similar to: Conditional correlation in R?

Dear all When using Sweave, I'm always hitting the same bump: I want to group repetitive calls in a function, but I want both the results and the function calls in the printed output. Let me explain myself. Consider the following computation in an Sweave document: summary(iris[,1:2]) cor(iris[,1:2]) When using these two calls directly, I obtain the following output: > summary(iris[,1:2])

In using cor(data.frame), it is annoying that you have to explicitly filter out non-numeric columns, and when you don't, the error message is misleading: > cor(iris) Error in cor(iris) : missing observations in cov/cor In addition: Warning message: In cor(iris) : NAs introduced by coercion It would be nicer if stats:::cor() did the equivalent *itself* of the following for a data.frame:

I have a data set where I want the correlations between 2 variables conditional on a students grade level. This code works just fine. by(tmp[,c('mtsc07', 'DCBASmathscoreSPRING')], tmp$Grade, cor, use='complete', method='pearson') However, this generates an error by(tmp[,c('mtsc07', 'DCBASmathscoreSPRING')], tmp$Grade, cor, use='complete',

Hi, I've looked at the parameters available for the legend function and cannot find a way to change the distance between the top of the box surrounding a legend and the legend's title. I have a math expression that raises the height of my title. If you don't mind the non-sensical title I give to the legend for this plot (Figure 3.20 in R Graphics): with(iris,

The lattice package uses special logic to allow for multiple left-hand-side variables in a formula, e.g. y1 + y2 ~ x. Is there an elegant way to do this outside of lattice? I'm trying to implement a data summarization function that logically takes multiple dependent variables. The usual invocation of model.frame( ) causes R to try to do arithmetic addition to create a single dependent

Fellow R-helpers, Suppose we create a histogram as follows (although it could be any vector with zeroes in it): R> lenh <- hist(iris$Sepal.Length, br=seq(4, 8, 0.05)) R> lenh$counts [1] 0 0 0 0 0 1 0 3 0 1 0 4 0 2 0 5 0 6 0 10 0 9 0 4 0 [26] 1 0 6 0 7 0 6 0 8 0 7 0 3 0 6 0 6 0 4 0 9 0 7 0 5 [51] 0 2 0 8 0 3 0 4 0 1 0 1 0 3

Hello, I am puzzled by the behavior of hist() when generating multiple plots per page on the pdf device. In the following example two pdf files are generated. The first results in 4 plots on one pdf page as expected. However, the second, which swaps one of the plot() calls for hist(), results in a 4 page pdf with one plot per page. How might I get the histogram with 3 other scatter

Hello, I'll use part of the iris dataset for an example of what I want to do. > data(iris) > iris<-iris[1:10,1:4] > iris Sepal.Length Sepal.Width Petal.Length Petal.Width 1 5.1 3.5 1.4 0.2 2 4.9 3.0 1.4 0.2 3 4.7 3.2 1.3 0.2 4 4.6 3.1 1.5

Hi list, One of my co-workers found this problem with 'cor' in his code and I confirm it too (see below). He's using R 2.2.1 under Win 2K and I'm using R 2.3.0 under Win XP. =========================================== > R.Version() $platform [1] "i386-pc-mingw32" $arch [1] "i386" $os [1] "mingw32" $system [1] "i386, mingw32" $status

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

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

Por si alguno pudiera ayudarme. Al realizar el t.test para una muestra, junto con el valor de t y el p-valor, la función proporciona la estimación de la media y su INTERVALO DE CONFIANZA. Desde el punto de vista de la estadística de rangos esto se puede hacer mediante: > iris$MEDIANA <- with(iris, 2.95) > median(iris$Sepal.Width - iris$MEDIANA, na.rm=TRUE) # median difference [1]

Dear R community I am trying to assign alphabets to integer values 1, 2, 3 etc. in y given below. Can any body suggest some simple way to do the same job? ds=iris; dl=nrow(ds) c1=ds[,1]; c2=ds[,2]; c3=ds[,3]; c4=ds[,4]; c5=ds[,5]; iris=cbind(c1,c2,c3,c4,c5) y=iris[,5] y1=which(y==1); y[y1] <- c("a"); y2=which(y==2); y[y2] <- c("b"); y3=which(y==3); y[y3] <-

I would like to create an xyplot with varying y-axis limits and horizontal labels at the y-axis tickmarks. The following does not seem to work, although I think it should, going by the documentation for par. R version 2.5.1, Windows XP Prof. Thanks for a clue. Andreas Krause library(lattice) # axis labels for y-axis are horizontal xyplot(Sepal.Length ~ Sepal.Width | Species, data=iris) #

Hi Dieter and R community: I tried both of these three versions with ylim as suggested, none work: I am getting only single (pch = 16) not overlayed (pch =3) everytime. *vs 1* require(lattice) xyplot(Sepal.Length ~ Sepal.Width | Species , data= iris, panel= function(x, y, subscripts) { panel.xyplot(x, y, pch=16, col = "green4", ylim = c(0, 10)) panel.lmline(x, y, lty=4, col =

Hi, The below very strange: # a) aov function av <- aov(Sepal.Length ~ Species, data=iris) # Error in parse(text = x) : # unexpected symbol in "Sepal(Sepal.Length+Species)Length" av <- aov(iris[, 1] ~ iris[, 5]) # summary(av) # Df Sum Sq Mean Sq F value Pr(>F) # iris[, 5] 2 63.2 31.6 119 <2e-16 *** # Residuals 147 39.0 0.3 # ---

Dear all Is there a simpler method to achieve the following: When I obtain an empty data.frame after subsetting, I need for it to contain one line of NAs. Here's a dummy example: > (.xb <- iris[ iris$Species=='zz', ]) [1] Sepal.Length Sepal.Width Petal.Length Petal.Width Species <0 rows> (or 0-length row.names) > dim(.xb) [1] 0 5 > (.xa <-

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

Hallo, I'd like to add grid lines to a lattice graph having 2 series of Y data. See these 2 examples: data(iris) [1] xyplot(Sepal.Length + Sepal.Width ~ Petal.Length , data = iris, allow.multiple = TRUE, scales = "same",type="l", ) [2] xyplot(Sepal.Length + Sepal.Width ~ Petal.Length , data = iris, allow.multiple = TRUE, scales =

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