similar to: help needed to find zero areas in a vector

Dear everyone in the Mailing list:   It is easy to get the unique elements in a column. But I would like to get rid of those rows that the elements of this column are redundant. Or sometimes, to have a look at the rows that the elements of this column are redundant is also important. I guess it boils down to throw out the index of the redundant elements.   With millions of rows, how can I

Dear Helpers: If you know how to fit a Weibull model using R, you are my savior for my project. The hazard ramda is specified as t^alpha * (1+a*dose*exp(-b*dose)). Or, can we modify the Cox model to capture the proportional part above? Here t is the survial time, dose is a subject received in a exposed group. Thank you a lot. Boshao

I have the following data (a subset of the real data set): structure(list(Date = structure(c(1209625080, 1209641460, 1209652500, 1209676800, 1209682860, 1209692100, 1209706980, 1209722580, 1209726300, 1209739620, 1209762780, 1209765720, 1209770520, 1209791040, 1209812580, 1209829920, 1209837180, 1209848160, 1209854640, 1209859440, 1209870780, 1209887760, 1209901080, 1209921660, 1209929280,

Hi. I have the following code in a loop. It splits a vector into subvectors of equal size. But if the size of the original vector is not an exact multiple of the desired subvector size, then the first few subvectors have one more element than the last few. I know that the cut function could be used to determine where to break up the vector, but it doesn't seem to provide control over

Dear all, I'm trying to optimize code and want to avoid for-loops as much as possible. I'm applying a calculation on subvectors from a big one, and I get the subvectors by using a vector of starting positions: x <- 1:10 pos <- c(1,4,7) n <- length(x) I try to do something like this : pos2 <- c(pos, n+1) out <- c() for(i in 1:n){ tmp <- x[pos2[i]:pos2[i+1]]

Hi. Suppose I have a vector that I partition into disjoint, contiguous subvectors. For example, let v = c(1,4,2,6,7,5), partition it into three subvectors, v1 = v[1:3], v2 = v[4], v3 = v[5:6]. I want to find the maximum element of each subvector. In this example, max(v1) is 4, max(v2) is 6, max(v3) is 7. If I knew that the successive subvector maxima would never decrease, as in the example,

Hi, I've got a list of ~20000 elements that look like this: [1] "A00096:A00096:A00096:A00096:A02178:A02178:A07776" [2] "A00046:A00076:A01101:A04146:A05671:A07169" [3] "A00038:A00932:A02185:A02370:A02818:A02818:A02818:A02818:A04732:A07142:A07142" [4] "A00096:A01352:A01352:A02023:A05001:A05001:A07776" [5]

hi, All. Date Dtime Hour Min Second Rep App_dur 9 2006-02-22 14:36:11 14 36 11 4 1 10 2006-02-22 14:36:12 14 36 12 3 86 11 2006-02-22 14:37:38 14 37 38 0 58 14 2006-02-22 14:38:36 14 38 36 3 1 15 2006-02-22 14:38:37 14 38 37 4 1 16 2006-02-22 14:38:38 14 38 38 1 9 18 2006-02-22 14:38:47 14 38

Hi David, Responses below. -Graham On 11 Jun 2018, at 22:19, David A. Greene <dag at cray.com<mailto:dag at cray.com>> wrote: Graham Hunter <Graham.Hunter at arm.com<mailto:Graham.Hunter at arm.com>> writes: ======== 1. Types ======== To represent a vector of unknown length a boolean `Scalable` property has been added to the `VectorType` class, which indicates that

R-help readers, I''m having a problem with an R script (see below), which regularly generates the error message, Error in start:(start + (sample.length - 1)) : NA/NaN argument , for which I am unsure of the cause. In essence, the script (below) generates the start and end points for random subsamples from along a vector (in reality a transect (of a given length,

Hi, everyone I am trying to estimate 3 parameters for my survival function. It's very complicated. The negative loglikelihood function is: l<- function(m1,m2,b) -sum( d*( log(m1) + log(m2) + log(1- exp(-(b + m2)*t)) ) + (m1/b - d)*log(m2 + b*exp(-(b + m2)*t) ) + m1*t - m1/b*log(b+m2) ) here d and t are given, "sum" means sum over these two vairables. the parameters

Hi, I have the following sequence: in <- c(0, 0, 0, 2, 1, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 1, 0, 1, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 2, 0, 2, 0, 0, 2) >From this sequence I would like to get to the following sequence: out <- c(0, 0, 0, 3, 3, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 1, 0, 1, 0, 0, 0, 0, 11, 11, 11, 11, 11, 11, 11, 11, 11, 11, 11, 0, 2, 0, 2, 0, 0, 2) Basically,

Please tell me whether the following implementation is correct..... My target supports 64 bit mask means immediate(0-2^63) I have implemented it but i dont know whether its correct or not. Please see the changes below that i have made in x86isellowering.cpp static SDValue lower2048BitVectorShuffle(const SDLoc &DL, ArrayRef<int> Mask, MVT VT,

Hi all, I have a dataset. Each time I want to sample N(i) elements from it and I want to repeated sampling M times. N(i) is varied from time to time. For example, dataset = 1:50; a = list(); M = 1000; I want to do something like for(i in 1:M) { a[[i]] = sample(dataset,

Hello, My out-of-tree target features some high latency instructions (let's call them FXLV). When an FXLV issues, it reserves its destination register and execution continues; if a subsequent instruction attempts to read or write that register, the pipline will stall until the FXLV completes. I have attempted to encode this constraint in the machine scheduler (excerpt at bottom of email).

Hi everyone, Please forgive me if my question is simple and my code terrible, I'm new to R. I am not looking for a ready-made answer, but I would really appreciate it if someone could share conceptual hints for programming, or point me toward an R function/package that could speed up my processing time. Thanks a lot for your help! ## My dataframe includes the variables 'year',

Hi! Does anyone know hot to save (to png, pdf, ...) only plot area. That is without box around, titles, axis, ... I tried: par(mar=c(0,0,0,0)) and it works for internal viewer (linux and windows). But when I save the image in png it adds that annoying margin around plot area. Any ideas? by, Tine

Hello dear R-helpers, I haven't been able to figure out of find a solution in the R-help archives about how to delete all the characters contained in groups of parenthesis. I have a vector that looks more or less like this: myvector<-c("something (80 km/h, sd) & more (6 kg/L,sd)", "somethingelse (48 m/s, sd) & moretoo (50g/L , sd)") I want to extract all

I generate random vector from Weibull distribution sampWB <-urweibull(sampleSize, shape=shape.true, scale=scale.true, lb=0, ub=Inf) how can I create subvector containing 30% of samplesize of sampWB which should be assigned as Censored data? The probability for each value in sampWB can be uniform to be included in the subvector.

The code I'm using is shown below. I would like to have a larger median at the top of the plot so that I can show the entirity of "title_text". Several times I tried messing with "par(mar", but that seemed to make matters worse. By any chance can anyone provide any insight as to the best way to increase the top/upper/northern margin, so the entirity of the title is