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Teste Comparação - Pacote Agricolae
by Evelyn Alecrim 19 Jan '12

19 Jan '12

Olá a todos, Alguém sabe me informar qual o nome do teste de comparação feito após o Kruskal Wallis no pacote Agricolae? Obrigada! -- Evelyn da Fonseca Alecrim Bragion Bióloga/Universidade Federal de Lavras - UFLA Mestranda em Ecologia Aplicada - UFLA Lavras - MG/Brasil

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Distribuição de pareto
by Vinicius Brito Rocha 19 Jan '12

19 Jan '12

Senhores, estou querendo gerar números a partir da distribuição de Pareto. Baixei o pacote VGAM. Meu problea é gerar dados de acordo com a proposta de que 80% da riqueza está na mão de 20% da sociciedade (no brasil é 90% em 2% rs) Alguem tem alguma noção de qual é a paremetrização para isso? No Pacote VGAM tem varias opões da distribuição de Pareto. Abs -- Vinicius Brito Rocha. Estatístico e Atuário, Mestre em Pesquisa Operacional www.aplicademic.blogspot.com http://twitter.com/viniciusbritor "Não se preocupe muito com as suas dificuldades em Matemática, posso assegurar-lhe que as minhas são ainda maiores." - Albert Einstein.

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Re: [R-br] Dúvida teste
by Ivan Bezerra Allaman 18 Jan '12

18 Jan '12

Bom dia Peterson! Sinceramente eu não entendi o seu problema. Ficou meio confuso. Você tem 4 linhas certo? Deu a entender que: Linha 1: Época 1 Linha 2: Época 2 Linha 3: Época 3 Linha 4: Época 4 É isso? Se for isso, cada planta constitui uma unidade experimental. Em outro momento, deu entender: Linha 1,2,3,4: Época 1 Linha 1,2,3,4: Época 2 Linha 1,2,3,4: Época 3 Linha 1,2,3,4: Época 4 Nesse caso, cada linha seria uma repetição. Talvez outros colegas tenham entedido, mas para agilizar a ajuda, se você "desenhar" fica melhor para ajudá-lo. Abraço! (S,f,P) Allaman \begin{signature} <<>>= Prof. Dr. Ivan Bezerra Allaman Universidade Estadual de Santa Cruz Departamento de Ciências Exatas e Tecnológicas Ilhéus/BA - Brasil Fone: +55 73 3680-5076 E-mail: ivanalaman@yahoo.com.br/ivanalaman@gmail.com @ \end{signature}

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Acessar funções de pacotes!
by Ivan Bezerra Allaman 18 Jan '12

18 Jan '12

Bom dia senhores! Gostaria de saber como faço para acessar as funções dos pacotes do R. Em um tempo remoto, bastava acessar a pastar "R", que lá estava todas as funções. Nos dias atuais, aparece aquirvos com extensões '.rdb' e '.rdx' que eu não consigo abrir. Será que as funções estão nestes arquivos? Caso sim, qual programa uso para abrí-los? Utilizo o ubuntu 10.04 lts. Desde já, agradeço a atenção de todos! (S,f,P) Allaman \begin{signature} <<>>= Prof. Dr. Ivan Bezerra Allaman Universidade Estadual de Santa Cruz Departamento de Ciências Exatas e Tecnológicas Ilhéus/BA - Brasil Fone: +55 73 3680-5076 E-mail: ivanalaman@yahoo.com.br/ivanalaman@gmail.com @ \end{signature}

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Dúvida teste
by Peterson Gardin 18 Jan '12

18 Jan '12

Pessoal Eu tenho um parreiral de uvas cabernet sauvignon. Nesse, tenho 4 linhas com 60 plantas cada. Fiz 4 tratamentos, que correspondem a época de colocação de uma cobertura plástica. Dependendo da época que coloquei a cobertura, ao final do ciclo, ocorrem diferenças de ataques de doenças, durante a maturação das uvas. Não existem repetições destas coberturas plásticas, pois cada linha custa mil reais, portanto, cada linha é um tratamento (ou seja, época de colocação da lona), pois se eu fizesse 4 repetições o experimento me comeria R$ 16.000,00, só de cobertura. Penso que 60 plantas sejam suficientes para realizar uma amostragem. Pergunto: Posso realizar uma amostragem aleatória em cada tratamento e fazer um teste de média como se fosse um DIC? Como faço se quiser avaliar uma doença que aumenta conforme transcorre o tempo de maturação, por exemplo, quanto mais madura a uva mais ataca. Coloco no modelo a variável tempo? Tenho dúvidas de analisar variáveis que evoluem com o tempo. Se tiverem alguma dica de bibliografia, para que eu veja as análises, tente fazer e retorne com as dúvidas eu agradeço. Obrigado. João Peterson Pereira Gardin Pesquisador Epagri - Videira - SC Doutor em Fisiologia Vegetal Engenheiro Agrônomo Fones: (49) 3566-0054

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Problema na instalação do pacote geepck
by Olympio Neto 18 Jan '12

18 Jan '12

Pessoal, bom dia. Ontem tentei instalar o pacote 'geepack' no R.2.14.1 rodando no Ubuntu 11.10, mas ao terminar de compilar o pacote dá um erro que impossibilita a finalização da instalação. Alguém já passou por isso? No Windows instala corretamente. Alguém teve o mesmo problema? Abs Olympio

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Gerar gráficos diferentes e em 3D
by Mauro Sznelwar 18 Jan '12

18 Jan '12

Tenho esta situação: f=rmultinom(30,3,prob=c(0.5,0.3,0.8))*dmultinom(x=1:3,prob=c(0.5,0.3,0.8)) n=5 for(i in 1:n){ g=f contour(g) } #preciso que gere 5 gráficos diferentes (mas não parece que gera gráficos #diferentes) jpeg(filename = "MeusGraficos%03d.jpg", width = 480, height = 480) for (i in 1:5) contour(g) dev.off() E como adapto o caso da normal para o multinomial para fazer gráfico 3D, já tentei mas acusa medidas diferentes. z <- matrix(rnorm(50*10),50,10) image(z) persp(z)

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Dúvida equação GLM e Anova
by Caroline Figueiredo 17 Jan '12

17 Jan '12

Olá pessoal, Estou com uma dúvida, que na verdade deve ser bem ridícula, mas realmente não sei a resposta. A questão é a seguinte: Tenho dados binários como variável resposta ( 1 – vivo; 0 - morto), e variáveis contínuas como explicativas, e vi no The R book do Crawley que poderia usar GLM para construir um modelo, tendo binomial como família. Usei o seguinte código (os valores são fictícios, pois li meus dados direto da tabela, pois são muitas linhas). status<-c(0,1,0,1,1,1,1,1,0,1) altura<-c(2.2,1.3,4.5,6.7,1.3,4.5,1.2,2.0,7.8,1.2) newy<-c(5.5,1.2,7.8,1.4,1.3,2.3,3.2,1.3,6.7,1.4) m1<-glm(status~altura+newy+altura*newy, binomial) E o summary deu isso: Call: glm(formula = status ~ altura + newy + altura * newy, family = binomial) Deviance Residuals: Min 1Q Median 3Q Max -3.3122 0.1586 0.1969 0.2486 0.6394 Coefficients: Estimate Std. Error z value Pr(>|z|) (Intercept) 0.24827 0.52797 0.470 0.63818 altura 0.80654 0.20006 4.032 5.54e-05 *** newy 0.23918 0.05101 4.689 2.74e-06 *** altura:newy -0.04578 0.01753 -2.611 0.00903 ** --- Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1 (Dispersion parameter for binomial family taken to be 1) Null deviance: 1705.0 on 7191 degrees of freedom Residual deviance: 1610.4 on 7188 degrees of freedom AIC: 1618.4 Number of Fisher Scoring iterations: 7 Minha dúvida é a seguinte: Qual é a equação ajustada para esse modelo? P. ex., se fosse uma regressão linear o modelo seria y~x, e a equação y=a + bx, mas e para esse modelo? A outra dúvida é se poderia fazer uma anova para comparar esse modelo com o modelo nulo, e qual teste poderia usar, pois no exemplo do livro ele usa o test= chi, porém está comparando dois modelos diferentes, e queria saber se posso usar para comparar com o modelo nulo também. Bem é isso pessoal, espero que possam me ajudar. Obrigada, Carol Figueiredo

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Operações com datas
by ASANTOS 17 Jan '12

