Nsryjdtyk

I am looking for the fastest way to find document distance. How can I improve this?

Split each document into words. replaceAll() is faster than replace()

String.prototype.replaceAll = function(search, replacement) {

    return this.replace(new RegExp(search, 'g'), replacement);

};

Creating word array here:

const wordsFromDocument = doc => {

    //we care only about alphanumeric characters and space

    return doc.replaceAll(/[^a-zA-Z0-9 ]/g, ``).split(` `);

};

count word frequencies (document vectors)

const wordFrequencies = words => {

    const wordFrequencyMap = new Map();

    for (const word of words) {

        const lowerCaseWord = word.toLowerCase(); //We don't care about case sensitivity. What is faster way to achieve this?

        if (!wordFrequencyMap.has(lowerCaseWord)) {

            wordFrequencyMap.set(lowerCaseWord, 1);

        } else {

            wordFrequencyMap.set(lowerCaseWord, wordFrequencyMap.get(lowerCaseWord) + 1);

        }

    }

    return wordFrequencyMap;

};

Compute dot product (& divide).

const dotProduct = (wordFrequencyMapOne, wordFrequencyMapTwo) => {

    let sum = 0;

    const wordOneSize = wordFrequencyMapOne.size;

    const wordTwoSize = wordFrequenciesTwo.size;



    if (wordOneSize < wordTwoSize) {

        for (let key of wordFrequencyMapOne.keys()) {

            if (wordFrequencyMapTwo.has(key)) {

                sum = sum + wordFrequencyMapOne.get(key) *

                    wordFrequencyMapTwo.get(key);

            }

        }

    } else {

        for (let key of wordFrequenciesTwo.keys()) {

            if (wordFrequenciesOne.has(key)) {

                sum = sum + wordFrequencyMapOne.get(key) *

                    wordFrequencyMapTwo.get(key);

            }

        }

    }

    return sum;

};

vector angle = (dot(1, 2)) / Sqrt (dot(1, 1) * dot(2, 2))

const vectorAngle = (wordFrequencyMapOne, wordFrequencyMapTwo) => {

    const numerator = dotProduct(wordFrequencyMapOne, wordFrequencyMapTwo);

    const denominator = Math.sqrt(dotProduct(wordFrequencyMapOne, wordFrequencyMapOne) * (dotProduct(wordFrequenciesTwo, wordFrequenciesTwo)))

    return numerator / denominator;

}