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?
What is the English pronunciation of "pain au chocolat"?
How does electrical safety system work on ISS?
What (the heck) is a Super Worm Equinox Moon?
Did the UK lift the requirement for registering SIM cards?
Is this part of the description of the Archfey warlock's Misty Escape feature redundant?
Unable to get dependencies from jcenter with a new project
C++ copy constructor called at return
Stack Interview Code methods made from class Node and Smart Pointers
What to do when eye contact makes your coworker uncomfortable?
Is there a nicer/politer/more positive alternative for "negates"?
Does grappling negate Mirror Image?
How to make money from a browser who sees 5 seconds into the future of any web page?
Review your own paper in Mathematics
Is it ethical to recieve stipend after publishing enough papers?
What do you call a word that can be spelled forward or backward forming two different words
Giving feedback to someone without sounding prejudiced
Permission on Database
Does the reader need to like the PoV character?
15% tax on $7.5k earnings. Is that right?
Is it allowed to activate the ability of multiple planeswalkers in a single turn?
Why should universal income be universal?
How to preserve electronics (computers, iPads and phones) for hundreds of years
Make a Bowl of Alphabet Soup
Why Shazam when there is already Superman?
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 yesterday
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 yesterday
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
yesterday
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
yesterday
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
yesterday
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 yesterday
pushkin
4,323112954
4,323112954
asked yesterday
mph85mph85
1229
1229
marked as duplicate by Jared Smith, pushkin, BlueRaja - Danny Pflughoeft, the_lotus, Moira yesterday
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 yesterday
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
yesterday
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
yesterday
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
yesterday
add a comment |
If you're using lodash, you can use_.uniq()
– BlueRaja - Danny Pflughoeft
yesterday
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
yesterday
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
yesterday
If you're using lodash, you can use
_.uniq()
– BlueRaja - Danny Pflughoeft
yesterday
If you're using lodash, you can use
_.uniq()
– BlueRaja - Danny Pflughoeft
yesterday
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
yesterday
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
yesterday
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
yesterday
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
yesterday
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
yesterday
Yes @Mukyuu, that would also be useful
– Jack Bashford
yesterday
5
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
yesterday
1
Most voted with multiple array#filter, array#sort and array#indexOf... That is not performant
– Yosvel Quintero
yesterday
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 thefinalResult
variable in the question though.)
– Florrie
yesterday
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
yesterday
@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
yesterday
could you remind me why we need thespread operator
? what happens if we don't have it? @Maheer Ali
– mph85
yesterday
Is it because if we don't have it, it'll just log an object?
– mph85
yesterday
@mph85 No if we will not have it. We will have aSet()
inside array. We use it convertSet()
toArray
– Maheer Ali
yesterday
|
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)
thank for that, for theresult1
what is going on with thea[i - 1] !== v
?
– mph85
23 hours ago
a
is the array,i
is the actual index, andv
is the actual value. it takes the value from the index before of the actual index and looks if the values are not equal.
– Nina Scholz
23 hours ago
ah ok, and sorry, it takes thevalue from the index before of the actual index
? Not sure what you mean by that
– mph85
23 hours ago
for example if you have the value9
of the array asv
, then the value before is8
.
– Nina Scholz
23 hours ago
ah ok, thank you for that
– mph85
23 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
yesterday
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
yesterday
add a comment |
You can use Array.prototype.reduce() create a hash
object where the keys
are the numbers in the array and the values
are going to be the the repeated occurrence of numbers in the arr
array variable..
Then using Object.keys():
- Remove all duplicates
Object.keys(hash)
- Remove all duplicates but filtering with Array.prototype.filter() to get the numbers with only one occurrence
Code:
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 yesterday
answered yesterday
Aswin KumarAswin Kumar
899115
899115
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
yesterday
Yes @Mukyuu, that would also be useful
– Jack Bashford
yesterday
5
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
yesterday
1
Most voted with multiple array#filter, array#sort and array#indexOf... That is not performant
– Yosvel Quintero
yesterday
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 thefinalResult
variable in the question though.)
– Florrie
yesterday
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
yesterday
Yes @Mukyuu, that would also be useful
– Jack Bashford
yesterday
5
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
yesterday
1
Most voted with multiple array#filter, array#sort and array#indexOf... That is not performant
– Yosvel Quintero
yesterday
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 thefinalResult
variable in the question though.)
– Florrie
yesterday
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 yesterday
Mukyuu
1,79831123
1,79831123
answered yesterday
Jack BashfordJack Bashford
12.5k31847
12.5k31847
You could also add.sort()
to sort them by numerical order:.sort(function(a, b)return a - b)
on finalresult
– Mukyuu
yesterday
Yes @Mukyuu, that would also be useful
– Jack Bashford
yesterday
5
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
yesterday
1
Most voted with multiple array#filter, array#sort and array#indexOf... That is not performant
– Yosvel Quintero
yesterday
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 thefinalResult
variable in the question though.)
