similar to: configuring a package for own personal needs

Hi All, I get the message failed with message ‘Dimension out of range’ when using cuhre in package R2Cuba. Does anyone know what this mean? Or would I need to email the package author? The funny thing is it does give a result and comparing it to "adaptIntegrate" in package cubature, the two numbers are very close. Thanks, Sachin [[alternative HTML version deleted]]

Hi all, What would be an efficient way to match rows of a matrix to a vector? ex: m<-matrix(1:9, nrow=3) m [,1] [,2] [,3] [1,] 1 4 7 [2,] 2 5 8 [3,] 3 6 9 ################################# which(m==c(2,5,8)) # I want this to return 2 ###################### Thanks, Sachin [[alternative HTML version deleted]]

Hi all, It seems like I cannot use normal 'if' for data frames. What would be the best way to do the following. if data$col1='high' data$col2='H' else if data$col1='Neutral' data$col2='N' else if data$col='low' data$col2='L' else #chuch a warning? Note that col2 was not an existing column and was newly assigned for this

I am using a similar dataset to the following: a= c("Fruits", "Adam","errorA", "steve", "errorS", "apples", 17.1,2.22, 3.2,1.1, "oranges", 3.1,2.55, 18.1,3.2 ) a_table=data.matrix(t(matrix(a,nrow=5))) I would like to plus minus every second column starting from errorA (using xtable/ hmisc) example output (ignoring

Hi all, I want to do the following: data<-data.frame(col1=c(1,2,3,4,5)) getcol2<-function(data){ data$col2[data$col1<=2]="L" } getcol2(data) Unfortunately in the above col2 does not appear in the final data. So how would you pass this by reference such that you would get it back? Thanks, Sachin [[alternative HTML version deleted]]

Hi all, I have a feeling the most efficient way to do the following is to use apply, but I'm still wrapping my head around the function. k=matrix(1:6,nrow=3) div=1:2 Questions is how do I get R to divide the first column by 1 (div[1]) and the second column by 2 (div[2]) k/div treats k as a vector and does the following (not what I want) >k/div [,1] [,2] [1,] 1 2 [2,]

Hi all, I have a data frame that has the columns OFB1, OFB2, OFB3,... OFB10. How do I select the first 8 columns efficiently without typing each and every one of them. i.e. I want something like: a<-data.frame(initial_data$OFB1-10) #i know this is wrong, what would be the correct syntax? Thanks, Sachin [[alternative HTML version deleted]]

Hi all, Is there a way to get cran R to set the working directory to be wherever the source file is? Each time I work on a project on different computers I keep having to set the working directory which is getting quite annoying. Thanks, Sachin [[alternative HTML version deleted]]

my=function(x){ len=1 for(i in 1:len){ y[i]=x[i] } g=1 w=NULL t=NULL for(i in 1:len)w[i]=x[i+len] for(i in 1:len)t[i]=x[i+2*len] for(i in 1:len)g=g*dnorm(y[i])*dnorm(w[i])*dnorm(z[i]) return(g) } cuhre(6,1,my,rep(-100,6),rep(100,6)) Error in crff(match.call(), integrand, "cuhre", libargs, ...) : Additional argument not expected in the integrand function function change to

Hi All, I have written a function (say) called foo, saved in a file called foo.R. Just going by Matlab syntax I usually just change my folder path and therefore can call it at will. When it comes to R, how is the usual way of calling/loading it? because R doesnt seem to automatically find the function from a folder (which might be stupid to attempt in the first place). Thanks, Sachin

Hi All, I want to do something along the lines of: for (i in 1:n){ for (j in 1:n){ A[i,j]<-myfunc(x[i], x[j]) } } The question is what would be the most efficient way of doing this. Would using functions such as sapply be more efficient that using a for loop? Note that n can be a few thousand. Thus atleast a 1000x1000 matrix. Thanks, Sachin [[alternative HTML version

Hi all, Suppose I have a data frame with mixed content (name age and address). a<-"Name: John Smith Age: 35 Address: 32, street, sub, something" b<-data.frame(a) 1. The question is I want to extract the name age and address separately from this data frame (containing potentially more people). 2. Also just incase I have to deal with it how would the syntax change if I had

Hi All, I have the following problem (read the commented bit below): a<-matrix(1:9,nrow=3) a [,1] [,2] [,3] [1,] 1 4 7 [2,] 2 5 8 [3,] 3 6 9 div<-1:3 apply(a,2,function(x)x/div) ##want to divide each column by div- instead each row is divided## [,1] [,2] [,3] [1,] 1 4.0 7 [2,] 1 2.5 4 [3,] 1 2.0 3

If this is posted elsewhere I cannot find it. I need to perform multiple integration where some of the variables are in the bounds of the other variables. I was trying to use R2Cuba function but cannot set the upper and lower bounds. My code so far is : int <- function(y){ u2 = y[1] z2 = y[2] u1 =y[3] z1 = y[4] ff <- u1*(z1-u1)*u2*(z2-u2)*exp(-0.027*(12-z2)) return(ff) }

Hi all, I have two dataframes. The first (A) contains all the stock prices for today including today. So the first column is the stock Symbol and the second column is the stock price. The second (B) is the symbol list in the top 100 stocks. I want to pick out from dataframe A only the rows containing the symbols from B. i.e. something like: prices <- A[A[,1]==B,2] is there any

Hi All, So I figured out how to do multiple outputs, but whats the best/ recommended way of assigning them. f<-function{a=1; b=1; list(a,b)} I want to be able to say assign into a and b straight away rather that doing a=f()[[1]] and b=f()[[2]]. It would be best if I can get around this without having to assigning this to a third dummy variable. Thanks, Sachin [[alternative HTML version

Hi All, I want to have an array/ matrix that looks this x<- 0 0 1 1 1 3 5 4 4 7 -1 8 9 10 6 I hope this makes sense. So basically if I want x[1,3] it will access 0 and similarly x[4,2], -1. Thanks in advance, Sachin p.s. sorry about the corporate notice. --- Please consider the environment before printing this email --- Allianz - Best General Insurance Company of the Year 2010*

Hi All, I am trying to convert the following piece of matlab code to R: XX1 = sum(w(:,ones(1,N1)).*X1.*X1,1); #square the elements of X1, weight it and repeat this vector N1 times XX2 = sum(w(:,ones(1,N2)).*X2.*X2,1); #square the elements of X2, weigh and repeat this vector N2 times X1X2 = (w(:,ones(1,N1)).*X1)'*X2; #get the weighted 'covariance'

Hi all, Say I have the following data: a<-data.frame(col1=c(rep("a",5),rep("b",7)),col2=runif(12)) a_aov<-aov(a$col2~a$col1) summary(aov) Note that there are 5 observations for a and 7 for b, thus is unbalanced. What would be the correct way of doing anova for this set? Thanks, Sachin [[alternative HTML version deleted]]

Hi all, Is it possible to get the latitude and longitude of the location of a tweet? If I do tweets<- searchTwitter("#obama", n=200) #get tweets df <- twListToDF(tweets) #converts to data frame for ease of viewing it does not seem to be getting the location of where that tweet was posted. >From what I read from the twitter API this is possible?