I have two arrays:
var arr1 = [1,2,3,4,5]
var arr2 = [7,1,8,2,12,3,4,28,5]
I need to go through arr2 looking for matches to arr1, but it has to be in order (1,2,3,4,5). As you can see in arr2, the order does exists, but there are some numbers in between.
[7,1,8,2,12,3,4,28,5]
I have about 50 arrays similar to arr2, so I need to look through each one, and when I find a match, push it out to a "results" object. Small issue though is that some arrays will not have the entire match, may only have 1,2,3 or any variation of the search. Also, if the array I'm searching in is NOT in order, (IE: starts at 2,3,4) skip over it entirely.
The idea is to loop through these arrays, and when I find a match, add a count to the results array.
For example, using arr1 as the search, go through these arrays:
[7,1,8,2,12,3,4,28,5],
[7,1,8,2,12,3,4],
[7,8,1,2],
[1,2,3]
and have a result that looks like this (a dictionary of what was searched for, and a count of what was found) :
{1:4, 2:4, 3:3, 4:2, 5:1}
I tried doing a bunch of for-loops, but I can't figure out how to skip over a number that I'm not looking for, and continue onto the next iteration, while saving the results into a dictionary object.
let list = [[7,1,8,2,12,3,4,28,5], [7,1,8,2,12,3,4], [7,8,1,2], [1,2,3]];
let search = [1, 2, 3, 4, 5];
// Initialize result with zeros:
let result = search.reduce((result, next) => {
result[next] = 0;
return result;
}, {});
// Increment result for items found:
list.forEach(array => {
for (let i = 0, j = 0; i < array.length && j < search.length; ++i) {
if (array[i] == search[j]) {
++result[search[j]];
++j;
}
}
});
console.log(result);
Essentially this:
var needle = [1,2,3,4,5]
var collection = [[7,1,8,2,12,3,4,28,5], [7,1,8,2,12,3,4], [7,8,1,2], [1,2,3]]
// start with an object
var results = {}
// populate object with zeros
needle.forEach(function (i) { results[i] = 0 })
// define an index to iterate through collection
var i = 0
// define an index to conditionally iterate through "arr1"
var j = 0
// define an index to iterate through collection arrays
var k = 0
// define surrogate for the arrays in the collection
var arr
while (i < collection.length) {
// get collection array
arr = collection[i]
// reset the indices
j = 0
k = 0
while (k < arr.length) {
// if same element on needle is in a collection array
if (needle[j] === arr[k]) {
// save it in an object starting at 1
results[needle[j]]++
j++ // increment needle
}
k++ // increment array in collection
}
i++ // increment collection
}
console.log(results) // {1:4, 2:4, 3:3, 4:2, 5:1}
I hope that helps!
var arr1 = [1,2,3,4,5];
var arr2 = [7,1,8,2,12,3,4,28,5];
function givenTwoArrays(a,b, obj){
var obj = obj || {};
var cond = true;
function otherMatch(indexFound,elementFound){
var indexOnA = a.indexOf(elementFound);
return a.some(function(ele, idx){
if(idx > indexOnA)
return b.some(function(bele,bidx){
return ele == bele && bidx < indexFound;
});
});
}
a.map(function(aele,idx){
if(cond){
var indexFound = b.findIndex(function(bele){
return aele == bele;
});
if(typeof indexFound !== 'undefined'){
if(!otherMatch(indexFound,aele)){
if(typeof obj[aele] !== 'undefined')
obj[aele]++;
else{
obj[aele] = 1;
}
} else {
cond = false;
}
}else
cond = false;
}
});
return obj;
}
console.log("first pass");
console.log(givenTwoArrays(arr1,arr2))
console.log("second pass");
console.log(givenTwoArrays(arr1,arr2,{
"1": 1,
"2": 1,
"3": 1,
"4": 1,
"5": 1
}));
I think this will work, just need to add a little recursion!
