I'm trying to build a collaborative doc editor and implement operational transformation. Imagine we have a string that is manipulated simultaneously by 2 users. They can only add characters, not remove them. We want to incorporate both of their changes.
The original string is: catspider
The first user does this: cat<span id>spider</span>
The second user does this: c<span id>atspi</span>der
I'm trying to write a function that will produce: c<span id>at<span id>spi</span>der</span>
The function I've written is close, but it produces c<span id>at<span i</span>d>spider</span> codepen here
String.prototype.splice = function(start, newSubStr) {
return this.slice(0, start) + newSubStr + this.slice(start);
};
function merge(saved, working, requested) {
if (!saved || !working || !requested) {
return false;
}
var diffSavedWorking = createDiff(working, saved);
var diffRequestedWorking = createDiff(working, requested);
var newStr = working;
for (var i = 0; i < Math.max(diffRequestedWorking.length, diffSavedWorking.length); i++) {
//splice does an insert `before` -- we will assume that the saved document characters
//should always appear before the requested document characters in this merger operation
//so we first insert requested and then saved, which means that the final string will have the
//original characters first.
if (diffRequestedWorking[i]) {
newStr = newStr.splice(i, diffRequestedWorking[i]);
//we need to update the merge arrays by the number of
//inserted characters.
var length = diffRequestedWorking[i].length;
insertNatX(diffSavedWorking, length, i + 1);
insertNatX(diffRequestedWorking, length, i + 1);
}
if (diffSavedWorking[i]) {
newStr = newStr.splice(i, diffSavedWorking[i]);
//we need to update the merge arrays by the number of
//inserted characters.
var length = diffSavedWorking[i].length;
insertNatX(diffSavedWorking, length, i + 1);
insertNatX(diffRequestedWorking, length, i + 1);
}
}
return newStr;
}
//arr1 should be the shorter array.
//returns inserted characters at their
//insertion index.
function createDiff(arr1, arr2) {
var diff = [];
var j = 0;
for (var i = 0; i < arr1.length; i++) {
diff[i] = "";
while (arr2[j] !== arr1[i]) {
diff[i] += arr2[j];
j++;
}
j++;
}
var remainder = arr2.substr(j);
if (remainder) diff[i] = remainder;
return diff;
}
function insertNatX(arr, length, pos) {
for (var j = 0; j < length; j++) {
arr.splice(pos, 0, "");
}
}
var saved = 'cat<span id>spider</span>';
var working = 'catspider';
var requested = 'c<span id>atspi</span>der';
console.log(merge(saved, working, requested));
Would appreciate any thoughts on a better / simpler way to achieve this.
Related
I have a string with repeated letters. I want letters that are repeated more than once to show only once.
Example input: aaabbbccc
Expected output: abc
I've tried to create the code myself, but so far my function has the following problems:
if the letter doesn't repeat, it's not shown (it should be)
if it's repeated once, it's show only once (i.e. aa shows a - correct)
if it's repeated twice, shows all (i.e. aaa shows aaa - should be a)
if it's repeated 3 times, it shows 6 (if aaaa it shows aaaaaa - should be a)
function unique_char(string) {
var unique = '';
var count = 0;
for (var i = 0; i < string.length; i++) {
for (var j = i+1; j < string.length; j++) {
if (string[i] == string[j]) {
count++;
unique += string[i];
}
}
}
return unique;
}
document.write(unique_char('aaabbbccc'));
The function must be with loop inside a loop; that's why the second for is inside the first.
Fill a Set with the characters and concatenate its unique entries:
function unique(str) {
return String.prototype.concat.call(...new Set(str));
}
console.log(unique('abc')); // "abc"
console.log(unique('abcabc')); // "abc"
Convert it to an array first, then use Josh Mc’s answer at How to get unique values in an array, and rejoin, like so:
var nonUnique = "ababdefegg";
var unique = Array.from(nonUnique).filter(function(item, i, ar){ return ar.indexOf(item) === i; }).join('');
All in one line. :-)
Too late may be but still my version of answer to this post:
function extractUniqCharacters(str){
var temp = {};
for(var oindex=0;oindex<str.length;oindex++){
temp[str.charAt(oindex)] = 0; //Assign any value
}
return Object.keys(temp).join("");
}
You can use a regular expression with a custom replacement function:
function unique_char(string) {
return string.replace(/(.)\1*/g, function(sequence, char) {
if (sequence.length == 1) // if the letter doesn't repeat
return ""; // its not shown
if (sequence.length == 2) // if its repeated once
return char; // its show only once (if aa shows a)
if (sequence.length == 3) // if its repeated twice
return sequence; // shows all(if aaa shows aaa)
if (sequence.length == 4) // if its repeated 3 times
return Array(7).join(char); // it shows 6( if aaaa shows aaaaaa)
// else ???
