Counting the frequency of elements in an array in JavaScript - javascript

how do I count the frequency of the elements in the array, I'm new to Javascript and completely lost, I have looked at other answers here but can't get them to work for me. Any help is much appreciated.
function getText() {
var userText;
userText = document.InputForm.MyTextBox.value; //get text as string
alphaOnly(userText);
}
function alphaOnly(userText) {
var nuText = userText;
//result = nuText.split("");
var alphaCheck = /[a-zA-Z]/g; //using RegExp create variable to have only alphabetic characters
var alphaResult = nuText.match(alphaCheck); //get object with only alphabetic matches from original string
alphaResult.sort();
var result = freqLet(alphaResult);
document.write(countlist);
}
function freqLet(alphaResult) {
count = 0;
countlist = {
alphaResult: count
};
for (i = 0; i < alphaResult.length; i++) {
if (alphaResult[i] in alphaResult)
count[i] ++;
}
return countlist;
}


To count frequencies you should use an object which properties correspond to the letters occurring in your input string.
Also before incrementing the value of the property you should previously check whether this property exists or not.
function freqLet (alphaResult) {
var count = {};
countlist = {alphaResult:count};
for (i = 0; i < alphaResult.length; i++) {
var character = alphaResult.charAt(i);
if (count[character]) {
count[character]++;
} else {
count[character] = 1;
}
}
return countlist;
}


If you can use a third party library, underscore.js provides a function "countBy" that does pretty much exactly what you want.
_.countBy(userText, function(character) {
return character;
});
This should return an associative array of characters in the collection mapped to a count.
Then you could filter the keys of that object to the limited character set you need, again, using underscore or whatever method you like.


Do as below:
var __arr = [6,7,1,2,3,3,4,5,5,5]
function __freq(__arr){
var a = [], b = [], prev
__arr.sort((a,b)=>{return a- b} )
for(let i = 0; i<__arr.length; i++){
if(__arr[i] !== prev){
a.push(__arr[i])
b.push(1)
}else{
b[b.length - 1]++
}
prev = __arr[i]
}
return [a , b]
}

Related

How can i solve this exercixe with String.prototype?

How can i create this to console log like this?
String.prototype.sheldonize = function () {
return `knock ${this}`
}
'Penny'.sheldonize(3)
I have this code at the moment, but I dont know how to repeat knock more times

Use the repeat method to establish a number of 'knocks' in the line and to establish how many times the line should repeat
String.prototype.sheldonize = function (repeats) {
const line = `${'knock '.repeat(repeats)}${this}, `.repeat(repeats)
return `${line.substring(0,line.length-2)}.`;
}
console.log('Penny'.sheldonize(3));

Using For Loop and repeat method
String.prototype.sheldonize = function(count) {
let ans = "";
for (let i = 0; i < count; i++) {
ans += "knock ";
}
ans = `${ans}${this}, `.repeat(count)
ans = ans.substring(0, ans.length - 2) + "."
return ans;
}
console.log('Penny'.sheldonize(3))
console.log('Penny'.sheldonize(2))

Create an array and on each iteration up to n - 1 add the string to it, finally joining it up and returning the string from the function.
// If you're adding to a prototype it's always best
// to double check to see if the method already exists
// no matter (in this case) how unlikely
if (!('sheldonize' in String.prototype)) {
String.prototype.sheldonize = function (n) {
// Create the array
const out = [];
// Create the string
const knock = 'knock '.repeat(n);
// Loop until `n - 1` has been reached
// pushing the string into the array
// on each iteration
for (let i = 0; i < n; i++) {
out.push(`${knock}${this}`);
}
// Finally return the joined array
return out.join(', ');
}
}
console.log('Penny'.sheldonize(3));
console.log('Penny'.sheldonize(2));
console.log('John'.sheldonize(4));
Additional documentation
repeat

