Develop Reference

Break each new anagram of an array item to a new line

I am working an anagram generator and am trying to break off each new item in the array into a new line. The way this works is it slices each array item and loops through each character.
The output needs to be:
cat, cta, act, atc, tca, tac,
bat, bta, abt, atb, tba, tab,
rat, rta, art, atr, tra, tar,
But it is:
cat, cta, act, atc, tca, tac, bat, bta, abt, atb, tba, tab, rat, rta, art, atr, tra, tar, splat, splta, spalt, spatl,...
So far the code I have is this:
HTML:
<div id="anagrams"></div>
JS:
var arr = ['cat', 'bat', 'rat', 'splat'];
var allAnagrams = function(arr) {
var anagrams = {};
arr.forEach(function(str) {
var recurse = function(ana, str) {
if (str === '')
anagrams[ana] = 1;
for (var i = 0; i < str.length; i++)
recurse(ana + str[i], str.slice(0, i) + str.slice(i + 1));
};
recurse(' ', str);
});
return Object.keys(anagrams);
}
document.getElementById("anagrams").innerHTML = (allAnagrams(arr));
To accomplish a new line per array item I basically want to check if the amount of the characters exceeds the amount of characters in the string/array item and if it does, insert a break into the HTML. I tried doing that by:
var arr = ['cat', 'bat', 'rat', 'splat'];
var allAnagrams = function(arr) {
var anagrams = {};
arr.forEach(function(str) {
var recurse = function(ana, str) {
if (str === '')
anagrams[ana] = 1;
for (var i = 0; i < str.length; i++)
recurse(ana + str[i], str.slice(0, i) + str.slice(i + 1));
// check if string length is greater than the count and
// if it is, insert a break between the string
if (i > str.length) {
recurse(' <br>', str);
}
};
recurse(' ', str);
});
return Object.keys(anagrams);
}
document.getElementById("anagrams").innerHTML = (allAnagrams(arr));
However it still prints across a single line. Am I approaching this the correct way? I also tried using ana in place of i but I think I need to use i since that's the actual count - is that correct?
A jsfiddle can be seen here: https://jsfiddle.net/4eqhd1m4/1/

I would slightly restructure the anagram creation.
Anagrams is now a string.
Recurse no longer takes care of adding break lines. Considering you want a break per element, it's cleaner to add it in the Array.forEach
jsfiddle
Edit
Adding a second jsfiddle to demonstrate the same behavior, except instead of using strings directly it returns an array (which gets split and rejoined using breaklines). It may be preferable to have the anagrams returned as an array.
jsfiddle

Is this is what you need ? All anagrams in different lines
var arr = ['cat', 'bat', 'rat', 'splat'];
var allAnagrams = function(arr) {
var anagrams = {};
arr.forEach(function(str) {
var recurse = function(ana, str) {
if (str === '')
anagrams[ana] = 1;
for (var i = 0; i < str.length; i++)
recurse(ana + str[i], str.slice(0, i) + str.slice(i + 1));
// check if string length is greater than the count and
// if it is, insert a break between the string
if (i > str.length) {
recurse(' <br \/>', str);
}
};
recurse(' <br \/>', str);
});
return Object.keys(anagrams);
}
document.getElementById("anagrams").innerHTML = (allAnagrams(arr));
<div id="anagrams"></div>

How can I avoid counting triplicates as pairs while iterating through an ordered series of letters within an array?

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));

Testing for a common word between 2 strings in javascript

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

Remove several words from an array - Javascript

I've got an array of words I need to sort by frequency. Before I do that, I need to strip out words like 'the,' 'it,' etc (anything less than three letters, really), as well as all numerals and any words beginning with # (the array of words is pulled from Twitter, although the example below is just a random paragraph from Wikipedia).
I can remove one word, but have been going crazy trying to remove more than one, or a range. Any suggestions? Thank you!
http://jsfiddle.net/9NzAC/6/
HTML:
<div id="text" style="background-color:Teal;position:absolute;left:100px;top:10px;height:500px;width:500px;">
Phrenology is a pseudoscience primarily focused on measurements of the human skull, based on the concept that the brain is the organ of the mind, and that certain brain areas have localized, specific functions or modules. The distinguishing feature of phrenology is the idea that the sizes of brain areas were meaningful and could be inferred by examining the skull of an individual.
</div>
JS:
//this is the function to remove words
<script type="text/javascript">
function removeA(arr){
var what, a= arguments, L= a.length, ax;
while(L> 1 && arr.length){
what= a[--L];
while((ax= arr.indexOf(what))!= -1){
arr.splice(ax, 1);
}
}
return arr;
}
</script>
//and this does the sorting & counting
<script type="text/javascript">
var getMostFrequentWords = function(words) {
var freq={}, freqArr=[], i;
// Map each word to its frequency in "freq".
for (i=0; i<words.length; i++) {
freq[words[i]] = (freq[words[i]]||0) + 1;
}
// Sort from most to least frequent.
for (i in freq) freqArr.push([i, freq[i]]);
return freqArr.sort(function(a,b) { return b[1] - a[1]; });
};
var words = $('#text').get(0).innerText.split(/\s+/);
//Remove articles & words we don't care about.
var badWords = "the";
removeA(words,badWords);
var mostUsed = getMostFrequentWords(words);
alert(words);
</script>

