similar to: handling of missing values in aov/lm

Displaying 20 results from an estimated 5000 matches similar to: "handling of missing values in aov/lm"

2002 Jul 11
0
another aov question: unbalanced multiple responses
Hi, This question is related to the bwplot issue I reported yesterday. I have a 3 factors (2x3x2) dataset that I collapsed into a 2 factors dataset (3x2 = sizexModality). For size==small, I have 2 observations per subject (Snr), for the other sizes only 1. I reckoned that aov (and underneath, lm) might handle this as it should, since the subjects are idendified, when I do > aov(
2002 Jun 10
2
Crashing R (PR#1651)
Concerns: R 1.5.0 gui version, Windows (downloaded binary) and Linux (installed from sources). # Load the data from the attached file: kk<-read.table("__filename__", header=1) # attach the data: attach(kk) Snr<-factor(Snr) # fool around with a call to anova.glm(): anova.glm( aov( nFD~Type+size+Modality+Error(Snr/(Type+size+Modality)) ) ) # Error: object nFD not found # Well, I
2005 Jun 17
0
another aov results interpretation question
I commend you to (a) the recent article by Doug Bates on "Fitting nonlinear mixed models in R" pp. 27-30 in the latest issue of "R News" available from "www.r-project.org" -> Newsletter and (b) Doug's book with Pinheiro (2000) Mixed-Effects Models in S and S-PLUS (Springer). I suggest you try the same analysis using in "lmer", library(lme4), and
2002 Jul 08
0
factanal results interpretation - am I right here?
Hi, I have obtained some results with factanal that seem to support a hypothesis I already had, and I'd like to verify that I can indeed conclude this from this new analysis. We had subjects reproduce perceived trajectories with a device that allowed us to measure spatial position (the path) and the device's orientation at any of those positions. From this, we calculated the rotation of
2007 Aug 30
1
(no subject)
Dear R users,I am looking for an easy (i.e., direct) way in obtaining the F and p values from the intercept in anovas with within-subject designs. Data are from a psychophysics expeirment where I am using d' (d-prime) values obtained from 3 modalities of presentation in each subject.I would like to know not only whether there is an effect of modality, but also wether the main effect is
2007 Aug 30
1
How to obtain intercept statistics in anova with within-subject factors?
Dear R users, I am looking for an easy (i.e., direct) way of obtaining the F and p values from the intercept in anovas with within-subject designs. My data are from a psychophysics experiment where I am using d' (d-prime) values obtained from 3 modalities of presentation in each subject. I would like to know not only whether there is an effect of modality, but also whether the main effect is
2013 Mar 05
0
Agreement and Consistency of 2D data
Hi, I have two different imaging modalities (for the identification of areas of infarcted myocardium) that I need to compare regarding agreement and consistency. However, I don't think that methods like Cohen's Kappa, PCC, Bland-Altmann or ICC are sufficient here as there is not just a pairwise but also a spatial relationship between measured data points. For example if the results of the
2005 Jun 30
2
Linear Models with mean as Intercept.
Dear advanced statisticians, *******Objectif******** I try to set up linear models with mean as intercept: Answer: y Variable: x, as factor of two modalities: x(1), x(2). I would like to have a model as: y = mean(y)+A(i)+residuals, with i in (1,2) and A(1) coefficient for x(1) and A(2) coefficient for x(2). *******Trials in R******* ## Firstly: I write in R: >Model<-lm(y~x,Data)
2007 Feb 28
2
lm ANOVA vs. AOV
Why would someone use lm and ANOVA (anova(lm(x))) instead of AOV (or the other way around)? The mean squares and sum of squares are the same, but the F values and p-values are slightly different. I am modeling a dependent~independent1*independent2. Thanks, Matt Bridgman [[alternative HTML version deleted]]
2003 Mar 22
1
extracting the names of the dataframe and variables in aov or lm
Dear R Users, I want to write a function that applies to the dataframe and variables that were used in a previous call to lm or aov. In order to do this, I need to write a function that applies to the output of lm or aov, and yields the names of the dataframe and variables that were used in the lm or aov analysis. For example, suppose that I give the command: aov.out <- aov( Rt ~
2011 Sep 30
1
different results aov vs. lm
Hi, I currently running regression models on an experimental dataset. The model contains one independent continuous variable and two independent experimental conditions (one with two factors, the other with three factors) and several covariates. Now I get different results for a covariate in this model when I run aov(modell) vs. lm(modell). In the ancova model, one of the covariates seems to
2011 Apr 18
1
Comparing two lines - Ancova: lm or aov?
