Remove all of the duplicate numbers in an array of numbers [duplicate]Get all unique values in a JavaScript array (remove duplicates)Get all non-unique values (i.e.: duplicate/more than one occurrence) in an arrayRemove occurrences of duplicate words in a stringremove all elements that occur more than once from arrayCreate ArrayList from arrayHow do I remove a property from a JavaScript object?How do I check if an array includes an object in JavaScript?How to append something to an array?How to replace all occurrences of a string in JavaScriptLoop through an array in JavaScriptHow to check if an object is an array?How do I remove a particular element from an array in JavaScript?Remove duplicate values from JS arrayFor-each over an array in JavaScript?
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Remove all of the duplicate numbers in an array of numbers [duplicate]
Get all unique values in a JavaScript array (remove duplicates)Get all non-unique values (i.e.: duplicate/more than one occurrence) in an arrayRemove occurrences of duplicate words in a stringremove all elements that occur more than once from arrayCreate ArrayList from arrayHow do I remove a property from a JavaScript object?How do I check if an array includes an object in JavaScript?How to append something to an array?How to replace all occurrences of a string in JavaScriptLoop through an array in JavaScriptHow to check if an object is an array?How do I remove a particular element from an array in JavaScript?Remove duplicate values from JS arrayFor-each over an array in JavaScript?
This question already has an answer here:
Get all unique values in a JavaScript array (remove duplicates)
66 answers
I received this question for practice and the wording confused me, as I see 2 results that it might want.
And either way, I'd like to see both solutions.
For example, if I have an array:
let arr = [1, 2, 4, 2, 3, 3, 4, 5, 5, 5, 8, 8, 9, 10];
I'm taking this as wanting the final result as either:
let finalResult = [1, 2, 3, 4, 5, 8, 9, 10];
OR:
let finalResult = [1, 9, 10];
The difference between the two being, one just removes any duplicate numbers and leaves the rest and the second just wants any number that isn't a duplicate.
Either way, I'd like to write two functions that does one of each.
This, given by someone else gives my second solution.
let elems = ,
arr2 = arr.filter(function (e)
if (elems[e] === undefined)
elems[e] = true;
return true;
return false;
);
console.log(arr2);
I'm not sure about a function for the first one (remove all duplicates).
javascript arrays duplicates
marked as duplicate by Jared Smith, pushkin, BlueRaja - Danny Pflughoeft, the_lotus, Moira 25 mins ago
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
add a comment |
This question already has an answer here:
Get all unique values in a JavaScript array (remove duplicates)
66 answers
I received this question for practice and the wording confused me, as I see 2 results that it might want.
And either way, I'd like to see both solutions.
For example, if I have an array:
let arr = [1, 2, 4, 2, 3, 3, 4, 5, 5, 5, 8, 8, 9, 10];
I'm taking this as wanting the final result as either:
let finalResult = [1, 2, 3, 4, 5, 8, 9, 10];
OR:
let finalResult = [1, 9, 10];
The difference between the two being, one just removes any duplicate numbers and leaves the rest and the second just wants any number that isn't a duplicate.
Either way, I'd like to write two functions that does one of each.
This, given by someone else gives my second solution.
let elems = ,
arr2 = arr.filter(function (e)
if (elems[e] === undefined)
elems[e] = true;
return true;
return false;
);
console.log(arr2);
I'm not sure about a function for the first one (remove all duplicates).
javascript arrays duplicates
marked as duplicate by Jared Smith, pushkin, BlueRaja - Danny Pflughoeft, the_lotus, Moira 25 mins ago
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
If you're using lodash, you can use_.uniq()
– BlueRaja - Danny Pflughoeft
9 hours ago
1
Further, this is asking for the inverse of Get all non-unique values (i.e.: duplicate/more than one occurrence) in an array. Finally, this post is asking two separate questions and both have good answers elsewhere already.
– Søren D. Ptæus
9 hours ago
To answer the question "which one is it" in a comment-answer: if you're asked to remove duplicates, I believe you should understand the first variant. The second variant removes all element that have duplicates, meaning the "original" value AND its duplicates.
– Pierre Arlaud
6 hours ago
add a comment |
This question already has an answer here:
Get all unique values in a JavaScript array (remove duplicates)
66 answers
I received this question for practice and the wording confused me, as I see 2 results that it might want.
And either way, I'd like to see both solutions.
For example, if I have an array:
let arr = [1, 2, 4, 2, 3, 3, 4, 5, 5, 5, 8, 8, 9, 10];
I'm taking this as wanting the final result as either:
let finalResult = [1, 2, 3, 4, 5, 8, 9, 10];
OR:
let finalResult = [1, 9, 10];
The difference between the two being, one just removes any duplicate numbers and leaves the rest and the second just wants any number that isn't a duplicate.
Either way, I'd like to write two functions that does one of each.
This, given by someone else gives my second solution.
let elems = ,
arr2 = arr.filter(function (e)
if (elems[e] === undefined)
elems[e] = true;
return true;
return false;
);
console.log(arr2);
I'm not sure about a function for the first one (remove all duplicates).
javascript arrays duplicates
This question already has an answer here:
Get all unique values in a JavaScript array (remove duplicates)
66 answers
I received this question for practice and the wording confused me, as I see 2 results that it might want.
And either way, I'd like to see both solutions.
For example, if I have an array:
let arr = [1, 2, 4, 2, 3, 3, 4, 5, 5, 5, 8, 8, 9, 10];
I'm taking this as wanting the final result as either:
let finalResult = [1, 2, 3, 4, 5, 8, 9, 10];
OR:
let finalResult = [1, 9, 10];
The difference between the two being, one just removes any duplicate numbers and leaves the rest and the second just wants any number that isn't a duplicate.
Either way, I'd like to write two functions that does one of each.
This, given by someone else gives my second solution.
let elems = ,
arr2 = arr.filter(function (e)
if (elems[e] === undefined)
elems[e] = true;
return true;
return false;
);
console.log(arr2);
I'm not sure about a function for the first one (remove all duplicates).
This question already has an answer here:
Get all unique values in a JavaScript array (remove duplicates)
66 answers
javascript arrays duplicates
javascript arrays duplicates
edited 1 hour ago
pushkin
4,263112954
4,263112954
asked 12 hours ago
mph85mph85
1089
1089
marked as duplicate by Jared Smith, pushkin, BlueRaja - Danny Pflughoeft, the_lotus, Moira 25 mins ago
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
marked as duplicate by Jared Smith, pushkin, BlueRaja - Danny Pflughoeft, the_lotus, Moira 25 mins ago
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
If you're using lodash, you can use_.uniq()
– BlueRaja - Danny Pflughoeft
9 hours ago
1
Further, this is asking for the inverse of Get all non-unique values (i.e.: duplicate/more than one occurrence) in an array. Finally, this post is asking two separate questions and both have good answers elsewhere already.
– Søren D. Ptæus
9 hours ago
To answer the question "which one is it" in a comment-answer: if you're asked to remove duplicates, I believe you should understand the first variant. The second variant removes all element that have duplicates, meaning the "original" value AND its duplicates.
