search for: sachinthaka

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, 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]]

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, 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

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, 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 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, 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, 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 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 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've reproduced the example from Prof. Friedrich Leisch's webpage. When I write sweave("Example-1.Snw") OR sweave("Example-1.Rnw"), (yes, I renamed them). I get the following error: Writing to file example-1.tex Processing code chunks ... 1 : echo term verbatim Error: chunk 1 Error in library(ctest) : there is no package called 'ctest' Also while

Hey all, Suppose a=b^2 for starters. I want to be able to create a graph that displays a initially and if i was to click on 'a' to show 'b' on the chart itself. Does anyone know if this is possible in R? Also as an extension (not necessary as yet) to output the above into a 'html' file. Thanks, Sachin --- Please consider the environment before printing this email ---

Hi All, I'm trying to create labels to plot such that it doesn't show up as scientific notation. So for example how do I get R to show 1e06 as $1,000,000. I was wondering if there was a single function which allows you to do that, the same way that as.Date() allows you to show in date format on a plot. Thanks, Sachin --- Please consider the environment before printing this email ---

Hi R, In the following code my x-axis is formatted in month format. Which Im happy with. The y-axis is what I want to re-format with something else. My question is, is it possible just to switch off the xaxis in plot function (see below). If not how do you get the months to show up as FEB-YYYY, MAR-YYYY and so on, so I could fit a label on x-axis. Thanks, Sachin p.s. sorry about corporate

Hi All, I am trying to run R scripts on a server rather than my own machine. The biggest reason being that the data can be 3GB+; more than my RAM can handle. Anyway is there a way to do this. I am trying to find a SAS alternative. In SAS you can do (keyword) rsubmit; and get things running remotely on a server. Is there something similar on R. Couldn't find a comprehensive answer online.

Hi All, Is it possible to write a program such that it downloads a csv from a given web address? Would be great if this could be done at a particular time during the day as well. Say 9AM monday-friday. Incase you are curious Im just trying to analyse some stocks data. Thanks, Sachin p.s. sorry about corporate notice --- Please consider the environment before printing this email --- Allianz -