I am struggling to create a simple function that will taking an array such as
["chicken","10 lbs","hot sauce","4 cups","celery","1 stalk"]
and turn it into a multidimensional array that looks like this
[["chicken","10 lbs"],["hot sauce","4 cups"],["celery","1 stalk"]]
essentially index 0 and 1 merge into a new sub array then 2 and 3 merge and so on...
i am trying for loops where i increases +2 but i can't get it quite right, i thought this would be a simple solution but the loops i've made so far are almost crashing my browser...any help would be appreciated. thanks
EDIT: WOW Thanks for all the rapid responses! I looked at everything and learned a lot!
Using Array#reduce(), Array#concat() & Array#slice()
var data = ["chicken", "10 lbs", "hot sauce", "4 cups", "celery", "1 stalk"];
var res = data.reduce((a, c, i) => {
return i % 2 === 0 ? a.concat([data.slice(i, i + 2)]) : a;
}, []);
console.log(res)
Hope this helps...
I added some comments but feel free to ask if you need more explanation.
There are definitely more advanced (and much shorter!) ways of doing this, but this is the most intuitive and easiest to understand, in my opinion.
Right now, if there is an odd amount of elements, the last element will end up being "undefined." Let me know if you'd like this to work differently.
var array = ["chicken","10 lbs","hot sauce","4 cups","celery","1 stalk"];
function combineTwo(inputArray) {
//Starting with the beginning of the array, this function combines index 0 with 1, 2 with 3, and so on for the entire length of the array
var result = []; //this will the variable that we store our result in
for (var i = 0; i < inputArray.length; i+=2) {
//This for loop iterates through every other index of the array... for example: 0, 2, 4, etc.
result.push([inputArray[i], inputArray[i+1]]); //Adds i and i+1 as a new array to the result array
}
return result;
}
console.log(combineTwo(array));
There are many ways to do this. Here, we are using the Array.map() method to return items into a new array and then cleaning out any undefined elements using Array.filter().
var startArray = ["chicken","10 lbs","hot sauce","4 cups","celery","1 stalk"];
// Create a new array by iterating the first one...
var resultArray = startArray.map(function(item, index, arry){
// If we are on an even index...
if(index % 2 === 0){
// Send the item and the one that follows it into the new array
return [item, arry[index + 1]];
}
});
// The resulting array will contain undefined for each enumeration that didn't
// return anything, so we can clean that up by filtering out the undefined elements
resultArray = resultArray.filter(function( element ) {
return element !== undefined;
});
console.log(resultArray);
However, it seems that you want to make an array of ingredients and ingredients combine to make a final product, so creating an object would be more appropriate, since objects are really nothing more than groupings of key/value pairs and can be more versatile if you plan on doing OOP operations. The result here is an array of recipe objects:
// Create a new array that will store recipe objects
var cookbook = [];
function addRecipeToCookbook(ary){
// Create a "recipe" object
var recipe = {};
// Iterate the passed array...
ary.forEach(function(item, index, array){
// If we are on an even index...