var orign = [1,2,3,4,5];
var arr = [[7,1,8,2,12,3,4,28,5], [7,1,8,2,12,3,4], [7,8,1,2], [1,2,3]];
//temp result
var arrTmp = [];
for (var x in arr){
var match = 0;
var mis = 1;
var curIndex = 0;
var cur = orign[curIndex];
var arrTmpX = [];
for(var y in arr[x]){
if(arr[x][y] !== cur){
mis=1;
}else{
//add match after mismatch
arrTmpX.push(cur);
curIndex++
cur = orign[curIndex];
}
}
arrTmp.push(arrTmpX);
}
//calc result
var result = {};
for (var x in orign){
result[orign[x]] = 0;
for(var y in arrTmp){
if(arrTmp[y].length>x)result[orign[x]]++;
}
}
console.log(result);
this works
Related
This question already has answers here:
Cartesian product of multiple arrays in JavaScript
(35 answers)
Closed 1 year ago.
I'm having trouble coming up with code to generate combinations from n number of arrays with m number of elements in them, in JavaScript. I've seen similar questions about this for other languages, but the answers incorporate syntactic or library magic that I'm unsure how to translate.
Consider this data:
[[0,1], [0,1,2,3], [0,1,2]]
3 arrays, with a different number of elements in them. What I want to do is get all combinations by combining an item from each array.
For example:
0,0,0 // item 0 from array 0, item 0 from array 1, item 0 from array 2
0,0,1
0,0,2
0,1,0
0,1,1
0,1,2
0,2,0
0,2,1
0,2,2
And so on.
If the number of arrays were fixed, it would be easy to make a hard coded implementation. But the number of arrays may vary:
[[0,1], [0,1]]
[[0,1,3,4], [0,1], [0], [0,1]]
Any help would be much appreciated.
Here is a quite simple and short one using a recursive helper function:
function cartesian(...args) {
var r = [], max = args.length-1;
function helper(arr, i) {
for (var j=0, l=args[i].length; j<l; j++) {
var a = arr.slice(0); // clone arr
a.push(args[i][j]);
if (i==max)
r.push(a);
else
helper(a, i+1);
}
}
helper([], 0);
return r;
}
Usage:
cartesian([0,1], [0,1,2,3], [0,1,2]);
To make the function take an array of arrays, just change the signature to function cartesian(args) instead of using rest parameter syntax.
I suggest a simple recursive generator function:
// JS
function* cartesianIterator(head, ...tail) {
const remainder = tail.length ? cartesianIterator(...tail) : [[]];
for (let r of remainder) for (let h of head) yield [h, ...r];
}
// get values:
const cartesian = items => [...cartesianIterator(items)];
console.log(cartesian(input));
// TS
function* cartesianIterator<T>(items: T[][]): Generator<T[]> {
const remainder = items.length > 1 ? cartesianIterator(items.slice(1)) : [[]];
for (let r of remainder) for (let h of items.at(0)!) yield [h, ...r];
}
// get values:
const cartesian = <T>(items: T[][]) => [...cartesianIterator(items)];
console.log(cartesian(input));
You could take an iterative approach by building sub arrays.
var parts = [[0, 1], [0, 1, 2, 3], [0, 1, 2]],
result = parts.reduce((a, b) => a.reduce((r, v) => r.concat(b.map(w => [].concat(v, w))), []));
console.log(result.map(a => a.join(', ')));
.as-console-wrapper { max-height: 100% !important; top: 0; }
After doing a little research I discovered a previous related question:
Finding All Combinations of JavaScript array values
I've adapted some of the code from there so that it returns an array of arrays containing all of the permutations:
function(arraysToCombine) {
var divisors = [];
for (var i = arraysToCombine.length - 1; i >= 0; i--) {
divisors[i] = divisors[i + 1] ? divisors[i + 1] * arraysToCombine[i + 1].length : 1;
}
function getPermutation(n, arraysToCombine) {
var result = [],
curArray;
for (var i = 0; i < arraysToCombine.length; i++) {
curArray = arraysToCombine[i];
result.push(curArray[Math.floor(n / divisors[i]) % curArray.length]);
}
return result;
}
var numPerms = arraysToCombine[0].length;
for(var i = 1; i < arraysToCombine.length; i++) {
numPerms *= arraysToCombine[i].length;
}
var combinations = [];
for(var i = 0; i < numPerms; i++) {
combinations.push(getPermutation(i, arraysToCombine));
}
return combinations;
}
I've put a working copy at http://jsfiddle.net/7EakX/ that takes the array you gave earlier ([[0,1], [0,1,2,3], [0,1,2]]) and outputs the result to the browser console.
