Filtering duplicate hashes from array of hashes - Javascript
9
I have an array of hashes, like this:
[{id: "4bf58dd8d48988d110941735", name: "italy"},
{id: "4bf58dd8d48988d1c6941735", name: "skandi"},
{id: "4bf58dd8d48988d147941735", name: "diner"},
{id: "4bf58dd8d48988d110941735", name: "italy"},
{id: "4bf58dd8d48988d1c4941735", name: "resto"},
{id: "4bf58dd8d48988d14a941735", name: "vietnam"},
{id: "4bf58dd8d48988d1ce941735", name: "fish"},
{id: "4bf58dd8d48988d1c4941735", name: "resto"},
{id: "4bf58dd8d48988d1c4941735", name: "resto"}]
I want to throw out duplicate hashes. Set doesn't work because hashes are unique objects.
I feel stuck and need a kick to think. Please advise!
javascript arrays hash
edited Dec 21 at 11:07
Kamil Kiełczewski
8,86085788
asked Dec 21 at 9:40
Anna Kizilova
8819
|
show 3 more comments
9
I have an array of hashes, like this:
[{id: "4bf58dd8d48988d110941735", name: "italy"},
{id: "4bf58dd8d48988d1c6941735", name: "skandi"},
{id: "4bf58dd8d48988d147941735", name: "diner"},
{id: "4bf58dd8d48988d110941735", name: "italy"},
{id: "4bf58dd8d48988d1c4941735", name: "resto"},
{id: "4bf58dd8d48988d14a941735", name: "vietnam"},
{id: "4bf58dd8d48988d1ce941735", name: "fish"},
{id: "4bf58dd8d48988d1c4941735", name: "resto"},
{id: "4bf58dd8d48988d1c4941735", name: "resto"}]
I want to throw out duplicate hashes. Set doesn't work because hashes are unique objects.
I feel stuck and need a kick to think. Please advise!
javascript arrays hash
edited Dec 21 at 11:07
Kamil Kiełczewski
8,86085788
asked Dec 21 at 9:40
Anna Kizilova
8819
2
Reduce the array to an object, using the id as key, and then convert back to array usingObject.values().
– Ori Drori
Dec 21 at 9:44
What you tried?
– ZiTAL
Dec 21 at 9:45
You would have to loop through the array for each possible hash and check, remove. Or change the data format, use an object with the hash string as a key and the name and whatever else as a separate object for the value
– Patrick Evans
Dec 21 at 9:46
Only pure JS allowed?
– hindmost
Dec 21 at 9:49
@hindmost, yes, this is part of my react app
– Anna Kizilova
Dec 21 at 9:52
|
show 3 more comments
9
9
9
1
I have an array of hashes, like this:
[{id: "4bf58dd8d48988d110941735", name: "italy"},
{id: "4bf58dd8d48988d1c6941735", name: "skandi"},
{id: "4bf58dd8d48988d147941735", name: "diner"},
{id: "4bf58dd8d48988d110941735", name: "italy"},
{id: "4bf58dd8d48988d1c4941735", name: "resto"},
{id: "4bf58dd8d48988d14a941735", name: "vietnam"},
{id: "4bf58dd8d48988d1ce941735", name: "fish"},
{id: "4bf58dd8d48988d1c4941735", name: "resto"},
{id: "4bf58dd8d48988d1c4941735", name: "resto"}]
I want to throw out duplicate hashes. Set doesn't work because hashes are unique objects.
I feel stuck and need a kick to think. Please advise!
javascript arrays hash
edited Dec 21 at 11:07
Kamil Kiełczewski
8,86085788
asked Dec 21 at 9:40
Anna Kizilova
8819
I have an array of hashes, like this:
[{id: "4bf58dd8d48988d110941735", name: "italy"},
{id: "4bf58dd8d48988d1c6941735", name: "skandi"},
{id: "4bf58dd8d48988d147941735", name: "diner"},
{id: "4bf58dd8d48988d110941735", name: "italy"},
{id: "4bf58dd8d48988d1c4941735", name: "resto"},
{id: "4bf58dd8d48988d14a941735", name: "vietnam"},
{id: "4bf58dd8d48988d1ce941735", name: "fish"},
{id: "4bf58dd8d48988d1c4941735", name: "resto"},
{id: "4bf58dd8d48988d1c4941735", name: "resto"}]
I want to throw out duplicate hashes. Set doesn't work because hashes are unique objects.
I feel stuck and need a kick to think. Please advise!
javascript arrays hash
javascript arrays hash
edited Dec 21 at 11:07
Kamil Kiełczewski
8,86085788
asked Dec 21 at 9:40
Anna Kizilova
8819
edited Dec 21 at 11:07
Kamil Kiełczewski
8,86085788
asked Dec 21 at 9:40
Anna Kizilova
8819
edited Dec 21 at 11:07
Kamil Kiełczewski
8,86085788
edited Dec 21 at 11:07
Kamil Kiełczewski
8,86085788
edited Dec 21 at 11:07
Kamil Kiełczewski
8,86085788
8,86085788
asked Dec 21 at 9:40
Anna Kizilova
8819
asked Dec 21 at 9:40
Anna Kizilova
8819
asked Dec 21 at 9:40
Anna Kizilova
8819
8819
2
Reduce the array to an object, using the id as key, and then convert back to array usingObject.values().
– Ori Drori
Dec 21 at 9:44
What you tried?
– ZiTAL
Dec 21 at 9:45
You would have to loop through the array for each possible hash and check, remove. Or change the data format, use an object with the hash string as a key and the name and whatever else as a separate object for the value
– Patrick Evans
Dec 21 at 9:46
Only pure JS allowed?
– hindmost
Dec 21 at 9:49
@hindmost, yes, this is part of my react app
– Anna Kizilova
Dec 21 at 9:52
|
show 3 more comments
2
Reduce the array to an object, using the id as key, and then convert back to array usingObject.values().
– Ori Drori
Dec 21 at 9:44
What you tried?
– ZiTAL
Dec 21 at 9:45
You would have to loop through the array for each possible hash and check, remove. Or change the data format, use an object with the hash string as a key and the name and whatever else as a separate object for the value
– Patrick Evans
Dec 21 at 9:46
Only pure JS allowed?
