search for: aas

Dear All I have a data.frame like that: structure(list(name = c("Gga_rs10722041", "Gga_rs10722249", "Gga_rs10722565", "Gga_rs10723082", "Gga_rs10723993", "Gga_rs10724555", "Gga_rs10726238", "Gga_rs10726461", "Gga_rs10726774", "Gga_rs10726967", "Gga_rs10727581", "Gga_rs10728004",

I was wondering if somebody could explain why I get different results here: >treats[,2]<-as.factor(treats[,2]) >treats[,5]<-as.factor(treats[,5]) >treats[,4]<-as.factor(treats[,4]) #there are 'c' on more days than I have 'h2o2', where treats[,4] is the day. I only want 'c' that correspond to the same days that I have a 'h2o2' also.

Dear mailing list I have a data that have 20,000 rows and 20 columns. Io wonted to extract the 10th row only. Example the 10th, 20th, 30th 40th…..20000 th. can you please help me how do I do that.Than kyou. Example is below. Inpute: AG GG GG AG CC CC CC CC CT CC CT CT GG GG GG GG CC CC CC CC GG GG GG GG CC CC CC CC GG CG CG GG GG GG GG GG *CC CC CC CC* AA AG AG AA AA AA AA AA GG AG AG GG GG AG AG

Hi, I am trying to split multiple columns. One column works just fine, but I want to do it for multiple columns??? Example > a ID V2 V3 V4 V5 V6 V7 V8 V9 V10 1 PBBA0644 -- GG AA -- AA -- AA GG GG 2 PBBA1010 -- GG AA -- AA -- AA GG GG 3 0127ATPR -- GG AA -- AA -- AA GG GG 4 0128EHAB -- GG AA -- AG -- AA AG GG 5 PBBA0829 -- GG AA -- AA -- AA GG AG

...es get a mix of Anders and Basic AA. Am I to conclude that the Anderson's Alias Analysis is only applied to LICM? and other passes which require alias analysis use BasicAA or NoAA depending which on which one is specified on the command line? Is this the reason I am seeing calls to different AAs when although I specify an AA which neither Basic nor No AA? I read on llvm docs that it is not possible to actually specify one user-created alias analysis on the command line which is applied to all subsequent passes on the command line. Does this mean that there is a constraint on the command...

Hola a todos. Estoy atascado en un sitio que no me he encontrado antes, debe ser fácil y rápido, a ver si alguien me puede ayudar. Puesto fácil, tengo: > aa <- matrix(11:19,3,3) > aa [,1] [,2] [,3] [1,] 11 14 17 [2,] 12 15 18 [3,] 13 16 19 Quiero la fila cuyo primer elemento es 12 y el segundo 15. > aa[aa[,1:2] == (c(12,15)),] Error: (subscript) logical

Hi, I am using the lda function from the MASS library. I would to find the decision boundaries of each class and subsequently plot them. I wonder if anybody can offer any help on this topic? Below I applied the lda function on a small dataset of mine. Any help will be much appreciated. > library(MASS) >

Full_Name: Robert Denham Version: R-2.4.1 OS: Windows Xp Submission from: (NULL) (61.88.57.1) R gui exits without warning when I run a function which has an argument with a default that is not found. This was a result of an error in a function I wrote, but I thought that it should exit more gracefully than it does. Here is an example: testfun <- function(aa=aa) { aa <-

Hi All, I have the following problem, that's driving me mad. I have a dataframe of factors, from a genetic scan of SNPs. I DO have NAs in the dataframe, which would look like: V4 V5 V6 V7 V8 V9 V10 1 TT GG TT AC AG AG TT 2 AT CC TT AA AA AA TT 3 AT CC TT AC AA <NA> TT 4 TT CC TT AA AA AA TT 5 AT CG TT CC AA AA TT 6 TT CC TT AA AA AA TT 7 AT CC

Dear Prof Azzalini, You have an interesting example of recursive use of uniroot(). [re-cited at the end] However, note that R currently has the same problem as S-plus: Uniroot() doesn't work reliably, recursively. When you found it to be better, you were just lucky. The relevant file in R's source, src/main/optimize.c says /* WARNING : As things stand, these routines should