if(index % 2 === 0){
// Add the item as a new property of the recipe object with the quantity as the value
recipe[item] = array[index + 1];
}
});
// Add the object to the new array
cookbook.push(recipe);
}
// Now you can add as many recipies as you like:
var recipeArray1 = ["chicken","10 lbs","hot sauce","4 cups","celery","1 stalk"];
var recipeArray2 = ["pork","2 lbs","BBQ sauce","2 cups","onion","3 ounces"];
var recipeArray3 = ["beef","5 lbs","worchestishire sauce","4 ounces","garlic","1 clove"];
// Just call the reusable function and pass a recipe array:
addRecipeToCookbook(recipeArray1);
addRecipeToCookbook(recipeArray2);
addRecipeToCookbook(recipeArray3);
console.log(cookbook);
Another option is to take in a second argument that dictates how many items you would like in each subarray. For instance:
function squash(arr, num) {
let results = [];
let start, end;
for (let i = 1; i < arr.length; i++) {
start = (i - 1) * num;
end = i * num;
let sliced = arr.slice(start, end);
if (start >= arr.length) { // need to make sure we don't go passed the end
break;
}
results.push(sliced);
}
return results;
}
Just another way to do this using .reduce():
var array = ["chicken","10 lbs","hot sauce","4 cups","celery","1 stalk"];
var array2d = array.reduce((array, element, index) => {
if (index % 2 > 0) {
array.push(array.pop().concat([element]));
} else {
array.push([element]);
}
return array;
}, []);
console.log(array2d);
A more generic method that could be used might look like the following:
function group(amount) {
return (array) => array
.reduce((array, element, index) => {
if (index % amount > 0) {
array.push(array.pop().concat([element]));
} else {
array.push([element]);
}
return array;
}, []);
}
var a = ["chicken","10 lbs","hot sauce","4 cups","celery","1 stalk"];
var b = ["chicken","10","lbs","hot sauce","4","cups","celery","1","stalk"];
var group2 = group(2);
var group3 = group(3);
var a2d = group2(a);
var b2d = group3(b);
console.log(a2d);
console.log(b2d);
most intuitive for me, just keep yanking items off the original array until it is empty:
let array = ["chicken", "10 lbs", "hot sauce", "4 cups", "celery", "1 stalk"],
newArray = [];
while (array.length) {
newArray.push([array.shift(), array.shift()]);
}
console.log(newArray);
Related
It goes something like this where I have a london array containing more than 10 million data
london = ['dwig7xmW','gIzbnHNI' ...]
And now I have a userTraveled which also contains millions of data
userTraveled = ['ntuJV09a' ...]
Now what's the most efficient way to split userTraveled into inLondon and notInLondon.
My attempt.
inLondon = []
notInLondon = []
userTraveled.forEach((p) => london.includes(p) ? inLondon.push(p) : notInLondon.push(p))
london.includes(p) will do a linear search over the array. Doing that for every userTraveled is horribly inefficient. Use a Set instead:
const usersInLondon = [], usersNotInLondon = [];
const lookup = new Set(london);
for (const p of usersTraveled) {
(lookup.has(p) ? usersInLondon : usersNotInLondon).push(p);
}
I can offer a O(n*log(n)) solution instead of your O(n^2), first order the passwords and later use the binary search on it instead of the include to search for an item
Hope it helps =)
const london = ['dwig7xmW','gIzbnHNI']
const userTraveled = ['ntuJV09a', 'dwig7xmW']
let inLondon = []
let notInLondon = []
const sortedlondon=london.sort();
userTraveled.forEach((p) => (binarySearch(sortedlondon,p)!=-1 ? inLondon.push(p) : notInLondon.push(p)))
//https://www.htmlgoodies.com/javascript/how-to-search-a-javascript-string-array-using-a-binary-search/
function binarySearch(items, value){
var startIndex = 0,
stopIndex = items.length - 1,
middle = Math.floor((stopIndex + startIndex)/2);
while(items[middle] != value && startIndex < stopIndex){
//adjust search area
if (value < items[middle]){
stopIndex = middle - 1;
} else if (value > items[middle]){
startIndex = middle + 1;
}
//recalculate middle
middle = Math.floor((stopIndex + startIndex)/2);
}
//make sure it's the right value
return (items[middle] != value) ? -1 : middle;
}
I hope you are not using these data in a wrong way.
const passwords = ['a', 'b']
const rawPasswords = ['c', 'b'];
const setPasswords = new Set(passwords)
const uniquePassword = [];
const usedPassword = [];
rawPasswords.forEach(rp => {
if (setPasswords.has(rp)) {
usedPassword.push(rp)
} else {
uniquePassword.push(rp)
}
})
console.log(uniquePassword, usedPassword)
Referring to this answer for performance tests: Get all unique values in a JavaScript array (remove duplicates) the best solution in your case would be to use an Object. Since you require to know about the duplicates and not just remove them.
function uniqueArray( ar ) {
var j = {};
var k = [];
var unique;
ar.forEach( function(v) {
if(j.hasOwnProperty(v)){
k.push(v);
} else {
j[v] = v;
}
});
unique = Object.keys(j).map(function(v){
return j[v];
});
return [unique, k];
}
var arr = [1, 1, 2, 3, 4, 5, 4, 3];
console.log(uniqueArray(arr));
First it loops through the input array and checks if the value is already existing as a key on the object. If that's not the case, it adds it. If it is, it pushes the value to another array. Since objects use a hash, the Javascript engine can work faster with it.
