search for: iris2

...[1:4], main = "Anderson's Iris Data -- 3 species",pch = "+", col = c("black", "red", "green3", "blue")[ 1+ unclass(iris$Species)]) One very kludgy work-around is to define a new level 1, say "foo" in the first row of iris: iris2=iris iris2$Species = as.character(iris2$Species) iris2$Species[1]="foo" iris2$Species = factor(iris2$Species) pairs(iris2[1:4], main = "Anderson's Iris Data -- 3 species", pch = "+", col = c( "black","red", "green3","blue")...

Hi list, not sure if this is the wanted behavior, but running the following code: > version platform i386-pc-mingw32 arch i386 os mingw32 system i386, mingw32 status major 2 minor 1.1 year 2005 month 06 day 20 language R > n <- 500 > d <- 4 > m <- matrix(runif(n*d, -1, 1), ncol=d) > c <- hsv(apply(m, 1, function(x) {sum(x*x)/d}),

Hi list, not sure if this is the wanted behavior, but running the following code: > version platform i386-pc-mingw32 arch i386 os mingw32 system i386, mingw32 status major 2 minor 1.1 year 2005 month 06 day 20 language R > n <- 500 > d <- 4 > m <- matrix(runif(n*d, -1, 1), ncol=d) > c <- hsv(apply(m, 1, function(x) {sum(x*x)/d}),

Given a data set with a group factor, I want to translate the numeric variables to their centroid, by subtracting out the group means (adding back the grand means). The following gives what I want, but there must be an easier way using sweep or apply or some such. iris2 <- iris[,c(1,2,5)] means <- colMeans(iris2[,1:2]) pooled <- lm(cbind(Sepal.Length, Sepal.Width) ~ Species, data=iris2)$residuals pooled[,1] <- pooled[,1] + means[1] pooled[,2] <- pooled[,2] + means[2] pooled <- as.data.frame(pooled) pooled$Species <- iris2$Species -- Michae...

...50 50 50 > nrow(iris) [1] 150 > iris1 <- iris[iris$Species == 'setosa',] > nrow(iris1) [1] 50 > summary(iris1$Species) setosa versicolor virginica 50 0 0 boxplot(Petal.Width ~ Species, data = iris1, plot=1) > iris2 <- subset(iris, Species == 'setosa') > nrow(iris2) [1] 50 > summary(iris2$Species) setosa versicolor virginica 50 0 0 > boxplot(Petal.Width ~ Species, data = iris2, plot=1) -- View this message in context: http://r.789695.n4.nabble.com/re...

I try to perform a clustering using an existing dissimilarity matrix that I calculated using distance (analogue) I tried two different things. One of them worked and one not and I don`t understand why. Here the code: not working example library(cluster) library(analogue) iris2<-as.data.frame(iris) str(iris2) 'data.frame': 150 obs. of 5 variables: $ Sepal.Length: num 5.1 4.9 4.7 4.6 5 5.4 4.6 5 4.4 4.9 ... $ Sepal.Width : num 3.5 3 3.2 3.1 3.6 3.9 3.4 3.4 2.9 3.1 ... $ Petal.Length: num 1.4 1.4 1.3 1.5 1.4 1.7 1.4 1.5 1.4 1.5 ... $ Petal.Width : num 0....

Hello! I'm a newcomer to R hoping to replace some convoluted database code with an R script. Unfortunately, I haven't been able to figure out how to implement the following logic. Essentially, we have a database of transactions that are coded with a geographic locale and a type. These are being loaded into a data.frame with named variables city, type, and price. E.g., trans$city

Hi all, I'd be grateful for your help. I am a new user struggling with a barplot issue. I am plotting categories (X axis) and their mean count (Y axies) with barplot(). The first call to barplot works fine. I remove records from the dataframe using final=[!final$varname == "some value",] I echo the dataframe and the records are no longer in the dataframe. When I call plot again

The diffobj package (https://cran.r-project.org/package=diffobj) is really helpful here. It provides "diff" functions diffPrint(), diffStr(), and diffChr() to compare two object 'x' and 'y' and provide neat colorized summary output. Example: > iris2 <- iris > iris2[122:125,4] <- iris2[122:125,4] + 0.1 > diffobj::diffPrint(iris2, iris) < iris2 > iris @@ 121,8 / 121,8 @@ ~ Sepal.Length Sepal.Width Petal.Length Petal.Width Species 120 6.0 2.2 5.0 1.5 virginica 121 6.9...

