similar to: list multiplied by a factor / mapply

Hi folk, I tried this and it works just perfectly tapply(iris[,1],iris[5],mean) but, how to obtain a single table from multiple variables? In tapply x is an atomic object so this code doesn't work tapply(iris[,1:4],iris[5],mean) Thanx and great summer holidays Gianandrea -- View this message in context: http://www.nabble.com/multiple-tapply-tp18868063p18868063.html Sent from the R help

Hi there: I have a question about generating mean value of a data.frame. Take iris data for example, if I have a data.frame looking like the following: --------------------- Sepal.Length Sepal.Width Petal.Length Petal.Width Species 1 5.1 3.5 1.4 0.2 setosa 2 4.9 3.0 1.4 0.2

Thanks Duncan. Awesome ideas! I think we're getting closer! I tried what you suggested and got a possibly better error... . . . rConnection.assign("boxMVariable", myDf); String resultBV = "str(boxMVariable)"; // your suggestion. RESULTING ERROR: Error in format.default(nam.ob, width = max(ncn), justify = "left") : invalid 'width' argument (No idea

Hey Duncan, Hard to debug? That's an understatement. Eyes bleeding.... In any case, I tried all your suggestions. To get "integer" for the final column, I had to change the code to get integers instead of strings. double[] d1 = ((REXPVector) ((RList) tableRead).get(0)).asDoubles(); double[] d2 = ((REXPVector) ((RList) tableRead).get(1)).asDoubles(); double[] d3 = ((REXPVector)

I'm not sure what you mean. Could you please be more specific? If I print the string, I get: boxM(boxMVariable[, -5], boxMVariable[, 5]) From this code: . . . // assign the data to a variable.rConnection.assign("boxMVariable", myDf); // create a string command with that variable name.String boxVariable = "boxM(boxMVariable[, -5], boxMVariable[, 5])";

Just out of curiosity, I took the default "iris" example in the RF helpfile... but seeing the admonition against using the formula interface for large data sets, I wanted to play around a bit to see how the various options affected the output. Found something interesting I couldn't find documentation for... Just like the example... > set.seed(12) # to be sure I have

Thanks Duncan. I can't tell you how helpful all your terrific replies have been. I think the biggest surprise is that nobody appears to be using Java and R together like I"m trying to do. I suppose it should be a surprise since there are no books on the subject and almost no technical documentation other than a few sites here and there. ----- I originally had the "int" as the

Hi, I was wondering if it is possible to get the rowindices without using the function "which" because I don't have a restriction criteria. Here's an example of what I mean: # take 10 randomly selected instances iris[sample(1:nrow(iris), 10),] # output Sepal.Length Sepal.Width Petal.Length Petal.Width Species 76 6.6 3.0 4.4 1.4

In the examples below, the first loses the name attached by foo(), the second retains names attached by bar(). Is this an intentional difference? I?d prefer that the names be retained in both cases. foo <- function(x) { c(mean = base::mean(x)) } bar <- function(x) { c(mean = base::mean(x), sd = stats::sd(x))} aggregate(iris$Sepal.Length, by = list(iris$Species), FUN = foo) #>

Dear all, I'd like to get a percentage variable based on a group, but without creating a new data frame. For example: data(iris) iris$percent <-unlist(tapply(iris$Sepal.Length,iris$Species,function(x) x/sum(x, na.rm=TRUE))) This does not work, I should have only three standard values, respectively for setosa, versicolor, and virginica. How can I do this? MANY THANKS, Karine

I'm confused, hope someone can point out what is not obvious to me. I thought I was creating a new data frame by 'deleting' rows from an existing dataframe - I've tried 2 methods. But this new data frame seems to remember values from its parent - even though there are no occurences. Where does it get the values versicolor and virginica from and give then a count of 0? What

Dear R users, I have a question on the confusion matrix generated by function randomForest. I used the entire data set to generate the forest, for example: > print(iris.rf) Call: randomForest(formula = Species ~ ., data = iris, importance = TRUE, keep.forest = TRUE) confusion setosa versicolor virginica class.error setosa 50 0 0 0.00

If do: > library("e1071") > example(svm) I get: svm> data(iris) svm> attach(iris) svm> ## classification mode svm> # default with factor response: svm> model <- svm(Species ~ ., data = iris) svm> # alternatively the traditional interface: svm> x <- subset(iris, select = -Species) svm> y <- Species svm> model <- svm(x, y) svm>

Dear R list, I'm trying to calculate Mahalanobis distances for 'Species' of 'iris' data as obtained below: Squared Distance to Species From Species: Setosa Versicolor Virginica Setosa 0 89.86419 179.38471 Versicolor 89.86419 0 17.20107 Virginica 179.38471 17.20107 0 This distances above were obtained with proc

Dear RRRRRrrrrrrrrlist! I?ve got two lists which contain sets of DNA-sequences. They look something like this: List of 33 $ Cunonia_atrorubens : chr [1:247] "t" "t" "n" "t" ... $ Cunonia_balansae : chr [1:254] "t" "c" "c" "c" ... $ Cunonia_capensis : chr

On Fri, Mar 23, 2018 at 6:43 PM, Rui Barradas <ruipbarradas at sapo.pt> wrote: > Hello, > > Not exactly an answer but here it goes. > If you use the formula interface the names will be retained. Also if you pass named arguments: aggregate(iris["Sepal.Length"], by = iris["Species"], FUN = foo) # Species Sepal.Length # 1 setosa 5.006 # 2

Hi R People: Several of you pointed out that using "tapply" on a data frame will work on the iris data frame. I'm still having a problem. The iris data frame has 150 rows, 5 variables. The first 4 are numeric, while the last is a factor, which has the Species names. I can use tapply for 1 variable at a time: >tapply(iris[,1],iris[,5],mean) setosa versicolor virginica

Dear R-help forum members, Suppose I have a data-frame having two variables and single data for each of them, as described below. variable_1           variable_2         10                          20 I have written a function, say, 'fun' which uses input 10 and 20 and gives me desired result. fun = function(X, Y)          {          X + Y              #( I am just giving an example of

Hopefully I will not be flamed for this on the list, but I am starting out with R and having some trouble with combining plots. I am playing with the famous iris dataset (checking out example dataset in R while reading through Introduction to datamining) What I would like to do is create three graphs (combined boxplots) besides each other for each of the three species (Setosa, Versicolour and

Hi, Since I've had no replies on my previous post about my problem I am posting it again in the hope someone notice it. The problem is that the randomForest function doesn't take datasets which has instances only containing a subset of all the classes. So the dataset with instances that either belong to class "a" or "b" from the levels "a", "b" and