const charSet = [["A", "B"],["C", "D", "E"],["F", "G", "H", "I"]];
console.log(charSet.reduce((a,b)=>a.flatMap(x=>b.map(y=>x+y)),['']))
Just for fun, here's a more functional variant of the solution in my first answer:
function cartesian() {
var r = [], args = Array.from(arguments);
args.reduceRight(function(cont, factor, i) {
return function(arr) {
for (var j=0, l=factor.length; j<l; j++) {
var a = arr.slice(); // clone arr
a[i] = factor[j];
cont(a);
}
};
}, Array.prototype.push.bind(r))(new Array(args.length));
return r;
}
Alternative, for full speed we can dynamically compile our own loops:
function cartesian() {
return (cartesian.cache[arguments.length] || cartesian.compile(arguments.length)).apply(null, arguments);
}
cartesian.cache = [];
cartesian.compile = function compile(n) {
var args = [],
indent = "",
up = "",
down = "";
for (var i=0; i<n; i++) {
var arr = "$"+String.fromCharCode(97+i),
ind = String.fromCharCode(105+i);
args.push(arr);
up += indent+"for (var "+ind+"=0, l"+arr+"="+arr+".length; "+ind+"<l"+arr+"; "+ind+"++) {\n";
down = indent+"}\n"+down;
indent += " ";
up += indent+"arr["+i+"] = "+arr+"["+ind+"];\n";
}
var body = "var res=[],\n arr=[];\n"+up+indent+"res.push(arr.slice());\n"+down+"return res;";
return cartesian.cache[n] = new Function(args, body);
}
var f = function(arr){
if(typeof arr !== 'object'){
return false;
}
arr = arr.filter(function(elem){ return (elem !== null); }); // remove empty elements - make sure length is correct
var len = arr.length;
var nextPerm = function(){ // increase the counter(s)
var i = 0;
while(i < len)
{
arr[i].counter++;
if(arr[i].counter >= arr[i].length){
arr[i].counter = 0;
i++;
}else{
return false;
}
}
return true;
};
var getPerm = function(){ // get the current permutation
var perm_arr = [];
for(var i = 0; i < len; i++)
{
perm_arr.push(arr[i][arr[i].counter]);
}
return perm_arr;
};
var new_arr = [];
for(var i = 0; i < len; i++) // set up a counter property inside the arrays
{
arr[i].counter = 0;
}
while(true)
{
new_arr.push(getPerm()); // add current permutation to the new array
if(nextPerm() === true){ // get next permutation, if returns true, we got them all
break;
}
}
return new_arr;
};
Here's another way of doing it. I treat the indices of all of the arrays like a number whose digits are all different bases (like time and dates), using the length of the array as the radix.
So, using your first set of data, the first digit is base 2, the second is base 4, and the third is base 3. The counter starts 000, then goes 001, 002, then 010. The digits correspond to indices in the arrays, and since order is preserved, this is no problem.
I have a fiddle with it working here: http://jsfiddle.net/Rykus0/DS9Ea/1/
and here is the code:
// Arbitrary base x number class
var BaseX = function(initRadix){
this.radix = initRadix ? initRadix : 1;
this.value = 0;
this.increment = function(){
return( (this.value = (this.value + 1) % this.radix) === 0);
}
}
function combinations(input){
var output = [], // Array containing the resulting combinations
counters = [], // Array of counters corresponding to our input arrays
remainder = false, // Did adding one cause the previous digit to rollover?
temp; // Holds one combination to be pushed into the output array
// Initialize the counters
for( var i = input.length-1; i >= 0; i-- ){
counters.unshift(new BaseX(input[i].length));
}
// Get all possible combinations
// Loop through until the first counter rolls over
while( !remainder ){
temp = []; // Reset the temporary value collection array
remainder = true; // Always increment the last array counter
// Process each of the arrays
for( i = input.length-1; i >= 0; i-- ){
temp.unshift(input[i][counters[i].value]); // Add this array's value to the result
// If the counter to the right rolled over, increment this one.
if( remainder ){
remainder = counters[i].increment();
}
}
output.push(temp); // Collect the results.
