similar to: splitting and saving a large dataframe

Hi there, From the vector X of integers, X = c(11999, 122000, 81997) I would like to make these two vectors: Z= c(1999, 2000, 1997) Y =c(1 , 12 , 8) That is, each entry of vector Z receives the four last digits of each entry of X, and Y receives "the rest". Any suggestions? Thanks in advance, Dimitri [[alternative HTML version deleted]]

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

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

Hello! I have larger and a smaller data frame with 1 factor in each - it's the same factor: large.frame<-data.frame(myfactor=LETTERS[1:10]) small.frame<-data.frame(myfactor=LETTERS[c(9,7,5,3,1)]) levels(large.frame$myfactor) levels(small.frame$myfactor) table(large.frame$myfactor) table(small.frame$myfactor) myfactor has 10 levels in large.frame and 5 levels in small.frame. All 5

Hi folks, I am working with repeated measures data and I ran into issues where the paired t-test results did not match those obtained by employing glht() contrasts on a lme model. While the lme model itself appears to be fine, there seems to be some discrepancy with using glht() on the lme model (unless I am missing something here). I was wondering if someone could help identify the issue. On

Hi, Does tapply change the order when applied on a factor? Below is the code I tried. > mylevels<-c("IN0020020155","IN0019800021","IN0020020064") >

I have a data frame with some factor columns. I want to drop the rows with rare factor values (and remove the factor values from the factors). E.g., frame$MyFactor takes values A 1,000 times, B 2,000 times, C 30 times and D 4 times. I want to remove all rows which assume rare values (<1%), i.e., C and D. i.e., frame <- frame[[! (frame$MyFactor %in% c("A","B"))]] except

Dear R help group, I am teaching myself linear mixed models with missing data since I would like to analyze a stats design with these kind of models. The textbook example is for the procedure "proc MIXED" in SAS, but I would like to know if there is an equivalent in R. This example only includes two time-measurements across subjects (a t-test "with missing values"), but I

Hello, I have a list of data with multiple elements, and each element in the list has multiple variables in it. Here's an example: ### Make the fake data dv <- c(1,3,4,2,2,3,2,5,6,3,4,4,3,5,6) subject <- factor(c("s1","s1","s1","s2","s2","s2","s3","s3","s3",

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))

Dear everybody, I have the following challenge. I have a data set with 2 subgroups, dates (days), and corresponding values (see example code below). Within each subgroup: I need to aggregate (sum) the values by week - for weeks that start on a Monday (for example, 2008-12-29 was a Monday). I find it difficult because I have missing dates in my data - so that sometimes I don't even have the

I have 3 questions (below). Background: I am teaching an introductory statistics course in which we are covering (among other things) repeated measures anova. This time around teaching it, we are using R for all of our computations. We are starting by covering the univariate approach to repeated measures anova. Doing a basic repeated measures anova (univariate approach) using aov() seems

R factors are the natural way to represent factors -- and should be efficient since they use small integers. But in fact, for many (but not all) operations, R factors are considerably slower than integers, or even character strings. This appears to be because whenever a factor vector is subsetted, the entire levels vector is copied. For example: > i1 <- sample(1e4,1e6,replace=T) > c1

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 R-experts! Every once in a while I need to convert a factor to a vector of numeric values. as.numeric(myfactor) of course returns a nice numeric vector of the indexes of the levels which is usually not what I had in mind: > v <- c(25, 3.78, 16.5, 37, 109) > f <- factor(v) > f [1] 25 3.78 16.5 37 109 Levels: 3.78 16.5 25 37 109 > as.numeric(f) [1] 3 1 2 4 5 > What I

Full_Name: jean coursol Version: 1.7.1, 1.8.0 OS: linux & Windows-XP Submission from: (NULL) (129.175.52.7) Binary MS-Windows akima module from CRAN (1.8.0 version) produces wrong results with some data. Installing akima source in linux, with same data: -with gcc-2.95.3 -O2 : give correct results (under R 1.7.1); -with gcc-3.2.3 -O2 : give wrong results (under R-1.7.1 and R-1.8.0); -with

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), +