similar to: Select rows from Data Frame with conditions

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, I am using Ubuntu and I am having quite a bit of difficulty trying to get R to use the system proxy settings. Any suggestions? Thanks, Sachin

Hi all, I am trying to install this NBPSeq package, but I'm now getting the error shown below with ANY package that I try to install. installation of package ‘NBPSeq’ had non-zero exit status This started happening ever since I tried to install a package that I downloaded from a personal website (for some reason only worked on my windows partition). I'm using Ubuntu. Any thoughts?

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 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

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]]

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 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, 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, There is a function in package "R2Cuba" called Cuhre that I need to use. It keeps spitting out a new-line which I really dont want it to do. So I was wondering what is the best way of configuring the package. I tried copying and pasting the code into Cuhre2 and getting rid of the newline command BUT that didn't work since it seems to have some private functions which I do

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, 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 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?

Hi all, I have a matrix that has many NaN values. As soon as one of the columns has a missing (NaN) value the covariance estimation gets thrown off. Is there a robust way to do this? Thanks, Sachin a=array(rnorm(9),dim=c(3,3))> a [,1] [,2] [,3] [1,] -0.79418236 0.7813952 0.855881 [2,] -1.65347906 -1.9462446 -0.376325 [3,] -0.03144987 0.6756862 -1.879801>