}
return output;
}
// Input is an array of arrays
console.log(combinations([[0,1], [0,1,2,3], [0,1,2]]));
You can use a recursive function to get all combinations
const charSet = [["A", "B"],["C", "D", "E"],["F", "G", "H", "I"]];
let loopOver = (arr, str = '', final = []) => {
if (arr.length > 1) {
arr[0].forEach(v => loopOver(arr.slice(1), str + v, final))
} else {
arr[0].forEach(v => final.push(str + v))
}
return final
}
console.log(loopOver(charSet))
This code can still be shorten using ternary but i prefer the first version for readability 😊
const charSet = [["A", "B"],["C", "D", "E"],["F", "G", "H", "I"]];
let loopOver = (arr, str = '') => arr[0].map(v => arr.length > 1 ? loopOver(arr.slice(1), str + v) : str + v).flat()
console.log(loopOver(charSet))
Another implementation with ES6 recursive style
Array.prototype.cartesian = function(a,...as){
return a ? this.reduce((p,c) => (p.push(...a.cartesian(...as).map(e => as.length ? [c,...e] : [c,e])),p),[])
: this;
};
console.log(JSON.stringify([0,1].cartesian([0,1,2,3], [[0],[1],[2]])));
This question already has answers here:
Cartesian product of multiple arrays in JavaScript
(35 answers)
Closed 1 year ago.
I'm having trouble coming up with code to generate combinations from n number of arrays with m number of elements in them, in JavaScript. I've seen similar questions about this for other languages, but the answers incorporate syntactic or library magic that I'm unsure how to translate.
Consider this data:
[[0,1], [0,1,2,3], [0,1,2]]
3 arrays, with a different number of elements in them. What I want to do is get all combinations by combining an item from each array.
For example:
0,0,0 // item 0 from array 0, item 0 from array 1, item 0 from array 2
0,0,1
0,0,2
0,1,0
0,1,1
0,1,2
0,2,0
0,2,1
0,2,2
And so on.
If the number of arrays were fixed, it would be easy to make a hard coded implementation. But the number of arrays may vary:
[[0,1], [0,1]]
[[0,1,3,4], [0,1], [0], [0,1]]
Any help would be much appreciated.
Here is a quite simple and short one using a recursive helper function:
function cartesian(...args) {
var r = [], max = args.length-1;
function helper(arr, i) {
for (var j=0, l=args[i].length; j<l; j++) {
var a = arr.slice(0); // clone arr
a.push(args[i][j]);
if (i==max)
r.push(a);
else
helper(a, i+1);
}
}
helper([], 0);
return r;
}
Usage:
cartesian([0,1], [0,1,2,3], [0,1,2]);
To make the function take an array of arrays, just change the signature to function cartesian(args) instead of using rest parameter syntax.
I suggest a simple recursive generator function:
// JS
function* cartesianIterator(head, ...tail) {
const remainder = tail.length ? cartesianIterator(...tail) : [[]];
for (let r of remainder) for (let h of head) yield [h, ...r];
}
// get values:
const cartesian = items => [...cartesianIterator(items)];
console.log(cartesian(input));
// TS
function* cartesianIterator<T>(items: T[][]): Generator<T[]> {
const remainder = items.length > 1 ? cartesianIterator(items.slice(1)) : [[]];
for (let r of remainder) for (let h of items.at(0)!) yield [h, ...r];
}
// get values:
const cartesian = <T>(items: T[][]) => [...cartesianIterator(items)];
console.log(cartesian(input));
You could take an iterative approach by building sub arrays.