– Florrie
yesterday
add a comment |
You could also add.sort()
to sort them by numerical order:.sort(function(a, b)return a - b)
on finalresult
– Mukyuu
yesterday
Yes @Mukyuu, that would also be useful
– Jack Bashford
yesterday
5
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
yesterday
1
Most voted with multiple array#filter, array#sort and array#indexOf... That is not performant
– Yosvel Quintero
yesterday
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 thefinalResult
variable in the question though.)
– Florrie
yesterday
You could also add
.sort()
to sort them by numerical order: .sort(function(a, b)return a - b)
on finalresult– Mukyuu
yesterday
You could also add
.sort()
to sort them by numerical order: .sort(function(a, b)return a - b)
on finalresult– Mukyuu
yesterday
Yes @Mukyuu, that would also be useful
– Jack Bashford
yesterday
Yes @Mukyuu, that would also be useful
– Jack Bashford
yesterday
5
5
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
yesterday
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
yesterday
1
1
Most voted with multiple array#filter, array#sort and array#indexOf... That is not performant
– Yosvel Quintero
yesterday
Most voted with multiple array#filter, array#sort and array#indexOf... That is not performant
– Yosvel Quintero
yesterday
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
yesterday
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
yesterday
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
yesterday
@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
yesterday
could you remind me why we need thespread operator
? what happens if we don't have it? @Maheer Ali
– mph85
yesterday
Is it because if we don't have it, it'll just log an object?
– mph85
yesterday
@mph85 No if we will not have it. We will have aSet()
inside array. We use it convertSet()
toArray
– Maheer Ali
yesterday
|
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
yesterday
@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
yesterday
could you remind me why we need thespread operator
? what happens if we don't have it? @Maheer Ali
– mph85
yesterday
Is it because if we don't have it, it'll just log an object?
– mph85
yesterday
@mph85 No if we will not have it. We will have aSet()
inside array. We use it convertSet()
toArray
– Maheer Ali
yesterday
|
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 yesterday
answered yesterday
Maheer AliMaheer Ali
6,836518
6,836518
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
yesterday
@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
yesterday
could you remind me why we need thespread operator
? what happens if we don't have it? @Maheer Ali
– mph85
yesterday
Is it because if we don't have it, it'll just log an object?
– mph85
yesterday
@mph85 No if we will not have it. We will have aSet()
inside array. We use it convertSet()
toArray
– Maheer Ali
yesterday
|
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
yesterday
@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
yesterday
could you remind me why we need thespread operator
? what happens if we don't have it? @Maheer Ali
– mph85
yesterday
Is it because if we don't have it, it'll just log an object?
– mph85
yesterday
@mph85 No if we will not have it. We will have aSet()
inside array. We use it convertSet()
toArray
– Maheer Ali
yesterday
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
yesterday
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
yesterday
@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
yesterday
@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
yesterday
could you remind me why we need the
spread operator
? what happens if we don't have it? @Maheer Ali– mph85
yesterday
could you remind me why we need the
spread operator
? what happens if we don't have it? @Maheer Ali– mph85
yesterday
Is it because if we don't have it, it'll just log an object?
– mph85
yesterday
Is it because if we don't have it, it'll just log an object?
– mph85
yesterday
@mph85 No if we will not have it. We will have a
Set()
inside array. We use it convert Set()
to Array
– Maheer Ali
yesterday
@mph85 No if we will not have it. We will have a
Set()
inside array. We use it convert Set()
to Array
– Maheer Ali
yesterday
|
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)
thank for that, for theresult1
what is going on with thea[i - 1] !== v
?
– mph85
23 hours ago
a
is the array,i
is the actual index, andv
is the actual value. it takes the value from the index before of the actual index and looks if the values are not equal.
– Nina Scholz
23 hours ago
ah ok, and sorry, it takes thevalue from the index before of the actual index
? Not sure what you mean by that
– mph85
23 hours ago
for example if you have the value9
of the array asv
, then the value before is8
.
– Nina Scholz
23 hours ago
ah ok, thank you for that
– mph85
23 hours ago
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)
thank for that, for theresult1
what is going on with thea[i - 1] !== v
?
– mph85
23 hours ago
a
is the array,i
is the actual index, andv
is the actual value. it takes the value from the index before of the actual index and looks if the values are not equal.
– Nina Scholz
23 hours ago
ah ok, and sorry, it takes thevalue from the index before of the actual index
? Not sure what you mean by that
– mph85
23 hours ago
for example if you have the value9
of the array asv
, then the value before is8
.
