similar to: definition of "dffits"

Displaying 20 results from an estimated 200 matches similar to: "definition of "dffits""

2008 Aug 29
1
lm() and dffits
All - My question is a bit involved, so bear with me. I have some data that looks like: Lake LL LW 81 2.176091259 1.342422681 81 2.176091259 1.414973348 81 2.176091259 1.447158031 81 2.181843588 1.414973348 81 2.181843588 1.447158031 81 2.184691431 1.462397998 81 2.187520721 1.447158031 81 2.187520721 1.477121255 81 2.187520721 1.505149978 ... [truncated] I'm trying to: 1) fit a simple
2006 Aug 31
1
NaN when using dffits, stemming from lm.influence call
Hi all I'm getting a NaN returned on using dffits, as explained below. To me, there seems no obvious (or non-obvious reason for that matter) reason why a NaN appears. Before I start digging further, can anyone see why dffits might be failing? Is there a problem with the data? Consider: # Load data dep <-
2006 Feb 17
1
Transforming results of the summary function
Hi all, I have a question about transforming the data from summary function. Let's say I have a data frame like this: > x = data.frame(a = c(rep("lev1", 5), rep("lev2", 5)), b = c(rnorm(5)+2, rnorm(5))) > x a b 1 lev1 1.5964765 2 lev1 2.2945609 3 lev1 3.5285787 4 lev1 1.4439838 5 lev1 2.2948826 6 lev2 1.7063506 7 lev2 -0.4042742 8 lev2
2008 Jan 02
3
Find missing days
Hi, I have a data.frame like this: y <- rnorm(60) lev <- gl(3,20, labels=paste("lev", 1:3, sep="")) date1 <- as.Date(seq(ISOdate(2007,9,1), ISOdate(2007,11,5), by=60*60*24)) date1 <- date1[-c(3,4,15,34,38,40)] df <- data.frame(lev=lev, date1=date1, y=y) I would like to produce a new data.frame with missing days in df$date1 in each df$lev, like this: lev
2007 Aug 07
1
Naming Lists
Hi Im pretty new to R and I have run in to a problem. How do I name all the levels in this list. Lev1 <- c("A1","A2") Lev2 <- c("B1","B2") Lev3 <- c("C1","C2") MyList <- lapply(Lev1,function(x){ lapply(Lev2,function(y){ lapply(Lev3,function(z){ paste(unlist(x),unlist(y),unlist(z)) })})}) I would like to name the different
2007 Aug 23
1
How to merge string to DF
#Hi R-users, #I have an example DF like this: y1 <- rnorm(10) + 6.8 y2 <- rnorm(10) + (1:10*1.7 + 1) y3 <- rnorm(10) + (1:10*6.7 + 3.7) y <- c(y1,y2,y3) x <- rep(1:3,10) f <- gl(2,15, labels=paste("lev", 1:2, sep="")) g <- seq(as.Date("2000/1/1"), by="day", length=30) DF <- data.frame(x=x,y=y, f=f, g=g) DF$wdays <- weekdays(DF$g)
2007 Aug 07
2
Interaction factor and numeric variable versus separate regressions
Dear list members, I have problems to interpret the coefficients from a lm model involving the interaction of a numeric and factor variable compared to separate lm models for each level of the factor variable. ## data: y1 <- rnorm(20) + 6.8 y2 <- rnorm(20) + (1:20*1.7 + 1) y3 <- rnorm(20) + (1:20*6.7 + 3.7) y <- c(y1,y2,y3) x <- rep(1:20,3) f <- gl(3,20,
2005 Jun 27
1
delta-beta's
Hi there I have created a multivariate logistic regression model looking at the presence/absence of disease on farms. I would like to plot the diagnostic plots recommended by Hosmer & Lemeshow to look particularly for any points of high influence. In order to do this I need to extract values for delta-beta. The function dfbeta gives a value for change in each coefficient but I am looking
2007 Aug 16
1
Trim trailng space from data.frame factor variables
Hi folks, I would like to trim the trailing spaces in my factor variables using lapply (described in this post by Marc Schwartz: http://tolstoy.newcastle.edu.au/R/e2/help/07/08/22826.html) but the code is not functioning (in this example there is only one factor with trailing spaces): y1 <- rnorm(20) + 6.8 y2 <- rnorm(20) + (1:20*1.7 + 1) y3 <- rnorm(20) + (1:20*6.7 + 3.7) y <-
2010 Feb 21
1
tests for measures of influence in regression
influence.measures gives several measures of influence for each observation (Cook's Distance, etc) and actually flags observations that it determines are influential by any of the measures. Looks good! But how does it discriminate between the influential and non- influential observations by each of the measures? Like does it do a Bonferroni-corrected t on the residuals identified by