var parts = [[0, 1], [0, 1, 2, 3], [0, 1, 2]],
result = parts.reduce((a, b) => a.reduce((r, v) => r.concat(b.map(w => [].concat(v, w))), []));
console.log(result.map(a => a.join(', ')));
.as-console-wrapper { max-height: 100% !important; top: 0; }
After doing a little research I discovered a previous related question:
Finding All Combinations of JavaScript array values
I've adapted some of the code from there so that it returns an array of arrays containing all of the permutations:
function(arraysToCombine) {
var divisors = [];
for (var i = arraysToCombine.length - 1; i >= 0; i--) {
divisors[i] = divisors[i + 1] ? divisors[i + 1] * arraysToCombine[i + 1].length : 1;
}
function getPermutation(n, arraysToCombine) {
var result = [],
curArray;
for (var i = 0; i < arraysToCombine.length; i++) {
curArray = arraysToCombine[i];
result.push(curArray[Math.floor(n / divisors[i]) % curArray.length]);
}
return result;
}
var numPerms = arraysToCombine[0].length;
for(var i = 1; i < arraysToCombine.length; i++) {
numPerms *= arraysToCombine[i].length;
}
var combinations = [];
for(var i = 0; i < numPerms; i++) {
combinations.push(getPermutation(i, arraysToCombine));
}
return combinations;
}
I've put a working copy at http://jsfiddle.net/7EakX/ that takes the array you gave earlier ([[0,1], [0,1,2,3], [0,1,2]]) and outputs the result to the browser console.
const charSet = [["A", "B"],["C", "D", "E"],["F", "G", "H", "I"]];
console.log(charSet.reduce((a,b)=>a.flatMap(x=>b.map(y=>x+y)),['']))
Just for fun, here's a more functional variant of the solution in my first answer:
function cartesian() {
var r = [], args = Array.from(arguments);
args.reduceRight(function(cont, factor, i) {
return function(arr) {
for (var j=0, l=factor.length; j<l; j++) {
var a = arr.slice(); // clone arr
a[i] = factor[j];
cont(a);
}
};
}, Array.prototype.push.bind(r))(new Array(args.length));
return r;
}
Alternative, for full speed we can dynamically compile our own loops:
function cartesian() {
return (cartesian.cache[arguments.length] || cartesian.compile(arguments.length)).apply(null, arguments);
}
cartesian.cache = [];
cartesian.compile = function compile(n) {
var args = [],
indent = "",
up = "",
down = "";
for (var i=0; i<n; i++) {
var arr = "$"+String.fromCharCode(97+i),
ind = String.fromCharCode(105+i);
args.push(arr);
up += indent+"for (var "+ind+"=0, l"+arr+"="+arr+".length; "+ind+"<l"+arr+"; "+ind+"++) {\n";
down = indent+"}\n"+down;
indent += " ";
up += indent+"arr["+i+"] = "+arr+"["+ind+"];\n";
}
var body = "var res=[],\n arr=[];\n"+up+indent+"res.push(arr.slice());\n"+down+"return res;";
return cartesian.cache[n] = new Function(args, body);
}
var f = function(arr){
if(typeof arr !== 'object'){
return false;
}
arr = arr.filter(function(elem){ return (elem !== null); }); // remove empty elements - make sure length is correct
var len = arr.length;
var nextPerm = function(){ // increase the counter(s)
var i = 0;
while(i < len)
{
arr[i].counter++;
if(arr[i].counter >= arr[i].length){
arr[i].counter = 0;
i++;
}else{
return false;
}
}
return true;
};
var getPerm = function(){ // get the current permutation
var perm_arr = [];
for(var i = 0; i < len; i++)
{
perm_arr.push(arr[i][arr[i].counter]);
}
return perm_arr;
};
var new_arr = [];
for(var i = 0; i < len; i++) // set up a counter property inside the arrays
{
arr[i].counter = 0;
}
while(true)
{
new_arr.push(getPerm()); // add current permutation to the new array
if(nextPerm() === true){ // get next permutation, if returns true, we got them all
break;
}
}
return new_arr;
};
Here's another way of doing it. I treat the indices of all of the arrays like a number whose digits are all different bases (like time and dates), using the length of the array as the radix.