– Nina Scholz
23 hours ago
ah ok, thank you for that
– mph85
23 hours ago
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 yesterday
Nina ScholzNina Scholz
192k15104177
192k15104177
thank for that, for theresult1
what is going on with thea[i - 1] !== v
?
– mph85
23 hours ago
a
is the array,i
is the actual index, andv
is the actual value. it takes the value from the index before of the actual index and looks if the values are not equal.
– Nina Scholz
23 hours ago
ah ok, and sorry, it takes thevalue from the index before of the actual index
? Not sure what you mean by that
– mph85
23 hours ago
for example if you have the value9
of the array asv
, then the value before is8
.
– Nina Scholz
23 hours ago
ah ok, thank you for that
– mph85
23 hours ago
add a comment |
thank for that, for theresult1
what is going on with thea[i - 1] !== v
?
– mph85
23 hours ago
a
is the array,i
is the actual index, andv
is the actual value. it takes the value from the index before of the actual index and looks if the values are not equal.
– Nina Scholz
23 hours ago
ah ok, and sorry, it takes thevalue from the index before of the actual index
? Not sure what you mean by that
– mph85
23 hours ago
for example if you have the value9
of the array asv
, then the value before is8
.
– Nina Scholz
23 hours ago
ah ok, thank you for that
– mph85
23 hours ago
thank for that, for the
result1
what is going on with the a[i - 1] !== v
?– mph85
23 hours ago
thank for that, for the
result1
what is going on with the a[i - 1] !== v
?– mph85
23 hours ago
a
is the array, i
is the actual index, and v
is the actual value. it takes the value from the index before of the actual index and looks if the values are not equal.– Nina Scholz
23 hours ago
a
is the array, i
is the actual index, and v
is the actual value. it takes the value from the index before of the actual index and looks if the values are not equal.– Nina Scholz
23 hours ago
ah ok, and sorry, it takes the
value from the index before of the actual index
? Not sure what you mean by that– mph85
23 hours ago
ah ok, and sorry, it takes the
value from the index before of the actual index
? Not sure what you mean by that– mph85
23 hours ago
for example if you have the value
9
of the array as v
, then the value before is 8
.– Nina Scholz
23 hours ago
for example if you have the value
9
of the array as v
, then the value before is 8
.– Nina Scholz
23 hours ago
ah ok, thank you for that
– mph85
23 hours ago
ah ok, thank you for that
– mph85
23 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
yesterday
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
yesterday
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
yesterday
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
yesterday
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 yesterday
AZ_AZ_
569210
569210
1
Nice one mate +1
– Maheer Ali
yesterday
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
yesterday
add a comment |
1
Nice one mate +1
– Maheer Ali
yesterday
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
yesterday
1
1
Nice one mate +1
– Maheer Ali
yesterday
Nice one mate +1
– Maheer Ali
yesterday
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
yesterday
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
yesterday
add a comment |
You can use Array.prototype.reduce() create a hash
object where the keys
are the numbers in the array and the values
are going to be the the repeated occurrence of numbers in the arr
array variable..
Then using Object.keys():
- Remove all duplicates
Object.keys(hash)
- Remove all duplicates but filtering with Array.prototype.filter() to get the numbers with only one occurrence
Code:
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 Array.prototype.reduce() create a hash
object where the keys
are the numbers in the array and the values
are going to be the the repeated occurrence of numbers in the arr
array variable..
Then using Object.keys():
- Remove all duplicates
Object.keys(hash)
- Remove all duplicates but filtering with Array.prototype.filter() to get the numbers with only one occurrence
Code:
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 Array.prototype.reduce() create a hash
object where the keys
are the numbers in the array and the values
are going to be the the repeated occurrence of numbers in the arr
array variable..
Then using Object.keys():
- Remove all duplicates
Object.keys(hash)
- Remove all duplicates but filtering with Array.prototype.filter() to get the numbers with only one occurrence
Code:
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 use Array.prototype.reduce() create a hash
object where the keys
are the numbers in the array and the values
are going to be the the repeated occurrence of numbers in the arr
array variable..
Then using Object.keys():
- Remove all duplicates
Object.keys(hash)
- Remove all duplicates but filtering with Array.prototype.filter() to get the numbers with only one occurrence
Code:
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 18 hours ago
answered yesterday
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 yesterday
birdbird
892620
892620
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 yesterday
JollyJokerJollyJoker
21115
21115
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 yesterday
JanspeedJanspeed
1,3231315
1,3231315
add a comment |
add a comment |
If you're using lodash, you can use
_.uniq()
– BlueRaja - Danny Pflughoeft
yesterday
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
yesterday
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
yesterday