17 Jan '12

Boa tarde pessoal, Tenho dados de temperatura e precipitação diárias de 1994 a 2001 em uma determinada fazenda. Preciso retira as médias do dia 1º ao 15º dia e do 15º ao último dia de cada mês, porém estou tendo problemas em trabalhar com as datas, exemplificando através de uma rotina venho fazendo: ### Criando o objeto com datas e a variável climática x <- as.Date("1994-01-01") times <- 1:2865 datas <- x+times datas<-as.data.frame(datas)##Cria o vetor data de 01/01/1994 a 5/11/2001 variavel<-rnorm(2865)## Simula a variável medida no dia dados<-cbind(datas,variavel)##Cria o objeto ## Cria um vetor temporal a cada 15 dias e adiciona aos dados t<-1:191 for(i in 1:length(t)){xv<-sort(rep(t,15))} xv<-as.data.frame(xv) xv<-xv[1:2865,] tp.1994_2001<-cbind(dados,xv) head(tp.1994_2001) ## Calcula as médias a cada 15 dias md<-1:2865 res<-NULL for(k in 1:length(md)){Tm<-mean(tp.1994_2001[md[k],2]) res<-rbind(res,c(md[k],Tm)) } Porém minha abordagem esta incorreta, quando na verdade eu desejo a média do 1° ao 15º dia e do 15º ao último dia do mês, e não a cada 15 dias, peguei alguns materiais de séries temporais usando o R, mas todos tratam diretamente das análises e não da manipulação de dados. Alguém saberia que abordagem eu poderia empregar para obter os resultados corretos ou alguma literatura que envolva a manipulação de datas, Obrigado -- Alexandre DOS SANTOS Engenheiro Florestal, Msc. Laboratório de Entomologia Florestal Departamento de Entomologia Universidade Federal de Lavras Caixa Postal 3037 37200-000 - Lavras - Minas Gerais - Brasil Tel: +55 35 92230304

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Como eu retiro so os 3 primeiro caracteres dos objetos de um Vetor?
by Augusto Ribas 17 Jan '12

17 Jan '12

Ola: #Vamos supor que eu tenha um vetor: teste<-c("Stomoxys.calcitrans","Musca.domestica","Syritta.flaviventris") Eu gostaria de estrair so as 3 primeiras letras, pra ter um resultado [1] "Sto" "Mus" "Syr" Meu interesse é em usar esse segundo vetor extraido para plotar um grafico, mas os nomes de especies inteiro ficam muito grandes e fica muito poluido, mas se der pra extrair os nomes assim fica mais facil fazer o grafico ja que são muitos nomes e além do trabalho é perigoso errar ordem de especies e tal ou escrever errado. Então se alguem souber como fazer. Outra coisa não sei se é muito complicado, mas todo nome de especie é o "Genero.especie" Daria pra mim tirar a primeira letra do Genero, e a primeira e segunda letra da especies, pra ter um resultado do tipo: [1] "Sca" "Mdo" "Sfl" E por ultimo, se o vetor for mistura especies e familias tipo: teste2<-c("Stomoxys.calcitrans","Musca.domestica","Syritta.flaviventris","Clusiidae") Neste caso o ultimo termo é uma familia, então ele é uma palavra so, e nao ta no esquema "Genero.especie" Daria pra nele ficar so as 3 primeiras letra, e chegar a um resultado final assim: [1] "Sca" "Mdo" "Sfl" "Clu" PS: "Genero.especie" Ta com um ponto separando os 2 mesmo. Não é espaço não. -- Grato Augusto C. A. Ribas Site Pessoal: http://augustoribas.heliohost.org Lattes: http://lattes.cnpq.br/7355685961127056

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dadaos de batimetria para o Maptools
by Rodrigo Plei 17 Jan '12

17 Jan '12

Prezados, Alguém sabe como extrair dados de batimetria (shape files) do GEBCO para plotar mapas com isóbatas usando o pacote maptools? Desde já, grato. Rodrigo -- =8-->=8-->=8-->=8-->=8-->=8-->=8-->=8-->=8-->=8-->=8-->=8--> Rodrigo Silvestre Martins, PhD Bolsista FAPESP Pós-Doutorado Laboratório de Ecossistemas Pesqueiros (LabPesq) Universidade de São Paulo, Instituto Oceanográfico Praça do Oceanográfico, 191. Cidade Universitaria (sala 107-A/B) Butantã - São Paulo/SP, Brasil 05508-900 Tel: +55 11 3091 6549 Email: rodrigo.plei(a)gmail.com ; ocersm(a)lycos.com; rsmartins(a)usp.br CV Lattes: http://lattes.cnpq.br/5350064124902777

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Tinn-R
by Joaquim Pereira Neto 17 Jan '12

17 Jan '12

Por que quando vou executar um programa com o Tinn-R versão 2.37.1, a execução para em algum ponto do programa, isto é, não executa totalmente o programa. Grato, Joaquim P. Neto.

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Transformar valores em datas
by ASANTOS 17 Jan '12

17 Jan '12

Bom dia Pessoal, Gostaria de saber se existe alguma função que transforma valores em datas para se trabalhar com séries temporais, no meu caso tenho um vetor que vai de 0 a 2865, sendo: times<-seq(0,2865,15)## Valores quinzenais de tempo onde o valor 0 representa 1/1/1994 e o valor 2865 representa 15/12/2001. Sendo que esta função teria de considerar os anos bissextos e as variações no número do dia dos meses. Obrigado, -- Alexandre DOS SANTOS Engenheiro Florestal, Msc. Laboratório de Entomologia Florestal Departamento de Entomologia Universidade Federal de Lavras Caixa Postal 3037 37200-000 - Lavras - Minas Gerais - Brasil Tel: +55 35 92230304

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Re: [R-br] ANOVA em regressão não linear!
by Thiago de Paula Protásio 17 Jan '12

17 Jan '12

Walmes e demais membros da lista; Vejam até onde consegui desenvolver com base no exemplo proposto na página das "Rídiculas". Tentei aplicar a função do Walmes para obter a anova do modelo não linear e o R², mas não obtive sucesso. Desde já agradeço a ajuda de todos! A <- 6.25551 > B <- -6.85223 > C <- -0.24253 > model8 <- nls(log(Carbtotal)~A+B*exp(C*Idade), data=citriodora, + start=list(A=A, B=B, C=C)) > summary(model8) Formula: log(Carbtotal) ~ A + B * exp(C * Idade) Parameters: Estimate Std. Error t value Pr(>|t|) A 6.25552 0.26527 23.58 1.14e-12 *** B -6.85224 0.27415 -25.00 5.14e-13 *** C -0.24253 0.02415 -10.04 8.88e-08 *** --- Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1 Residual standard error: 0.2841 on 14 degrees of freedom Number of iterations to convergence: 1 Achieved convergence tolerance: 2.429e-06 > F <- attr(model8$m$fitted(), "gradient") ## extraindo a matriz gradiente avaliada nas estimativas dos parâmetros > m0 <- lm(log(Carbtotal)~-1+F, data=citriodora)# passando a matriz gradiente para a lm(), importante remover intercepto > anova(m0) # partição da soma de quadrados total Analysis of Variance Table Response: log(Carbtotal) Df Sum Sq Mean Sq F value Pr(>F) F 3 271.26 90.422 1119.9 < 2.2e-16 *** Residuals 14 1.13 0.081 --- Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1 > anova(model8, lm(log(Carbtotal)~1, citriodora)) # SQ do modelo não linear corrigido para intercepto Analysis of Variance Table Model 1: log(Carbtotal) ~ A + B * exp(C * Idade) Model 2: log(Carbtotal) ~ 1 Res.Df Res.Sum Sq Df Sum Sq F value Pr(>F) 1 14 1.130 2 16 82.335 -2 -81.204 502.88 9.192e-14 *** --- Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1 > R2 <- function(nls.obj){ + da <- eval(nls.obj$data) + resp.name <- all.vars(summary(nls.obj)$formula)[1] + form <- paste(resp.name, "~1", sep="") + m0 <- lm(form, da) + an <- anova(nls.obj, m0) + sqn <- deviance(nls.obj) + sqe <- deviance(m0) + r2 <- 1-(sqn/sqe) + aov <- data.frame(fv=c("regression","residuals"), + gl=c(-an$Df[2],an$Res.Df[1]), + sq=c(-an$Sum[2],an$Res.Sum[1])) + aov$qm <- aov$sq/aov$gl + aov$F <- c(aov$qm[1]/aov$qm[2], NA) + aov$"Pr(>F)" <- c(1-pf(aov$F[1], df1=aov$gl[1], df2=aov$gl[2]), NA) + names(aov) <- c(" ","Df", "Sum Sq", "Mean Sq", "F value", "Pr(>F)") + return(list(anova=aov, R2=r2)) + } > R2(model8) Erro em anovalist.nls(object, ...) : 'anova' is only defined for sequences of "nls" objects Além disso: Warning message: In anovalist.nls(object, ...) : models with response "Carbtotal" removed because response differs from model 1 -- Thiago de Paula Protásio Acadêmico de Engenharia Florestal Universidade Federal de Lavras Departamento de Ciências Florestais Ciência e Tecnologia da Madeira (035) 9183 - 2246

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Re: [R-br] Fazer filme
by Ivan Bezerra Allaman 17 Jan '12

17 Jan '12

Estou estudando isso a algum tempo. Segue alguns links interessantes: http://www.drewconway.com/zia/?p=2800 http://www.stattler.com/article/add-animated-r-plot-your-latex-document http://www.r-bloggers.com/animated-plots-with-r/ http://robjhyndman.com/researchtips/animations/ http://r.yihui.name/ http://animation.yihui.name/prob:flipping_coins (S,f,P) Allaman \begin{signature} <<>>= Prof. Dr. Ivan Bezerra Allaman Universidade Estadual de Santa Cruz Departamento de Ciências Exatas e Tecnológicas Ilhéus/BA - Brasil Fone: +55 73 3680-5076 E-mail: ivanalaman@yahoo.com.br/ivanalaman@gmail.com @ \end{signature}

