I know this question was answered before multiple times.
but i didn't find any solution that helped me out.
I got an array of objects with a Name property. I only want to get the objects with the same name.
How my Array looks like:
[
{
Name: 'test',
coolProperty: 'yeahCool1'
},
{
Name: 'test1',
coolProperty: 'yeahCool2'
},
{
Name: 'test2',
coolProperty: 'yeahCool3'
},
{
Name: 'test3',
coolProperty: 'yeahCool4'
},
{
Name: 'test',
coolProperty: 'yeahCool5'
}
]
so I only want to get:
[
{
Name: 'test',
coolProperty: 'yeahCool1'
},
{
Name: 'test',
coolProperty: 'yeahCool5'
}
]
I hope someone can help me out :)
For an O(N) solution, first reduce the array into an object that counts the number of occurrences of each name, and then filter the input by the occurrence count being 2:
const arr = [
{
Name: 'test',
coolProperty: 'yeahCool1'
},
{
Name: 'test1',
coolProperty: 'yeahCool2'
},
{
Name: 'test2',
coolProperty: 'yeahCool3'
},
{
Name: 'test3',
coolProperty: 'yeahCool4'
},
{
Name: 'test',
coolProperty: 'yeahCool5'
}
];
const counts = arr.reduce((a, { Name }) => {
a[Name] = (a[Name] || 0) + 1;
return a;
}, {});
console.log(arr.filter(({ Name }) => counts[Name] === 2));
You could use reduce() and filter() method to get the required result.
Using filter() method you need to check if length is greater than 2 then need it will be push in new array inside of reduce() method
DEMO
const arr =[{"Name":"test","coolProperty":"yeahCool1"},{"Name":"test1","coolProperty":"yeahCool2"},{"Name":"test2","coolProperty":"yeahCool3"},{"Name":"test3","coolProperty":"yeahCool4"},{"Name":"test","coolProperty":"yeahCool5"}];
let getCount = (name)=>{
return arr.filter(o => o.Name == name).length;
}
console.log(arr.reduce((r,item) => {
let len = getCount(item.Name);
return r.concat(len>1?item:[]);
}, []));
I see that you have already got an answer. So I thought of adding another way using map.
var counts = {};
var repeats = {};
arr.map(i => {
counts[i['Name']] = (counts[i['Name']] || []);
counts[i['Name']].push(i);
if (counts[i['Name']].length > 1) {
repeats[i['Name']] = counts[i['Name']];
}
});
console.log(repeats);
Not the best solution considering the performance. Just wanted to add an alternative method.
Hope it helps!!
Related
I have this array in my next.js app
arr1
[
{
identifier: "60a17722225f2918c445fd19",
name: "Ben Awad",
_id: "60c94480b8d43c28d0a6eb73
},
{
identifier: "60a455d11fa62a1510b408f8",
name: "dev ed"
_id: "60bf62cede309f1a30fe88ab"
}
]
And i have this another big array
arr2
[
{
name: "Ben Awad",
_id: "60a17722225f2918c445fd19
},
{
name: "dev ed",
_id: "60a455d11fa62a1510b408f8"
},
{
name: "Katlyn",
_id: "60a52500ce96f30c14fdaff9"
},
{
name: "Mike",
_id: "60c95deeb8d43c28d0a6eb74"
},
{
name: "Kassandra",
_id: "60c960ddb8d43c28d0a6eb7a"
}
]
I want a new array who should have all users except for those who have similar ids with arr1
So this is the logic i did (Notice that arr1 and arr2 will change constantly)
Me = arr1
AllUsers = arr2
const LookFriends =
Me &&
AllUsers.filter(({ _id }) => {
return Me.friends.indexOf(_id) === -1;
});
console.log(LookFriends);
The output should be Katlyn, Mike and Kassandra, but the console.log says...
[
{
name: "Ben Awad",
_id: "60a17722225f2918c445fd19
},
{
name: "dev ed",
_id: "60a455d11fa62a1510b408f8"
},
{
name: "Katlyn",
_id: "60a52500ce96f30c14fdaff9"
},
{
name: "Mike",
_id: "60c95deeb8d43c28d0a6eb74"
},
{
name: "Kassandra",
_id: "60c960ddb8d43c28d0a6eb7a"
}
]
I'm really having a hard time trying to filter an array based on another array, what can i do ?
