similar to: more on apply on data frame

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

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

Colleagues ---------------------------------- System info: R version rw1020 on NT ESS using emacs ver. 20.4 ---------------------------------- I need to get the means for each column of a dataframe in the list created by splitting a data frame. At present, I am getting the mean of all columns in aggregate. The structure of the unsplit data is: > shuttle.tr1[1:10,] juliandate

Greetings R experts, I'm having some difficulty recovering lda objects that I've saved within sublists using the save.image() function. I am running a script that exports a variety of different information as a list, included within that list is an lda object. I then take that list and create a list of that with all the different replications I've run. Unfortunately I've been

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

Hello all, I apologize for a terribly simple question. I'm used to using Stata and am trying to `switch' over to R. I would like to recode a vector in a data frame when the value of it meets the following condition: if (dataframe$factor=='3'){dataframe$thevector<-(an arithmetic mean). What I would like to result is the creation of a new variable within the data frame for

I use by() to generate a summary statistics like so: Lbys <- by(dat[Nidx], dat$LipTest, summary) where Nidx is an index vector with names picking out the columns in the data frame dat. This returns a list of character arrays (see below for str() output) where the columns are named correctly but the rownames are empty strings and the values are strings prepended with the summary

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) #>

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

Dear all Suppose the object below: > require(Hmisc) > require(plyr) > x <- dlply(iris, .(Species), describe) How can I print the object without displaying the attributes? I inspected ?print and ?print.default with no luck. > x $setosa x[, "Sepal.Length"] n missing unique Mean .05 .10 .25 .50 .75 50 0 15 5.006 4.40 4.59

Hi, I was wondering if it's possible to get the row numbers from a filtering. Here's an example: # give me the rows with sepal.length == 6.2 iris[(iris[,1]==6.2),] # output Sepal.Length Sepal.Width Petal.Length Petal.Width Species 69 6.2 2.2 4.5 1.5 versicolor 98 6.2 2.9 4.3 1.3 versicolor 127 6.2

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

Hi For first question, maybe I am completely wrong but cor(swiss[,-1], swiss[,1]) should give you what you want in one step. Second question Without an example it is hard to say but maybe aggregate is the way forward. > aggregate(iris[,1:4], list(iris$Species), function (x) c(mean=mean(x), sd=sd(x))) Group.1 Sepal.Length.mean Sepal.Length.sd Sepal.Width.mean Sepal.Width.sd 1

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)

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

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

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

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

I require some help in debugging this code  library(neuralnet) ir<-read.table(file="iris_data.txt",header=TRUE,row.names=NULL) ir1 <- data.frame(ir[1:100,2:6]) ir2 <- data.frame(ifelse(ir1$Species=="setosa",1,ifelse(ir1$Species=="versicolor",0,""))) colnames(ir2)<-("Output") ir3 <- data.frame(rbind(ir1[1:4],ir2))

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