similar to: Adding values to the end of a data frame

I have a really long functions, and at the end of the function, I am using a if statement to tag certain keywords based on whether they have certain values contained in them. However, the if statement doesn't seem to work. When I had split up the commands into various functions, it worked fine, but I'm not sure what going on now that it's combined into a single function. myfunc

I'm writing a function and keep getting the following error message. myfunc <- function(lst) { lst <- list(roots = c("car insurance", "auto insurance"), roots2 = c("insurance"), prefix = c("cheap", "budget"), prefix2 = c("low cost"), suffix = c("quote", "quotes"), suffix2 = c("rate",

I'm trying to run a function inside a function but get an error message. lst <- list(roots = c("car insurance", "auto insurance"), roots2 = c("insurance"), prefix = c("cheap", "budget"), prefix2 = c("low cost"), suffix = c("quote", "quotes"), suffix2 = c("rate", "rates"), suffix3 =

I have a repetative task in R and i'm trying to find a more efficient way to perform the following task. lst <- list(roots = c("car insurance", "auto insurance"), roots2 = c("insurance"), prefix = c("cheap", "budget"), prefix2 = c("low cost"), suffix = c("quote", "quotes"),

I have the following data: prefix <- c("cheap", "budget") roots <- c("car insurance", "auto insurance") suffix <- c("quote", "quotes") prefix2 <- c("cheap", "budget") roots2 <- c("car insurance", "auto insurance") roots3 <- c("car insurance", "auto

I'm working with some data, and am trying to generate it in the following format. state city zipcode I like pizza 0 0 0 I live in Denver 0 1 0 All the fun stuff is in Alaska 1 0 0 he lives in 66062

I'm running a linear model in R using the car package. I have a variable education, which i have recoded and regrouped to my wishes. However, R seems to place each element of that variable in alphabetical order. When I am running the model, don't I need the model order from lowest to highest to make an inference that a one unit change in one variable produced a one unit change in

So I have the following data frame and I want to know how I can remove all "NA" values from each string, and also remove all "|" values from the START of the string. So they should something like "auto|insurance" or "auto|insurance|quote" one = data.frame(keyword=c("|auto", "NA|auto|insurance|quote", "NA|auto|insurance",

I'm trying to find the total number of letters in a row of a data frame. Let's say I have the following data frame. f1 <- data.frame(keyword=c("I live in Denver", I live in Kansas City, MO", "Pizza is good")) The following function gives me the number of characters in each string. So for "I live in Denver", I get 1, 4, 2, and 6. However, I want to

I didn't learn about data tables until recently. (They're never covered in any intro R books). In any case, I'm not sure what (if any) is the difference between a data frame and a data table. Can anyone provide a brief explanation? Is one preferred over another or is it just dependent on the task at hand? Thanks, Abraham M. [[alternative HTML version deleted]]

Villmow, excuse me about the previous reply. :$ So, I'm thinking about OpenGL, or other generic graphical libraries. 2011/8/26 Villmow, Micah <Micah.Villmow at amd.com> > In what context? Also, please have all replies go to the mailing list and > not to the contributor directly.**** > > ** ** > > Thanks,**** > > Micah**** > > ** ** > > *From:*

I'm running R 2.13 on Ubuntu 10.10 I have a data set which is comprised of character strings. site = readLines('http://www.census.gov/tiger/tms/gazetteer/zips.txt') dat <- c("01, 35004, AL, ACMAR, 86.51557, 33.584132, 6055, 0.001499") dat I want to loop through the data and construct a data frame with the zip code, state abbreviation, and city name in seperate columns.

I'm trying to develop a stacked bar plot in R with ggplot2. My data: conv = c(10, 4.76, 17.14, 25, 26.47, 37.5, 20.83, 25.53, 32.5, 16.7, 27.33) click = c(20, 42, 35, 28, 34, 48, 48, 47, 40, 30, 30) date = c("July 7", "July 8", "July 9", "July 10", "July 11", "July 12", "July 13", "July 14", "July 15",

Hi all, I would like to factorize the entries in a dataframe given some groupings. E.g: mydf = data.frame( a = rnorm(100,10), b = rnorm(100,10), c = rgamma(100, 1, scale=1)) group = hist(mydf$c, breaks="FD") group$breaks The idea is to create a factor "mydf$d" with levels corresponding to the ranges in group$breaks. There must be an easy way to do this that I

On converting character variables to ordered factors, regression result has strange names. Is it possible to obtain same variable names with and without intercept? Thanks, Naresh mydf <- data.frame(date = seq.Date(as.Date("2024-01-01"), as.Date("2024-03-31"), by = 1)) mydf[, "wday"] <- weekdays(mydf$date, abbreviate = TRUE) mydf.work <- subset(mydf, !(wday

Hello! I have a data frame with dates. I need to create a new "month" that starts on the 20th of each month - because I'll need to aggregate my data later by that "shifted" month. I wrote the code below and it works. However, I was wondering if there is some ready-made function in some package - that makes it easier/more elegant? Thanks a lot! # Example data:

Hi, as you come to speak of it, i have implemented an OpenCL-Backend for LLVM as part of my bachelor thesis (and for GLSlang as well, see http://www.cdl.uni-saarland.de/publications/theses/moll_bsc.pdf ). However, the code is currently unreleased. But that could be arranged, if you are interested in using it. Regards, Simon Am Freitag, den 26.08.2011, 20:11 -0500 schrieb llvmdev-request at

Hi All, I've been successfully using the with function for analyses and the transform function for multiple transformations. Then I thought, why not use "with" for both? I ran into problems & couldn't figure them out from help files or books. So I created a simplified version of what I'm doing: rm( list=ls() ) x1<-c(1,3,3) x2<-c(3,2,1) x3<-c(2,5,2)

Hi all, Can anyone explain why the following use of the subset() function produces a different outcome than the use of the "[" extractor? The subset() function as used in density(subset(mydf, ht >= 150.0 & wt <= 150.0, select = c(age))) appears to me from documentation to be equivalent to density(mydf[mydf$ht >= 150.0 & mydf$wt <= 150.0, "age"])

Dear all, I wanted to make a two-way-table of two variables with a counting variable stored in another column of a dataframe. In version 1.9.1, the behavior is as expected as shown in the simplified example code. > sex <- rep(c("F", "M"), 5) > income <- c(rep("low", 5), rep("high", 5)) > count <- 1:10 > mydf <-