Hello! I have measurements (length and volume) of fish collected in two years. I want to know if the the relationship between length and volume is the same for both years. The number of fish measured is different for each year. I don't know whether lm or aov is more appropriate to use. Here are the two output options: Call: lm(formula = Volume ~ Length * Year) Residuals: Min 1Q
2006 Mar 01
0
Help - lm, glm, aov results inconsistent with other statistical package
Hello, I 'm sure there must a be a simple explanation for what I'm doing wrong but I am stumped. I am a novice R user and this has shaken my confidence in what I'm doing! I am trying to run a simple ANCOVA using the model y~A*x, where y and x are continuous and A has two levels. Everything seems fine until I compare the output with what I get from another statistical package
2003 Dec 06
2
Difference between summary.lm() and summary.aov()
I have a simple linear model (fitted with lm()) with 2 independant variables : one categorical and one integer. When I run summary.lm() on this model, I get a standard linear regression summary (in which one categorical variable has to be converted into many indicator variables) which looks like : Estimate Std. Error t value Pr(>|t|) (Intercept) -3595.3 2767.1 -1.299
2017 Nov 07
0
New vcov(*, complete=TRUE) etc -- coef(<lm>) vs coef(<aov>)
Dear Martin, I think that your plan makes sense. It's too bad that aov() behaved differently in this respect from lm(), and thus created more work, but it's not be a bad thing that the difference is now explicit and documented. I expect that that other problems like this will surface, particularly with contributed packages (and I know that you're aware that this has already happened
2009 Apr 08
2
Doubt about aov and lm function... bug?
Hi, The below very strange: # a) aov function av <- aov(Sepal.Length ~ Species, data=iris) # Error in parse(text = x) : # unexpected symbol in "Sepal(Sepal.Length+Species)Length" av <- aov(iris[, 1] ~ iris[, 5]) # summary(av) # Df Sum Sq Mean Sq F value Pr(>F) # iris[, 5] 2 63.2 31.6 119 <2e-16 *** # Residuals 147 39.0 0.3 # ---
2005 Sep 12
0
grepping and splitting (with R 2.1.1)
Hi R experts I have the following regular expression problem. I am writing a basic corpus retrieval program, i.e. a concordancer/function where a user enters - a set or a directory of text files to search; - a regular expression to search for in these files. I want to provide an output in which the matches of the regular expression are listed in one central column and the neighboring columns
2002 Jul 10
0
lattice::bwplot with unbalanced design
Hi, I have an experimental design with 3 factors (Type, size and Modality), where the stimulus for Type==B, size==small was identical to that for Type==A, size==small. Since I am in the end only interested in Type==A, I "relabeled" all Type==B,size==small responses as Type==A,size==small. This gives me an unbalanced design, with a certain number of missing combinations, and others that
2009 Feb 20
1
lm and aov produce different results for nested fixed-factor anova
Dear R users, I have trouble obtaining the same results for nested Anova with two fixed factors when using lm and aov functions. The formulas are: > e1=aov(y~x/z) > e2=lm(y~x/z) summary(e1) Df Sum Sq Mean Sq F value Pr(>F) x 47 260.0 5.5 18.0088 < 2.2e-16 *** x:z 195 169.6 0.9 2.8318 < 2.2e-16 *** Residuals 14425
2015 May 08
0
[RFC PATCH v1 0/8] Ne10 fft fixed and previous
Hello Jean-Marc, **Resending.. not sure why subject got removed earlier** Below are the results that show test_unit_dft passes, but test_unit_mdct fails (only for nfft=480, 960, 1920) Note: Tested on BeagleboneBlack(Cortex-A8) fixed point on branch [1] ./test_unit_dft nfft=32 inverse=0,snr = 88.394372 nfft=32 inverse=1,snr = 93.896470 nfft=128 inverse=0,snr = 89.185895 nfft=128 inverse=1,snr =