...huge RDSs row by row. The diffobj package (https://cran.r-project.org/package=diffobj) is really helpful here. It provides "diff" functions diffPrint(), diffStr(), and diffChr() to compare two object 'x' and 'y' and provide neat colorized summary output. Example: > iris2 <- iris > iris2[122:125,4] <- iris2[122:125,4] + 0.1 > diffobj::diffPrint(iris2, iris) < iris2 > iris @@ 121,8 / 121,8 @@ ~ Sepal.Length Sepal.Width Petal.Length Petal.Width Species 120 6.0 2.2 5.0 1.5 virginica 121 6.9...

Dear R-help, This is about randomForest's handling of NA and NaNs in test set data. Currently, if the test set data contains an NA or NaN then predict.randomForest will skip that row in the output. I would like to change that behavior to outputting an NA. Can this be done with flags to randomForest? If not can some sort of wrapper be built to put the NAs back in? thanks, Clayton

...830 with the commit message "correct the work of dput() on the row names of a data frame with compact representation." Is there a problem / better way to use the result of a hefty dput than source()ing it? This seems to work rather robustly: data(iris) source(textConnection(paste0("iris2 <- ", capture.output(dput(iris))))) identical(iris, iris2) Cheers, Michael

The anti_join from the package dplyr might also be handy. install.package("dplyr") library(dplyr) anti_join (x1, x2) You can get help on the different functions by ?function.name(), so ?anti_join() will bring you help - and examples - on the anti_join function. It might be worth testing your approach on a small subset of the data. That makes it easier for you to follow what happens

I have many very large dataframes with 20 columns each. In order to conserve memory, I wish to separate the data frame into 20 vectors, each named the name of the dataframe followed by .1,.2,.3 .20. (For example purposes, one data frame is named ?testa?.) e.g. testa.1, testa.2, testa.3 I have written the code to do this (see below). I am trying to convert this into a function that I can reuse.

Has anyone sucessfully used the maxnodes feature in randomForest? I tried setting it, but when it is non-NULL I always get back a forest in which all trees have size 1. I am using a continuous response (regression). Any help would be appreciated. Thanks. [[alternative HTML version deleted]]

I have a dataset and I wish to use two different models to predict. Both models are SVM. The reason for two different models is based on the sex of the observation. I wish to be able to make predictions and have the results be in the same order as my original dataset. To illustrate I will use iris: # Take Iris and create a dataframe of just two Species, setosa and versicolor, shuffle them

Probably a stupidly simple question, but I wouldn't know how to google it: xyplot(neuro ~ time | UserID, data=data_sub) creates a proper plot. However, if I add type = "l" the lines do not go first through time1, then time2, then time3 etc but in about 50% of all subjects the lines go through points seemingly random (e.g. from 1 to 4 to 2 to 5 to 3). The lines always start at time

Hi list, I want to write a general function so that it would take an lm object, extract its data element, then use the data at another R function (eg, glm). I searched R-help list, and found this would do the trick of the first part: a.lm$call$data this would return a name object but could not be recognized as a data.frameby glm. I also tried call(as.character(a.lm$call$data)) or

I try tu use mob() with my data.frame ('data.frame': 288 obs. of 81 variables; factors, numerics and ordered factors) My response is a binary variable and I should use for modelling a logistic regression (family=binomial). I read in the "MOB" Vignette that I could use a formula like this if I would like to have only partitioning variables apart from the response.

Hello List, I am looking for efficient code to produce the Cartesian product of two or more data.frames. I'd like to be able to do this without resorting to looping. I have searched the FAQ, web, etc without luck. That being said, the help page for merge says that the function can produce what I'm looking for if the by vectors are of zero length. Would someone be so kind as to