– hindmost
Dec 21 at 9:49
@hindmost, yes, this is part of my react app
– Anna Kizilova
Dec 21 at 9:52
2
2
Reduce the array to an object, using the id as key, and then convert back to array using
Object.values().– Ori Drori
Dec 21 at 9:44
Reduce the array to an object, using the id as key, and then convert back to array using
Object.values().– Ori Drori
Dec 21 at 9:44
What you tried?
– ZiTAL
Dec 21 at 9:45
What you tried?
– ZiTAL
Dec 21 at 9:45
You would have to loop through the array for each possible hash and check, remove. Or change the data format, use an object with the hash string as a key and the name and whatever else as a separate object for the value
– Patrick Evans
Dec 21 at 9:46
You would have to loop through the array for each possible hash and check, remove. Or change the data format, use an object with the hash string as a key and the name and whatever else as a separate object for the value
– Patrick Evans
Dec 21 at 9:46
Only pure JS allowed?
– hindmost
Dec 21 at 9:49
Only pure JS allowed?
– hindmost
Dec 21 at 9:49
@hindmost, yes, this is part of my react app
– Anna Kizilova
Dec 21 at 9:52
@hindmost, yes, this is part of my react app
– Anna Kizilova
Dec 21 at 9:52
|
show 3 more comments
5 Answers
5
active
oldest
votes
7
You can use reduce too
//I added comma to each object
const data= [{id: "4bf58dd8d48988d110941735", name: "italy"},
{id: "4bf58dd8d48988d1c6941735", name: "skandi"},
{id: "4bf58dd8d48988d147941735", name: "diner"},
{id: "4bf58dd8d48988d110941735", name: "italy"},
{id: "4bf58dd8d48988d1c4941735", name: "resto"},
{id: "4bf58dd8d48988d14a941735", name: "vietnam"},
{id: "4bf58dd8d48988d1ce941735", name: "fish"},
{id: "4bf58dd8d48988d1c4941735", name: "resto"},
{id: "4bf58dd8d48988d1c4941735", name: "resto"}]
const result= data.reduce((current,next)=>{
if(!current.some(a=> a.name === next.name)){
current.push(next);
}
return current;
},)
console.log(result);answered Dec 21 at 10:04
Just code
8,57142966
add a comment |
4
Try this
h.filter(( t={}, a=>!(t[a.id]=++t[a.id]|0) ))
Input array in h, time complexity O(n), explanation here.
let h = [{id: "4bf58dd8d48988d110941735", name: "italy"},
{id: "4bf58dd8d48988d1c6941735", name: "skandi"},
{id: "4bf58dd8d48988d147941735", name: "diner"},
{id: "4bf58dd8d48988d110941735", name: "italy"},
{id: "4bf58dd8d48988d1c4941735", name: "resto"},
{id: "4bf58dd8d48988d14a941735", name: "vietnam"},
{id: "4bf58dd8d48988d1ce941735", name: "fish"},
{id: "4bf58dd8d48988d1c4941735", name: "resto"},
{id: "4bf58dd8d48988d1c4941735", name: "resto"}]
let t; // declare t to avoid use global (however works without it too)
let r= h.filter(( t={}, a=>!(t[a.id]=++t[a.id]|0) ))
console.log(JSON.stringify(r));answered Dec 21 at 9:45
Kamil Kiełczewski
8,86085788
add a comment |
3
Here. A little bit messy, but gets the job done.
let array = [{id: "4bf58dd8d48988d110941735", name: "italy"},
{id: "4bf58dd8d48988d1c6941735", name: "skandi"},
{id: "4bf58dd8d48988d147941735", name: "diner"},
{id: "4bf58dd8d48988d110941735", name: "italy"},
{id: "4bf58dd8d48988d1c4941735", name: "resto"},
{id: "4bf58dd8d48988d14a941735", name: "vietnam"},
{id: "4bf58dd8d48988d1ce941735", name: "fish"},
{id: "4bf58dd8d48988d1c4941735", name: "resto"},
{id: "4bf58dd8d48988d1c4941735", name: "resto"}];
let result = ;
let duplicate;
array.forEach(function(a){
duplicate = false;
result.forEach(function(r){
if (a.name === r.name){
duplicate = true;
}
});
if (!duplicate) {
result.push(a);
}
});
console.log(result);answered Dec 21 at 9:51
holydragon
1,2351618
add a comment |
3
Space for time
let arr = [
{ id: '4bf58dd8d48988d110941735', name: 'italy' },
{ id: '4bf58dd8d48988d1c6941735', name: 'skandi' },
{ id: '4bf58dd8d48988d147941735', name: 'diner' },
{ id: '4bf58dd8d48988d110941735', name: 'italy' },
{ id: '4bf58dd8d48988d1c4941735', name: 'resto' },
{ id: '4bf58dd8d48988d14a941735', name: 'vietnam' },
{ id: '4bf58dd8d48988d1ce941735', name: 'fish' },
{ id: '4bf58dd8d48988d1c4941735', name: 'resto' },
{ id: '4bf58dd8d48988d1c4941735', name: 'resto' }
]
let map = {};
let rest = arr.filter((item) => {
if(map[item.id] === void 0) {
map[item.id] = item.id;
return true;
}
});
map = null;
console.log(rest);answered Dec 21 at 10:03
zheng li
547
add a comment |
2
I would suggest an approach with associative arrays, this makes duplicate removal easier. If you can, you should build your array as an associative array in the first place, so that you don't have to convert it. Here is how you do it:
var array = [{
id: "4bf58dd8d48988d110941735",
name: "italy"
},
{
id: "4bf58dd8d48988d1c6941735",
name: "skandi"
}, {
id: "4bf58dd8d48988d147941735",
name: "diner"
}, {
id: "4bf58dd8d48988d110941735",
name: "italy"
}, {
id: "4bf58dd8d48988d1c4941735",
name: "resto"
}, {
id: "4bf58dd8d48988d14a941735",
name: "vietnam"
}, {
id: "4bf58dd8d48988d14a941735",
name: "fish"
}, {
id: "4bf58dd8d48988d1c4941735",
name: "resto"
}, {
id: "4bf58dd8d48988d1c4941735",
name: "resto"
}
];
// you can access the array with arrayAssociative[id], where the id is the real id like "4bf58dd8d48988d110941735"
var arrayAssociative = {};
for (item in array) {
// first get the unique id's
var addedNode = arrayAssociative[array[item].id] = arrayAssociative[array[item].id] || {};
if (addedNode.names == null)
addedNode.names = {};
// now get the unique names
var addedName = arrayAssociative[array[item].id].names[array[item].name] = arrayAssociative[array[item].id].names[array[item].name] || {};
}
console.log(arrayAssociative);I don't know the exact reason, why the line
var element = arrayAssociative[id] =arrayAssociative[id] || {};
works for this, but let's just accept the funcitonality as it is :)
answered Dec 21 at 9:52
Stephan T.
