similar to: Permutation Test

Any help with this problem would be greatly appreciated: I need to produce a custom plot i haven't come across in R. Basically, I want to show means, 1st standard deviation and 5th and 95th percentiles visually, using something resembling a boxplot. Is it possible to completely customize a boxplot so that it shows means as the bar (instead of, not as well as medians), standard deviations at

I've been comparing variables among objects (taxa) related by known trees, using phylogentically independent contrasts in the ape package, and want to move on to more complex models e.g. by using gls with appropriate correlation terms. My trees contain lots of (hard) polytomies and information about ancestors, which I've been including- creating fully dichotomous trees by using zero branch

Hello all, I have a *.csv file that looks like this (actual file is orders of magnitude larger): Site taxa no.ind forest LMA 1 forest LCY 1 forest SCO 1 meadow LMA 2 meadow LCY 1 meadow PNT

Dear all, I was wondering if you could help me with the following. I want to do calculate this equation: Ps(t)= (?_(r=1)^N Xtr* 1/?drs)/(?_(r=1)^N 1/?drs) Ps(t) ? Probability that taxon t will occur at site s (300 sites s) N ? Number of sites = from 1 to 49 Xtr ? Value for taxon t at site r (I have this information in a table, 40 taxa (columns) t, 49 sites r) drs ? Distance from site r to

Hello I apologise for the length of this entry but please bear with me. In short: I need a way of subsampling communities from all possible communities of n taxa taken 1:n at a time without having to calculate all possible combinations (because this gives me a memory error - using combn() or expand.grid() at least). Does anyone know of a function? Or can you help me edit the combn or

Dear R users, I've had no joy finding a solution to this online or in any of my R books. Many thanks in advance for any help you can give. I'm seeking to output a data frame (or matrix - it doesn't matter which for my purposes) to a .txt file, but omit any row or column names. The data frame that I'm using doesn't actually have column or row names to start with as it has been

Hi, How can I list all possilities of Distinguishable Permutation by R? For example, N=6, n1=n2=n3=2, the total possible answers are 6!/(2!2!2!)=90. Please help me. Thank you So much. ____________________________________________________________________________________¡A

Getting the through items from a has_many :through :uniq relationship In a nutshell, I have been using the has_many :through :uniq relationship, and I want a DRY way to list the join model objects. Currently, my objects include taxon objects, toxin objects, data_object objects, and taxon_toxin_citation objects ("citation" is another word for data_object). A taxon has many data objects,

Hi, I have probably overlooked something obvious, but could anybody help me with following, please? Trying to make regression analysis. I have a huge dataframe with results from National Opinion Survey on Crime and Justice (www.abacon.com/fox/) with two variables G5 and N3 which are imported to R as ordered factors: > levels(noscj$G5) [1] "Strongly agree" "Agree"

Hi Quentin, > It sounds like you only need the earlyclobber description for the N, N > variant. > In other words, as long as you use different opcodes for widen-op NN and > widen-op WN, you model exactly what you want. > > What am I missing? > we are using different opcodes for widen-op NN and widen-op WN. My understanding is that not setting earlyclobber to the W, N

> On Jun 6, 2017, at 4:01 AM, Jim Lemon <drjimlemon at gmail.com> wrote: > > Hi Bogdan, > Kinda messy, but: > > N <- data.frame(N=c("n1","n2","n3","n4")) > M <- data.frame(M=c("m1","m2","m3","m4","m5")) > C <-

Thank you David. Using xtabs operation simplifies the code very much, many thanks ;) On Tue, Jun 6, 2017 at 7:44 AM, David Winsemius <dwinsemius at comcast.net> wrote: > > > On Jun 6, 2017, at 4:01 AM, Jim Lemon <drjimlemon at gmail.com> wrote: > > > > Hi Bogdan, > > Kinda messy, but: > > > > N <-

Simple matrix indexing suffices without any fancier functionality. ## First convert M and N to character vectors -- which they should have been in the first place! M <- sort(as.character(M[,1])) N <- sort(as.character(N[,1])) ## This could be a one-liner, but I'll split it up for clarity. res <-matrix(NA, length(M),length(N),dimnames = list(M,N)) res[as.matrix(C[,2:1])] <-

Hi, I run this code to get the power of the test for modified bartlett's test..but I'm not really sure that my coding is right.. #normal distribution unequal variance asim<-5000 pv<-rep(NA,asim) for(i in 1:asim) {print(i) set.seed(i) n1<-20 n2<-20 n3<-20 mu<-0 sd1<-sqrt(25) sd2<-sqrt(50) sd3<-sqrt(100) g1<-rnorm(n1,mu,sd1) g2<-rnorm(n2,mu,sd2)

Here's another approach: N <- data.frame(N=c("n1","n2","n3","n4")) M <- data.frame(M=c("m1","m2","m3","m4","m5")) C <- data.frame(n=c("n1","n2","n3"), m=c("m1","m1","m3"), I=c(100,300,400)) # Rebuild the factors using M and N C$m <-

So I wanted to export a contour3d object into VTK and it worked well enough, except the top of the ball (the contour) was missing. Example below, or gist here: https://gist.github.com/muschellij2/5126544 rm(list=ls()) require(misc3d) f <- function(x, y, z) x^2+y^2+z^2 x <- seq(-2,2,len=20) ball <- contour3d(f,4,x,x,x, draw=FALSE) drawScene.rgl(ball) ### get the triangles allids

Hi all, I'm working on a target whose registers have equal-sized subregisters and all of those subregisters can be named (or the other way round: registers can be grouped into super registers). So for instance we've got 16 registers W (as in wide) W0..W15 and 32 registers N (as in narrow) N0..N31. This way, W0 is made by grouping N0 and N1, W1 is N2 and N3, W2 is N4 and N5, ..., W15 is

Hi Bogdan, Kinda messy, but: N <- data.frame(N=c("n1","n2","n3","n4")) M <- data.frame(M=c("m1","m2","m3","m4","m5")) C <- data.frame(n=c("n1","n2","n3"), m=c("m1","m1","m3"), I=c(100,300,400))

I am trying to solve one of my graduate level physics problems with ruby... Here is what I have so far... a6=0.0 for n1 in -10..10 for n2 in -10..10 for n3 in -10..10 if n1!=0 and n2!=0 and n3!=0 p=Math.sqrt(n1**2+n2**2+n3**2+n1*n2/1.414+n1*n3/1.41+n2*n3/1.414) a6+=(1/p)**6 end end end end puts a6 What I''ve got here is a 10x10x10 face-centered

Hi, how can I create an array of lists of three components? This approach does not work: n1 <- 2 n2 <- 4 n3 <- 5 res <- array(rep(vector("list",3), n1*n2*n3), dim = c(n1,n2,n3)) res[1,1,1] # is not a list with three components... The goal is that res[1,1,1] is a list with three components. Also, appending the components didn't work. For example, I tried: component