similar to: help with simple function

Hi all, Why cov(y, y) only gives one value, and cov(t(y), t(y)) gives 3x3 NA matrix? Here my y is listed below and it is a 3x1 matrix. I am expecting that if I have a random vector y=[y1 y2 y3]', here " ' " denotes a transposition so that y is a column vector, where y1, y2, y3 are independent random variables... then cov(y, y) should be E[ y * y' ] - E[y] * E[y] ',

Dear all: Assume I have 3 distributions, x1, x2, and x3. x1 ~ normal(mu1, sd1) x2 ~ normal(mu2, sd2) x3 ~ normal(mu3, sd3) y1 = x1 + x2 y2 = x1 + x3 Now that the data I can observed is only y1 and y2. It is easy to estimate (mu1+m2), (mu1+mu3), (sd1^2+sd2^2) and (sd1^2+sd3^2) by EM algorithm since y1 ~ normal(mu1+mu2, sqrt(sd1^2+sd2^2)) and y2 ~ normal(mu1+mu3, sqrt(sd1^2+sd3^2)) However, I want

Dear all, I am quite new to R; facing certain problems: Say, I have a text file( named as "try"): Year C1 C2 C3 C4 C5 C6 Y1 3.5 13.8 9.5 6.8 0.4 24.2 Y2 3.8 13.9 9.9 7.6 0.7 12.8 Y3 4.5 14.5 14.2 9.2 0.6 14.5 Y4 5.9 16.2 24.6 12.7 0.2 24.3 Y5 7.2 20.4 40.6 18.2 0.8 28.2 Y6 5.9 18.6 37.4 14.5 0.3 36.9 Y7 8.0 16.1 88.6 24.1 0.1 34.6 Y8 13.6 21.1 56.3 19.0 0.7 33.3 I wish to import the

Hi all, For the simple linear regression, I want to find the input "x" value so that the lower confidnece limit is a specific number, say 0.2. In other words, I want to find the value of x so that the lower confidence bound crosses the horizontal line 0.2. Is there a simple way (an R function) that can do this? Thanks. Hanna [[alternative HTML version deleted]]

I started to check what I thought I knew with autocovariance and it doesn’t jive with the the calculations given by ‘R’. I was wondering if there is some scaling or something that I am not aware of. Take the example Ø d <- 1:10 Ø (a <- acf(d, type="covariance", demean=FALSE, plot=FALSE)) Autocovariances of series ‘d’, by lag 0 1 2 3 4 5 6

All, I've been using aggregate() to compute means and standard deviations at time/treatment combinations for a longitudinal dataset, using na.rm = TRUE for missing data. This was working fine before, but now when I re-run some old code it isn't. I've backtracked my steps and can't seem to find out why it was working before but not now. In any event, below is a reproducible

Dear all, Has someone implemented in R (or any other language) Knol DL, ten Berge JMF. Least-squares approximation of an improper correlation matrix by a proper one. Psychometrika, 1989, 54, 53-61. or any other similar algorithm? Best regards Jens Oehlschl?gel Background: I want to factanal() matrices of polychoric correlations which have negative eigenvalue. I coded Highham 2002

Bueno, realmente no es necesaria que la serie esté centrada en este caso, ya que estoy sumando un ruído blanco Un saludo From: fjroar en hotmail.com To: caaperezan en gmail.com; r-help-es en r-project.org Date: Wed, 5 Nov 2014 13:00:49 +0000 Subject: Re: [R-es] Agregar ruido a una serie de tiempo Hola buenos d?as: Yo cuando he tenido que hacer estos trabajos, lo que hac?a era coger la serie

Hi, I have measured two response variables (y1, y2) at each treatment level (x = 0, 1.5 or 3). Now I would like to show the y1 and y2 against x in a bar plot. However, y1 and y2 differ in scale so I need two y-axises, one on the left side and one on the right side (and I dont want to standardize my responses). This is fairly easy if you want to show points,lines etc, but gets more complicated

Hi I would like to shorten mod1 <- nls(ColName2 ~ ColName1, data = table, ...) mod2 <- nls(ColName3 ~ ColName1, data = table, ...) mod3 <- nls(ColName4 ~ ColName1, data = table, ...) ... is there something like cols = c(ColName2,ColName3,ColName4,...) for i in ... mod[i-1] <- nls(ColName[i] ~ ColName1, data = table, ...) I am looking forward to help Christof