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Gráfico 3D
by Mauro Sznelwar 17 Jan '12

17 Jan '12

Como faço gráfico 3D desta função? f=rmultinom(30,3,prob=c(0.5,0.3,0.8))*dmultinom(x=1:3,prob=c(0.5,0.3,0.8)) round(f,1) contour(f) Mas nâo vai funcionar com estes comandos e x<-seq(-10,10,length= 30) y<-x z<-c(x, y, f) persp(z) Nem com estes require(lattice) wireframe(z)

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Fazer filme
by Mauro Sznelwar 16 Jan '12

16 Jan '12

Tenho uma sequência de vários gráficos gerados pelos comandos abaixo, e gostaria de fazer uma sequência, como um filme. Dá para fazer em qualquer editor de imagem como o Picasa, mas queria saber se tem algum comando ou uma sequência de comandos que o faz. jpeg("exemplohist.jpg") hist(rexp(150, rate=5)) dev.off()

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Ajuda na compilação: R-dev Linux
by Jose Claudio Faria 16 Jan '12

16 Jan '12

Pessoal, Estou tendo um problema localizado na compilação do R-dev relativo a função savePlot(grDevices). Tenho muitos scripts que uso esta função e minha extensão preferida sempre foi png. #----------------------------------------------- plot(1) savePlot('test.png', type='png') Erro em savePlot("test.png", type = "png") : somente possível copiar dos dispositivos 'X11(type="*cairo")' #----------------------------------------------- Na tentiva de contornar o problema usei o script Rpatch.sh (disponível no LEG: http://leg.ufpr.br/doku.php/dicas:r-compila) que faz a checagem de dependências, tem boas directivas de configuração, etc, mas não resolveu. Tenho duas versões do R em meu Linux: 2.13.1 que baixei pré-compilada e a R-dev sendo compilada na máquina. Na 2.13.1 a função savePlot funciona perfeitamente. Portanto, concluo que o problema etá mesmo na compilação da R-dev ou algum possível bug. Alguma dica? Abs, -- ///\\\///\\\///\\\///\\\///\\\///\\\///\\\///\\\ Jose Claudio Faria Estatistica - Prof. Pleno UESC/DCET/Brasil joseclaudio.faria at gmail.com ///\\\///\\\///\\\///\\\///\\\///\\\///\\\///\\\

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[OFF-TOPIC] Processador para rodar o R no Linux
by Fernando Mayer 16 Jan '12

16 Jan '12

Amigos do R, estou montando um computador para uso pessoal e gostaria de otimizá-lo para o uso do R no Linux. Por isso gostaria de saber a opinião de vocês sobre algumas coisas: 1) AMD x Intel Os processadores AMD são bem mais em conta do que os da Intel, mas pelos benchmarks que encontrei por aí eles sempre saem perdendo. O custo benefício às vezes é melhor que os da Intel, mas isso se deve aos seus preços muito mais baixos. Os processadores AMD que olhei são estes (do mais barato para o mais caro): http://www.cpubenchmark.net/cpu.php?cpu=AMD+Phenom+II+X6+1100T http://www.cpubenchmark.net/cpu.php?cpu=AMD+FX-6100+Six-Core http://www.cpubenchmark.net/cpu.php?cpu=AMD+FX-8120+Eight-Core Embora só tenha encontrado reclamações e problemas para estes mais novos (da série FX), principalmente no Linux. O que acontece é que o kernel (mesmo os mais novos) ainda não está com todos os módulos otimizados para estes processadores. Isso me levaria aos da Intel: http://www.cpubenchmark.net/cpu.php?cpu=Intel+Core+i5-2500K+%40+3.30GHz http://www.cpubenchmark.net/cpu.php?cpu=Intel+Core+i7-2600+%40+3.40GHz que parecem funcionar perfeitamente com os kernels mais novos e são superior em processamento quando comparados com os da AMD da mesma categoria. Com isso estou inclinado a ficar com a Intel, o que me leva a segunda pergunta. 2) Intel i7 ou i5 A principal diferença entre os dois é que o i5 possui 4 núcleos verdadeiros, enquanto que o i7 tem 4 núcleos verdadeiros + 4 núcleos "virtuais", gerados pela tecnologia Hyper-Threading (HT) da Intel. Sei que o R funciona basicamente utilizando um processador apenas, mas podemos otimizá-lo para fazer uso dos demais processadores através dos pacotes paralell, snow, etc. O que gostaria de saber de quem tem experiência com estes processadores é se vale a pena gastar ~ R$ 200,00 a mais no i7 para ter mais núcleos, e se realmente é possível que o R utilize os 4 núcleos virtuais da tecnologia HT. Pelos benchmarks parece que o i5 2500K tem o melhor custo/benefício (e com os 200 de diferença posso investir em mais RAM :) Agradeço desde já quem quiser dar uma opinião. []s, --- Fernando Mayer Universidade Federal de Santa Catarina - UFSC Departamento de Ecologia e Zoologia - ECZ/CCB URL: http://sites.google.com/site/fernandomayer e-mail: fernandomayer [@] gmail.com

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Como utilizar familia e ligação da classe glm
by Bruno Santos 15 Jan '12

15 Jan '12

Olá a todos, Estou tentando reutilizar a família e a função de ligação de um certo objeto da classe glm para ajustar um novo modelo, mas não estou conseguindo. O que eu quero fazer é gerar o modelo3, como no seguinte exemplo, mas recebo uma mensagem de erro. Obrigado desse já, Bruno ### CMR x <- rgamma(20, shape=1) y <- rgamma(20, shape=1) modelo <- glm(y~x, family=Gamma(link=identity)) fam <- modelo$family$family fam ligacao <- modelo$family$link ligacao modelo2 <- glm(y~x, family=fam) modelo3 <- glm(y~x, family=fam(link=ligacao)) # Error in glm(y ~ x, family = fam(link = ligacao)) : # could not find function "fam"

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Valor de coluna onde o registro tem valor máximo em outra coluna
by Roberto Pinho 15 Jan '12

15 Jan '12

Caros, Não consegui achar uma função que retorne o conteúdo de uma coluna que pertença a uma linha onde o valor de outra coluna é máximo. Exemplificando: data(iris) attach(iris) m = max(Sepal.Length) iris[Sepal.Length==m,"Species"] que retorna a espécie ou espécies do maior Sepal.Length encontrado; imagino algo na linha de valueatmax(data=iris, maxcol = "Sepal.Lenght", valuecol = "Species) Existe algo feito que pareça com isto? obrigado -- Roberto Pinho robertodepinho(a)gmail.com msn: robertopinho(a)yahoo.com.br Skype: roberto.pinho http://www.ascoisas.com

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ajuda em teste para comparação de frequências
by Flávia Machado 15 Jan '12

