similar to: spliting an integer

Hi, Suppose I I wish to run lm( y ~ x + z + log(w) ) where w assumes non-negative values. A problem arises when w=0, as log(0) = -Inf, and R doesn't accept that (as it "accepts" NA). Is there a way to tell R to do with -Inf the same it does with NA, i.e, to ignore it? ( Otherwise I have to do something like w[w==0] <- NA which doesn't hurt, but might be a bit

Hi, x and y are (numeric) vectors. I wonder if one of the following is more efficient than the other: x%*%y or sum(x*y) ? Thanks, Dimitri Szerman

Hi, I searched for this in the mailing list, but found no results. I have a large dataframe ( dim(mydata)= 1297059 16, object.size(mydata= 145280576) ) , and I want to perform some calculations which can be done by a factor's levels, say, mydata$myfactor. So what I want is to split this dataframe into nlevels(mydata$myfactor) = 80 levels. But I must do this efficiently, that is, I

Dear all, I have a list with three (named) numeric vectors: > lst = list(a=c(A=1,B=8) , b=c(A=2,B=3,C=0), c=c(B=2,D=0) ) > lst $a A B 1 8 $b A B C 2 3 0 $c B D 2 0 Now, I'd love to use this list to create the following data frame: > dtf = data.frame(a=c(A=1,B=8,C=NA,D=NA), + b=c(A=2,B=3,C=0,D=NA), + c=c(A=NA,B=2,C=NA,D=0) ) > dtf a b

Hi, I have two data frames. One is like > dtf = data.frame(y=c(rep(2002,4), rep(2003,5)), + m=c(9:12, 1:5), + def=c(.74,.75,.76,.78,.80,.82,.85,.85,.87)) and the other dtf2 = data.frame(y=rep( c(2002,2003),20), m=c(trunc(runif(20,1,5)),trunc(runif(20,9,12))), inc=rnorm(40,mean=300,sd=150) ) What I want is to divide

Dear R-mates, # Here's what I am trying to do. I have a dataset like this: id = c(rep(1,8), rep(2,8)) dur1 <- c( 17,18,19,18,24,19,24,24 ) est1 <- c( rep(1,5), rep(2,3) ) dur2 <- c(1,1,3,4,8,12,13,14) est2 <- rep(1,8) mydata = data.frame(id, estat=c(est1, est2), durat=c(dur1, dur2)) # I want to one have this: id = c(rep(1,8), rep(2,8))

Hi, I wonder if someone has already figured out a way of making summary(mylm) # where mylm is an object of the class lm() to print the "(Intercept)" at the last line, rather than the first line of the output. I don't know about, say, biostatistics, but in economics the intercept is usually the least interesting of the parameters of a regression model. That's why, say, Stata

Hi there, I am trying to compute Gini coefficients for vectors containing income classes. The data I possess look loke this: yit <- c(135, 164, 234, 369) piit <- c(367, 884, 341, 74 ) where yit is the vector of income classes, and fit is the vector of associated frequencies.(This data is from Rustichini, Ichino and Checci (Journal of Public Economics, 1999) ). In ineq pacakge, Gini( )

Dear all, A few months ago, I asked for your help on the following problem: I have a list with three (named) numeric vectors: > lst = list(a=c(A=1,B=8) , b=c(A=2,B=3,C=0), c=c(B=2,D=0) ) > lst $a A B 1 8 $b A B C 2 3 0 $c B D 2 0 Now, I'd love to use this list to create the following data frame: > dtf = data.frame(a=c(A=1,B=8,C=NA,D=NA), +

Hello, This is what I am trying to do: I wrote a little function that takes addresses (coordinates) as input, and returns the road distance between every two points using Google Maps. Catch is, there are 2000 addresses, so I have to get around 2x10^6 addresses. On my first go, this is what I did: ######################################### getRoadDist = function(X,complete=F){ # X must be a

I am trying to rasterize a SpatialPolygon object by a RasterBrick object. The documentation of the raster::rasterize function explicitly says this is allowed. Here's what I am doing # load the raster package library("raster") # create a raster brick object using the example from the brick function documentation b <- brick(system.file("external/rlogo.grd",

Hi everyone, I have a vector of strings, each string made up by different number of words. I want to get a new vector which has only the first word of each string in the first vector. I came up with this: str <- c('aaa bbb', 'cc', 'd eee aa', 'mmm o n') str1 <- rep(1, length(str)) for (i in 1:length(str)) { str1[i] <- strsplit(str, "

Dear R users, I have a huge vector that I would like to split into unequal slices. However, the only way I can do this is to create another huge vector to define the groups that are used to split the original vector, e.g. # my vector is this a.vector <- seq(2, by=5, length=100) # indices where I would like to slice my vector cut.values <- c(30, 50, 100, 109, 300, 601, 803) # so I have to

Currently as the first column in a data frame I have string values in the format xx_yy - I want to create a new column with just the substring xx (for each row in turn). Three possible ways to do this might be (1) split the string by '_' using strsplit and paste the first of the resulting variables into a new column, but I have been unable to do this for each row of my data frame in turn

Hello, I was wondering if there is a function in R that imports tables directly from a HTML document. I know there are functions (say, getURL() from {RCurl} ) that download the entire page source, but here I refer to something like google document's function importHTML() (if you don't know this function, go check it, it's very useful). Anyway, if someone of something that does this

Hi there, I am trying to export a regression output to Latex. I am using the xtable function in the xtable library. Doing myfit <- lm(myformula, mydata) print.xtable(xtable(myfit), file="myfile") only returns the estimated coefficients and the correspondent standard erros, t-statiscs and p-values. But I wish to get a bit more, say, the number of observations used in the

Dear all, A few weeks ago, I asked this list why small Stata files became huge R files. Thomas Lumley said it was because "Stata uses single-precision floating point by default and can use 1-byte and 2-byte integers. R uses double precision floating point and four-byte integers." And it seemed I couldn't do anythig about it. Is it true? I mean, isn't there a (more or less

Hi, I am trying to estimate the effects of covariates on the hazard function, rather than on the survival. I know this is actually the same thing. For example, using the survival package, and doing: > myfit <- survreg( Surv(time, event) ~ mymodel ) all I have to do to get the quantities of my interest is > -myfit$coefficients/myfit$scale The standard erros are easily worked out, as

Hi, I'm running a regression using lm(), in which one of the right-hand side variables is factor with many levels (say, 80). I am not intersted in the estimates of the resulting dummies, but I have to include them in my regression equation. So, I don't want the estimates associated with theses dummies to be printed by summary.lm( ). Is there an easy way to do this? Thank you, Dimitri

Hi, I have a new computer with Windows Vista and I am trying to use RWinEdt, which I have always used. I am using R version 2.5. The installation of the RWinEdt library is funny. First, it didn't install at all. Then, I uninstalled/reinstalled both R and WinEdt, downloaded the package again from the CRAN repositary, got some error messages, but RWinEdt initialized. I closed R and WinEdt,