Problems with finding unique values in regEx

I have a javascript regEx that is supposed to find all values with curly brackets around them eg {} and return a list of the unique values. I thought that it was working perfectly but I found that it doesn't work depending on the sequence of values.
For example: If the target document contains {lorem}{lorem}{ipsem}{ipsem} the script logs what's wanted [lorem, ipsem] but {lorem}{ipsem}{ipsem}{lorem} the script logs [lorem, ipsem,lorem]. What am I doing wrong!?
function getVariables() {
var doc = DocumentApp.getActiveDocument();
var str = doc.getText(); //get the text of the document
var result = str.match(/{.*?}/g).map(function(val) {
return val.replace(/[\])}[{(]/g, "");
//return val.replace(/(^.*\[|\].*$)/g,'');
});
//The purpose of sort_unique is to find one of every value or string represented in an array
function sort_unique(arr) {
if (result.length === 0) return arr;
arr = arr.sort(function(a, b) {
return a * 1 - b * 1;
});
var ret = [arr[0]];
for (var i = 1; i < arr.length; i++) {
if (arr[i - 1] !== arr[i]) {
ret.push(arr[i]);
}
}
for (var index = 0; index < ret.length; index++) {
Logger.log(ret[index]);
}
return ret;
}
result = sort_unique(result);
Logger.log("Getting final result for front end....");
Logger.log(result);
return result;
}

I believe part of your problem is the sort method. If you replace
arr = arr.sort(function(a, b) {
return a * 1 - b * 1;
});
with
arr = arr.sort();
Then the function appears to work, at least on my side.
This will run in O(n log n) time. You can do better without sorting, if you store the values you've found so far in a map instead of an array. This would run in linear time.
(Also you'll want to replace if (result.length === 0) return arr; with if (arr.length === 0) return arr; just to make your sort_unique function completely independent of the surrounding function.)

The simplest method would be to use a Set. Store each of the regex matches in a set, then return Array.from(mySet).
var mySet = new Set();
str.match(/{.*?}/g).forEach(function(val) {
mySet.add(val.replace(/[\])}[{(]/g, ""));
});
return Array.from(mySet);
A set's add() function is O(1) so the total running time is O(n) where n is the number of matches in your string. Though, realistically, the regex search will be where most of the processing time occurs.

You check if the subsequent items are the same and those that are not subsequent land in the resulting array.
Check if the found value is in the result, and if not add the match, else, ignore.
Use the code like
function getVariables() {
var doc = DocumentApp.getActiveDocument();
var str = doc.getText(); //get the text of the document
var m, result=[], rx = /{([^{}]*)}/g;
while (m=rx.exec(str)) {
if (result.indexOf(m[1]) == -1) {
result.push(m[1]);
}
}
result.sort(); // If you really want to sort use this
// Logger.log(result); // View the result
}
The /{([^{}]*)}/g regex matches {, then captures into Group 1 zero or more chars other than { and }. So, the value you need is in m[1]. The if (result.indexOf(m[1]) == -1) checks if the value is in result.

