I found a variation on this code elsewhere in StackOverflow. It takes all words from a textarea and converts them into a regular expression. It then tests an array to see if all the words in the regex are contained in the array:
<textarea id="inputtext" type="text"></textarea>
<input id="searchbutton" type="button" value="Click me" />
var links = new Array("taxi","Alpha","runway");
$("#searchbutton").click(function () {
var query = $("#inputtext").val();
var querywords = query.split(',');
for (var i = 0; i < querywords.length; i++) {
var regex = new RegExp('(?=.*\\b' + querywords[i].split(' ').join('\\b)(?=.*\\b') + '\\b)', 'i', 'g');
for (var j = 0; j < links.length; j++) {
if (regex.test(links[j])) {
console.log("Correct");
}
}
}
});
How can I reverse the process so the program returns "true" if the textarea words includes all of the keywords within the array? For example, if the textarea had the sentence "Taxi to the runway via taxiway alpha," and the array named "links" contained the keywords "taxi" "alpha" and "runway", the program would return "true".
That script you have seems to check if any of the words appears somewhere in the array. What you want is the every Array method:
var text = "Taxi to the runway via taxiway alpha",
links = ["taxi", "alpha", "runway"];
console.log( links.every(function(word) {
return new RegExp("\\b"+word+"\\b", "i").test(text);
}) ); // true
The methods provided by other answers are simple, but they could be more efficient.
It's almost always better to use an object as a map to speed up lookups instead of having to search the entiry array everytime.
var words = ['word1', 'word2'],
wordsMap = 'text area content, word1 and word2'.split(/\W+/).reduce(function (obj, word) {
obj[word] = true;
return obj;
}, {}),
areWordsAllContained = words.every(function (word) {
return wordsMap[word.toLowerCase()];
});
console.log(areWordsAllContained); //true
EDIT: I've changed the splitting regex from \s+ to \W+ to make sure that it splits on every non-word characters.
A non-regex way would be:
var arr = ['word1', 'word2'], haystack = textArea.value.toLowerCase().split(/\s+/);
var result = true, i = 0;
for(i=0; i<arr.length; i++) {
if(haystack.indexOf(arr[i].toLowerCase()) === -1) {
result = false;
break;
}
}
Related
I am trying to match all typed phrases(characters) with main array. I have a big array, when user types something i want to search my array and show only matching entries, for example: if user types puzzle then from my array 'my puzzle starts','enter your puzzle' these both should match.
I have created Fiddle here, here is my html code
Search: <input autocomplete="off" type="text" id="search" /><br />
Results: <input type="text" id="results_count" /><br />
<textarea id="results_text" style="width:300px; height:500px"></textarea>
My js code
var string =['1234sud','man in puzzle','rocking roll','1232 -powernap','all_broswers'];
$('#search').keyup(function() {
var search = $(this).val();
var resultsText = $('#results_text');
var resultsCount = $('#results_count');
if (!search) {
resultsText.val('');
resultsCount.val('0');
return;
}
var j = 0, results = [],result;
for (var i=0;i<string.length;i++){
var rx = new RegExp('"([*'+search+'*])"','gi');
if (rx.exec(string[i])) {
results.push(string[i]);
j += 1;
if (j >=100)
break;
}
}
//results=string;
resultsText.val(results);
console.log(results)
resultsCount.val(j);
});
I am not able to write that regular expression which matches types phrases (characters) with array and pushes matching element to result array. Any help will be appreciated thanks in advance.
var string = ['1234sud', 'man in puzzle', 'rocking roll', '1232 -powernap', 'all_broswers'];
$('#search').keyup(function() {
var search = $(this).val();
var resultsText = $('#results_text');
var resultsCount = $('#results_count');
if (!search) {
resultsText.val('');
resultsCount.val('0');
return;
}
var j = 0,
results = [],
result;
for (var i = 0; i < string.length; i++) {
if (string[i].indexOf(search) > -1) {
results.push(string[i]);
j += 1;
if (j >= 100)
break;
}
}
//results=string;
resultsText.val(results);
console.log(results)
resultsCount.val(j);
});
<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.0/jquery.min.js"></script>
Search: <input autocomplete="off" type="text" id="search" /><br /> Results: <input type="text" id="results_count" /><br />
<textarea id="results_text" style="width:300px; height:500px"></textarea>
Use String.prototype.indexOf() function, it's exactly what you want to accomplish.
