similar to: Interpreting Rprof output

Hi, >From reading ?Rprof, I checked R CMD Rprof --help and learned that there are options to specify the min % to print. This is currently (R-devel r52975) displayed with the --help option as --min%total minimum % to print for 'by total' --min%self minimum % to print for 'by self' So I tried R CMD Rprof --min%total 5 and got an error. After looking at

Hello, Appending information to the profiler's output seems to generate problems. Here is a small example of code : <code r> require(boot) Rprof( memory.profiling = TRUE) Rprof(NULL) for(i in 1:2){ Rprof( memory.profiling = TRUE, append = TRUE) example(boot) Rprof(NULL) } </code> The problem is that the file Rprof.out contains more than once the header information: $ grep

Rprof seems to ignore all time spent inside system() calls. E.g., this simple example actually takes about 10 seconds, but Rprof thinks the total time is only 0.12 seconds: > Rprof("sleep-system.out") ; system.time(system(command="sleep 10")) ; Rprof(NULL) user system elapsed 0.000 0.004 10.015 > summaryRprof("sleep-system.out")$by.total

Hello there, I am not quite sure how to interpret the output of Rprof (in the following the output I was staring at). I was poking around the web a little bit for documentation but without much success. I guess if I want to figure out what takes so long in my code the 2nd table $by.total and the total.pct column (pct = percent) is the most helpful. What does it mean that [ or [.data.frame is

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",

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

I have some trouble interpreting the output from profiling. I have read the help pages Rprof, summaryRprof and consult the R extensions manual, but I still have problems understanding the output. Basically the output consist of self.time and total.time. I have the understanding that total.time is the time spent in a given function including any subcalls or child functions or whatever the

Hello, (This is follow up from this thread: http://www.nabble.com/execution-time-of-.packages-td22304833.html but with a different focus) I am often confused by the result of the profiler, when a loop is involved. Consider these two scripts: script1: Rprof( ) x <- numeric( ) for( i in 1:10000){ x <- c( x, rnorm(10) ) } Rprof( NULL ) print( summaryRprof( ) ) script2:

Yes. It should have something to do with read/write permissions, but it is not clear how it happens. I can write file to C drive using R. I usually write my results matrix to a txt file in C drive. For Rprof(), the boot.out file can be created, but only with one line sample.interval=20000 The situation is the same even if I specify the directory to the D drive,where I have the full

Dear R-Users, I am working on an Hierarchical Bayes model and tried to replace the inner for-loop (which loops over a list with n.observations elements) with truely vectorized code (where I calculated everything based on ONE dataset over all respondents). However, when comparing the performance of the two alternatives, I found out that the code with the for-loop actually was faster! In order to

Hello, This is about the new "line profiling" feature in R 3.0.0. As I was testing it, I find the results somewhat disappointing so I'd like to get your opinion. I put some poorly written code in a test.R file, here are the contents: double <- function(x) { out <- c() for (i in x) { out <- c(out, 2*i) # line 4 } return(out) } Then this how I source the file

Short: get rid of the loops I use and optimize runtime Dear all, I want to calculate for each row the amount of the month ago. I use a matrix with 2100 rows and 22 colums (which is still a very small matrix. nrows of other matrixes can easily be more then 100000) Table before Year month quarter yearmonth Service ... Amount 2009 9 Q3 092009 A ...

Sorry, this might be a really basic question, but I'm trying to interpret the results from memory profiling, and I have a few questions (marked by *Q#*). From the summaryRprof() documentation, it seems that the four columns of statistics that are reported when setting memory.profiling=TRUE are - vector memory in small blocks on the R heap - vector memory in large blocks (from malloc) - memory

Hello, I am looking for more information about lazy loading introduced in R 2.0.0. Doing ?lazyLoad I got some and there is a 'see also' section that points to 'makeLazyLoading'... But I cannot reach this page. My problem is: I recompiled a library that uses a lot of functions from other libraries (of course I can give details if needed). I load it in my computer: library(svGUI),

Dear R-devel: I could use some advice about matrix calculations and steps that might make for faster computation of generalized inverses. It appears in some projects there is a bottleneck at the use of svd in calculation of generalized inverses. Here's some Rprof output I need to understand. > summaryRprof("Amelia.out") $by.self self.time self.pct

Hi folks, I'm attempting to use the EMD package to analyze some neuroimaging data (timeseries with 64 channels sampled across 1 million time points within each of 20 people). I found that processing a single channel of data using EMD::emd() took about 8 hours. Exploration using Rprof() suggested that most of the compute time was spent in EMD::extrema(). Looking at the code for EMD:extrema(),

Dear R helpers, I have what another member on this forum described as an embarrassingly parallel problem. I am trying to fit models on subsets of some data based on unique combinations of two id factors in the dataset. Total number of combinations is 30^5, and this takes a long time. So, I would like fit models for each of the datasets produced by subsetting on the unique combinations, splitting

Full_Name: John Garvin Version: 1.9.0 OS: Linux Submission from: (NULL) (128.42.129.78) This may or may not technically be a bug, but it's certainly an annoyance. Rprof only takes into account computation that occurs inside functions. If a time-consuming operation occurs outside a function, it doesn't record the time it takes. Consider this program 'array.r': Rprof() foo <-

I'm trying to configure Version 2.12.0 or R to do memory profiling. I've reconfigured the code: % ./compile --enable-memory-profiling=YES and verified that it's configured correctly by examining the output. I then rebuild R: % make Then I fire up R and run a script, using Rprof with the memory-profiling switch set to TRUE: Rprof("output", memory.profiling=TRUE); # a

Dear R's, I have been using R lately to perform some statistical analysis and, based on them, simulations to be exported in flat text files to other programs. These text files are nowadays of about 30MB in size, but they could finally be of up to 300MB. Writing these files with either write.table or write.matrix was desperately slow and the bottleneck of the whole process. Besides, the