search for: thisorthat

...ns'', ignore => ''.svn'', recurse => true, replace => true, owner => root, group => root, purge => false, mode => 755, } It happens every night around 9pm (4am UDT). EVERY night. Begin forwarded message: > From: "report@thisorthat.example.com" <report@thisorthat.example.com> > Date: October 25, 2011 9:02:50 PM PDT > To: puppet-info <puppet-info@example.com> > Subject: Puppet Report for thisorthat.example.com > > Wed Oct 26 04:02:35 +0000 2011 /File[/usr/lib64/nagios/plugins/check_clamd]/conte...

...]] <- "Other" educ2 = factor(educ2) levels(educ2) The above code is how I regrouped the variable. How can I regroup it so that it's levels are from lowest to highest. What if they're numeric values" -- *Abraham Mathew Statistical Analyst This or That Media, Inc. abraham@thisorthat.com 720-648-0108 @abmathewks www.amathew.com * [[alternative HTML version deleted]]

Hello Folks, I'm working on trying to scrape my first web site and ran into a issue because I'm really don't know anything about regular expressions in R. library(XML) library(RCurl) site <- "http://thisorthat.com/leader/month" site.doc <- htmlParse(site, ?, xmlValue) At the ?, I realize that I need to insert a regex command which will decipher the contents of the web page...right? First, I'm not sure if the contents of the site would be considered a table and I'm also not sure how to...

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",

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]]

Let's say that I'm trying to write a functions that will allow me to automate a process where I examine all possible combinations of various string groupings. Each time I run the one function, I want to include the new values to the end of a data frame. The data frame will basically be one column with a lot of rows. roots <- c("car insurance", "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'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 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 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 have a data frame in R with the following values. cars autocar cars info what is that donna drive car telephone i need car... I want to select all values which contain 'car', values with three words, and those keywords with car that contain three words. The first part is done with : sqldf("SELECT Keyword FROM dat WHERE Keyword like '%car%'") However, I'm not

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 have a data frame that looks as follows. df <- data.frame(city=c("Houston", "Houston", "El Paso", "Waco", Houston", "Plano", "Plano") What I want to do is get a list of the city values. Currently, when I run df$city, I get all the values. I just want to know the four cities that appear. So instead of: "Houston",

Lets say I have the following data frame. df = data.frame(word = c("David", "James", "Sara", "Jamie", "Jon")) df I was trying to place brackets , [ ] , around each string. I'll be exporting it with write.table and quotes=FALSE, so it will eventually look like: [David] [James] [Sara] .... Can anyone help with this task? I'd like to

I have a number of strings in a vector, and want the output to be seperated by commas. > t [1] "35004" "35005" "35006" "35007" "35010" "35014" "35016" So I want want it to look like: "35004", 35005", "35006", "35007",... Can anyone help? I initially thought strsplit would be the

This is a very basic question, so please bear with me. I've been learning about AB Testing, which is largely used in internet marketing to examine the effectiveness of certain aspects of ads, websites, etc. Here's a couple links to people who want to know more about AB Testing:

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"),