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How to sort an array of objects by multiple fields?
(38 answers)
Closed 1 year ago.
I want to sort Below array based on name and is_closed.
this.rawDataListDup = [
{id: 1, name: 'john','is_closed':true},
{id: 2, name: 'james','is_closed':true},
{id: 3, name: 'jane','is_closed':false},
{id: 4, name: 'alex','is_closed':false},
{id: 5, name: 'david','is_closed':true},
];
As of now i can only sort using any one of the attribute using below code.
let colName = 'name'
this.rawDataListDup.sort((b, a) => a[colName] < b[colName] ? 1 : a[colName] > b[colName] ? -1 : 0)
I want array objects with is_closed = false on top of array and also it should be in Alphabetical order. like this
this.rawDataListDup = [
{id: 4, name: 'alex','is_closed':false},
{id: 3, name: 'jane','is_closed':false},
{id: 5, name: 'david','is_closed':true},
{id: 2, name: 'james','is_closed':true},
{id: 1, name: 'john','is_closed':true},
];
how to do this?
You can do this with sort with multiple conditions inside sort:
const rawDataListDup = [
{id: 1, name: 'john','is_closed':true},
{id: 2, name: 'james','is_closed':true},
{id: 3, name: 'jane','is_closed':false},
{id: 4, name: 'alex','is_closed':false},
{id: 5, name: 'david','is_closed':true},
];
const sortedList = rawDataListDup.sort((a,b)=>(a.is_closed-b.is_closed) || a.name.localeCompare(b.name));
console.log(sortedList);
It just write simple code.
const rawDataListDup = [
{id: 1, name: 'john','is_closed':true},
{id: 2, name: 'james','is_closed':true},
{id: 3, name: 'jane','is_closed':false},
{id: 4, name: 'alex','is_closed':false},
{id: 5, name: 'david','is_closed':true},
];
rawDataListDup.sort(function(a, b) {
return a.is_closed - b.is_closed || a.name - b.name;
});
//OR
var obj = rawDataListDup.sort((a, b) => a.is_closed - b.is_closed || a.name - b.name);
You could also use lodash (sortBy).
Example:
_.sortBy(rawDataListDup, ['is_closed', 'name']);
I have a json array like this:
(3) [{…}, {…}, {…}]
0: {Id: 1, Name: "bask"}
1: {Id: 2, Name: "voll"}
2: {Id: 3, Name: "badminton"}
I want to turn it into something like this:
{1:"bask",2:"voll",3:"badminton"}
You can use reduce to loop through array and build a object of desired key/value pair
let data = [{Id: 1, Name: "bask"},{Id: 2, Name: "voll"},{Id: 3, Name: "badminton"}]
let output = data.reduce((op, {Id, Name}) => {
op[Id] = Name
return op
},{})
console.log(output)
You could take Object.fromEntries with the maped key/value pairs.
var array = [{ Id: 1, Name: "bask" }, { Id: 2, Name: "voll" }, { Id: 3, Name: "badminton" }],
object = Object.fromEntries(array.map(({ Id, Name }) => [Id, Name]));
console.log(object);
You can check out the reduce() function!
let array = [
{Id: 1, Name: "bask"},
{Id: 2, Name: "voll"},
{Id: 3, Name: "badminton"}
];
console.log(_.reduce(array, function(result, obj){
result[obj.Id] = obj.Name;
return result;
}, {}));
You can checkout lodash an awesome library with many other such utilities!
You can do this with reduce():
var a = [
{Id: 1, Name: "bask"},
{Id: 2, Name: "voll"},
{Id: 3, Name: "badminton"}
]
b = a.reduce((acc, item) => {
acc[item.Id] = item.Name;
return acc;
}
console.log(b);
You can do it in different ways, here one of them.