– Pierre Arlaud
6 hours ago
add a comment |
If you're using lodash, you can use_.uniq()
– BlueRaja - Danny Pflughoeft
9 hours ago
1
Further, this is asking for the inverse of Get all non-unique values (i.e.: duplicate/more than one occurrence) in an array. Finally, this post is asking two separate questions and both have good answers elsewhere already.
– Søren D. Ptæus
9 hours ago
To answer the question "which one is it" in a comment-answer: if you're asked to remove duplicates, I believe you should understand the first variant. The second variant removes all element that have duplicates, meaning the "original" value AND its duplicates.
– Pierre Arlaud
6 hours ago
If you're using lodash, you can use
_.uniq()– BlueRaja - Danny Pflughoeft
9 hours ago
If you're using lodash, you can use
_.uniq()– BlueRaja - Danny Pflughoeft
9 hours ago
1
1
Further, this is asking for the inverse of Get all non-unique values (i.e.: duplicate/more than one occurrence) in an array. Finally, this post is asking two separate questions and both have good answers elsewhere already.
– Søren D. Ptæus
9 hours ago
Further, this is asking for the inverse of Get all non-unique values (i.e.: duplicate/more than one occurrence) in an array. Finally, this post is asking two separate questions and both have good answers elsewhere already.
– Søren D. Ptæus
9 hours ago
To answer the question "which one is it" in a comment-answer: if you're asked to remove duplicates, I believe you should understand the first variant. The second variant removes all element that have duplicates, meaning the "original" value AND its duplicates.
– Pierre Arlaud
6 hours ago
To answer the question "which one is it" in a comment-answer: if you're asked to remove duplicates, I believe you should understand the first variant. The second variant removes all element that have duplicates, meaning the "original" value AND its duplicates.
– Pierre Arlaud
6 hours ago
add a comment |
9 Answers
9
active
oldest
votes
Using Set and Array.from()
let arr = [1, 2, 4, 2, 3, 3, 4, 5, 5, 5, 8, 8, 9, 10];
console.log(Array.from(new Set(arr)));Alternate using regex
regex explanation here
let arr = [1, 2, 4, 2, 3, 3, 4, 5, 5, 5, 8, 8, 9, 10];
let res = arr
.join(',')
.replace(/(b,w+b)(?=.*1)/ig, '')
.split(',')
.map(Number);
console.log(res);Alternate using objects
let arr = [1, 2, 4, 2, 3, 3, 4, 5, 5, 5, 8, 8, 9, 10];
let obj = arr.reduce((acc, val) => Object.assign(acc,
[val]: val
), );
console.log(Object.values(obj));add a comment |
Just use a simple array.filter one-liner:
let arr = [1, 2, 4, 2, 3, 3, 4, 5, 5, 5, 8, 8, 9, 10];
let finalResult = arr.filter((e, i, a) => a.indexOf(e) == i).sort(function(a, b)return a - b);
console.log(finalResult);You could use another filter statement if you wanted the second result:
let arr = [1, 2, 4, 2, 3, 3, 4, 5, 5, 5, 8, 8, 9, 10];
let finalResult = arr.filter((e, i, a) => a.filter(f => f == e).length == 1).sort(function(a, b)return a - b);
console.log(finalResult);
You could also add.sort()to sort them by numerical order:.sort(function(a, b)return a - b)on finalresult
– Mukyuu
12 hours ago
Yes @Mukyuu, that would also be useful
– Jack Bashford
12 hours ago
2
Worth pointing out that the run time of this approach will be quadratic on the size of the input, which is probably not great unless the input arrays are known to be always fairly small.
– Joe Lee-Moyet
7 hours ago
Most voted with multiple array#filter, array#sort and array#indexOf... That is not performant
– Yosvel Quintero
6 hours ago
Do note that the.sort()is not necessary - the end result without the sort is still an array without any duplicate items, just in the same order as the original array. (It does make it exactly match thefinalResultvariable in the question though.)
– Florrie
4 hours ago
add a comment |
For the first part you can use Set() and Spread Syntax to remove duplicates.
let arr = [1, 2, 4, 2, 3, 3, 4, 5, 5, 5, 8, 8, 9, 10];
let res = [...new Set(arr)]
console.log(res)For the second part you can use reduce()
let arr = [1, 2, 4, 2, 3, 3, 4, 5, 5, 5, 8, 8, 9, 10];
//to get the object with count of each number in array.
let obj = arr.reduce((ac,a) =>
//check if number doesnot occur before then set its count to 1
if(!ac[a]) ac[a] = 1;
//if number is already in object increase its count
else ac[a]++;
return ac;
,)
//Using reduce on all the keys of object means all numbers.
let res = Object.keys(obj).reduce((ac,a) =>
//check if count of current number 'a' is `1` in the above object then add it into array
if(obj[a] === 1) ac.push(+a)
return ac;
,[])
console.log(res)
nice appreciate that, that 2nd one looks crazy. I'm assuming the time complexity for would be less than ideal compared to other results?
– mph85
11 hours ago
@mph85 Yes its a little complex because it doesnot go through the array again and again instead it just store all the result obj and then filter it.Its better regarding performance
– Maheer Ali
11 hours ago
could you remind me why we need thespread operator? what happens if we don't have it? @Maheer Ali
– mph85
11 hours ago
Is it because if we don't have it, it'll just log an object?
– mph85
11 hours ago
@mph85 No if we will not have it. We will have aSet()inside array. We use it convertSet()toArray
– Maheer Ali
11 hours ago
|
show 1 more comment
You could sort the array before and filter the array by checking only one side for duplicates or both sides.
var array = [1, 2, 4, 2, 3, 3, 4, 5, 5, 5, 8, 8, 9, 10],
result1,
result2;
array.sort((a, b) => a - b);
result1 = array.filter((v, i, a) => a[i - 1] !== v);
result2 = array.filter((v, i, a) => a[i - 1] !== v && a[i + 1] !== v);
console.log(...result1);
console.log(...result2)add a comment |
You can create both arrays in One Go
let arr = [1, 2, 4, 2, 3, 3, 4, 5, 5, 5, 8, 8, 9, 10];
let unique = new Set();
let repeated = Array.from(arr.reduce((acc, curr) =>
acc.has(curr) ? unique.delete(curr) : acc.add(curr) && unique.add(curr);
return acc;
, new Set()));
console.log(Array.from(unique))
console.log(repeated)
1
Nice one mate +1
– Maheer Ali
11 hours ago
I wonder if the ternary plus&&could be unclear though (x?y:z&&w)? It's not obvious to me how JS's order of operations would handle that, and I wonder if you could get across the same logic and reasoning by usingif/else?