655620
1
arrayAssociative[array[item].id] || {}this line gives value in left of||if it is not null/undefined/0/false or value in the right if opposit - stackoverflow.com/q/2100758/860099
– Kamil Kiełczewski
Dec 21 at 10:03
thanks for the explanation <3
– Stephan T.
Dec 21 at 10:05
add a comment |
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5 Answers
5
active
oldest
votes
5 Answers
5
active
oldest
votes
active
oldest
votes
active
oldest
votes
7
You can use reduce too
//I added comma to each object
const data= [{id: "4bf58dd8d48988d110941735", name: "italy"},
{id: "4bf58dd8d48988d1c6941735", name: "skandi"},
{id: "4bf58dd8d48988d147941735", name: "diner"},
{id: "4bf58dd8d48988d110941735", name: "italy"},
{id: "4bf58dd8d48988d1c4941735", name: "resto"},
{id: "4bf58dd8d48988d14a941735", name: "vietnam"},
{id: "4bf58dd8d48988d1ce941735", name: "fish"},
{id: "4bf58dd8d48988d1c4941735", name: "resto"},
{id: "4bf58dd8d48988d1c4941735", name: "resto"}]
const result= data.reduce((current,next)=>{
if(!current.some(a=> a.name === next.name)){
current.push(next);
}
return current;
},)
console.log(result);answered Dec 21 at 10:04
Just code
8,57142966
add a comment |
7
You can use reduce too
//I added comma to each object
const data= [{id: "4bf58dd8d48988d110941735", name: "italy"},
{id: "4bf58dd8d48988d1c6941735", name: "skandi"},
{id: "4bf58dd8d48988d147941735", name: "diner"},
{id: "4bf58dd8d48988d110941735", name: "italy"},
{id: "4bf58dd8d48988d1c4941735", name: "resto"},
{id: "4bf58dd8d48988d14a941735", name: "vietnam"},
{id: "4bf58dd8d48988d1ce941735", name: "fish"},
{id: "4bf58dd8d48988d1c4941735", name: "resto"},
{id: "4bf58dd8d48988d1c4941735", name: "resto"}]
const result= data.reduce((current,next)=>{
if(!current.some(a=> a.name === next.name)){
current.push(next);
}
return current;
},)
console.log(result);answered Dec 21 at 10:04
Just code
8,57142966
add a comment |
7
7
7
You can use reduce too
//I added comma to each object
const data= [{id: "4bf58dd8d48988d110941735", name: "italy"},
{id: "4bf58dd8d48988d1c6941735", name: "skandi"},
{id: "4bf58dd8d48988d147941735", name: "diner"},
{id: "4bf58dd8d48988d110941735", name: "italy"},
{id: "4bf58dd8d48988d1c4941735", name: "resto"},
{id: "4bf58dd8d48988d14a941735", name: "vietnam"},
{id: "4bf58dd8d48988d1ce941735", name: "fish"},
{id: "4bf58dd8d48988d1c4941735", name: "resto"},
{id: "4bf58dd8d48988d1c4941735", name: "resto"}]
const result= data.reduce((current,next)=>{
if(!current.some(a=> a.name === next.name)){
current.push(next);
}
return current;
},)
console.log(result);answered Dec 21 at 10:04
Just code
8,57142966
You can use reduce too
//I added comma to each object
const data= [{id: "4bf58dd8d48988d110941735", name: "italy"},
{id: "4bf58dd8d48988d1c6941735", name: "skandi"},
{id: "4bf58dd8d48988d147941735", name: "diner"},
{id: "4bf58dd8d48988d110941735", name: "italy"},
{id: "4bf58dd8d48988d1c4941735", name: "resto"},
{id: "4bf58dd8d48988d14a941735", name: "vietnam"},
{id: "4bf58dd8d48988d1ce941735", name: "fish"},
{id: "4bf58dd8d48988d1c4941735", name: "resto"},
{id: "4bf58dd8d48988d1c4941735", name: "resto"}]
const result= data.reduce((current,next)=>{
if(!current.some(a=> a.name === next.name)){
current.push(next);
}
return current;
},)
console.log(result);//I added comma to each object
const data= [{id: "4bf58dd8d48988d110941735", name: "italy"},
{id: "4bf58dd8d48988d1c6941735", name: "skandi"},
{id: "4bf58dd8d48988d147941735", name: "diner"},
{id: "4bf58dd8d48988d110941735", name: "italy"},
{id: "4bf58dd8d48988d1c4941735", name: "resto"},
{id: "4bf58dd8d48988d14a941735", name: "vietnam"},
{id: "4bf58dd8d48988d1ce941735", name: "fish"},
{id: "4bf58dd8d48988d1c4941735", name: "resto"},
{id: "4bf58dd8d48988d1c4941735", name: "resto"}]
const result= data.reduce((current,next)=>{
if(!current.some(a=> a.name === next.name)){
current.push(next);
}
return current;
},)
console.log(result);//I added comma to each object
const data= [{id: "4bf58dd8d48988d110941735", name: "italy"},
{id: "4bf58dd8d48988d1c6941735", name: "skandi"},
{id: "4bf58dd8d48988d147941735", name: "diner"},
{id: "4bf58dd8d48988d110941735", name: "italy"},
{id: "4bf58dd8d48988d1c4941735", name: "resto"},
{id: "4bf58dd8d48988d14a941735", name: "vietnam"},
{id: "4bf58dd8d48988d1ce941735", name: "fish"},
{id: "4bf58dd8d48988d1c4941735", name: "resto"},
{id: "4bf58dd8d48988d1c4941735", name: "resto"}]
const result= data.reduce((current,next)=>{
if(!current.some(a=> a.name === next.name)){
current.push(next);
}
return current;
},)
console.log(result);answered Dec 21 at 10:04
Just code
8,57142966
edited Dec 21 at 11:50
answered Dec 21 at 10:04
Just code
8,57142966
answered Dec 21 at 10:04
Just code
8,57142966
answered Dec 21 at 10:04
Just code
8,57142966
8,57142966
add a comment |
add a comment |
4
Try this
h.filter(( t={}, a=>!(t[a.id]=++t[a.id]|0) ))
Input array in h, time complexity O(n), explanation here.