return sequence;
});
}
Using lodash:
_.uniq('aaabbbccc').join(''); // gives 'abc'
Per the actual question: "if the letter doesn't repeat its not shown"
function unique_char(str)
{
var obj = new Object();
for (var i = 0; i < str.length; i++)
{
var chr = str[i];
if (chr in obj)
{
obj[chr] += 1;
}
else
{
obj[chr] = 1;
}
}
var multiples = [];
for (key in obj)
{
// Remove this test if you just want unique chars
// But still keep the multiples.push(key)
if (obj[key] > 1)
{
multiples.push(key);
}
}
return multiples.join("");
}
var str = "aaabbbccc";
document.write(unique_char(str));
Your problem is that you are adding to unique every time you find the character in string. Really you should probably do something like this (since you specified the answer must be a nested for loop):
function unique_char(string){
var str_length=string.length;
var unique='';
for(var i=0; i<str_length; i++){
var foundIt = false;
for(var j=0; j<unique.length; j++){
if(string[i]==unique[j]){
foundIt = true;
break;
}
}
if(!foundIt){
unique+=string[i];
}
}
return unique;
}
document.write( unique_char('aaabbbccc'))
In this we only add the character found in string to unique if it isn't already there. This is really not an efficient way to do this at all ... but based on your requirements it should work.
I can't run this since I don't have anything handy to run JavaScript in ... but the theory in this method should work.
Try this if duplicate characters have to be displayed once, i.e.,
for i/p: aaabbbccc o/p: abc
var str="aaabbbccc";
Array.prototype.map.call(str,
(obj,i)=>{
if(str.indexOf(obj,i+1)==-1 ){
return obj;
}
}
).join("");
//output: "abc"
And try this if only unique characters(String Bombarding Algo) have to be displayed, add another "and" condition to remove the characters which came more than once and display only unique characters, i.e.,
for i/p: aabbbkaha o/p: kh
var str="aabbbkaha";
Array.prototype.map.call(str,
(obj,i)=>{
if(str.indexOf(obj,i+1)==-1 && str.lastIndexOf(obj,i-1)==-1){ // another and condition
return obj;
}
}
).join("");
//output: "kh"
<script>
uniqueString = "";
alert("Displays the number of a specific character in user entered string and then finds the number of unique characters:");
function countChar(testString, lookFor) {
var charCounter = 0;
document.write("Looking at this string:<br>");
for (pos = 0; pos < testString.length; pos++) {
if (testString.charAt(pos) == lookFor) {
charCounter += 1;
document.write("<B>" + lookFor + "</B>");
} else
document.write(testString.charAt(pos));
}
document.write("<br><br>");
return charCounter;
}
function findNumberOfUniqueChar(testString) {
var numChar = 0,
uniqueChar = 0;
for (pos = 0; pos < testString.length; pos++) {
var newLookFor = "";
for (pos2 = 0; pos2 <= pos; pos2++) {
if (testString.charAt(pos) == testString.charAt(pos2)) {
numChar += 1;
}
}
if (numChar == 1) {
uniqueChar += 1;
uniqueString = uniqueString + " " + testString.charAt(pos)
}
numChar = 0;
}
return uniqueChar;
}
var testString = prompt("Give me a string of characters to check", "");
var lookFor = "startvalue";
while (lookFor.length > 1) {
if (lookFor != "startvalue")
alert("Please select only one character");
lookFor = prompt(testString + "\n\nWhat should character should I look for?", "");
}
document.write("I found " + countChar(testString, lookFor) + " of the<b> " + lookFor + "</B> character");
document.write("<br><br>I counted the following " + findNumberOfUniqueChar(testString) + " unique character(s):");
document.write("<br>" + uniqueString)
</script>
Here is the simplest function to do that
function remove(text)
{
var unique= "";
for(var i = 0; i < text.length; i++)
{
if(unique.indexOf(text.charAt(i)) < 0)
{
unique += text.charAt(i);
}
}
return unique;
}
The one line solution will be to use Set. const chars = [...new Set(s.split(''))];
If you want to return values in an array, you can use this function below.
const getUniqueChar = (str) => Array.from(str)
.filter((item, index, arr) => arr.slice(index + 1).indexOf(item) === -1);
console.log(getUniqueChar("aaabbbccc"));
Alternatively, you can use the Set constructor.
const getUniqueChar = (str) => new Set(str);
console.log(getUniqueChar("aaabbbccc"));
Here is the simplest function to do that pt. 2
const showUniqChars = (text) => {
let uniqChars = "";
for (const char of text) {
if (!uniqChars.includes(char))
uniqChars += char;
}
return uniqChars;
};
const countUnique = (s1, s2) => new Set(s1 + s2).size
a shorter way based on #le_m answer
let unique=myArray.filter((item,index,array)=>array.indexOf(item)===index)
I'm trying to shorten a text, if it is longer than a specified number.
So shorten_text_easy(text, 30), should return "We believe. In the future".
The first loop is working and displaying the correct total length of the strings in the array, but the second is where the error is, but I can't seem to find the issue.
var text = "We believe. In the future. The future is here. This is a test. We are testing.";
function shorten_text_easy(text, number) {
var text_array = text.split('. ').join('.///').split('///'); // Splitting the text into an array
var text_array_length = text_array.length;
var total_text_array_length = 0; // Predefining value to zero; The total length of all strings in the array
for (var i = 0; i < text_array_length; i++) { // To run while i is short than the length of the array
total_text_array_length += text_array[i].length; //
}
total_text_array_length = total_text_array_length + text_array_length - 1; // To exclude spaces which are omitted in array
console.log(total_text_array_length); // To show in console the first value of total_array_length
for (total_text_array_length; total_text_array_length > number; text_array.pop(-1), text_array_length--) { // Trying to remove the last item of an array if the total length of all strings in the array is larger than the number
for (var i = 0; i < text_array_length; i++) {
total_text_array_length += text_array[i].length;
console.log(total_text_array_length);
}
total_text_array_length = total_text_array_length + text_array_length - 1;
}
return text_array // This should return the end text array, when the total length of all strings is lower than 'number'
};
console.log(
shorten_text_easy(text, 30)
);
Slicing on last full stop that fits
const shorten_text_easy = (text, number) => {
let res = text.slice(0,number);
const pos = res.lastIndexOf(".");
return pos > 0 ? res.slice(0,pos+1) : res;
}
var text = "We believe. In the future. The future is here. This is a test. We are testing.";
console.log(
shorten_text_easy(text, 30)
);
I ended up doing this:
function shortenTextNextDot (text, length) {
var firstDot = text.indexOf(".") + 1;
if (length <= firstDot) {
return text.slice(0, firstDot);
} else {
var textLength = text.slice(0, length);
var restText = text.slice(length);
var nextDot = restText.indexOf(".") + 1;
var finalSentence = textLength + restText.slice(0, nextDot);
return finalSentence;
}
}
function shortenTextPrevDot (text, length) {
var firstDot = text.indexOf(".") + 1;
if (length <= firstDot) {
return text.slice(0, firstDot);
} else {
var textLength = text.slice(0, length);
var lastDot = textLength.lastIndexOf(".") + 1;
var finalSentence = textLength.slice(0, lastDot);
return finalSentence;
}
}
I have written this js code for palindrome, I know there are better and more efficient palindrome methods online but I want to know why I am unable to get my palindrome function to work properly?
CODE:
var pal = function(str) {
var len = str.length;
for (var i = 0; i < len; i++) {
var comp1 = str.substring(i, i + 1);
for (var j = len; j > 0; j--) {
var comp2 = str.substring(j - 1, j);
}
if (comp1 != comp2) {
console.log("not palindrome")
break;
} else {
console.log('palindrome')
}
}
}
pal('maddog');
OUTPUT :
palindrome
not palindrome
There are lot of better algorithms to check Palindrome. Let use the similar algorithm that you are using.
We basically use two pointers - left and right, and move to middle at the same time. In the original question, left pointer and right pointer doesn't move at the same time.
Pointers should move like this -
a b c b a
^ ^
a b c b a
^ ^
a b c b a
^
var isPalindrome = function (str) {
for (var i = 0, j = str.length-1; i < j; i++ , j--) {
if (str[i] != str[j]) {
return false;
}
}
return true;
}
console.log('maddog : ' + isPalindrome('maddog'));
console.log('abcba : ' + isPalindrome('abcba'));
console.log('deed : ' + isPalindrome('deed'));
console.log('a : ' + isPalindrome('a'));
Try the following code. It works by dividing the string length by 2, and then iterating up, checking mirroring characters against each other:
var pal = function(str){
var len = str.length;
for(var i = 0; i < Math.floor(len/2); i++){
if(str[i] != str[(len-1)-i]){
return false;
}
}
return true;
}
console.log(pal("bunny"));
console.log(pal("amoreroma"));
The inner loop is totally unnecessary. It does the same thing every time -- it loops through the whole string, starting from the end, repeatedly setting comp2 to the character; when it's done, comp2 always contains the first character. So your function just tests whether every character in the string is the same as the first character.
To test if something is a palindrome, you need to compare each character with the corresponding character from the other end of the string. You don't need two loops for this. You also only need to loop through the first half of the string, not the whole string.
Finally, you should only echo Palindrome at the end of the loop. Inside the loop you only know that one character matches, not all of them.
var pal = function(str) {
var len = str.length;
var half = Math.floor(len / 2);
var isPal = true;
for (var i = 0; i < half; i++) {
var comp1 = str[i];
var comp2 = str[len - i - 1];
if (comp1 != comp2) {
console.log("not palindrome")
isPal = false;
break;
}
}
if (isPal) {
console.log('palindrome')
}
}
pal('maddog');
pal('maddam');
You don't really need the nested loops, you can just loop backwards through the string to invert the string and then compare it to the original string. I updated the Snippet to work.
Before, your code was not inverting the string but rather just iterating through the characters and assigning them to the comp1 and comp1 variables. You need to concatenate the strings in order to build the new string backwards comp = comp + str.substring(j-1, j);
var pal = function(str) {
var len = str.length;
var comp = '';
for (var j = len; j > 0; j--) {
comp = comp + str.substring(j - 1, j);
}
if (str !== comp) {
console.log("not palindrome")
return;
}
console.log('palindrome')
}
pal('arepera');
I am caching longitude and latitude (plus a bit more info) for possibly 1000s of locations, currently using a JavaScript hash, a {}. e.g.
var cache = {};
cache['Boston, MA'] = { id: someid, latlon: [lat, long] };
cache['Someotherplace, TX'] = { id: someotherid, latlon: [itslat, itslong]};
Everytime a new location comes up I do a geocode and add the results to the cache. I don't think Boston's latitude will change anytime soon...
Will lookups be reasonably fast? I don't need blazing fast, I'm not running Amazon, but as this data grows to, say 2000 locations, will it bog down? If so, what might be a good alternative?
Much of the performance of the entire javascript engine is based on property lookups on objects so I'm quite sure that significant effort has been paid to the performance of that in the basic JS engine.
But, as with all things related to performance you should measure yourself. It would only take a few minutes to build a test harness in jsperf and either compare it to an alternative or just see if regular JS lookup appears fast enough for you.
Here's a [little test harness][1] that shows more than 20,000 key lookups per ms on my computer. I would think that's fast enough for you.
function log(args) {
var str = "";
for (var i = 0; i < arguments.length; i++) {
if (typeof arguments[i] === "object") {
str += JSON.stringify(arguments[i]);
} else {
str += arguments[i];
}
}
var div = document.createElement("div");
div.innerHTML = str;
document.body.appendChild(div);
}
function addCommas(str) {
var amount = str + "";
var parts = amount.split(".");
amount = parts[0].split("").reverse();
var output = "";
for (var i = 0; i < amount.length; i++) {
output = amount[i] + output;
if ((i+1) % 3 == 0 && (amount.length-1) !== i) {
output = ',' + output;
}
}
if (parts.length > 1) {
output += "." + parts[1];
}
return output;
}
function now() {
return new Date().getTime();
}
// now fill the cache with a random set of keys
// the keys will var in length between minKeyLen and maxKeyLen
function createRandomKeys(num, minKeyLen, maxKeyLen, obj) {
function rand(min, max) {
return Math.floor(Math.random() * (max - min)) + min;
}
var chars = "abcdefghijlkmnopqrstuvwzyz";
var len, key, numKeys = 0;
while (numKeys < num) {
// generate random key length
len = rand(minKeyLen, maxKeyLen);
key = "";
// now select len random chars and combine into a string
for (var j = 0; j < len; j++) {
key += chars.charAt(rand(0, chars.length))
}
// put this key into our object, only count it if it's not already there
if (!Object.prototype.hasOwnProperty.call(obj, key)) {
++numKeys;
obj[key] = true;
}
}
}
var cache = {};
// put all the keys into our object
createRandomKeys(200000, 3, 15, cache);
// now get the list of keys, just so we know what to fetch in our test
var keys = Object.keys(cache);
// now time getting every key
var total = 0;
var start = now();
for (var i = 0; i < keys.length; i++) {
if (cache[keys[i]]) {
++total;
}
}
var end = now();
var elapsed = end - start;
log("Elapsed time = " + elapsed + "ms for " + addCommas(keys.length) + " key lookups - found " + addCommas(total));
log(elapsed/keys.length + "ms per lookup");
log(addCommas((keys.length / elapsed).toFixed(2)) + " key lookups per ms");
// show some sample keys
log("<hr>Sample keys (first 100 keys):<br>");
log(keys.slice(0, 100).join(", "));
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]).