similar to: FW: creating a custom background

Consider the following missing data problem: y = c(1, 2, 2, 2, 3) a = factor(c(1, 1, 1, 2, 2)) b = factor(c(1, 2, 3, 1, 2)) fit = lm(y ~ a + b) anova(fit) Analysis of Variance Table Response: y Df Sum Sq Mean Sq F value Pr(>F) a 1 0.83333 0.83333 1.3637e+33 < 2.2e-16 *** b 2 1.16667 0.58333 9.5461e+32 < 2.2e-16 *** Residuals 1 0.00000 0.00000 ---

Hi. I know that R computes sums of squares based on the diagonal of t(Q) %*% y %*% t(y) %*% Q, where Q comes from the QR-decomposition of the model matrix. Does anyone know where I can find a proof for this result? All Best and Happy New Year, Ethan [[alternative HTML version deleted]]

temp.ttt <- "ttt <- 1\nttt" conn.ttt <- textConnection(temp.ttt) source(conn.ttt, echo=TRUE) ## name of variable is echoed close(conn.ttt) cat(file="c:/temp/temp.R", temp.ttt) ## name of variable not echoed source("c:/temp/temp.R", echo=TRUE) temp.abc <- "abc <- 1\nabc" conn.abc <- textConnection(temp.abc) source(conn.abc, echo=TRUE)

Hi. I want to use the text() command to produce graphics text in the (Windows) font Tahoma, but I don't know how to do this. I appreciate any suggestions/ideas you may have. All Best, Ethan [[alternative HTML version deleted]]

Excellent! Worked like a charm. Thanks. --Chris Ryan On Tue, May 1, 2018 at 4:33 PM, William Dunlap <wdunlap at tibco.com> wrote: > The ts method for plot() is quirky. You can use the default method: > > plot(as.vector(time(ttt)), as.vector(ttt), type = "p", col=ifelse(ttt<8, > "black", "red")) > > > Bill Dunlap > TIBCO Software

How would I color points conditional on their value in a plot of a time series. Something like this: ## demonstration data ttt <- ts(rpois(12, lambda = 8), start = c(2000, 1), freq = 4) ttt plot(ttt, type = "p") ## doesn't work--all points the same color plot(ttt, type = "p", col = ifelse(ttt < 8, "black", "red")) ## also doesn't work--all

You may also want to check this out: plot(ttt, type = "p") points(ttt, col = ifelse(ttt < 8, "black", "red")) Eivind K. Dovik Bergen, NO On Tue, 1 May 2018, Christopher W Ryan wrote: > Excellent! Worked like a charm. Thanks. > > --Chris Ryan > > On Tue, May 1, 2018 at 4:33 PM, William Dunlap <wdunlap at tibco.com> wrote: > >> The

Hello, everyone, I don't quite understand the following message: > TTT <- t.test(1:10, y=c(7:20)) > class(TTT) [1] "htest" > TTT@p.value Error: Trying to get slot "p.value" from an object whose class ("htest") is not defined > TTT$p.value [1] 1.855282e-05 Why the message says the class of TTT is not defined while class(TTT) gets

The ts method for plot() is quirky. You can use the default method: plot(as.vector(time(ttt)), as.vector(ttt), type = "p", col=ifelse(ttt<8, "black", "red")) Bill Dunlap TIBCO Software wdunlap tibco.com On Tue, May 1, 2018 at 1:17 PM, Christopher W Ryan <cryan at binghamton.edu> wrote: > How would I color points conditional on their value in a plot

Dear all, I have looked for an answer for a couple of days, but can't come with any solution. I have a set of functions, say: > t0 <- function(x) {1} > t1 <- function(x) {x} > t2 <- function(x) {x^2} > t3 <- function(x) {x^3} I would like to find a way to add up the previous 4 functions and obtain a new function: > rrr <- function(x) {1+x+x^2+x^3} without,

Hello R users I have a question with Kaplan-Meier Curve with respect to my research. We have done a retrospective study on fillings in the tooth and their survival in relation to the many influencing factors. We had a long follow-up time (upto 8yrs for some variables). However, we decided to stop the analysis at the 6year follow up time, so that we can have uniform follow-up time for all the

Hi: I have some problems when I use the function "solve" function in a loop. In the following code, I have a diagonal martix "ttt" whose elements change in every iteration in a loop. I defined a "dpoMatrix"class before the loop so I do not need to define this class every time in the loop. The reason is to save some computing time. The code is below. The inverse

I have a timeseries object, ts, and want to get the first date in the series into a string so I can concatenate it with a SQL query. Input and output are shown below. I must be missing something very basic, but I can't seem to pry the data ("2008-07-01") into a string variable. Any suggestions would be appreciated. Thank you, Andrew =====script: class(ts) (ts[1,0]) #returns

Hello, I am forwarding this from ESS mailing list, as it's a failure of internal R completion system: This fails: utils:::.assignLinebuffer('iris[iris$Spec') utils:::.assignEnd(15) utils:::.guessTokenFromLine() utils:::.completeToken() utils:::.retrieveCompletions() ## -> [1] "iris[iris$Spec" This works utils:::.assignLinebuffer('iris[ iris$Spec') # note the

I am trying to copy some digital images to a directory on my Redhat 7.3 Samba 2.2.3a-6 server and am getting extra files in the directory. I have included my SMB.Conf file contents below in case that is relevant. The files being copied reside on a Windows 2000 workstation and were created by Jasc Image robot. These files would have been accessed by both a Windows 2000 machine as well as a Windows

Why fails nls with "singular gradient" here? I post a minimal example on the bottom and would be very happy if someone could help me. Kind regards, ########### # define some constants smallc <- 0.0001 t <- seq(0,1,0.001) t0 <- 0.5 tau1 <- 0.02 # generate yy(t) yy <- 1/2 * ( 1- tanh((t - t0)/smallc) * exp(-t / tau1) ) + rnorm(length(t))*0.01 # show the curve

Dear all, I'd like to subset a df within a function, and use select for choosing the variable. Something like (simplified example): mydf <- data.frame(a= 0:9, b= 10:19) ttt <- function(vv) { tmpdf <- subset(mydf, select= vv) mean(tmpdf$vv) } ttt(mydf$b) But this is not the correct way. Any help? Thanks in advance Juli

At times we want to convert a two-sided formula to a one-sided formula. In S we can do this by dropping the second entry in the formula. In R that object no longer has a formula class. R> ttt <- score ~ age | Infant R> class(ttt) [1] "formula" R> length(ttt) [1] 3 R> ttt[-2] [[1]] ~ [[2]] age | Infant R> class(ttt[-2]) NULL R> do.call("~",

Dear R Wizards, After much frustration and days of confusion I have finally broken down and am asking for help, which I don’t like doing, but I just can’t figure this one out on my own. I’ve conducted a laboratory experiment testing the effects of temperature and salinity on whether or not a biological event will occur (Go or NoGo). I’ve coded the factors temperature and salinity as factors for

Hi, I encountered a problem of not being able to read in an excel spreadsheet using read.xls() in the xlsReadWrite package. can anyone help? Here is an example code write.xls(matrix(1:9,nrow=3),"ttt.xls") read.xls("ttt.xls") Error in read.xls("ttt.xls") : Unexpected error. Message: Can't find the file "ttt.xls" The "ttt.xls" file was