similar to: Data Gaps

Hi, For this example: O <- c(0 0 0 2 0 0 2 0) I want to create an array every time O[i] > 0. The array should be in the form; R[j] <- array(-1, dim=c(2,O[i])) i.e. if O[i] > 0 4 times I want 4 R arrays. Does anyone have any suggestions? Thanks, Doug -- View this message in context: http://r.789695.n4.nabble.com/Create-Arrays-tp2996706p2996706.html Sent from the R help

Hi, I am having some problems using the if statement correctly. I have used it many times previously so I dona't know what is different with this case. Here is my problem: I have a 1X10 matrix of values as follows: > H.MC [,1] [1,] 4.257669 [2,] 7.023242 [3,] 4.949857 [4,] 5.107000 [5,] 4.257669 [6,] 4.257669 [7,] 4.257669 [8,] 4.257669 [9,] 4.257669 [10,] 4.257669

I am trying to convert an array from numeric values back to date and time format. The code I have used is as follows; for (i in 0:(length(DateTime3)-1)) { DateTime3[i] <- (strptime(start, "%m/%d/%Y %H:%M")+ i*interval) where start <- [1] "1/1/1981 00:00" However the created array (DateTime3) contains [1,] 347156400 347157600 347158800 347160000 347161200 347162400

R community, I am creating a bivariate return level plot by adding calculated return period values as lines onto an existing plot using the following code with the points representing the return periods. plot(H2,D2,pch="+",axes=TRUE) points(H.10,D.10, type="l",col="blue") points(H.20,D.20, type="l",col="green") points(H.50,D.50,

Is there a way to estimate a nominal response model? To be more specific let's say I want to calibrate: \pi_{v}(\theta_j)=\frac{e^{\xi_{v}+\lambda_{v}\theta_j}}{\sum_{h=1}^m e^{\xi_{h}+\lambda_{h}\theta_j}} Where $\theta_j$ is a the dependent variable and I need to estimate $\xi_{h}$ and $\lambda_{h}$ for $h \in {1...,m}$. Thank you, Mauricio Romero Quantil S.A.S. Cel: 3112231150

Looking for a little help figuring out what's driving gaps in data after merging two xts objects (msci.m and x2). The merge statement I'm using is ... y <-merge(x2,msci.m, all=FALSE). Here's info on the output , y: head(y) t-bill msci Sep 1985 7.310 316.963 Mar 1986 6.560 463.471 Jun 1986 6.180 498.791 Jul 1987 6.200 778.898 Aug 1987 6.400 833.519 Nov 1987

Dear R users, I have a dataset with two variables: $esan - a grouping factor with 8 levels and $reus. I'd like to do wilcox.test on this dataset as sugested Weiwei here: https://stat.ethz.ch/pipermail/r-help/2007-July/136627.html. I tried to adapt his recommendation but no succes. Can anyone help me? Regards, Iurie Malai, Senior Lecturer Department of Psychology Faculty of Psychology and

Hi, I am trying to find m breakpoints of a linear regression model. I used the segmented package. It works fine for small number of predicators and breakpoints.(3 r.v. 3 points). However, my model has 14 variables it even would not work even for just one breakpoints!. The error message is always estimated breakpoints are out of range. Since my problem is time related problem. So I

Hi, Sorry, but how did you bring it out? Thanks On Fri, Dec 28, 2012 at 8:48 AM, arun kirshna [via R] < ml-node+s789695n4654093h10@n4.nabble.com> wrote: > Hi, > bp.sub<- structure(list(CODEA = c(1L, 3L, 4L, 7L, 8L, 9L, 10L, 11L, 12L, > 13L, 14L, 16L, 17L), C45 = c(NA, 2L, 2L, 2L, 2L, 1L, NA, 1L, > 1L, 2L, 1L, 2L, 1L), ragek = c(3L, 3L, 3L, 4L, 4L, 3L, 3L, 3L, > 3L, 3L,

R community, I am trying to create an array of the time differences between datapoints for a very large set. For some reason for 4 of the values the difference has been calculated as NA. Looking at the individual points two of them are "1981-03-29 01:40:00" and "1981-03-29 02:00:00" This is the exact same format as the other points that return values of 20 mins as expected.