You need to use 'findIndex' in this case, then compare the _id field itself:
const LookFriends =
Me &&
AllUsers.filter(({ _id }) => {
return Me.friends.findIndex(friend => friend._id === _id) === -1;
});
If you use "indexOf", it will compare the entire object to just that _id value.
You can extract the identifiers from the first array and filter the second array if the element doesn't match any id in our object or our Set
// with an object
const ids = {}
arr1.forEach(user => {
ids[user.identifier] = true
})
const filtered = arr2.filter(user => !ids[user._id])
// or with a Set
const ids = new Set(arr1.map(user => user.identifier))
const filtered = arr2.filter(user => !ids.has(user._id))
Please have a try with this code.
Me &&
AllUsers.filter(({ _id }) => {
let bExist = false
Me.friends.map( (friend) => {
if ( friend._id === _id )
bExist = true
})
return bExist
});
Let me know if it works or not.
I have an array of duplicated objects in Javascript. I want to create an array of unique objects by adding the index of occurrence of the individual value.
This is my initial data:
const array= [
{name:"A"},
{name:"A"},
{name:"A"},
{name:"B"},
{name:"B"},
{name:"C"},
{name:"C"},
];
This is expected end result:
const array= [
{name:"A-0"},
{name:"A-1"},
{name:"A-2"},
{name:"B-0"},
{name:"B-1"},
{name:"C-0"},
{name:"C-1"},
];
I feel like this should be fairly simple, but got stuck on it for a while. Can you please advise how I'd go about this? Also if possible, I need it efficient as the array can hold up to 1000 items.
EDIT: This is my solution, but I don't feel like it's very efficient.
const array = [
{ name: "A" },
{ name: "A" },
{ name: "C" },
{ name: "B" },
{ name: "A" },
{ name: "C" },
{ name: "B" },
];
const sortedArray = _.sortBy(array, 'name');
let previousItem = {
name: '',
counter: 0
};
const indexedArray = sortedArray.map((item) => {
if (item.name === previousItem.name) {
previousItem.counter += 1;
const name = `${item.name}-${previousItem.counter}`;
return { name };
} else {
previousItem = { name: item.name, counter: 0};
return item;
}
});
Currently you are sorting it first then looping over it, which may be not the most efficient solution.
I would suggest you to map over it with a helping object.
const a = [{name:"A"},{name:"A"},{name:"A"},{name:"B"},{name:"B"},{name:"C"},{name:"C"},], o = {};
const r = a.map(({ name }) => {
typeof o[name] === 'number' ? o[name]++ : o[name] = 0;
return { name: `${name}-${o[name]}` };
});
console.log(r);
Keep a counter, and if the current name changes, reset the counter.
This version mutates the objects. Not sure if you want a copy or not. You could potentially sort the array by object name first to ensure they are in order (if that's not already an existing precondition.)
const array = [
{ name: "A" },
{ name: "A" },
{ name: "A" },
{ name: "B" },
{ name: "B" },
{ name: "C" },
{ name: "C" },
];
let name, index;
for (let i in array) {
index = array[i].name == name ? index + 1 : 0;
name = array[i].name;
array[i].name += `-${index}`;
}
console.log(array);
Another way, if you don't want to sort, and don't want to mutate any objects, is to use a map and keep track of the current index for each object.
const array = [
// NOTE: I put the items in mixed up order.
{ name: "A" },
{ name: "C" },
{ name: "A" },
{ name: "B" },
{ name: "A" },
{ name: "C" },
{ name: "B" },
];
let index = {};
let next = name => index[name] = index[name] + 1 || 0;
let result = array.map(obj => ({ ...obj, name: obj.name + '-' + next(obj.name) }));
console.log(result);
I am working on lodash ...I want to find particular array if matching id mind in that particular array...This is my collection:-
[ { name: 'Thor',
id:
[ '5a1676da509a93571c15501f',
'59ffb40f62d346204e09c9ad',
'5a0a92e01bb28a276abfa548' ] },
{ name: 'Loki',
id:
[ '59ffb40f62d346204e09c9ad',
'59ffb41b62d346204e0a03ed',
'59ffb40e62d346204e09c298',
'5a65853af431924e73c401fe' ] } ]
What I am trying to find i have given id if it should find it above collection then it should return that collection...for example..if i have id:59ffb40e62d346204e09c298,then it should return 2 collection.....
b = _.find(upperArray,(candidate) => {
// console.log(candidate)
return _.find(candidate.id,id)//here actor_id is given id
})
but it does not return 2 collection.If it does not found any matching element it should return blank array or if found it should return collection of array.Where I am doing wrong??
