create dummy variable from string variable

Question

I am trying to create a dummy variable from a column variable from an existing data set. The variable I am interested in is a title in this format:

CHEMICALS - Commission Delegated Directive (EU) 2015/863 of 31 March 2015 amending Annex II to Directive 2011/65/EU of the European Parliament and of the Council as regards the list of restricted substances (Text with EEA relevance)

or

Commission Implementing Directive (EU) 2015/2392...

I want to create a dummy variable indicating that the Title is either implementing or delegated. In other words, when the word "delegated" is in my title variable, this will be labeled 1 and everything else will be labeled 0.

Can anyone help me with this? It is very appreciated. So far, I have used this code:

infringements$delegated <- ifelse(infringements$Title=="Delegated", 1, 0)
table(infringements$delegated, infringements$Title)  
summary(infringements$delegated)

When I run the code, I get 0 matches, even though I know that there are 41 matches.

You can use str_detect() from the package stringr instead of == because == will only check if your string is equal to "Delegated" and what you're trying to do is to detect a pattern in your title. — Mbr Mbr
– Mbr Mbr, Commented Mar 23, 2017 at 10:00
Use grepl, i.e. as.integer(grepl('Delegated', infringements$Title)) — Sotos
– Sotos, Commented Mar 23, 2017 at 10:06
great, thank you! I used the grepl suggestion because I have already been working with the grep package, and this worked. — reveraert
– reveraert, Commented Mar 23, 2017 at 10:28

akrun · Accepted Answer · 2017-03-23 10:19:26Z

4

We can do

+(grepl('Delegated', infringements$Title))

answered Mar 23, 2017 at 10:19

akrun

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2 Comments

Aleksandr Over a year ago

What is "+" sign is stands for?

akrun Over a year ago

@AleksandrVoitov It is a hacky way to coerce the logical to binary. Better approach is as.integer(grepl(...

Mbr Mbr · Accepted Answer · 2017-03-23 10:09:59Z

3

Use str_detect() from the package stringr

library(stringr)

as.integer(str_detect(infringements$Title,"Delegated"))

answered Mar 23, 2017 at 10:09

Mbr Mbr

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Comments

Aleksandr · Accepted Answer · 2017-03-23 10:24:42Z

1

infringements = data.frame(lapply(data.frame(Title=c("CHEMICALS - Commission Delegated Directive (EU) 2015/863 of 31 March 2015 amending Annex II to Directive 2011/65/EU of the European Parliament and of the Council as regards the list of restricted substances (Text with EEA relevance)","No Text","Text3Delegated")), as.character), stringsAsFactors=FALSE)
infringements$delegated = lapply(infringements$Title, function(x) ifelse(length(grep("Delegated", x))!=0, 1, 0))

edited Mar 23, 2017 at 10:24

answered Mar 23, 2017 at 10:19

Aleksandr

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