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

Description Usage Examples

View source: R/stringtools.R

Description

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

Usage

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

Examples

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## 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 March 20, 2018, 3:23 p.m.