Normalize the case and use Array#filter()
$('#search').keyup(function() {
var search = $(this).val().toLowerCase(),
// add validation that search isn't empty, return if it is
// create results array
results = string.filter(function(s){
return s.toLowerCase().indexOf(search)>-1;
})
// do something with results
})
Your simplest solution: don't use regex.
For this particular problem, you're best off just looking for the index of the substring, using indexOf.
const words = ['hello', 'jello', 'cake', 'pie'];
const term = 'ello';
const result = words.filter(word => word.indexOf(term) > -1);
console.log(result);
You don't specify, but if you want word only matches (as oppose to partial word matches like I have above), you could pad it with a space.
If you insist on using regex, then all you need is the word itself:
const words = ['hello', 'jello', 'cake', 'pie'];
const term = 'ello';
const result = words.filter(word => (new RegExp(term, 'gi')).test(word));
console.log(result);
You don't need any fancy symbols or groupings or anything.
A regex of /ello/ just means "find this word anywhere in the string". Since there isn't a ^ or $, it isn't limited by start or end, so no need for wildcards.
Also, use test() instead of exec(), since test() resolves to a boolean, it's more accurate for an if statement.
Again, if you want whole words, you could just wrap it in spaces with the start and end having /[^\s]word[\s$]/ which would means "either a space or the start of the phrase" and "either a space or the end of the phrase".
Maybe you want to use a filter() for that task.
var words = ['spray', 'limit', 'elite', 'exuberant', 'destruction', 'present'];
const result = words.filter(word => word.length > 6);
// const result = words.filter(word => word.indexOf(words.search) >= 0);
console.log(result);
// expected output: Array ["exuberant", "destruction", "present"]
https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/Array/filter
Hope this helps.
Use String.match if you need exact match:
if (search.match(string[i])) {
results.push(string[i]);
j += 1;
}
Or dynamically created regex for case of not exact match:
var rxs = '.*' + search + '.*';
var rx = new RegExp(rxs,'gi');
const words = ['spray', 'limit', 'elite', 'exuberant', 'destruction', 'present'];
let checkWord = "elite"
words.forEach(x=>{
checkWord.match(new RegExp(x)
})
How can I implement javascript function to calculate frequency of each word in a given sentence.
this is my code:
function search () {
var data = document.getElementById('txt').value;
var temp = data;
var words = new Array();
words = temp.split(" ");
var uniqueWords = new Array();
var count = new Array();
for (var i = 0; i < words.length; i++) {
//var count=0;
var f = 0;
for (j = 0; j < uniqueWords.length; j++) {
if (words[i] == uniqueWords[j]) {
count[j] = count[j] + 1;
//uniqueWords[j]=words[i];
f = 1;
}
}
if (f == 0) {
count[i] = 1;
uniqueWords[i] = words[i];
}
console.log("count of " + uniqueWords[i] + " - " + count[i]);
}
}
am unable to trace out the problem ..any help is greatly appriciated.
output in this format:
count of is - 1
count of the - 2..
input: this is anil is kum the anil
Here is a JavaScript function to get the frequency of each word in a sentence:
function wordFreq(string) {
var words = string.replace(/[.]/g, '').split(/\s/);
var freqMap = {};
words.forEach(function(w) {
if (!freqMap[w]) {
freqMap[w] = 0;
}
freqMap[w] += 1;
});
return freqMap;
}
It will return a hash of word to word count. So for example, if we run it like so:
console.log(wordFreq("I am the big the big bull."));
> Object {I: 1, am: 1, the: 2, big: 2, bull: 1}
You can iterate over the words with Object.keys(result).sort().forEach(result) {...}. So we could hook that up like so:
var freq = wordFreq("I am the big the big bull.");
Object.keys(freq).sort().forEach(function(word) {
console.log("count of " + word + " is " + freq[word]);
});
Which would output:
count of I is 1
count of am is 1
count of big is 2
count of bull is 1
count of the is 2
JSFiddle: http://jsfiddle.net/ah6wsbs6/
And here is wordFreq function in ES6:
function wordFreq(string) {
return string.replace(/[.]/g, '')
.split(/\s/)
.reduce((map, word) =>
Object.assign(map, {
[word]: (map[word])
? map[word] + 1
: 1,
}),
{}
);
}
JSFiddle: http://jsfiddle.net/r1Lo79us/
I feel you have over-complicated things by having multiple arrays, strings, and engaging in frequent (and hard to follow) context-switching between loops, and nested loops.
Below is the approach I would encourage you to consider taking. I've inlined comments to explain each step along the way. If any of this is unclear, please let me know in the comments and I'll revisit to improve clarity.
