populate column in r based on regex in another column

Question

I was doing some data wrangling of wine reviews in R and could not find an elegant way to do what I wanted.
My goal was to look at the title column of the wine reviews which usually contain the year of the wine and put that year in a different column. Kernal: https://www.kaggle.com/kieroneil/data-wrangling-wine-reviews-in-r

This is the code that did what I wanted but I'm hoping someone can show me a better way:

# Create the year columns and assign an arbitrary value.
library(tidyverse)
wine_04$year <- 1900
year_2000 <- unlist(str_detect(wine_04$title, "2000"))
year_2001 <- unlist(str_detect(wine_04$title, "2001"))
year_2002 <- unlist(str_detect(wine_04$title, "2002"))
year_2003 <- unlist(str_detect(wine_04$title, "2003"))
year_2004 <- unlist(str_detect(wine_04$title, "2004"))
year_2005 <- unlist(str_detect(wine_04$title, "2005"))
year_2006 <- unlist(str_detect(wine_04$title, "2006"))
year_2007 <- unlist(str_detect(wine_04$title, "2007"))
year_2008 <- unlist(str_detect(wine_04$title, "2008"))
year_2009 <- unlist(str_detect(wine_04$title, "2009"))
year_2010 <- unlist(str_detect(wine_04$title, "2010"))
year_2011 <- unlist(str_detect(wine_04$title, "2011"))
year_2012 <- unlist(str_detect(wine_04$title, "2012"))
year_2013 <- unlist(str_detect(wine_04$title, "2013"))
year_2014 <- unlist(str_detect(wine_04$title, "2014"))
year_2015 <- unlist(str_detect(wine_04$title, "2015"))
year_2016 <- unlist(str_detect(wine_04$title, "2016"))
year_2017 <- unlist(str_detect(wine_04$title, "2017"))

wine_04[year_2000 == TRUE, 15] <- 2000
wine_04[year_2001 == TRUE, 15] <- 2001
wine_04[year_2002 == TRUE, 15] <- 2002
wine_04[year_2003 == TRUE, 15] <- 2003
wine_04[year_2004 == TRUE, 15] <- 2004
wine_04[year_2005 == TRUE, 15] <- 2005
wine_04[year_2006 == TRUE, 15] <- 2006
wine_04[year_2007 == TRUE, 15] <- 2007
wine_04[year_2008 == TRUE, 15] <- 2008
wine_04[year_2009 == TRUE, 15] <- 2009
wine_04[year_2010 == TRUE, 15] <- 2010
wine_04[year_2011 == TRUE, 15] <- 2011
wine_04[year_2012 == TRUE, 15] <- 2012
wine_04[year_2013 == TRUE, 15] <- 2013
wine_04[year_2014 == TRUE, 15] <- 2014
wine_04[year_2015 == TRUE, 15] <- 2015
wine_04[year_2016 == TRUE, 15] <- 2016
wine_04[year_2017 == TRUE, 15] <- 2017

Thanks for the help.

You would like wine_04 to contain a column labelled year containing the year of the wine? Would this work? wine_04$year <- sub('.*(\\d{4}).*', '\\1', wine_04$title) — C. Braun
– C. Braun, Commented Feb 26, 2018 at 20:39

YOLO · Accepted Answer · 2018-02-26 20:40:37Z

5

This works.

library(stringr)
df <- data.table(text = c('the wine is from 1898','the wine is since 2008'))
df[,year := str_extract(string = text, pattern = '\\d{4}')]

                     text year
1:  the wine is from 1898 1898
2: the wine is since 2008 2008

answered Feb 26, 2018 at 20:40

YOLO

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1 Comment

Kier Over a year ago

Worked perfect. Thanks Manish.

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