Hi Herve, Inline. On Thu, May 16, 2019 at 4:45 PM Pages, Herve <hpages at fredhutch.org> wrote: > Hi Gabe, > > ncol(data.frame(aa=c("a", "b", "c"), AA=c("A", "B", "C"))) > # [1] 2 > > ncol(data.frame(aa="a", AA="A")) > # [1] 2 > > ncol(data.frame(aa=character(0),

Hi, all, When running > AB.fit=adaboost(ylearn, xlearn, xtest, presel=0) I got the following error: Error in `[[<-.data.frame`(`*tmp*`, preds, value = c(4L, 6L, 6L, 6L, 3L, : replacement has 186 rows, data has 62 The data structure is attached below: [1] "ylearn" [1] 0 0 0 0 0 0 0 0 0 1 1 1 1 1 1 0 0 0 0 0 0 0 0 0 1 1 1 1 1 1 0 0 0 0 0 0 0 0 0 [40] 1 1 1 1 1 1 1 0

Hi everybody, I have a list, where every element of this list is a data frame. An example: Mylist<-list(A=data.frame, B=data.frame, C=data.frame, D=data.frame) I want to rbind some elements of this list. As an example: Output<-list(AB=data.frame, CD=data.frame) Where AB=rbind(A,B) CD=rbind(C,D) I’ve tried: f<-function(x){ for (i in

On Wed, Jan 4, 2012 at 12:10 PM, David Gardner <daveg at xmos.com> wrote: > Jianzhou Zhao <jianzhou <at> seas.upenn.edu> writes: >> The documents say that all the aa analysis are chained, and give an >> example like opt -basicaa -ds-aa -licm. In this case, does ds-aa >> automatically call basicaa for the case when ds-aa can only return >> MayAlias? This

In a number of different situations I'm trying to remove factor levels that are represented by less than a certain number of rows, e.g. if I had the dataset aa below and wanted to remove the species that are represented in less than 2 rows: data(iris) aa <- iris[1:101,] In this case, since I can see that the species virginica only has one row, I can write: table(aa$Species) setosa

Hi there, I am trying to understand how AliasAnalysis works out in LLVM. I used the following simple test-case (test4.c): -- void test() { int *jj, *kk; int aa = 100; jj = &aa; *jj = 300; } int main() { test(); return 0; } -- Then I did "llvm-gcc -emit-llvm -c -o test4.bc test4.c" to get bc. I tried the following 2 ways to get what I expect to get as

Hi, I am sure there is an easy way to do it, but I can't find it. I have a data frame that has 15 columns and 7000 rows. The only values inside the data.frame are "aa", "ab", "bb" as you can see an example bellow. 1 2 3 1 aa ab ab 2 ab ab ab 3 aa aa aa 4 bb bb bb What I would like to do, is to generate a vector (or another data.frame) with 7000 rows, and 3

I'm not sure I'm doing this right. I have two identical machines, A and B.: ----------------------- CONSOLE ----------------------- A ~$: gluster peer probe domU-BB-BB-BB-BB.compute-1.internal Probe successful A ~$: gluster peer status Number of Peers: 1 Hostname: domU-BB-BB-BB-BB.compute-1.internal Uuid: 8d5d9af4-6a92-4d56-b063-c8fc9ac17a45 State: Peer in Cluster (Connected) B ~$:

Hi, I met with this error message with the following data set. Do you know how to resolve it? Thanks. > data<-read.table("c://temp3//abc.csv", sep = ",", header=T) > classwt<-c( 0.5806452, 0.4193548) > y<-data[,1] > x<-data[,2:ncol(data)] > print(y) [1] 1 1 1 1 1 1 1 1 1 2 2 2 2 2 2 1 1 1 1 1 1 1 1 1 2 2 2 2 2 2 1 1 1 1 1 [36] 1 1 1 1 2 2 2 2 2 2

It's found more and more like "get CFL-AA turned on by default" might be a viable GSoC project for the right student. It would require someone with existing knowledge of AA and a willingness to debug nasty problems, but it sounds like there's definitely interest in the community in seeing this happen. If the student finished early (unlikely), they could start on SCEV-AA as