similar to: Naming rows/columns in a 3 dimensional array/dataframe

Hi there, I am trying to run a liner regression using lm with na.action = NULL, but I am getting an error message. Any ideas as to why this may be happening? Please see code and error message below: > reg_test<-lm(yy~.,data=test,na.action=NULL) Error in lm.fit(x, y, offset = offset, singular.ok = singular.ok, ...) : NA/NaN/Inf in foreign function call (arg 4) This message and

Hello list members, I am working with simulated data for landscape pattern analysis. I have 1000 replicates of binary (2 colour) gridded landscapes at each combination of 9 levels of class proportion and 11 levels of spatial autocorrelation. The results are stored in an array as follows: > dim(surfaces) [1] 38 9 11 1000 The dimensions are defined as follows: [x,,,] 1:38, integers

foo <- rnorm(30*34*12) dim(foo) <- c(30, 34, 12) I want to make a data.frame out of this three-dimensional array. Each dimension will be a variabel (column) in the data.frame. I know how this can be done in a very slow way using for loops, like this: x <- rep(seq(from = 1, to = 30), 34) y <- as.vector(sapply(1:34, function(x) {rep(x, 30)})) month <- as.vector(sapply(1:12,

If I have a two column data frame like: > dat <- cbind("x"=c(1:100),"y"=c(100:1)) How can I create an array that splits every ten rows of that data frame into a third dimension of an array so that: > newarray[,,1] ,,1 x y 1 100 2 99 3 98 ... ... 10 91 ,,2 x y 11 90 12 89 ... ... ... Thanks. [[alternative HTML version deleted]]

Dear R-Helpers, I wonder whether there is a function which cuts a multiple dimensional array along a chosen dimension and then store each piece (still an array of one dimension less) into a list. For example, arr <- array(seq(1*2*3*4),dim=c(1,2,3,4)) # I made a point to set the length of the first dimension be 1to test whether I worry about drop=F option. brkArrIntoListAlong <-

Dear all! After hours of trying around, I gave up: I have a 2-dimensional array, and I know how to split it into its rows and how to get the mean for every row using 'sapply'. But what I want is to calculate the mean over the first n rows, and then the second n rows, etc., so that I get a vector like: v == mean1(row 1:5), mean2(row6:10),... (trivial, you might say. I find it rather

Hi, Suppose I have a multidimensional array: tmp <- array(1:8, c(2,2,2)) is there a function out there that, given a one-dimensional array index, will return the separate indices for each array dimension? for instance, tmp[8] is equivalent to tmp[2,2,2]. I'd like to derive the vector (2,2,2) from the index 8. thanks, Brad Buchsbaum

Hi everyone! Thank you for the help you have been given to me, and here I'm with another problem with my dataframes: I have two dataframes (with much more observations), like these: Dataframe1 Firm Year cash 500400200 2007 100 500400200 2006 200 500400200 2005 400 500400300 2007 300 500400300 2006 240 500400300 2005 120 500400400

Is there any way to rotate a plot generated by the wireframe function (lattice)? Thank you, Fabian Garavito

Hi everyone: I need to count the number of banks of each firm in my data. The firm is identified by the fiscal number. The banks of each firm appears like this: Firm Banks 500600700 Citybank 500600700 CGD 500600700 BES 500600800 Citybank 500600800 Bank1 500600900 CGD I want to obtain the following dataframe: Firm

I'd like to have a three dimensional array of matrices. I thought I could construct a five dimensional array to have the three dimensional array of matrices. However, not all of the matrices in the array have the same dimensions, which seems to mean I can't use a five dimensional array. What I'd like is this: A = matrix(1:4,2,2) B = matrix(1:25,5,5) C = matrix(1,3,3) D =

If I have 2 n-dimensional arrays, how do I compose them into a n+1-dimension array? Is there a standard R function that's something like the following, but that gives clean errors, handles all the edge cases, etc. abind <- function(a,b) structure( c(a,b), dim = c(dim(a), 2) ) m1 <- array(1:6,c(2,3)) m2 <- m1 + 10 abind(m1,m2) ==> , , 1 [,1] [,2] [,3] [1,] 1 3 5

hello everyone. Look at the following R idiom: a <- array(1:30,c(3,5,2)) M <- (matrix(1:15,c(3,5)) %% 4) < 2 a[M,] <- 0 Now, I think that "a[M,]" has an unambiguous meaning (to a human). However, the last line doesn't work as desired, but I expected it to...and it recently took me an indecent amount of time to debug an analogous case. Just to be explicit, I would

Colleagues I want to convert a 10x2 array: # create a 10x2 matrix. datavals <- matrix(nrow=10,ncol=2) datavals[,] <- rep(c(1,2),10)+c(rnorm(10),rnorm(10)) datavals into a 10x3 array, ThreeDArray, dim(10,2,10). The values storede in ThreeDArray's first dimensions will be the data stored in datavalues. ThreeDArray[i,,] <- datavals[i,] The values storede in ThreeDArray's second

Hi R helpers! I have the following dataframe ?choose? choose<-data.frame(firm=c(1,1,2,2,2,2,3,3,4,4,4,4,4,4), year=c(2000,2001,2000,2001,2002,2003,2000,2003,2001,2002,2003,2004,2005,2006),code=c(10,10,11,11,11,11,12,12,13,13,13,13,13,13)) choose I want to subset it to obtain another one with those observations for which there more than 2 observations in the column ?code?. So I want a

Dear list after quite a bit of research in the archive, I gave up. This seems to be a simple problem: I would like to aggregate a (3-dimensional) array, either by another array, or by a vector, indicating the dimension which should be aggregated. I don't think I have to provide an example, it's really the 3-dimensional equivalent for the standard aggregate command. I am sure

Hello: I would like to sum every x columns of a dataframe for each row. For instance, if x is 10, then for dataframe df, this function will sum the first ten elements together and then the next ten: sapply(list(colnames(df)[1:10], colnames(df)[11:20]),function(x)apply( df[,x], 1, sum)) If the number of columns is quite large (1000's), then manually entering the list above is not practical.

how to implement multi dimensional array in ruby in ruby multi dimension array look like tree structure plz help me & explain with code have a pleasant day thx narayana -- Posted via http://www.ruby-forum.com/.

maybe we've already diskussed this before, but Kurt and I can't remember ... is.vector() of an one-dimensional array returns FALSE. this is also the behavior of Splus, but totally counter-intuitive for me ... IMO an array of dimension 1 is exactly the definition of a vector ... it also breaks our current plot.factor, which is simply a barplot(table(x)) table() returns an

Hello, How would you export a 3-dimensional array to an SQL database? a<-array(1:24, 2:4) Is there an open source DB that would be more adequate for this type of operation? Is there a way to reshape/flatten a 3-dimensional array? Regards, Pierre Lapointe ************************************************** AVIS DE NON-RESPONSABILITE: Ce document transmis par courrie...{{dropped}}