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  1. df <- structure(list(ADDRESS = c("401 300th st", "40120 30 street", 
  2. "30 401 plz", "401 30th st", "401 e gibbsborro rd, 305", "401 e 30th street, shelter", 
  3. "401 east 30st", "401 e30th street, 3", "77-02 30th ave, 3rd fl", 
  4. "401 e30 st."), Outcome = c(FALSE, FALSE, TRUE, TRUE, FALSE, 
  5. TRUE, TRUE, TRUE, FALSE, TRUE)), class = "data.frame", row.names = c(NA, 
  6. -10L))
  7. grepl("^(?=.*\\b401\\b)(?=.*?\\be?30(?:th|st)?\\b)", df$ADDRESS, perl=TRUE)
  8.  
Success #stdin #stdout 0.22s 38920KB
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 [1] FALSE FALSE  TRUE  TRUE FALSE  TRUE  TRUE  TRUE FALSE  TRUE
https://ideone.com/o3cO72
language:
R (R 3.5.2)
created:
3 years ago
visibility:
secret

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