NA while in the last two is not. I basically want to get rid of column 5 and 6 and row 4 as they do not contain any relevant information. If i do a is.na.data.frame(t1): is.na.data.frame(t1) A BX D X.1 X.2 [1,] FALSE FALSE TRUE FALSE TRUE FALSE [2,] TRUE FALSE TRUE FALSE TRUE FALSE [3,] FALSE FALSE TRUE FALSE TRUE FALSE [4,] FALSE FALSE TRUE FALSE TRUE FALSE does not give me the result I hoped to get. It seems that NA and NA are treated as NA but in t1[4,6] it is treated as FALSE because if I do as.character(t1[4,6]) [1] NA Hi. I do not know, how NA appeared, however, it is possible to change them to real NA as follows. # some data frame df - structure(list(a = c(NA, 2L, 3L, 4L), b = c(a, NA, c, NA ), c = structure(c(1L, 2L, NA, 4L), .Label = c(e, f, g, h), class = factor)), .Names = c(a, b, c), row.names = c(NA, -4L), class = data.frame) df abc 1 NAae 2 2 NAf 3 3c NA 4 4 NA h df[4, 2] # this is not NA, but NA [1] NA # replace all NA by NA in column 2 df[which(df[,2] == NA ), 2] - NA df abc 1 NAae 2 2 NAf 3 3c NA 4 4 NAh Hope this helps. Petr Savicky. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Error in apply(x2, 1, diff) : dim(X) must have a positive length
What package is the bootstrap() function from? There are many
functions by that name
My hunch is that it takes a vector at a time and puts it through
dif.mns so that leads to the error in the apply() call but I can't
verify.
Michael
On Mon, Feb 13, 2012 at 10:34 AM, hithit168 ccchri...@live.com wrote:
Thank you for your help, David. I was trying to run bootrapping on the
dataset shoes from MASS package. But I still have some problem here.
as.matrix(data.frame(shoes),nrows=10,ncols=2,byrow=T)
A B
[1,] 13.2 14.0
[2,] 8.2 8.8
[3,] 10.9 11.2
[4,] 14.3 14.2
[5,] 10.7 11.8
[6,] 6.6 6.4
[7,] 9.5 9.8
[8,] 10.8 11.3
[9,] 8.8 9.3
[10,] 13.3 13.6
y=as.matrix(data.frame(shoes),nrows=10,ncols=2,byrow=T)
class(y)
[1] matrix
apply(y,1,diff)
[1] 0.8 0.6 0.3 -0.1 1.1 -0.2 0.3 0.5 0.5 0.3
dif.mns - function(x2,tr1=.1,tr2=.2){
+ diffs=apply(x2,1,diff)
+ mn1=mean(diffs)
+ mn2=mean(diffs,tr=.1)
+ mn3=mean(diffs,tr=.2)
+ mn4=median(diffs)
+ mns=c(mn1,mn2,mn3,mn4)
+ list(mnds=round(mns,3))}
dif.mns(y,tr1=.1,tr2=.2)
$mnds
[1] 0.410 0.400 0.417 0.400
bootstrap(y,nboot=1000,theta=dif.mns)
Error in apply(x2, 1, diff) : dim(X) must have a positive length
I am not sure what I am doing wrong. When I run apply(y,1,diff) I am able to
get output, and when I run dif.mns with just the original dataset, I am also
able to get output.
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[R] Including .exe files in an R package
I am in the process of creating a package in R which calls pre-compiled C code i.e. '.exe' files In reading the manual, I came across this: A source package if possible should not contain binary executable files: they are not portable, and a security risk if they are of the appropriate architecture. R CMD check will warn about them2 unless they are listed (one filepath per line) in a file ‘BinaryFiles’ at the top level of the package. Note that CRAN will no longer accept submissions containing binary files even if they are listed. From my understanding, the .exe files are a subset of binary files, which are no longer accepted by CRAN. If this is the case for my situation, is there any other way to include the C code, so that CRAN will accept the package? 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] pairwise comparisons with multcomp package
Hi, I've got this model and following Hothorn et al advices, I used glht for a post hoc comparison modezqM-glm(rojos~estacion*zona3,quasipoisson,subset=(edadysexo==M)) anova(modezqM,test=F) Df Deviance Resid. Df Resid. Dev F Pr(F) NULL293 41148 estacion 1 3647.0 292 37501 23.6250 1.927e-06 *** zona3 2 931.4 290 36569 3.0167 0.050510. estacion:zona3 2 2421.7 28834148 7.84380.000482 *** summary(glht(modezqM,linfct=mcp(zona3=Tukey,interaction_average=T))) Estimate Std. Error z-value Pr(|z|) norte - turist == 0 -0.051820.23701 -0.219 0.9727 occid - turist == 0 -1.059700.41618 -2.546 0.0271 * occid - norte == 0 -1.007880.45830 -2.199 0.0665 . I'm trying to understanding the way this command made pairwise comparisons, but, where does this z-value comes from? I'm suspected this not come from Tukey's test, but i don't know it's Wald or something i've never use I know it is not an habitual doubt. Sorry and thanks in advance I suspect it - Mario Garrido Escudero PhD student Dpto. de Biología Animal, Ecología, Parasitología, Edafología y Qca. Agrícola Universidad de Salamanca -- View this message in context: http://r.789695.n4.nabble.com/pairwise-comparisons-with-multcomp-package-tp4384499p4384499.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] Cumulative density (kernel smoothing)
Hi, in R there is the function density which computes kernel density estimates. Is there a cumulative version of it? Something like they have in Matlab: http://www.mathworks.nl/help/toolbox/stats/ksdensity.html I know there is ecdf, but I'm not sure it's based on kernel density smoothing. Thanks -- View this message in context: http://r.789695.n4.nabble.com/Cumulative-density-kernel-smoothing-tp4384652p4384652.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] Deleting rows and columns containing NA's and only
Hi, thanks for you suggestion. I finally solved it in a different way using apply and is.na for TRUE NA's and if(as.character(x) == NA) etc. However, I just spotted that read.xls seems to have problems reading in special characters such as or . Is there any workaround for that? -- View this message in context: http://r.789695.n4.nabble.com/Deleting-rows-and-columns-containing-NA-s-and-only-tp4384173p4384663.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] Including .exe files in an R package
On 13/02/2012 17:15, sahir bhatnagar wrote: I am in the process of creating a package in R which calls pre-compiled C code i.e. '.exe' files Which are only used on Windows, so presumably you forgot to mention that OS. In reading the manual, I came across this: A source package if possible should not contain binary executable files: they are not portable, and a security risk if they are of the appropriate architecture. R CMD check will warn about them2 unless they are listed (one filepath per line) in a file ‘BinaryFiles’ at the top level of the package. Note that CRAN will no longer accept submissions containing binary files even if they are listed. From my understanding, the .exe files are a subset of binary files, which are no longer accepted by CRAN. If this is the case for my situation, is there any other way to include the C code, so that CRAN will accept the package? CRAN will neither accept packages with binary code files, nor will it accept packages tied to one architecture and OS (as such files would be). You can do what you like provided you do not want to distribute the package: at that point you need to meet the conditions of the license and of the repository. There are several examples of CRAN packages which compile up executables: arulesSequences is one. But note that this is tricky as platforms such as Windows and OS X support multiple architectures in the same package installation. 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. And that does make clear that this is (or would rapidly become) a R-devel topic. -- Brian D. Ripley, rip...@stats.ox.ac.uk Professor of Applied Statistics, http://www.stats.ox.ac.uk/~ripley/ University of Oxford, Tel: +44 1865 272861 (self) 1 South Parks Road, +44 1865 272866 (PA) Oxford OX1 3TG, UKFax: +44 1865 272595 __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Logistic regression with non-gaussian random effects
Hi, does anyone know of an implementation of generalized linear models with random effects, where the random effects are non-gaussian? Actually, what I need is to do a logistic regression (or binomial regression) where the linear predictor in addition to fixed effects and gaussian random effects also has a term b*z where z is a random effect variable with p(z=1)=p(z=-1)=0.5 and b is a parameter of interest. I am very grateful to you for any help or comments. Sincerely, Line __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Error in apply(x2, 1, diff) : dim(X) must have a positive length
Thanks. I use the bootstrap() function from the bootstrap package. Let me try it again to see if I get any luck :) -- View this message in context: http://r.789695.n4.nabble.com/Error-in-apply-x2-1-diff-dim-X-must-have-a-positive-length-tp4382435p4384608.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] non-isomorphic sequences
Dear All,
Sorry for the typoes earlier, let me repost the question.
