similar to: boxplot of two variables

I need to generate output files as .csv file in a loop: let's say, I want to generate a sequence of files according to the loop number "i". the first file genearted should be file1.csv the second should be file2.csv I use: write.csv(temp, "file$i$.csv") in the loop but it did not work. can anyone give suggestions? thanks a lot. Jack

Hi all, I wondered whether anyone has some advice on a stats-related 'sanity check', as I ran a nonparametric multivariate test (mulrank function as decribed by R. Wilcox, 2005) on both systems, but got different results (please see below for the system-specific outputs)! The functions I used are attached as well. Any advice would be much appreciated! Thanks in advance for getting back to

Hi all, I suspect the answer to this query will be the tongue-in-cheek "use a quantum computer", but I thought my understanding might be sufficiently limited that I'm missing a simpler option. I'm looking for a way to cycle through all possible combinations of 2 groups of data. For 10 data points, that's 2^10 combinations, for 20 data points it's 2^20, etc. combn() from

Hola, mi pregunta es la siguiente: Tengo un modelo genético aditivo, donde los individuos estan emparentados. Quiero que la matriz de parentesco sea la matriz de coeficientes de las correlaciones entre genotipos. En SAS, con PROC MIxed podia agregarla como un archivo y usarla en la sentencia random...No encuentro como hacerlo en R, alguna idea?? En resumen, puedo imponerle a los efectos

The ez package was developed to aid those that are new to statistical programming. Over the course of several years of helping colleagues and students learn R, I observed that folks are often initially turned off R because they have difficulty obtaining SPSS-like results quickly (SPSS is the dominant environment in my field, psychology). ez attempts to fill this gap, providing quick and easy

The ez package was developed to aid those that are new to statistical programming. Over the course of several years of helping colleagues and students learn R, I observed that folks are often initially turned off R because they have difficulty obtaining SPSS-like results quickly (SPSS is the dominant environment in my field, psychology). ez attempts to fill this gap, providing quick and easy

Dear R users, I have a matrix "final" which looks like this: final oSO4 oNO3 mSO4 mNO3 [1,] 3.3728 0.2110 1.9517421 1.01883602 [2,] 0.8249 0.0697 1.5970292 0.11368781 [3,] 0.2636 0.1004 0.6012445 0.24356332 [4,] 8.0072 0.3443 6.1016998 3.63207149 [5,] 13.5079 0.6593 12.4011068 1.55323386 [6,] 6.1293 0.1989 5.7620926 0.12884845 [7,] 0.6004 0.0661

Dear list, I'm having trouble to see how my elements on a 4 dimensional array are listed. For example, I generated the following array: junk.melt=melt(occ.data,id.var=c("Especie", "Site", "Rep", "Año"), measure.var="Pres") y=cast(junk.melt, Site ~ Rep ~ Especie ~ Año) Now, I want to be able to look at how my species (Especie) are listed, in

> > Hola, a todos! Estoy ajustando un modelo componentes de varianza utilizando el paquete lme4, quiero extraer luego los valores estimados de varianza de cada factor aleatorio de este objeto lmer y guardarlo en un vector. Existe la función VarCorr, Esto es lo que hago: Mi ejemplo: tengo 3 factores NDVIaño; Máx.IH; TV # Ajusto el modelos lmer(peligro100 ~ 1 + (1|NDVIaño) + (1|Máx.IH) +

Hola a todos, quisiera saber como hacer para cambiar la escala del eje x, por ejemplo en un gráfico de puntos. Saludos _________________________________________________________________ Windows Live Messenger GRATIS: lo que faltaba en tu BlackBerry http://www.messengerentublackberry.com?ocid=WL_BB_LandPage_TagLine [[alternative HTML version deleted]]

Hi all, I'm a newbie R developer, am trying to dotplot a few graphs using a for loop. The following code works fine but once I wanna plot inside a loop, nothing happens. > for(i in 1:1){dotplot(y~x)} > y <- c(1,2,3) > x <- c('a','b','c') > dotplot(y~x) > for (i in 1:3) {dotplot(y~x)} (y and x depends on I in actual case) Nothing happens. I

How do you fit a Weibull distribution in R?

Dear All: I am new here. Please, advise how can I reach the R-archives to look through the libraries available for download.   Thanks   Xao Ping R&R Pharmakinetics Taiwan [[alternative HTML version deleted]]

Hi, there: I have a dataset with 50 states and for each state, I have its associated mean estimate (for some parameters) and the lower and upper bound of the 95% CI. The data look like below: state ami_mean ami_low ami_up 1 MS -0.58630 -0.90720 -0.29580 2 KY -0.48100 -0.75990 -0.19470 3 FL -0.47900 -0.62930 -0.32130 I would like to have a plot the 95% CI (characterized by

Dear R users, Does anyone know how to write function for Kumaraswamy distribution in R? Since I cannot write dkumar, pkumar, etc. in R. Please help. Thanks a lot, Debbie _________________________________________________________________ [[elided Hotmail spam]] [[alternative HTML version deleted]]

Hello, I'm trying to do a comparsion on a large scale say 10L bottle of liquid and a small scale bottle of liquid 0.5L, I have 5 different samples from each and they are measured over the space of 8 days as % viability and the % viability decreases over time. However not all 10 samples got measured every day. How would I do a two-way anova on this in R? Thanks for any help. Regards, Al

Dear R user, I am trying to understand this following code. Basically it's using a permutation method to calculate p value. But I would like to know exactly how the permutation works. #calculates null statistics tt0 <- 0 set.seed(123) B <- 100 for(i in 1:B) { v <- sample(y) tt0 <- c(tt0,ttest(dat,v)$tt) } tt0 <- tt0[-1] #form p-values att <- abs(tt) att0 <- abs(tt0) v

I have a list of numbers with NAs as below: > A[,1] [1] NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA [19] NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA [37] NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA [55] NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA [73] NA NA NA 62 78 98

Hi, In the example dataset below - how can I cahnge "gray20", to "blue" # data black <- rep(c("black","red"),10) gray <- rep(c("gray10","gray20"),10) black_gray <- data.frame(black,gray) # none of this desperate things works # replace(black_gray$gray, gray=="gray20","red") #

All, Can someone help me create a skewed distribution, mean = 30, with probability of selecting a random number from the distribution greater than or equal 60 equal to 10%? I need the probability density function to equal zero at zero, and have a maximum height at or near 30. Is this possible? And if it is possible, how can I adjust the distribution so that the probability of selecting a