– Florrie
4 hours ago
add a comment |
You can do:
const arr = [1, 2, 4, 2, 3, 3, 4, 5, 5, 5, 8, 8, 9, 10];
const hash = arr.reduce((a, c) => (a[c] = (a[c] || 0) + 1, a), );
// [1, 2, 3, 4, 5, 8, 9, 10];
const finalResultOne = Object.keys(hash);
// [1, 9, 10];
const finalResultTwo = Object.keys(hash).filter(k => hash[k] === 1);
console.log('finalResultOne:', ...finalResultOne);
console.log('finalResultTwo:', ...finalResultTwo);add a comment |
You can use closure and Map
let arr = [1, 2, 4, 2, 3, 3, 4, 5, 5, 5, 8, 8, 9, 10];
const build = ar =>
const mapObj = ar.reduce((acc, e) =>
acc.has(e) ? acc.set(e, true) : acc.set(e, false)
return acc
, new Map())
return function(hasDup = true)
if(hasDup) return [...mapObj.keys()]
else return [...mapObj].filter(([key, val]) => !val).map(([k, v])=> k)
const getArr = build(arr)
console.log(getArr())
console.log(getArr(false))add a comment |
As many other have said, the first one is just [...new Set(arr)]
For the second, just filter out those that occur more than once:
const arr = [1, 2, 4, 2, 3, 3, 4, 5, 5, 5, 8, 8, 9, 10];
const count = (arr, e) => arr.filter(n => n == e).length
const unique = arr => arr.filter(e => count(arr, e) < 2)
console.log(unique(arr));add a comment |
var arr = [1, 2, 4, 2, 3, 3, 4, 5, 5, 5, 8, 8, 9, 10];
var map = ;
var finalResult = [];
for(var i = 0; i < arr.length; i++)
if(!map.hasOwnProperty(arr[i]))
map[arr[i]] = true;
finalResult.push(arr[i]);
//if you need it sorted otherwise it will be in order
finalResult.sort(function(a, b)return a-b);
add a comment |
9 Answers
9
active
oldest
votes
9 Answers
9
active
oldest
votes
active
oldest
votes
active
oldest
votes
Using Set and Array.from()
let arr = [1, 2, 4, 2, 3, 3, 4, 5, 5, 5, 8, 8, 9, 10];
console.log(Array.from(new Set(arr)));Alternate using regex
regex explanation here
let arr = [1, 2, 4, 2, 3, 3, 4, 5, 5, 5, 8, 8, 9, 10];
let res = arr
.join(',')
.replace(/(b,w+b)(?=.*1)/ig, '')
.split(',')
.map(Number);
console.log(res);Alternate using objects
let arr = [1, 2, 4, 2, 3, 3, 4, 5, 5, 5, 8, 8, 9, 10];
let obj = arr.reduce((acc, val) => Object.assign(acc,
[val]: val
), );
console.log(Object.values(obj));add a comment |
Using Set and Array.from()
let arr = [1, 2, 4, 2, 3, 3, 4, 5, 5, 5, 8, 8, 9, 10];
console.log(Array.from(new Set(arr)));Alternate using regex
regex explanation here
let arr = [1, 2, 4, 2, 3, 3, 4, 5, 5, 5, 8, 8, 9, 10];
let res = arr
.join(',')
.replace(/(b,w+b)(?=.*1)/ig, '')
.split(',')
.map(Number);
console.log(res);Alternate using objects
let arr = [1, 2, 4, 2, 3, 3, 4, 5, 5, 5, 8, 8, 9, 10];
let obj = arr.reduce((acc, val) => Object.assign(acc,
[val]: val
), );
console.log(Object.values(obj));add a comment |
Using Set and Array.from()
let arr = [1, 2, 4, 2, 3, 3, 4, 5, 5, 5, 8, 8, 9, 10];
console.log(Array.from(new Set(arr)));Alternate using regex
regex explanation here
let arr = [1, 2, 4, 2, 3, 3, 4, 5, 5, 5, 8, 8, 9, 10];
let res = arr
.join(',')
.replace(/(b,w+b)(?=.*1)/ig, '')
.split(',')
.map(Number);
console.log(res);Alternate using objects
let arr = [1, 2, 4, 2, 3, 3, 4, 5, 5, 5, 8, 8, 9, 10];
let obj = arr.reduce((acc, val) => Object.assign(acc,
[val]: val
), );
console.log(Object.values(obj));Using Set and Array.from()
let arr = [1, 2, 4, 2, 3, 3, 4, 5, 5, 5, 8, 8, 9, 10];
console.log(Array.from(new Set(arr)));Alternate using regex
regex explanation here
let arr = [1, 2, 4, 2, 3, 3, 4, 5, 5, 5, 8, 8, 9, 10];
let res = arr
.join(',')
.replace(/(b,w+b)(?=.*1)/ig, '')
.split(',')
.map(Number);
console.log(res);Alternate using objects
let arr = [1, 2, 4, 2, 3, 3, 4, 5, 5, 5, 8, 8, 9, 10];
let obj = arr.reduce((acc, val) => Object.assign(acc,
[val]: val
), );
console.log(Object.values(obj));let arr = [1, 2, 4, 2, 3, 3, 4, 5, 5, 5, 8, 8, 9, 10];
console.log(Array.from(new Set(arr)));let arr = [1, 2, 4, 2, 3, 3, 4, 5, 5, 5, 8, 8, 9, 10];
console.log(Array.from(new Set(arr)));let arr = [1, 2, 4, 2, 3, 3, 4, 5, 5, 5, 8, 8, 9, 10];
let res = arr
.join(',')
.replace(/(b,w+b)(?=.*1)/ig, '')
.split(',')
.map(Number);
console.log(res);let arr = [1, 2, 4, 2, 3, 3, 4, 5, 5, 5, 8, 8, 9, 10];
let res = arr
.join(',')
.replace(/(b,w+b)(?=.*1)/ig, '')
.split(',')
.map(Number);
console.log(res);let arr = [1, 2, 4, 2, 3, 3, 4, 5, 5, 5, 8, 8, 9, 10];
let obj = arr.reduce((acc, val) => Object.assign(acc,
[val]: val
), );
console.log(Object.values(obj));let arr = [1, 2, 4, 2, 3, 3, 4, 5, 5, 5, 8, 8, 9, 10];
let obj = arr.reduce((acc, val) => Object.assign(acc,
[val]: val
), );
console.log(Object.values(obj));edited 11 hours ago
answered 12 hours ago
Aswin KumarAswin Kumar
782115
782115
add a comment |
add a comment |
Just use a simple array.filter one-liner:
let arr = [1, 2, 4, 2, 3, 3, 4, 5, 5, 5, 8, 8, 9, 10];
let finalResult = arr.filter((e, i, a) => a.indexOf(e) == i).sort(function(a, b)return a - b);
console.log(finalResult);You could use another filter statement if you wanted the second result:
let arr = [1, 2, 4, 2, 3, 3, 4, 5, 5, 5, 8, 8, 9, 10];
let finalResult = arr.filter((e, i, a) => a.filter(f => f == e).length == 1).sort(function(a, b)return a - b);
console.log(finalResult);
You could also add.sort()to sort them by numerical order:.sort(function(a, b)return a - b)on finalresult
– Mukyuu
12 hours ago
Yes @Mukyuu, that would also be useful
– Jack Bashford
12 hours ago
2
Worth pointing out that the run time of this approach will be quadratic on the size of the input, which is probably not great unless the input arrays are known to be always fairly small.
