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Dear All I want to store matrix in an array Suppose s<-array(0,4) for(i in 1:4) s[i] <- read_matrix(a,2,2) But the error - number of items to replace is not a multiple of replacement length. Can you suggest me any alternative method for storing a matrix in an array. Thanks In advance. Kind Regards Wesley C Mathew [[alternative HTML version deleted]]

I have big .csv file. I would like to filter that file into a new table. For example, I have .csv file as below: f1 f2 f3 f4 f5 f6 f7 f9 f10 f11 t1 1 0 1 0 1 0 0 0 0 1 t2 1 0 0 0 0 1 1 1 1 1 t3 0 0 0 0 0 0 0 0 0 0 t4 1 0 0 0 1 0 0 0 0 0 t5 0 0 0 0 0 0 0 0 0 0 t6 0 0 0 0 0 0

Hi, I have big .csv file. I would like to filter that file into a new table. For example, I have .csv file as below: f1 f2 f3 f4 f5 f6 f7 f9 f10 f11 t1 1 0 1 0 1 0 0 0 0 1 t2 1 0 0 0 0 1 1 1 1 1 t3 0 0 0 0 0 0 0 0 0 0 t4 1 0 0 0 1 0 0 0 0 0 t5 0 0 0 0 0 0 0 0 0 0 t6 0 0 0 0 0 0 0 0 0 0

R Help: I have read a number of tables into R with identical headings and I would now like to make a single table that has all the data appended under this single heading line. for example: t1 <- read.csv("f1",header=TRUE) t2 <- read.csv("f2",header=TRUE) all <- c(t1,t2) #all is now twice as wide as t1 or t2 with the same number of row!!!! #I need to know how

i m using a package NbClust for cluster analysis. in the following algorithm ->NbClust(m, diss="NULL", distance = "euclidean", min.nc=2, max.nc=15, method = "ward", index = "all", alphaBeale = 0.1) i want to define my own dissimilarity matrix of dimension 38*38. my original data "m" is a matrix of 365*38. whenever i define my own dissimilarity

Hi, I wonder whether the mechanism of parsing parameters to functions has changed between 0.63.1 and 0.63.3? The following code yeils different results in R 0.63.1 (Version 0.63.1 (Dec 5, 1998)) and R 0.63.3. cave<-function(x,a,b) { return(c(mean(x[a],na.rm=T),mean(x[b],na.rm=T))) } datx <- data.frame(rbind(c(1,2,3,4),c(4,5,6,7)))

Hi, I wonder whether the mechanism of parsing parameters to functions has changed between 0.63.1 and 0.63.3? The following code yeils different results in R 0.63.1 (Version 0.63.1 (Dec 5, 1998)) and R 0.63.3. cave<-function(x,a,b) { return(c(mean(x[a],na.rm=T),mean(x[b],na.rm=T))) } datx <- data.frame(rbind(c(1,2,3,4),c(4,5,6,7)))

Hi, all together. I have - a maybe trivial - problem with aggregating a list of weights. Here is the problem: - At first I have set of nodes (X/Y-coordinates) and associated weights, where the set of nodes is typically not unique - I want to get a set of unique nodes and the sum of associated weights I am grateful for any help See for example: # weights: w <- c(1, 1, 1, 1, 1) #

Hi, I want to encode Loongson ISA initially https://gist.github.com/xiangzhai/8ae6966e2f02a94e180dd16ff1cd60ac gslbx           $2,0($3,$4) It is equivalent to: dadd $1, $3, $4 lb $2,0($1) I just use  mem_simmptr  as the default value of  DAGOperand MO , because  MipsMemAsmOperand  use  parseMemOperand  to parse general  MemOffset  and only *one*  AnyRegister , for example: 0($1) But 

Dear sir, I have a matrix like a<-matrix(c(0,2,0,4,0,6,5,8,0),nrow=3) colnames(a)<-c("F1","F2","F3") rownames(a)<-c("A1","A2","A3") a F1 F2 F3 A1 0 4 5 A2 2 0 8 A3 0 6 0 I want to extract all pairs (rownames, columnames) from which the value in the matrix is 0 The result should be something like this A1, F1 A2,

Dear All I am doing one distributed data mining program. So I selected GridR package for the distributed programming. *grid.init(service="local",debug=FALSE, localTmpDir="GridRTmp/") grid.apply("x",UCS, wait=TRUE )* UCS is a function. When I execute this statement, I have some errors like : *cannot load local function/variable: xmlRoot extracted from line: : no

On Tue, Mar 12, 2013 at 7:21 PM, Krzysztof Parzyszek <kparzysz at codeaurora.org> wrote: > On 3/12/2013 12:13 PM, Manman Ren wrote: >> >> Given >> struct A { >> int x; >> int y; >> }; >> struct B { >> A a; >> int z; >> }; >> struct C { >> B b1; >> B b2; >> };

I am trying to add a constant to the previous value of a variable based on certain conditions. Maybe there is a simple way to do this that I am missing completely. I have given an example below: df <- data.frame(x = c(1,2,3,4,5), y = c(10,20,30,NA,NA)) > df x y 1 1 10 2 2 20 3 3 30 4 4 NA 5 5 NA I want to add 2 to the previous value of y, if x exceeds 3 (also will have to handle NAs in

ear folks ? I have a question, though it is more of a logic- or a good practices-question than a programming question per se. I am working with data from the American Community Survey summary file. It is mainly categorical count data. Currently I am working with about 40 tables covering about 35 variables, mainly in two-way tables, with some 3-way and a handful of four-way tables. I am going to

Someone privately asked me to explain this example, so here goes ... > There are simpler examples of this kind for C++, because placement > new can change the dynamic type of the object (I actually haven't > looked to see if they changed this in 2012, but it was definitely > legal in C++98): > > #include <new> > struct Foo { long i; }; > struct Bar { void *p; };

I have a data frame with columns which draw on the same underlying universe, so I want them to be factors with the same level set: --8<---------------cut here---------------start------------->8--- > z <- data.frame(a=c("a","b","c"),b=c("b","c","d"),stringsAsFactors=FALSE) > str(z) 'data.frame': 3 obs. of 2

Dear list, I am trying to choose the scale parameter for the cgt transform but I don't know how to do it. In time I would like to be able to separate points 30 samples apart, and in frequency I would like to separate bands 0.04 Hz apart. I tried the two approaches described below and they gave me different results. I would appreciate advise on how to do this. The Rwave Gabor transform uses

Hi all I'm doing some things with a colleague comparing different sorts of models. My colleague has fitted a number of glms in Genstat (which I have never used), while the glm I have been using is only available for R. He has a spreadsheet of fitted means from each of his models obtained from using the Genstat "predict" function. For example, suppose we fit the model of the type

Hi, table1D() below can be up to 60x faster than base::table() for the 1D case. Here are the detailed speedups compared to base::table(). o With a logical vector of length 5M: 11x faster (or more if 'useNA="always"') o With factor/integer/numeric/character of length 1M and 9 levels (or 9 distinct values for non-factors):

Dear All I was trying to install R-2.10.0-2.fc11.X86 in Fedora-13-i386 but it makes error. Could you please tell me which is the exact version of R for Fedora 13. Thanks in advance Kind Regards Wesley [[alternative HTML version deleted]]