Create column based on presence of string pattern and ifelse

Question

I would like to fill in a new column with one of two values if a pattern is matched.

Here is my data frame:

df <- structure(list(loc_01 = c("ap         


        

Answer 1

To check if a string contains a certain substring, you can't use == because it performs an exact matching (i.e. returns true only if the string is exactly "non").
You could use for example grepl function (belonging to grep family of functions) that performs a pattern matching:

df$loc01 <- ifelse(grepl("non",df$loc_01),'outside','inside')

Result :

> df
     loc_01 loc01_land   loc01
1      apis  165730500  inside
2      indu   62101800  inside
3      isro  540687600  inside
4      miss  161140500  inside
5  non_apis 1694590200 outside
6  non_indu 1459707300 outside
7  non_isro 1025051400 outside
8  non_miss 1419866100 outside
9  non_piro 2037064500 outside
10 non_sacn 2204629200 outside
11 non_slbe 1918840500 outside
12 non_voya  886299300 outside
13     piro  264726000  inside
14     sacn  321003900  inside
15     slbe  241292700  inside
16     voya  530532000  inside

Answer 2

You only need one line of code:

library(dplyr)
library(stringr)


df %>% 
  mutate(loc01 = if_else(str_starts(loc_01, "non_"), "outside", "inside"))

For using more complex regex-pattern you can use str_detect instead str_starts:

df %>% 
  mutate(loc01 = if_else(str_detect(loc_01, "^(non_)"), "outside", "inside"))

Output:

   loc_01   loc01_land loc01  
   <chr>         <dbl> <chr>  
 1 apis      165730500 inside 
 2 indu       62101800 inside 
 3 isro      540687600 inside 
 4 miss      161140500 inside 
 5 non_apis 1694590200 outside
 6 non_indu 1459707300 outside
 7 non_isro 1025051400 outside
 8 non_miss 1419866100 outside
 9 non_piro 2037064500 outside