– Joe Lee-Moyet
7 hours ago
Most voted with multiple array#filter, array#sort and array#indexOf... That is not performant
– Yosvel Quintero
6 hours ago
Do note that the.sort()is not necessary - the end result without the sort is still an array without any duplicate items, just in the same order as the original array. (It does make it exactly match thefinalResultvariable in the question though.)
– Florrie
4 hours ago
add a comment |
Just use a simple array.filter one-liner:
let arr = [1, 2, 4, 2, 3, 3, 4, 5, 5, 5, 8, 8, 9, 10];
let finalResult = arr.filter((e, i, a) => a.indexOf(e) == i).sort(function(a, b)return a - b);
console.log(finalResult);You could use another filter statement if you wanted the second result:
let arr = [1, 2, 4, 2, 3, 3, 4, 5, 5, 5, 8, 8, 9, 10];
let finalResult = arr.filter((e, i, a) => a.filter(f => f == e).length == 1).sort(function(a, b)return a - b);
console.log(finalResult);
You could also add.sort()to sort them by numerical order:.sort(function(a, b)return a - b)on finalresult
– Mukyuu
12 hours ago
Yes @Mukyuu, that would also be useful
– Jack Bashford
12 hours ago
2
Worth pointing out that the run time of this approach will be quadratic on the size of the input, which is probably not great unless the input arrays are known to be always fairly small.
– Joe Lee-Moyet
7 hours ago
Most voted with multiple array#filter, array#sort and array#indexOf... That is not performant
– Yosvel Quintero
6 hours ago
Do note that the.sort()is not necessary - the end result without the sort is still an array without any duplicate items, just in the same order as the original array. (It does make it exactly match thefinalResultvariable in the question though.)
– Florrie
4 hours ago
add a comment |
Just use a simple array.filter one-liner:
let arr = [1, 2, 4, 2, 3, 3, 4, 5, 5, 5, 8, 8, 9, 10];
let finalResult = arr.filter((e, i, a) => a.indexOf(e) == i).sort(function(a, b)return a - b);
console.log(finalResult);You could use another filter statement if you wanted the second result:
let arr = [1, 2, 4, 2, 3, 3, 4, 5, 5, 5, 8, 8, 9, 10];
let finalResult = arr.filter((e, i, a) => a.filter(f => f == e).length == 1).sort(function(a, b)return a - b);
console.log(finalResult);Just use a simple array.filter one-liner:
let arr = [1, 2, 4, 2, 3, 3, 4, 5, 5, 5, 8, 8, 9, 10];
let finalResult = arr.filter((e, i, a) => a.indexOf(e) == i).sort(function(a, b)return a - b);
console.log(finalResult);You could use another filter statement if you wanted the second result:
let arr = [1, 2, 4, 2, 3, 3, 4, 5, 5, 5, 8, 8, 9, 10];
let finalResult = arr.filter((e, i, a) => a.filter(f => f == e).length == 1).sort(function(a, b)return a - b);
console.log(finalResult);let arr = [1, 2, 4, 2, 3, 3, 4, 5, 5, 5, 8, 8, 9, 10];
let finalResult = arr.filter((e, i, a) => a.indexOf(e) == i).sort(function(a, b)return a - b);
console.log(finalResult);let arr = [1, 2, 4, 2, 3, 3, 4, 5, 5, 5, 8, 8, 9, 10];
let finalResult = arr.filter((e, i, a) => a.indexOf(e) == i).sort(function(a, b)return a - b);
console.log(finalResult);let arr = [1, 2, 4, 2, 3, 3, 4, 5, 5, 5, 8, 8, 9, 10];
let finalResult = arr.filter((e, i, a) => a.filter(f => f == e).length == 1).sort(function(a, b)return a - b);
console.log(finalResult);let arr = [1, 2, 4, 2, 3, 3, 4, 5, 5, 5, 8, 8, 9, 10];
let finalResult = arr.filter((e, i, a) => a.filter(f => f == e).length == 1).sort(function(a, b)return a - b);
console.log(finalResult);edited 12 hours ago
Mukyuu
1,79531123
1,79531123
answered 12 hours ago
Jack BashfordJack Bashford
12.2k31847
12.2k31847
You could also add.sort()to sort them by numerical order:.sort(function(a, b)return a - b)on finalresult
– Mukyuu
12 hours ago
Yes @Mukyuu, that would also be useful
– Jack Bashford
12 hours ago
2
Worth pointing out that the run time of this approach will be quadratic on the size of the input, which is probably not great unless the input arrays are known to be always fairly small.
– Joe Lee-Moyet
7 hours ago
Most voted with multiple array#filter, array#sort and array#indexOf... That is not performant
– Yosvel Quintero
6 hours ago
Do note that the.sort()is not necessary - the end result without the sort is still an array without any duplicate items, just in the same order as the original array. (It does make it exactly match thefinalResultvariable in the question though.)
– Florrie
4 hours ago
add a comment |
You could also add.sort()to sort them by numerical order:.sort(function(a, b)return a - b)on finalresult
– Mukyuu
12 hours ago
Yes @Mukyuu, that would also be useful
– Jack Bashford
12 hours ago
2
Worth pointing out that the run time of this approach will be quadratic on the size of the input, which is probably not great unless the input arrays are known to be always fairly small.
– Joe Lee-Moyet
7 hours ago
Most voted with multiple array#filter, array#sort and array#indexOf... That is not performant
– Yosvel Quintero
6 hours ago
Do note that the.sort()is not necessary - the end result without the sort is still an array without any duplicate items, just in the same order as the original array. (It does make it exactly match thefinalResultvariable in the question though.)
– Florrie
4 hours ago
You could also add
.sort() to sort them by numerical order: .sort(function(a, b)return a - b) on finalresult– Mukyuu
12 hours ago
You could also add
.sort() to sort them by numerical order: .sort(function(a, b)return a - b) on finalresult– Mukyuu
12 hours ago
Yes @Mukyuu, that would also be useful
– Jack Bashford
12 hours ago
Yes @Mukyuu, that would also be useful
– Jack Bashford
12 hours ago
2
2
Worth pointing out that the run time of this approach will be quadratic on the size of the input, which is probably not great unless the input arrays are known to be always fairly small.
– Joe Lee-Moyet
7 hours ago
Worth pointing out that the run time of this approach will be quadratic on the size of the input, which is probably not great unless the input arrays are known to be always fairly small.
– Joe Lee-Moyet
7 hours ago
Most voted with multiple array#filter, array#sort and array#indexOf... That is not performant
– Yosvel Quintero
6 hours ago
Most voted with multiple array#filter, array#sort and array#indexOf... That is not performant
– Yosvel Quintero
6 hours ago
Do note that the
.sort() is not necessary - the end result without the sort is still an array without any duplicate items, just in the same order as the original array. (It does make it exactly match the finalResult variable in the question though.)– Florrie
4 hours ago
Do note that the
.sort() is not necessary - the end result without the sort is still an array without any duplicate items, just in the same order as the original array. (It does make it exactly match the finalResult variable in the question though.)– Florrie
4 hours ago
add a comment |
For the first part you can use Set() and Spread Syntax to remove duplicates.
let arr = [1, 2, 4, 2, 3, 3, 4, 5, 5, 5, 8, 8, 9, 10];
let res = [...new Set(arr)]
console.log(res)For the second part you can use reduce()
let arr = [1, 2, 4, 2, 3, 3, 4, 5, 5, 5, 8, 8, 9, 10];
//to get the object with count of each number in array.
let obj = arr.reduce((ac,a) =>
//check if number doesnot occur before then set its count to 1
if(!ac[a]) ac[a] = 1;
//if number is already in object increase its count
else ac[a]++;
return ac;
,)
//Using reduce on all the keys of object means all numbers.
let res = Object.keys(obj).reduce((ac,a) =>
//check if count of current number 'a' is `1` in the above object then add it into array
if(obj[a] === 1) ac.push(+a)
return ac;
,[])
console.log(res)
nice appreciate that, that 2nd one looks crazy. I'm assuming the time complexity for would be less than ideal compared to other results?
– mph85
11 hours ago
@mph85 Yes its a little complex because it doesnot go through the array again and again instead it just store all the result obj and then filter it.Its better regarding performance
– Maheer Ali
11 hours ago
could you remind me why we need thespread operator? what happens if we don't have it? @Maheer Ali
– mph85
11 hours ago
Is it because if we don't have it, it'll just log an object?
– mph85
11 hours ago
@mph85 No if we will not have it. We will have aSet()inside array. We use it convertSet()toArray
– Maheer Ali
11 hours ago
|
show 1 more comment
For the first part you can use Set() and Spread Syntax to remove duplicates.
let arr = [1, 2, 4, 2, 3, 3, 4, 5, 5, 5, 8, 8, 9, 10];
let res = [...new Set(arr)]
console.log(res)For the second part you can use reduce()
let arr = [1, 2, 4, 2, 3, 3, 4, 5, 5, 5, 8, 8, 9, 10];
//to get the object with count of each number in array.
let obj = arr.reduce((ac,a) =>
//check if number doesnot occur before then set its count to 1
if(!ac[a]) ac[a] = 1;
//if number is already in object increase its count
else ac[a]++;
return ac;
,)
//Using reduce on all the keys of object means all numbers.
let res = Object.keys(obj).reduce((ac,a) =>
//check if count of current number 'a' is `1` in the above object then add it into array
if(obj[a] === 1) ac.push(+a)
return ac;
,[])
console.log(res)
nice appreciate that, that 2nd one looks crazy. I'm assuming the time complexity for would be less than ideal compared to other results?
– mph85
11 hours ago
@mph85 Yes its a little complex because it doesnot go through the array again and again instead it just store all the result obj and then filter it.Its better regarding performance
– Maheer Ali
11 hours ago
could you remind me why we need thespread operator? what happens if we don't have it? @Maheer Ali
– mph85
11 hours ago
Is it because if we don't have it, it'll just log an object?
– mph85
11 hours ago
@mph85 No if we will not have it. We will have aSet()inside array. We use it convertSet()toArray
– Maheer Ali
11 hours ago
|
show 1 more comment
For the first part you can use Set() and Spread Syntax to remove duplicates.
let arr = [1, 2, 4, 2, 3, 3, 4, 5, 5, 5, 8, 8, 9, 10];
let res = [...new Set(arr)]
console.log(res)For the second part you can use reduce()
let arr = [1, 2, 4, 2, 3, 3, 4, 5, 5, 5, 8, 8, 9, 10];
//to get the object with count of each number in array.
let obj = arr.reduce((ac,a) =>
//check if number doesnot occur before then set its count to 1
if(!ac[a]) ac[a] = 1;
//if number is already in object increase its count
else ac[a]++;
return ac;
,)
//Using reduce on all the keys of object means all numbers.
let res = Object.keys(obj).reduce((ac,a) =>
//check if count of current number 'a' is `1` in the above object then add it into array
if(obj[a] === 1) ac.push(+a)
return ac;
,[])
console.log(res)For the first part you can use Set() and Spread Syntax to remove duplicates.
let arr = [1, 2, 4, 2, 3, 3, 4, 5, 5, 5, 8, 8, 9, 10];
let res = [...new Set(arr)]
console.log(res)For the second part you can use reduce()
let arr = [1, 2, 4, 2, 3, 3, 4, 5, 5, 5, 8, 8, 9, 10];
//to get the object with count of each number in array.
let obj = arr.reduce((ac,a) =>
//check if number doesnot occur before then set its count to 1
if(!ac[a]) ac[a] = 1;
//if number is already in object increase its count
else ac[a]++;
return ac;
,)
//Using reduce on all the keys of object means all numbers.
let res = Object.keys(obj).reduce((ac,a) =>
//check if count of current number 'a' is `1` in the above object then add it into array
if(obj[a] === 1) ac.push(+a)
return ac;
,[])
console.log(res)let arr = [1, 2, 4, 2, 3, 3, 4, 5, 5, 5, 8, 8, 9, 10];
let res = [...new Set(arr)]
console.log(res)let arr = [1, 2, 4, 2, 3, 3, 4, 5, 5, 5, 8, 8, 9, 10];
let res = [...new Set(arr)]
console.log(res)let arr = [1, 2, 4, 2, 3, 3, 4, 5, 5, 5, 8, 8, 9, 10];
//to get the object with count of each number in array.
let obj = arr.reduce((ac,a) =>
//check if number doesnot occur before then set its count to 1
if(!ac[a]) ac[a] = 1;
//if number is already in object increase its count
else ac[a]++;
return ac;
,)
//Using reduce on all the keys of object means all numbers.
let res = Object.keys(obj).reduce((ac,a) =>
//check if count of current number 'a' is `1` in the above object then add it into array
if(obj[a] === 1) ac.push(+a)
return ac;
,[])
console.log(res)let arr = [1, 2, 4, 2, 3, 3, 4, 5, 5, 5, 8, 8, 9, 10];
//to get the object with count of each number in array.
let obj = arr.reduce((ac,a) =>
//check if number doesnot occur before then set its count to 1
if(!ac[a]) ac[a] = 1;
//if number is already in object increase its count
else ac[a]++;
return ac;
,)
//Using reduce on all the keys of object means all numbers.
let res = Object.keys(obj).reduce((ac,a) =>
//check if count of current number 'a' is `1` in the above object then add it into array
if(obj[a] === 1) ac.push(+a)
return ac;
,[])
console.log(res)edited 3 hours ago
answered 12 hours ago
Maheer AliMaheer Ali
6,765417
6,765417
nice appreciate that, that 2nd one looks crazy. I'm assuming the time complexity for would be less than ideal compared to other results?
