similar to: Combinations

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]]

For a list say; list1<-{1,2,3,4,5,2,1} How do I remove the duplicates please? My real list is 20,000 obs long of dates with many duplicates 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]]

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]]

Hello all, I am trying to create a matrix of 1s and -1s without any repetitions for a specified number of columns. e.g. 1s and -1s for 3 columns can be done uniquely in 2^3 ways. -1 -1 -1 -1 -1 1 -1 1 -1 -1 1 1 1 -1 -1 1 -1 1 1 1 -1 1 1 1 and for 4 columns in 2^4 ways and so on. I finally used the function combn([0 1],3) that I found at the following link

I can get the interactions between factors like this: > idx=c(1,3,6,9) > jdx=idx > levels(interaction(idx,jdx,lex.order=TRUE)) [1] "1.1" "1.3" "1.6" "1.9" "3.1" "3.3" "3.6" "3.9" "6.1" "6.3" "6.6" "6.9" [13] "9.1" "9.3" "9.6"

Very quick one please team; How do I return the number of possible unique subsets in a data set (not combinations i.e. not (1,2,3,4) and (4,3,2,1)) please ? e.g. In a list of integers 1 to 12, there is 495 sets with 4 elements in regards glenn [[alternative HTML version deleted]]

Let say, I have a matrix with 8 rows and 6 columns:  >  df1  <- matrix(NA, 8, 4)  > df1       [,1] [,2] [,3] [,4]  [1,]   NA   NA   NA   NA  [2,]   NA   NA   NA   NA  [3,]   NA   NA   NA   NA  [4,]   NA   NA   NA   NA  [5,]   NA   NA   NA   NA  [6,]   NA   NA   NA   NA  [7,]   NA   NA   NA   NA  [8,]   NA   NA   NA   NA  Now I want to get **all possible** ways to fetch 6 cells at a

I have a list (length 750), each element containing a vector of unique strings (unique gene ids), with length up to ~40 (median 15). I want to compile a matrix of all possible triplets and their frequency within gene elements. Using combn and a lot of looping, I am accomplishing this but it is VERY slow. I've tried to figure out a way to vectorize this, using "match" and

Hello I apologise for the length of this entry but please bear with me. In short: I need a way of subsampling communities from all possible communities of n taxa taken 1:n at a time without having to calculate all possible combinations (because this gives me a memory error - using combn() or expand.grid() at least). Does anyone know of a function? Or can you help me edit the combn or

Hello! I have a regular data frame (DATA) with 10 people and 1 column ('variable'). Its cases are people with names ('a', 'b', 'c', 'd', 'e', 'f', etc.). I would like to write a function that would sum up the values on 'variable' of all possible combinations of people, i.e. 1. I would like to write a loop - in such a way that it

Hi, I am trying to generate all unordered combinations of a set of numbers / characters, and I can only find a (very) clumsy way of doing this using expand.grid. For example, all unordered combinations of the numbers 0, 1, 2 are: 0, 0, 0 0, 0, 1 0, 0, 2 0, 1, 1 0, 1, 2 0, 2, 2 1, 1, 1 1, 1, 2 1, 2, 2 2, 2, 2 (I have not included, for example, 1, 0, 0, since it is equivalent to 0, 0, 1). I have

R-2.9.1, Windows7 Dear list, I have a question to you that seems very simple to me, but I just can't figure it out. I have a dataframe called "ratings" which contains the following variables: evalR1, evalR2, evalR3, evalR4, scoreR1, scoreR2, scoreR3, scoreR4, opinionR1, opinionR2, opinionR3, opinionR4. (there are more variables, but this gives an idea of the data structure). What

Hi again, I am exploring if R can help me to get all possible combinations of members in a group. Let say I have a group with 5 members : A, B, C, D, E Now I want to generate all possible unique combinations with all possible lengths from that group e.g. 1st combination : A 2nd combination : B ..... 5th combination : E 6th combination : A, B 7th combination : B, C .... last combination: A, B,

hello list, i've got 8 objects and i want to calculate a list with all combinations of them. the output should contain the object names. for example: object1 & 2, 1 $ 3, ... best regards mirca -- "Ein Herz f?r Kinder" - Ihre Spende hilft! Aktion: www.deutschlandsegelt.de Unser Dankesch?n: Ihr Name auf dem Segel der 1. deutschen America's Cup-Yacht!