Suppose I want to generate sequences of length 3 from two symbols {1,2}, we get
the following 8 sequences
1 1 1
1 1 2
1 2 1
1 2 2
2 1 1
2 1 2
2 2 1
2 2 2
However, I do not want all these 8 sequences. I call two sequencs to be
isomorphic if one sequence could be obtained from the other by relabelling the
symbols. For example, 111 is isomorphic to 222, 112 is isomorphic to 221. By
eliminating all these isomorphic ones, what I want is the following
1 1 1
1 1 2
1 2 1
2 1 1
In general, I need to generate non-isomorphic sequences of length p from t
distinct symbols. For example, when p=3, t=3 we have
matrix(c(1,2,3,1,1,2,2,1,1,1,2,1,1,1,1),3,5)
[1,] 1 1 2 1 1
[2,] 2 1 1 2 1
[3,] 3 2 1 1 1
When p=4, t=4 we have
matrix(c(1,2,3,4,1,1,2,3,1,2,1,3,1,2,3,1,2,1,1,3,2,3,1,1,2,1,3,1,1,1,2,2,1,2,1,2,1,2,2,1,1,1,1,2,1,1,2,1,1,2,1,1,2,1,1,1,1,1,1,1),4,15)
[1,] 1 1 1 1 2 2 2 1 1 1 1 1 1
2 1
[2,] 2 1 2 2 1 3 1 1 2 2 1 1 2
1 1
[3,] 3 2 1 3 1 1 3 2 1 2 1 2 1
1 1
[4,] 4 3 3 1 3 1 1 2 2 1 2 1 1
1 1
In general, I need to do this for arbitrary choices of p and t.
Thaks a lot,
Wei
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Re: [R] best subset selection on random effects model
The question is where do your models come from? Passing nested models to ?anova.lme in nlme package or lme4 results in a likelihood ratio test. Are you looking for something else/more ? On Sun, Feb 12, 2012 at 8:02 PM, Tao Zhang zt020...@gmail.com wrote: Hi, I know leaps() computes the best subset selection for linear model, and the bestglm() computes the best subset selection for generalized linear model. Is there any package for best subset selection on random effects model, or mixed effects model? Thank you so much. Tao [[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] Cumulative density (kernel smoothing)
On Feb 13, 2012, at 1:01 PM, francogrex wrote: Hi, in R there is the function density which computes kernel density estimates. Is there a cumulative version of it? Something like they have in Matlab: I'm not aware of one, but you could use `integrate`. You will need to limit your range or use the version in package logspline if your range is fixed. You will also need to account for the fact that the inteegral will only be approximately == 1. Searching the Archives will bring up several similar questions from prior years.\ http://www.mathworks.nl/help/toolbox/stats/ksdensity.html I know there is ecdf, but I'm not sure it's based on kernel density smoothing. Thanks `ecdf` is not based on density. -- David Winsemius, MD West Hartford, CT __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] for loop
Hi guys, This is a very beginner question. Anybody willing to help? for(i in 1:1000) x=29.5 + i/500 y=2x plot(y,x) The idea is to produce 1000 values of x and y then plot them. Cheers, Eddie [[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] for loop
x - 29.5 + (1:1000)/500 y - 2 * x plot(y,x) On Mon, Feb 13, 2012 at 1:34 PM, eddie smith eddie...@gmail.com wrote: Hi guys, This is a very beginner question. Anybody willing to help? for(i in 1:1000) x=29.5 + i/500 y=2x plot(y,x) The idea is to produce 1000 values of x and y then plot them. Cheers, Eddie [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Jim Holtman Data Munger Guru What is the problem that you are trying to solve? Tell me what you want to do, not how you want to do it. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Deleting rows and columns containing NA's and only
On Feb 13, 2012, at 1:05 PM, syrvn wrote: Hi, thanks for you suggestion. I finally solved it in a different way using apply and is.na for TRUE NA's and if(as.character(x) == NA) etc. However, I just spotted that read.xls seems to have problems reading in special characters such as or . Or could it be that you have problems in not reading the help pages carefully? read.table(text=NA) V1 1 NA read.table(text=NA, na.strings=c(NA,NA)) V1 1 NA -- David Winsemius, MD West Hartford, CT __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] meboot - can it handle outliers and missing values?
Hi everyone, I would like to use your meboot package in R in a power simulation study, where meboot stands for Maximum Entropy Bootstrap. In this study, each time series that will be bootstrapped includes both missing values and outliers. Can meboot accomodate these two features, which are the hallmark of many real time series? If yes, I would very much appreciate your thoughts on how this can be achieved in R. Many thanks and kind regards, Isabella Isabella R. Ghement, Ph.D. Ghement Statistical Consulting Company 301-7031 Blundell Road, Richmond, B.C., Canada, V6Y 1J5 Tel: 604-767-1250 Fax: 604-270-3922 E-mail: isabe...@ghement.ca Web: www.ghement.ca [[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] only 0s may be mixed with negative subscripts
I'd like to get the sum of every other row in a data.frame. When I
actually set about doing this, I get the error in the subject line of
this message. A sample of my data is below, followed by the function
call that should give me the results I want:
dput(head(sens2))
structure(list(Time = c(1328565067, 1328565067.05, 1328565067.1,
1328565067.15, 1328565067.2, 1328565067.25), Y = c(0.0963890795246276,
0.227296347215609, 0.240972698811569, 0.221208948983498, 0.230898231782485,
0.203282153087549), X = c(0.0245045248243853, 0.0835679411703579,
0.0612613120609633, 0.058568910563872, 0.0511868450318788, 0.0557714205674231
), rownumber = 1:6), .Names = c(Time, Y, X, rownumber
), row.names = c(NA, 6L), class = data.frame)
speedX - sapply(sens2[sens2$rownumber %% 2 == 0,], function(row) {
cumsum(c(sens2[row+1,3], sens2[row,3]))}, simplify=TRUE)
Error in xj[i] : only 0's may be mixed with negative subscripts
Help?
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Re: [R] Getting codebook data into R
This is how I get a whole SPSS data files into R. You specifically asked about the codebook, so this may not be exactly what you are after. spssFileInfo - spss.system.file ( file = path to my SPSS file ) spssDataSet - as.data.set ( spssFileInfo) spssDataFrame - as.data.frame ( spssDataSet ) (Not tested. Adapted from working code.) memisc documentation has more info about doing this and how it works. eRic - Original message - From: barny garyb.dav...@btinternet.com To: r-help@r-project.org Date: Sat, 11 Feb 2012 10:04:16 -0800 (PST) Subject: Re: [R] Getting codebook data into R Hi Eric - after seeing the difficulty of inputting this kind of data into R I decided to use your method. It was rather painless using PSPP to do what I wanted - however, how do I now create an SPSS file and then use the memisc package to read it in? -- View this message in context: http://r.789695.n4.nabble.com/Getting-codebook-data-into-R-tp4374331p4379433.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Deleting rows and columns containing NA's and only
Hi David, I am using read.xls not read.table. -- View this message in context: http://r.789695.n4.nabble.com/Deleting-rows-and-columns-containing-NA-s-and-only-tp4384173p4384866.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] Puzzling... puzzling... puzzling...
Hi all, I made sure that it's env$sRes1$nPositionsOptimizedM that's correct... not the env$sRes1$nPositionsOptimized... But it seems both point to the same memory area... This is very dangerous because I have used naming conventions such as: MyLongVariableNameForA MyLongVariableNameForB MyLongVariableNameForC ... ... Then if internally they are actually the same thing then all my programs messed up... Any thoughts? env=new.env() load(MyResults.rData, env) identical(env$sRes1$nPositionsOptimized, env$sRes1$nPositionsOptimizedM) [1] TRUE [[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] Deleting rows and columns containing NA's and only
On Feb 13, 2012, at 1:57 PM, syrvn wrote: Hi David, I am using read.xls not read.table. Please read the help page for read.xls more carefully. -- View this message in context: http://r.789695.n4.nabble.com/Deleting-rows-and-columns-containing-NA-s-and-only-tp4384173p4384866.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. David Winsemius, MD West Hartford, CT __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] for loop
In addition to Jim's neat solution (see also below),
some comments on your original code.
Your for loop executes x=29.5 + i/500 100 times,
producing a single value each time and replacing the
previous value which was in x. So, at the end of the loop,
you have a single value of x. Then you compute y=2x;
that, as it stands, would prokoke an error:
Error: unexpected symbol in y=2x
since variable names cannort start with a digit. You
need, of course, the mutltiplaction operator * as
in Jim's y - 2 * x.
The scope of your for loop (i.e. the set of commands
that is executed for each round of the loop) is solely
the command x=29.5 + i/500. The y - 2 * x is not
part of the scope of the loop, and would only be executed
once, when the loop was finished. You would need
for(i in 1:1000) {
commands
}
to cause the execution of several commands in each round
of the loop.
Finally, even if you did think that your entire series of
commands (re-written):
for(i in 1:1000)
x=29.5 + i/500
y=2*x
plot(y,x)
would all be executed (down to and including the plot() command)
in each round of the loop, nevertheless each call to plot()
creates a new graph, discarding the previous one, so only a single
point would be plotted each time.
The solution (as in Jim's suggestion) is to create the full vector
of x-values and y-values, and then use plot(x,y) where x and y are
now vectors.
There are all sorts of little details about how R puts things
together, which will become familiar as you use R. However,
you do need to get hold of the basics of how R operates, so
I would suggest having R for Beginners
http://cran.r-project.org/doc/contrib/Paradis-rdebuts_en.pdf
to hand while you learn R. It is very good about how the basics
work. The next step up would be the more systematic exposition
of how R works in An Introduction to R:
http://cran.r-project.org/doc/manuals/R-intro.html
http://cran.r-project.org/doc/manuals/R-intro.pdf
Hoping this helps!
Ted.
On 13-Feb-2012 jim holtman wrote:
x - 29.5 + (1:1000)/500
y - 2 * x
plot(y,x)
On Mon, Feb 13, 2012 at 1:34 PM, eddie smith eddie...@gmail.com wrote:
Hi guys,
This is a very beginner question. Anybody willing to help?
for(i in 1:1000)
x=29.5 + i/500
y=2x
plot(y,x)
The idea is to produce 1000 values of x and y then plot them.
Cheers,
Eddie
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--
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Data Munger Guru
What is the problem that you are trying to solve?
Tell me what you want to do, not how you want to do it.
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-
E-Mail: (Ted Harding) ted.hard...@wlandres.net
Date: 13-Feb-2012 Time: 19:12:47
This message was sent by XFMail
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Re: [R] Puzzling... puzzling... puzzling...
On 13/02/2012 2:00 PM, Michael wrote: Hi all, I made sure that it's env$sRes1$nPositionsOptimizedM that's correct... not the env$sRes1$nPositionsOptimized... But it seems both point to the same memory area... How did you determine that? The test below just shows that they contain the same thing. This is very dangerous because I have used naming conventions such as: MyLongVariableNameForA MyLongVariableNameForB MyLongVariableNameForC ... ... Then if internally they are actually the same thing then all my programs messed up... Any thoughts? env=new.env() load(MyResults.rData, env) identical(env$sRes1$nPositionsOptimized, env$sRes1$nPositionsOptimizedM) [1] TRUE Changing one of them and seeing the other one change would show that they point to the same memory area. This can happen with environments: if you create env1 and set env2 - env1, then changes to either environment will affect the other, because that's how environments work. That's not true of most of the other kinds of objects in R. (The exceptions are fairly exotic things that you are unlikely to use.) 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] R's AIC values differ from published values
Using the Cement hardening data in Anderson (2008) Model Based Inference in the Life Sciences. A Primer on Evidence, and working with the best model which is lm ( y ~ x1 + x2,data = cement ) the AIC value from R is model-lm ( formula = y ~ x1 + x2 , data = cement ) AIC ( model ) 64.312 which can be converted to AICc by adding the bias correction factor 2*K*(K+1)/(n-K-1) to give the AICc value of 69.312 (addition of 5, where n=13 and K=4). This same value, 69.31, can be obtained using R package AICcmodavg library ( AICcmodavg ) data (cement) cement Cand.models - list( ) Cand.models[[1]] - lm ( y ~ x1 + x2,data = cement ) Cand.models[[2]] - lm ( y~ x3 + x4, data = cement ) Cand.models[[3]] - lm ( y ~ x1 + x2 + x1 * x2, data = cement ) Cand.models[[4]] - lm ( y ~ x3 + x4 + x3 * x4, data = cement ) ## vector of model names Modnames - paste(MODEL, 1:4, sep= ) ## AICc aictab ( cand.set = Cand.models, modnames = Modnames ) However, the AICc value reported by Anderson (2008) is 32.41. The AICc value obtained using RSS value (i.e., calculating AICc manually from the output of linear regression) is 32.41. Thanks for any help. David New R user, minimal familiarity with statistics. [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Question scatterplot axis cut point
Hi everybody, i made a scatterplot using the command plot (datafile1, xlim=c(0,10), ylim=c(0.001, 1), log=y, xlab=x Achse, ylab=y Achse, frame.plot=FALSE, axes = TRUE). Now i have a problem. There is a gap between the x and the y axis. I want that the x and y axis cut at 0 and 0.001 without having this gap. Is this possible? thanks for your help! nice greetings michi __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help 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's AIC values differ from published values
This is answered in ?AIC. Have you read it? -- Bert On Mon, Feb 13, 2012 at 10:22 AM, david hamer j.david.ha...@gmail.com wrote: Using the Cement hardening data in Anderson (2008) Model Based Inference in the Life Sciences. A Primer on Evidence, and working with the best model which is lm ( y ~ x1 + x2, data = cement ) the AIC value from R is model - lm ( formula = y ~ x1 + x2 , data = cement ) AIC ( model ) 64.312 which can be converted to AICc by adding the bias correction factor 2*K*(K+1)/(n-K-1) to give the AICc value of 69.312 (addition of 5, where n=13 and K=4). This same value, 69.31, can be obtained using R package AICcmodavg library ( AICcmodavg ) data (cement) cement Cand.models - list( ) Cand.models[[1]] - lm ( y ~ x1 + x2, data = cement ) Cand.models[[2]] - lm ( y ~ x3 + x4, data = cement ) Cand.models[[3]] - lm ( y ~ x1 + x2 + x1 * x2, data = cement ) Cand.models[[4]] - lm ( y ~ x3 + x4 + x3 * x4, data = cement ) ## vector of model names Modnames - paste(MODEL, 1:4, sep= ) ## AICc aictab ( cand.set = Cand.models, modnames = Modnames ) However, the AICc value reported by Anderson (2008) is 32.41. The AICc value obtained using RSS value (i.e., calculating AICc manually from the output of linear regression) is 32.41. Thanks for any help. David New R user, minimal familiarity with statistics. [[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. -- Bert Gunter Genentech Nonclinical Biostatistics Internal Contact Info: Phone: 467-7374 Website: http://pharmadevelopment.roche.com/index/pdb/pdb-functional-groups/pdb-biostatistics/pdb-ncb-home.htm __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Puzzling... puzzling... puzzling...
