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Array, return original index from a sorted array in javascript

'I wish to sort an array in numerical order but once it is sorted I wish to be able to find the original index.
For example the original array:
ptsGP = [3,8,2,5,6,9,8,4]
I am using the following code below to sort the array:
arr = ptsGP;
var arr2 = arr.map(function(o, i){return {idx: i, obj: o}; }).sort(function(a, b) {
return b.obj - a.obj;
});
for(var i = 1, j = arr2.length; i <= j; i++){
document.write('i:' + i + ' = arr2[i].obj: PTS: ', arr2[i-1].obj+"<br/>");
}`
This is fine as the sorted array is :
arr = [2,3,4,5,6,8,8,9];
How can I find the index of sorted number in the original array? In this case it would be :
Index on original array would be = [2,0,7,3,4,1,6,5]
I know I could use map on the original array but how can I deal with duplicate numbers i.e, in this case I have two number 8's within the array?

You can achieve it by following below steps :
Creating a deep copy of an original array by using spread operator. So that we can get proper indexing.
Now we can iterate deep copy array to get the index of the elements from an original array.
Regarding duplicate values we can check via .indexOf() and .lastIndexOf() methods.
via and then via comparison. For fetching the correct index of duplicate values I wrote a logic based on the count of duplicate value.
Working Demo :
// Original array.
const originalArray = [3, 8, 2, 5, 6, 9, 8, 4];
// Creating a deep copy of an original array.
const deepCopy = [...originalArray].sort(function(a, b){
return a-b
});
// result array
const arr = [];
// count to get the index based on duplicate values.
let count = 0;
// Iterating deepCopy array to get the actual index.
deepCopy.forEach((elem) => {
// Checking for duplicate value in an array
if (originalArray.indexOf(elem) === originalArray.lastIndexOf(elem)) {
// This line of code execute if there is no duplicates in an array.
arr.push(originalArray.indexOf(elem))
} else {
// This line of code execute if there is duplicate values in an array.
count++;
// Inserting the index one by one.
arr.push(originalArray.indexOf(elem, count))
}
});
// Result array.
console.log(arr);

Add all the elements in an array without using loops [closed]

Closed. This question needs details or clarity. It is not currently accepting answers.
Want to improve this question? Add details and clarify the problem by editing this post.
Closed 1 year ago.
Improve this question
How to add elements without using loops and the built-in method of the array like for , foreach etc) and built in methods(like reduce,map...) of array
How can we do the addition?
var arr =[1,2,2,3,4...];
arr length also dynamic.

You can use Spread ... operator like this
var arr1 = [1, 2, 3, 4];
var arr2 = [5,6,7,8];
console.log([...arr1, ...arr2]);

Addition as in sum?
const arr = [1,2,2,3,4];
// normal way
let sum = arr.reduce((a,b) => a+b)
console.log(sum)
// weird way not using loops but still using a built-in method
sum = eval(arr.toString().replace(/,/g,"+"))
console.log(sum)

You can use concat for that whice creates a new array and returns the result so arr1 and arr2 will remain unchanged.
const arr1 = [1, 2];
const arr2 = [3, 4,5,6,7,8];
const combined = arr1.concat(arr2);
console.log(combined);
console.log(arr1);
console.log(arr2);
If you want to mutate the original array you can use push and the spread operator (...):
const a = [1,2]
const b = [3,4,5,6]
a.push(...b)
console.log(a);

Compare values of two different arrays [duplicate]

This question already has answers here:
Check whether an array exists in an array of arrays?
(11 answers)
Closed 3 years ago.
My problem is about comparing the values of two different arrays, not to know if they are equal, but if a value in array A exists in array B.
Includes doesn't work, and I don't compare the length, only the values.
I spend many hours looking for an answer but found nothing precisely about this problem.
firstArray = [0,1];
secondArray = [[0,1],[0,2],[0,3],[1,1],[1,2],[1,3]];
How can I can compare firstArray and secondArray, to know if secondArray has the value of firstArray. It is like an equal comparison but only if the value of firstArray is in secondArray.
If it is, then the player can move on the board. The idea is that as long as firstArray value is one of secondArray value, the player can move. If not, no move possible.

You can stringify array items and compare:
var firstArray = [0,1];
var secondArray = [[0,1],[0,2],[0,3],[1,1],[1,2],[1,3]];
var res = secondArray.some(x => JSON.stringify(x) == JSON.stringify(firstArray));
console.log(res);

You can use Array.prototype.some() combined with Array.prototype.toString():
const firstArray = [0,1];
const secondArray = [[0,1],[0,2],[0,3],[1,1],[1,2],[1,3]];
const isFound = secondArray.some(a => a.toString() === firstArray.toString());
console.log(isFound);

Simply use JSON.stringify with some and sort():
var firstInSecond = secondArray.some(e => JSON.stringify(e.sort()) == JSON.stringify(firstArray.sort()));

Best way to filter for duplicates? [duplicate]

This question already has answers here:
Get all unique values in a JavaScript array (remove duplicates)
(91 answers)
Closed 4 years ago.
I posted about this before trying to use a regular expression, but this time is a bit different.
So I have this list of 500 phone numbers. Here is a little sample of them:
{
8664665844
8885444636
8664604776
8776434327
8887441938
8882642882
8888097429
8668943258
8777711234
8669894327
}
It looks a bit different, this is on my mongoDB server, so its more like an array of objects with a unique uid... so for example:
[
{
_id: kasjf234lkj109euf9023u4n,
field1: 8669894327,
}
]
something like that right.
So basically, my webapp is using a csv file of numbers that you want to compare to your base numbers.
My first thought was using a for, while, for loop kind of thing but im not sure how well it'll work:
for(var i = 0; i < basenums.length; i++){
while ( i >= 0 ){
for(var j = 0; j < comparing.length; j++){
if comparing[j] == basenums.field1[i]{
push that number to a 'dupes' array
}else{
break or something?
}
}
}
}
...this logic is starting to hurt my head...
I know theres an 'includes()' method, but I havn't really used it and when I tried it in this case, It gave me everything as false, even tho the list I was using to compare is just a copy of my 'basenums' list on my server.
What would be the more 'correct' way of doing this?
I'm trying to compare an array with an array. Not a value in its specific array to its own array.
So, rather than marking this as a duplicate, maybe you should re-read:
I have 2 arrays of numbers: [1, 2, 3, 4, 5, 6] and [1, 2, 3, 4, 5, 6] How do I take array a index 0 and compare it to array b every index, then take index 1 of array a and compare to every index in array b, so on and so forth until i > arrayA.length.

Don't try to do everything in one single step. Break your problem into smaller chunks:
Format mongo's response into a hash map that you can efficiently query (O(1))
Test each input number against this map
Step 1:
var hashmap = {}
basenums.forEach(function(basenum) {
hashmap[basenum.field1] = true;
})
Step 2:
var dupes = [];
comparing.forEach(function(comparingNum) {
if(hashmap[comparingNum]) {
dupes.push(comparingNum);
}
})
EDIT: I am not sure, whether you want to have your dupes as a set (unique array). If so, you could use an alternate Step 2:
var dupesMap = {};
comparing.forEach(function(comparingNum) {
if(hashmap[comparingNum]) {
dupesMap[comparingNum] = true;
}
})
var dupes = Object.keys(dupesMap);

You can push values into an array and create a new set with the array. Sets do not keep redundant elements in them. Try this;
let a = ["a","b","a","b","a","b","a","b","a","c"];
console.log(a);
let uniqueA = [...new Set(a)];
console.log(uniqueA);
By the way, for piece of code I got a help from this page: https://medium.com/front-end-hacking/getting-unique-values-in-javascript-arrays-17063080f836

Undo sort on sorted array in javascript

I have an array. I sort it.
I get a second array which is already sorted based on the first one.
I need to reverse the sorting on the second array.
For example, if the first array (unsorted) is: [9, 5, 3, 0, 2] then I want to to sort it, so that it becomes [0, 2, 3, 5, 9].
Then I receive the second array sorted based on the first one, for example ["home", "car", "train", "pc", "mouse"]. I need it to become ["mouse, "pc", "train", "home", "car"].
I can't make a copy of the array.
I have the following code:
//data_r is an array with values
var i = 0;
var sort_order = new Array();
data_r.sort(function (a,b) {
var res = a[0] - b[0];
sort_order[i] = res;
i++;
return res;
});
In the end, the the sort_order array will contain the actions performed when we sorted items. If I want to sort a second array exactly the same way as the first then I can do the following:
//data_x is an array with values
var i = 0;
data_x.sort(function (a,b) {
i++;
return sort_order[i-1];
});
Now the data_x array is sorted exactly the same way as the data_r array.
How can I undo sort on the data_r array?
The following code is incorrect:
var unsort = new Array();
for(var i = 0; i < data_r.length; i++)
unsort[i] = sort_order[i]*(-1);//-1 so we perfom the oposite action

Your premise here is flawed.
In the end, the sort_order array contains the actions performed when we sorted items.
No, it doesn't; it contains a log of the comparisons performed by the Javascript Array.sort function. The actions it took in response to those comparison results are private to it.
If I want to sort a second array exactly the same way as the first then I can do the following:
This is not guaranteed to work. Even if the two arrays are the same size, Array.sort may not always compare the same elements in the same order each time it's called - it's possible that it's using a randomized algorithm, that it performs comparisons based on other data that are internal to the interpreter, or that it switches between multiple entirely different sort algorithms under some circumstances.
While this code may work for you, right now, in your current web browser, it is likely to fail in surprising ways in other circumstances (possibly in future browsers). Do not use this technique in production code.
The question is, how can i unsort the data_r array?
Make a copy of the array before you sort it.

Storing res[i] = a - b is like journaling the sort() algorithm - but what if it used a random pivot?
This code is inherently unreliable unless you write sort() yourself. It's also inefficient.
A better approach, one that will solve both your needs, is to create an array of indices and sort that. This is trivial to invert. Then you can implement a permute function that takes an array of indices, and it achieves a sort or unsort, depending on the input.
If x is from 0:n-1, create an array sort_i of same size, then initialize each sort_i[i] = i.
for(var i = 0; i < n; i++)
sort_i[i] = i;
Then
sort_i.sort(function (a,b) { return x[a] - x[b]; });
Now you have the indices. To apply to x:
for(var i = 0; i < n; i++)
sort_x[i] = x[sort_i[i]];
To unsort it, first invert the indices
for(var i = 0; i < n; i++)
unsort_i[sort_i[i]] = i;
Then apply the indices. Exercise left to question asker.
This approach of sorting an array of integer indices is needed when you don't want to move the original elements around in memory (maybe they are big objects), and many other circumstances. Basically you are sorting pointers. The result is an index to the data, and a reverse index.

See #duskwuff's answer on why your approach doesn't work.
Instead, just introduce a mapping between the original data and the sorted data.
{0:2, 1:3, 2:1, 3:0}
Which means the first element became the third, the second became the last and so on. Below we'll use an array instead of an object.
Why does this map help? You can sort it like another dataset by just using the indizes in it as pointers to the data you're going to compare. And you can apply the mapping easily on other datasets. And you can even reverse that mapping very easily. See it in the code:
// data_r, data_x are arrays with values
var l = data_r.length;
var sort_order = new Array(l);
for (var i=0; i<l; i++) sort_order[i] = i; // initialised as 1-1 mapping
// change the sort_order first:
sort_order.sort(function (a,b) {
// a and b being indices
return data_r[a] - data_r[b];
});
// Making a new, sorted array
var data_x_sorted = new Array(l);
for (var i=0; i<l; i++)
data_x_sorted[ sort_order[i] ] = data_x[i]; // put it to sorted position
If you want to sort the data_x array itself, just use the "apply" algorithm which I showed for data_r.
The question is, how can I undo sort on the data_r array?
Either don't sort it at all, and just make a copy of it which gets sorted (or do nothing at all).
Or use the sort_order to reverse it. You just would need to swap i and newIndex (sortOrder[i]) everywhere. Example for building a new, "unsorted" (old-order) array:
var unsorted = new Array(l);
for (var i=0; i<l; i++)
unsorted[i] = data_r[ sort_order[i] ]; // take it from its new position

While this question is 8 years old at this point, I came across it when trying to find the same solution to the problem and I was unable to find a suitable, performant, and intuitive way of doing so, so I wrote one myself.
Please take a look at the sort-unwind library. If ranks is a list of indexes that would rank an array in order...
import unwind from 'sort-unwind'
const suits = ['♥', '♠', '♣', '♦']
const ranks = [2, 0, 3, 1]
const [sortedSuits, tenet] = unwind(ranks, suits)
// sortedSuits <- ['♠', '♦', '♥', '♣']
// unwind <- [1, 3, 0, 2]
You can then use the tenet variable that's returned to unsort an array and restore the original ordering.
const names = ['spades', 'diamonds', 'hearts', 'clubs']
const [tenetNames, tenetRanks] = unwind(tenet, names)
// tenetNames <- ['hearts', 'spades', 'clubs', 'diamonds']
// tenetRanks <- [2, 0, 3, 1]

The sort function just returns a number which can be positive,zero, or negative telling it if the current element goes before,has same weight, or goes after the element it is comparing it too. I would imagine your sort order array is longer than your data_r array because of the number of comparisons you make. I would just make a copy of data_r before you sort it and then set data_r equal to that array when you want it unsorted.

If you have a lot of these arrays to maintain, it might be as well to
convert array1 into an array of objects, each one containing the value
and its original position in the array. This keeps everything together
in one array.
var array1 = [9, 5, 3, 0, 2];
var array2 = ["home", "car", "train", "pc", "mouse"];
var sort = function(array){
var indexed_objects = array.map(function(value, index){
return {index: index, value: value};
});
indexed_objects.sort(function(a,b){
return a.value <= b.value ? -1 : 1;
});
return indexed_objects;
};
var sorted1 = sort(array1);
sorted1; // [{index: 3, value:0}, {index: 4, value: 2}, ...]
And now, given an array of sorted objects, we can write a function to
unsort any other array accordingly:
var unsort = function(array, sorted_objects){
var unsorted = [];
sorted_objects.forEach(function(item, index){
unsorted[item.index] = array[index];
});
return unsorted;
};
var array2_unsorted = unsort(array2, sorted1);
array2_unsorted; // ["mouse", "pc", "train", "home", "car"]

v1 = [0,1,2,3,4,5,6]
q = v1.length
b = []
for(i=0;i<q;i++){
r = parseInt(Math.random()*v1.length)
b.push(v1[r])
a = v1.indexOf(v1[r])
v1.splice(a,1)
}