has.substr: Returns for every element of str whether there is a match...

In skranz/stringtools: Tools for working with strings in R

Returns for every element of str whether there is a match with pattern works similar than grepl

Description

Returns for every element of str whether there is a match with pattern works similar than grepl

Usage

has.substr(str, pattern, fixed = TRUE, perl = FALSE, ignore = NULL)

Examples

## Not run: 
  str = c("12347382709")
  pattern = c("a","4","56","34","766","b")
  has.substr(str,pattern)  

## End(Not run)

skranz/stringtools documentation built on May 11, 2022, 4:48 a.m.