So, using your first set of data, the first digit is base 2, the second is base 4, and the third is base 3. The counter starts 000, then goes 001, 002, then 010. The digits correspond to indices in the arrays, and since order is preserved, this is no problem.
I have a fiddle with it working here: http://jsfiddle.net/Rykus0/DS9Ea/1/
and here is the code:
// Arbitrary base x number class
var BaseX = function(initRadix){
this.radix = initRadix ? initRadix : 1;
this.value = 0;
this.increment = function(){
return( (this.value = (this.value + 1) % this.radix) === 0);
}
}
function combinations(input){
var output = [], // Array containing the resulting combinations
counters = [], // Array of counters corresponding to our input arrays
remainder = false, // Did adding one cause the previous digit to rollover?
temp; // Holds one combination to be pushed into the output array
// Initialize the counters
for( var i = input.length-1; i >= 0; i-- ){
counters.unshift(new BaseX(input[i].length));
}
// Get all possible combinations
// Loop through until the first counter rolls over
while( !remainder ){
temp = []; // Reset the temporary value collection array
remainder = true; // Always increment the last array counter
// Process each of the arrays
for( i = input.length-1; i >= 0; i-- ){
temp.unshift(input[i][counters[i].value]); // Add this array's value to the result
// If the counter to the right rolled over, increment this one.
if( remainder ){
remainder = counters[i].increment();
}
}
output.push(temp); // Collect the results.
}
return output;
}
// Input is an array of arrays
console.log(combinations([[0,1], [0,1,2,3], [0,1,2]]));
You can use a recursive function to get all combinations
const charSet = [["A", "B"],["C", "D", "E"],["F", "G", "H", "I"]];
let loopOver = (arr, str = '', final = []) => {
if (arr.length > 1) {
arr[0].forEach(v => loopOver(arr.slice(1), str + v, final))
} else {
arr[0].forEach(v => final.push(str + v))
}
return final
}
console.log(loopOver(charSet))
This code can still be shorten using ternary but i prefer the first version for readability 😊
const charSet = [["A", "B"],["C", "D", "E"],["F", "G", "H", "I"]];
let loopOver = (arr, str = '') => arr[0].map(v => arr.length > 1 ? loopOver(arr.slice(1), str + v) : str + v).flat()
console.log(loopOver(charSet))
Another implementation with ES6 recursive style
Array.prototype.cartesian = function(a,...as){
return a ? this.reduce((p,c) => (p.push(...a.cartesian(...as).map(e => as.length ? [c,...e] : [c,e])),p),[])
: this;
};
console.log(JSON.stringify([0,1].cartesian([0,1,2,3], [[0],[1],[2]])));
Write a function that takes in a list and returns a list with all of the duplicates removed (list will only have unique numbers).
Here's what I have so far:
var lista = [1,4,5,1,1,3,5,6,4,4,3];
function dupRemove (lista) {
//Sort the array in case it isn't sorted
lista.sort();
//Object to store duplicates and unique numbers
var listNumbers = {
"Duplicate Numbers": [],
"Unique Numbers": []
};
for (var i = 0; i < lista.length; i++) {
//check if it is not equal to the index of the array before it and after. if it isn't, that means its unique, push it in the uniques array.
if (lista[i] !== lista[i-1] && lista[i] !== lista[i+1]) {
listNumbers["Unique Numbers"].push(lista[i]);
} else {
listNumbers["Duplicate Numbers"].push(lista[i]);
}
}
return listNumbers;
}
Currently, my solution returns an object with keys with the values of "Duplicates": 1, 1, 1, 3, 3, 4, 4, 4, 5, 5 and "Uniques": 6.
How do I remove the duplicates from duplicates and then join these two keys into a single array?