Secondly it goes through the object's keys to turn it back into an array and finally returns both. I didn't add this explanation because the provided reference already explained it.
The result will be an array containing 2 arrays. First the array with unique values, second the array with duplicates.
I've searched SO for a way to do this but most questions only support two arrays (I need a solution for multiple arrays).
I don't want to compare exact objects, I want to compare objects by their ID, as their other parameters may differ.
So here's the example data:
data1 = [{'id':'13','name':'sophie'},{'id':'22','name':'andrew'}, etc.]
data2 = [{'id':'22','name':'mary'},{'id':'85','name':'bill'}, etc.]
data3 = [{'id':'20','name':'steve'},{'id':'22','name':'john'}, etc.]
...
I'd like to return all objects whose ID appears in all arrays, and I don't mind which of the set of matched objects is returned.
So, from the data above, I'd expect to return any one of the following:
{'id':'22','name':'andrew'}
{'id':'22','name':'mary'}
{'id':'22','name':'john'}
Thanks
First, you really need an array of arrays - using a numeric suffix is not extensible:
let data = [ data1, data2, ... ];
Since you've confirmed that the IDs are unique within each sub array, you can simplify the problem by merging the arrays, and then finding out which elements occur n times, where n is the original number of sub arrays:
let flattened = data.reduce((a, b) => a.concat(b), []);
let counts = flattened.reduce(
(map, { id }) => map.set(id, (map.get(id) || 0) + 1), new Map()
);
and then you can pick out those objects that did appear n times, in this simple version they'll all come from the first sub array:
let found = data[0].filter(({ id }) => counts.get(id) === data.length);
Picking an arbitrary (unique) match from each sub array would be somewhat difficult, although picking just one row of data and picking the items from that would be relatively easy. Either would satisfy the constraint from the question.
If you want the unique object by Name
data1 = [{'id':'13','name':'sophie'},{'id':'22','name':'mary'}]
data2 = [{'id':'26','name':'mary'},{'id':'85','name':'bill'}]
data3 = [{'id':'29','name':'sophie'},{'id':'22','name':'john'}]
flattened = [ ...data1, ...data2, ...data3 ];
counts = flattened.reduce(
(map, { name }) => map.set(name, (map.get(name) || 0) + 1), new Map()
);
names = []
found = flattened.filter(({ name }) => {
if ((counts.get(name) > 1) && (!names.includes(name))) {
names.push(name);
return true
}
return false
});
its too many loops but , if u can find the common id which is present in all the arrays then it would make your finding easier i think .you can have one array value as reference to find the common id
var global = [];
for(var i = 0;i<data1.length;i++){
var presence = true;
for(var j=0;j<arrays.length;j++){
var temp = arrays[j].find(function(value){
return data1[i].id == value.id;
});
if(!temp){
presence = false;
break;
}
}
if(presence){
global.push(data1[i].id)
}
}
console.log(global);
var data1 = [{'id':'13','name':'sophie'},{'id':'22','name':'andrew'}];
var data2 = [{'id':'22','name':'mary'},{'id':'85','name':'bill'}];
var data3 = [{'id':'20','name':'steve'},{'id':'22','name':'john'}];
var arrays = [data1, data2, data3];
var global = [];
for(var i = 0;i<data1.length;i++){
var presence = true;
for(var j=0;j<arrays.length;j++){
var temp = arrays[j].find(function(value){
return data1[i].id == value.id;
});
if(!temp){
presence = false;
break;
}
}
if(presence){
global.push(data1[i].id)
}
}
console.log(global);
There's mention you you need n arrays, but also, given that you can:
put all the arrays into an array called data
you can:
combine your arrays
get a list of duplicated IDs (via sort by ID)
make that list unique (unique list of IDs)
find entries in the combined list that match the unique IDs
where the count of those items match the original number of arrays
Sample code:
// Original data
var data1 = [{'id':'13','name':'sophie'},{'id':'22','name':'andrew'}]
