similar to: Transpose array

Hi listers, I have the following code... data<-as.matrix(c(1,2,3,4,5,6,7,8,9,10),nrow=10,ncol=1,byrow=TRUE) n<-nrow(data) m<-n-1 boot<-data[sample(m,replace=T),] So, I need to count the number of times each observation was selected at the sample with replacement... Suppose I sampled... 4 5 1 3 7 5 1 4 7 So, I would count... x count 1 2 2 0 3 1 4 2 5 2 6 0 7 2 8 0 9 0 10 0

Hi listers, I have a data frame with 5 variables... And I want reorder the data frame according a specific variable X... I checked the sort and order functions... In my case I think that the ORDER function is applicable... But I did not understand at all... The idea is simple... Data order by X... I know that this is very simple... Anyways... How I would do that... Thanks in advance, Marcio --

Hi listers, I have a simple doubt... I need to manipulate the data base as: var1 var2 7 0.1 7 0.1 8 0.12 10 0.15 12 0.18 20 0.31 I would like to aggregate the equal cases as var1 making the sum of var2... I would get... var1 var2 7 0.2 8 0.12 10 0.15 12 0.18 20 0.31 Thanks in advance, Marcio -- View this message in context:

Hi listers, I am having some trouble in a matrix multiplication... I have already checked some posts, but I didn't find my problem... I have the following code... But I am not getting the right multiplication... I checked the dimension and they are fine... id_y <- array(1:10,dim=c(2,1,5)) id_yt<-aperm(id_y,c(2,1,3)) m_id<-array(dim=c(dim(id_y)[1],dim(id_y)[1],dim(id_y)[3])) for (i in

Hi All, I want to ask if there is a transpose function for data frame like the procedure of transpose in SAS? Because I want to partially transpose a data frame which contains 5 columns (siteid, date, time, obs, mod), what I want to do is to put time as the column variables along with siteid, and date, and put obs and mod in the row names. specifically to transpose a data frame: siteid date

Dear list, We have a number of files containing similarly structured data: file1: A B C 1 2 3 4 5 6 file2: A B C 7 8 9 10 11 12 ... etc My part of R receives all these data as an array: 1,2,3... 12 together with info about dimensions (row,col,fileN) . ( Converting the data into 3D cannot simply done by: array(x, c(2,3,2)) because breaks the structure (e.g. 1,3,5 is type mismatch)

Hi listers, I am working on a simulation... But I am having some troubles... Suppose I have a function A which produces two results (mean and variance)... Then I would like to simulate this function A with a function B many times using the results from function A For example: #Function A boot<-function(a,b,c){ mean_boot<-(a+b)/2 var_boot<-c #list(a=a,b=b,c=c) return(a) } Then I would

Thanks for both the mapply and array approaches! However, although intended to generate the same result, they don't: # mapply approach n = 3 sa <- rnorm(n,0.8,0.1) so <- rnorm(n,0.5,0.1) m <- rnorm(n,1.2,0.1) mats = mapply(function(sa1, so1, m1) matrix(c(0,sa1*m1,so1,sa1),2,2,byrow=T), sa, so, m, SIMPLIFY = FALSE) print(mats) [[1]] ????????? [,1]????? [,2] [1,] 0.0000000

Sure -- thanks -- only took me 3-4 attempts to get aperm to work (as opposed to really thinking hard about how it works ;-) On 9/28/2017 11:55 AM, Duncan Murdoch wrote: > On 28/09/2017 9:10 AM, Evan Cooch wrote: >> Thanks for both the mapply and array approaches! However, although >> intended to generate the same result, they don't: >> >> # mapply approach

Hi listers, I am working on an canonical discriminant analysis, but I am having some trouble to use the CANDISC function... I just installed the last R version and installed the package CANDISC... But, I am getting the following message because about a permission: The downloaded packages are in C:\Users\Marcio\AppData\Local\Temp\Rtmpz2kFUm\downloaded_packages updating HTML package descriptions

On 28/09/2017 9:10 AM, Evan Cooch wrote: > Thanks for both the mapply and array approaches! However, although > intended to generate the same result, they don't: > > # mapply approach > > n = 3 > sa <- rnorm(n,0.8,0.1) > so <- rnorm(n,0.5,0.1) > m <- rnorm(n,1.2,0.1) > mats = mapply(function(sa1, so1, m1) > matrix(c(0,sa1*m1,so1,sa1),2,2,byrow=T),

I've dug around but not been able to find anything, am probably missing something obvious. How can I fill a three-dimensional (or higher dimension) array by rows instead of columns. eg new1 <- array(c(1:125), c(5,5,5)) works fine for me but fills it by columns and new2 <- array(c(1:125), c(5,5,5), byrow=TRUE) throws an error. Am I missing something obvious? I also tried

Hi Listers, I am producing some graphics that the commands are in a FUNCTION... The problem is that I end up viewing just last graphic and in my FUNCTION there are 4 graphics with the PAR command function... Like those below... How do I view/get the other 3 graphics? Any help? Thanks in advance... histogram<-par(mfrow=c(1,2)) hist(rw_mean_app,main='Bootstrap Method

I have the following: Time A1 A1 B1 B1 C1 C2 x y x y x y 0 5 6 6 7 7 9 1 3 4 4 3 9 9 2 5 2 6 4 7 4 I want to change it to the following: 0 1 2 x y x y x y A1 5 6 3 4 5 2 B1

Hi listers, I am trying to arrange my data and I didn't find any information how to do it! I have a data with 3 variables: X Y Z 1-I would like to multiplicate de information of X according to the number I have for my Y variable... 2-Then I want to identify with a dicotomic variable by the number according my variable Z from X... I can do the first part by... z<-rep(x,y) But I don't

Hello everybody When one is working with complex matrices, "transpose" very nearly always means *Hermitian* transpose, that is, A[i,j] <- Conj(A[j,i]). One often writes A^* for the Hermitian transpose. I have only once seen a "real-life" case where transposition does not occur simultaneously with complex conjugation. And I'm not 100% sure that that wasn't a

Hi listers, Does it necessary to install any package in order to use the GINI or INEQ functions. If I use the following command the R tells me that didn't find the GINI function. x<-c(541, 1463, 2445, 3438, 4437, 5401, 6392, 8304, 11904, 22261) G<-gini(x) Thanks in advance, Marcio -- View this message in context:

Suppose this is my list of transactions: set.seed(200) tran=random.transactions(100,3) inspect(tran) items transactionID 1 {item80} trans1 2 {item8, item20} trans2 3 {item28} trans3 I want to get the 'transpose' of the data, i.e. transactionID items 1 {trans2} item8 2 {trans2} item20 3 {trans3} item28 4 {trans1} item80 I

I want to transpose a dataframe like level 2006 2007 2008 A .... B ..... C ........ into level year A 2006 A 2007 A 2008 B 2006 B 2007 ...... There is a procedure in SAS can do this, is there any function in R can do this? -- View this

Hi guys, I need your help!! My goal is to make a csv file from ncdf file. This is the code i've used : > hyo=open.ncdf("C:/CRUTEM3.nc") > hyo [1] "file C:/CRUTEM3.nc has 4 dimensions:" [1] "longitude Size: 72" [1] "latitude Size: 36" [1] "unspecified Size: 1" [1] "t Size: 1916" [1] "------------------------"