Javascript sort and order

So i have this array
[ 'vendor/angular/angular.min.js',
'vendor/angular-nice-bar/dist/js/angular-nice-bar.min.js',
'vendor/angular-material/modules/js/core/core.min.js',
'vendor/angular-material/modules/js/backdrop/backdrop.min.js',
'vendor/angular-material/modules/js/dialog/dialog.min.js',
'vendor/angular-material/modules/js/button/button.min.js',
'vendor/angular-material/modules/js/icon/icon.min.js',
'vendor/angular-material/modules/js/tabs/tabs.min.js',
'vendor/angular-material/modules/js/content/content.min.js',
'vendor/angular-material/modules/js/toolbar/toolbar.min.js',
'vendor/angular-material/modules/js/input/input.min.js',
'vendor/angular-material/modules/js/divider/divider.min.js',
'vendor/angular-material/modules/js/menu/menu.min.js',
'vendor/angular-material/modules/js/select/select.min.js',
'vendor/angular-material/modules/js/radioButton/radioButton.min.js',
'vendor/angular-material/modules/js/checkbox/checkbox.min.js',
'vendor/angular-material/modules/js/switch/switch.min.js',
'vendor/angular-material/modules/js/tooltip/tooltip.min.js',
'vendor/angular-material/modules/js/toast/toast.min.js',
'vendor/angular-clipboard/angular-clipboard.js',
'vendor/angular-animate/angular-animate.min.js',
'vendor/angular-aria/angular-aria.min.js',
'vendor/angular-messages/angular-messages.min.js',
'vendor/angular-ui-router/release/angular-ui-router.js',
'src/app/about/about.js',
'src/app/hekate.cfg.js',
'src/app/hekate.ctrl.js',
'src/app/hekate.module.js',
'src/app/home/home.js',
'src/app/user/dialog/user.signIn.ctrl.js',
'src/app/user/dialog/user.signIn.module.js',
'src/app/user/user.cfg.js',
'src/app/user/user.ctrl.js',
'src/app/user/user.module.js',
'src/common/services/toast.service.js',
'templates-common.js',
'templates-app.js'
]
And taking the following part from the above array as example:
[
'src/app/hekate.cfg.js',
'src/app/hekate.ctrl.js',
'src/app/hekate.module.js',
]
I want to sort it like
[
'src/app/hekate.module.js',
'src/app/hekate.cfg.js',
'src/app/hekate.ctrl.js',
]
So more specific of what i want is to find in that array where string is duplicated and after check if has at the end [.cfg.js, .ctrl.js, .module.js] and automatic order them to [.module.js, .cfg.js, .ctrl.js]
Can anyone please help me with that?

A single sort proposal.
var array = ['src/app/about/about.js', 'src/app/hekate.cfg.js', 'src/app/hekate.ctrl.js', 'src/app/hekate.module.js', 'src/app/home/home.js', 'src/app/user/dialog/user.signIn.ctrl.js', 'src/app/user/dialog/user.signIn.module.js', 'src/app/user/user.cfg.js', 'src/app/user/user.ctrl.js', 'src/app/user/user.module.js'];
array.sort(function (a, b) {
function replaceCB(r, a, i) { return r.replace(a, i); }
var replace = ['.module.js', '.cfg.js', '.ctrl.js'];
return replace.reduce(replaceCB, a).localeCompare(replace.reduce(replaceCB, b));
});
document.write('<pre>' + JSON.stringify(array, 0, 4) + '</pre>');
To prevent so much replaces, i suggest to have a look to sorting with map.

You can try something like this:
Algo:
Group based on path and store file names as value.
Check for existence of one of special file ".cfg.js"
Sort following list based on custom sort.
Loop over object's property and join key with values to form full path again.
If you wish to sort full array, you can sort keys itself and then merge path with names. I have done this. If you do not wish to do this, just remove sort function from final loop.
Sample
var data=["vendor/angular/angular.min.js","vendor/angular-nice-bar/dist/js/angular-nice-bar.min.js","vendor/angular-material/modules/js/core/core.min.js","vendor/angular-material/modules/js/backdrop/backdrop.min.js","vendor/angular-material/modules/js/dialog/dialog.min.js","vendor/angular-material/modules/js/button/button.min.js","vendor/angular-material/modules/js/icon/icon.min.js","vendor/angular-material/modules/js/tabs/tabs.min.js","vendor/angular-material/modules/js/content/content.min.js","vendor/angular-material/modules/js/toolbar/toolbar.min.js","vendor/angular-material/modules/js/input/input.min.js","vendor/angular-material/modules/js/divider/divider.min.js","vendor/angular-material/modules/js/menu/menu.min.js","vendor/angular-material/modules/js/select/select.min.js","vendor/angular-material/modules/js/radioButton/radioButton.min.js","vendor/angular-material/modules/js/checkbox/checkbox.min.js","vendor/angular-material/modules/js/switch/switch.min.js","vendor/angular-material/modules/js/tooltip/tooltip.min.js","vendor/angular-material/modules/js/toast/toast.min.js","vendor/angular-clipboard/angular-clipboard.js","vendor/angular-animate/angular-animate.min.js","vendor/angular-aria/angular-aria.min.js","vendor/angular-messages/angular-messages.min.js","vendor/angular-ui-router/release/angular-ui-router.js","src/app/about/about.js","src/app/hekate.cfg.js","src/app/hekate.ctrl.js","src/app/hekate.module.js","src/app/home/home.js","src/app/user/dialog/user.signIn.ctrl.js","src/app/user/dialog/user.signIn.module.js","src/app/user/user.cfg.js","src/app/user/user.ctrl.js","src/app/user/user.module.js","src/common/services/toast.service.js","templates-common.js","templates-app.js"];
// Create groups based on path
var o = {};
data.forEach(function(item) {
var lastIndex = item.lastIndexOf('/') + 1;
var path = item.substring(0, lastIndex);
var fname = item.substring(lastIndex);
if (!o[path]) o[path] = [];
o[path].push(fname);
});
var manualOrder= [".module.js", ".cfg.js", ".ctrl.js"];
Array.prototype.fuzzyMatch = function(search){
return this.some(function(item){
return item.indexOf(search)>-1;
});
}
Array.prototype.fuzzySearchIndex = function(search){
var pos = -1;
this.forEach(function(item, index){
if(search.indexOf(item)>-1){
pos = index;
}
});
return pos;
}
function myCustomSort(a,b){
var a_pos = manualOrder.fuzzySearchIndex(a);
var b_pos = manualOrder.fuzzySearchIndex(b);
return a_pos > b_pos ? 1 : a_pos < b_pos ? -1 : 0;
}
// Check for ".cfg.js" and apply custom sort
for (var k in o) {
if (o[k].fuzzyMatch(".cfg.js")) {
o[k].sort(myCustomSort);
}
}
// Merge Path and names to create final value
var final = [];
Object.keys(o).sort().forEach(function(item) {
if (Array.isArray(o[item])) {
final = final.concat(o[item].map(function(fn) {
return item + fn
}));
} else
final = final.concat(o[item]);
});
console.log(final);

