I have a huge data frame. One of the columns in the data frame is an email address. In addition, I have a vector with domain extensions (for example: c(".ac",".ad",".ae",".af",".ag",".ai") – a total length of 259 extensions.) I want to filter my data frame to contain records whose email ends with one of the strings… Read More Filtering a tibble in R by list of strings and returning all records that end with the strings in the list

I have a dataset in which my unit of analysis and variable are together in the same column. I would like to split this up into two variables, but not sure how to. Say I have the following data: de <- data.frame(year = c(2010, 2010, 2010, 2010, 2010, 2010, 2010, 2010, 2010, 2012, 2012, 2012,… Read More Take string from variable and split the variable up in two variables r

What is the best method to check to see if a text is in another variable in R? Below is an example: a= ‘homer simpson the third’ b= c(‘homer simpson’,’marge’,’bart’) grepl(a,b) # this is what it returns FALSE FALSE FALSE I am wanting it to return TRUE FALSE FALSE thanks! >Solution : Perhaps, use agrep… Read More check if text vector contains a certain text in R

I have a vector of strings either starting with \n1. or \ntext and I wish to filter all those starting with \n1. Sample: [1] "\n1. Morgenhanen matter" [2] "\n1. Morgenstund har guld" [3] "\nMorgensange for børn be" but I can’t seem to grap those sentences starting with \n1. Here’s where I’m at: grepl("^[‘\\\\’]n1", df$text) but… Read More Grapping \n1 with grepl in r

I want to create a list that contains vectors that display either a 1 or a 0 depending on whether a certain character is present. The data has the following form: list <- list(‘a’, ‘b’, ‘c’, ‘d’, ‘e’) col1 <- c(‘ta’, ‘ta’, ‘tb’, ‘tb’, ‘tb’, ‘tc’, ‘td’, ‘te’) What I want is a list for… Read More How do I loop over a list of strings in R?

I have example data as follows: library(data.table) dat <- fread("q1 q2 …1 ..2 q3..1 ..1 NA response other else response other 1 4 NA NA 1 NA") I wanted to filter out all columns that are automatically named when reading in an Excel file with missing column names, which have names like ..x. I thought… Read More Getting grepl to return TRUE only if there is a match with the full string

I’m trying to extract the character string "segments" separated by an "_" and make them their own columns. I know how to combine many columns to a single one, but I’m curious how going backwards works (single column to many). My data: structure(list(REF = c("M120", "M70", "M100", "M94", "M74", "M54", "M99", "M09", "M111", "M17"), SEAS… Read More Extract and create columns from character strings separated by "_" in R