let h = [{id: "4bf58dd8d48988d110941735", name: "italy"},
{id: "4bf58dd8d48988d1c6941735", name: "skandi"},
{id: "4bf58dd8d48988d147941735", name: "diner"},
{id: "4bf58dd8d48988d110941735", name: "italy"},
{id: "4bf58dd8d48988d1c4941735", name: "resto"},
{id: "4bf58dd8d48988d14a941735", name: "vietnam"},
{id: "4bf58dd8d48988d1ce941735", name: "fish"},
{id: "4bf58dd8d48988d1c4941735", name: "resto"},
{id: "4bf58dd8d48988d1c4941735", name: "resto"}]
let t; // declare t to avoid use global (however works without it too)
let r= h.filter(( t={}, a=>!(t[a.id]=++t[a.id]|0) ))
console.log(JSON.stringify(r));answered Dec 21 at 9:45
Kamil Kiełczewski
8,86085788
add a comment |
4
Try this
h.filter(( t={}, a=>!(t[a.id]=++t[a.id]|0) ))
Input array in h, time complexity O(n), explanation here.
let h = [{id: "4bf58dd8d48988d110941735", name: "italy"},
{id: "4bf58dd8d48988d1c6941735", name: "skandi"},
{id: "4bf58dd8d48988d147941735", name: "diner"},
{id: "4bf58dd8d48988d110941735", name: "italy"},
{id: "4bf58dd8d48988d1c4941735", name: "resto"},
{id: "4bf58dd8d48988d14a941735", name: "vietnam"},
{id: "4bf58dd8d48988d1ce941735", name: "fish"},
{id: "4bf58dd8d48988d1c4941735", name: "resto"},
{id: "4bf58dd8d48988d1c4941735", name: "resto"}]
let t; // declare t to avoid use global (however works without it too)
let r= h.filter(( t={}, a=>!(t[a.id]=++t[a.id]|0) ))
console.log(JSON.stringify(r));answered Dec 21 at 9:45
Kamil Kiełczewski
8,86085788
add a comment |
4
4
4
Try this
h.filter(( t={}, a=>!(t[a.id]=++t[a.id]|0) ))
Input array in h, time complexity O(n), explanation here.
let h = [{id: "4bf58dd8d48988d110941735", name: "italy"},
{id: "4bf58dd8d48988d1c6941735", name: "skandi"},
{id: "4bf58dd8d48988d147941735", name: "diner"},
{id: "4bf58dd8d48988d110941735", name: "italy"},
{id: "4bf58dd8d48988d1c4941735", name: "resto"},
{id: "4bf58dd8d48988d14a941735", name: "vietnam"},
{id: "4bf58dd8d48988d1ce941735", name: "fish"},
{id: "4bf58dd8d48988d1c4941735", name: "resto"},
{id: "4bf58dd8d48988d1c4941735", name: "resto"}]
let t; // declare t to avoid use global (however works without it too)
let r= h.filter(( t={}, a=>!(t[a.id]=++t[a.id]|0) ))
console.log(JSON.stringify(r));answered Dec 21 at 9:45
Kamil Kiełczewski
8,86085788
Try this
h.filter(( t={}, a=>!(t[a.id]=++t[a.id]|0) ))
Input array in h, time complexity O(n), explanation here.
let h = [{id: "4bf58dd8d48988d110941735", name: "italy"},
{id: "4bf58dd8d48988d1c6941735", name: "skandi"},
{id: "4bf58dd8d48988d147941735", name: "diner"},
{id: "4bf58dd8d48988d110941735", name: "italy"},
{id: "4bf58dd8d48988d1c4941735", name: "resto"},
{id: "4bf58dd8d48988d14a941735", name: "vietnam"},
{id: "4bf58dd8d48988d1ce941735", name: "fish"},
{id: "4bf58dd8d48988d1c4941735", name: "resto"},
{id: "4bf58dd8d48988d1c4941735", name: "resto"}]
let t; // declare t to avoid use global (however works without it too)
let r= h.filter(( t={}, a=>!(t[a.id]=++t[a.id]|0) ))
console.log(JSON.stringify(r));let h = [{id: "4bf58dd8d48988d110941735", name: "italy"},
{id: "4bf58dd8d48988d1c6941735", name: "skandi"},
{id: "4bf58dd8d48988d147941735", name: "diner"},
{id: "4bf58dd8d48988d110941735", name: "italy"},
{id: "4bf58dd8d48988d1c4941735", name: "resto"},
{id: "4bf58dd8d48988d14a941735", name: "vietnam"},
{id: "4bf58dd8d48988d1ce941735", name: "fish"},
{id: "4bf58dd8d48988d1c4941735", name: "resto"},
{id: "4bf58dd8d48988d1c4941735", name: "resto"}]
let t; // declare t to avoid use global (however works without it too)
let r= h.filter(( t={}, a=>!(t[a.id]=++t[a.id]|0) ))
console.log(JSON.stringify(r));let h = [{id: "4bf58dd8d48988d110941735", name: "italy"},
{id: "4bf58dd8d48988d1c6941735", name: "skandi"},
{id: "4bf58dd8d48988d147941735", name: "diner"},
{id: "4bf58dd8d48988d110941735", name: "italy"},
{id: "4bf58dd8d48988d1c4941735", name: "resto"},
{id: "4bf58dd8d48988d14a941735", name: "vietnam"},
{id: "4bf58dd8d48988d1ce941735", name: "fish"},
{id: "4bf58dd8d48988d1c4941735", name: "resto"},
{id: "4bf58dd8d48988d1c4941735", name: "resto"}]
let t; // declare t to avoid use global (however works without it too)
let r= h.filter(( t={}, a=>!(t[a.id]=++t[a.id]|0) ))
console.log(JSON.stringify(r));answered Dec 21 at 9:45
Kamil Kiełczewski
8,86085788
edited Dec 23 at 17:49
answered Dec 21 at 9:45
Kamil Kiełczewski
8,86085788
answered Dec 21 at 9:45
Kamil Kiełczewski
8,86085788
answered Dec 21 at 9:45
Kamil Kiełczewski
8,86085788
8,86085788
add a comment |
add a comment |
3
Here. A little bit messy, but gets the job done.
