similar to: indexing into a list

I didn't explain myself well. You are right about not needing a list on the right hand side of the equation below. What I end up with is a list of arrays. Now what i would like to do is to access the individual elements in the arrays in the list. For example, when I type, y[[1]], I just get the whole first array. How do I get the first element in the first array for instance? Is

Please excuse me if this is a very basic question and I am just misunderstanding the basic structure of R. Right now, I am writing everything in a buffer (emacs) and then copying it into R when it is correct. Is there a way to save a file of commands and then just type the file name or something in R and have the commands executed within R? Anna

When I am trying to put a legend on a plot where I am using "lines", R just ignores it. I can do it with boxplot or plot, but just not with lines. Am I doing something wrong? Maybe I am just making a mistake? Anna

I would like to do a one way anova and then a tukeyHSD. I have three vectors A,B and C. In a previous help message, I was told to do the following for the anova: y = c(A,B,C) group = factor(rep(a:3,c(7,9,13))) #provided there a 7 elements in A,9 in B and 13 in C and then anova(lm(y~group)) Looking at the tukeyHSD method it looks like it wants the aov method which I don't understand.

Does anyone know of some methods already programmed up ( ie freeware : ) ) in R for the post hoc methods in anova. Particularly Scheffe's method or Tukey's? Anna

I've tried looking at ?colors and ?palette and if I'm understanding it correctly, I'm supposed to type in (for example) palette(rainbow(13)) before I type in my plot (of 13 lines) if I want 13 different colors. But this does not work. Other things that i have tried besides rainbow give me errors. Am I just doing something completely wrong? Anna

Is there a simpler way then the solution to the one that was posted here? I'm not very proficient with legend, and I don't understand this solution. All I have is two or more lines on one plot that I want to put a legend on and I can't figure out how to do it from the examples. Can you give a very simple example? It does not have to be fancy!! I have never worked with a

Is there a modulus operator in R? Anna

try "%%" as the modulus operator and "%/%" for integer division eg 5%%2 equals 1 John Marsland > > -----Original Message----- > > From: Anna H. Pryor [mailto:anna at ptolemy.arc.nasa.gov] > > Sent: 19 May 2003 16:56 > > To: R Help > > Subject: [R] modulus operator? > > > > > > > > Is there a modulus operator in R?

I know how to print my own labels along the x axis: plot(1:10,xlab = "My label", axes = FALSE) axis(1,at=seq(1,10,by=2),labels=c("first","second","third","fourth","fifth")) Is there a way to make the labels print vertically along the axis rather than horizontally? (I'm not sure if this is even going to look all right, but I

I am totally confused as to how to use anova. I have three vectors and would like to use anova on them but I don't understand how lm or glm comes into play. In matlab, you just give the three vectors. Why isn't it the same in R? Any help would be greatly appreciated. Anna

I have a plot with 13 lines and R repeats colors. Is there any way to make R not repeat colors, or is this just a fundamental limitation on R's part to say have only 8 colors that it can use? Anna

I have two separate questions. The first is when I use xlab, the labels on my tic marks disappear. Is there a way to make them reappear? Also, is there a way to make my own labels for the tick marks? My second question is, after I have spent a lot of time getting my plot to look just right, I want to save it to a file, but the only way I seem to know how to do it is to first use

I've been programming in one directory and recently switched to another directory. It appears that in doing so I've uncovered a problem. My environment was saving something so that my code would work and now I don't know how to fix it. I have the following bit of code: for(i in 1:index){ indexList = lst[i] for(j in 2:jobs-1){ indexList =

I don't know how to retain my row labels. I read in a table with row labels, ie: Price Floor Area Rooms 52. 111. 830 5 54. 128. 710 5 47.4 101 1000 4 ... but then I need to transform the data raggedRidzeros <- function(x){ y = list(ridzeros(x[[1]])) for(i in 2:length(x)){ y = c(y,list(ridzeros(x[[i]]))) } y } (where ridzeros is a function

Hello everyone, I need some help with lists inside lists (a good way to emulate a struct)\ Assume that there is a small list called fred: fred <- list(happy = 1:10, name = "squash") and a big list called bigfred that includes fred list 5 times bigfred <- rep(fred,5) Is it possible somehow to index all these sublists(fred) inside bigfred with a more direct way like this:

Hi everyone (again), before you all start screaming that the reordering of factors has been discusse on several threads and is not particular to ggplot2, hear me out. I can easily reorder my x-axis factor in facet.grid() in ggplot2. What I cannot reorder are the factors represented on the strips. I can see that the graphs are changing, so I am afraid of what it is I am doing. Why is ggplot2

I'm seeing a lot of activity over the last two days with what looks to be a kiddie script. Mostly trying to access several of our servers with the username anna. All failed... in fact I don't think we have a user anna on any of our servers. Meanwhile... I'm running Sendmail. This pertains to Centos 4 and 5 servers. I'm also running fail2ban on some and Ossec on others. So far,

Hi guys, here is a list of names that I have: MyList: > myList<-list("A", "B","C","D") > myList [[1]] [1] "A" [[2]] [1] "B" [[3]] [1] "C" [[4]] [1] "D" I want to turn this list into a matrix of 1 row and 4 columns with those four components (A, B, C, D) so here is what I do: myDataFrame <-

Hello guys, I have a list L1 of matrix. I have another list L2 with the same number of elements representing the row of the L matrix that I want to delete (L1[[i]][-L2[[i]],]) but I can't do this with lapply as it iterates through L1 (first argument) and not L2. Any idea? ----- Anna Lippel -- View this message in context: