search for: 0.02

Dear Rs: Hi, I am trying to read a big text file (nrows=243440, ncols=144). It seems the computational time of all the read methods (scan,readtable,read.delim) is not linear to the number of rows I want to read in: things became really slow once I tried to read in 100000 lines compare to 10000 lines). If I am reading the profiling result right, I guess scan wouldn't help either. My

Hi Stepan, Sorry for the delay. It's great that you are working on MergeFunctions as well and I agree, we should definitely try to combine our efforts to improve MergeFunctions. Just to give you some context, the pass (with the similar function merging patch) is already being used in a production setting. From my point of view, it would be better if we focus on improving its capability

Hello Sean and Tobias, Sean, Thank you. Could you describe Nick's ideas in few words or give me links to your discussion, so I could adapt my ideas to it. Tobias, Your patch fails on several modules in my benchmark (73 of ~1800 tests). I have sent one as attachment. See statistics files for more details, all the .ll files you could simply find in test-suite object directory (after

Hello, One more question.. I have the data.frame "pop": xloc yloc gonad ind Ene W Area 1 23 20 516.74 1 0.02 20.21 1 2 23 20 1143.20 1 0.02 20.21 1 3 23 20 250.00 1 0.02 20.21 1 4 22 15 251.98 1 0.02 18.69 2 5 22 15 598.08 1 0.02 18.69 2 6 21 19 250.00 1 0.02 20.21 3 7 22 20 251.98 1

hello, i'm pete ,how can i order rows of matrix by max to min value? I have a matrix of membership degrees, with 82 (i) rows and K coloumns, K are clusters. I need first and second largest elements of the i-th row. for example 1 0.66 0.04 0.01 0.30 2 0.02 0.89 0.09 0.00 3 0.06 0.92 0.01 0.01 4 0.07 0.71 0.21 0.01 5 0.10 0.85 0.04 0.01 6 0.91 0.04 0.02 0.02 7 0.00 0.01 0.98 0.00 8 0.02

This looks like a bug? Unable to use scan(pipe()) while profiling. I have no idea whether this version of R violates the "do not use `Rprof' in an executable built for profiling" warning in ?Rprof. Thanks -Don > version _ platform powerpc-apple-darwin1.4 arch powerpc os darwin1.4 system powerpc, darwin1.4 status Patched major 1 minor 3.1 year

Hello everyone,           I have a basic question regarding logical operators. > x<-seq(-1,1,by=0.02) > x   [1] -1.00 -0.98 -0.96 -0.94 -0.92 -0.90 -0.88 -0.86 -0.84 -0.82 -0.80 -0.78  [13] -0.76 -0.74 -0.72 -0.70 -0.68 -0.66 -0.64 -0.62 -0.60 -0.58 -0.56 -0.54  [25] -0.52 -0.50 -0.48 -0.46 -0.44 -0.42 -0.40 -0.38 -0.36 -0.34 -0.32 -0.30  [37] -0.28 -0.26 -0.24 -0.22 -0.20 -0.18 -0.16

Hello, I'm trying to create a matrix (95x55) with data from a data.frame pop: xloc yloc go ind Ene totW 1 23 20 516 1 0.02 20.21 2 23 20 1143 1 0.02 20.21 3 23 20 250 1 0.02 20.21 4 22 15 251 1 0.02 18.69 5 22 15 598 1 0.02 18.69 6 21 19 250 1 0.02 20.21 7 22 20 251 1

Hi all. I'm developing a function, which must return a square matrix. Here is the code: http://pastebin.com/THzEW9N7 These functions implement an analog of two embedded for cycles. The first variant creates the resulting matrix by columns, cbind()-ing them one by one. The second variant creates the matrix with two columns, which rows contain all possible variants of i and j and calls apply

Hi there, I would like to draw a scatter plot and fit a smooth line by loess. Below is the data. However, the curve line started from 0, which my "resid" list doesn't consist of 0 value. It returned some warnings which I don't know if this is the reason affecting such problem. Here I also attached the warning messages. Please let me know if there is a solution to fix this. Thank

Hello! I have run Rprof on a function of mine and the results look very strange, to say the least. At the end I of this email is an output of summaryRprof. Can someone help me interpret this output? I have read the appropriate section in the manual "Writing R Extensions" and help pages. If I understand this output correctly, it is saying that "unlist" has been active in

Dear Jim, Here is again exactly what I did and with the output of Rprof (with this reduced dataset and with a simpler function, it is here much faster than in real life). Thanks you again for your help! ## CODE ## mydata1<- structure(list(species = structure(1:8, .Label = c("alsen","gogor", "loalb", "mafas", "pacyn", "patro",

Hi All, I have a data.frame: >pop consumed ind recruits gonad cell 1 516.74 1 0.02 20.21 0.25 2 1143.20 1 0.02 20.21 0.50 3 250.00 1 0.02 20.21 0.25 4 251.98 1 0.02 18.69 0.25 5 598.08 1 0.02 18.69 0.25 6 437.38 1 0.02 18.69 0.25 7 250.00 1 0.02 17.39 0.25 8 250.00 1 0.02 17.39 0.25 9

I have 4 years of data and for each year, I have initialize the value so now for fitting the model, I want to remove the initial value and get the model based on remaining data set. Could anyone can help on this? I want to get linear model based on fourth column and 13th column but need to remove the initial value for each year and each treatment ( the second column I have 1:36) . Thank you,

Hi all, I'm porting a library I wrote in S-plus over to R. All but one of the functions work fine. The only one that doesn't is really giving me a headache. I'm admitting defeat. Can anybody help? Object example3 has been read in using read.table. The statement works fine in S-Plus. When I run the same statement in R the column Prefix, which is read in using the as.character

With the ATI bug I was hitting earlier fixed, only my btrfs partition continues to show high cpu usage for some operations. Rsync, git pull, git checkout and svn up are typicall operations which trigger the high cpu usage. As an example, this perf report is from using git checkout to change to a new branch; the change needed to checkout 208 files out of about 1600 total files. du(1) reports

Hello, A collegue of mine has compared the runtime of a linear model + anova in SAS and S+. He got the same results, but SAS took a bit more than a minute whereas S+ took 17 minutes. I've tried it in R (1.9.0) and it took 15 min. Neither machine run out of memory, and I assume that all machines have similar hardware, but the S+ and SAS machines are on windows whereas the R machine is Redhat

Dear expeRts, I would like to color a certain region in a levelplot. The region for z <= 0.02 should have a dark gray color. Below is a minimal example. It almost does what I want, but The region between z=0.02 and z=1 is also colored in dark gray (instead of just the region for z <= 0.02). How can I solve this? Cheers, Marius ## z values for given x and y f <- function(x)

Dear R users, I am trying to have an xyplot of a data set which has the following variables: case (n=10,20,30) parameter (parm=a,b) group (grp=g1,g2) y (y values) x (x=2,4,8) My plot should be parameter by case such that I have 2 rows (each row= each parameter) and 3 columns (each column=each case). My R-code is as follows but I am not able to get what I want to: tp1.sim <- xyplot(y~ x | case

** Disclaimer: I'm looking for general suggestions ** I'm sorry, but can't send out the file I'm using, so there is no reproducible example. I'm using read.table and it's taking over 30 seconds to read a tiny file. The strange thing is that it takes roughly the same amount of time if the file is 100 times larger. After re-reviewing the data Import / Export manual I think