var data2 = [{'id':'22','name':'mary'},{'id':'85','name':'bill'}]
var data3 = [{'id':'13','name':'steve'},{'id':'22','name':'john'}]
var arraycount = 3;
// Combine data into a single array
// This might be done by .pushing to an array of arrays and then using .length
// https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/Array/sort?v=control
var data = [].concat(data1).concat(data2).concat(data3);
//console.log(data)
// Sort array by ID
// http://stackoverflow.com/questions/840781/easiest-way-to-find-duplicate-values-in-a-javascript-array
var sorted_arr = data.slice().sort(function(a, b) {
return a.id - b.id;
});
//console.log(sorted_arr)
// Find duplicate IDs
var duplicate_arr = [];
for (var i = 0; i < data.length - 1; i++) {
if (sorted_arr[i + 1].id == sorted_arr[i].id) {
duplicate_arr.push(sorted_arr[i].id);
}
}
// Find unique IDs
// http://stackoverflow.com/questions/1960473/unique-values-in-an-array
var unique = duplicate_arr.filter(function(value, index, self) {
return self.indexOf(value) === index;
});
//console.log(unique);
// Get values back from data
//https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/Array/filter?v=control
var matches = [];
for (var i = 0; i < unique.length; ++i) {
var id = unique[i];
matches.push(data.filter(function(e) {
return e.id == id;
}))
}
//console.log(matches)
// for data set this will be 13 and 22
// Where they match all the arrays
var result = matches.filter(function(value, index, self) {
return value.length == arraycount;
})
//console.log("Result:")
console.log(result)
Note: There's very likely to be more efficient methods.. I've left this in the hope part of it might help someone
var arr1 = ["558", "s1", "10"];
var arr2 = ["55", "s1", "103"];
var arr3 = ["55", "s1", "104"];
var arr = [arr1, arr2, arr3];
console.log(arr.reduce((p, c) => p.filter(e => c.includes(e))));
// output ["s1"]
I am pushing element in the array using javascript .
and i am getting output like this
["api", "management", "provision", "tenant", "external", "provider"] => Correct
after pushing the data into array now i want append two element from the array management and provision and want output like this
["api", "management/provision", "tenant", "external", "provider"]
how would i do this ?
any example would be good for me
thanx in advance
The simplest thing you can do is to remove them both, join with / and put at the right place. You can do this in a few lines of code:
const input = ["api", "management", "provision", "tenant", "external", "provider"];
input.splice(1, 3, input.slice(1, 3).join('/'));
// now input contains ["api", "management/provision", "external", "provider"]
This can be further made more readable and reusable:
function mergeTerms(startIndex, endIndex, array) {
const input = array.slice(0); // make a shallow copy of the array in order not to mutate it
input.splice(startIndex, endIndex, input.slice(startIndex, endIndex).join('/'));
return input;
}
and then:
mergeTerms(1, 3, input);
Lets say its the var array that holds the values.
var temp = array.splice(2,3);
array[1] = array[1]+"/"+temp;
function mergeArrayEntries(e1, e2, arr) {
var index = -1;
for (var i = 0; i < arr.length; i++) {
if (arr[i] == e1) {
index = i;
}
if (arr[i] == e2 && index > -1) {
arr[index] = arr[index]+"/"+arr[i];
arr.splice(i, 1);
}
}
}
this function searches for e1 and e2 in the arr and once found, it merges them to position of e1 and deletes e2 entry in arr.
hope it helps
Not the nicest code, and not at all reusable, but will give you what you are looking for --- just another approach
var array = ["api", "management", "provision", "tenant", "external", "provider"];
var subArray = [];
subArray.push(array[1]);
subArray.push(array[2])
array.splice(1, 2);
subArray = subArray.join("/")
array.splice(1, 0, subArray);
I have an array like [1,4,3,1,6,5,1,4,4]
Here Highest element frequency is 3 ,I need to select all elements from array that have 3 frequency like [1,4] in above example.
I have tried with this
var count = {},array=[1,4,3,1,6,5,1,4,4],
value;
for (var i = 0; i < array.length; i++) {
value = array[i];
if (value in count) {
count[value]++;
} else {
count[value] = 1;
}
}
console.log(count);
this will output array element with their frequency , now i need all elements that have highest frequency.
I'd approach this problem as follows.
First, write down how you think the problem can be solved IN ENGLISH, or something close to English (or your native language of course!). Write down each step. Start off with a high-level version, such as:
Count the frequency of each element in the input.
Find the highest frequency.
and so on. At this point, it's important that you don't get bogged down in implementation details. Your solution should be applicable to almost any programming language.
Next flesh out each step by adding substeps. For instance, you might write:
Find the highest frequency.
a. Assume the highest frequency is zero.
b. Examine each frequency. If it is higher than the current highest frqeuency, make it the current highest frequency.
Test your algorithm by executing it manually in your head.
Next, convert what you have written about into what is sometimes called pseudo-code. It is at this point that our algorithm starts to look a little bit like a computer program, but is still easily human-readable. We may now use variables to represent things. For instance, we could write "max_freq ← cur_freq". We can refer to arrays, and write loops.
Finally, convert your pseudo-code into JS. If all goes well, it should work the first time around!
In recent years, a lot of people are jumping right into JavaScript, without any exposure to how to think about algorithms, even simple ones. They imagine that somehow they need to be able to, or will magically get to the point where they can, conjure up JS out of thin air, like someone speaking in tongues. In fact, the best programmers do not instantly start writing array.reduce when confronted with a problem; they always go through the process--even if only in their heads--of thinking about the approach to the problem, and this is an approach well worth learning from.
If you do not acquire this skill, you will spend the rest of your career posting to SO each time you can't bend your mind around a problem.
A proposal with Array.prototype.reduce() for a temporary object count, Object.keys() for getting the keys of the temporary object, a Array.prototype.sort() method for ordering the count results and Array.prototype.filter() for getting only the top values with the most count.
Edit: Kudos #Xotic750, now the original values are returned.
var array = [1, 4, 3, 1, 6, 5, 1, 4, 4],
result = function () {
var temp = array.reduce(function (r, a, i) {
r[a] = r[a] || { count: 0, value: a };
r[a].count++;
return r;
}, {});
return Object.keys(temp).sort(function (a, b) {
return temp[b].count - temp[a].count;
}).filter(function (a, _, aa) {
return temp[aa[0]].count === temp[a].count;
}).map(function (a) {
return temp[a].value;
});
}();
document.write('<pre>' + JSON.stringify(result, 0, 4) + '</pre>');
Bonus with a different attempt
var array = [1, 4, 3, 1, 6, 5, 1, 4, 4],
result = array.reduce(function (r, a) {
r.some(function (b, i) {
var p = b.indexOf(a);
if (~p) {
b.splice(p, 1);
r[i + 1] = r[i + 1] || [];
r[i + 1].push(a);
return true;
}