(function () {
/* Below is a regular expression that finds alphanumeric characters
Next is a string that could easily be replaced with a reference to a form control
Lastly, we have an array that will hold any words matching our pattern */
var pattern = /\w+/g,
string = "I I am am am yes yes.",
matchedWords = string.match( pattern );
/* The Array.prototype.reduce method assists us in producing a single value from an
array. In this case, we're going to use it to output an object with results. */
var counts = matchedWords.reduce(function ( stats, word ) {
/* `stats` is the object that we'll be building up over time.
`word` is each individual entry in the `matchedWords` array */
if ( stats.hasOwnProperty( word ) ) {
/* `stats` already has an entry for the current `word`.
As a result, let's increment the count for that `word`. */
stats[ word ] = stats[ word ] + 1;
} else {
/* `stats` does not yet have an entry for the current `word`.
As a result, let's add a new entry, and set count to 1. */
stats[ word ] = 1;
}
/* Because we are building up `stats` over numerous iterations,
we need to return it for the next pass to modify it. */
return stats;
}, {} );
/* Now that `counts` has our object, we can log it. */
console.log( counts );
}());
const sentence = 'Hi my friend how are you my friend';
const countWords = (sentence) => {
const convertToObject = sentence.split(" ").map( (i, k) => {
return {
element: {
word: i,
nr: sentence.split(" ").filter(j => j === i).length + ' occurrence',
}
}
});
return Array.from(new Set(convertToObject.map(JSON.stringify))).map(JSON.parse)
};
console.log(countWords(sentence));
Here is an updated version of your own code...
<!DOCTYPE html>
<html>
<head>
<title>string frequency</title>
<style type="text/css">
#text{
width:250px;
}
</style>
</head>
<body >
<textarea id="txt" cols="25" rows="3" placeholder="add your text here"> </textarea></br>
<button type="button" onclick="search()">search</button>
<script >
function search()
{
var data=document.getElementById('txt').value;
var temp=data;
var words=new Array();
words=temp.split(" ");
var unique = {};
for (var i = 0; i < words.length; i++) {
var word = words[i];
console.log(word);
if (word in unique)
{
console.log("word found");
var count = unique[word];
count ++;
unique[word]=count;
}
else
{
console.log("word NOT found");
unique[word]=1;
}
}
console.log(unique);
}
</script>
</body>
I think your loop was overly complicated. Also, trying to produce the final count while still doing your first pass over the array of words is bound to fail because you can't test for uniqueness until you have checked each word in the array.
Instead of all your counters, I've used a Javascript object to work as an associative array, so we can store each unique word, and the count of how many times it occurs.
Then, once we exit the loop, we can see the final result.
Also, this solution uses no regex ;)
I'll also add that it's very hard to count words just based on spaces. In this code, "one, two, one" will results in "one," and "one" as being different, unique words.
While both of the answers here are correct maybe are better but none of them address OP's question (what is wrong with the his code).
The problem with OP's code is here:
if(f==0){
count[i]=1;
uniqueWords[i]=words[i];
}
On every new word (unique word) the code adds it to uniqueWords at index at which the word was in words. Hence there are gaps in uniqueWords array. This is the reason for some undefined values.
Try printing uniqueWords. It should give something like:
["this", "is", "anil", 4: "kum", 5: "the"]
Note there no element for index 3.
Also the printing of final count should be after processing all the words in the words array.
Here's corrected version:
function search()
{
var data=document.getElementById('txt').value;
var temp=data;
var words=new Array();
words=temp.split(" ");
var uniqueWords=new Array();
var count=new Array();
for (var i = 0; i < words.length; i++) {
//var count=0;
var f=0;
for(j=0;j<uniqueWords.length;j++){
if(words[i]==uniqueWords[j]){
count[j]=count[j]+1;
//uniqueWords[j]=words[i];
f=1;
}
}
if(f==0){
count[i]=1;
uniqueWords[i]=words[i];
}
}
for ( i = 0; i < uniqueWords.length; i++) {
if (typeof uniqueWords[i] !== 'undefined')
console.log("count of "+uniqueWords[i]+" - "+count[i]);
}
}
I have just moved the printing of count out of the processing loop into a new loop and added a if not undefined check.