Hi, I am using the uniroot function in order to carry out a bivariate Monte Carlo simulation using the logistics model. I have defined the function as: BV.FV <- function(x,y,a,A) (((x^(-a^-1)+y^(-a^-1))^(a-1))*(y^(a-1/a))*(exp(-((1^(-a^-1)+y^(-a^-1))^a)+y^-1)))-A and the procedure is as follows: Randomly generate values of A~(0,1), y0 = -(lnA)^-1 Where: A=Pr{X<xi|Y=yi-1} and a is the

Hey, i want to define 3 ideal breaks (bin) for each variable one of those variables is attached in the previous email, i don't want to consider quartile method because quartile is not working ideally for that data set because data distribution is non normal. so i want you to suggest another method so that i can define 3 breaks with the ideal interval for Recency, frequency and monetary to

#is there a way to get NA in the table of descriptive statistics instead of the function stopping Thank you in advance #data x.f <- structure(list(Site = structure(c(9L, 9L, 9L, 9L, 9L, 9L, 9L, 9L, 9L, 9L, 9L, 9L, 9L, 9L, 9L, 9L, 9L, 9L, 9L, 9L, 9L, 9L, 9L, 9L, 9L, 9L, 9L), .Label = c("BC", "HC", "RM119", "RM148", "RM179", "RM185",

On Wed, May 29, 2013 at 03:25:30PM +0200, Duncan Sands wrote: > Hi Jack, I pulled the loop vectorizer and fast math changes into the 3.3 branch, > so hopefully they will be part of 3.3 rc3 (and 3.3 final!). It would be great > if you could redo the benchmarks rc3. > Duncan, As requested, appended are the updated Polyhedron 2005 benchmark results with both RC1 and RC3 llvm 3.3

(Apologies to those Toronto Asterisk Users' Group folks who have seen this message... I figured I'd have more success with a wider audience) I'm trying to boot a Cisco 7960 from an ISC DHCPD server (3.0.3 on FreeBSD 4.11), so far unsuccessful, and getting some odd behaviour on the wire. I wonder if anyone has done this before and therefore can validate whether or not the traffic I am

Estimados usuarios de R: Soy nuevo en R, le ruego disculpa por mi *Ignorancia*. Estoy corriendo la siguiente instruccion, ggsave(paste('img/',plotName,'.png', sep='') y resulta el error: ggsave(paste('img/',plotName,'.png', sep='') Saving 7.32 x 4.87 in image Error in grDevices::png(..., width = width, height = height, res = dpi, :

Hello! Given is an Excel-Sheet with actually 11,000 rows and 9 columns. I want to work with the data in R. The contents are similar to my following example. I have a list with ID-number, personal name and two kinds of loan-values. I want to aggregate the list, that for each person only one row remains and where the loan-values are added. First I tried some commands with tapply but had no

Below are the results for the Polyhedron 2005 benchmarks compiled with llvm/compiler-rt/dragonegg 3.3svn at r182439 against current FSF gcc 4.7.3svn and 4.8.1svn. The only major bug remaining in the dragonegg 3.3svn support for gcc 4.8.x is http://llvm.org/bugs/show_bug.cgi?id=15980 which results in unresolved symbols for _iround and _iroundf in the aermod and rnflow testcases. Note that this

On 03/17/2010 10:12 PM, Tanya Lattner wrote: > The 2.7 binaries are available for testing: > http://llvm.org/pre-releases/2.7/pre-release1/ > > You will also find the source tarballs there as well. > > We rely on the community to help make our releases great, so please help > test 2.7 if you can. Please follow these instructions to test 2.7: > > /To test llvm-gcc:/

OK, so maybe last night was a little too much at one throw, so I have reduced the data to two stations- one that has precipitation and one that does not. This is going to be in the context of a larger data set. I would like to be able to issue a ggplot command and have cum sum just act on the facets (factors) to apply this. library(chron) library(ggplot2) DF <- structure(list(date_time =