Replace the syntax List$Name with List[[Name]] and see if things work better. '[[' does not do the partial matching that '$' does. E.g., x - list(AB=10, BC=20, CD=30) x$A # returns 10 because A is the initial part of exactly one name in x, AB x[[A]] # returns NULL However, if you have y - list(AB=1, AC=2, AD=3) then y$A will return NULL because there is not a unique partial match to A among the names of y. Bill Dunlap Spotfire, TIBCO Software wdunlap tibco.com -Original Message- From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On Behalf Of Michael Sent: Monday, February 13, 2012 11:00 AM To: r-help Subject: [R] Puzzling... puzzling... puzzling... Hi all, I made sure that it's env$sRes1$nPositionsOptimizedM that's correct... not the env$sRes1$nPositionsOptimized... But it seems both point to the same memory area... This is very dangerous because I have used naming conventions such as: MyLongVariableNameForA MyLongVariableNameForB MyLongVariableNameForC ... ... Then if internally they are actually the same thing then all my programs messed up... Any thoughts? env=new.env() load(MyResults.rData, env) identical(env$sRes1$nPositionsOptimized, env$sRes1$nPositionsOptimizedM) [1] TRUE [[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] R's AIC values differ from published values
hi: the definition of AIC can vary a lot from paper to paper and textbook to textbook because some people keep the multiplicative constants and other's don't. all that matters when using AIC is COMPARISON. the value itself means nothing. So, you'll be fine no matter what you use as long as you're consistent. On Mon, Feb 13, 2012 at 1:22 PM, david hamer j.david.ha...@gmail.comwrote: Using the Cement hardening data in Anderson (2008) Model Based Inference in the Life Sciences. A Primer on Evidence, and working with the best model which is lm ( y ~ x1 + x2,data = cement ) the AIC value from R is model-lm ( formula = y ~ x1 + x2 , data = cement ) AIC ( model ) 64.312 which can be converted to AICc by adding the bias correction factor 2*K*(K+1)/(n-K-1) to give the AICc value of 69.312 (addition of 5, where n=13 and K=4). This same value, 69.31, can be obtained using R package AICcmodavg library ( AICcmodavg ) data (cement) cement Cand.models - list( ) Cand.models[[1]] - lm ( y ~ x1 + x2,data = cement ) Cand.models[[2]] - lm ( y~ x3 + x4, data = cement ) Cand.models[[3]] - lm ( y ~ x1 + x2 + x1 * x2, data = cement ) Cand.models[[4]] - lm ( y ~ x3 + x4 + x3 * x4, data = cement ) ## vector of model names Modnames - paste(MODEL, 1:4, sep= ) ## AICc aictab ( cand.set = Cand.models, modnames = Modnames ) However, the AICc value reported by Anderson (2008) is 32.41. The AICc value obtained using RSS value (i.e., calculating AICc manually from the output of linear regression) is 32.41. Thanks for any help. David New R user, minimal familiarity with statistics. [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] nice report generator?
On 06/02/2012 4:12 PM, Hadley Wickham wrote: 2. It's more flexible to construct the language object as a language object, rather than pasting something together and parsing it. For one thing, that allows non-syntactic variable names; I think it's also easier to read. So your code txt- paste(tabular(value*v*, LEFT , ~ ,RIGHT ,, data = m_xx, suppressLabels = 2,...), sep = ) eval(parse(text = txt )) could be rewritten as formula- substitute( value*v*LEFT ~ RIGHT, list(LEFT=LEFT, RIGHT=RIGHT)) tabular(formula, data = m_xx, suppressLabels = 2, ...) To be strictly correct, shouldn't that be: formula- eval(substitute( value*v*LEFT ~ RIGHT, list(LEFT=LEFT, RIGHT=RIGHT))) ? It might make sense to put something like this into the tables package, but I don't want to have a dependency on reshape. Would you consider making tabular generic? I have now made tabular() into a generic function, but because of the problems at R-forge, can't commit the changes immediately. The old tabular() function is now the tabular.formula() method; the default method tries to coerce the object to a formula to call that. I think both my suggestion and yours would likely have problems in the new system (as they did in the old one) because the environment associated with the formula would be wrong. It's a little tricky, but now tabular() works a lot more like model.frame(), which I think has to be considered to be the standard way to do 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.
Re: [R] non-isomorphic sequences
On Mon, Feb 13, 2012 at 10:05:02AM -0800, zheng wei wrote:
Dear All,
Sorry for the typoes earlier, let me repost the question.
Suppose I want to generate sequences of length 3 from two symbols {1,2}, we
get the following 8 sequences
1 1 1
1 1 2
1 2 1
1 2 2
2 1 1
2 1 2
2 2 1
2 2 2
However, I do not want all these 8 sequences. I call two sequencs to be
isomorphic if one sequence could be obtained from the other by relabelling
the symbols. For example, 111 is isomorphic to 222, 112 is isomorphic to
221.?By eliminating all these isomorphic ones, what I want is the following
1 1 1
1 1 2
1 2 1
2 1 1
Eliminating isomorphic sequences may be done differently,
if we select different representatives of each equivalence
class. The following also eliminates isomorphic 1,2 sequences
1 1 1
1 1 2
1 2 1
1 2 2
Is this solution OK?
In general, I need to generate non-isomorphic sequences of length p from t
distinct symbols. For example, when p=3, t=3 we have
matrix(c(1,2,3,1,1,2,2,1,1,1,2,1,1,1,1),3,5)
[1,]??? 1??? 1??? 2??? 1??? 1
[2,]??? 2??? 1??? 1??? 2??? 1
[3,]??? 3??? 2??? 1??? 1??? 1
When p=4, t=4 we have
matrix(c(1,2,3,4,1,1,2,3,1,2,1,3,1,2,3,1,2,1,1,3,2,3,1,1,2,1,3,1,1,1,2,2,1,2,1,2,1,2,2,1,1,1,1,2,1,1,2,1,1,2,1,1,2,1,1,1,1,1,1,1),4,15)
[1,]??? 1??? 1??? 1??? 1??? 2??? 2??? 2??? 1??? 1 1 1 1 1
2 1
[2,]??? 2??? 1??? 2??? 2??? 1??? 3??? 1??? 1??? 2 2 1 1 2
1 1
[3,]??? 3??? 2??? 1??? 3??? 1??? 1??? 3??? 2??? 1 2 1 2 1
1 1
[4,]??? 4??? 3??? 3??? 1??? 3??? 1??? 1??? 2??? 2 1 2 1 1
1 1
In general, I need to do this for arbitrary choices of p and t.
