从字符串中删除字符串中的每个单词

Question

So I'm in a situation where I want to remove dynamically generated text from another piece of text..

  var string1 = "Hello, my name is irrelevant";
  var string2 = "Hello irrelevant, my name is blank";

So what I want is that when string1 is "removed" from string2 the result is blank because string1 had all the same words as string2 but even though they were in a different pattern they still get removed.

Is that possible? If I have a very large string would it be slow?

Answer 1

Generate a regular expression from string1 and use that to eliminate words from string2 :

var reg = new RegExp('\\b('+string1.match(/\w+/g).join('|')+')\\b','gi');
var unmatched_words = string2.replace(reg,'').match(/\w+/g);
alert(unmatched_words);

This should produce an array of every word that appears in string2 but not in string1 .

EDIT: As Guffa pointed out, the regular expression will also match partial words unless it is bookended with word boundary match strings ( \\b ). I've updated the answer to include this change.

 var string1 = "Hello, my name is irrelevant"; var string2 = "Hello irrelevant, my name is blank"; var reg = new RegExp('\\\\b('+string1.match(/\\w+/g).join('|')+')\\\\b','gi'); var unmatched_words = string2.replace(reg,'').match(/\\w+/g); alert(unmatched_words); 

Answer 2

You can make a map of the words from the first string, then loop through the words in the second string and check which ones are not in the map. Example:

 var string1 = "Hello, my name is irrelevant"; var string2 = "Hello irrelevant, my name is blank"; var map = {}; var words1 = string1.replace(/,/g, '').split(' '); for (var i = 0; i < words1.length; i++) { map[words1[i]] = 1; } var result = []; var words2 = string2.replace(/,/g, '').split(' '); for (var i = 0; i < words2.length; i++) { if (!(words2[i] in map)) { result.push(words2[i]); } } document.write(JSON.stringify(result)); 

This will be an O(n) solution, ie the performance will be linear (or close to linear) to the length of the strings, so that will perform very well for large strings.