Hi R lovers I would like to know how to step to the next line in the R console editor without breaking the continuity of my code more clearly : if for example I write a function, so far i have to write the all code inside on the same line wich may become obscure as the function is more and more complex. I would like to do like in the example of the manuels: >twosam <- function(y1, y2) {

Dear all I have a code like x<-1:10 y1<-x+runif(10)*2 y2<-seq(0,50,length.out=10)+rnorm(10)*10 par(mfrow=c(1,2)) plot(y1~x) plot(y2~x) Now I would like to plot y1 and y2 on the same graph, with its two scales (y1 on left and y2 on rigth side). Any help are welcome. Kind regards Miltinho Brazil [[alternative HTML version deleted]]

Hi R, y1 <- zoo(matrix(1:10, ncol = 2), 1:5) colnames(y1)=c("a","b") y2 <- zoo(matrix(rnorm(10), ncol = 2), 6:10) colnames(y2)=c("b","a") > y1 a b 1 1 6 2 2 7 3 3 8 4 4 9 5 5 10 > y2 b a 6 0.9070204 0.3527630 7 1.2405943 0.8275001 8 -0.1690653 -0.1724976 9 -0.6905223 -1.1127670 10

Hi - I'm up against a complicated reshape problem. I have data of the form X1,Y1,hr1,hr2,hr3 X1,Y2,hr1,hr2,hr3 X1,Y3,hr1,hr2,hr3 X2,Y1,hr1,hr2,hr3 X2,Y2,hr1,hr2,hr3 X2,Y3,hr1,hr2,hr3 where X and Y are factors and the hr(1,2,3) are values. I need it as ,X1, X2 Y1,hr1,hr1 Y1,hr2,hr2 Y1,hr3,hr3 Y2,hr1,hr1 Y2,hr2,hr2 Y2,hr3,hr3 .., Any hints? I've been at it for hours. p -- View

Hi all, I'm extracting the name of the term in a regression model that dropterm specifies as the least significant one, and I'm assigning this name to an object. However, when I use update(), it ignores this object. Is there a way I can make it not ignore it? A reproducible example is below: > lm(x1~1+y1*y2+y3+y4,data=anscombe)->my.lm >

Although this should be trivial, I'm having a spot of trouble. I want to make a lattice plot of the format y1+y2 ~ x | grp but then fit a lm to each y variable and add an abline of those models in different colors. If the xyplot followed y~x|grp I would write a panel function as below, but I'm unsure of how to do that with y1 and y2 without reshaping the data before hand. Thoughts

Hello all, I have data in the form of a table: X Y1 Y2 0.1 3 2 0.2 2 1 And I would like to transform in the form: X Y 0.1 Y1 0.1 Y1 0.1 Y1 0.1 Y2 0.1 Y2 0.2 Y1 0.2 Y1 0.2 Y2 Any ideas how? Thanks in advance, IOanna [[alternative HTML version deleted]]

Dear All, I would be very appreciative of your help with the following 1). I am running multivariate multiple regression through the manova() function (kindly suggested by Professor Venables) and getting two different answers for test=c("Wilks","Roy","Pillai") and tests=c("Wilks","Roy",'"Pillai") as shown below. In the

Dear List, I am trying to find each time a value changes in a dataset. The numbers are variables for day vs. night values, so what I am really getting is the daily sunrise and sunset. A simplified example is the following: x<-seq(1:100) y1<-rep(1,10) y2<-rep(2,10) y<-c(y1,y2,y1,y1,y1,y2,y1,y2,y1,y2) xy<-cbind(x,y) I would like to know each time the numbers change. Correct

Hi all, Using xyplot I want to print to Y variables (y1, y2) versus X, conditional on the group. How can I obtain a line (type="l") for one relationship (ie. y1 ~ x) and points (type="p") for the other (y2 ~ x) ? library(lattice) # create some sample data df<-data.frame(group=as.factor(c(rep("a",4), rep("b",4))), # grouping variable for conditional