}) || (
r[1] = r[1] || [],
r[1].push(a)
);
return r;
}, []).pop();
document.write('<pre>' + JSON.stringify(result, 0, 4) + '</pre>');
you can try this
var input = [1,4,3,1,6,5,1,4,4];
var output = {};
for ( var counter = 0; counter < input.length; counter++ )
{
if ( !output[ input[ counter ] ] )
{
output[ input[ counter ] ] = 0;
}
output[ input[ counter ] ]++;
}
var outputArr = [];
for (var key in output)
{
outputArr.push([key, output[key]])
}
outputArr = outputArr.sort(function(a, b) {return b[1] - a[1]})
now initial values of outputArr are the ones with highest frequency
Here is the fiddle
Check this updated fiddle (this will give the output you want)
var input = [1,4,3,1,6,5,1,4,4];
var output = {}; // this object holds the frequency of each value
for ( var counter = 0; counter < input.length; counter++ )
{
if ( !output[ input[ counter ] ] )
{
output[ input[ counter ] ] = 0; //initialized to 0 if value doesn't exists
}
output[ input[ counter ] ]++; //increment the value with each occurence
}
var outputArr = [];
var maxValue = 0;
for (var key in output)
{
if ( output[key] > maxValue )
{
maxValue = output[key]; //find out the max value
}
outputArr.push([key, output[key]])
}
var finalArr = []; //this array holds only those keys whose value is same as the highest value
for ( var counter = 0; counter < outputArr.length; counter++ )
{
if ( outputArr[ counter ][ 1 ] == maxValue )
{
finalArr.push( outputArr[ counter ][ 0 ] )
}
}
console.log( finalArr );
I would do something like this. It's not tested, but it's commented for helping you to understand my approach.
// Declare your array
var initial_array = [1,4,3,1,6,5,1,4,4];
// Declare an auxiliar counter
var counter = {};
// Loop over the array
initial_array.forEach(function(item){
// If the elements is already in counter, we increment the repetition counter.
if counter.hasOwnProperty(item){
counter[item] += 1;
// If the element is not in counter, we set the repetitions to one
}else{
counter[item] = 1;
}
});
// counter = {1 : 3, 4 : 3, 3 : 1, 6 : 1, 5 : 1}
// We move the object keys to an array (sorting it's more easy this way)
var sortable = [];
for (var element in counter)
sortable.push([element, counter[element]]);
// sortable = [ [1,3], [4,3], [3,1], [6,1], [5,1] ]
// Sort the list
sortable.sort(function(a, b) {return a[1] - b[1]})
// sortable = [ [1,3], [4,3], [3,1], [6,1], [5,1] ] sorted, in this case both are equals
// The elements in the firsts positions are the elements that you are looking for
// This auxiliar variable will help you to decide the biggest frequency (not the elements with it)
higgest = 0;
// Here you will append the results
results = [];
// You loop over the sorted list starting for the elements with more frequency
sortable.forEach(function(item){
// this condition works because we have sorted the list previously.
if(item[1] >= higgest){
higgest = item[1];
results.push(item[0]);
}
});
I'm very much with what #torazaburo had to say.
I'm also becoming a fan of ES6 as it creeps more and more into my daily browser. So, here is a solution using ES6 that is working in my browser now.
The shims are loaded to fix browser browser bugs and deficiencies, which is recommended in all environments.
'use strict';
// Your array of values.
const array = [1, 4, 3, 1, 6, 5, 1, 4, 4];
// An ES6 Map, for counting the frequencies of your values.
