similar to: [External] Re: Speeding up a loop

General problem: I have 20 projects that can be invested in and I need to decide which combinations meet a certain set of standards. The total possible combinations comes out to 2^20. However I know for a fact that the number of projects must be greater than 5 and less than 13. So far the the code below is the best I can come up with for iteratively creating a set to check against my set of

It looks from str(SA) that Response IPS1 is a data.frame of class "anova", which probably cannot be coerced to vector. Maybe you can use unlist() instead of as.vector() Or something like SA[["Response IPS1"]]["as.factor(WSD)",] ## to select the first row only, even maybe with unlist() Without a better REPRODUCIBLE example, I cannot tell more (maybe some others

I have 4 columns and 56 rows of made up data that I want to plot as a series of bar graphs. The idea is to create one bar graph for each of the 4 columns using a for loop. I tried the following command in RStudio and when I type x in the console I get just the 4th graph instead of all four graphs. I did not define what x is before hand. I was not sure what it would be. Any suggestions on how you

Dear R users, I want to check matrices when i change the order of the rows or/and the order of the columns or/and the combination of them i will give an example what i want  1  -1  1  1      1  1   1  1 -1  -1 -1 -1    -1 -1  -1  -1   1  1    1  1     1 -1    1   1 these 2 matrices are identical because i change the first row and make it third   1  -1  1  1      -1  1   1  1 -1  -1 -1

library(xlsReadWrite) mydata<-read.xls("file path", header=TRUE) however if I change xls to csv it works just fine. Any ideas what I'm doing wrong? I have have also using the package gdata with the exact same error. Below is the error that pops up. Error in findPerl(verbose = verbose) : perl executable not found. Use perl= argument to specify the correct path. Error in

Hi again, I've got few very good options from the list and since they were not posted to the list, I will provide a summary. Thank you very much to all who answered and I hope this summary will benefit others interested in solving similar problems like that. Yasir Kaheil re-wrote my original code in a more streamlined way: pr<-array(0,c(dim(x)[2],dim(x)[2])); for (i in 1:dim(x)[2]) for

I am just learning to use qplot and can't get the fill/colour to work. Below is the R code for a scatter plot and bar graph. library(ggplot2) x<-c(1,2,3,4,5,6,7) y<-c(1,2,3,2,5,6,3) qplot(x,y, main="Scatter Plot Test", xlab="X Label Test", ylab="Y Label Test", colour="blue")z<-c("van", "van", "van",

I wish to create a matrix of all possible percentages with two decimal place percision. I then want each row to sum to 100%. I started with the code below with the intent to then subset the data based on the row sum. This works great for 2 or 3 columns, but if I try 4 or more columns the number of rows become to large. I would like to find a way to break it down into some kind of for loop, so

Hi R-listers, I run R version 2.13.0, on a i386-pc-mingw32/i386 platform under Windows XP. I perform daily update of my installed packages. I've got the most recent Java for this platform installed, as I am told by the JavaUpdater when checking. (and I've read the posting guide, so that I'm not going to get fired by Uwe as poor wwreith [joke])... But : since one of the last daily

I am trying to print out a list of strings of length 11 based on integers 0 through 10. The rules as given to me for the ordering are: The first digit must be 0. The 2nd digit must be 0 or 1. The 3rd digit must equal the 2nd digit or the 2nd digit +1. ... Given the final digit, n, all digits 0 through n must appear in a given sequence. So the final 1024 item list should look like 0 1 2 3 4 5

It seems people are reinventing the wheel here: The goal is to generate all combinations of 1:n of size k. This (typically) results in a matrix of size k * choose(n,k) i.e. needs O(n ^ k) space, hence is only applicable to relatively small k. Then alternatives have been devised to generate the combinations "one by one", and I think I remember there has been a quiz/challenge about 20

I am just starting to experiment with writing a function and have run into what seems like a limitation or more likely a lack of understanding on my part. Very Simple Example: I want to define a function that does 1+1=2. z<-1 ADD<-function(x) { x<-x+1 } ADD(z) z output for z is 1 not the expected 2. Now if I were to do print(x+1) instead of x<-x+1 it does return 2, so the

I have several thousand rows of shipment data imported into R as a data frame, with two columns of particular interest, col 1 is the entry date, and col 2 is the tracking number (colname is REQ.NR). Tracking numbers should be unique but on occassion aren't because they get entered more than once. This creates two or more rows of with the same tracking number but different dates. I wrote a for

Hi, all When I am using mle.cv(wle), I find a interesting problem: I can't do leave-one-out cross-validation with mle.cv(wle). I will illustrate the problem as following: > xx=matrix(rnorm(20*3),ncol=3) > bb=c(1,2,0) > yy=xx%*%bb+rnorm(20,0,0.001)+0 > summary(mle.cv(yy~xx,split=nrow(xx)-1,monte.carlo=2*nrow(xx),verbose=T), num.max=1)[[1]] mle.cv: dimension of the split subsample

Hi, On 12/06/12 13:09, Abdul Wahid Memon wrote: > Thanks alot. How can we view those optimization which are enabled when we do > > llc -O3 comb.ll llc -O3 comb.ll -debug-pass=Arguments > > Are these the same as the ones produced by the following command line no, as I mentioned before they are of a completely different nature to the optimizations that opt does. Ciao, Duncan.

Using R 2.6.2, say I have the following list of lists, "comb": data1 <- list(a = 1, b = 2, c = 3) data2 <- list(a = 4, b = 5, c = 6) data3 <- list(a = 3, b = 6, c = 9) comb <- list(data1 = data1, data2 = data2, data3 = data3) So that all names for the lowest level list are common. How can I most efficiently access all of the sublist items "a" indexed by the outer

Hi, > I executed the following command line > > llc -O3 comb.ll.bc -debug-pass=Arguments > > and got > > Pass Arguments: -targetdata -targetpassconfig -no-aa -tbaa > -targetlibinfo -basicaa -collector-metadata -machinemoduleinfo > -machine-branch-prob -preverify -domtree -verify -loops -loop-simplify > -scalar-evolution -loop-simplify -iv-users -loop-reduce

Thanks again. I executed the following command line llc -O3 comb.ll.bc -debug-pass=Arguments and got Pass Arguments: -targetdata -targetpassconfig -no-aa -tbaa -targetlibinfo -basicaa -collector-metadata -machinemoduleinfo -machine-branch-prob -preverify -domtree -verify -loops -loop-simplify -scalar-evolution -loop-simplify -iv-users -loop-reduce -gc-lowering -unreachableblockelim

Thanks alot. How can we view those optimization which are enabled when we do llc -O3 comb.ll Are these the same as the ones produced by the following command line llvm-as < /dev/null | opt -O3 -disable-output -debug-pass=Arguments Regards Shahzad On Tue, Jun 12, 2012 at 1:01 PM, Duncan Sands <baldrick at free.fr> wrote: >> Yes. But how exactly code generation (optimized one)

I need to save data in fixed-width format without headers and reading the help archive leads me to believe that sprintf is pretty much the only way to do this. My question is, is there anyway to change the output so the text in each column is left justified instead of right justified? My code sample is below where comb is the data frame. TIA, Roger out <- sprintf("%6s %22s %8s %15s