If p and t are not too large, try the following
check.row - function(x)
{
y - unique(x)
all(y == seq.int(along=y))
}
p - 4
tt - 4 # do not rewrite t() for transpose
elem - lapply(as.list(pmin(1:p, tt)), function(x) seq.int(length=x))
a - as.matrix(rev(expand.grid(rev(elem
ok - apply(a, 1, check.row)
out - a[ok, ]
out
Var4 Var3 Var2 Var1
[1,]1111
[2,]1112
[3,]1121
[4,]1122
[5,]1123
[6,]1211
[7,]1212
[8,]1213
[9,]1221
[10,]1222
[11,]1223
[12,]1231
[13,]1232
[14,]1233
[15,]1234
This solution differs from yours, for example, in
the row c(1, 2, 3, 3), which is in your solution
represented by c(2, 3, 1, 1). This a different choice
of the representatives. Is the choice important?
A related thread started at
https://stat.ethz.ch/pipermail/r-help/2012-January/301489.html
There was an additional requirement that each of t symbols
has at least one occurrence.
Hope this helps.
Petr Savicky.
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[R] Singling out observations
Greetings I am attempting to plot observations of a cave aquatic invertebrate dating from 1901-2004. I can come up with a nice lattice plot of the eight sites from which I have data easily enough. However, I'd like to be able to highlight the 0 observations on the plots, i.e., attempts to find it at the site were unsuccessful. I'd like to be able to highlight these observations with either a symbol or a color. Here's a sample of the data (Reach=site): (Embedded image moved to file: pic17437.jpg) Here's my code for the plot: xyplot(No.Invert~Year|Reach, data=Invert.Coll.1901.2004, layout=c(4,2), ylab=No. Invert 'collected', scales=list(x=list(rot=45))) *** Kurt Lewis Helf, Ph.D. Ecologist EEO Counselor National Park Service Cumberland Piedmont Network P.O. Box 8 OR NPS Warehouse 61 Maintenance Rd. Mammoth Cave, KY 42259 Ph: 270-758-2163 Lab: 270-758-2151 Fax: 270-758-2609 Cumberland Piedmont Network (CUPN) Homepage: http://tiny.cc/e7cdx *** Science, in constantly seeking real explanations, reveals the true majesty of our world in all its complexity. -Richard Dawkins [Scientific] theories are passed on not as dogmas but rather with the challenge to discuss them and improve upon them. -Karl Popper ...consider yourself a guest in the home of other creatures as significant as yourself. -Wayside at Wilderness Threshold in McKittrick Canyon, Guadalupe Mountains National Park, TX CUPN Forest Pest Monitoring Website: http://bit.ly/9rhUZQ CUPN Cave Cricket Monitoring Website: http://tiny.cc/ntcql CUPN Cave Aquatic Biota Monitoring Website: http://tiny.cc/n2z1o__ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] access R basic libraries
Hello Everyone, I am trying to access all R basic C/Fotran libraries to see how some functions are written in fotran or C so that in case I need to change some stuff I can do that. Any help would be great, nitin [[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] MCMCglmm with cross-classified random effects
Agostino Moro agostino.moro99 at gmail.com writes: I would like to fit a glmm with cross-classified random effects with the function MCMCglmm. Something along the lines: model1-MCMCglmm(response~pred1, random=~re1+re2, data=data) where re1 and re2 should be crossed random effects. [snip to make gmane happy. Thank you for trying to find the answer before posting ...] It would probably be best to post this one to the r-sig-mixed-models at r-project.org mailing list instead ... Ben Bolker __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] access R basic libraries
On 13-02-2012, at 21:47, nitin kumar wrote: Hello Everyone, I am trying to access all R basic C/Fotran libraries to see how some functions are written in fotran or C so that in case I need to change some stuff I can do that. BTW; it's fortran. Well, goto CRAN (http://cran.r-project.org/) and download the sources (link R-2.14.1.tar.gz). Unpack and browse. Berend __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] entropy package: how to compute mutual information?
suppose I have two factor vectors: x - as.factor(c(a,b,a,c,b,c)) y - as.factor(c(b,a,a,c,c,b)) I can compute their entropies: entropy(table(x)) [1] 1.098612 using library(entropy) but it is not clear how to compute their mutual information directly. I can compute the joint entropy as entropy(table(paste(x,y,sep=))) [1] 1.791759 and then mutual information will be h(x) + h(y) - h(x,y) = 1.098612 + 1.098612 - 1.791759 0.405465 but I was wondering whether there was a better way (without creating a fresh factor vector and a fresh factor class, both of which are immediately discarded). -- Sam Steingold (http://sds.podval.org/) on Ubuntu 11.10 (oneiric) X 11.0.11004000 http://www.childpsy.net/ http://iris.org.il http://ffii.org http://camera.org http://americancensorship.org http://dhimmi.com http://pmw.org.il There is Truth, and its value is T. Or just non-NIL. So 0 is True! __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] for loop
Dear Ted,
Thank you very much for your details explanation!
Cheers.
On Mon, Feb 13, 2012 at 7:12 PM, Ted Harding ted.hard...@wlandres.netwrote:
In addition to Jim's neat solution (see also below),
some comments on your original code.
Your for loop executes x=29.5 + i/500 100 times,
producing a single value each time and replacing the
previous value which was in x. So, at the end of the loop,
you have a single value of x. Then you compute y=2x;
that, as it stands, would prokoke an error:
Error: unexpected symbol in y=2x
since variable names cannort start with a digit. You
need, of course, the mutltiplaction operator * as
in Jim's y - 2 * x.
The scope of your for loop (i.e. the set of commands
that is executed for each round of the loop) is solely
the command x=29.5 + i/500. The y - 2 * x is not
part of the scope of the loop, and would only be executed
once, when the loop was finished. You would need
for(i in 1:1000) {
commands
}
to cause the execution of several commands in each round
of the loop.
Finally, even if you did think that your entire series of
commands (re-written):
for(i in 1:1000)
x=29.5 + i/500
y=2*x
plot(y,x)
would all be executed (down to and including the plot() command)
in each round of the loop, nevertheless each call to plot()
creates a new graph, discarding the previous one, so only a single
point would be plotted each time.
The solution (as in Jim's suggestion) is to create the full vector
of x-values and y-values, and then use plot(x,y) where x and y are
now vectors.
There are all sorts of little details about how R puts things
together, which will become familiar as you use R. However,
you do need to get hold of the basics of how R operates, so
I would suggest having R for Beginners
http://cran.r-project.org/doc/contrib/Paradis-rdebuts_en.pdf
to hand while you learn R. It is very good about how the basics
work. The next step up would be the more systematic exposition
of how R works in An Introduction to R:
http://cran.r-project.org/doc/manuals/R-intro.html
http://cran.r-project.org/doc/manuals/R-intro.pdf
Hoping this helps!
Ted.
On 13-Feb-2012 jim holtman wrote:
x - 29.5 + (1:1000)/500
y - 2 * x
plot(y,x)
On Mon, Feb 13, 2012 at 1:34 PM, eddie smith eddie...@gmail.com wrote:
Hi guys,
This is a very beginner question. Anybody willing to help?
for(i in 1:1000)
x=29.5 + i/500
y=2x
plot(y,x)
The idea is to produce 1000 values of x and y then plot them.
Cheers,
Eddie
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--
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Data Munger Guru
What is the problem that you are trying to solve?
Tell me what you want to do, not how you want to do it.
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-
E-Mail: (Ted Harding) ted.hard...@wlandres.net
Date: 13-Feb-2012 Time: 19:12:47
This message was sent by XFMail
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Re: [R] entropy package: how to compute mutual information?
