similar to: Linear regression on several groups

Dear all, This question may be too basic quesition for this list, but if someone has time to answer I will be happy. I have tried to find out, but haven't found a consice answer. As an example I use "Pinheiro, J. C. & Bates, D. M. 2000. Mixed-effects models in S and S-PLUS. Springer, New York." page 225, where rats are fed by 3 different diets over time, which body mass has

Hi, In the examples from the ReShape package there is a simple example of using melt followed by cast that produces a smallish amount of output about the chicks database. Here's the code: library(reshape) names(ChickWeight) <- tolower(names(ChickWeight)) chick_m <- melt(ChickWeight, id=2:4, na.rm=TRUE) DietResults <- cast(chick_m, diet + chick ~ time) DietResults My challenge

An excerpt from dataset ChickWeight: ??? weight Time Chick Diet 1?????? 42??? 0???? 1??? 1 2?????? 51??? 2???? 1??? 1 3?????? 59??? 4???? 1??? 1 I am interested in the residuals of the dataset.? Specifically in saving them to another format. I have been creating text files with sink. CW.lm <- lm(weight ~ Diet, ChickWeight) resid.CW.lm <- resid(CW.lm) But when I call: resid.CW.lm The data

I've noticed that R has a number of very useful functions for obtaining descriptive statistics on groups of variables, including summary {stats}, describe {Hmisc}, and describe {psych}, but none that I have found is able to provided weighted descriptives of subsets of a data set (ex. descriptives for both males and females for age, where accurate results require use of sampling

according to the documentation of the cast function in the reshape function, I would expect this bit of code from the examples to calculate marginal means over only the 'diet' variable. #Chick weight example names(ChickWeight) <- tolower(names(ChickWeight)) chick_m <- melt(ChickWeight, id=2:4, na.rm=TRUE) cast(chick_m, diet + chick ~ time, mean, margins="diet") But,

Estimado Oliver Nuñez Envío un ejemplo reproducible. Javier Marcuzzi # de donde tomo datos, y tiene el modelo (en el pdf) library(MCMCglmm) # librería con las funciónes que voy a usar library(nlme) datos0<-ChickWeight # creo algunos datos que agrego a los origonales Factor<-as.numeric(datos0$Chick) Factor[Factor > 0 & Factor <= 10] <- 'A' Factor[Factor > 10

Les consulto por lo siguiente: Se me ocurrió usar datos de MCMCglmm en otros procedimientos de R, comparar los resultados y el tiempo como facilidad de uso, para lo cual use LME4, los resultados fueron parecidos, entendiendo que internamente los algoritmos tienen diferencias, es lógico que los resultados no coincidan exactamente, pero aproximadamente tienen que ser compatibles, en este caso

Estimados Tengo un error que me desconcierta, es un código que simplifiqué de otro trabajo donde no hay problemas, sin embargo me da un error. Una diferencia es que en mi otro trabajo uso spline y ahora polinomio, este es de segundo grado y se encuentra tanto en efectos fijos como aleatorios, el modelo es correcto, corre con MCMCglmm pero no con nlme. grid <- expand.grid(Tiempo=4:6,

When I try to run the example from Variogram with an lme object, I get an error (although summary works): R : Copyright 2005, The R Foundation for Statistical Computing Version 2.2.1 (2005-12-20 r36812) ISBN 3-900051-07-0 ... > fm1 <- lme(weight ~ Time * Diet, BodyWeight, ~ Time | Rat) Error: couldn't find function "lme" > Variogram(fm1, form = ~ Time | Rat, nint =

Is there anyway I can convert a vectors objectname to a string to be used in fx: Monkey<-c(0,0,0,1,1,1) Wax<-c(1,0,1,0,1,0) f<-function(x,y){ table(x,y) } f(Monkey,Wax) so that the printout is not y x 0 1 0 1 2 1 2 1 but Wax Monkey 0 1 0 1 2 1 2 1 -- View this message in context:

Dear R communitiy, I am trying to use multiple functions for aggregation within a function call for dcast. However this seems to result in an error. Also I have not managed to make dcast() work with fun.aggregate=sd. Please find attached some example code using the ChickWeight data. Many thanks for your help! Jokel #Chick weight example names(ChickWeight) <- tolower(names(ChickWeight))

Dear Dr. Bates, and R-help, I've tried the anova function to test the difference between two coefficients, as shown on page 225 of your book "Mixed Effects Models in S and S-Plus (Statistics and Computing)". When I type: anova( fm2BW.lme, L = c(TimeDiet2 = 1, TimeDiet3 = -1) ) I got the following error message: Error: unexpected '=' in "anova( fm2BW.lme, L =

Hi, I'm having difficulty with getting a table to show with multiple rows and columns. Below is the commands that I've typed in and errors that I am getting. Thank you. Laura Table trying to enter: Diet: Binger-yes: Binger-No: Total: None 24 134 158 Healthy 9 52 61 Unhealthy 23 72 95 Dangerous 12 15 27 >

Otras variantes con y sin paquetes adicionales... > sapply(split(datIn$Gain, as.factor(datIn$Diet)), mean) d1 d2 d3 280 278 312 > by(datIn$Gain, datIn$Diet, mean) datIn$Diet: d1 [1] 280 -------------------------------------------------------------- datIn$Diet: d2 [1] 278 -------------------------------------------------------------- datIn$Diet: d3 [1] 312 > > library(dplyr) >

Me gusta la respuesta uqe has dado, pero si por ejemplo, alguno de los datos tiene datos faltantes, entonces devuelve NA. He probado con: sapply(split(datos$uno, as.factor(datos$dos)), mean(na.rm=TRUE)) pero da fallo. ¿Cómo se podría hacer para que devolviera además la media obviando los NA y que contara el numero de NA por categoria? > Date: Wed, 28 Oct 2015 00:13:45 +0100 > From: cof

Estimados Cuando existia epicalc, había una manera muy fácil de determinar la media de una variable (en esta caso Gain) por grupos, en este caso (Diet). ?Como se puede hacer ahora? Diet Gain 1 d1 270 2 d1 300 3 d1 280 4 d1 280 5 d1 270 6 d2 290 7 d2 250 8 d2 280 9 d2 290 10 d2 280 11 d3 290 12 d3 340 13 d3 330 14 d3 300 15 d3 300

H all, I have relative abundance data from >100 sites. This is from acoustic monitoring and usually the data is for 2-3 nights but in some cases my be longer like months or years for each location.. The data output from my management data base is proved by species by night for each location so data frame would look like this below. What I need to do is sum the Survey_time by Spec_Code for

Hello This is Elaine. I am using package lattice to generate boxplots. Using Richard's code, the display was almost perfect except the outlier shape. Based on the following code, the outliers are vertical lines. However, I want the outliers to be empty circles. Please kindly help how to modify the code to change the outlier shapes. Thank you. code package (lattice) dataN <-

Hi all , I have a dataframe (Hypertension) with following headers :- > Hypertension ID Hypertension(before drug A) Hypertension(On drug A) On drug B? Healthy diet? 1 160 90 True True 2 190

Hello I ran the code below but it said: no object "'panel.xyplot.intermediate.hh'" Please kindly advise how to modify the code. thank you. (It works with panel.bwplot.intermediate.hh) Elaine code library(HH) # data input dataN <-read.csv("H:/R_data/Mig_bird_586.csv",header=T, row.names=1) dim(dataN) dataN[1,] str(dataN) diet.code <-