let array = [{id: "4bf58dd8d48988d110941735", name: "italy"},
{id: "4bf58dd8d48988d1c6941735", name: "skandi"},
{id: "4bf58dd8d48988d147941735", name: "diner"},
{id: "4bf58dd8d48988d110941735", name: "italy"},
{id: "4bf58dd8d48988d1c4941735", name: "resto"},
{id: "4bf58dd8d48988d14a941735", name: "vietnam"},
{id: "4bf58dd8d48988d1ce941735", name: "fish"},
{id: "4bf58dd8d48988d1c4941735", name: "resto"},
{id: "4bf58dd8d48988d1c4941735", name: "resto"}];
let result = ;
let duplicate;
array.forEach(function(a){
duplicate = false;
result.forEach(function(r){
if (a.name === r.name){
duplicate = true;
}
});
if (!duplicate) {
result.push(a);
}
});
console.log(result);answered Dec 21 at 9:51
holydragon
1,2351618
add a comment |
3
Here. A little bit messy, but gets the job done.
let array = [{id: "4bf58dd8d48988d110941735", name: "italy"},
{id: "4bf58dd8d48988d1c6941735", name: "skandi"},
{id: "4bf58dd8d48988d147941735", name: "diner"},
{id: "4bf58dd8d48988d110941735", name: "italy"},
{id: "4bf58dd8d48988d1c4941735", name: "resto"},
{id: "4bf58dd8d48988d14a941735", name: "vietnam"},
{id: "4bf58dd8d48988d1ce941735", name: "fish"},
{id: "4bf58dd8d48988d1c4941735", name: "resto"},
{id: "4bf58dd8d48988d1c4941735", name: "resto"}];
let result = ;
let duplicate;
array.forEach(function(a){
duplicate = false;
result.forEach(function(r){
if (a.name === r.name){
duplicate = true;
}
});
if (!duplicate) {
result.push(a);
}
});
console.log(result);answered Dec 21 at 9:51
holydragon
1,2351618
add a comment |
3
3
3
Here. A little bit messy, but gets the job done.
let array = [{id: "4bf58dd8d48988d110941735", name: "italy"},
{id: "4bf58dd8d48988d1c6941735", name: "skandi"},
{id: "4bf58dd8d48988d147941735", name: "diner"},
{id: "4bf58dd8d48988d110941735", name: "italy"},
{id: "4bf58dd8d48988d1c4941735", name: "resto"},
{id: "4bf58dd8d48988d14a941735", name: "vietnam"},
{id: "4bf58dd8d48988d1ce941735", name: "fish"},
{id: "4bf58dd8d48988d1c4941735", name: "resto"},
{id: "4bf58dd8d48988d1c4941735", name: "resto"}];
let result = ;
let duplicate;
array.forEach(function(a){
duplicate = false;
result.forEach(function(r){
if (a.name === r.name){
duplicate = true;
}
});
if (!duplicate) {
result.push(a);
}
});
console.log(result);answered Dec 21 at 9:51
holydragon
1,2351618
Here. A little bit messy, but gets the job done.
let array = [{id: "4bf58dd8d48988d110941735", name: "italy"},
{id: "4bf58dd8d48988d1c6941735", name: "skandi"},
{id: "4bf58dd8d48988d147941735", name: "diner"},
{id: "4bf58dd8d48988d110941735", name: "italy"},
{id: "4bf58dd8d48988d1c4941735", name: "resto"},
{id: "4bf58dd8d48988d14a941735", name: "vietnam"},
{id: "4bf58dd8d48988d1ce941735", name: "fish"},
{id: "4bf58dd8d48988d1c4941735", name: "resto"},
{id: "4bf58dd8d48988d1c4941735", name: "resto"}];
let result = ;
let duplicate;
array.forEach(function(a){
duplicate = false;
result.forEach(function(r){
if (a.name === r.name){
duplicate = true;
}
});
if (!duplicate) {
result.push(a);
}
});
console.log(result);let array = [{id: "4bf58dd8d48988d110941735", name: "italy"},
{id: "4bf58dd8d48988d1c6941735", name: "skandi"},
{id: "4bf58dd8d48988d147941735", name: "diner"},
{id: "4bf58dd8d48988d110941735", name: "italy"},
{id: "4bf58dd8d48988d1c4941735", name: "resto"},
{id: "4bf58dd8d48988d14a941735", name: "vietnam"},
{id: "4bf58dd8d48988d1ce941735", name: "fish"},
{id: "4bf58dd8d48988d1c4941735", name: "resto"},
{id: "4bf58dd8d48988d1c4941735", name: "resto"}];
let result = ;
let duplicate;
array.forEach(function(a){
duplicate = false;
result.forEach(function(r){
if (a.name === r.name){
duplicate = true;
}
});
if (!duplicate) {
result.push(a);
}
});
console.log(result);let array = [{id: "4bf58dd8d48988d110941735", name: "italy"},
{id: "4bf58dd8d48988d1c6941735", name: "skandi"},
{id: "4bf58dd8d48988d147941735", name: "diner"},
{id: "4bf58dd8d48988d110941735", name: "italy"},
{id: "4bf58dd8d48988d1c4941735", name: "resto"},
{id: "4bf58dd8d48988d14a941735", name: "vietnam"},
{id: "4bf58dd8d48988d1ce941735", name: "fish"},
{id: "4bf58dd8d48988d1c4941735", name: "resto"},
{id: "4bf58dd8d48988d1c4941735", name: "resto"}];
let result = ;
let duplicate;
array.forEach(function(a){
duplicate = false;
result.forEach(function(r){
if (a.name === r.name){
duplicate = true;
}
});
if (!duplicate) {
result.push(a);
}
});
console.log(result);answered Dec 21 at 9:51
holydragon
1,2351618
answered Dec 21 at 9:51
holydragon
1,2351618
answered Dec 21 at 9:51
holydragon
1,2351618
answered Dec 21 at 9:51
holydragon
1,2351618
1,2351618
add a comment |
add a comment |
3
