I tried to run your code. There were formatting issues. The code below runs for me. I have no csv to test it with but it may work for you. library(shiny) library(car) runApp( list( ui = pageWithSidebar( headerPanel('Analysis of Variance'), sidebarPanel( fileInput("file1", "CSV File", accept=c("text/csv", "text/comma-separated-values,text/plain", ".csv")), checkboxInput("header", "Header", TRUE),...

You probably should be treating "subject" as a categorical variable rather than numeric. You can make this clear to R using subject = factor(rep(subject,6)) in your above example....

You can change your data for 0 to NA if you also want to exclude it: df$Workout[df$Workout == "0"] <- NA or filter your data: df <- df[df$Workout!=0, ] ...

It appears that you somehow changed the name of the data and/or the variables between computing your fit and calling glht. Your code has V1 but the error has trt. It's hard to say more in detail because your example is not fully reproducible (the computation of fit is missing)....

r,transformation,interaction,anova,sqrt

Just to close the question: interaction.plot(Gastropods$Zone, Gastropods$Species, Gastropods$sqrtAbundance, main= "Gastropod Interaction Plot", xlab = "Gastropod Zone", ylab= "Mean of SQRT Gastropod Abundance", legend = TRUE) ...

When you use the ezANOVA function from the ez package, the eta-squared is automatically calculated: library(ez) m1 <- ezANOVA(data = data.ex4, dv = Recall, wid = Subject, within = .(Task,Valence), type = 3, detailed = TRUE) m1$ANOVA This gives the following output: Effect DFn DFd SSn SSd F p p<.05...

Using base lapply: lapply(split(df, df$Letter), aov, formula=Question ~ Number) Alternatively using dplyr: library(dplyr) obj <- df %>% group_by(Letter) %>% do(model = aov(Question~Number, data = .)) obj$model Using data.table: library(data.table) df <- as.data.table(df) df[, list(Model = list(aov(Question ~ Number))), keyby = Letter]$Model ...

The page you are learning from has an error, in that it doesn't tell you how to enter the data correctly. The ses variable is supposed to be a factor, as we can see from the data they give us, it is read in as numeric: str(hsb2$ses) If we convert...

Okay, first off, your code is unnecessarily complicated. Check out several simplifications below that will hopefully make things simpler. Then, to your core question, you want a split-apply-combine strategy. You need to split the data by the relevant grouping variables, then do a t.test in each subset of the data....

You could try either Map(function(x) summary(aov(Obs~Treat,data=x)) , mget(Names)) Or lapply(mget(Names), function(x) summary(aov(Obs~Treat, data=x))) If you need a function that also changes the "data" in the "Call" formula. f1 <- function(formula= as.formula('Obs~Treat'), data, ...){ Call <- match.call(expand.dots=TRUE) Call[[1]] <- as.name('aov') Call$formula <- as.formula(terms(formula)) Call$data <- as.symbol(substitute(nm, list(nm=data))) eval(Call) } Map(function(x,y) f1(data=y),...

I think some of your cv lags have all missing values (at the landcover level of detail). This code requires that the cv.lags (within the landcover class) have at least 3 observations. #Create fake dataset in your same mold set.seed(1) d <- data.frame(row.names=1:40,timestamp=rep(1:10,4),station=c(rep("A",10),rep("B",10),rep("C",10),rep("D",10)), year=rep(c(2000,2000,2000,2001,2001,2001,2001,2002,2002,2002),4), month=rep(c("jan","feb","mar","april","may","jun","jul","aug","sep","oct"),4),ndvi=runif(40,min=0.3,max=0.6),...

You can specify the model with random terms and how to divide to get the F ratios by using the Error term in the formula (I changed Machine and Operator to factors in the data): > summary(aov(Measure~Machine + Error(interaction(Machine,Operator)),data=mydat)) Error: interaction(Machine, Operator) Df Sum Sq Mean Sq F value Pr(>F)...

You could use the function lm, which runs a linear regression (in the end, ANOVA and ANCOVA are just restricted versions of the linear model). mod <- lm(output ~ input + category, data=data) You can view the output with the summary function summary(mod) If you really need to have the...

Got the clue from this post: Why do R and statsmodels give slightly different ANOVA results? When the data is being read from CSV file, the Diet column is becoming an ordinary numeric column, but for ANOVA it has to be a factor variable (I am still not clear why...

I was trying and thinking and thinking and saying how could this be possible that the degrees of freedom are not calculated correctly? But sometimes we only think about complicated things and forget about the easy things. I found what the problem is: data <- read.table(header=T,text='Runs I A B C...

Your issue is that when you try to do something like boxplot(temp~Tissue, data=allgenedata), R is using the variable temp you created rather than referring to that column in allgenedata. For the boxplot and aov commands, just change temp to allgenedata[,temp] to make sure R knows what you are referencing. Also,...

The answer is in the FAQ, FAQ 7.22 in fact. However this is not obvious until you realize that the plot.lme function from the nlme package uses lattice/trellis graphics to do the actual plotting (there are references on the help page for plot.lme, but not obvious). The short form of...

r,for-loop,foreach,statistics,anova

Well, you never actually made an attempt to use the i variable that you are looping over. I went ahead and changed the for loop to use lapply instead. How about this res<-lapply(levels(data$metabolite), function(i) { met<-data[data$metabolite==i, ] #subset for metabolite i part1<-with(met, aov(area ~ treatment)) print(part1) list( part1 = part1,...

Example data (which may or may not fulfil the criteria for an ANOVA) X <- rnorm(100) Z <- rnorm(100) dependent1 <- rnorm(100) dependent2 <- rnorm(100) dependent3 <- rnorm(100) dataset <- cbind(data.frame(X, Z, dependent1, dependent2, dependent3)) The following script would work, you need to put in the row column numbers of...

I kind of do the same with you. I want to feel in control so whenever I have time I design the dummies myself with the following: d = data.frame( Temperature = c(rep("Cool", 6), rep("Warm", 6)), Bact = c(rep("Bact 1", 2), rep("Bact 2", 2), rep("Bact 3", 2), rep("Bact 1", 2),...

As suggested by Jaap and Andrew Taylor, the problem was formatting a linear regression. This was achieved through the 'stack' and 'cbind' functions. df1 <- as.data.frame(matrix(sample(0:1000, 36*10, replace=TRUE), ncol=1)) df2 <- as.data.frame(matrix(sample(0:500, 36*10, replace=TRUE), ncol=1)) df3 <- as.data.frame(matrix(sample(0:200, 36*10, replace=TRUE), ncol=1)) Example <- cbind(df2,df3) Stacked <- stack(Example) Combined <- cbind(df1,Stacked)...

The p values are in the summary. If you try: summary(aov(score~teacher, data = mydata)) You'll get: Df Sum Sq Mean Sq F value Pr(>F) teacher 3 108.0 36.01 4.114 0.0192 * Residuals 21 183.8 8.75 --- Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1...