similar to: Removing duplicates from a list

Hi All, would appreciate an answer on this if you have a moment; Is there a function (before I try and write it !) that allows the input of a covariance or correlation matrix to calculate PCA, rather than the actual data as in princomp() Regards Glenn [[alternative HTML version deleted]]

Hi All, I have a small dataframe [dates, values) I am plotting with plot(df,type=²l²) And the date date covers a year. The graph only have marks at Œ2008¹ and Œ2009¹. How do I get the months labeled at the bottom please Thanks as always Glenn [[alternative HTML version deleted]]

Hi all - has anyone ot and experienced problems with R+. Have downloaded the trail and it will not work -does it need vista (like it says on the box!) Thanks glenn [[alternative HTML version deleted]]

Hi All, some help on this would be appreciated; Understood combinations(7,4) returns all the possible 4 part combinations out of 7. Is is possible to substitute the ³7² for a list of stuff you would like to see the mix of; c(³a²,...,²g²) say ? Thanks Glenn [[alternative HTML version deleted]]

Dear All; Is it possible to create a list of lists (I am sure it is) along these lines; I have a dataframe data02 that holds a lot of information, and the first column is ³date² I have a list of dates in; data03<-c(date1,.....,daten) And would like to create a list; data04 <- subset(data02, date == data03[1,]) Ie. data04 holds the data from data02 that matches a date in data03 How do

Help with this much appreciated I have a large dataframe that I would like to subset where the constraint Test1 <- subset(df, date == uniques[[1]]), where uniques is a list of dates that must be matched to create Test1. I would like to perform an operation on Test1 that results in a single column of data. So far so good. How do loop through all values in the uniques list (say

Very simple questions if anyone can help: (1) what is the value in saving workspaces, which is offered at every close? I thought it might save the set up of the GUI but I cant work out what it does ­ I run a script that loads the packages I need at the start of every session. (2) mathematica has a term (%) that when typed means the last returned value, so > % + 1 say would return the last

Newbie question sorry (have tried the help pages I promise) I have a dataframe (date,stockprice) say and looking how I might get the return of: dataframe (difference in days, change in stock price) using sapply - I require a very simple function and don't really want to go down the zoo and quant mod route Regards glenn [[alternative HTML version deleted]]

Very quick newbie question sorry; How do I standardize a list of data please (other than do the calculation of course). Is there a quick way please ? glenn [[alternative HTML version deleted]]

In a quest to remove ?duplicate? messages sent to both me and lists I subscribe to I came up with this, which I think should clean out my Archive folder, but I?ve been unable to get it to work for scanning all on my list-user email. $ doveadm -f table fetch -u kremels 'hdr.message-id guid uid hdr.x-listname' mailbox "Archive" | sort| awk 'cnt[$1]++{if (cnt[$1]==2) print

Dear list I have a response variable coded 0/1 i.e. a binary response. There are 20,000 individual responses that I would like to aggregate into numbers of each category (i.e. 0/1) by group called dn (350 different groups) and by month mth (there are several hundred responses per month. What is the simplest way to perform this operation in R?

Hello R Help Team, I have created graph from weighted adjecency matrix .Is there a way I can find highly connected nodes in Igraph like the Package RBGL does. nathan -- View this message in context: http://www.nabble.com/Highly-Connected-Nodes-in-Igraph-tp22377522p22377522.html Sent from the R help mailing list archive at Nabble.com.

Dear colleagues, I'm trying to have a look at the Assortative and Disassortative ( http://en.wikipedia.org/wiki/Assortative_mixing) of the network I have. But it seems that the igraph hasn't mentioned that yet. I have to get the in/out degree of the vertices of each edge and calculate the Pearson's Correlation coefficient which seems to be quite a huge task for me. :( So I wonder if

I'm trying to draw an arrow with a curved shaft on the graph as a straight one looks messy on a detailed graph. I've looked in arrows but it doesn't seem to give an option. larrows doesn't look much more promising. I had a look in the archive and couldn't find anything. Any thoughts? Thanks Paul -- View this message in context:

Dear R helpers, I'm trying to write a numerical scheme for a boundary integral method to solve an electromagnetic problem. This requires the computation of the distance between points at the surface of an object (a sphere, in my example). Here is my code, > require(rgl) > r<-1 > size<-10 > theta<-seq(0,2*pi,length=size*2) > phi<-seq(0,pi,length=size) > pc

Hi all, I'm very new to R and read a few tutorials, however I'm having difficulty trying to figure out how to plot a minimum spanning tree. I have a csv file that contains an n-by-n matrix of distances between strains of bacteria called matrix.csv. Looks like: id,strain1, strain2,strain3 strain1,0,.2,.8 strain2,.3,0,.7 strain3,.4,.6,0 I've been messing around with some information

Dear R users, A co-worker and I are writing a function to facilitate graph plotting in R. The function makes use of a lot of lists in its defaults. However, we discovered that R does not necessarily preserve the defaults if we were to input them in the form of list() when initializing the function. For example, if you feed the function codes below into R: myfunction=function( list1=list

Dear all, "Writing R Extensions" explicitly says that A method must have all the arguments of the generic, including ... if the generic does. A method must have arguments in exactly the same order as the generic. If the generic specifies defaults, all methods should use the same defaults. This is clear. R CMD check even checks for this. But then how is it possible that

Hi, Given a simple example, test <- matrix(c(0.1, 0.2, 0.1, 0.2, 0.1, 0.1, 0.3, 0.1, 0.1), 3, 3) How to generate row indexes for which their corresponding row values are less than or equal to 0.2 ? For this example, row 2 and 3 are the correct ones. Thanks [[alternative HTML version deleted]]

Hi R users: I got this code to generate a graphic for each member of a lists. list1<-list(A=data.frame(x=c(1,2),y=c(5,6)),B=data.frame(x=c(8,9),y=c(12,6))) names1<-names(list1) sapply(1:length(list1),function(i) with(list1[[i]],plot(x,y,type="l",main=paste("Graphic of",names1[i])))) Is there a more elegant solution for not to use two separate lists? I would like to