* Sam Steingold f...@tah.bet [2012-02-13 16:14:36 -0500]: suppose I have two factor vectors: x - as.factor(c(a,b,a,c,b,c)) y - as.factor(c(b,a,a,c,c,b)) I can compute their entropies: entropy(table(x)) [1] 1.098612 using library(entropy) but it is not clear how to compute their mutual information directly. I can compute the joint entropy as entropy(table(paste(x,y,sep=))) this can be simplified to entropy(table(x,y)) [1] 1.791759 and then mutual information will be h(x) + h(y) - h(x,y) = 1.098612 + 1.098612 - 1.791759 0.405465 but I was wondering whether there was a better way (without creating a fresh factor vector and a fresh factor class, both of which are immediately discarded). -- Sam Steingold (http://sds.podval.org/) on Ubuntu 11.10 (oneiric) X 11.0.11004000 http://www.childpsy.net/ http://ffii.org http://mideasttruth.com http://thereligionofpeace.com http://americancensorship.org http://memri.org If money were measured in piles, I would have had a pit of 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] Writing R-scripts
Hello,
This is my first attempt to write a script in R. The program below
is intended to do some parametric tests on group data. There are subroutines
for each type of test. The call to the parametric.tests, routine sets the
argument testtype for the test to be used. How can I transfer the
calculated values (in result below) in each routine to the calling
parametric.tests routine?
Cem
## testtype : 1: Duncan
## 2: Dunnett
## resp: response variable, must be numeric and vector
## group: group id for resp, numeric or character
## alpha: CL 0.05 or 0.01
## vehicle: Control group name for Dunnett
parametric.tests-function(testtype, resp, group, vehicle, alpha)
{
if (testtype==1){
## resp: response variable, must be numeric and vector
## group: group id for resp, numeric or character
## alpha: CL 0.05 or 0.01
duncan.test - function (resp, group, alpha) {
.
result - data.frame(label=label, estimate=Estimate, alpha=alpha,
lower=Lower, upper=Upper, p.value=pval, significance=sig)
return(result)
}
}
else if (testtype==2){
dunnett.test - function(resp, group, vehicle, alpha)
{
.
result - data.frame(label=label, estimate=Estimate, alpha=alpha,
lower=Lower, upper=Upper, p.value=pval, significance=sig)
return(result)
}
}
}
Cem
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[R] Change dataframe-structure
Hello everybody, I have the following problem and have no idea how to solve it: In my dataframe I have six columns representing six societal problems (p1, p2, ..., p6). The values are ranks between 1 (worst problem) and 6 (best problem) p1 p2 p3 p4 p5 p6 1 3 2 5 4 6 2 3 1 6 4 5 1 2 3 4 6 5 but I'd like the dataframe the other way round: 123456 p1 p3 p2 p4 p4 p6 p3 p1 p2 p5 p6 p4 p1 p2 p3 p4 p6 p5 Can anyone help? Thanks! [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Error Message Comes from the Vuong Function
I want to compare the poisson and the zero-inflated poisson distribution on describing the data. So, after using the GLM and the ZEROINFL function, I used the Voung function to compare them. Here is my code: library(pscl) glm1 - glm(nmer9_1[, 1] ~ 1, family = poisson) zip - zeroinfl(nmer9_1[, 1] ~ 1) vuong(glm1, zip) However, R returns the following error message: Error: cannot allocate vector of size 3.3 Gb. Would you please tell me what was happening there? Please note that I am not trying to do a regression, but only want to compare which of the distributions explain the data better. Thank you. -- View this message in context: http://r.789695.n4.nabble.com/Error-Message-Comes-from-the-Vuong-Function-tp4385497p4385497.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] If (x 0)
Hi,
I am new to R. I was trying to get a very simple program to run. Take one
number from the command line. If the number 0 return -1. If number 0 return
1 and if the number == 0 return 0. The code is in a file called test1.R
The code:
#useage: R --no-save --args 5 test1.R
args = (commandArgs(TRUE))
x = as.numeric(args[1])
print(x)
res - conditional1(x)
cat(result= ,res,\n)
conditional1 - function(x){
result - 0
if (x 0) {
result - 1
} else if (x 0) {
result - -1
}
return(result)
}
The output:
R --no-save --slave --args 1 test1.R
[1] 1
result= 1
R --no-save --slave --args -1 test1.R
[1] -1
result= -1
] R --no-save --slave --args 0 test1.R
[1] 0
result= -1
The problem:
For arguments 1 and -1 it works as intended. For 0 (zero) it does not. If the 0
value is passed into the function named conditional1 I would expect both
if-statements to evaluate to false and 0 being return. From what I can tell (0
0) evaluates to true since -1 is returned. Hm...
What is going on? What am I doing wrong? Why is this happening? I am baffled!
Any help would be appreciated.
Thanks
Mike
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Re: [R] non-isomorphic sequences
Dear Petr,
This is fantastic!
I have one more question, when p=4, tt=4. We have 15 non-isomorphic sequences
as you have generated. Among these 15, I selected 2 sequences. How do I recover
all the members of the equivalent classes corresponding to these 2 sequences?
For example, corresponding to the sequence of , I would like to recover
,,, from this sequence.
Thanks,
Wei
From: Petr Savicky savi...@cs.cas.cz
To: r-help@r-project.org
Sent: Monday, February 13, 2012 2:53 PM
Subject: Re: [R] non-isomorphic sequences
On Mon, Feb 13, 2012 at 10:05:02AM -0800, zheng wei wrote:
Dear All,
Sorry for the typoes earlier, let me repost the question.
Suppose I want to generate sequences of length 3 from two symbols {1,2}, we
get the following 8 sequences
1 1 1
1 1 2
1 2 1
1 2 2
2 1 1
2 1 2
2 2 1
2 2 2
However, I do not want all these 8 sequences. I call two sequencs to be
isomorphic if one sequence could be obtained from the other by relabelling
the symbols. For example, 111 is isomorphic to 222, 112 is isomorphic to
221.?By eliminating all these isomorphic ones, what I want is the following
1 1 1
1 1 2
1 2 1
2 1 1
Eliminating isomorphic sequences may be done differently,
if we select different representatives of each equivalence
class. The following also eliminates isomorphic 1,2 sequences
1 1 1
1 1 2
1 2 1
1 2 2
Is this solution OK?
In general, I need to generate non-isomorphic sequences of length p from t
distinct symbols. For example, when p=3, t=3 we have
matrix(c(1,2,3,1,1,2,2,1,1,1,2,1,1,1,1),3,5)
[1,]??? 1??? 1??? 2??? 1??? 1
[2,]??? 2??? 1??? 1??? 2??? 1
[3,]??? 3??? 2??? 1??? 1??? 1
When p=4, t=4 we have
matrix(c(1,2,3,4,1,1,2,3,1,2,1,3,1,2,3,1,2,1,1,3,2,3,1,1,2,1,3,1,1,1,2,2,1,2,1,2,1,2,2,1,1,1,1,2,1,1,2,1,1,2,1,1,2,1,1,1,1,1,1,1),4,15)
[1,]??? 1??? 1??? 1??? 1??? 2??? 2??? 2??? 1??? 1 1 1 1 1
2 1
[2,]??? 2??? 1??? 2??? 2??? 1??? 3??? 1??? 1??? 2 2 1 1 2
1 1
[3,]??? 3??? 2??? 1??? 3??? 1??? 1??? 3??? 2??? 1 2 1 2 1
1 1
[4,]??? 4??? 3??? 3??? 1??? 3??? 1??? 1??? 2??? 2 1 2 1 1
1 1
In general, I need to do this for arbitrary choices of p and t.
If p and t are not too large, try the following
check.row - function(x)
{
y - unique(x)
all(y == seq.int(along=y))
}
p - 4
tt - 4 # do not rewrite t() for transpose
elem - lapply(as.list(pmin(1:p, tt)), function(x) seq.int(length=x))
a - as.matrix(rev(expand.grid(rev(elem
ok - apply(a, 1, check.row)
out - a[ok, ]
out
Var4 Var3 Var2 Var1
[1,] 1 1 1 1
[2,] 1 1 1 2
[3,] 1 1 2 1
[4,] 1 1 2 2
[5,] 1 1 2 3
[6,] 1 2 1 1
[7,] 1 2 1 2
[8,] 1 2 1 3
[9,] 1 2 2 1
[10,] 1 2 2 2
[11,] 1 2 2 3
[12,] 1 2 3 1
[13,] 1 2 3 2
[14,] 1 2 3 3
[15,] 1 2 3 4
This solution differs from yours, for example, in
the row c(1, 2, 3, 3), which is in your solution
represented by c(2, 3, 1, 1). This a different choice
of the representatives. Is the choice important?
A related thread started at
https://stat.ethz.ch/pipermail/r-help/2012-January/301489.html
There was an additional requirement that each of t symbols
has at least one occurrence.
Hope this helps.
Petr Savicky.
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Re: [R] non-isomorphic sequences
On Mon, Feb 13, 2012 at 02:04:51PM -0800, zheng wei wrote:
Dear Petr,
This is fantastic!
I have one more question, when p=4, tt=4. We have 15 non-isomorphic sequences
as you have generated. Among these 15, I selected 2 sequences. How do I
recover all the members of the equivalent classes corresponding to these 2
sequences? For example, corresponding to the sequence of , I would like
to recover ,,, from this sequence.
Dear Wei:
Try the following.
getEquivalent - function(a, tt)
{
b - as.matrix(rev(expand.grid(rep(list(1:tt), times=max(a)
ok - apply(b, 1, function(x) length(unique(x))) == ncol(b)
b - b[ok, , drop=FALSE]
dimnames(b) - NULL
t(apply(b, 1, function(x) x[a]))
}
getEquivalent(c(1, 1, 1, 1), 4)
[,1] [,2] [,3] [,4]
[1,]1111
[2,]2222
[3,]3333
[4,]4444
getEquivalent(c(1, 1, 1, 2), 4)
[,1] [,2] [,3] [,4]
[1,]1112
[2,]1113
[3,]1114
[4,]2221
[5,]2223
[6,]2224
[7,]3331
[8,]3332
[9,]3334
[10,]4441
[11,]4442
[12,]4443
Hope this helps.
Petr.
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[R] matrix subsetting
if I have a vector, I can find the indexes which satisfy a condition: x - rnorm(10) [1] 0.4751132 -0.5442322 -0.1979854 -0.2455521 0.8349336 -0.4283345 [7] 0.6108130 2.0576160 1.1251716 -1.3933637 x[x0] [1] 0.4751132 0.8349336 0.6108130 2.0576160 1.1251716 (1:10)[x0] [1] 1 5 7 8 9 how about a matrix? 0G 0Q 2O 35 37 3A 0G NA 0.008226002 0.005631718 0.0002625585 0.0010673235 0.0045310915 0Q NA NA 0.003714951 0.0002007877 0.0016983426 0.0011083925 2O NA NA NA 0.0024233691 0.0069849678 0.0024510792 35 NA NA NA NA 0.0006499707 0.0008420414 37 NA NA NA NA NA 0.0005448872 3A NA NA NA NA NA NA 3B NA NA NA NA NA NA 3G NA NA NA NA NA NA 3H NA NA NA NA NA NA 3I NA NA NA NA NA NA 3B 3G 3H 3I 0G 0.0007254316 0.0093954826 3.420231e-04 0.003974556 0Q 0.0013883606 0.0019778247 3.609791e-05 0.001552725 2O 0.0027340806 0.0082790646 2.561862e-03 0.012297821 35 0.0006025767 0.0004534397 1.053946e-03 0.001681780 37 0.0005501974 0.0021137871 1.733880e-03 0.003712675 3A 0.0008969849 0.0043855527 2.786194e-06 0.002492954 3B NA 0.0019791290 8.301787e-04 0.002816774 3G NA NA 3.181892e-07 0.010608717 3H NA NA NA 0.001066141 3I NA NA NA NA how do I find the cells whose values are, e.g., 0.01 ?? thanks! -- Sam Steingold (http://sds.podval.org/) on Ubuntu 11.10 (oneiric) X 11.0.11004000 http://www.childpsy.net/ http://pmw.org.il http://americancensorship.org http://ffii.org http://iris.org.il http://truepeace.org http://dhimmi.com The difference between genius and stupidity is that genius has its limits. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Change dataframe-structure
On Mon, Feb 13, 2012 at 5:07 PM, David Studer stude...@gmail.com wrote: Hello everybody, I have the following problem and have no idea how to solve it: In my dataframe I have six columns representing six societal problems (p1, p2, ..., p6). The values are ranks between 1 (worst problem) and 6 (best problem) p1 p2 p3 p4 p5 p6 1 3 2 5 4 6 2 3 1 6 4 5 1 2 3 4 6 5 but I'd like the dataframe the other way round: 1 2 3 4 5 6 p1 p3 p2 p4 p4 p6 p3 p1 p2 p5 p6 p4 p1 p2 p3 p4 p6 p5 First we read the data and then rearrange it into long form (DF) and turn that into a 2d matrix (tapply): Lines - p1 p2 p3 p4 p5 p6 1 3 2 5 4 6 2 3 1 6 4 5 1 2 3 4 6 5 DF0 - read.table(text = Lines, header = TRUE) DF - as.data.frame.table(as.matrix(DF0), stringsAsFactors = FALSE, responseName = Ranks) tapply(DF[[Var2]], DF[-2], c) The result of the last statement is: Ranks Var1 123456 A p1 p3 p2 p5 p4 p6 B p3 p1 p2 p5 p6 p4 C p1 p2 p3 p4 p6 p5 -- Statistics Software Consulting GKX Group, GKX Associates Inc. tel: 1-877-GKX-GROUP email: ggrothendieck at gmail.com __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Change dataframe-structure
P - paste(P,1:6,sep=) t(apply(yourdataframe,1,function(x)P[order(x)])) ## result is a mtrix, though. -- Bert On Mon, Feb 13, 2012 at 2:07 PM, David Studer stude...@gmail.com wrote: Hello everybody, I have the following problem and have no idea how to solve it: In my dataframe I have six columns representing six societal problems (p1, p2, ..., p6). The values are ranks between 1 (worst problem) and 6 (best problem) p1 p2 p3 p4 p5 p6 1 3 2 5 4 6 2 3 1 6 4 5 1 2 3 4 6 5 but I'd like the dataframe the other way round: 1 2 3 4 5 6 p1 p3 p2 p4 p4 p6 p3 p1 p2 p5 p6 p4 p1 p2 p3 p4 p6 p5 Can anyone help? 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. -- Bert Gunter Genentech Nonclinical Biostatistics Internal Contact Info: Phone: 467-7374 Website: http://pharmadevelopment.roche.com/index/pdb/pdb-functional-groups/pdb-biostatistics/pdb-ncb-home.htm __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Change dataframe-structure
There is probably a more ellegant way, but: df - data.frame(p1=c(1,2,1),p2=c(3,3,2),p3=c(2,1,3),p4=c(5,6,4),p5=c(4,4,6),p6=c(6,5,5)) as.data.frame(t(apply(df,1,function(x) names(x)[match(1:6,x)]))) V1 V2 V3 V4 V5 V6 1 p1 p3 p2 p5 p4 p6 2 p3 p1 p2 p5 p6 p4 3 p1 p2 p3 p4 p6 p5 On Mon, Feb 13, 2012 at 2:07 PM, David Studer stude...@gmail.com wrote: Hello everybody, I have the following problem and have no idea how to solve it: In my dataframe I have six columns representing six societal problems (p1, p2, ..., p6). The values are ranks between 1 (worst problem) and 6 (best problem) p1 p2 p3 p4 p5 p6 1 3 2 5 4 6 2 3 1 6 4 5 1 2 3 4 6 5 but I'd like the dataframe the other way round: 123456 p1 p3 p2 p4 p4 p6 p3 p1 p2 p5 p6 p4 p1 p2 p3 p4 p6 p5 Can anyone help? 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. [[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] survey package svystat objects from predict()
Hello,
I'm running R 2.14.1 on OS X (x86_64-apple-darwin9.8.0/x86_64 (64-bit)), with
version 3.28 of Thomas Lumley's survey package. I was using predict() from
svyglm(). E.g.:
data(api)
dstrat-svydesign(id=~1,strata=~stype, weights=~pw, data=apistrat, fpc=~fpc)
out - svyglm(sch.wide~ell+mobility, design=dstrat,
family=quasibinomial())
pred.df - expand.grid(ell=c(20,50,80), mobility=20)
out.pred - predict(out, pred.df)
From the console out.pred looks like this:
class(out.pred)
[1] svystat
print(out.pred) # or just 'out.pred'
link SE
1 1.8504 0.2414
2 1.6814 0.3033
3 1.5124 0.5197
From here I wanted to conveniently access the predicted values and SEs. I
thought that I might be able to do this using either ftable() or
as.data.frame(), as methods for these exist for the objects of class
svystat. But while the predicted values come through fine, the SE only gets
calculated for the first element and is then repeated:
ftable(out.pred)
A B
1 1.8504120 0.2413889
2 1.6814293 0.2413889
3 1.5124466 0.2413889
as.data.frame(out.pred)
linkSE
1 1.850412 0.2413889
2 1.681429 0.2413889
3 1.512447 0.2413889
I think what's happening is that as.data.frame.svystat() method in the survey
package ends up calling the wrong function to calculate the standard errors.