let dataArray = [
{id: 1, name: 'bask'},
{id: 2, name: 'voll'},
{id: 3, name: 'badminton'}
]
let ouputObject = {}
dataArray.map(data => {
ouputObject[`${data.id}`] = data.name
})
console.log(ouputObject)
outputObject will be
Object {
1: "bask",
2: "voll",
3: "badminton"
}
Using Array.reduce() :
var arr = [{
Id: 1,
Name: "bask"
}, {
Id: 2,
Name: "voll"
}, {
Id: 3,
Name: "badminton"
}];
var reduceObj = arr.reduce(function(result, currentElement) {
result[currentElement.Id] = currentElement.Name;
return result;
}, {});
console.log(reduceObj);
Using Array.map() :
var arr = [{
Id: 1,
Name: "bask"
}, {
Id: 2,
Name: "voll"
}, {
Id: 3,
Name: "badminton"
}];
var mapObject = {}
arr.map(obj => {
mapObject[obj.Id] = obj.Name
})
console.log(mapObject);
I'm trying to convert my data from API to my needs. Would like to create a nested array from plain array. I would like to group elements by parentId property, if parentId would not exist I would put it as a root. id value is unique. Like so (raw data):
[
{id: 1, name: 'sensor'},
{id: 2, name: 'sensor', parent: 1},
{id: 3, name: 'sensor', parent: 1},
{id: 4, name: 'sensor', parent: 3},
{id: 5, name: 'sensor'},
{id: 6, name: 'sensor', parent: 5}
]
Converted Data:
const results = [
{
id: 1,
name: "sensor",
children: [
{ id: 2, name: "sensor", parent: 1 },
{
id: 3,
name: "sensor",
parent: 1,
children: [{ id: 4, name: "sensor", parent: 3 }]
}
]
},
{ id: 5, name: "sensor", children: [{ id: 6, name: "sensor", parent: 5 }] }
];
I found this recursive method but it assumes that the parent property exist for every element in an array. In my example root level element would not have parent property.
function getNestedChildren(arr, parent) {
var out = []
for(var i in arr) {
if(arr[i].parent == parent) {
var children = getNestedChildren(arr, arr[i].id)
if(children.length) {
arr[i].children = children
}
out.push(arr[i])
}
}
return out
}
You could take an approach which uses both relations, one from children to parent and vice versa. At the end take the children of the root node.
This approach works for unsorted data.
var data = [{ id: 1, name: 'sensor' }, { id: 2, name: 'sensor', parent: 1 }, { id: 3, name: 'sensor', parent: 1 }, { id: 4, name: 'sensor', parent: 3 }, { id: 5, name: 'sensor' }, { id: 6, name: 'sensor', parent: 5 }],
tree = function (data, root) {
var t = {};
data.forEach(o => {
Object.assign(t[o.id] = t[o.id] || {}, o);
t[o.parent] = t[o.parent] || {};
t[o.parent].children = t[o.parent].children || [];
t[o.parent].children.push(t[o.id]);
});
return t[root].children;
}(data, undefined);
console.log(tree);
.as-console-wrapper { max-height: 100% !important; top: 0; }
Given the limited amount of information (will update if more info is added).
The algorithm would be, given an array of data entries, check if entry has a parent and if that parent exists, in which case we want to add the entry to the array of children of the parent entry otherwise add the entry as a parent.
var dataFromAPI = [
{id: 1, name: 'sensor'},
{id: 2, name: 'sensor', parent: 1},
{id: 3, name: 'sensor', parent: 1},
{id: 4, name: 'sensor', parent: 3},
{id: 5, name: 'sensor'},
{id: 6, name: 'sensor', parent: 5}
];
var transformedData = { };
dataFromAPI.forEach(function(entry){
if(entry.parent !== undefined && entry.parent in transformedData) {
transformedData[entry.parent].children.push(entry);
} else {
entry["children"] = [];
transformedData[entry.id] = entry;
}
});
console.log(transformedData);
Please note:
there are a couple assumptions made within this algorithm/code. It assumes that all parent entries exist before their child entry. It also only accounts for two levels (parent or child), meaning a child cannot act as the parent (otherwise you'd have to store the children as an object and not an array)
use a for loop to go through each item.
check if parent property exists (or has value).
If not its a child item. Attach it to appropriate parent.
to check if property exists:
var myProp = 'prop';
if (myObj.hasOwnProperty(myProp)) {
alert("yes, i have that property");
}
try
let h={}, r=[]; // result in r
d.forEach(x=> (h[x.id]=x, x.children=[]) );
d.forEach(x=> x.parent ? h[x.parent].children.push(x) : r.push(x) );
let d = [
{id: 1, name: 'sensor'},
{id: 2, name: 'sensor', parent: 1},
{id: 3, name: 'sensor', parent: 1},
{id: 4, name: 'sensor', parent: 3},
{id: 5, name: 'sensor'},
{id: 6, name: 'sensor', parent: 5}
];
let h = {},r = []; // result in r
d.forEach(x => (h[x.id] = x, x.children = []));
d.forEach(x => x.parent ? h[x.parent].children.push(x) : r.push(x));
console.log(r);
If you want that parent element should not have a parent , than you can manually check and remove fields of an object in an array that has null parent. than you can make a tree... here is an example...
