[R] The contrary of command %in%
Dear r-list, I've got a data base: HData[1:10,] NumTree Site Species Date Age DBHH IdentTree 11 Queige Spruce 2002 184 49 33.5 Queige 1 22 Queige Fir 2002 NA 5 4.6 Queige 2 33 Queige Fir 2002 25 8 6.6 Queige 3 44 Queige Spruce 2002 198 47 32.5 Queige 4 55 Queige Fir 2002 200 59 35.3 Queige 5 66 Queige Spruce 2002 80 16 9.4 Queige 6 77 Queige Fir 2002 NA 5 4.2 Queige 7 88 Queige Fir 2002 200 44 32.5 Queige 8 99 Queige Fir 2002 NA 5 3.4 Queige 9 10 10 Queige Spruce 2002 167 48 32.8 Queige 10 ... I want to remove particular points determined by HDataPartP-HData[H1.30,] and HDataPartP2-HData[H8DBH20,] That's why I want to use subset in a close form to: HData2-subset(HData,HData$H1.30HData$IdentTree not%in%HDataPartP2$IdentTree) How should I do that ? Is there any R-syntax saying not element of that object ? Thanks for your help. Ghislain. -- Ghislain Vieilledent 06 24 62 65 07 [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
[R] Question on glm for Poisson distribution.
Good afternoon, I REALLY try to answer to my question as an autonomous student searching in the huge pile of papers on my desk and on the Internet but I can't find out the solution. Would you mind giving me some help? Please. # I'm trying to use glm with factors: Pyr.1.glm-glm(Pyrale~Trait,DataRav,family=poisson) If I have correctly payed attention to my cyber professor explanations I have, for the variable Pyrale which I suppose Poisson-distributed, the following mathematical expression: P(Pyrale=k)=exp(-m).[(m^k)/k!] with log(m)=Intercept+Trait(i) (link function is log for Poisson distribution) Then I test the significativity of Trait: anova(Pyr.1.glm,test=Chisq) Analysis of Deviance Table Model: poisson, link: log Response: Pyrale Terms added sequentially (first to last) Df Deviance Resid. Df Resid. Dev P(|Chi|) NULL 19 49.813 Trait 3 31.281 16 18.532 7.419e-07 Which means that variable Trait is significant for determining the value of P(Pyrale=k). I tried to order the effects of the modalities of my variable Trait using: summary(Pyr.1.glm) Call: glm(formula = Pyrale ~ Trait, family = poisson, data = DataRav) Deviance Residuals: Min 1Q Median 3Q Max -1.7117 -0.8944 -0.6237 0.6390 1.5224 Coefficients: Estimate Std. Error z value Pr(|z|) (Intercept) 1.3350 0.2294 5.819 5.92e-09 *** TraitInsFong -2.9444 1.0259 -2.870 0.00410 ** TraitInsecticide -2.2513 0.7434 -3.028 0.00246 ** TraitTemoin -0.2364 0.3454 -0.684 0.49372 --- Signif. codes: 0 `***' 0.001 `**' 0.01 `*' 0.05 `.' 0.1 ` ' 1 (Dispersion parameter for poisson family taken to be 1) Null deviance: 49.813 on 19 degrees of freedom Residual deviance: 18.532 on 16 degrees of freedom AIC: 61.85 Number of Fisher Scoring iterations: 5 ## I have therefore two questions: - Considering the values of estimated coefficients for Trait(i), does it mean that the bigger is the coefficient, the lower is the probability considering the mathematical expression (exp(-m))? - How can I check that coefficients are significatively different one from each other (as with function TukeyHSD for other models)? Thanks for you help. Regards Ghislain. -- Ghislain Vieilledent 30, rue Bernard Ortet 31 500 TOULOUSE 06 24 62 65 07 [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
[R] Significant difference of coefficients in glm with factors?
Hi and sorry to distur, ### Setting # I'm trying to use glm with factors: Pyr.1.glm-glm(Pyrale~Trait,DataRav,family=poisson) summary(Pyr.1.glm) Call: glm(formula = Pyrale ~ Trait, family = poisson, data = DataRav) Deviance Residuals: Min 1Q Median 3Q Max -1.7117 -0.8944 -0.6237 0.6390 1.5224 Coefficients: Estimate Std. Error z value Pr(|z|) (Intercept) 1.3350 0.2294 5.819 5.92e-09 *** TraitInsFong -2.9444 1.0259 -2.870 0.00410 ** TraitInsecticide -2.2513 0.7434 -3.028 0.00246 ** TraitTemoin -0.2364 0.3454 -0.684 0.49372 --- Signif. codes: 0 `***' 0.001 `**' 0.01 `*' 0.05 `.' 0.1 ` ' 1 (Dispersion parameter for poisson family taken to be 1) Null deviance: 49.813 on 19 degrees of freedom Residual deviance: 18.532 on 16 degrees of freedom AIC: 61.85 Number of Fisher Scoring iterations: 5 anova(Pyr.1.glm,test=Chisq) Analysis of Deviance Table Model: poisson, link: log Response: Pyrale Terms added sequentially (first to last) Df Deviance Resid. Df Resid. Dev P(|Chi|) NULL 19 49.813 Trait 3 31.281 16 18.532 7.419e-07 ## Question ## I would like to know if the coefficients for the levels of the factors are significatively different from each other and not only from the intercept. Do you know what function can I use and in which package can I found it? Thanks very much for you help. Regards. Ghislain V. -- Ghislain Vieilledent 30, rue Bernard Ortet 31 500 TOULOUSE 06 24 62 65 07 [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
[R] Name for factor's levels with contr.sum
