similar to: options(digits) (PR#1879)

Hi Gents, I need to get the control limits from qcc function. As follows: qcc(MDI, type = "xbar.one") Call: qcc(data = MDI, type = "xbar.one") xbar.one chart for MDI Summary of group statistics: Min. 1st Qu. Median Mean 3rd Qu. Max. 0.3266 0.4249 0.4371 0.4333 0.4451 0.4858 Group sample size: 1 Number of groups: 383 Center

Dear Sir or Madam? I am a doctor of urology,and I am engaged in developing a nomogram of bladder cancer. May I ask for your help on below issue? I set up a dataset which include 317 cases. I got the Binary Logistic Regression model by SPSS.And then I try to reconstruct the model ?lrm(RECU~Complication+T.Num+T.Grade+Year+TS)? by R-Project,and try to internal validate the model through

Hi! Do not understand following behavior. > summary(test$dif) Min. 1st Qu. Median Mean 3rd Qu. Max. 0.7389 0.9713 0.9850 0.9818 1.0000 1.0000 length(test$dif) [1] 85879 tmp <- hist(test$dif,breaks=100,freq=FALSE) The density on the Y axis in the plot are in the range 0-200. Thought that the density should be in the range 0-1 (something like

I'm having trouble with a simple application with metRology. I need to estimate the uncertainty of the density thickness of seven sheets of film. This is calculated from measurements of mass, length and width of rectangular samples of film. It's not too hard to calculate the whole thing with a little Monte Carlo loop. I get about 0.07 with this: #sample area

Hi, I'm sure this is simple, but I haven't been able to find this in TFM, say I have some data in R like this (pasted here: http://pastebin.com/raw.php?i=sjS9Zkup): > head(df) gender age smokes disease Y 1 female 65 ever control 0.18 2 female 77 never control 0.12 3 male 40 state1 0.11 4 female 67 ever control 0.20 5 male 63 ever state1 0.16

Hello all, I am running into a problem with garbage collection not being able to free up all memory. Unfortunately I am unable to provide a minimal self-contained example, although I can provide a self contained example if anyone feels like wading through some 600 lines of code. I would love to isolate the relevant parts from the code but whenever I try to run a simpler example, the problem does

Hello, Is this a bug in the lmer routine? > library(lme4) > ### test case based on rats data from Crawley > a<-rnorm(36);b<-rep(1:3,each=12);c<-rep(1:2,each=6,3);d<-rep (1:3,each=2,6) > > ### mixed model works when c & d are numeric, lmer assumes they are factors > m <- lmer(a ~ b + (1|c/d)) > > ### but bails out when they are actually

CentOS Errata and Bugfix Advisory 2016:1879 Upstream details at : https://rhn.redhat.com/errata/RHBA-2016-1879.html The following updated files have been uploaded and are currently syncing to the mirrors: ( sha256sum Filename ) x86_64: a59f9402df8b2a324097eb3ebdc7a645dd104f78aba503b98449ee0762b9b11c bmc-snmp-proxy-1.8.13-9.el7_2.noarch.rpm

Hi All, I would like to ask two questions about printCoefmat(). First, I found a behavior of printCoefmat() that looks strange to me, but I am not sure whether this is an intended behavior: ``` r set.seed(5689417) n <- 10000 x1 <- rnorm(n) x2 <- rnorm(n) y <- .5 * x1 + .6 * x2 + rnorm(n, -0.0002366, .2) dat <- data.frame(x1, x2, y) out <- lm(y ~ x1 + x2, dat) out_summary <-

Dear All, I?m using R package ?msm? to fit a multi state model to infection history data (counts of infections per month upto diagnosis of a particular disease (sink state is state 11). The observed transitions are as follows: to from 1 2 3 4 5 6 7 8 10 11 1 35192 3806 899 233 46 11 3 0 1 534 2 3801 790 249 69 15

>>>>> Shu Fai Cheung >>>>> on Thu, 6 Jul 2023 17:14:27 +0800 writes: > Hi All, > I would like to ask two questions about printCoefmat(). Good... this function, originally named print.coefmat(), is 25 years old (in R) now: -------------------------------------------------------------------- r1902 | maechler | 1998-08-14 19:19:05 +0200 (Fri,

Hi, I wonder if anyone could advise me with this: I've been trying to make a standard curve in R with lm() of some standards from a spectrophotometer, so as I can express the curve as a formula, and so obtain values from my treated samples by plugging in readings into the formula, instead of trying to judge things by eye, with a curve drawn by hand. It is a curve and so I used the

Hola:   Tengo un modelo de regresión lineal en el cual las componentes son cosenoidales, y lo construyo del siguiente modo:     modelo = "y ~ I(t) + I(t^2) + I(t^3) + x1[, 1] + x2[, 1]" x1[, 1] = cos(2 * pi * t / periods[1]) x2[, 1] = sin(2 * pi * t / periods[1]) for (i in 2:nComp) {   x1[, i] = cos(2 * pi * t / periods[i])   x2[, i] = sin(2 * pi * t / periods[i])   modelo =

Dear all I'm familiarising myself with Ridge Regressions in R and the following is bugging me: How does one get p-values for the coefficients obtained from MASS::lm.ridge() output (for a given lambda)? Consider the example below (adapted from PRA [1]): > require(MASS) > data(longley) > gr <- lm.ridge(Employed ~ .,longley,lambda = seq(0,0.1,0.001)) > plot(gr) > select(gr)

The test is manova. I tried to use manova() function, I used the code below:fit <- manova(Y ~ X)summary(fit, test="Wilks")but I get p values for intercept and regression coefficient as in anova() function, not for the hull model. Date: Mon, 7 Feb 2011 00:57:43 -0800 Subject: Re: [R] FW: multivariate regression From: djmuser@gmail.com To: denizsigirli@hotmail.com CC:

R Help - I'm trying to run a correlation matrix with a covariate of "age" and will at some point will also want to covary other variables concurrently. I'm using the "psych" package and have tried other methods such as writing a loop to extract semi-partial correlations, but it does not seem to be working. How can I accomplish this? library(psych) > set.cor(y =

Dear R user, I am new with split-plot designs and I have problems with multiple comparisons. This data correspond to an split-plot experiment with two replications (bloque).(Hoshmand, 2006 pp 138). Briefly, the whole-plot factor is Nitrogen concentration ("nitrogeno") and the subplot factor is the variety of corn ("hibrido"). The aim is to determine if major differences

How can I run multivariate linear regression in R (I have got 3 dependent variables and only 1 independent variable)? I tried lm function, but it gave different R2 and p values for every dependent variable. I need one R2 and p value for the model. [[alternative HTML version deleted]]

Hi, I have a fairly simple linear regression using the lm function. There are about 100 variables and 30,000 rows of data. It runs fine and produces a decent looking R2 value. I'm interested in performing a stepwise variable selection to see if things can be cleaned up a bit. Calling the step function returns ONE iteration (all the variables) and then stops. No errors are reported.

Dear R-helpers, this mail is intended to mention a rather trange result and generate potential useful comments on it. I am not aware of another posts on this issue ( RSiteSearch("quasipoisson lmer version dispersion")). MUsing the exemple in the reference of the lmer function (in lme4 library) and turning it into a quasi-poisson or quasi-binomial analysis, we get different results,