– mph85
11 hours ago
@mph85 Yes its a little complex because it doesnot go through the array again and again instead it just store all the result obj and then filter it.Its better regarding performance
– Maheer Ali
11 hours ago
could you remind me why we need thespread operator? what happens if we don't have it? @Maheer Ali
– mph85
11 hours ago
Is it because if we don't have it, it'll just log an object?
– mph85
11 hours ago
@mph85 No if we will not have it. We will have aSet()inside array. We use it convertSet()toArray
– Maheer Ali
11 hours ago
|
show 1 more comment
nice appreciate that, that 2nd one looks crazy. I'm assuming the time complexity for would be less than ideal compared to other results?
– mph85
11 hours ago
@mph85 Yes its a little complex because it doesnot go through the array again and again instead it just store all the result obj and then filter it.Its better regarding performance
– Maheer Ali
11 hours ago
could you remind me why we need thespread operator? what happens if we don't have it? @Maheer Ali
– mph85
11 hours ago
Is it because if we don't have it, it'll just log an object?
– mph85
11 hours ago
@mph85 No if we will not have it. We will have aSet()inside array. We use it convertSet()toArray
– Maheer Ali
11 hours ago
nice appreciate that, that 2nd one looks crazy. I'm assuming the time complexity for would be less than ideal compared to other results?
– mph85
11 hours ago
nice appreciate that, that 2nd one looks crazy. I'm assuming the time complexity for would be less than ideal compared to other results?
– mph85
11 hours ago
@mph85 Yes its a little complex because it doesnot go through the array again and again instead it just store all the result obj and then filter it.Its better regarding performance
– Maheer Ali
11 hours ago
@mph85 Yes its a little complex because it doesnot go through the array again and again instead it just store all the result obj and then filter it.Its better regarding performance
– Maheer Ali
11 hours ago
could you remind me why we need the
spread operator? what happens if we don't have it? @Maheer Ali– mph85
11 hours ago
could you remind me why we need the
spread operator? what happens if we don't have it? @Maheer Ali– mph85
11 hours ago
Is it because if we don't have it, it'll just log an object?
– mph85
11 hours ago
Is it because if we don't have it, it'll just log an object?
– mph85
11 hours ago
@mph85 No if we will not have it. We will have a
Set() inside array. We use it convert Set() to Array– Maheer Ali
11 hours ago
@mph85 No if we will not have it. We will have a
Set() inside array. We use it convert Set() to Array– Maheer Ali
11 hours ago
|
show 1 more comment
You could sort the array before and filter the array by checking only one side for duplicates or both sides.
var array = [1, 2, 4, 2, 3, 3, 4, 5, 5, 5, 8, 8, 9, 10],
result1,
result2;
array.sort((a, b) => a - b);
result1 = array.filter((v, i, a) => a[i - 1] !== v);
result2 = array.filter((v, i, a) => a[i - 1] !== v && a[i + 1] !== v);
console.log(...result1);
console.log(...result2)add a comment |
You could sort the array before and filter the array by checking only one side for duplicates or both sides.
var array = [1, 2, 4, 2, 3, 3, 4, 5, 5, 5, 8, 8, 9, 10],
result1,
result2;
array.sort((a, b) => a - b);
result1 = array.filter((v, i, a) => a[i - 1] !== v);
result2 = array.filter((v, i, a) => a[i - 1] !== v && a[i + 1] !== v);
console.log(...result1);
console.log(...result2)add a comment |
You could sort the array before and filter the array by checking only one side for duplicates or both sides.
var array = [1, 2, 4, 2, 3, 3, 4, 5, 5, 5, 8, 8, 9, 10],
result1,
result2;
array.sort((a, b) => a - b);
result1 = array.filter((v, i, a) => a[i - 1] !== v);
result2 = array.filter((v, i, a) => a[i - 1] !== v && a[i + 1] !== v);
console.log(...result1);
console.log(...result2)You could sort the array before and filter the array by checking only one side for duplicates or both sides.
var array = [1, 2, 4, 2, 3, 3, 4, 5, 5, 5, 8, 8, 9, 10],
result1,
result2;
array.sort((a, b) => a - b);
result1 = array.filter((v, i, a) => a[i - 1] !== v);
result2 = array.filter((v, i, a) => a[i - 1] !== v && a[i + 1] !== v);
console.log(...result1);
console.log(...result2)var array = [1, 2, 4, 2, 3, 3, 4, 5, 5, 5, 8, 8, 9, 10],
result1,
result2;
array.sort((a, b) => a - b);
result1 = array.filter((v, i, a) => a[i - 1] !== v);
result2 = array.filter((v, i, a) => a[i - 1] !== v && a[i + 1] !== v);
console.log(...result1);
console.log(...result2)var array = [1, 2, 4, 2, 3, 3, 4, 5, 5, 5, 8, 8, 9, 10],
result1,
result2;
array.sort((a, b) => a - b);
result1 = array.filter((v, i, a) => a[i - 1] !== v);
result2 = array.filter((v, i, a) => a[i - 1] !== v && a[i + 1] !== v);
console.log(...result1);
console.log(...result2)answered 11 hours ago
Nina ScholzNina Scholz
192k15104177
192k15104177
add a comment |
add a comment |
You can create both arrays in One Go
let arr = [1, 2, 4, 2, 3, 3, 4, 5, 5, 5, 8, 8, 9, 10];
let unique = new Set();
let repeated = Array.from(arr.reduce((acc, curr) =>
acc.has(curr) ? unique.delete(curr) : acc.add(curr) && unique.add(curr);
return acc;
, new Set()));
console.log(Array.from(unique))
console.log(repeated)
1
Nice one mate +1
– Maheer Ali
11 hours ago
I wonder if the ternary plus&&could be unclear though (x?y:z&&w)? It's not obvious to me how JS's order of operations would handle that, and I wonder if you could get across the same logic and reasoning by usingif/else?
– Florrie
4 hours ago
add a comment |
You can create both arrays in One Go
let arr = [1, 2, 4, 2, 3, 3, 4, 5, 5, 5, 8, 8, 9, 10];
let unique = new Set();
let repeated = Array.from(arr.reduce((acc, curr) =>
acc.has(curr) ? unique.delete(curr) : acc.add(curr) && unique.add(curr);
return acc;
, new Set()));
console.log(Array.from(unique))
console.log(repeated)
1
Nice one mate +1
– Maheer Ali
11 hours ago
I wonder if the ternary plus&&could be unclear though (x?y:z&&w)? It's not obvious to me how JS's order of operations would handle that, and I wonder if you could get across the same logic and reasoning by usingif/else?