15 Jan '12

Galera, preciso muito da ajuda de vocês, tenho umas análises pra fazer mas não sei quais testes aplicar para responder as minhas perguntas. Tenho um conjunto de dados que correspondem a frequência com que indivíduos de uma população realizam determinados comportamentos ao longo do período de um ano. Quero verificar se há diferença significativa nas frequências dos comportamentos entre os indivíduos neste período. Além disso, se as frequências dos comportamentos de um determinado indivíduo diferem da média dos comportamentos dos demais indivíduos. Ou seja, quero ver: os indivíduos possuem mesmo comportamento? Se não, como é essa diferença e como ela varia ao longo do ano? Essa diferença aumenta, diminui ou se mantêm estável ao longo do ano? Os comportamentos do indivíduo selecionado difere dos demais? Caso haja diferença, verificar o quão diferente e se essa diferença nas freq. diminuem, aumentam, ou se mantêm constantes ao longo do ano. Imagino que para saber se existe diferença nas frequências eu poderia fazer um qui-quadrado, mas não sei fazer. Agora para o outro questionamento nem faço ideia de que teste aplicar. Segue o CMR de parte dos meus dados para vocês verem como eles estão. OBS: as letras antes do "." na função se referem ao indivíduos e as depois dele aos comportamentos. Rep - Repouso; Ali - Alimentação; D - deslocamento; S - Social; O - outros comportamentos. Cada linha corresponde a um dia amostral (71 no total). São seis indivíduos e o que eu quero comparar com os demais está nomeado "RE". # Essas são as frequências dos comportamentos de cada indivíduo: Rep.RE<-c(0.000,0.714,0.500,1.000,0.500,0.636,1.000,0.700,0.600,0.000,0.789,0.417,0.000,0.941,0.200,0.500,0.462,1.000,0.500,0.500,0.273,0.643,0.467,0.353,0.000,0.364,0.250,0.316,0.273,0.429,0.000,0.000,0.258,0.182,0.421,0.667,0.250,0.688,0.167,0.400,0.400,0.692,0.500,0.333,0.400,0.091,0.360,0.333,0.882,0.583,0.400,0.724,0.000,0.636,0.522,0.563,0.368,0.667,0.633,0.640,0.500,0.571,0.621,0.375,0.522,0.000,0.500,0.800,0.375,0.000,0.375) Ali.RE<-c(0.000,0.000,0.500,0.000,0.000,0.182,0.000,0.300,0.200,1.000,0.105,0.250,0.000,0.000,0.600,0.188,0.462,0.000,0.000,0.321,0.091,0.143,0.200,0.353,0.400,0.364,0.450,0.579,0.727,0.357,0.000,0.800,0.484,0.727,0.316,0.333,0.500,0.188,0.556,0.450,0.500,0.154,0.000,0.333,0.400,0.818,0.480,0.444,0.000,0.333,0.500,0.172,0.333,0.242,0.348,0.188,0.421,0.222,0.000,0.200,0.100,0.143,0.172,0.375,0.304,0.000,0.167,0.000,0.500,0.000,0.375) D.RE <-c(1.000,0.286,0.000,0.000,0.500,0.182,0.000,0.000,0.200,0.000,0.105,0.250,1.000,0.059,0.200,0.313,0.077,0.000,0.250,0.179,0.545,0.143,0.333,0.235,0.600,0.136,0.200,0.105,0.000,0.214,1.000,0.200,0.226,0.091,0.211,0.000,0.250,0.125,0.222,0.150,0.100,0.077,0.500,0.333,0.200,0.091,0.160,0.222,0.118,0.083,0.100,0.103,0.667,0.121,0.130,0.188,0.211,0.111,0.100,0.160,0.300,0.286,0.172,0.250,0.174,0.500,0.333,0.200,0.125,0.000,0.125) S.RE <-c(0.000,0.000,0.000,0.000,0.000,0.000,0.000,0.000,0.000,0.000,0.000,0.083,0.000,0.000,0.000,0.000,0.000,0.000,0.250,0.000,0.091,0.071,0.000,0.059,0.000,0.136,0.100,0.000,0.000,0.000,0.000,0.000,0.032,0.000,0.000,0.000,0.000,0.000,0.056,0.000,0.000,0.000,0.000,0.000,0.000,0.000,0.000,0.000,0.000,0.000,0.000,0.000,0.000,0.000,0.000,0.063,0.000,0.000,0.267,0.000,0.100,0.000,0.034,0.000,0.000,0.500,0.000,0.000,0.000,0.000,0.125) O.RE <-c(0.000,0.000,0.000,0.000,0.000,0.000,0.000,0.000,0.000,0.000,0.000,0.000,0.000,0.000,0.000,0.000,0.000,0.000,0.000,0.000,0.000,0.000,0.000,0.000,0.000,0.000,0.000,0.000,0.000,0.000,0.000,0.000,0.000,0.000,0.053,0.000,0.000,0.000,0.000,0.000,0.000,0.077,0.000,0.000,0.000,0.000,0.000,0.000,0.000,0.000,0.000,0.000,0.000,0.000,0.000,0.000,0.000,0.000,0.000,0.000,0.000,0.000,0.000,0.000,0.000,0.000,0.000,0.000,0.000,0.000,0.000) Rep.SO<-c(0.667,0.833,0.750,0.500,0.000,0.789,0.889,0.500,0.250,0.625,0.000,0.733,0.727,0.500,0.563,0.429,0.455,0.750,0.500,0.600,0.857,0.000,0.500,0.550,0.444,0.308,0.200,0.500,0.300,0.200,0.333,0.000,0.235,0.000,0.278,0.154,0.429,0.333,0.091,0.393,0.619,0.692,0.522,0.500,0.000,0.333,0.154,0.429,0.824,0.619,0.143,0.385,0.000,0.368,0.182,0.714,0.471,0.464,0.400,0.500,0.000,0.545,0.560,0.643,0.167,0.200,0.833,0.000,0.615,0.250,0.000) Ali.SO<-c(0.000,0.083,0.000,0.357,0.000,0.158,0.000,0.429,0.500,0.125,0.000,0.067,0.091,0.375,0.250,0.429,0.091,0.000,0.500,0.200,0.000,0.333,0.250,0.200,0.222,0.308,0.600,0.250,0.550,0.467,0.333,0.889,0.706,0.000,0.333,0.308,0.571,0.400,0.364,0.393,0.286,0.231,0.217,0.000,0.333,0.000,0.538,0.214,0.059,0.143,0.429,0.385,0.000,0.421,0.727,0.000,0.353,0.321,0.000,0.000,0.500,0.273,0.160,0.143,0.389,0.400,0.000,0.667,0.154,0.250,0.000) D.SO<-c(0.333,0.000,0.250,0.143,1.000,0.053,0.000,0.071,0.250,0.125,0.000,0.200,0.182,0.125,0.125,0.143,0.455,0.250,0.000,0.133,0.143,0.667,0.250,0.250,0.333,0.385,0.200,0.250,0.150,0.333,0.333,0.111,0.059,1.000,0.333,0.538,0.000,0.267,0.545,0.179,0.095,0.077,0.261,0.500,0.667,0.667,0.231,0.357,0.118,0.190,0.429,0.231,0.000,0.211,0.091,0.286,0.176,0.214,0.067,0.500,0.500,0.182,0.280,0.214,0.444,0.200,0.167,0.333,0.231,0.500,0.000) S.SO<-c(0.000,0.083,0.000,0.000,0.000,0.000,0.111,0.000,0.000,0.125,0.000,0.000,0.000,0.000,0.063,0.000,0.000,0.000,0.000,0.067,0.000,0.000,0.000,0.000,0.000,0.000,0.000,0.000,0.000,0.000,0.000,0.000,0.000,0.000,0.000,0.000,0.000,0.000,0.000,0.036,0.000,0.000,0.000,0.000,0.000,0.000,0.000,0.000,0.000,0.000,0.000,0.000,0.000,0.000,0.000,0.000,0.000,0.000,0.533,0.000,0.000,0.000,0.000,0.000,0.000,0.200,0.000,0.000,0.000,0.000,0.000) O.SO<-c(0.000,0.000,0.000,0.000,0.000,0.000,0.000,0.000,0.000,0.000,0.000,0.000,0.000,0.000,0.000,0.000,0.000,0.000,0.000,0.000,0.000,0.000,0.000,0.000,0.000,0.000,0.000,0.000,0.000,0.000,0.000,0.000,0.000,0.000,0.056,0.000,0.000,0.000,0.000,0.000,0.000,0.000,0.000,0.000,0.000,0.000,0.077,0.000,0.000,0.048,0.000,0.000,0.000,0.000,0.000,0.000,0.000,0.000,0.000,0.000,0.000,0.000,0.000,0.000,0.000,0.000,0.000,0.000,0.000,0.000,0.000) Rep.LA<-c(0.333,0.857,0.333,0.833,0.667,0.750,0.538,0.600,0.000,0.429,0.765,0.786,1.000,0.818,0.286,0.833,0.500,0.857,0.500,0.583,0.600,0.833,0.636,0.417,0.800,0.000,0.375,0.476,0.143,0.357,0.200,0.333,0.304,0.385,0.688,0.143,0.167,0.375,0.250,0.167,0.682,0.000,0.000,0.500,0.000,0.385,0.435,0.333,0.826,0.000,0.417,0.842,0.643,0.750,0.429,0.286,0.000,0.409,0.444,0.000,0.000,0.536,0.667,0.636,0.632,0.500,0.600,0.000,0.444,1.000,0.000) Ali.LA<-c(0.667,0.000,0.333,0.083,0.111,0.083,0.077,0.300,0.000,0.000,0.000,0.000,0.000,0.000,0.429,0.000,0.000,0.000,0.250,0.000,0.200,0.083,0.091,0.333,0.000,1.000,0.375,0.429,0.714,0.357,0.400,0.400,0.478,0.231,0.063,0.286,0.667,0.125,0.583,0.667,0.182,0.714,0.000,0.167,0.000,0.308,0.435,0.500,0.130,0.000,0.417,0.053,0.000,0.167,0.500,0.143,0.750,0.227,0.000,0.333,0.000,0.214,0.111,0.182,0.158,0.000,0.000,0.000,0.222,0.000,0.000) D.LA <-c(0.000,0.000,0.222,0.000,0.000,0.167,0.077,0.100,0.000,0.571,0.176,0.214,0.000,0.091,0.286,0.167,0.500,0.143,0.250,0.333,0.200,0.083,0.273,0.250,0.200,0.000,0.188,0.095,0.143,0.286,0.400,0.267,0.217,0.308,0.250,0.571,0.167,0.500,0.167,0.167,0.091,0.286,0.000,0.167,0.000,0.308,0.130,0.167,0.043,1.000,0.167,0.105,0.071,0.083,0.071,0.571,0.250,0.364,0.167,0.667,1.000,0.214,0.222,0.182,0.211,0.000,0.400,1.000,0.333,0.000,0.000) S.LA <-c(0.000,0.143,0.111,0.083,0.222,0.000,0.308,0.000,0.000,0.000,0.059,0.000,0.000,0.091,0.000,0.000,0.000,0.000,0.000,0.083,0.000,0.000,0.000,0.000,0.000,0.000,0.063,0.000,0.000,0.000,0.000,0.000,0.000,0.077,0.000,0.000,0.000,0.000,0.000,0.000,0.000,0.000,0.000,0.167,0.000,0.000,0.000,0.000,0.000,0.000,0.000,0.000,0.071,0.000,0.000,0.000,0.000,0.000,0.389,0.000,0.000,0.036,0.000,0.000,0.000,0.500,0.000,0.000,0.000,0.000,0.000) O.LA <-c(0.000,0.000,0.000,0.000,0.000,0.000,0.000,0.000,0.000,0.000,0.000,0.000,0.000,0.000,0.000,0.000,0.000,0.000,0.000,0.000,0.000,0.000,0.000,0.000,0.000,0.000,0.000,0.000,0.000,0.000,0.000,0.000,0.000,0.000,0.000,0.000,0.000,0.000,0.000,0.000,0.045,0.000,0.000,0.000,0.000,0.000,0.000,0.000,0.000,0.000,0.000,0.000,0.214,0.000,0.000,0.000,0.000,0.000,0.000,0.000,0.000,0.000,0.000,0.000,0.000,0.000,0.000,0.000,0.000,0.000,0.000) Rep.PE<-c(0.000,0.000,0.000,0.000,0.429,0.600,0.667,0.650,0.909,0.600,0.688,0.875,0.636,0.929,0.200,0.300,1.000,0.000,0.286,0.375,0.000,0.818,0.500,0.500,0.250,0.000,0.308,0.000,0.500,0.000,0.400,0.231,0.125,0.333,0.182,0.500,0.600,0.286,0.182,0.692,0.800,0.182,0.750,0.778,0.556,0.125,0.238,0.000,0.846,0.000,0.357,0.565,0.333,0.583,0.350,0.667,0.273,0.407,0.300,0.684,0.143,0.368,0.429,0.316,0.600,0.000,1.000,0.000,0.667,0.579,0.500) Ali.PE<-c(0.000,0.000,0.000,0.000,0.143,0.050,0.000,0.300,0.000,0.000,0.000,0.125,0.227,0.071,0.600,0.600,0.000,0.000,0.143,0.250,0.500,0.091,0.333,0.200,0.250,1.000,0.308,0.000,0.500,0.500,0.600,0.538,0.250,0.444,0.545,0.500,0.000,0.429,0.545,0.154,0.133,0.455,0.000,0.111,0.222,0.250,0.476,0.667,0.077,1.000,0.429,0.217,0.000,0.167,0.400,0.000,0.455,0.333,0.000,0.158,0.571,0.263,0.333,0.368,0.250,0.333,0.000,0.000,0.222,0.211,0.250) D.PE <-c(0.000,0.000,0.000,0.000,0.143,0.050,0.000,0.000,0.091,0.200,0.000,0.000,0.000,0.000,0.200,0.100,0.000,0.000,0.429,0.375,0.250,0.091,0.167,0.200,0.500,0.000,0.154,0.000,0.000,0.500,0.000,0.154,0.375,0.222,0.000,0.000,0.200,0.286,0.273,0.154,0.067,0.364,0.250,0.000,0.222,0.625,0.190,0.333,0.077,0.000,0.214,0.174,0.000,0.250,0.200,0.333,0.273,0.259,0.000,0.158,0.286,0.316,0.190,0.211,0.150,0.333,0.000,0.667,0.111,0.211,0.000) S.PE <-c(0.000,0.000,0.000,0.000,0.000,0.300,0.333,0.050,0.000,0.200,0.313,0.000,0.136,0.000,0.000,0.000,0.000,0.000,0.143,0.000,0.250,0.000,0.000,0.100,0.000,0.000,0.231,0.000,0.000,0.000,0.000,0.077,0.250,0.000,0.273,0.000,0.200,0.000,0.000,0.000,0.000,0.000,0.000,0.111,0.000,0.000,0.000,0.000,0.000,0.000,0.000,0.043,0.000,0.000,0.050,0.000,0.000,0.000,0.700,0.000,0.000,0.053,0.048,0.105,0.000,0.333,0.000,0.333,0.000,0.000,0.250) O.PE <-c(0.000,0.000,0.000,0.000,0.286,0.000,0.000,0.000,0.000,0.000,0.000,0.000,0.000,0.000,0.000,0.000,0.000,0.000,0.000,0.000,0.000,0.000,0.000,0.000,0.000,0.000,0.000,0.000,0.000,0.000,0.000,0.000,0.000,0.000,0.000,0.000,0.000,0.000,0.000,0.000,0.000,0.000,0.000,0.000,0.000,0.000,0.095,0.000,0.000,0.000,0.000,0.000,0.667,0.000,0.000,0.000,0.000,0.000,0.000,0.000,0.000,0.000,0.000,0.000,0.000,0.000,0.000,0.000,0.000,0.000,0.000) Rep.BR<-c(0.000,0.000,0.000,0.364,0.000,0.000,0.900,0.619,0.591,0.500,0.000,0.600,0.500,0.000,0.333,0.700,0.750,0.000,0.333,0.364,0.333,0.000,0.368,0.474,0.300,0.000,0.333,0.471,0.368,0.273,0.429,0.407,0.143,0.500,0.000,0.714,0.000,0.200,0.500,0.400,0.846,0.000,0.000,0.000,0.571,0.111,0.357,0.333,0.000,0.000,0.467,0.333,0.000,0.421,0.333,0.200,0.000,0.909,0.333,0.600,0.000,0.438,0.111,0.750,0.667,0.000,0.556,0.000,1.000,1.000,0.000) Ali.BR<-c(0.000,0.000,0.000,0.182,0.333,1.000,0.000,0.286,0.182,0.000,0.000,0.100,0.083,0.000,0.467,0.200,0.125,0.000,0.167,0.409,0.556,0.000,0.368,0.474,0.400,1.000,0.333,0.294,0.368,0.455,0.429,0.444,0.714,0.167,0.750,0.143,0.000,0.400,0.375,0.400,0.154,0.500,0.000,0.000,0.214,0.222,0.571,0.333,0.000,0.000,0.467,0.333,0.727,0.421,0.500,0.400,0.000,0.000,0.000,0.200,0.000,0.188,0.667,0.000,0.200,0.818,0.000,0.000,0.000,0.000,0.000) D.BR <-c(0.000,0.000,0.000,0.182,0.333,0.000,0.000,0.048,0.227,0.500,0.000,0.200,0.167,0.000,0.133,0.100,0.125,0.000,0.500,0.136,0.111,0.000,0.211,0.053,0.300,0.000,0.167,0.176,0.211,0.273,0.143,0.148,0.143,0.167,0.250,0.143,0.000,0.400,0.000,0.200,0.000,0.500,0.000,1.000,0.214,0.667,0.071,0.333,0.000,0.000,0.067,0.333,0.182,0.158,0.111,0.400,0.000,0.091,0.083,0.200,1.000,0.313,0.222,0.083,0.133,0.091,0.444,0.000,0.000,0.000,0.000) S.BR <-c(0.000,0.000,1.000,0.273,0.333,0.000,0.100,0.048,0.000,0.000,0.000,0.100,0.250,0.000,0.067,0.000,0.000,0.000,0.000,0.091,0.000,0.000,0.053,0.000,0.000,0.000,0.167,0.059,0.053,0.000,0.000,0.000,0.000,0.167,0.000,0.000,0.000,0.000,0.125,0.000,0.000,0.000,0.000,0.000,0.000,0.000,0.000,0.000,0.000,0.000,0.000,0.000,0.091,0.000,0.056,0.000,0.000,0.000,0.583,0.000,0.000,0.063,0.000,0.167,0.000,0.091,0.000,0.000,0.000,0.000,0.000) O.BR <-c(1.000,0.000,0.000,0.000,0.000,0.000,0.000,0.000,0.000,0.000,0.000,0.000,0.000,0.000,0.000,0.000,0.000,0.000,0.000,0.000,0.000,0.000,0.000,0.000,0.000,0.000,0.000,0.000,0.000,0.000,0.000,0.000,0.000,0.000,0.000,0.000,0.000,0.000,0.000,0.000,0.000,0.000,0.000,0.000,0.000,0.000,0.000,0.000,0.000,0.000,0.000,0.000,0.000,0.000,0.000,0.000,0.000,0.000,0.000,0.000,0.000,0.000,0.000,0.000,0.000,0.000,0.000,0.000,0.000,0.000,0.000) Rep.ZU<-c(0.000,0.000,0.000,0.000,0.000,0.800,0.556,0.250,0.000,0.000,0.700,0.500,0.000,0.000,0.000,0.750,0.800,0.000,0.000,0.375,0.565,0.000,0.111,0.000,0.000,0.000,0.333,0.000,0.000,0.375,0.000,0.143,0.333,0.500,0.333,0.250,0.000,0.000,0.500,0.000,0.643,0.375,0.875,0.250,0.000,0.000,0.000,0.167,0.818,0.429,0.000,0.600,0.000,0.000,0.000,0.667,0.000,0.000,0.000,0.000,0.000,0.250,0.000,0.231,0.833,0.000,0.000,0.000,0.938,0.000,0.000) Ali.ZU<-c(0.000,0.000,0.000,0.000,0.000,0.000,0.000,0.750,0.000,0.000,0.000,0.000,0.000,0.000,0.000,0.000,0.000,0.000,0.000,0.250,0.174,0.000,0.556,0.000,0.000,1.000,0.167,1.000,0.500,0.375,0.000,0.857,0.067,0.400,0.333,0.750,0.000,0.500,0.250,1.000,0.143,0.000,0.000,0.000,0.000,0.000,0.500,0.167,0.000,0.143,0.000,0.000,0.000,0.667,1.000,0.000,0.000,0.000,0.000,0.000,0.000,0.125,0.000,0.308,0.000,0.500,1.000,0.000,0.000,0.000,0.000) D.ZU<-c(0.000,0.000,0.000,0.000,0.000,0.100,0.111,0.000,0.000,0.000,0.100,0.250,0.000,0.000,1.000,0.250,0.200,0.000,0.000,0.375,0.217,0.000,0.333,0.000,0.000,0.000,0.167,0.000,0.500,0.125,1.000,0.000,0.533,0.000,0.333,0.000,1.000,0.500,0.250,0.000,0.143,0.375,0.125,0.750,1.000,0.000,0.500,0.667,0.182,0.429,0.000,0.400,0.000,0.333,0.000,0.333,0.000,0.000,0.000,0.000,1.000,0.625,1.000,0.231,0.167,0.500,0.000,0.000,0.063,0.000,0.000) S.ZU<-c(0.000,0.000,0.000,0.000,0.000,0.100,0.333,0.000,0.000,0.000,0.200,0.250,0.000,0.000,0.000,0.000,0.000,0.000,0.000,0.000,0.043,0.000,0.000,0.000,0.000,0.000,0.333,0.000,0.000,0.125,0.000,0.000,0.067,0.100,0.000,0.000,0.000,0.000,0.000,0.000,0.071,0.000,0.000,0.000,0.000,0.000,0.000,0.000,0.000,0.000,0.000,0.000,0.000,0.000,0.000,0.000,0.000,0.000,0.000,0.000,0.000,0.000,0.000,0.231,0.000,0.000,0.000,0.000,0.000,0.000,0.000) O.ZU<-c(0.000,0.000,0.000,0.000,0.000,0.000,0.000,0.000,0.000,0.000,0.000,0.000,0.000,0.000,0.000,0.000,0.000,0.000,0.000,0.000,0.000,0.000,0.000,0.000,0.000,0.000,0.000,0.000,0.000,0.000,0.000,0.000,0.000,0.000,0.000,0.000,0.000,0.000,0.000,0.000,0.000,0.250,0.000,0.000,0.000,0.000,0.000,0.000,0.000,0.000,0.000,0.000,0.000,0.000,0.000,0.000,0.000,0.000,0.000,0.000,0.000,0.000,0.000,0.000,0.000,0.000,0.000,0.000,0.000,0.000,0.000) # Funções para criação das tabelas dos comportamentos para cada um dos indivíduos. tabela.repouso<-cbind(Rep.RE,Rep.SO,Rep.LA,Rep.PE,Rep.BR,Rep.ZU) tabela.alimentacao<-cbind(Ali.RE,Ali.SO,Ali.LA,Ali.PE,Ali.BR,Ali.ZU) tabela.deslocamento<-cbind(D.RE,D.SO,D.LA,D.PE,D.BR,D.ZU) tabela.social<-cbind(S.RE,S.SO,S.LA,S.PE,S.BR,S.ZU) tabela.outros<-cbind(O.RE,O.SO,O.LA,O.PE,O.BR,O.ZU) Espero que tenha dado para entender as minhas dúvidas e os meus dados. Agradeço desde já. Abraços, -- Flávia Machado