Good morning, I used in R contr.sum for the contrast in a lme model: options(contrasts=c(contr.sum,contr.poly)) Septo5.lme-lme(Septo~Variete+DateSemi,Data4.Iso,random=~1|LieuDit) intervals(Septo5.lme)$fixed lower est. upper (Intercept) 17.0644033 23.106110 29.147816 Variete1 9.5819873 17.335324 25.088661 Variete2 -3.3794907 6.816101 17.011692 Variete3 -0.5636915 8.452890 17.469472 Variete4 -22.8923812 -10.914912 1.062558 Variete5 -10.7152821 -1.865884 6.983515 Variete6 0.2743390 9.492175 18.710012 Variete7 -23.7943250 -15.070737 -6.347148 Variete8 -21.7310554 -12.380475 -3.029895 Variete9 -27.9782575 -17.480555 -6.982852 DateSemi1 -5.7903419 -1.547875 2.694592 DateSemi2 3.6571596 8.428417 13.199675 attr(,label) [1] Fixed effects: How is it possible to obtain a return with the name of my factor's levels as with contr.treatment ? Thanks for you help. -- Ghislain Vieilledent 30, rue Bernard Ortet 31 500 TOULOUSE 06 24 62 65 07 [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
[R] Fwd: Plotting confidence intervals for lme
-- Forwarded message -- From: Ghislain Vieilledent [EMAIL PROTECTED] Date: 7 juil. 2005 11:38 Subject: Re: [R] Plotting confidence intervals for lme To: Spencer Graves [EMAIL PROTECTED] That's what I was looking for. Thanks a lot! That's true units are mixing and that values of coef depend on the contrasts but I want to use this graph in order to compare levels' coefficient of each of the factor taken independantly, not to compare factors' coefficients to eachothers. Does it mean something? Thanks again. Ghislain Vieilledent. 2005/7/6, Spencer Graves [EMAIL PROTECTED]: Consider the following extension of an example in ?lme: fm2 - lme(distance ~ age + Sex, data = Orthodont, random = ~ 1) int - intervals(fm2) class(int$fixed) kf - dim(int$fixed)[1] plot(int$fixed[,2], kf:1, xlab=x, ylab=, xlim=range(int$fixed), axes=FALSE) axis(1) axis(2, kf:1, dimnames(int$fixed)[[1]]) segments(int$fixed[,1], kf:1, int$fixed[,3], kf:1) abline(v=0) However, we are mixing units, i.e., the units for the intercept are distance, while for age are distance/time, and the interpretation of the coefficient of a factor like Sex depends on contrasts used. Thus, I don't know how much sense it makes to prepare plots like this. For similar plots that make more sense, see Pinheiro and Bates (2000 Mixed-Effects Models in S and S-PLUS (Springer). spencer graves Ghislain Vieilledent wrote: Hello and sorry to disturb. I'm trying to plot the confidence intervals for the fixed effects of a lme. I want to obtain graphically, if it is possible, a bar with Estimate, upper and lower CI for each level of the factors. I know how to do for a lm model but for a lme one, I tried with plot(intervals(...)) and plot(ci(...)) from the gmodels package but it doesn't work well. Thanks for you help and have a good day. -- Spencer Graves, PhD Senior Development Engineer PDF Solutions, Inc. 333 West San Carlos Street Suite 700 San Jose, CA 95110, USA [EMAIL PROTECTED] www.pdf.com http://www.pdf.com http://www.pdf.com Tel: 408-938-4420 Fax: 408-280-7915 -- Ghislain Vieilledent 30, rue Bernard Ortet 31 500 TOULOUSE 06 24 62 65 07 -- Ghislain Vieilledent 30, rue Bernard Ortet 31 500 TOULOUSE 06 24 62 65 07 [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
[R] Plotting confidence intervals for lme
Hello and sorry to disturb. I'm trying to plot the confidence intervals for the fixed effects of a lme. I want to obtain graphically, if it is possible, a bar with Estimate, upper and lower CI for each level of the factors. I know how to do for a lm model but for a lme one, I tried with plot(intervals(...)) and plot(ci(...)) from the gmodels package but it doesn't work well. Thanks for you help and have a good day. -- Ghislain Vieilledent 30, rue Bernard Ortet 31 500 TOULOUSE 06 24 62 65 07 [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
[R] 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) summary(Model) ... I've got the coefficients for: - the intercept (x(1) as been choosen) that we can call B(1) - the second modality: x(2) that we can call B(2) If I have well understood we have for the model and predictions: if x(1): y=B(1) if x(2): y=B(1)+B(2) which is quite different as y=mean(y)+A(i) ## Secondly I tried to skip the intercept Model2-lm(y~0+x,Data) summary(Model2) ... I've got the coefficients for: - the first modality: x(1) that we can call C(1) - the second modality: x(2) that we can call C(2) And the model and predictions, if I'm right, are: if x(1): y=C(1) if x(2): y=C(2) *** Questions *** How can I obtain a predictive model y=mean(y)+A(i) ? Is it possible to settle mean(y) as intercept? Thanks for your help. Ghislain V., retarded statistician. __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
[R] Selecting rows regarding the frequency of a factor variable.
Hi and sorry to disturb, I'll try to be as clear as possible: I want to select rows of a data frame called Data2.Iso regarding the frequency of a factor variable called Variete that I want =4. I used function table to have the frequency: FRAMEVARIETE-as.data.frame(table(Data2.Iso$Variete)) Then I selected the modalities with a frequency =4: FRAMEVARIETE2-FRAMEVARIETE[FRAMEVARIETE$Freq=4,] as.character(FRAMEVARIETE2[,Variete]) But then, how to select the rows with those modalities? Does anyone can help me? Thanks! Ghislain. __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