– Florrie
4 hours ago
add a comment |
You can create both arrays in One Go
let arr = [1, 2, 4, 2, 3, 3, 4, 5, 5, 5, 8, 8, 9, 10];
let unique = new Set();
let repeated = Array.from(arr.reduce((acc, curr) =>
acc.has(curr) ? unique.delete(curr) : acc.add(curr) && unique.add(curr);
return acc;
, new Set()));
console.log(Array.from(unique))
console.log(repeated)You can create both arrays in One Go
let arr = [1, 2, 4, 2, 3, 3, 4, 5, 5, 5, 8, 8, 9, 10];
let unique = new Set();
let repeated = Array.from(arr.reduce((acc, curr) =>
acc.has(curr) ? unique.delete(curr) : acc.add(curr) && unique.add(curr);
return acc;
, new Set()));
console.log(Array.from(unique))
console.log(repeated)let arr = [1, 2, 4, 2, 3, 3, 4, 5, 5, 5, 8, 8, 9, 10];
let unique = new Set();
let repeated = Array.from(arr.reduce((acc, curr) =>
acc.has(curr) ? unique.delete(curr) : acc.add(curr) && unique.add(curr);
return acc;
, new Set()));
console.log(Array.from(unique))
console.log(repeated)let arr = [1, 2, 4, 2, 3, 3, 4, 5, 5, 5, 8, 8, 9, 10];
let unique = new Set();
let repeated = Array.from(arr.reduce((acc, curr) =>
acc.has(curr) ? unique.delete(curr) : acc.add(curr) && unique.add(curr);
return acc;
, new Set()));
console.log(Array.from(unique))
console.log(repeated)answered 11 hours ago
AZ_AZ_
559210
559210
1
Nice one mate +1
– Maheer Ali
11 hours ago
I wonder if the ternary plus&&could be unclear though (x?y:z&&w)? It's not obvious to me how JS's order of operations would handle that, and I wonder if you could get across the same logic and reasoning by usingif/else?
– Florrie
4 hours ago
add a comment |
1
Nice one mate +1
– Maheer Ali
11 hours ago
I wonder if the ternary plus&&could be unclear though (x?y:z&&w)? It's not obvious to me how JS's order of operations would handle that, and I wonder if you could get across the same logic and reasoning by usingif/else?
– Florrie
4 hours ago
1
1
Nice one mate +1
– Maheer Ali
11 hours ago
Nice one mate +1
– Maheer Ali
11 hours ago
I wonder if the ternary plus
&& could be unclear though (x?y:z&&w)? It's not obvious to me how JS's order of operations would handle that, and I wonder if you could get across the same logic and reasoning by using if/else?– Florrie
4 hours ago
I wonder if the ternary plus
&& could be unclear though (x?y:z&&w)? It's not obvious to me how JS's order of operations would handle that, and I wonder if you could get across the same logic and reasoning by using if/else?– Florrie
4 hours ago
add a comment |
You can do:
const arr = [1, 2, 4, 2, 3, 3, 4, 5, 5, 5, 8, 8, 9, 10];
const hash = arr.reduce((a, c) => (a[c] = (a[c] || 0) + 1, a), );
// [1, 2, 3, 4, 5, 8, 9, 10];
const finalResultOne = Object.keys(hash);
// [1, 9, 10];
const finalResultTwo = Object.keys(hash).filter(k => hash[k] === 1);
console.log('finalResultOne:', ...finalResultOne);
console.log('finalResultTwo:', ...finalResultTwo);add a comment |
You can do:
const arr = [1, 2, 4, 2, 3, 3, 4, 5, 5, 5, 8, 8, 9, 10];
const hash = arr.reduce((a, c) => (a[c] = (a[c] || 0) + 1, a), );
// [1, 2, 3, 4, 5, 8, 9, 10];
const finalResultOne = Object.keys(hash);
// [1, 9, 10];
const finalResultTwo = Object.keys(hash).filter(k => hash[k] === 1);
console.log('finalResultOne:', ...finalResultOne);
console.log('finalResultTwo:', ...finalResultTwo);add a comment |
You can do:
const arr = [1, 2, 4, 2, 3, 3, 4, 5, 5, 5, 8, 8, 9, 10];
const hash = arr.reduce((a, c) => (a[c] = (a[c] || 0) + 1, a), );
// [1, 2, 3, 4, 5, 8, 9, 10];
const finalResultOne = Object.keys(hash);
// [1, 9, 10];
const finalResultTwo = Object.keys(hash).filter(k => hash[k] === 1);
console.log('finalResultOne:', ...finalResultOne);
console.log('finalResultTwo:', ...finalResultTwo);You can do:
const arr = [1, 2, 4, 2, 3, 3, 4, 5, 5, 5, 8, 8, 9, 10];
const hash = arr.reduce((a, c) => (a[c] = (a[c] || 0) + 1, a), );
// [1, 2, 3, 4, 5, 8, 9, 10];
const finalResultOne = Object.keys(hash);
// [1, 9, 10];
const finalResultTwo = Object.keys(hash).filter(k => hash[k] === 1);
console.log('finalResultOne:', ...finalResultOne);
console.log('finalResultTwo:', ...finalResultTwo);const arr = [1, 2, 4, 2, 3, 3, 4, 5, 5, 5, 8, 8, 9, 10];
const hash = arr.reduce((a, c) => (a[c] = (a[c] || 0) + 1, a), );
// [1, 2, 3, 4, 5, 8, 9, 10];
const finalResultOne = Object.keys(hash);
// [1, 9, 10];
const finalResultTwo = Object.keys(hash).filter(k => hash[k] === 1);
console.log('finalResultOne:', ...finalResultOne);
console.log('finalResultTwo:', ...finalResultTwo);const arr = [1, 2, 4, 2, 3, 3, 4, 5, 5, 5, 8, 8, 9, 10];
const hash = arr.reduce((a, c) => (a[c] = (a[c] || 0) + 1, a), );
// [1, 2, 3, 4, 5, 8, 9, 10];
const finalResultOne = Object.keys(hash);
// [1, 9, 10];
const finalResultTwo = Object.keys(hash).filter(k => hash[k] === 1);
console.log('finalResultOne:', ...finalResultOne);
console.log('finalResultTwo:', ...finalResultTwo);edited 7 hours ago
answered 11 hours ago
Yosvel QuinteroYosvel Quintero
11.8k42531
11.8k42531
add a comment |
add a comment |
You can use closure and Map
let arr = [1, 2, 4, 2, 3, 3, 4, 5, 5, 5, 8, 8, 9, 10];
const build = ar =>
const mapObj = ar.reduce((acc, e) =>
acc.has(e) ? acc.set(e, true) : acc.set(e, false)
return acc
, new Map())
return function(hasDup = true)
if(hasDup) return [...mapObj.keys()]
else return [...mapObj].filter(([key, val]) => !val).map(([k, v])=> k)
const getArr = build(arr)
console.log(getArr())
console.log(getArr(false))add a comment |
You can use closure and Map
let arr = [1, 2, 4, 2, 3, 3, 4, 5, 5, 5, 8, 8, 9, 10];
const build = ar =>
const mapObj = ar.reduce((acc, e) =>
acc.has(e) ? acc.set(e, true) : acc.set(e, false)
return acc
, new Map())
return function(hasDup = true)
if(hasDup) return [...mapObj.keys()]