Fiddle: https://jsfiddle.net/cdLgaq3a/
I had a similar assignment. This is what I did:
Assignment : Clean the following text and find the most frequent word (hint, use replace and regular expressions).
const sentence = '%I $am#% a %tea#cher%, &and& I lo%#ve %te#a#ching%;. The#re $is no#th#ing; &as& mo#re rewarding as educa#ting &and& #emp%o#weri#ng peo#ple. ;I found tea#ching m%o#re interesting tha#n any ot#her %jo#bs. %Do#es thi%s mo#tiv#ate yo#u to be a tea#cher!? %Th#is 30#Days&OfJavaScript &is al#so $the $resu#lt of &love& of tea&ching'
console.log(`\n\n 03.Clean the following text and find the most frequent word (hint, use replace and regular expressions) \n\n ${sentence} \n\n`)
console.log(`Cleared sentence : ${sentence.replace(/[.,\/#!$%\^&\*;:{}=\-_`~()#]/g, "")}`)
console.log(mostFrequentWord(sentence))
function mostFrequentWord(sentence) {
sentence = sentence.replace(/[.,\/#!$%\^&\*;:{}=\-_`~()#]/g, "").trim().toLowerCase()
let sentenceArray = sentence.split(" ")
let word = null
let count = 0
for (i = 0; i < sentenceArray.length; i++) {
word = sentenceArray[i]
count = sentence.match(RegExp(sentenceArray[i], 'gi')).length
if (count > count) {
count = count
word = word
}
}
return `\n Count of most frequent word "${word}" is ${count}`
}
I'd go with Sampson's match-reduce method for slightly better efficiency. Here's a modified version of it that is more production-ready. It's not perfect, but it should cover the vast majority of scenarios (i.e., "good enough").
function calcWordFreq(s) {
// Normalize
s = s.toLowerCase();
// Strip quotes and brackets
s = s.replace(/["“”(\[{}\])]|\B['‘]([^'’]+)['’]/g, '$1');
// Strip dashes and ellipses
s = s.replace(/[‒–—―…]|--|\.\.\./g, ' ');
// Strip punctuation marks
s = s.replace(/[!?;:.,]\B/g, '');
return s.match(/\S+/g).reduce(function(oFreq, sWord) {
if (oFreq.hasOwnProperty(sWord)) ++oFreq[sWord];
else oFreq[sWord] = 1;
return oFreq;
}, {});
}
calcWordFreq('A ‘bad’, “BAD” wolf-man...a good ol\' spook -- I\'m frightened!') returns
{
"a": 2
"bad": 2
"frightened": 1
"good": 1
"i'm": 1
"ol'": 1
"spook": 1
"wolf-man": 1
}
I wrote a simple program to analyze a string to find the word with the greatest amount of duplicate letters within it. It essentially takes a given string, breaks it up into an array of separated words, and then breaks up each separate word into alphabetically sorted groups of individual letters (which are then compared as prev and next, 2 at a time, as the containing array is iterated through). Any two adjacent and matching values found adds one tally to the hash-file next to the word in question, and the word with the most tallied pairs of duplicate letters is returned at the end as greatest. No matching pairs found in any word returns -1. This is what it's supposed to do.
Below, I've run into a problem: If I don't use a REGEXP to replace one of my matched characters, then my code gives false positives as it will count triplicates (eg, "EEE"), as two separate pairs, (eg, "EEE" = "EE & EE", instead of being viewed as "EE, E"). However, if I DO use the REGEXP below to prevent triplicate counts, then doing so breaks my loop mid-stride, and skips to the next word. Is there no way to make this way work? If not, would it be better to employ a REGEXP which deletes all chars EXCEPT the duplicate characters in question, and then perhaps I could divide the .length of each word by 2 to get the number of pairs remaining? Any ideas as to how to solve this would greatly help.
var str = "Helloo aplpplpp pie";
//var str = "no repting letrs";
//var str = "ceoderbyte";
function LetterCountI(str) {
var input = str.split(" ");
console.log(input);
console.log("\n")
var hashObject = {};
var word = "";
var count = 0;
for(var i = 0; i<input.length; i++) {
var currentItem = input[i];
var currentWordIntoChars = currentItem.split("").sort();
console.log(currentWordIntoChars);
var counter = 0;
for(var j=1; j<currentWordIntoChars.length; j++) {
console.log(currentWordIntoChars[j-1] + "=currentChar j-1");
console.log(currentWordIntoChars[j] + "=prev j");
console.log("-");
var final = currentItem;
if(currentWordIntoChars[j-1] == currentWordIntoChars[j]) {
counter++;
hashObject[final] = counter;
//currentWordIntoChars = currentWordIntoChars[j-1].replace(/[a-z]/gi, String.fromCharCode(currentItem.charCodeAt(0)+1));
//HERE REPLACE j-1 with random# or something
//to avoid 3 in a row being counted as 2 pair
//OR use regexp to remove all but pairs, and
//then divide .length/2 to get pairs.
