similar to: Problem about for loop

Hi, I' am interested in differences between sample's result when samples consist of full elements and consist of only distinct elements. When sample consist of full elements it take about 120 sec., but when consist of only distinct elements it take about 4.5 or 5 times more sec. I expected that opposite of this result, because unique(sample) has less elements than full sample. Code as

I am a beginner in R, so please don't step on me if this is too simple. I have two data sets datax and datay for which I created a qqplot qqplot(datax,datay) but now I want a line that indicates the perfect match so that I can see how much the plot diverts from the ideal. This ideal however is not normal, so I think qqnorm and qqline cannot be applied. Perhaps you can help? Ralf

Hi Everyone, When I have a load of functions which have various arguments passed via the ellipsis argument it keeps on assigning them as symbol making them unusable to the function. My current work around involves using do.call but this is rather cumbersome. Does anyone know why it suddenly changes the types to symbol and if there is a way to get the actual data pointed to by the symbol? (I

Dear all, I am trying to my dataframe for the PLS analysis using the PLS package. However I have some trouble generating the correct dataframe. The main problem is how to use one name to represent several columns in the dataframe. The example dataframe in PLS package is called "sensory". I cannot directly read the data file since it's a binary file. If I use

Actually what I meant in my original loop, that there is a dependency between every two consecutive iterations. So, how the loop vectorizer says 'we can vectorize this loop'? for(int k=20;k<50;k++) dataY[k] = dataY[k-1]; From: Henrique Santos [mailto:henrique.nazare.santos at gmail.com] Sent: Sunday, November 03, 2013 4:28 PM To: Sara Elshobaky Cc: <llvmdev at

Hi Sarah, the loop vectorizer runs not on the C code but on LLVM IR this c code was lowered to. Before the loop vectorizer runs many other optimization change the shape of this IR. You can see in the LLVM IR you referenced below, a preceding LLVM IR transformation has change your loop from: > for(int k=20;k<50;k++) > dataY[k] = dataY[k-1]; to > int a = d[19]; >

Notice that the code you provided, for globals and stack allocations, at least, is semantically equivalent to: int a = d[19]; for(int k = 20; k < 50; k++) dataY[k] = a; Like so, the load you see missing was redundant, probably hoisted by GVN/PRE and replaced with "%.pre". H. On Sun, Nov 3, 2013 at 11:26 AM, Sara Elshobaky <sara.elshobaky at gmail.com>wrote: >

Hello, I was trying to trace the Loop vectorizer of the LLVM, I wrote a simple loop with a clear dependency. But found that the debug shows that 'we can vectorize this loop' Here you are my loop with dependency: for(int k=20;k<50;k++) dataY[k] = dataY[k-1]; And the debug prints: LV: Checking a loop in "main" LV: Found a loop: for.body4 LV: Found an

Dear Sir / Madam, I am new for R coding. Kindly help me out in sorting out the following problem. There are 50 rows with six coloumns(you could see in the attached .txt file). I wish to go for filtering this 50 rows for any one of the six coloumns satisfying the value >= 64. I need to have a final table with rows having >= 64 value in any one of the six coloumns and the rest could be

Buenas tardes, quedarme con las muestras de una BD (data) que están presentes en otra (datax), cuando se tiene una variable que nunca se repite (Key) es fácil: data <- subset(data,data$Key %in% datax$Key). Mi problema es cuando la exclusividad viene dada por dos variables. P.e., las coordenadas de un mapa: lon y lat. ¿Como puedo quedarme con las muestras de una df cuya lon y lat son iguales a

Dear All, A library (PHYLOGR) that passed the usual tests in R-1.1.1 gives errors with R-devel; my (mis?)understanding of scoping rules is that it should have worked in both. The problems seem related to using the name of the data frame for extracting weights or subsets within a function call. The problems can be reproduced as follows: ********************** datai <- data.frame( y =

Dear addresses, I need perform a batch of 10 000 simulations for each of 4 options considered. (The idea is to obtain the parameter estimates in a heteroskedastic linear regression model - with additive or mixed heteroskedasticity - via the Kenward-Roger small-sample adjusted covariance matrix of disturbances). For this purpose I wrote an R program which would capture all possible options (true

I found out the other day that crossprod() will take a single matrix argument; crossprod(x) notionally returns crossprod(x,x). The two forms do not return identical matrices: x <- matrix(rnorm(3000000),ncol=3) M1 <- crossprod(x) M2 <- crossprod(x,x) R> max(abs(M1-M2)) [1] 1.932494e-08 But what really surprised me is that crossprod(x) is slower than crossprod(x,x): R>

Hi everyone, I do a lot of work with large variance matrices, and I like to use "crossprod" for speed and to keep everything symmetric, i.e. I often compute "crossprod(Q %*% t(A))" for "A %*% Sigma %*% t(A)", where "Sigma" decomposes as "t(Q) %*% Q". I notice in the code that "crossprod", current definition > crossprod function (x,

The following is at least as much out of intellectual curiosity as for practical reasons. On reviewing some code written by novices to R, I came across: crossprod(x, y)[1,1] I thought, "That isn't a very S way of saying that, I wonder what the penalty is for using 'crossprod'." To my surprise the penalty was substantially negative. Handily the client had S-PLUS as

Hi I have a square matrix Ainv of size N-by-N where N ~ 1000 I have a rectangular matrix H of size N by n where n ~ 4. I have a vector d of length N. I need X = solve(t(H) %*% Ainv %*% H) %*% t(H) %*% Ainv %*% d and H %*% X. It is possible to rewrite X in the recommended crossprod way: X <- solve(quad.form(Ainv, H), crossprod(crossprod(Ainv, H), d)) where quad.form() is a little

Dear R-users, I found a strange problem working with products of two matrices, say: a <- A[i, ] ; crossprod(a) where i is a set of integers selecting rows. When i is empty the result is in a sense random. After some trials the right answer (a matrix of zeros) appears. --------------- Illustration -------------------- R : Copyright 2003, The R Development Core Team Version 1.8.0

Hi the manpage says that crossprod(x,y) is formally equivalent to, but faster than, the call 't(x) %*% y'. I have a vector 'a' and a matrix 'A', and need to evaluate 't(a) %*% A %*% a' many many times, and performance is becoming crucial. With f1 <- function(a,X){ ignore <- t(a) %*% X %*% a } f2 <- function(a,X){ ignore <-

Hi all, I have been searching all sorts of documentation, reference cards, cheat sheets but can't find why R's crossprod(A, B) which is identical to A%*%B does not produce the same as Matlabs cross(A, B) Supposedly both calculate the cross product, and say so, or where do I go wrong? R is only doing sums in the crossprod however, as indicated by (z <- crossprod(1:4)) # = sum(1 +

Dear Rdevelopers The background for this email is that I was helping a PhD student to improve the speed of her R code. I suggested to replace calls like t(AA)%*% BB by crossprod(AA,BB) since I expected this to be faster. The surprising result to me was that this change actually made her code slower. > ## Examples : > > AA <- matrix(rnorm(3000*1000),3000,1000) > BB <-