#Barmar is correct you should use _.filter. Try the following
var upperArray = [
{
name: 'Thor',
id: [
'5a1676da509a93571c15501f',
'59ffb40f62d346204e09c9ad',
'5a0a92e01bb28a276abfa548'
]
},
{
name: 'Loki',
id: [
'59ffb40f62d346204e09c9ad',
'59ffb41b62d346204e0a03ed',
'59ffb40e62d346204e09c298',
'5a65853af431924e73c401fe'
]
}
];
function findById(idToFind) {
return _.filter(upperArray,function(candidate) {
return _.includes(candidate.id, idToFind);
});
}
var b = findById('59ffb40e62d346204e09c298');
console.log(b);
<script src="https://cdnjs.cloudflare.com/ajax/libs/lodash.js/4.17.4/lodash.js"></script>
Using can achieve the same using native array methods
var id = "59ffb40e62d346204e09c298";
var arr = [{
name: 'Thor',
id: ['5a1676da509a93571c15501f',
'59ffb40f62d346204e09c9ad',
'5a0a92e01bb28a276abfa548'
]
},
{
name: 'Loki',
id: ['59ffb40f62d346204e09c9ad',
'59ffb41b62d346204e0a03ed',
'59ffb40e62d346204e09c298',
'5a65853af431924e73c401fe'
]
}
]
function getMatchedCollection(id) {
// filter will return a new array of matched condition
return arr.filter(function(item) {
// indexOf is used to check if an element is present
if (item.id.indexOf(id) !== -1) {
return item
}
})
}
console.log(getMatchedCollection(id))
See jsfiddle here: https://jsfiddle.net/remenyLx/2/
I have data that contains objects that each have an array of images. I want only the first image of each object.
var data1 = [
{
id: 1,
images: [
{ name: '1a' },
{ name: '1b' }
]
},
{
id: 2,
images: [
{ name: '2a' },
{ name: '2b' }
]
},
{
id: 3
},
{
id: 4,
images: []
}
];
var filtered = [];
var b = data1.forEach((element, index, array) => {
if(element.images && element.images.length)
filtered.push(element.images[0].name);
});
console.log(filtered);
The output needs to be flat:
['1a', '2a']
How can I make this prettier?
I'm not too familiar with JS map, reduce and filter and I think those would make my code more sensible; the forEach feels unnecessary.
First you can filter out elements without proper images property and then map it to new array:
const filtered = data1
.filter(e => e.images && e.images.length)
.map(e => e.images[0].name)
To do this in one loop you can use reduce function:
const filtered = data1.reduce((r, e) => {
if (e.images && e.images.length) {
r.push(e.images[0].name)
}
return r
}, [])
You can use reduce() to return this result.
var data1 = [{
id: 1,
images: [{
name: '1a'
}, {
name: '1b'
}]
}, {
id: 2,
images: [{
name: '2a'
}, {
name: '2b'
}]
}, {
id: 3
}, {
id: 4,
images: []
}];
var result = data1.reduce(function(r, e) {
if (e.hasOwnProperty('images') && e.images.length) r.push(e.images[0].name);
return r;
}, [])
console.log(result);
All answers are creating NEW arrays before projecting the final result : (filter and map creates a new array each) so basically it's creating twice.
Another approach is only to yield expected values :
Using iterator functions
function* foo(g)
{
for (let i = 0; i < g.length; i++)
{
if (g[i]['images'] && g[i]["images"].length)
yield g[i]['images'][0]["name"];
}
}
var iterator = foo(data1) ;
var result = iterator.next();
while (!result.done)
{
console.log(result.value)
result = iterator.next();
}
This will not create any additional array and only return the expected values !