First make an array for names like 'hekate'.
Then make an array for final results.
We need 3 searching loops for ctrls, cfgs and modules.
If string contains arrayWithNames[0] + '.module' push the whole record to new array that you created. Same with ctrls and cfgs.
var allItems = []; //your array with all elements
var namesArray = [];
var finalResultsArray = [];
//fill name array here:
for(var i=0; i<=allItems.length; i++){
//you have to split string and find the module name (like 'hekate'). i hope you know how to split strings
}
//sort by modules, cfgs, ctrls:
for(var i=0; i<=namesArray.length; i++){
if(allItems[i].indexOf(namesArray[i] + '.module') > -1) {
finalResultsArray.push(allItems[i]);
}
}
for(var i=0; i<=namesArray.length; i++){
if(allItems[i].indexOf(namesArray[i] + '.cfg') > -1) {
finalResultsArray.push(allItems[i]);
}
}
for(var i=0; i<=namesArray.length; i++){
if(allItems[i].indexOf(namesArray[i] + '.ctrl') > -1) {
finalResultsArray.push(allItems[i]);
}
}
//now finalResultsArray have what you wanted

You can provide your own compare function to array.sort (see https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/Array/sort)
Write one that returns the correct order for modules, ctrls and cfgs:
It should first remove the suffixes, and if the rest is the same, use the correct logic to return the order according to the suffix. Otherwise return a value according to the alphabetical order.
Update
I didn't test this code (not is it finished), but it should look something like that:
arr.sort(function(a, b) {
if ((a.endsWith(".cfg.js") || a.endsWith(".ctrl.js") || a.endsWith(".module.js")) &&
(b.endsWith(".cfg.js") || b.endsWith(".ctrl.js") || b.endsWith(".module.js"))) {
var sortedSuffixes = {
".module.js": 0,
".cfg.js": 1,
".ctrl.js": 2
};
var suffixAIdx = a.lastIndexOf(".cfg.js");
if (suffixAIdx < 0) suffixAIdx = a.lastIndexOf(".ctrl.js");
if (suffixAIdx < 0) suffixAIdx = a.lastIndexOf(".module.js");
var suffixBIdx = b.lastIndexOf(".cfg.js");
if (suffixBIdx < 0) suffixBIdx = b.lastIndexOf(".ctrl.js");
if (suffixBIdx < 0) suffixBIdx = b.lastIndexOf(".module.js");
var prefixA = a.substring(0, suffixAIdx);
var prefixB = b.substring(0, suffixAIdx);
if (prefixA != prefixB)
{
return a.localeCompare(b);
}
var suffixA = a.substring(suffixAIdx);
var suffixB = b.substring(suffixBIdx);
return sortedSuffixes[suffixA] - sortedSuffixes[suffixB];
} else {
return a.localeCompare(b);
}
});
Update 2
Here is a fiddle (https://jsfiddle.net/d4fmc7ue/) that works.