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ANOVA em regressão não linear!
by Thiago de Paula Protásio 14 Jan '12

14 Jan '12

Prezados membros da R-Br; Recorro a valiosa ajuda de vcs! Estou trabalhando com os seguintes modelos não lineares para modelar o acúmulo de biomassa com o tempo: Yi = A*exp(B*X) ln(Yi) = A * (1 - exp(-B*X) ln (Yi) = A + B*exp(C*X) Nas "Rídiculas" eu encontrei uma função e explicação do Walmes: http://ridiculas.wordpress.com/2011/05/29/analise-de-residuos-em-regressao-… Mas gostaria de saber se posso aplicar essa teoria/função para os modelos que estou trabalhando. Na verdade, estou testando modelos lineares também e gostaria de saber se é fundamental utilizar o Teste F para os modelos não lineares à semelhança dos modelos lineares.?? Cheguei a fazer alguma coisa como apresentado abaixo para um dos modelos. Mas na ANOVA que foi realizada entre o modelo não linear e o modelo reduzido para o intercepto obtive uma soma de quadrado negativa, daí a minha dúvida. Muito obrigado pela valiosa ajuda! -- Thiago de Paula Protásio Acadêmico de Engenharia Florestal Universidade Federal de Lavras Departamento de Ciências Florestais Ciência e Tecnologia da Madeira (035) 9183 - 2246

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Ajuda para buscar coordenadas de um ponto qq em uma matriz com coordenadas.
by Maria Papa 13 Jan '12

13 Jan '12

Olá pessoal, boa tarde, Eu estou com outro problema! Eu tenho um conjunto de coordenadas, (X,Y,Z) uma matriz com uma coluna de valores para X, outra para Y e outra para Z. Dado uma coordenada (XYZ) qualquer, eu preciso buscar neste conjunto (matriz) os valores imediatamente maiores e menores que as coordenadas deste ponto qualquer. Eu pensei no comando ifelse, mas eu teria que fazer para cada coordenadas. Alguém saberia me dizer se existe alguma forma mais simples para fazer isso? Desde já agradeço a todos. célia

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Função para somar pontuação
by Carlos Andrade 12 Jan '12

12 Jan '12

Prezados amigos da lista Que 2012 seja repleto de sucesso. Minha dúvida. Elaborei um questionário e atribuí pontos para as diferentes respostas. Como codifico uma função para gerar uma variável que será o resultado da computação da soma dos pontos de cada indivíduo de acordo com suas respostas? Abaixo segue o esquema de pontuação das questões que entram no cálculo. Pergunta Resposta Pontuação p1 1 5 2 1 3 0 p2 1 5 2 3 3 3 4 3 5 1 6 0 p3.1 1 1 0 0 P3.2 1 3 0 0 P3.3 1 3 0 0 P3.4 1 5 0 0 P3.5 1 5 0 0 P3.6 1 3 0 0 P3.7 1 0 0 0 p4 1 1 2 3 3 3 4 3 5 5 6 0 P5.1 1 3 0 0 P5.2 1 1 0 0 P5.3 1 3 0 0 P5.4 1 1 0 0 P5.5 1 1 0 0 P5.6 1 5 0 0 P5.7 1 1 0 0 P5.8 1 1 0 0 P5.9 1 1 0 0 P5.10 1 0 0 0 p6 1 1 2 3 3 3 4 3 5 5 6 0 P7.1 1 5 0 0 P7.2 1 5 0 0 P7.3 1 5 0 0 P7.4 1 5 0 0 P7.5 1 5 0 0 P7.6 1 3 0 0 P7.7 1 0 0 0 P8.1 1 5 0 0 P8.2 1 3 0 0 P8.3 1 5 0 0 P8.4 1 1 0 0 P8.5 1 0 0 0 P9.1 1 5 0 0 P9.2 1 5 0 0 P9.3 1 5 0 0 P9.4 1 5 0 0 P9.5 1 5 0 0 P9.6 1 5 0 0 P9.7 1 0 0 0 P10.1 1 5 0 0 P10.2 1 5 0 0 P10.3 1 5 0 0 P10.4 1 5 0 0 P10.5 1 5 0 0 P10.6 1 0 0 0 P10.7 1 0 0 0 p11 1 5 2 3 3 1 4 0 5 0 p15 1 5 2 3 3 3 4 1 5 1 6 0 Também: como substituir os NAs por zeros de uma vez só? Agradeço antecipadamente qualquer ajuda. -- Atenciosamente, Prof. Carlos A. S. de Andrade LAPEA - Laboratório de Pesquisa em Economia Aplicada e Engenharia de Produção Universidade Federal de Campina Grande. Centro de Humanidades Unidade Acadêmica de Economia

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Gerar multinomial
by Mauro Sznelwar 11 Jan '12

11 Jan '12

Gostaria de saber onde posso achar comandos do R para geração de números aleatórios da distribuição multinomial. Não posso usar o comando rmultinom que existe, preciso implementar algo para três variáveis, teta1, teta2 e teta3. Desde já agradeço.