Thank you.
that answer is seriously over -engineered- all you need to to is push all values into a new array if they are not already in it.
function=removeDups()
{
var lista = [1,4,5,1,1,3,5,6,4,4,3];
var uniqueValues=[];
var duplicateValues=[];
for(i=0;i<lista.length;i++)
{
if(uniqueValues.indexof(lista[i] == -1){uniqueValues.push(lista[i]}else{duplicateValues.push(lista[i]}
}
}
You could just use the default filter method that is on all Arrays
You don't need the sort function either. If the item is already found using the indexOf method it will not be added to the newly returned array created by the filter method
var list = [1,4,5,1,1,3,5,6,4,4,3];
function removeDup (arr) {
return arr.filter(function(item, pos) {
return arr.indexOf(item) == pos;
})
}
var sortedList = removeDup(list).sort(function(a,b){
return a - b
})
document.getElementsByTagName('div')[0].textContent = sortedList
<div></div>
Kind of a non elegant solution but it gives you the two arrays: one with the duplicate values and one with the unique ones. Since you cannot rely on .sort() you can just count things.
Function checkList will give you back those two arrays.
var list = [1,4,5,1,1,3,5,6,4,4,3];
console.log(checkList(list));
function checkList(list) {
var uniques = []; // will be [6]
var dups = []; // will be [1, 4, 5, 3]
var checked = []; // save what you have already checked so far
for(i = 0; i < list.length; i++) {
if(notChecked(list[i], checked)) {
checked.push(list[i]);
if(count(list[i], list) > 1) {
dups.push(list[i]);
} else {
uniques.push(list[i]);
}
}
}
return {dups: dups, uniques: uniques}
}
// count how many num in arr
function count(num, arr) {
var count = 0;
var i;
for(i = 0; i < arr.length; i++) {
if(arr[i] == num) count++;
if(count > 1) return count;
}
return count;
}
// check if num has not been checked
function notChecked(num, arr) {
return (arr.indexOf(num) == -1) ? true : false;
}
I am trying to display the contents of an array in a more readable way, this is the array:
["malevolent", "pariah", "judicious", "pariah", "judicious"]
I'm simply adding that array to an HTML element to display it, but I want to display it like this:
malevolent, pariah x 2, judicious x2
How would I do this?
It's quite simple actually:
var myArray = new Array("a", "b", "c", "b", "a");
var newObject = {};
// Iterate over the array
for(var i = 0; i < myArray.length; i++){
// If the new object already contains the key (e.g. a, b, or c), increment value by one
if(myArray[i] in newObject){
newObject[myArray[i]]++;
} else {
// Otherwise add a key (e.g. a, b, or c) to the object and assign 1 to it (first occurence)
newObject[myArray[i]] = 1;
}
}
// Write the resulting object to console
window.console && console.log(newObject);
newObject contains a list of keys (a,b,c) and values (number of occurrences of each key). You can than use that data to output it in any format you like, but that's left up to you as an excercise.
You can try the following:
var myArray = ["malevolent", "pariah", "judicious", "pariah", "judicious"];
var resultArray = [];
var countArray = [];
for(index in myArray) {
var element = myArray[index];
var isInArray = resultArray.indexOf(element);
if(isInArray !== -1) {
var tmpCnt = countArray[isInArray];
tmpCnt++;
countArray[isInArray] = tmpCnt;
} else {
resultArray.push(element);
countArray.push(1);
}
}
console.log(resultArray);
console.log(countArray);
Felix Kling provided a Link to an answer on how to count your elements. I just shamelessly use the reduce method described there and then just iterate over the object to build a string.