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));

Javascript Function to split and return a value from a string

I am trying to grab a certain value. I am new to javascript and I can't figure out why this is not working.
If I parse "kid_2" I should get "kostas". Instead of "Kostas" I always get "02-23-2000". So I must have a logic problem in the loop but I am really stuck.
function getold_val(fieldname,str){
var chunks=str.split("||");
var allchunks = chunks.length-1;
for(k=0;k<allchunks;k++){
var n=str.indexOf(fieldname);
alert(chunks[k]);
if(n>0){
var chunkd=chunks[k].split("::");
alert(chunkd);
return chunkd[1];
}
}
}
var test = getold_val('kid_2','date_1::02-23-2000||date_2::06-06-1990||kid_1::George||kid_2::Kostas||');
alert(test);

A regex may be a little more appealing. Here's a fiddle:
function getValue(source, key){
return (new RegExp("(^|\\|)" + key + "::([^$\\|]+)", "i").exec(source) || {})[2];
}
getValue("date_1::02-23-2000||date_2::06-06-1990||kid_1::George||kid_2::Kostas||","kid_2");
But if you want something a little more involved, you can parse that string into a dictionary like so (fiddle):
function splitToDictionary(val, fieldDelimiter, valueDelimiter){
var dict = {},
fields = val.split(fieldDelimiter),
kvp;
for (var i = 0; i < fields.length; i++) {
if (fields[i] !== "") {
kvp = fields[i].split(valueDelimiter);
dict[kvp[0]] = kvp[1];
}
}
return dict;
}
var dict = splitToDictionary("date_1::02-23-2000||date_2::06-06-1990||kid_1::George||kid_2::Kostas||","||","::");
console.log(dict["date_1"]);
console.log(dict["date_2"]);
console.log(dict["kid_1"]);
console.log(dict["kid_2"]);​

This works, here's my fiddle.
function getold_val(fieldname,str) {
var chunks = str.split('||');
for(var i = 0; i < chunks.length-1; i++) {
if(chunks[i].indexOf(fieldname) >= 0) {
return(chunks[i].substring(fieldname.length+2));
}
}
}
alert(getold_val('kid_2', 'date_1::02-23-2000||date_2::06-06-1990||kid_1::George||kid_2::Kostas||'));
The issue with your code was (as #slebetman noticed as well) the fact that a string index can be 0 because it starts exactly in the first letter.
The code is almost the same as yours, I just didn't use the second .split('::') because I felt a .substring(...) would be easier.

There are two bugs. The first error is in the indexOf call:
var n = str.indexOf(fieldname);
This will always return a value greater than or equal to 0 since the field exists in the string. What you should be doing is:
var n = chunks[k].indexOf(fieldname);
The second error is in your if statement. It should be:
if(n >= 0) {
...
}
or
if(n > -1) {
...
}
The substring you are looking for could very well be the at the beginning of the string, in which case its index is 0. indexOf returns -1 if it cannot find what you're looking for.
That being said, here's a better way to do what you're trying to do:
function getold_val(fieldName, str) {
var keyValuePairs = str.split("||");
var returnValue = null;
if(/||$/.match(str)) {
keyValuePairs = keyValuePairs.slice(0, keyValuePairs.length - 1);
}
var found = false;
var i = 0;
while(i < keyValuePairs.length && !found) {
var keyValuePair = keyValuePairs[i].split("::");
var key = keyValuePair[0];
var value = keyValuePair[1];
if(fieldName === key) {
returnValue = value;
found = true;
}
i++;
}
return returnValue;
}

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