else return [...mapObj].filter(([key, val]) => !val).map(([k, v])=> k)
const getArr = build(arr)
console.log(getArr())
console.log(getArr(false))add a comment |
You can use closure and Map
let arr = [1, 2, 4, 2, 3, 3, 4, 5, 5, 5, 8, 8, 9, 10];
const build = ar =>
const mapObj = ar.reduce((acc, e) =>
acc.has(e) ? acc.set(e, true) : acc.set(e, false)
return acc
, new Map())
return function(hasDup = true)
if(hasDup) return [...mapObj.keys()]
else return [...mapObj].filter(([key, val]) => !val).map(([k, v])=> k)
const getArr = build(arr)
console.log(getArr())
console.log(getArr(false))You can use closure and Map
let arr = [1, 2, 4, 2, 3, 3, 4, 5, 5, 5, 8, 8, 9, 10];
const build = ar =>
const mapObj = ar.reduce((acc, e) =>
acc.has(e) ? acc.set(e, true) : acc.set(e, false)
return acc
, new Map())
return function(hasDup = true)
if(hasDup) return [...mapObj.keys()]
else return [...mapObj].filter(([key, val]) => !val).map(([k, v])=> k)
const getArr = build(arr)
console.log(getArr())
console.log(getArr(false))let arr = [1, 2, 4, 2, 3, 3, 4, 5, 5, 5, 8, 8, 9, 10];
const build = ar =>
const mapObj = ar.reduce((acc, e) =>
acc.has(e) ? acc.set(e, true) : acc.set(e, false)
return acc
, new Map())
return function(hasDup = true)
if(hasDup) return [...mapObj.keys()]
else return [...mapObj].filter(([key, val]) => !val).map(([k, v])=> k)
const getArr = build(arr)
console.log(getArr())
console.log(getArr(false))let arr = [1, 2, 4, 2, 3, 3, 4, 5, 5, 5, 8, 8, 9, 10];
const build = ar =>
const mapObj = ar.reduce((acc, e) =>
acc.has(e) ? acc.set(e, true) : acc.set(e, false)
return acc
, new Map())
return function(hasDup = true)
if(hasDup) return [...mapObj.keys()]
else return [...mapObj].filter(([key, val]) => !val).map(([k, v])=> k)
const getArr = build(arr)
console.log(getArr())
console.log(getArr(false))answered 11 hours ago
birdbird
822619
822619
add a comment |
add a comment |
As many other have said, the first one is just [...new Set(arr)]
For the second, just filter out those that occur more than once:
const arr = [1, 2, 4, 2, 3, 3, 4, 5, 5, 5, 8, 8, 9, 10];
const count = (arr, e) => arr.filter(n => n == e).length
const unique = arr => arr.filter(e => count(arr, e) < 2)
console.log(unique(arr));add a comment |
As many other have said, the first one is just [...new Set(arr)]
For the second, just filter out those that occur more than once:
const arr = [1, 2, 4, 2, 3, 3, 4, 5, 5, 5, 8, 8, 9, 10];
const count = (arr, e) => arr.filter(n => n == e).length
const unique = arr => arr.filter(e => count(arr, e) < 2)
console.log(unique(arr));add a comment |
As many other have said, the first one is just [...new Set(arr)]
For the second, just filter out those that occur more than once:
const arr = [1, 2, 4, 2, 3, 3, 4, 5, 5, 5, 8, 8, 9, 10];
const count = (arr, e) => arr.filter(n => n == e).length
const unique = arr => arr.filter(e => count(arr, e) < 2)
console.log(unique(arr));As many other have said, the first one is just [...new Set(arr)]
For the second, just filter out those that occur more than once:
const arr = [1, 2, 4, 2, 3, 3, 4, 5, 5, 5, 8, 8, 9, 10];
const count = (arr, e) => arr.filter(n => n == e).length
const unique = arr => arr.filter(e => count(arr, e) < 2)
console.log(unique(arr));const arr = [1, 2, 4, 2, 3, 3, 4, 5, 5, 5, 8, 8, 9, 10];
const count = (arr, e) => arr.filter(n => n == e).length
const unique = arr => arr.filter(e => count(arr, e) < 2)
console.log(unique(arr));const arr = [1, 2, 4, 2, 3, 3, 4, 5, 5, 5, 8, 8, 9, 10];
const count = (arr, e) => arr.filter(n => n == e).length
const unique = arr => arr.filter(e => count(arr, e) < 2)
console.log(unique(arr));answered 7 hours ago
JollyJokerJollyJoker
1915
1915
add a comment |
add a comment |
var arr = [1, 2, 4, 2, 3, 3, 4, 5, 5, 5, 8, 8, 9, 10];
var map = ;
var finalResult = [];
for(var i = 0; i < arr.length; i++)
if(!map.hasOwnProperty(arr[i]))
map[arr[i]] = true;
finalResult.push(arr[i]);
//if you need it sorted otherwise it will be in order
finalResult.sort(function(a, b)return a-b);
add a comment |
var arr = [1, 2, 4, 2, 3, 3, 4, 5, 5, 5, 8, 8, 9, 10];
var map = ;
var finalResult = [];
for(var i = 0; i < arr.length; i++)
if(!map.hasOwnProperty(arr[i]))
map[arr[i]] = true;
finalResult.push(arr[i]);
//if you need it sorted otherwise it will be in order
finalResult.sort(function(a, b)return a-b);
add a comment |
var arr = [1, 2, 4, 2, 3, 3, 4, 5, 5, 5, 8, 8, 9, 10];
var map = ;
var finalResult = [];
for(var i = 0; i < arr.length; i++)
if(!map.hasOwnProperty(arr[i]))
map[arr[i]] = true;
finalResult.push(arr[i]);
//if you need it sorted otherwise it will be in order
finalResult.sort(function(a, b)return a-b);
var arr = [1, 2, 4, 2, 3, 3, 4, 5, 5, 5, 8, 8, 9, 10];
var map = ;
var finalResult = [];
for(var i = 0; i < arr.length; i++)
if(!map.hasOwnProperty(arr[i]))
map[arr[i]] = true;
finalResult.push(arr[i]);
//if you need it sorted otherwise it will be in order
finalResult.sort(function(a, b)return a-b);
answered 5 hours ago
JanspeedJanspeed
1,3031315
1,3031315
add a comment |
add a comment |
If you're using lodash, you can use
_.uniq()– BlueRaja - Danny Pflughoeft
9 hours ago
1
Further, this is asking for the inverse of Get all non-unique values (i.e.: duplicate/more than one occurrence) in an array. Finally, this post is asking two separate questions and both have good answers elsewhere already.
– Søren D. Ptæus
9 hours ago
To answer the question "which one is it" in a comment-answer: if you're asked to remove duplicates, I believe you should understand the first variant. The second variant removes all element that have duplicates, meaning the "original" value AND its duplicates.
– Pierre Arlaud
6 hours ago