console.log(counter + " === # total char pairs");
}
if(count<hashObject[currentItem]) {
word = final;
count = hashObject[currentItem];
}
}
}
console.log(hashObject);
console.log("\n");
for (var o in hashObject) if (o) return word;
return -1;
}
console.log(LetterCountI(str));
An other way to do it, consists to replace duplicate characters in a sorted word:
var str = "Helloo aplpplpp pie";
function LetterCountI(str) {
var input = str.split(" ");
var count = 0;
var result = -1;
for(var i = 0; i<input.length; i++) {
var nb = 0;
var sortedItem = input[i].split("").sort().join("");
sortedItem.replace(/(.)\1/g, function (_) { nb++ });
if (nb > count) {
count = nb;
result = input[i];
}
}
return result;
}
console.log(LetterCountI(str));
Notes: The replace method is only a way to increment nb using a callback function. You can do the same using the match method and counting results.
if two words have the same number of duplicates, the first word will be returned by default. You can easily change this behaviour with the condition of the if statement.
Whenever you find a match within a word, increment j by 1 to skip comparing the next letter.
var str = "Helloo aplpplpp pie";
//var str = "no repting letrs";
//var str = "ceoderbyte";
function LetterCountI(str)
{
var input = str.split(" ");
console.log(input);
console.log("\n")
var hashObject = {};
var word = "";
var count = 0;
for(var i = 0; i<input.length; i++)
{
var currentItem = input[i];
var currentWordIntoChars = currentItem.split("").sort();
console.log(currentWordIntoChars);
var counter = 0;
for(var j=1; j<currentWordIntoChars.length; j++)
{
console.log(currentWordIntoChars[j-1] + "=currentChar j-1");
console.log(currentWordIntoChars[j] + "=prev j");
console.log("-");
var final = currentItem;
if(currentWordIntoChars[j-1] == currentWordIntoChars[j])
{
counter++;
hashObject[final] = counter;
j++; // ADD HERE
console.log(counter + " === # total char pairs");
}
if(count<hashObject[currentItem])
{
word = final;
count = hashObject[currentItem];
}
}
}
console.log(hashObject);
console.log("\n");
for (var o in hashObject) if (o) return word;
return -1;
}
console.log(LetterCountI(str));
How can I return the url part of the first page in pages for which the content part matches pattern (in the style of function 1, case insensitive) but return an empty string if no page is found?
e.g.
url1(pages,"GREAT") returns "www.xyz.ac.uk"
url1(pages,"xyz") returns ""
Here is my code so far:
var pg = [ "|www.cam.ac.uk|Cambridge University offers degree programmes and world class research." , "!www.xyz.ac.uk!An great University" , "%www%Yet another University" ];
var pt = "great";
function url1(pages, pattern) {
var result = "";
for (x in pages) {
current = pages[x].split(pattern);
result = current[1];
}
return result;
}
alert(url1(pg, pt));
Try this:
var pg = [ "|www.cam.ac.uk|Cambridge University offers degree programmes and world class research." , "!www.xyz.ac.uk!An great University" , "%www%Yet another University" ];
function find(pages, pattern) {
var i, l, page, arr;
pattern = pattern.toLowerCase();
for(i=0, l=pages.length; i<l; i++) {
page = pages[i];
arr = page.split(page[0]);
if(arr.slice(2).join(page[0]).toLowerCase().indexOf(pattern) >=0) {
return arr[1];
}
}
return '';
}
console.log(find(pg,'great')); // 'www.xyz.ac.uk'
console.log(find(pg,'xyz')); // ''
Fiddle here: http://jsbin.com/uGebeQi/2/edit
Here's another solution using a regular expression. I feel this is slightly better than blindly splitting on the first character since the content could always contain the first character as well.
Note: The accepted answer was corrected.
function url1(pages, pattern) {
var rx = /^(.)(.+?)\1(.+)/,
i = 0,
len = pages.length,
matches;
for (; i < len; i++) {
if ((matches = pages[i].match(rx)) && matches[3].indexOf(pattern) !== -1)
return matches[2];
}
return '';
}
//www.xyz.ac.uk
url1(['!www.xyz.ac.uk!An great University! The best in the world!'], 'best');
You can always lowercase the pattern and the content if you want case-insensitive searches.