Space for time
let arr = [
{ id: '4bf58dd8d48988d110941735', name: 'italy' },
{ id: '4bf58dd8d48988d1c6941735', name: 'skandi' },
{ id: '4bf58dd8d48988d147941735', name: 'diner' },
{ id: '4bf58dd8d48988d110941735', name: 'italy' },
{ id: '4bf58dd8d48988d1c4941735', name: 'resto' },
{ id: '4bf58dd8d48988d14a941735', name: 'vietnam' },
{ id: '4bf58dd8d48988d1ce941735', name: 'fish' },
{ id: '4bf58dd8d48988d1c4941735', name: 'resto' },
{ id: '4bf58dd8d48988d1c4941735', name: 'resto' }
]
let map = {};
let rest = arr.filter((item) => {
if(map[item.id] === void 0) {
map[item.id] = item.id;
return true;
}
});
map = null;
console.log(rest);answered Dec 21 at 10:03
zheng li
547
add a comment |
3
Space for time
let arr = [
{ id: '4bf58dd8d48988d110941735', name: 'italy' },
{ id: '4bf58dd8d48988d1c6941735', name: 'skandi' },
{ id: '4bf58dd8d48988d147941735', name: 'diner' },
{ id: '4bf58dd8d48988d110941735', name: 'italy' },
{ id: '4bf58dd8d48988d1c4941735', name: 'resto' },
{ id: '4bf58dd8d48988d14a941735', name: 'vietnam' },
{ id: '4bf58dd8d48988d1ce941735', name: 'fish' },
{ id: '4bf58dd8d48988d1c4941735', name: 'resto' },
{ id: '4bf58dd8d48988d1c4941735', name: 'resto' }
]
let map = {};
let rest = arr.filter((item) => {
if(map[item.id] === void 0) {
map[item.id] = item.id;
return true;
}
});
map = null;
console.log(rest);answered Dec 21 at 10:03
zheng li
547
add a comment |
3
3
3
Space for time
let arr = [
{ id: '4bf58dd8d48988d110941735', name: 'italy' },
{ id: '4bf58dd8d48988d1c6941735', name: 'skandi' },
{ id: '4bf58dd8d48988d147941735', name: 'diner' },
{ id: '4bf58dd8d48988d110941735', name: 'italy' },
{ id: '4bf58dd8d48988d1c4941735', name: 'resto' },
{ id: '4bf58dd8d48988d14a941735', name: 'vietnam' },
{ id: '4bf58dd8d48988d1ce941735', name: 'fish' },
{ id: '4bf58dd8d48988d1c4941735', name: 'resto' },
{ id: '4bf58dd8d48988d1c4941735', name: 'resto' }
]
let map = {};
let rest = arr.filter((item) => {
if(map[item.id] === void 0) {
map[item.id] = item.id;
return true;
}
});
map = null;
console.log(rest);answered Dec 21 at 10:03
zheng li
547
Space for time
let arr = [
{ id: '4bf58dd8d48988d110941735', name: 'italy' },
{ id: '4bf58dd8d48988d1c6941735', name: 'skandi' },
{ id: '4bf58dd8d48988d147941735', name: 'diner' },
{ id: '4bf58dd8d48988d110941735', name: 'italy' },
{ id: '4bf58dd8d48988d1c4941735', name: 'resto' },
{ id: '4bf58dd8d48988d14a941735', name: 'vietnam' },
{ id: '4bf58dd8d48988d1ce941735', name: 'fish' },
{ id: '4bf58dd8d48988d1c4941735', name: 'resto' },
{ id: '4bf58dd8d48988d1c4941735', name: 'resto' }
]
let map = {};
let rest = arr.filter((item) => {
if(map[item.id] === void 0) {
map[item.id] = item.id;
return true;
}
});
map = null;
console.log(rest);let arr = [
{ id: '4bf58dd8d48988d110941735', name: 'italy' },
{ id: '4bf58dd8d48988d1c6941735', name: 'skandi' },
{ id: '4bf58dd8d48988d147941735', name: 'diner' },
{ id: '4bf58dd8d48988d110941735', name: 'italy' },
{ id: '4bf58dd8d48988d1c4941735', name: 'resto' },
{ id: '4bf58dd8d48988d14a941735', name: 'vietnam' },
{ id: '4bf58dd8d48988d1ce941735', name: 'fish' },
{ id: '4bf58dd8d48988d1c4941735', name: 'resto' },
{ id: '4bf58dd8d48988d1c4941735', name: 'resto' }
]
let map = {};
let rest = arr.filter((item) => {
if(map[item.id] === void 0) {
map[item.id] = item.id;
return true;
}
});
map = null;
console.log(rest);let arr = [
{ id: '4bf58dd8d48988d110941735', name: 'italy' },
{ id: '4bf58dd8d48988d1c6941735', name: 'skandi' },
{ id: '4bf58dd8d48988d147941735', name: 'diner' },
{ id: '4bf58dd8d48988d110941735', name: 'italy' },
{ id: '4bf58dd8d48988d1c4941735', name: 'resto' },
{ id: '4bf58dd8d48988d14a941735', name: 'vietnam' },
{ id: '4bf58dd8d48988d1ce941735', name: 'fish' },
{ id: '4bf58dd8d48988d1c4941735', name: 'resto' },
{ id: '4bf58dd8d48988d1c4941735', name: 'resto' }
]
let map = {};
let rest = arr.filter((item) => {
if(map[item.id] === void 0) {
map[item.id] = item.id;
return true;
}
});
map = null;
console.log(rest);answered Dec 21 at 10:03
zheng li
547
answered Dec 21 at 10:03
zheng li
547
answered Dec 21 at 10:03
zheng li
547
answered Dec 21 at 10:03
zheng li
547
547
add a comment |
add a comment |
2
I would suggest an approach with associative arrays, this makes duplicate removal easier. If you can, you should build your array as an associative array in the first place, so that you don't have to convert it. Here is how you do it:
var array = [{
id: "4bf58dd8d48988d110941735",
name: "italy"
},
{
id: "4bf58dd8d48988d1c6941735",
name: "skandi"
}, {
id: "4bf58dd8d48988d147941735",
name: "diner"
}, {
