similar to: memory.size help

Consider a data frame (x) with 2 variables, x1 and x2, having equal values. It looks like: x1 x2 1 1 2 2 3 3 Now, consider a second data frame (xk): xk1 xk2 0.5 0.5 1.0 0.5 1.5 0.5 2.0 0.5 0.5 1 1.0 1 1.5 1 2.0 1 0.5 1.5 1.0 1.5 1.5 1.5 2.0 1.5 0.5 2 1.0 2 1.5 2 2.0 2 I have written code to calculate some differences between these

This is a continuation from a previous posting of mine: The following algorithm below is what I want to accomplish: Z(xk) = Average(Yi, i belongs to Ik), where Ik contains all i such that for each j, |Xi,j - xkj?? 2. Here, j = 1, 2 and i corresponds to the elements in each X and/or xk >data x1 x2 y 1 1 2 2 2 6 3 3 12 Now, consider a second data frame or matrix (xk):

useR's, Consider some variables and a data frame of points: x1 <- c(1,2,3) x2 <- c(3,4,5) xk1 <- seq(min(x1)-.5, max(x1)+.5,.5) xk2 <- seq(min(x2)-.5, max(x2)+.5,.5) expand.grid(xk1=xk1,xk2=xk2) xk1 xk2 1 0.5 2.5 2 1.0 2.5 3 1.5 2.5 4 2.0 2.5 5 2.5 2.5 6 3.0 2.5 7 3.5 2.5 ... 46 2.0 5.5 47 2.5 5.5 48 3.0 5.5 49 3.5 5.5 I want to restrict the data frame to only contain

useR's, Consider the variables defined below: yvals <- c(25,30,35) x1 <- c(1,2,3) x2 <- c(3,4,5) x3 <- c(6,7,8) x <- as.data.frame(cbind(x1,x2,x3)) delta <- c(2.5, 1.5, 0.5) h <- delta/2 vars <- 3 xk1 <- seq(min(x1)-0.5, max(x1)+0.5, 0.5) xk2 <- seq(min(x2)-0.5, max(x2)+0.5, 0.5) xk3 <- seq(min(x3)-0.5, max(x3)+0.5, 0.5) xks <- list(xk1,xk2,xk3) xk <-

useR's, I am trying trying to find out if there is a faster way to do a certain computation. I have successfully used FOR loops and the apply function to do this, but it can take some time to fully compute, but I was wondering if anyone may know of a different function or way to do this: > x [1] 1 2 3 4 5 > xk [1] 0.0 0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.0 4.5 5.0 5.5 6.0 I want to do:

Hello! With this email we release an extension to OpenSSH that was initially developed as project for our studies at the Univerity of Applied Sciences in Hagenberg. First we would like to describe the purpose of using Zero-Knowledge (ZK) for user authentication. Traditional authentication methods like challenge-response with passwords or public keys leak information about the credentials of

useR's, I would like to know if there is a way to avoid using FOR loops to perform the below calculation. Consider the following data: > x [,1] [,2] [,3] [1,] 4 11 1 [2,] 1 9 2 [3,] 7 3 3 [4,] 3 6 4 [5,] 6 8 5 > xk Var1 Var2 Var3 1 -0.25 1.75 0.5 2 0.75 1.75 0.5 3 1.75 1.75 0.5 4 2.75 1.75 0.5 5 3.75 1.75

I have the following problem. Consider a `classical' test which works for k .ge. 2 samples. Possible interfaces are e.g. xxx.test(x, g) x ... all data, g ... corresponding groups xxx.test(x1, ..., xk) xxx.test(list(x1, ..., xk)) etc etc. Clearly, the first and the second one are nice, but cannot be combined without making `g' (i.e., `group') a named argument. Hence, in

Is anybody using mod_cache/mod_disk_cache with Puppet? I found a post talking about it here (http://paperairoplane.net/?p=380) and I tried to implement it .. but I found that nothing was being cached. Near as I can tell, Apache refuses to cache any URL that has a query-string attached to it: (http://httpd.apache.org/docs/2.2/caching.html) > • If the URL included a query string (e.g. from a

I'm pulling my hair (and there's not much left!) on this one. Basically I'm not getting the same result t when I "step" through the program and evaluate each element separately than when I use the outer() function in the FindLikelihood() function below. Here's the functions: Dk<- function(xk,A,B) { n0 *(A*exp(-0.5*(xk/w)^2) + B) } FindLikelihood <-

Dear readers of the list, I have a problem a comparison of two data from a vector. The comparison yields FALSE but should be TRUE. I have checked for mode(), length() and attributes(). See the following code (R2.10.0): ----------------------------------------------- # data vector of 66 double data X =

I would like to be able to exert certain types of control over the plotting of dendrograms (representing hierarchical clusterings) that I think is best achieved by modifying the dendrogram object prior to plotting. I am using the "dendrogram" class and associated methods. Define the cluster number of each cluster formed as the corresponding row of the merge object. So, if you are

Dear all, Is there any method in R to find all possible nonnegative integer solutions to the linear equation with unit coefficients as follow: X1+X2+...+Xk=N Thank you, Amin Zollanvari

Dear all, I have a simple question which I unfortunately do not seem to be able to solve myself. I have a (NxK) matrix and want to generate a new matrix by multiplying each row with itself such that the new matrix has dimension ((N*K)xK) (or better, generate an array with dimension (K,K,N)). I tried apply, but that did not work. Any suggestions? Thanks! Stephan ## Here is a simple

Hey all I am having instances where voicemail boxes will have a 00001 message and no 00000 message this causes the user to be told that they have a message that they can't get at. If I renumber the messages manually to start with the 00000 numbering then the user can get their messages. What could be causing this and how can I get it out of the system. Is there a patch I can apply to the

Can somebody please take a look at this and tell me whats going wrong? It seems to be parsing wronly around the 'if' statement and gives me a directory listing. Thanks in advance Tom N.B. datan is an invented dataset xvals<-c(1,0.4,0.2) datan<-data.frame(s1=c(3,4,5),s2=c(5,5,5),s3=c(21,55,34),s4=c(5,3,2)) datan$sint<-NA datan$sgrad<-NA for(icount in 1:dim(datan)[1]) {

On Mon, Feb 26, 2018 at 3:25 PM, Gary Black <gwblack001 at sbcglobal.net> wrote: (Sorry to be a bit slow responding.) You have not supplied a complete example, which would be good in this case because what you are suggesting could be a serious bug in R or a package. Serious journals require reproducibility these days. For example, JSS is very clear on this point. To your question >

Thanks for the quick reply, I believe I am using MIT based on log file names; but is there a better way to tell? I'm not very knowledgeable about the distinction between MIT and Heimdal regarding KDC. Can you direct me to a resource that explains how to make the switch as I am just using the defaults in SUSE. Additionally, many of the domains experiencing this bug were working fine; before

Thank you, everybody, who replied! I appreciate your valuable advise! I will move the location of the set.seed() command to after all packages have been installed and loaded. Best regards, Gary Sent from my iPad > On Mar 4, 2018, at 12:18 PM, Paul Gilbert <pgilbert902 at gmail.com> wrote: > > On Mon, Feb 26, 2018 at 3:25 PM, Gary Black <gwblack001 at sbcglobal.net> >

Dear R experts, I'm trying to find a code to calculate the p-value from the score test for the logistic regression. My fit is like this: logit=beta0+beta1*x1+beta2*x2 +....+ betak* xk. And my H0 is beta1=beta2=...=betak =0. Any help will be highly appreciated. Thank you! Ying