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Re: [R-br] Superfície de Resposta (Resolvido)
by Evelyn Alecrim 11 Jan '12

11 Jan '12

Pessoal, obrigada pela ajuda. Consegui fazer o gráfico usando persp(). -- Evelyn da Fonseca Alecrim Bragion Mestranda em Ecologia Aplicada - UFLA Lavras - MG/Brasil

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Ajuda com NAs
by Daniel C Bezerra 11 Jan '12

11 Jan '12

Amigos do R, Eu tenho uma variável que foi medida três vezes no tempo com NAs em pontos diferentes de cada um dos momentos em que foi medida: Eu gostaria de fazer a média destas 3 medidas mas caso uma medida fosse NA eu não gostaria de ignorar esta linha e torná-la NA e sim usar a média das medidas restantes. Alguém teria uma sugestão? Exemplo: m1<-c(1,2,3,4,NA,NA) m2<-c(1,NA,3,4,5,NA) m3<-c(NA,2,3,NA,NA,NA) Se eu faço apply(cbind(m1,m2,m3),1,mean) Eu tenho: [1] NA NA 3 NA NA NA Mas eu gostaria de: [1] 1 2 3 4 5 NA Muito obrigado, D

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Anova de um modelo.
by Julia Maria 11 Jan '12

11 Jan '12

Pessoal, enviai um trabalho pra um Congresso e voltou pedindo algumas correções, dentre elas pedindo Anova dos modelos que eu utilizei. Como estou descobrindo o R agora, preciso da ajuda de voces. A partir do modelo: Y=a+b+c+ab+ac+bc, eu consigo recuperar uma anova? Tenho os valores de a, b e c e tb tenho os Y originais e os preditos. Bastaria escrever o modelo e pedir a anova?

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Superfície de Resposta
by Evelyn Alecrim 11 Jan '12

11 Jan '12

Olá Pessoal, Estou precisando produzir um gráfico de superfície para o seguinte modelo : lm(y=a+bx+cz+dx*z) Alguém poderia me dizer que pacote poderia usar e qual o código? Obrigada... -- Evelyn da Fonseca Alecrim Bragion Mestranda em Ecologia Aplicada - UFLA Lavras - MG/Brasil

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Resultado gráfico do teste de kruskal-Wallis
by Maria Papa 10 Jan '12

10 Jan '12

Olá Pessoal, boa tarde, Eu gostaria de saber da possibilidade de gerar um resultado gráfico do teste de Kruska-Wallis, assim como o Tukey apresenta o gráfico da diferença das médias. A comparação com o Tukey não tem sentido teórico, é apenas para exemplificar o tipo de resultado que eu preciso. Eu estou usando a função kruskal do pacote agricolae. Desde já agradeço a todos Maria Célia

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Criando um processo de pontos sobre uma área regular
by ASANTOS 10 Jan '12

10 Jan '12

Boa tarde pessoal, Estou tentando criar um processo de pontos aleatório sobre uma área regular representada por árvores regularmente espaçadas, sendo que o processo de pontos só pode ocorrer onde não existam árvores e estou quebrando a cabeça com isso, bom primeiro criei a área com as árvores: require(plotrix) ##Cria o limite da área limx<-c(2,2,20.5,20.5) limy<-c(2,20.5,20.5,2) plot(limx,limy) lim=cbind(limx,limy) lim<-as.matrix(lim) polygon(lim) ##Aqui criei a posição das árvores e com área igual ao seu diâmetro sq=seq(2.5,20,2.5) for (k in sq) { for(j in sq){ draw.circle(k,j,radius=0.3,col="red") }} Bom agora eu queria jogar sobre essa área o processo de pontos: ##Criando o processo de pontos x <- runif(n=500,min=0, max=20) y <- runif(n=500,min=0, max=20) xy<-cbind(x,y) points(xy) ## Porém, os pontos também caem nas mesmas posições ocupadas pelas árvores (obviamente), onde gostaria de saber se alguém conheceria alguma solução para eu fazer com que não haja sobreposição entre os raios que representam os diâmetros das árvores e os pontos criados? Na verdade, preciso definir uma maneira de fazer com que as coordenadas contidas nos diâmetros das árvores não sejam validas no momento da geração do processo de pontos, atualmente estou tentando soluções com o pacote spatstat mais sem sucesso, pois não consigo fazer com que a área criada seja um objeto ppp valido. Obrigado, -- Alexandre DOS SANTOS Engenheiro Florestal, Msc. Laboratório de Entomologia Florestal Departamento de Entomologia Universidade Federal de Lavras Caixa Postal 3037 37200-000 - Lavras - Minas Gerais - Brasil Tel: +55 35 92230304

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Fatorial com parcela subdivida
by Wenceslau Teixeira 10 Jan '12

10 Jan '12

Caros colegas da lista do R, um colega me pediu ajuda para analisar um experimento com um delineamento (complexo para meus conhecimentos) E um fatorial 2 x 3 x 3 com parcela subvididida (micronutrientes), feito em blocos com 3 repeticoes. ** Sera que alguem poderia nos indicar algum material do R sobre como fazer o modelo para este tipo de delineamento. Agradeco antecipadamente a ajuda Wenceslau Teixeira Embrapa Solos ** * * *FV GL .F* Blocos 2**** N 1**** P 2**** K 2**** NxP 2**** NxK 2* *** PxK 4**** NxPxK 4**** ** Erro a ?* *** ** ** Micro (?) 1 **** * * *Erro b (?) ? . * Total 53**** ** **

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Como faço para atualizar minha versão do R
by andre neves 10 Jan '12

10 Jan '12

Prezados, Como faço para atualizar minha versão do R. Um abraço a todos. André =========================== Aviso de confidencialidade: =========================== - Esta mensagem da Empresa Brasileira de Pesquisa Agropecuaria (Embrapa), empresa publica federal regida pelo disposto na Lei Federal n. 5.851, de 7 de dezembro de 1972, eh enviada exclusivamente a seu destinatario e pode conter informacoes confidenciais, protegidas por sigilo profissional. Sua utilizacao desautorizada eh ilegal e sujeita o infrator as penas da lei. Se voce a recebeu indevidamente, queira, por gentileza, reenvia-la ao emitente, esclarecendo o equivoco. ===================== Confidentiality note: ===================== - "This message from Empresa Brasileira de Pesquisa Agropecuaria (Embrapa), a government company established under Brazilian law (5.851/72), is directed exclusively to its addressee and may contain confidential data, protected under professional secrecy rules. Its unauthorized use is illegal and may subject the transgressor to the law's penalties. If you are not the addressee, please send it back, elucidating the failure".

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Linux: Duas versões do R
by Jose Claudio Faria 09 Jan '12

09 Jan '12

Pessoal, É possível ter 2 versões do R (no caso 2.13.1 e 2.14.0-dev) instaladas e funcionais no notebook? A 2.14.0 eu compilo dos fontes, a outra baixo já compilada. Sempre usei as dev no Linux, mais ela está apresentando conflito com alguns pacotes que necessito para rodar uma análise. Ando procurando essa informação mas ainda não encontrei: Qualquer dica é bem vinda. Ab, -- ///\\\///\\\///\\\///\\\///\\\///\\\///\\\///\\\ Jose Claudio Faria Estatistica - Prof. Pleno UESC/DCET/Brasil joseclaudio.faria at gmail.com ///\\\///\\\///\\\///\\\///\\\///\\\///\\\///\\\

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post: invocando R através do Stata
by Daniel Marcelino 09 Jan '12

09 Jan '12

A quem possa interessar, acabo de publicar um post sobre como rodar o R através do Stata em Mac OS. Assim, talvez o post sirva para quem use os dois pacotes statisticos em Mac OS. http://danielmarcelino.com/

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post: invocando o R através do Stata
by Daniel Marcelino 09 Jan '12

09 Jan '12

A quem possa interessar, acabo de publicar um post sobre como rodar o R através do Stata em Mac OS. Assim, talvez o post sirva para quem use os dois pacotes statisticos em Mac OS. http://danielmarcelino.com/

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Erro em shapiro.test
by Flávia Machado 06 Jan '12

06 Jan '12

Pessoal, fui fazer um novo teste de shapiro e encontrei outro erro. Fiz: > dados<-read.table(file.choose(),header=T) > attach(dados) > dados apareceu: Erro em shapiro.test(FrMad, FrImat) : argumento(s) não utilizado(s) (FrImat) Coloquei em anexo a planilha que estou usando. Dessa vez não esqueci de "attachar" o arquivo. Alguém saberia me dizer o motivo do erro? Att. -- Flávia Machado

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Software publico brasileiro
by Pedro Rafael 06 Jan '12

06 Jan '12

http://linuxmagazine.com.br/lm/noticia/software_publico_brasileiro_na_linux… -- Saudações, Pedro Rafael Diniz Marinho. Estatístico - Secretaria de Estado da Saúde - PB.