const arr2 = [
{id: 1, name: 'gender', parent: null, parent_id: null },
{id: 2, name: 'material', parent: null, parent_id: null },
{id: 3, name: 'male', parent: 1, parent_name: "gender" },
{ id: 5, name: 'female', parent: 1, parent_name: "gender" },
{ id: 4, name: 'shoe', parent: 3, parent_id: "male"},
]
let newarr=[];
for(let i=0 ; i< arr2.length; i++ ){
if(arr2[i].id){
if(newarr[i] != {} ){
newarr[i] = {}
}
newarr[i].id = arr2[i].id
}
if( arr2[i].name ){
newarr[i].name = arr2[i].name
}
if( arr2[i].parent ){
newarr[i].parent = arr2[i].parent
}
if( arr2[i].parent_id ){
newarr[i].parent_id = arr2[i].parent_id
}
}
console.log('newarr', newarr );
let tree = function (data, root) {
var obj = {};
data.forEach(i => {
Object.assign(obj[i.id] = obj[i.id] || {}, i);
obj[i.parent] = obj[i.parent] || {};
obj[i.parent].children = obj[i.parent].children || [];
obj[i.parent].children.push(obj[i.id]);
});
return obj[root].children;
}(newarr, undefined);
console.log('tree ', tree);
I want to return matching proprieties of two arrays of objects. But I got undefined from map function.
let fruits1 = [
{id: 1, name: "apple"},
{id: 2, name: "dragon fruit"},
{id: 3, name: "banana"},
{id: 4, name: "kiwi"},
{id: 5, name: "pineapple"},
{id: 6, name: "watermelon"},
{id: 7, name: "pear"},
]
let fruits2 = [
{id: 7, name: "pear"},
{id: 10, name: "avocado"},
{id: 5, name: "pineapple"},
]
fruits1.forEach((fruit1) => {
fruits2.filter((fruit2) => {
return fruit1.name === fruit2.name;
}).map((newFruit) => {
//console.log(newFruit.name);
return newFruit.name;
})
})
What are you looking for is an array intersection:
// Generic helper function that can be used for the three operations:
const operation = (list1, list2, isUnion = false) =>
list1.filter( a => isUnion === list2.some( b => a.name === b.name ) );
// Following functions are to be used:
const inBoth = (list1, list2) => operation(list1, list2, true),
inFirstOnly = operation,
inSecondOnly = (list1, list2) => inFirstOnly(list2, list1);
Usage:
console.log('inBoth:', inBoth(list1, list2));
Working Example:
// Generic helper function that can be used for the three operations:
const operation = (list1, list2, isUnion = false) =>
list1.filter( a => isUnion === list2.some( b => a.name === b.name ) );
// Following functions are to be used:
const inBoth = (list1, list2) => operation(list1, list2, true),
inFirstOnly = operation,
inSecondOnly = (list1, list2) => inFirstOnly(list2, list1);
let fruits1 = [
{id: 1, name: "apple"},
{id: 2, name: "dragon fruit"},
{id: 3, name: "banana"},
{id: 4, name: "kiwi"},
{id: 5, name: "pineapple"},
{id: 6, name: "watermelon"},
{id: 7, name: "pear"},
]
let fruits2 = [
{id: 7, name: "pear"},
{id: 10, name: "avocado"},
{id: 5, name: "pineapple"},
]
console.log('inBoth:', inBoth(fruits1, fruits2));
You could use a Set and filter the names.
const names = ({ name }) => name;
var fruits1 = [{ id: 1, name: "apple" }, { id: 2, name: "dragon fruit" }, { id: 3, name: "banana" }, { id: 4, name: "kiwi" }, { id: 5, name: "pineapple" }, { id: 6, name: "watermelon" }, { id: 7, name: "pear" }],
fruits2 = [{ id: 7, name: "pear" }, { id: 10, name: "avocado" }, { id: 5, name: "pineapple" }],
common = fruits1
.map(names)
.filter(Set.prototype.has, new Set(fruits2.map(names)));
console.log(common);
What you want to do is this:
/* first we filter fruits1 (arbitrary) */
let matchingFruits = fruits1.filter(f1 => {
/* then we filter the frut if it exists in frtuis2 */
return fruits2.find(f2 => f2.name === f1.name)
}).map(fruit => fruit.name) // and now we map if we only want the name strings
If you're not using a polyfill Array.find will not work in IE. The alternative would be using Array.indexOf (thanks for pointing this out #JakobE).
Be aware that Array.forEach return value is undefined and that, in order to actually use the Array.map correctly, one has to consume the returned value somehow or assign it to a variable, as we just did with matchingFruits.
I am trying to merge data from json that comes in array of objects. I was using the underscore solution from here merge two json object based on key value in javascript, but it turns out it doesnt override existing items which I need to do as well now.