I have to match 2 strings where at least one word is same, I need to give a success msg.
var str1 = "Hello World";
var str2 = "world is beautiful";
I need to match/compare these 2 strings, in both strings world is matching, So i need to print a success message. How do I go about it.
The following code will output all the matching words in the both strings:
var words1 = str1.split(/\s+/g),
words2 = str2.split(/\s+/g),
i,
j;
for (i = 0; i < words1.length; i++) {
for (j = 0; j < words2.length; j++) {
if (words1[i].toLowerCase() == words2[j].toLowerCase()) {
console.log('word '+words1[i]+' was found in both strings');
}
}
}
You can avoid comparing all the words in one list with all the words in the other by sorting each and eliminating duplicates. Adapting bjornd's answer:
var words1 = str1.split(/\s+/g),
words2 = str2.split(/\s+/g);
var allwords = {};
// set 1 for all words in words1
for(var wordid=0; wordid < words1.length; ++wordid) {
var low = words1[wordid].toLowerCase();
allwords[low] = 1;
}
// add 2 for all words in words2
for(var wordid=0; wordid < words2.length; ++wordid) {
var current = 0;
var low = words2[wordid].toLowerCase();
if(allwords.hasOwnProperty(low)) {
if(allwords[low] > 1) {
continue;
}
}
current += 2;
allwords[low] = current;
}
// now those seen in both lists have value 3, the rest either 1 or 2.
// this is effectively a bitmask where the unit bit indicates words1 membership
// and the 2 bit indicates words2 membership
var both = [];
for(var prop in allwords) {
if(allwords.hasOwnProperty(prop) && (allwords[prop] == 3)) {
both.push(prop);
}
}
This version should be reasonably efficient, because we are using a dictionary/hash structure to store information about each set of words. The whole thing is O(n) in javascript expressions, but inevitably dictionary insertion is not, so expect something like O(n log n) in practise. If you only care that a single word matches, you can quit early in the second for loop; the code as-is will find all matches.
This is broadly equivalent to sorting both lists, reducing each to unique words, and then looking for pairs in both lists. In C++ etc you would do it via two sets, as you could do it without using a dictionary and the comparison would be O(n) after the sorts. In Python because it's easy to read:
words1 = set(item.lower() for item in str1.split())
words2 = set(item.lower() for item in str2.split())
common = words1 & words2
The sort here (as with any set) happens on insertion into the set O(n log n) on word count n, and the intersection (&) is then efficent O(m) on the set length m.
I just tried this on WriteCodeOnline and it works there:
var s1 = "hello world, this is me";
var s2 = "I am tired of this world and I want to get off";
var s1s2 = s1 + ";" + s2;
var captures = /\b(\w+)\b.*;.*\b\1\b/i.exec(s1s2);
if (captures[1])
{
document.write(captures[1] + " occurs in both strings");
}
else
{
document.write("no match in both strings");
}
Just adapting #Phil H's code with a real bitmask:
var strings = ["Hello World", "world is beautiful"]; // up to 32 word lists
var occurrences = {},
result = [];
for (var i=0; i<strings.length; i++) {
var words = strings[i].toLowerCase().split(/\s+/),
bit = 1<<i;
for (var j=0, l=words.length; j<l; j++) {
var word = words[j];
if (word in occurrences)
occurrences[word] |= bit;
else
occurrences[word] = bit;
}
}
// now lets do a match for all words which are both in strings[0] and strings[1]
var filter = 3; // 1<<0 | 1<<1
for (var word in occurrences)
if ((occurrences[word] & filter) === filter)
result.push(word);
OK, the simple way:
function isMatching(a, b)
{
return new RegExp("\\b(" + a.match(/\w+/g).join('|') + ")\\b", "gi").test(b);
}
isMatching("in", "pin"); // false
isMatching("Everything is beautiful, in its own way", "Every little thing she does is magic"); // true
isMatching("Hello World", "world is beautiful"); // true
...understand?
I basically converted "Hello, World!" to the regular expression /\b(Hello|World)\b/gi
Something like this would also do:
isMatching = function(str1, str2) {
str2 = str2.toLowerCase();
for (var i = 0, words = str1.toLowerCase().match(/\w+/g); i < words.length; i++) {
if (str2.search(words[i]) > -1) return true;
}
return false;
};
var str1 = "Hello World";
var str2 = "world is beautiful";
isMatching(str1, str2); // returns true
isMatching(str1, 'lorem ipsum'); // returns false