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ajuda boxplo
by Ana Paula Sousa 06 Jan '12

06 Jan '12

Colegas, Fiz um gráfico boxplot de séries temporais ajustadas por mês. Na escala horizontal, ele entende os valores com 1,2...,12. Preciso que 1=jan;2=fev e assim por diante. Como faço? Atenciosamente,

3 2

Ajuda com pacote "seas"
by Inajara Welter Macedo 06 Jan '12

06 Jan '12

Ola, Estou trabalhando com dados de precipitacao, de 10 anos, e quero obter resultados de periodos de seca. Preciso contar quantos dias consecutivos sem chuva houve, ao todo, e por ano e depois por mes. Meus dados eram uma matriz de 4 colunas, chamado "mod1", cada coluna corresponde ao ano, mes, dia, e precipitacao, mas transformei para data.frame, para poder utilizar o pacote "seas": as.data.frame(mod1) Tentei usar a funcao "interarrival": CCS<-interarrival (mod1, var = "Precip", p.cut = 0, inv = TRUE) # CCS= casos consecutivos secos Mas, ocorre um erro: Erro em seas.df.check(x, orig, var) : a ‘date’ column must exist in ‘mod1’ Nao sei como colocar essa coluna de data que pede, porque existem as colunas de ano , mes, e dia. Como consigo converter essas colunas em uma unica coluna de data?? Obrigada, Inajara - UFPEL/RS

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Ajuda para nomear variáveis para uma função repetida
by Maria Papa 05 Jan '12

05 Jan '12

Olá pessoal, boa tarde, Gostaria da ajuda de vcs para nomear variáveis. Eu tenho este "for" que é aplicado para as três coordenadas de um ponto (XYZ). O procedimento é idêntico para os três, por isso, eu gostaria de saber se tem alguma forma de usar este mesmo conjunto mundando apenas o nome das coordenadas, no exemplo abaixo, eu faço com X e tenho os resutlados para X, depois e preciso para Y e Z. Eu fiz algumas tentativas com "for," mas não deu certo. for (j in 1:n) { X_e = rep(X[j], length(m))+ (runif(m, -U,U)) funX = function(X_e){mean(X_e)} X_boot = one.boot(X_e,funX,R) X_boot_ic= boot.ci(X_boot,conf = c(0.95),type = c("norm")) ic_inf_X [j]= X_boot_ic$normal[2] ic_sup_X [j] = X_boot_ic$normal[3] X_b[j]= X_boot$t0 } os resultados que eu preciso no final e portanto devem ser trocados o nome das coordenadas são os intervalos de confiança ic_inf_X, ic_sup_X e X_b Desde já agradeço a todos. Célia

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Re: [R-br] Ajuda com pacote "seas"
by Ivan Bezerra Allaman 05 Jan '12

05 Jan '12

A mensagem de erro é clara. Uma coluna com datas deve existir em mod1. Para uma ajuda mais efetiva, nos envie parte dos seus dados e um código reproduzível. Um ?interarrival ajuda a entender melhor o uso da função. (S,f,P) Allaman \begin{signature} <<>>= Prof. Dr. Ivan Bezerra Allaman Universidade Estadual de Santa Cruz Departamento de Ciências Exatas e Tecnológicas Ilhéus/BA - Brasil Fone: +55 73 3680-5076 E-mail: ivanalaman@yahoo.com.br/ivanalaman@gmail.com @ \end{signature}

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SOMAR VARIAVEIS
by Luiz André Zardo 05 Jan '12

05 Jan '12

PREZADOS Preciso criar uma variavel Y que é a soma de outras 3 variaveis (X1+X2+X3), porém quando uma delas é algum NA não consigo fazer a soma das outras duas. como resolvo isso, como nos casos 2 e 3 abaixo. caso *X1* *X2* *X3* *Y* 1 1 2 1 4 2 4 2 NA NA 3 NA 2 1 NA 4 2 0 3 5

2 1

Leitura de arquivos
by Emerson Cotta Bodevan 04 Jan '12

04 Jan '12

Prezados, tenho um arquivo .txt com uma sequência enorme de números. Por ex.: 0120304405560085673947684658495604932570932847509836536793586593420574092398564230562439065423095624359842350234567230945 7340127459042759834265093248560147237561092436509864230543609586723096756759867243670934675.... . . . ...87350274523475982734596239467358927609650972345792034759237459623767235967526 chego a ter milhares de números! Veja que não há espaço entre os dígitos da sequência! Gostaria de ler esse arquivo para um vetor mas, ao ler, quero que o R entenda que cada dígito deva ser armazenado numa posição diferente no vetor criado. Por ex. (no arquivo acima): 0 1 2 0 3 0 4 4 0 5 5 6 0 0 8... cada um pertencendo a uma posição no vetor criado. Agradeço qualquer ajuda, -- *Emerson*

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Re: [R-br] Teste de homocedasticidade e ANOVA
by Ivan Bezerra Allaman 04 Jan '12

04 Jan '12

Antes de sermos mecanicistas, vamos nos ater no problema como um todo. Se você tomou medidas em mesmo animal durante intervalos de tempo, essas medidas não são independentes, infrigindo um dos primeiros pressupostos da ANAVA. Se você utilizar as médias dessas medidas, obtendo um valor para cada animal, aí tudo bem, pois a unidade experimental será o animal. No entanto, você irá ter perda de informações. (S,f,P) Allaman \begin{signature} <<>>= Prof. Dr. Ivan Bezerra Allaman Universidade Estadual de Santa Cruz Departamento de Ciências Exatas e Tecnológicas Ilhéus/BA - Brasil Fone: +55 73 3680-5076 E-mail: ivanalaman@yahoo.com.br/ivanalaman@gmail.com @ \end{signature}

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Problema para reinstalar pacote (EspDes) no Linux Ubuntu 10.04
by Andre Oliveira 04 Jan '12

04 Jan '12

Pessoal estou com problemas para reinstalar o EspDes !! Alguém saberia dizer o que pode ser? install.packages("EspDes", dep=T) Installing package(s) into ‘/home/andre/R/i486-pc-linux-gnu-library/2.14’ (as ‘lib’ is unspecified) --- Please select a CRAN mirror for use in this session --- Carregando a interface Tcl/Tk ... pronto Mensagens de aviso perdidas: In getDependencies(pkgs, dependencies, available, lib) : package ‘EspDes’ is not available (for R version 2.14.1) André Oliveira Souza obrigado

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Teste de homocedasticidade e ANOVA
by Flávia Machado 04 Jan '12

04 Jan '12

Pessoal, trabalho com dieta de muriquis (primata da Mata Atlântica) e quero, com os dados que obtive, verificar qual foi o alimento mais consumido em um determinado período de tempo (todo período de estudo, estação seca e e. chuvosa). São quatro categorias alimentícias (Fruto, Folha, Broto e Flor) e calculei, para cada uma delas, as frequências média de consumo para os três períodos de tempo. Então, cada linha do meu arquivo corresponde a frequência de consumo de cada um desses em um determinado dia, no determinado período. Fiz uma ANOVA, mas não soube interpretar o valor do teste e o bloxplot. De qualquer forma, pode ter dado errado porque eu não sabia que tinha que fazer o teste de homocedasticidade. Não entendi o que preciso colocar para fazer o "levene's test". Na ajuda do R diz: levene.test(y, group, option = c("mean", "median","trim.mean"), trim.alpha = 1) Arguments your data groupfactor of the data optionThe option must be "mean" corresponding to classical Levene's test (default), "median" corresponding to the robust Brown-Forsythe Levene-type test or "trim.mean" corresponding to the robust Levene-type test using the trimmed mean. trim.alphathe fraction (0 to 0.5) of observations to be trimmed from each end of 'x' before the mean is computed. Porém, continuo sem entender... Alguém pode me dar um help? Abraços! Att. -- Flávia Machado

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Escala de y
by Fernanda Balbino 04 Jan '12

04 Jan '12

Olá, boa tarde, Gostaria de saber como alterar em um gráfico a escala em y. Por exemplo, no caso gostaria que ela fosse de 10 em 10 unidades. Obrigada. Fernanda Balbino.

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Alocação de Memória
by Edson Lira 04 Jan '12

04 Jan '12

Amigos, tenho um arquivo com milhões de registro. Uso a rotina abaixo. require(RODBC) channel<-odbcConnect("saddw",uid="saddw",case="tolower") #SOROLOGIA GERAL channel<-odbcConnect("saddw",uid="saddw",case="tolower") > soro<- sqlQuery(channel, "select doador_dw.sexo,doador_dw.idade,doador_dw.qtde_doacoes, + doador_dw.numerogente,doacao_dw.numerogente,sorologias_dw.numerogente, + sorologias_dw.exame,sorologias_dw.patologia,sorologias_dw.resultado,doacao_dw.tipodoacao, + sorologias_dw.metodo,sorologias_dw.basicasorologia,sorologias_dw.dataexamesorologia + from saddw.doador_dw, saddw.doacao_dw,saddw.sorologias_dw + where doador_dw.numerogente=sorologias_dw.numerogente + and doacao_dw.numerogente=doador_dw.numerogente + and to_char(sorologias_dw.dataexamesorologia,'DD/MM/YYYY') >= to_date('01/01/2010','DD/MM/YYYY') + and to_char(sorologias_dw.dataexamesorologia,'DD/MM/YYYY') <= to_date('31/12/2011','DD/MM/YYYY')") tenho esse erro, como resolver? Erro: não é possível alocar vetor de tamanho 26.8 Mb Meu computador tem 2 giga de ram. Uso o windows 7 32 bits. A versão do R é a 2.14 Edson Lira Estatístico Manaus-Amazonas

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