id: "4bf58dd8d48988d110941735",
name: "italy"
}, {
id: "4bf58dd8d48988d1c4941735",
name: "resto"
}, {
id: "4bf58dd8d48988d14a941735",
name: "vietnam"
}, {
id: "4bf58dd8d48988d14a941735",
name: "fish"
}, {
id: "4bf58dd8d48988d1c4941735",
name: "resto"
}, {
id: "4bf58dd8d48988d1c4941735",
name: "resto"
}
];
// you can access the array with arrayAssociative[id], where the id is the real id like "4bf58dd8d48988d110941735"
var arrayAssociative = {};
for (item in array) {
// first get the unique id's
var addedNode = arrayAssociative[array[item].id] = arrayAssociative[array[item].id] || {};
if (addedNode.names == null)
addedNode.names = {};
// now get the unique names
var addedName = arrayAssociative[array[item].id].names[array[item].name] = arrayAssociative[array[item].id].names[array[item].name] || {};
}
console.log(arrayAssociative);I don't know the exact reason, why the line
var element = arrayAssociative[id] =arrayAssociative[id] || {};
works for this, but let's just accept the funcitonality as it is :)
answered Dec 21 at 9:52
Stephan T.
655620
1
arrayAssociative[array[item].id] || {}this line gives value in left of||if it is not null/undefined/0/false or value in the right if opposit - stackoverflow.com/q/2100758/860099
– Kamil Kiełczewski
Dec 21 at 10:03
thanks for the explanation <3
– Stephan T.
Dec 21 at 10:05
add a comment |
2
I would suggest an approach with associative arrays, this makes duplicate removal easier. If you can, you should build your array as an associative array in the first place, so that you don't have to convert it. Here is how you do it:
var array = [{
id: "4bf58dd8d48988d110941735",
name: "italy"
},
{
id: "4bf58dd8d48988d1c6941735",
name: "skandi"
}, {
id: "4bf58dd8d48988d147941735",
name: "diner"
}, {
id: "4bf58dd8d48988d110941735",
name: "italy"
}, {
id: "4bf58dd8d48988d1c4941735",
name: "resto"
}, {
id: "4bf58dd8d48988d14a941735",
name: "vietnam"
}, {
id: "4bf58dd8d48988d14a941735",
name: "fish"
}, {
id: "4bf58dd8d48988d1c4941735",
name: "resto"
}, {
id: "4bf58dd8d48988d1c4941735",
name: "resto"
}
];
// you can access the array with arrayAssociative[id], where the id is the real id like "4bf58dd8d48988d110941735"
var arrayAssociative = {};
for (item in array) {
// first get the unique id's
var addedNode = arrayAssociative[array[item].id] = arrayAssociative[array[item].id] || {};
if (addedNode.names == null)
addedNode.names = {};
// now get the unique names
var addedName = arrayAssociative[array[item].id].names[array[item].name] = arrayAssociative[array[item].id].names[array[item].name] || {};
}
console.log(arrayAssociative);I don't know the exact reason, why the line
var element = arrayAssociative[id] =arrayAssociative[id] || {};
works for this, but let's just accept the funcitonality as it is :)
answered Dec 21 at 9:52
Stephan T.
655620
1
arrayAssociative[array[item].id] || {}this line gives value in left of||if it is not null/undefined/0/false or value in the right if opposit - stackoverflow.com/q/2100758/860099
– Kamil Kiełczewski
Dec 21 at 10:03
thanks for the explanation <3
– Stephan T.
Dec 21 at 10:05
add a comment |
2
2
2
I would suggest an approach with associative arrays, this makes duplicate removal easier. If you can, you should build your array as an associative array in the first place, so that you don't have to convert it. Here is how you do it:
var array = [{
id: "4bf58dd8d48988d110941735",
name: "italy"
},
{
id: "4bf58dd8d48988d1c6941735",
name: "skandi"
}, {
id: "4bf58dd8d48988d147941735",
name: "diner"
}, {
id: "4bf58dd8d48988d110941735",
name: "italy"
}, {
id: "4bf58dd8d48988d1c4941735",
name: "resto"
}, {
id: "4bf58dd8d48988d14a941735",
name: "vietnam"
}, {
id: "4bf58dd8d48988d14a941735",
name: "fish"
}, {
id: "4bf58dd8d48988d1c4941735",
name: "resto"
}, {
id: "4bf58dd8d48988d1c4941735",
name: "resto"
}
];
// you can access the array with arrayAssociative[id], where the id is the real id like "4bf58dd8d48988d110941735"
var arrayAssociative = {};
for (item in array) {
// first get the unique id's
var addedNode = arrayAssociative[array[item].id] = arrayAssociative[array[item].id] || {};
if (addedNode.names == null)
addedNode.names = {};
// now get the unique names
var addedName = arrayAssociative[array[item].id].names[array[item].name] = arrayAssociative[array[item].id].names[array[item].name] || {};
}
console.log(arrayAssociative);I don't know the exact reason, why the line
var element = arrayAssociative[id] =arrayAssociative[id] || {};
works for this, but let's just accept the funcitonality as it is :)
answered Dec 21 at 9:52
Stephan T.
