similar to: converting blank cells to NAs

Hi, Is there a way in llvm IR to emit multiple data elements within a single compilation unit that a) are guaranteed to appear sequentially in the final binary (in the order they appear in the llvm IR), and b) will be removed on an individual basis by the optimizers and/or linker in case they are not referenced from anywhere ? Defining them as "appending" puts them all into a

Is there no better way to merge one array into another than iterating over the array you wish to append with each() and push()ing the elements on to the other array? Here''s what I mean: var arr1 = [1, 2, 3, 4, 5]; var arr2 = [''a'', ''b'', ''c'']; $A(arr2).each(function(el) { arr1.push(el) }); Is there no better way to do it than this?

Dear Duncan, Everything worked out with your help. Thanks. I'm now looking at bound checks that an Ada program has, so I can remove all possible. It seams that Ada already does some optimization to remove bound checks. Do you know how are they done, and where? I tested two codes (below) and the first code did not produce any bound check and the second produced two bound checks. -- -- Simple

Hi everyone, Speed is the key here. I need to find the difference between a vector and its one-period lag (i.e. the difference between each value and the subsequent one in the vector). Let's say the vector contains 10 million random integers between 0 and 1,000. The solution vector will have 9,999,999 values, since their is no lag for the 1st observation. In R we have: #Set up input vector

Hello Chandler, > ``` > void f(int *arr1, int *arr2, int length) { > intptr_t rel_offset = arr2 - arr1; > int *arr1_end = arr1 + length; > for (int *p = arr1; p < arr1_end; p += sizeof(int)) { > do_something(p, p + rel_offset); > } > } > ``` > > For example, a common loop optimization technique is something like the following in *pseudocode* (note

Hi, I have been using R for a few months and I have this working code. Don't seen any problem but this takes a long time. So if I have about 30000 rows it takes a few minutes. If I have 100000 it does not seem to complete. Original Data: Proto Recv-Q Send-Q Local-Address Foreign-Address State tcp 0 0 172.20.100.2:60255

I'm seeing some inconsistent behavior when re-assigning values in a data frame. The first assignment turns all of the 0s in my data frame to 2s, the second fails to do so. > df1 <- data.frame(a = c(NA, 0, 3, 4)) > df2 <- data.frame(a = c(NA, 0, 0, 4)) > df1[df1 == 0] <- 2 ## Works > df2[df2 == 0] <- 2 Error: NAs are not allowed in subscripted assignments Checking an

Hello R specialists, I have a base vector called vBase with 102 elements. I have another vector vec1 which elements are only part of vBase but is shorter. I transform vec1 so I get a vector with the same size as vBase and with each common element on the same indexed row. If a value is missing in vec1 then I put a Na like this: vec1 vBase Amsterdam Amsterdam Na

Hallo all Please help me. I am lost and do not know what is the problem. I have a factor called kvartaly. > attributes(kvartaly) $levels [1] "1Q.04" "2Q.04" "3Q.04" "4Q.04" "1Q.05" "2Q.05" "3Q.05" "4Q.05" $class [1] "factor" > mode(kvartaly) [1] "numeric" > str(kvartaly) Factor w/ 8

On 04 Aug 2014, at 09:27, Reid Kleckner wrote: > On Sat, Aug 2, 2014 at 7:51 AM, Jonas Maebe > <jonas.maebe at elis.ugent.be> > wrote: > >> On 01/08/14 19:37, Reid Kleckner wrote: >> >>> What happens if you drop appending linkage? I think it will just >>> work, >>> since you are already using a custom section, which will ensure

This can be vectorized. Try ix <- seq_along(vec2) S_diff2 <- sapply(seq_len(N1-(N2-1)*ratio_sampling), \(j) sum((vec1[(ix-1)*ratio_sampling+j] - vec2[ix])**2)) On Sun, Jun 16, 2024 at 11:27?AM Laurent Rhelp <laurentRHelp at free.fr> wrote: > > Dear RHelp-list, > > I try to use the package comprehenr to replace a for loop by a list > comprehension. > > I

I would like to know whether there is a faster way to do the below operation (updating vec1). My objective is to update the elements of a vector (vec1), where a particular element i is dependent on the previous one. I need to do this on vectors that are 1 million or longer and need to repeat that process several hundred times. The for loop works but is slow. If there is a faster way, please let

Dear R-helpers: I have a question related to using do.call to call cbind and get. #the following works vec1 <- c(1,2) vec2 <- c(3,4) ColNameVec <- c('vec1','vec2') mat <- do.call("cbind",lapply(ColNameVec,get)) mat #put code above into a function then it does not work #before doing so, first remove vec1 and vec2 from global environment rm(vec1,vec2) test

Hello. Sorry I am not sure this question should go to llvm or mesa3d-dev mailing list, so I post it to both. I am writing a llvm backend for a modern graphics processor which has a ISA very similar to that of Direct 3D. I am reading the code in Gallium-3D driver in a mesa3d branch, which converts the shader programs (TGSI tokens) to LLVM IR. For the shader instruction also found in LLVM IR,

Is there any built-in way to lexicographically compare two vectors of the same length in R? The textbook algorithm could be coded as follows: lex.cmp <- function (vec1,vec2) { for (j in 1:length(vec1)) { if (vec1[j] < vec2[j]) { return(-1) } if (vec1[j] > vec2[j]) { return(1) } } return(0) } Thanks, Gabriel

Hello, I'm using the eclat function of the arules package (1.0-6) for the identification of frequent itemsets. I need the tidLists, but if I set in the function tidLists=TRUE R crashes (Windows XP Professional SP3, 32 bit, R version 2.12.1 (2010-12-16), reproducible on two different computers) with two different error messages or non at all. Minimum examples are: library(arules)

Hi, In the example you showed: m1<- matrix(0,length(vec),max(vec)) 1*!upper.tri(m1) #or ?m1[!upper.tri(m1)] <-? rep(rep(1,length(vec)),vec) #But, in a case like below, perhaps: vec1<- c(3,4,5) ?m2<- matrix(0,length(vec1),max(vec1)) ?indx <- cbind(rep(seq_along(vec1),vec1),unlist(tapply(vec1,list(vec1),FUN=seq),use.names=FALSE)) m2[indx]<- 1 ?m2 #???? [,1] [,2] [,3] [,4] [,5]

Emilio Ahora cuando quiero instalar los paquetes pdftools, magick y otros más me salen el siguiente error WARNING: Rtools is required to build R packages but is not currently installed. Please download and install the appropriate version of Rtools before proceeding: https://cran.rstudio.com/bin/windows/Rtools/ Installing package into ?C:/Users/bdominguez/Documents/R/win-library/3.6? (as ?lib?

HI Elisa, You could use ?cut() vec1<-c(33,18,13,47,30,10,6,21,39,25,40,29,14,16,44,1,41,4,15,20,46,32,38,5,31,12,48,27,36,24,34,2,35,11,42,9,8,7,26,22,43,17,19,28,23,3,49,37,50,45) label1<-unlist(lapply(mapply(c,lapply(seq(0,45,5),function(x) x),lapply(seq(5,50,5),function(x) x),SIMPLIFY=FALSE),function(i) paste(i[1],"<x<=",i[2],sep="")))

I'm not sure this list is the right place for this thing. I noticed some erratic behaviour in sammon(). Running sammon on two nearly identical sets of data results in very different results. Below is an example. I create an initial configuration with cmdscale() and store it into 'vec1'. I write this to file, and read it back in again to 'vec2'. According to cor() on the three