similar to: Ordering long vectors

I am trying to plot fitted lme values as a smooth line of a graph showing the exponential relationship between temperature and soil respiration. In the plot, the x-axis has temperature, and the y-axis has soil respiration. When I try to add a line showing temperature versus the fitted values, it is jagged and not smooth. Here is the code I used: lme.1<-lme(fixed=LnFlux~Temp,

library(survival) library(boot) data=NULL lambda=NULL result=NULL pat=rep(1:102,each=1) trt=rep(c(1,0),51) status=rep(1,102) site=rep(1:51, each=2) nr.datasets=100 seed=2006 beta=log(1/2) for (i in 1:51) { lambda[i]=1+((3-1)/50)*(i-1)} lambda1=rep(lambda, each=2) dummy=rep(c(exp(beta),1),51) elf=lambda1*dummy r=70 #the number of bootstrap replicates

Say, I just picked this up on my messages! There are a whole host of these requests! Anyone know whow there people are? Is there a way to report them? Any suggestions as to how to block them? [Aug 23 10:34:16] NOTICE[1010] chan_sip.c: Registration from '"912" <sip:1 at 41.1.1.1>' failed for '184.106.217.112' - Wrong password [Aug 23 10:34:16] NOTICE[1010]

Suppose one has x <- c(1, 2, 7, 9, 14) y <- c(71, 72, 77) How would one write an R function which alternates between elements of one vector and the next? In other words, one wants z <- c(x[1], y[1], x[2], y[2], x[3], y[3], x[4], y[4], x[5], y[5]) I couldn't think of a clever and general way to write this. I am aware of gdata::interleave() but it deals

Anybody else having problems accessing this site? Haven''t had much luck over the past couple of days. -- Posted via http://www.ruby-forum.com/.

H ello R-experts, I want to do ordination plots using vegan metaMDS. I have a where many cells have zero values. Data structure: X[1:10,1:14] Height.1 Height.2 Height.3 Height.4 Height.5 Height.6 Height.7 Height.8 Height.9 Height.10 Height.11 Height.12 Height.13 D30I1A 46 0 0 0 0 0 0 0 0 0 39 0 98 D30I1B

I am trying to sort a list and the data is obiously not in the right format. I am trying: x <- list() x[["A"]] <- 1 x[["B"]] <- 2 order(x) But am getting: Error in order(x) : unimplemented type 'list' in 'orderVector1' How should I change the list so that it can be sorted? What kinds of objects (classes of objects) can be sorted? Thank you. Kevin

Dear R-help members, Apologies - I am posting on behalf of a colleague, who is a little puzzled as STATA and R seem to be yielding different survival estimates for the same dataset when treating a variable as ordinal. Ordered() is used to represent an ordinal variable) I understand that R's coxph (by default) uses the Efron approximation, whereas STATA uses (by default) the Breslow. but we

Dear all, If I have a data frame x<-data.frame(x1=c(1,7),x2=c(4,6),x3=c(8,2)): x1 x2 x3 1 4 8 7 6 2 I want to sort the whole data and get this: x1 1 x3 2 x2 4 x2 6 x1 7 x3 8 If I do sort(X), R reports: Error in order(list(x1 = c(1, 7), x2 = c(4, 6), x3 = c(8, 2)), decreasing = FALSE) : unimplemented type 'list' in 'orderVector1' The only way

Hi, base::order main input arguments are defined as: a sequence of numeric, complex, character or logical vectors, all of the same length, or a classed R object When passing a list or a data.frame, the resuts seems to be a bit useless. Shouldn't that raise an error, or at least warning? Best Regards, Jan Gorecki

Dear R user, I am a biochemist/bioinformatician, at the moment working on protein clusterings by conformation similarity. I only started seriously working with R about a couple of months ago. I have been able so far to read my way through tutorials and set-up my hierarchical clusterings. My problem is that I cannot find a way to obtain information on the rooting of specific nodes, i.e. of

In the text file pressione2008.csv I have the following "Data","MAX","MIN","Note" "07-01-2008 08:00:00", 135, 90, "Eccessi feste, inizio dieta" "07-01-2008 18:00:00", 135, 85, "" "08-01-2008 08:00:00", 125, 75, "" which is a collection of blood pressure data at different time of the day. I would

I guess we could make it do the equivalent of do.call(order, df). On Mon, May 18, 2020 at 8:32 AM Rui Barradas <ruipbarradas at sapo.pt> wrote: > > Hello, > > There is a result with lists? I am getting > > > order(list(letters, 1:26)) > #Error in order(list(letters, 1:26)) : > # unimplemented type 'list' in 'orderVector1' > >

Hello, i would like to import this txt file: Giorno;PM10 2006-01-01 10:10;10.3 2006-02-02 20:22;50.3 2006-03-03 23:33;20.1 ......... As it's an irregular time series i use zoo as follow: require(zoo) z <- read.table("c:\\1.csv", sep=";", na.strings="-999", header=TRUE) q <- zoo(z$PM10, strptime(as.character(z$Giorno),"%Y-%m-%d %H:%M")) At this

If anyone can assist with this problem you have my great thanks: I am trying to establish and plot confidence intervals on a bootstrapped function. I have a more complicated function that has no problems with determining the confidence intervals using the quantile command. This is outside the bootstrap portion of the code that is working fine it is just determining everything for the more

Full_Name: Murray Smith Version: 2.2.1 OS: Windows XP Submission from: (NULL) (202.36.29.1) order() will not allow a complex vector for its first argument. This contrasts with sort() which will happily sort a complex vector. While this may not be regarded as a bug by some, it is annoying for anyone who makes frequent use of complex numbers. The problem occured in the Windows version 2.2.1 of

Dear Sir, In the functions orderVector1, orderVector1l (R-3.5.2/src/main/sort.c) there are two loops concerning nalast (lines 1096, 1105). I am not sure about the possibility of redefining them, so this function should be a little faster. The first one (line 1096) can be included in the previous 'switch' block (line 1079) (see below). And if you rewrite/duplicate this 'switch'

Dear R-helpers, We marked 6000 leaves from 5 SPECIES - 10 individuals/species - in two different TREATMENTs: a control and a dry-plot from which 50% of incoming precipitation was excluded. We followed those leaves for 42 months and noted the presence and absence at each visit. I then carried out a Cox Harzard model to see differences in leaf mortality between parcels and among species over time:

Hey all, I noticed that my puppet server running CentOS 6.5 was acting a little pokey. So I logged in and did what well just about anyone would've done. And ran the uptime command to have a look at the load. And it was astonishingly high! [root at puppet:~] #uptime 21:28:01 up 1:26, 3 users, load average: 107.37, 72.06, 75.52 So then I had a look at top and saw a LOT of processes

Hello, I’m trying to follow the syntax of a script from a journal website. In order to create a regression formula used later in the script, the regression matrix must have column names “X1”, “X2”, etc. I have tried to assign these column names to my matrix ScoutRSM.mat using a for loop, but I don’t know how to interpret the error message. Suggestions? Thanks, Paul