655620
I would suggest an approach with associative arrays, this makes duplicate removal easier. If you can, you should build your array as an associative array in the first place, so that you don't have to convert it. Here is how you do it:
var array = [{
id: "4bf58dd8d48988d110941735",
name: "italy"
},
{
id: "4bf58dd8d48988d1c6941735",
name: "skandi"
}, {
id: "4bf58dd8d48988d147941735",
name: "diner"
}, {
id: "4bf58dd8d48988d110941735",
name: "italy"
}, {
id: "4bf58dd8d48988d1c4941735",
name: "resto"
}, {
id: "4bf58dd8d48988d14a941735",
name: "vietnam"
}, {
id: "4bf58dd8d48988d14a941735",
name: "fish"
}, {
id: "4bf58dd8d48988d1c4941735",
name: "resto"
}, {
id: "4bf58dd8d48988d1c4941735",
name: "resto"
}
];
// you can access the array with arrayAssociative[id], where the id is the real id like "4bf58dd8d48988d110941735"
var arrayAssociative = {};
for (item in array) {
// first get the unique id's
var addedNode = arrayAssociative[array[item].id] = arrayAssociative[array[item].id] || {};
if (addedNode.names == null)
addedNode.names = {};
// now get the unique names
var addedName = arrayAssociative[array[item].id].names[array[item].name] = arrayAssociative[array[item].id].names[array[item].name] || {};
}
console.log(arrayAssociative);I don't know the exact reason, why the line
var element = arrayAssociative[id] =arrayAssociative[id] || {};
works for this, but let's just accept the funcitonality as it is :)
var array = [{
id: "4bf58dd8d48988d110941735",
name: "italy"
},
{
id: "4bf58dd8d48988d1c6941735",
name: "skandi"
}, {
id: "4bf58dd8d48988d147941735",
name: "diner"
}, {
id: "4bf58dd8d48988d110941735",
name: "italy"
}, {
id: "4bf58dd8d48988d1c4941735",
name: "resto"
}, {
id: "4bf58dd8d48988d14a941735",
name: "vietnam"
}, {
id: "4bf58dd8d48988d14a941735",
name: "fish"
}, {
id: "4bf58dd8d48988d1c4941735",
name: "resto"
}, {
id: "4bf58dd8d48988d1c4941735",
name: "resto"
}
];
// you can access the array with arrayAssociative[id], where the id is the real id like "4bf58dd8d48988d110941735"
var arrayAssociative = {};
for (item in array) {
// first get the unique id's
var addedNode = arrayAssociative[array[item].id] = arrayAssociative[array[item].id] || {};
if (addedNode.names == null)
addedNode.names = {};
// now get the unique names
var addedName = arrayAssociative[array[item].id].names[array[item].name] = arrayAssociative[array[item].id].names[array[item].name] || {};
}
console.log(arrayAssociative);var array = [{
id: "4bf58dd8d48988d110941735",
name: "italy"
},
{
id: "4bf58dd8d48988d1c6941735",
name: "skandi"
}, {
id: "4bf58dd8d48988d147941735",
name: "diner"
}, {
id: "4bf58dd8d48988d110941735",
name: "italy"
}, {
id: "4bf58dd8d48988d1c4941735",
name: "resto"
}, {
id: "4bf58dd8d48988d14a941735",
name: "vietnam"
}, {
id: "4bf58dd8d48988d14a941735",
name: "fish"
}, {
id: "4bf58dd8d48988d1c4941735",
name: "resto"
}, {
id: "4bf58dd8d48988d1c4941735",
name: "resto"
}
];
// you can access the array with arrayAssociative[id], where the id is the real id like "4bf58dd8d48988d110941735"
var arrayAssociative = {};
for (item in array) {
// first get the unique id's
var addedNode = arrayAssociative[array[item].id] = arrayAssociative[array[item].id] || {};
if (addedNode.names == null)
addedNode.names = {};
// now get the unique names
var addedName = arrayAssociative[array[item].id].names[array[item].name] = arrayAssociative[array[item].id].names[array[item].name] || {};
}
console.log(arrayAssociative);answered Dec 21 at 9:52
Stephan T.
655620
edited Dec 21 at 10:00
answered Dec 21 at 9:52
Stephan T.
655620
answered Dec 21 at 9:52
Stephan T.
655620
answered Dec 21 at 9:52
Stephan T.
655620
655620
1
arrayAssociative[array[item].id] || {}this line gives value in left of||if it is not null/undefined/0/false or value in the right if opposit - stackoverflow.com/q/2100758/860099
– Kamil Kiełczewski
Dec 21 at 10:03
thanks for the explanation <3
– Stephan T.
Dec 21 at 10:05
add a comment |
1
arrayAssociative[array[item].id] || {}this line gives value in left of||if it is not null/undefined/0/false or value in the right if opposit - stackoverflow.com/q/2100758/860099
– Kamil Kiełczewski
Dec 21 at 10:03
thanks for the explanation <3
– Stephan T.
Dec 21 at 10:05
1
1
arrayAssociative[array[item].id] || {} this line gives value in left of || if it is not null/undefined/0/false or value in the right if opposit - stackoverflow.com/q/2100758/860099– Kamil Kiełczewski
Dec 21 at 10:03
arrayAssociative[array[item].id] || {} this line gives value in left of || if it is not null/undefined/0/false or value in the right if opposit - stackoverflow.com/q/2100758/860099– Kamil Kiełczewski
Dec 21 at 10:03
thanks for the explanation <3
– Stephan T.
Dec 21 at 10:05
thanks for the explanation <3
– Stephan T.
Dec 21 at 10:05
add a comment |
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2
Reduce the array to an object, using the id as key, and then convert back to array using
Object.values().– Ori Drori
Dec 21 at 9:44
What you tried?
– ZiTAL
Dec 21 at 9:45
You would have to loop through the array for each possible hash and check, remove. Or change the data format, use an object with the hash string as a key and the name and whatever else as a separate object for the value
– Patrick Evans
Dec 21 at 9:46
Only pure JS allowed?
– hindmost
Dec 21 at 9:49
@hindmost, yes, this is part of my react app
– Anna Kizilova
Dec 21 at 9:52