From the survey package:
as.data.frame.svystat-function(x,...){
rval-data.frame(statistic=coef(x),SE=SE(x))
names(rval)[1]-attr(x,statistic)
if (!is.null(attr(x,deff)))
rval-cbind(rval,deff=deff(x))
rval
}
The relevant SE method seems to be:
SE.svrepstat-function(object,...){
if (is.list(object)){
object-object[[1]]
}
vv-as.matrix(attr(object,var))
if (!is.null(dim(object)) length(object)==length(vv))
sqrt(vv)
else
sqrt(diag(vv))
}
Instead of returning sqrt(vv) on each element, it calculates sort(diag(vv))
instead. At least I think this is what's happening.
I apologize in advance if all this is the result of some elementary error on my
part.
Thanks,
Kieran
--
Kieran Healy :: http://kieranhealy.org
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Re: [R] only 0s may be mixed with negative subscripts
The function you posted runs without error (on these 6 lines), but
does not return anything that looks remotely like a sum, or cumsum of
anything. Can you clarify what you are trying to do? I assume by sum
of every other row you don't mean summing Time, X and Y for rows
1,3,5,..., ?
For the sum of sens2[c(1,3,5,...),] for every column (assuming no NA's
in the data) you could
(1:nrow(sens2) %% 2) %*% as.matrix(sens2)
Time Y X rownumber
[1,] 3985695201 0.56826 0.1369527 9
Hope this helps
On Mon, Feb 13, 2012 at 11:56 AM, Hasan Diwan hasan.di...@gmail.com wrote:
I'd like to get the sum of every other row in a data.frame. When I
actually set about doing this, I get the error in the subject line of
this message. A sample of my data is below, followed by the function
call that should give me the results I want:
dput(head(sens2))
structure(list(Time = c(1328565067, 1328565067.05, 1328565067.1,
1328565067.15, 1328565067.2, 1328565067.25), Y = c(0.0963890795246276,
0.227296347215609, 0.240972698811569, 0.221208948983498, 0.230898231782485,
0.203282153087549), X = c(0.0245045248243853, 0.0835679411703579,
0.0612613120609633, 0.058568910563872, 0.0511868450318788, 0.0557714205674231
), rownumber = 1:6), .Names = c(Time, Y, X, rownumber
), row.names = c(NA, 6L), class = data.frame)
speedX - sapply(sens2[sens2$rownumber %% 2 == 0,], function(row) {
cumsum(c(sens2[row+1,3], sens2[row,3]))}, simplify=TRUE)
Error in xj[i] : only 0's may be mixed with negative subscripts
Help?
--
Sent from my mobile device
Envoyait de mon portable
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and provide commented, minimal, self-contained, reproducible code.
Re: [R] matrix subsetting
which(x0) or which(x0, arr.ind=TRUE) depending on your application. Michael On Feb 13, 2012, at 5:38 PM, Sam Steingold s...@gnu.org wrote: if I have a vector, I can find the indexes which satisfy a condition: x - rnorm(10) [1] 0.4751132 -0.5442322 -0.1979854 -0.2455521 0.8349336 -0.4283345 [7] 0.6108130 2.0576160 1.1251716 -1.3933637 x[x0] [1] 0.4751132 0.8349336 0.6108130 2.0576160 1.1251716 (1:10)[x0] [1] 1 5 7 8 9 how about a matrix? 0G 0Q 2O 35 37 3A 0G NA 0.008226002 0.005631718 0.0002625585 0.0010673235 0.0045310915 0Q NA NA 0.003714951 0.0002007877 0.0016983426 0.0011083925 2O NA NA NA 0.0024233691 0.0069849678 0.0024510792 35 NA NA NA NA 0.0006499707 0.0008420414 37 NA NA NA NA NA 0.0005448872 3A NA NA NA NA NA NA 3B NA NA NA NA NA NA 3G NA NA NA NA NA NA 3H NA NA NA NA NA NA 3I NA NA NA NA NA NA 3B 3G 3H 3I 0G 0.0007254316 0.0093954826 3.420231e-04 0.003974556 0Q 0.0013883606 0.0019778247 3.609791e-05 0.001552725 2O 0.0027340806 0.0082790646 2.561862e-03 0.012297821 35 0.0006025767 0.0004534397 1.053946e-03 0.001681780 37 0.0005501974 0.0021137871 1.733880e-03 0.003712675 3A 0.0008969849 0.0043855527 2.786194e-06 0.002492954 3B NA 0.0019791290 8.301787e-04 0.002816774 3G NA NA 3.181892e-07 0.010608717 3H NA NA NA 0.001066141 3I NA NA NA NA how do I find the cells whose values are, e.g., 0.01 ?? thanks! -- Sam Steingold (http://sds.podval.org/) on Ubuntu 11.10 (oneiric) X 11.0.11004000 http://www.childpsy.net/ http://pmw.org.il http://americancensorship.org http://ffii.org http://iris.org.il http://truepeace.org http://dhimmi.com The difference between genius and stupidity is that genius has its limits. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] only 0s may be mixed with negative subscripts
On 13 February 2012 14:46, ilai ke...@math.montana.edu wrote: The function you posted runs without error (on these 6 lines), but does not return anything that looks remotely like a sum, or cumsum of anything. Can you clarify what you are trying to do? I assume by sum of every other row you don't mean summing Time, X and Y for rows 1,3,5,..., ? I'm trying to get a piecewise sum of every n rows in a given column in this data set. For the sum of sens2[c(1,3,5,...),] for every column (assuming no NA's in the data) you could I do format checking well before getting to this stage in the analysis. (1:nrow(sens2) %% 2) %*% as.matrix(sens2) That does not do what I want... Again, what it should return is a list of the sum of the current and preceding row. -- Sent from my mobile device Envoyait de mon portable __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