// Capable of distinguishing all unique values except `+0` and `-0`
// i.e. SameValueZero (see ES6 specification for explanation)
const frequencies = new Map();
// Loop through all the `values` of `array`
for (let item of array) {
// If item exists in frequencies increment the count or set the count to `1`
frequencies.set(item, frequencies.has(item) ? frequencies.get(item) + 1 : 1);
}
// Array to group the frequencies into list of `values`
const groups = [];
// Loop through the frequencies
for (let item of frequencies) {
// The `key` of the `entries` iterator is the value
const value = item[0];
// The `value` of the `entries` iterator is the frequency
const frequency = item[1];
// If the group exists then append the `value`,
// otherwise add a new group containing `value`
if (groups[frequency]) {
groups[frequency].push(value);
} else {
groups[frequency] = [value];
}
}
// The most frequent values are the last item of `groups`
const mostFrequent = groups.pop();
document.getElementById('out').textContent = JSON.stringify(mostFrequent);
console.log(mostFrequent);
<script src="https://cdnjs.cloudflare.com/ajax/libs/es5-shim/4.4.1/es5-shim.min.js"></script>
<script src="https://cdnjs.cloudflare.com/ajax/libs/json3/3.3.2/json3.min.js"></script>
<script src="https://cdnjs.cloudflare.com/ajax/libs/es6-shim/0.34.0/es6-shim.js"></script>
<pre id="out"></pre>
you can do like this to find count occurrence each number
var array = [1, 4, 3, 1, 6, 5, 1, 4, 4];
var frequency = array.reduce(function(sum, num) {
if (sum[num]) {
sum[num] = sum[num] + 1;
} else {
sum[num] = 1;
}
return sum;
}, {});
console.log(frequency)
<script src="https://getfirebug.com/firebug-lite-debug.js"></script>
In a JavaScript array how can I get the index of duplicate strings?
Example:
MyArray = ["abc","def","abc"]; //----> return 0,2("abc");
Another example:
My Array = ["abc","def","abc","xyz","def","abc"]
//----> return 0,2,5("abc") and 1,4("def");
I have no idea how to do this.
Thanks in advance for your help!
Update 01/2022: It's not 2013 anymore, and many things have changed. I neither recommend modifying the prototype, nor is the approach in this answer the "best" as it requires several iterations over the array.
Here's an updated version of the original answer, retaining its spirit, as well as the original answer below.
function getDuplicates<T>(input: T[]): Map<T, number[]> {
return input.reduce((output, element, idx) => {
const recordedDuplicates = output.get(element);
if (recordedDuplicates) {
output.set(element, [...recordedDuplicates, idx]);
} else if (input.lastIndexOf(element) !== idx) {
output.set(element, [idx]);
}
return output;
}, new Map<T, number[]>());
}
Yet another approach:
Array.prototype.getDuplicates = function () {
var duplicates = {};
for (var i = 0; i < this.length; i++) {
if(duplicates.hasOwnProperty(this[i])) {
duplicates[this[i]].push(i);
} else if (this.lastIndexOf(this[i]) !== i) {
duplicates[this[i]] = [i];
}
}
return duplicates;
};
It returns an object where the keys are the duplicate entries and the values are an array with their indices, i.e.
["abc","def","abc"].getDuplicates() -> { "abc": [0, 2] }
Another less sophisticated approach:
Iterate over the whole array and keep track of the index of each element. For this we need a string -> positions map. An object is the usual data type to use for this. The keys are the elements of the array and the values are arrays of indexes/positions of each element in the array.
var map = {};
for (var i = 0; i < arr.length; i++) {
var element = arr[i]; // arr[i] is the element in the array at position i
// if we haven't seen the element yet,
// we have to create a new entry in the map
if (!map[element]) {
map[element] = [i];
}
else {
// otherwise append to the existing array
map[element].push(i);
}
// the whole if - else statement can be shortend to
// (map[element] || (map[element] = [])).push(i)
}
Now you can iterate over the map and remove all entries where the array value has a length of one. Those are elements that appear only once in an array:
for (var element in map) {
if (map[element].length === 1) {
delete map[element];
}
}
Now map contains a string -> positions mapping of all duplicate elements of the array. For example, if you array is ["abc","def","abc","xyz","def","abc"], then map is an object of the form
var map = {
'abc': [0,2,5],
'def': [1,4]
};
and you can process it further in any way you like.
Further reading:
Eloquent JavaScript - Data structures: Objects and Arrays
MDN - Working with objects
MDN - Predefined core objects, Array object
This covers finding the indices efficiently:
var inputArray = [1, 2, 3, 4, 5, 6, 6, 7, 8, 9];
var encounteredIndices = {};
for(var i = 0; i < inputArray.length; i++)
if (encounteredIndices[inputArray[i]])
console.log(i); // Or add to some array if you wish
else
encounteredIndices[inputArray[i]] = 1;