var a = ["malevolent", "pariah", "judicious", "pariah", "judicious"].reduce(function (acc, curr) {
if (typeof acc[curr] == 'undefined') {
acc[curr] = 1;
} else {
acc[curr] += 1;
}
return acc;
}, {});
var out = "";
for (var k in a) {
out += k + " x " + a[k] + "; ";
}
console.log(out);
try this
var arr = new Array("malevolent", "pariah", "judicious", "pariah", "judicious");
var new_arr1 = new Array(); // for containing distinct values like pariah
var new_arr2 = new Array(); // for containing distinct values count like 2 for pariah
// both will be on same index in new_arr1 and new_arr2
for(var i=0; i<arr.length; i++)
{
// fetch every value of arr
var indx = $.inArray(arr[i], new_arr1); // check value is exists in new_arr1 and get index
if(indx > -1) // values exists in new_arr1
{
var v = new_arr2[indx]+1; // increment the previous count +1
new_arr2[indx] = v; // update it on the index of new_arr2
}
else
{
// if value not in new_arr1 means the value comes first time
var l = new_arr1.length;
new_arr1[l] = arr[i]; // insert value in new_arr1
new_arr2[l] = 1; // initate count 1 for the same index of new value in new_arr2
}
}
// now new_arr1 will contains the distinct values
// and new_arr2 contains the count for distinct values
// eg new_arr1[0] = 'malevolent';
// new_arr2[0] = 1;
// new_arr1[1] = 'pariah';
// new_arr2[1] = 2;
// now you can fetch distinct values and their count like given below
for(var i=0; i<new_arr1.length; i++)
{
var str = new_arr1[i]+" X "+new_arr2[i];
alert(str);
}
See FIDDLE
I need to know if one or more duplicates exist in a list. Is there a way to do this without travelling through the list more than once?
Thanks guys for the suggestions. I ended up using this because it was the simplest to implement:
var names = [];
var namesLen = names.length;
for (i=0; i<namesLen; i++) {
for (x=0; x<namesLen; x++) {
if (names[i] === names[x] && (i !== x)) {alert('dupe')}
}
}
Well the usual way to do that would be to put each item in a hashmap dictionary and you could check if it was already inserted. If your list is of objects they you would have to create your own hash function on the object as you would know what makes each one unique. Check out the answer to this question.
JavaScript Hashmap Equivalent
This method uses an object as a lookup table to keep track of how many and which dups were found. It then returns an object with each dup and the dup count.
function findDups(list) {
var uniques = {}, val;
var dups = {};
for (var i = 0, len = list.length; i < len; i++) {
val = list[i];
if (val in uniques) {
uniques[val]++;
dups[val] = uniques[val];
} else {
uniques[val] = 1;
}
}
return(dups);
}
var data = [1,2,3,4,5,2,3,2,6,8,9,9];
findDups(data); // returns {2: 3, 3: 2, 9: 2}
var data2 = [1,2,3,4,5,6,7,8,9];
findDups(data2); // returns {}
var data3 = [1,1,1,1,1,2,3,4];
findDups(data3); // returns {1: 5}
Since we now have ES6 available with the built-in Map object, here's a version of findDups() that uses the Map object:
function findDups(list) {
const uniques = new Set(); // set of items found
const dups = new Map(); // count of items that have dups
for (let val of list) {
if (uniques.has(val)) {
let cnt = dups.get(val) || 1;
dups.set(val, ++cnt);
} else {
uniques.add(val);
}
}
return dups;
}
var data = [1,2,3,4,5,2,3,2,6,8,9,9];
log(findDups(data)); // returns {2 => 3, 3 => 2, 9 => 2}
var data2 = [1,2,3,4,5,6,7,8,9];
log(findDups(data2)); // returns empty map
var data3 = [1,1,1,1,1,2,3,4];
log(findDups(data3)); // returns {1 => 5}
// display resulting Map object (only used for debugging display in snippet)
function log(map) {
let output = [];
for (let [key, value] of map) {
output.push(key + " => " + value);
}
let div = document.createElement("div");
div.innerHTML = "{" + output.join(", ") + "}";
document.body.appendChild(div);
}
If your strings are in an array (A) you can use A.some-
it will return true and quit as soon as it finds a duplicate,
or return false if it has checked them all without any duplicates.
has_duplicates= A.some(function(itm){
return A.indexOf(itm)===A.lastIndexOf(itm);
});
If your list was just words or phrases, you could put them into an associative array.
var list=new Array("foo", "bar", "foobar", "foo", "bar");
var newlist= new Array();
for(i in list){
if(newlist[list[i]])
newlist[list[i]]++;
else
newlist[list[i]]=1;
}
Your final array should look like this:
"foo"=>2, "bar"=>2, "foobar"=>1