2007 Aug 08
1
tapply grand mean
Hi R-users, I have a data.frame like this (modificated from https://stat.ethz.ch/pipermail/r-help/2007-August/138124.html). y1 <- rnorm(20) + 6.8 y2 <- rnorm(20) + (1:20*1.7 + 1) y3 <- rnorm(20) + (1:20*6.7 + 3.7) y <- c(y1,y2,y3) x <- rep(1:5,12) f <- gl(3,20, labels=paste("lev", 1:3, sep="")) d <- data.frame(x=x,y=y, f=f) and this is how I can
2013 Dec 02
1
pamer.fnc y la nueva versión de R
Hace unos meses os escribir para comunicaros que había un fallo en esta función. Como os prometí os comento la respuesta por si alguno está interesado en utilizar el paquete LMERconvenientsfucntions Dear Javier, The package has been updated and should work for you fine now. Note that function mcp.fnc does not return the fourth plot (dffits) anymore. We still have to figure out a way to compute
2004 Jan 05
1
lda() called with data=subset() command
Hi I have a data.frame with a grouping variable having the levels C, mild AD, mod AD, O and S since I want to compute a lda only for the two groups 'C' and 'mod AD' I call lda with data=subset(mydata.pca,GROUP == 'mod AD' | GROUP == 'C') my.lda <- lda(GROUP ~ Comp.1 + Comp.2 + Comp.3 + Comp.4+ Comp.5 + Comp.6 + Comp.7 + Comp.8 ,
2009 Jun 22
1
The gradient of a multivariate normal density with respect to its parameters
Does anybody know of a function that implements the derivative (gradient) of the multivariate normal density with respect to the *parameters*? It?s easy enough to implement myself, but I?d like to avoid reinventing the wheel (with some bugs) if possible. Here?s a simple example of the result I?d like, using numerical differentiation: library(mvtnorm) library(numDeriv) f=function(pars, xx, yy)
2013 May 17
1
Error with adehabitatHR and kernelbb
Dear all, I'm trying to get a Brownian bridge kernel (kernelbb) for each combination of two consecutive animal locations (see commands below) and put them, with a loop, inside a list. It works well at the beginning but after 42 runs, it appears the following warning : >Error in seq.default(yli[1], yli[2], by = diff(xg[1:2])) :  >  invalid (to - from)/by in seq(.) I looked at the
2005 Nov 09
2
About: Error in FUN(X[[1]], ...) : symbol print-name too long
Hi, I??m trying to use the Win2BUGS package from R and I have a similar problem that reurns with the message: Error in FUN(X[[1]], ...) : symbol print-name too long But, there is no stray ` character in the file ( Sugestions given by: Duncan Temple Lang <duncan> Date: Mon, 26 Sep 2005 07:31:08 -0700 ) The progam in R is: library(R2WinBUGS) library(rbugs) dat <-
2007 Dec 01
2
How to cbind DF:s with differing number of rows?
#Hi R-users, #Suppose that I have a data.frame like this: y1 <- rnorm(10) + 6.8 y2 <- rnorm(10) + (1:10*1.7 + 1) y3 <- rnorm(10) + (1:10*6.7 + 3.7) y <- c(y1,y2,y3) x <- rep(1:3,10) f <- gl(2,15, labels=paste("lev", 1:2, sep="")) g <- seq(as.Date("2000/1/1"), by="day", length=30) DF <- data.frame(x=x,y=y, f=f, g=g) DF$g[DF$x == 1]
2011 Feb 11
6
linear models with factors
i am trying to fit a linear model with both continuous covariates and factors. When fitted with the intercept term the first level of the factor is treated by R as intercept and the estimate of the effects of remaining levels(say i th level) are given as true estimate of i th level - estimate of 1st level.can any please help me? thanks in advance..... -- View this message in context:
2013 Mar 11
2
how to convert a data.frame to tree structure object such as dendrogram
I have a data.frame object like: > data.frame(x=c('A','A','B','B'), y=c('Ab','Ac','Ba','Bd')) x y 1 A Ab 2 A Ac 3 B Ba 4 B Bd how could I create a tree structure object like this: |---Ab A---| _| |---Ac | | |---Ba B---| |---Bb Thanks, Zech [[alternative HTML version deleted]]
2012 Feb 15
1
influence.measures()
Hi dear all, I'm wondering about the question that; Does the influence.measures(model) for linear models valid for general linear models such as logistic regression models? That is; If I fit the model like model <- glm( y~X1+X2, family=binomial) Then, if i apply the function "influence.measures(model), i will get the result of influence measures. These result are valid for