similar to: Probem with argument "append" in "Rprof"

Displaying 20 results from an estimated 2000 matches similar to: "Probem with argument "append" in "Rprof""

2009 Jun 12
1
Rprof loses all system() time
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
2007 Aug 23
2
read big text file into R
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
2009 Mar 03
1
profiler and loops
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:
2013 Apr 05
2
line profiling
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
2010 Sep 23
0
R CMD Rprof --help suggestion
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
2011 Feb 11
1
Help optimizing EMD::extrema()
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(),
2004 Jul 16
3
interpreting profiling output
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
2005 Jan 20
0
Interpreting Rprof output
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
2007 Oct 22
2
Help interpreting output of Rprof
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
2017 May 18
1
Interpreting R memory profiling statistics from Rprof() and gc()
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
2009 Oct 19
2
how to get rid of 2 for-loops and optimize runtime
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 ...
2010 Jan 05
1
Naming functions for the purpose of profiling
Hi all, I have some long-running code that I'm trying to profile. I am seeing a lot of time spent inside the <Anonymous> function. Of course, this can in fact be any of several functions, but I am unable to see how I could use the information from Rprof.out to discern which function is taking the most time. An example line from my Rprof.out is: rbernoulli <Anonymous>
2010 Nov 19
1
memory profiling
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
2004 Oct 19
0
Question on Rprof(); was: Re: sapply and loop
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
2009 Nov 10
1
standardGeneric seems slow; any way to get around it?
Hi, I'm running some routines with standard matrix operations like solve() and diag(). When I do a profile, the lead item under total time is standardGeneric(). Furthermore, solve() and diag() have much greater total time than self time. ??? I assume there is some time-consuming decision going on in the usual functions; is there any way to avoid that and go straight to the calculaions? Thanks
2012 Jul 21
1
alternative to rbind for data.table
Hi I want to add a row to a "data.table" in each round of a for loop. "rbind" seems to be a inefficient way to implement this. How would you do this? The "slow" solution: library(data.table) Rprof("test.out") dt <- data.table() for (i in (1:10000)) { # algorithm that generates a list with different values, # but same key-names, each round, for
2004 Oct 16
3
Lazy loading... advices
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),
2011 Feb 28
0
Fwd: Re: speed up process
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",
2008 Aug 26
1
Dramatic slowdown of R 2.7.2?
Dear R users/developers, simple comparison of code execution time of R 2.7.1 and R 2.7.2 shows a dramatic slowdown of the newer version. Rprof() identifies .Call function as a main cause (see the code below). What happened with R 2.7.2? Kind regards Marek Wielgosz Bayes Consulting ######### Probably useful info ############### ### CPU: Core2Duo T 7300, 2 GB RAM ### WIN XP ### both standard
2012 Dec 05
1
Understanding svd usage and its necessity in generalized inverse calculation
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