Add extra columns based on two conditions, with the possibility for NA values-CodePudding

I have a dataset that kind of look like this:

NAME         JOB TITLE           ROLE DESCRIPTION
Mark         Cyber specialist    N/A
Peter        Manager             Manager with focus on cyber
Steven       CFO                 N/A
Lucas        VP                  General manager technology safety

Now I want to create an extra column which indicates if the job of the person is cyber related. But there have to be a 1 (cyber related) at Mark, Peter and Lucas. So, there should be looked at 2 columns instead of one. If the condition is only based on 1 column I know that the code should look like this:

pattern <- paste(c("cyber", "Cyber", "technology", "Technology", "computer", "Computer"), collapse = "|")
df <- df %>% 
  mutate(`Cyber Job` = ifelse(str_detect('column', pattern), 1, 0))

But i don't know how the code should look like if the condition is based on two columns

CodePudding user response：

You can use the following code which checks whether there is any string that matches the pattern using grepl, across the columns:

library(dplyr)
df %>% 
  rowwise() %>%
  mutate(`Cyber Job` = any(grepl(pattern,across(everything())))*1)

Output:

# A tibble: 4 × 4
# Rowwise: 
  NAME   JOB.TITLE        ROLE.DESCRIPTION                  `Cyber Job`
  <chr>  <chr>            <chr>                                   <dbl>
1 Mark   Cyber specialist N/A                                         1
2 Peter  Manager          Manager with focus on cyber                 1
3 Steven CFO              N/A                                         0
4 Lucas  VP               General manager technology safety           1

Data

df <- data.frame(NAME = c("Mark", "Peter", "Steven", "Lucas"),
                 `JOB TITLE` = c("Cyber specialist", "Manager", "CFO", "VP"),
                 `ROLE DESCRIPTION` = c("N/A", "Manager with focus on cyber", "N/A", "General manager technology safety"))

CodePudding user response：

We may also use if_any

library(dplyr)
library(stringr)
df %>%
  mutate(`Cyber Job` =  (if_any(everything(), str_detect, pattern = pattern)))
    NAME        JOB.TITLE                  ROLE.DESCRIPTION Cyber Job
1   Mark Cyber specialist                               N/A         1
2  Peter          Manager       Manager with focus on cyber         1
3 Steven              CFO                               N/A         0
4  Lucas               VP General manager technology safety         1