similar to: Zoo Data

I want to count attributes of IDs: --8<---------------cut here---------------start------------->8--- z <- data.frame(id=c(10,20,10,30,10,20), a1=c("a","b","a","c","b","b"), a2=c("x","y","x","z","z","y"),

hi, i'm trying to figure out if there's any possibility to write a whole vector into a matrix or data.frame or something like that. i don't mean transormation. Here an example: [,1] [,2] [1,] "a" "d" [2,] "b" "e" [3,] "c" "f" where e.g. a is a<-c(0,1) vector of length 2, b a vector of length 4,... (i know that

Hi, I am trying to get the maximum likelihood estimator for lognormal distribution with censored data;when we have left, interval and right censord. I built my code in R, by writing the deriving of log likelihood function and using newton raphson method but my estimators were too high " overestimation", where the values exceed the 1000 in some runing of my code. is there any one can

Hi I have to time series with a different time base. The first has only sporadic datapoints: 2011-02-01 15.29130 2011-02-08 17.60278 2011-02-15 17.99737 2011-02-22 25.43690 The other has a daily datapoint: 2011-02-01 342.34 2011-02-02 68.45 2011-02-03 130.47 2011-02-04 129.86 2011-02-05 81.98 2011-02-06 77.30 2011-02-07 81.38 2011-02-08 139.95 2011-02-09 124.40 ...etc. In Excel, it is fairly

Hi , I am trying to get some estimator based on lognormal distribution when we have left,interval, and right censored data. Since, there is now avalible pakage in R can help me in this, I had to write my own code using Newton Raphson method which requires first and second derivative of log likelihood but my problem after runing the code is the estimators were too high. with this email ,I provide

Hello, I have a problem shutting down a domU with xen-4.1.2, which doesn''t terminate the corresponding blktap2 process, since one (other) VM uses a image file, which contains spaces in its file name. /var/log/xen/xend-debug.log has the following information: Unhandled exception in thread started by Traceback (most recent call last): File

hello I make a subroutine that give-me a (mathematical) function in string format. I would like transform this string into function ( R function ). thanks for any tips. cleber #e.g. fun_String = "-100*x1 + 0*x2 + 100*x3" fun <- function(x1,x2,x3){ return( ############ evaluation( fun_String ) ############ ) True String mathematical function :-( :-( > nomes [1]

Hi, R users, I'm trying to transform a matrix A into B (see below). Anyone knows how to do it in R? Thanks. Matrix A (zone to zone travel time) zone z1 z2 z3 z1 0 2.9 4.3 z2 2.9 0 2.5 z3 4.3 2.5 0 B: from to time z1 z1 0 z1 z2 2.9 z1 z3 4.3 z2 z1 2.9 z2 z2 0 z2 z3 2.5 z3 z1 4.3 z3 z2 2.5 z3 z3 0 The real matrix I have is much larger, with more than 2000 zones. But I think it should

Hi list, I am unsuccessfully trying to produce a serious of trellis barcharts from within a for-loop. The barcharts work outside the loop. What am I missing? Example attached. Thanks Herry #XXXXXXXXXXXXXXXXXXXXXX trellis.device(bg="white") trellis.par.get("fontsize")->fontsize fontsize$default<-16 trellis.par.set("fontsize",fontsize)

Hi Sarah, Thank you so much!! I got your good ideas. Ding -----Original Message----- From: Sarah Goslee [mailto:sarah.goslee at gmail.com] Sent: Friday, May 11, 2018 11:40 AM To: Ding, Yuan Chun Cc: r-help mailing list Subject: Re: [R] add one variable to a data frame [Attention: This email came from an external source. Do not open attachments or click on links from unknown senders or

Sarah's solutions are good, and here's another, even more basic: tmp1 <- unique(dat1$B) tmp2 <- seq_along(tmp1) dat1$C <- tmp2[ match( dat1$B, tmp1) ] > dat1 N B C 1 1 29_log 1 2 2 29_log 1 3 3 29_log 1 4 4 27_cat 2 5 5 27_cat 2 6 6 1_log 3 7 7 1_log 3 8 8 1_log 3 9 9 1_log 3 10 10 1_log 3 11 11 3_cat 4 12 12 3_cat 4 As a single line

Dear R users, I have posted this question several days ago and received not a single suggestion. I believe I have provided sufficient information for at least some help. Here I repost the question with several modifications. I want to run nested fixed-factor Anova in R on different experiments. I have 48 levels of the main factor x1 and 242 levels of the nested factor z1, and continuous response

Hi Team, I am trying to very simple plot with command plot. Question : I am trying to plot (x,y) based on the value of Column e. If column e value is greater than 0 then plot(x,y) otherwise do not plot it. Data : structure(list(x = c(1, 1, 1, 2, 2, 2, 3, 3, 3), y = c(1, 2, 3, 1, 2, 3, 1, 2, 3), e = c(0, -1, -2, 1, 0, -1, 2, 1, 0)), row.names = c(NA, -9L), .Names = c("x",

Rather than “fixing” it, it might be better to support a separate method for signed extension. My reasoning is as follows: int x = 7; llvm::PointerIntPair<double*, 3, int> pip; pip.setInt(x); There could be code out there that expects pip.getInt() to return 7 and not -1. So if you really want to set a negative and return a negative value, a separate method setSignedInt and getSignedInt

Dear all, I have a dataset. I want to make barplot from this data. Zero1 <- " V1 V2 V3 V4 V5 V6 V7 V8 V9 1 1 0 0 0 1 0 0 0 Positive 2 0 0 1 0 1 0 1 1 Negative 3 0 0 1 0 0 0 1 1 Positive 4 0 1 0 1 1 1 0 1 Negative 5 0 0 1 0 1 1 0 0 Positive 6 0 1 0 0 1 1 1 1 Negative 7 1 0 1 1 1 1 1 1 Negative 8 0 0 0 0 1 0 0 1

Hi All, I have a data frame dat1: dat1 <-data.frame(N=seq(1, 12,1), B=c("29_log","29_log", "29_log", "27_cat", "27_cat", "1_log", "1_log", "1_log", "1_log", "1_log",

> Hello, > > On 01.08.2014 19:22, Ady wrote:>> On 24.07.2014 17:32, Gene Cumm wrote: > >>> On Thu, Jul 24, 2014 at 10:29 AM, Philipp Hahn <hahn at univention.de> > wrote: > >>>> On 24.07.2014 13:52, Gene Cumm wrote: > ... > >>>> Your special build gets the gateway working, but the downloaded InitRD > >>>> still

Hey All, I have a dataset(dat1) like this: ObsNumber ID Weight 1 0001 12 2 0001 13 3 0001 14 4 0002 16 5 0002 17 And another dataset(dat2) like this: ID Height 0001 3.2 0001 2.6 0001

Dear Rusers, I would like some comments about the following results (under R-2.2.0) > m = matrix(1:6 , 2 , 3) > m [,1] [,2] [,3] [1,] 1 3 5 [2,] 2 4 6 > z1 = m[(m[,1]==2),] > z1 [1] 2 4 6 > is.matrix(z1) [1] FALSE > z2 = m[(m[,1]==0),] > z2 [,1] [,2] [,3] > is.matrix(z2) [1] TRUE Considered together, I'm a bit surprised about

Colleagues, I am in the process of learning R. I've been able to import my dataset (from Stata) and do some simple coding. I have now come to coding situation that requires some assistance. This is some code in SPSS that I would like to be able to execute in R: if (race eq 1 and usborn=0) confused=1 . if (race eq 2 and usborn=0) confused=1 . if (race eq 1 and usborn=1) confused=0 . if (race