However if you must return an array , rather than to do something with the actual values , then use other solutions suggested here.
https://jsfiddle.net/remenyLx/7/
I have an array of objects and I want to get a new array from it that is unique based only on a single property, is there a simple way to achieve this?
Eg.
[ { id: 1, name: 'bob' }, { id: 1, name: 'bill' }, { id: 1, name: 'bill' } ]
Would result in 2 objects with name = bill removed once.
Use the uniq function
var destArray = _.uniq(sourceArray, function(x){
return x.name;
});
or single-line version
var destArray = _.uniq(sourceArray, x => x.name);
From the docs:
Produces a duplicate-free version of the array, using === to test object equality. If you know in advance that the array is sorted, passing true for isSorted will run a much faster algorithm. If you want to compute unique items based on a transformation, pass an iterator function.
In the above example, the function uses the objects name in order to determine uniqueness.
If you prefer to do things yourself without Lodash, and without getting verbose, try this uniq filter with optional uniq by property:
const uniqFilterAccordingToProp = function (prop) {
if (prop)
return (ele, i, arr) => arr.map(ele => ele[prop]).indexOf(ele[prop]) === i
else
return (ele, i, arr) => arr.indexOf(ele) === i
}
Then, use it like this:
const obj = [ { id: 1, name: 'bob' }, { id: 1, name: 'bill' }, { id: 1, name: 'bill' } ]
obj.filter(uniqFilterAccordingToProp('abc'))
Or for plain arrays, just omit the parameter, while remembering to invoke:
[1,1,2].filter(uniqFilterAccordingToProp())
If you want to check all the properties then
lodash 4 comes with _.uniqWith(sourceArray, _.isEqual)
A better and quick approach
var table = [
{
a:1,
b:2
},
{
a:2,
b:3
},
{
a:1,
b:4
}
];
let result = [...new Set(table.map(item => item.a))];
document.write(JSON.stringify(result));
Found here
You can use the _.uniqBy function
var array = [ { id: 1, name: 'bob' }, { id: 2, name: 'bill' }, { id: 1, name: 'bill' },{ id: 2, name: 'bill' } ];
var filteredArray = _.uniqBy(array,function(x){ return x.id && x.name;});
console.log(filteredArray)
<script src="https://cdnjs.cloudflare.com/ajax/libs/lodash.js/4.17.5/lodash.js"></script>
In the above example, filtering is based on the uniqueness of combination of properties id & name.
if you have multiple properties for an object.
then to find unique array of objects based on specific properties, you could follow this method of combining properties inside _.uniqBy() method.
I was looking for a solution which didn't require a library, and put this together, so I thought I'd add it here. It may not be ideal, or working in all situations, but it's doing what I require, so could potentially help someone else:
const uniqueBy = (items, reducer, dupeCheck = [], currentResults = []) => {
if (!items || items.length === 0) return currentResults;
const thisValue = reducer(items[0]);
const resultsToPass = dupeCheck.indexOf(thisValue) === -1 ?
[...currentResults, items[0]] : currentResults;
return uniqueBy(
items.slice(1),
reducer,
[...dupeCheck, thisValue],
resultsToPass,
);
}
const testData = [
{text: 'hello', image: 'yes'},
{text: 'he'},
{text: 'hello'},
{text: 'hell'},
{text: 'hello'},
{text: 'hellop'},
];
const results = uniqueBy(
testData,
item => {
return item.text
},
)
console.dir(results)
In case you need pure JavaScript solution:
var uniqueProperties = {};
var notUniqueArray = [ { id: 1, name: 'bob' }, { id: 1, name: 'bill' }, { id: 1, name: 'bill' } ];
for(var object in notUniqueArray){
uniqueProperties[notUniqueArray[object]['name']] = notUniqueArray[object]['id'];
}
var uniqiueArray = [];
for(var uniqueName in uniqueProperties){
uniqiueArray.push(
{id:uniqueProperties[uniqueName],name:uniqueName});
}
//uniqiueArray
unique array by id property with ES6:
arr.filter((a, i) => arr.findIndex(b => b.id === a.id) === i); // unique by id
replace b.id === a.id with the relevant comparison for your case