search for: anas

Hi, I'm having this Sendmail configuration problem and stuck. I've been doing googling/reading posts and none of the solutions matched my problem. I'm doing some testing, and have been getting this 550 5.1.1 ....User unknown error when I send the email to my company's email address. When I send it to an external email address, such as yahoo, the message delivered successfully.

Hello. I have a hudge problem, don't know how to deal with it... I'm supposed to implement a tree structure in R, without using rpart library... I don't have any clue about appropriate approach... Anyway, I can't use pointers under R...so how to to it? lists, vectors... Anyway, does anybody knows how the regression tree in rpart was implemented... Someone please help! Thanks, ana

Hi to all, I initiate in R - Linux and I've some problems to find an editor with R interface as like RWinEdt for WinEdt. Anyone know one? Thanks in advance for your kind cooperation. Best regards Atenciosamente Ana Patricia Martins ------------------------------------------- Serviço Métodos Estatísticos Departamento de Metodologia Estatística Telef: 218 426 100 - Ext: 3210

In this real example (below), all four of the replicates in one treatment combination had zero failures, and this produced a very high standard error in the summary.lm. =20 Just adding one failure to one of the replicates produced a well-behaved standard error. =20 I don't know if this is a bug, but it is certainly hard for users to understand. =20 I would value your comments=20 =20 Thanks =20

Hi folks, Moving the discussion to llvm.dev. None of the changes we talked earlier help. Find attached the C source code that you can use to reproduce the issue. clang --target=aarch64-linux-gnu -c -mcpu=cortex-a57 -Ofast -fno-math-errno test.c -S -o test.s -mllvm -debug-only=licm LICM hoisting to while.body.lr.ph: %21 = load double** %arrayidx8, align 8, !tbaa !5 LICM hoisting to

Can you send me actual LLVM IR or a preprocessed source from using -E? I don't have a machine handy that has headers that target that arch. On Tue Jan 13 2015 at 4:33:29 PM Daniel Berlin <dberlin at dberlin.org> wrote: > Anything other than noalias or mustalias should be getting passed down the > stack, so either that is not happening or CFL aa is giving better answers > and

Oh, sorry, i didn't rebase it when i changed the fix, you would have had to apply the first on top of the second. Here is one against HEAD On Wed, Jan 14, 2015 at 12:32 PM, Ana Pazos <apazos at codeaurora.org> wrote: > Daniel, your patch does not apply cleanly. Are you on the tip? > > The code I see there is no line if (QueryResult == MayAlias|| QueryResult == PartialAlias)

Hola a todos/as: Tengo la siguiente duda: Construyo una lista de tamaño 2, del siguiente modo: miLista<-list() miLista[[1]]<-list(nombre="Ana", apellido1="Pérez", apellido2="Sánchez") miLista[[2]]<-list(nombre="Carlos", apellido1="Núñez", apellido2="Sierra") Le aplico la función turnoLaboratorio: miResul<-lapply(miLista,

Inline - George > On Jan 14, 2015, at 10:49 AM, Daniel Berlin <dberlin at dberlin.org> wrote: > > > >> On Tue, Jan 13, 2015 at 11:26 PM, Nick Lewycky <nlewycky at google.com> wrote: >>> On 13 January 2015 at 22:11, Daniel Berlin <dberlin at dberlin.org> wrote: >>> This is caused by CFLAA returning PartialAlias for a query that BasicAA can

On 13 January 2015 at 22:11, Daniel Berlin <dberlin at dberlin.org> wrote: > This is caused by CFLAA returning PartialAlias for a query that BasicAA > can prove is NoAlias. > One of them is wrong. Which one? I'm not sure from your description that this is a chaining issue. PartialAlias doesn't chain and isn't supposed to, it's a final answer just like NoAlias and

Yes. I've attached an updated patch that does the following: 1. Fixes the partialalias of globals/arguments 2. Enables partialalias for cases where nothing has been unified to a global/argument 3. Fixes that select was unifying the condition to the other pieces (the condition does not need to be processed :P). This was causing unnecessary aliasing. 4. Adds a regression test to

The directory that I am trying to clean up is huge . every time get this error msg -bash: /usr/bin/find: Argument list too long Please advise Anas -------------- next part -------------- An HTML attachment was scrubbed... URL: <http://lists.centos.org/pipermail/centos/attachments/20100123/f6534851/attachment-0001.html>

Eva, ¿Algo asi o te entiendo mal? l <- list()l$nombre <- "Ana"l$apellido <- "Pérez"parametro <- "nombre"if (exists(paste('nombre', 'OnValidate', sep=''))) nombreOnValidate(1:5) # cuidado con el enviroment nombreOnValidateif (exists(paste('s', 'um', sep=''))) sum(1:5) # esto lo agrego como ejemplo[1] 15l[

Hola, tengo un script para usar la función POSIXct y lo corro en otras computadoras con otras versiones de R y no tengo problemas ahora cuando lo corro en mi maquina donde tengo la versión 3.1.3 no me da ningun error pero genera una nueva columna con NA puede ser sólo problema de la versión de R? gracias ana -- Ana Marina Srur Departamento de Dendrocronología e Historia Ambiental

Hi, Is there any way to estimate a DEPENDENT variable through a GLM/LM model? Suppose I have the linear model: y=a0+a1*x1+a2*x2 (a0=1, a1=0.6, a2=0.8, x1~N(1,1), x2~N(0,1)). The alphas and the auxiliary variables are given and I have to estimate y. The point is if I estimate it, let¹s say algebraically, I get high variances that do not decrease as sample sizes increases... Is the any other way

Recientemente me intalaron la plataforma Platform: i686-pc-linux-gnu (32-bit). He intentado instalar el paquete dplR tanto desde el cran como desde un archivo y me da el siguiente error: ERROR: dependency ?XML? is not available for package ?dplR? Esto se debe a la versión de Linux que tengo instalada? gracias -- Ana Marina Srur Departamento de Dendrocronología e Historia Ambiental

Buen día estimados, Tengo el siguiente código: df_1 <- data.frame(ana = c(15, 20, 30), maria = c(15,20,30), jose = c(15, 20, 30)) df_2 <- data.frame(nombre = c("jose", "ana", "maria"), valor = c(1,2,3)) # Find the corresponding columns in df_1 based on the values in df_2$nombre cols <- match(df_2$nombre, names(df_1)) # Subtract the values of df_2$valor

Hi everybody. I'm a new R user and i've been searching a tool for construction of regression tree... I found function "tree()" written by a certain Mr. Ripley, and seems to be just what i'm looking for, but when i try to use it in R replies me:"Object not found". So I was wandering if I should include one special library or something like that? Thanks a lot, ana

Algo asi? l <- list()l$nombre <- "Ana"l$apellido <- "Pérez"parametro <- "nombre"paste(l[ parametro][[1]], 'OnValidate', sep='')[1] "AnaOnValidate"l[ parametro][[1]][1] "Ana"x <- 1:5AnaOnValidate <- function(x) print(x)if (exists(paste(l[ parametro][[1]], 'OnValidate', sep=''))) do.call(paste(l[

Hi, if i just want a vector filled with names which has length(index) > 0. For example if nombreC <- c("Juan", "Carlos", "Ana", "Mar?a") nombreL <- c("Juan Campo", "Carlos Gallardo", "Ana Iglesias", "Mar?a Bacaldi", "Juan Grondona", "Dario Grandineti", "Jaime Acosta",