Instead of removing from the original array, just push to a new one, it's simpler, and it'll make your code shorter and more readable.
var words = ['the', 'it', '12', '#twit', 'aloha', 'hello', 'bye']
var filteredWords = []
for (var i = 0, l = words.length, w; i < l; i++) {
w = words[i]
if (!/^(#|\d+)/.test(w) && w.length > 3)
filteredWords.push(w)
}
console.log(filteredWords) // ['aloha', 'hello']
Demo: http://jsfiddle.net/VcfvU/

I recommend you to do array[i] = null (or "") and then just clean up your arrays empty nodes. You can easily achieve that using Array#filter
Test: http://jsfiddle.net/6LPep/
Code:
var FORGETABLE_WORDS = ',the,of,an,and,that,which,is,was,';
var words = text.innerText.split(" ");
for(var i = 0, word; word = words[i++]; ) {
if (FORGETABLE_WORDS.indexOf(',' + word + ',') > -1 || word.length < 3) {
words[i-1] = "";
}
}
// falsy will get deleted
words.filter(function(e){return e});
// as example
output.innerHTML = words.join(" ");
// just continue doing your stuff with "words" array.
// ...​
I think it's cleaner than the way you're doing it currently. If you need anything else I will update this answer.

console.log(
['🍇','🍈','🍌','🍉','🍊','🍋'].filter(a => !['🍌','🍊'].includes(a))
)

What's the best way to count keywords in JavaScript?

What's the best and most efficient way to count keywords in JavaScript? Basically, I'd like to take a string and get the top N words or phrases that occur in the string, mainly for the use of suggesting tags. I'm looking more for conceptual hints or links to real-life examples than actual code, but I certainly wouldn't mind if you'd like to share code as well. If there are particular functions that would help, I'd also appreciate that.
Right now I think I'm at using the split() function to separate the string by spaces and then cleaning punctuation out with a regular expression. I'd also want it to be case-insensitive.

Cut, paste + execute demo:
var text = "Text to be examined to determine which n words are used the most";
// Find 'em!
var wordRegExp = /\w+(?:'\w{1,2})?/g;
var words = {};
var matches;
while ((matches = wordRegExp.exec(text)) != null)
{
var word = matches[0].toLowerCase();
if (typeof words[word] == "undefined")
{
words[word] = 1;
}
else
{
words[word]++;
}
}
// Sort 'em!
var wordList = [];
for (var word in words)
{
if (words.hasOwnProperty(word))
{
wordList.push([word, words[word]]);
}
}
wordList.sort(function(a, b) { return b[1] - a[1]; });
// Come back any time, straaanger!
var n = 10;
var message = ["The top " + n + " words are:"];
for (var i = 0; i < n; i++)
{
message.push(wordList[i][0] + " - " + wordList[i][1] + " occurance" +
(wordList[i][1] == 1 ? "" : "s"));
}
alert(message.join("\n"));
Reusable function:
function getTopNWords(text, n)
{
var wordRegExp = /\w+(?:'\w{1,2})?/g;
var words = {};
var matches;
while ((matches = wordRegExp.exec(text)) != null)
{
var word = matches[0].toLowerCase();
if (typeof words[word] == "undefined")
{
words[word] = 1;
}
else
{
words[word]++;
}
}
var wordList = [];
for (var word in words)
{
if (words.hasOwnProperty(word))
{
wordList.push([word, words[word]]);
}
}
wordList.sort(function(a, b) { return b[1] - a[1]; });
var topWords = [];
for (var i = 0; i < n; i++)
{
topWords.push(wordList[i][0]);
}
return topWords;
}

Once you have that array of words cleaned up, and let's say you call it wordArray:
var keywordRegistry = {};
for(var i = 0; i < wordArray.length; i++) {
if(keywordRegistry.hasOwnProperty(wordArray[i]) == false) {
keywordRegistry[wordArray[i]] = 0;
}
keywordRegistry[wordArray[i]] = keywordRegistry[wordArray[i]] + 1;
}
// now keywordRegistry will have, as properties, all of the
// words in your word array with their respective counts
// this will alert (choose something better than alert) all words and their counts
for(var keyword in keywordRegistry) {
alert("The keyword '" + keyword + "' occurred " + keywordRegistry[keyword] + " times");
}
That should give you the basics of doing this part of the work.

Try to split you string on words and count the resulting words, then sort on the counts.

This builds upon a previous answer by insin by only having one loop:
function top_words(text, n) {
// Split text on non word characters
var words = text.toLowerCase().split(/\W+/)
var positions = new Array()
var word_counts = new Array()
for (var i=0; i<words.length; i++) {
var word = words[i]
if (!word) {
continue
}
if (typeof positions[word] == 'undefined') {
positions[word] = word_counts.length
word_counts.push([word, 1])
} else {
word_counts[positions[word]][1]++
}
}
// Put most frequent words at the beginning.
word_counts.sort(function (a, b) {return b[1] - a[1]})
// Return the first n items
return word_counts.slice(0, n)
}
// Let's see if it works.
var text = "Words in here are repeated. Are repeated, repeated!"
alert(top_words(text, 3))
The result of the example is: [['repeated',3], ['are',2], ['words', 1]]

I would do exactly what you have mentioned above to isolate each word. I would then probably add each word as the index of an array with the number of occurrences as the value.
For example:
var a = new Array;
a[word] = a[word]?a[word]+1:1;
Now you know how many unique words there are (a.length) and how many occurrences of each word existed (a[word]).