The result should be all items of array 1 in the same order, overriden by array 2 where id = id. Items in array 2 that does not exist in array 1 should be pushed to the end of the result.
First array:
[
{id: 8, category: "A"}
{id: 2, category: "D"}
{id: 5, category: "C"}
{id: 9, category: "B"}
]
Second array:
[
{id: 1, category: "X"}
{id: 2, category: "Y"}
]
Expected result:
[
{id: 8, category: "A"}
{id: 2, category: "Y"}
{id: 5, category: "C"}
{id: 9, category: "B"}
{id: 1, category: "X"}
]
Use filter, find and concat
Given that
var arr1 = [
{id: 8, category: "A"},
{id: 2, category: "D"},
{id: 5, category: "C"},
{id: 9, category: "B"}
];
var arr2 = [
{id: 12, category: "X"},
{id: 2, category: "Y"}
];
If the order is not important
var output = arr2.concat(
arr1.filter( s =>
!arr2.find( t => t.id == s.id )
)//end filter
);//end concat
Demo
var arr1 = [{
id: 8,
category: "A"
},
{
id: 2,
category: "D"
},
{
id: 5,
category: "C"
},
{
id: 9,
category: "B"
}
];
var arr2 = [{
id: 12,
category: "X"
},
{
id: 2,
category: "Y"
}
];
var output = arr2.concat(
arr1.filter(s =>
!arr2.find(t => t.id == s.id)
) //end filter
); //end concat
console.log(output);
If the order is important
var output = arr1.map(
s => arr2.find(
t => t.id == s.id ) || s
).concat( //end map of arr1
arr2.filter(
s => !arr1.find( t => t.id == s.id )
) //end filter
);//end concat
Demo
var arr1 = [{
id: 8,
category: "A"
},
{
id: 2,
category: "D"
},
{
id: 5,
category: "C"
},
{
id: 9,
category: "B"
}
];
var arr2 = [{
id: 12,
category: "X"
},
{
id: 2,
category: "Y"
}
];
var output = arr1.map(
s => arr2.find(
t => t.id == s.id) || s
).concat( //end map of arr1
arr2.filter(
s => !arr1.find(t => t.id == s.id)
) //end filter
); //end concat
console.log(output);
You could use a Map as closure and store the index of the result array for this id.
var first = [{ id: 8, category: "A" }, { id: 2, category: "D" }, { id: 5, category: "C" }, { id: 9, category: "B" }],
second = [{ id: 12, category: "X" }, { id: 2, category: "Y" }],
result = [first, second].reduce((m => (r, a) => {
a.forEach(o => {
if (m.has(o.id)) {
r[m.get(o.id)] = o;
return;
}
m.set(o.id, r.push(o) - 1);
});
return r;
})(new Map), []);
console.log(result);
You can set a loop on your secondArray and check each object with id value against object with id of firstArray. If you find a match then simply replace the object else push the object:
var firstArray = [
{id: 8, category: "A"},
{id: 2, category: "D"},
{id: 5, category: "C"},
{id: 9, category: "B"}
];
var secondArray = [
{id: 12, category: "X"},
{id: 2, category: "Y"}
];
secondArray.forEach((obj)=>{
var match = false;
for(var i=0; i<firstArray.length; i++){
if(firstArray[i].id === obj.id){
match = true;
firstArray[i] = obj;
break;
}
}
if(!match){
firstArray.push(obj);
}
});
console.log(firstArray);
You can use a forEach to iterate through second array. For each object with the same id in the first array, update the category otherwise push in the new array.
const first = [{id: 8, category: "A"},{id: 2, category: "D"},{id: 5, category: "C"},{id: 9, category: "B"}],
second = [{id: 12, category: "X"},{id: 2, category: "Y"}],
merged = [...first];
second.forEach(o => {
let obj = first.find(({id,category}) => id === o.id);
obj ? obj.category = o.category : merged.push({...o});
});
console.log(merged);
I think reduce is better
first.reduce((res, item) => res.filter(i => i.id !== item.id).concat(item), second);
Using underscore I managed to come up with this answer my own question. It is probably not the most efficent
const newarr = _.map(arr1, obj1 => {
const r = _.find(arr2, obj2 => {
return obj1[match] === obj2[match]
})
if (typeof r === 'undefined') {
return obj1
} else {
return r
}
})
_.each(arr2, obj => {
if (_.indexOf(arr1, _.findWhere(arr1, {id: obj.id})) === -1) {
newarr.push(obj)
}
})
This should work:
const newArr = second.reduce((res, item) => res.filter(i => i.id !== item.id).concat(item), first);