I have an array of objects, each with an 'id' and a 'name'. I'm retrieving an 'id' from the server and need to reorder the array starting from this id.
Example code:
var myList = [
{
id: 0,
name: 'Joe'
},
{
id: 1,
name: 'Sally'
},
{
id: 2,
name: 'Chris'
},
{
id: 3,
name: 'Tiffany'
},
{
id: 4,
name: 'Kerry'
}
];
Given an 'id' of 2, how can I reorder the array so my output is as follows:
var newList = [
{
id: 2,
name: 'Chris'
},
{
id: 3,
name: 'Tiffany'
},
{
id: 4,
name: 'Kerry'
},
{
id: 0,
name: 'Joe'
},
{
id: 1,
name: 'Sally'
}
];
Try this:
function orderList(list, id){
return list.slice(id).concat(list.slice(0,id));
}
Link to demo
You could slice the array at given index and return a new array using spread syntax.
const myList = [{id:0,name:'Joe'},{id:1,name:'Sally'},{id:2,name:'Chris'},{id:3,name:'Tiffany'},{id:4,name:'Kerry'}];
const slice = (arr, num) => [...arr.slice(num), ...arr.slice(0, num)];
console.log(slice(myList, 2));
myList.sort(function(a,b){
return a.id>2===b.id>2?a.id-b.id:b.id-a.id;
});
newList=myList;
http://jsbin.com/kenobunali/edit?console
You could splice the wanted part and use splice to insert it at the end of the array.
var myList = [{ id: 0, name: 'Joe' }, { id: 1, name: 'Sally' }, { id: 2, name: 'Chris' }, { id: 3, name: 'Tiffany' }, { id: 4, name: 'Kerry' }],
id = 2;
myList.splice(myList.length, 0, myList.splice(0, myList.findIndex(o => o.id === id)));
console.log(myList);
using es6 spread syntax
var myList = [{ id: 0, name: 'Joe' }, { id: 1, name: 'Sally' }, { id: 2, name: 'Chris' }, { id: 3, name: 'Tiffany' }, { id: 4, name: 'Kerry' }],
id = 2;
var index = myList.findIndex(o => o.id == id);
var arr = myList.splice(0, index);
var result = [...myList, ...arr];
console.log(result);
Related
I am having two arrays
const selected = [];
const current = [
{ id: 1, name: "abc" },
{ id: 2, name: "def" }
];
const result = []
I need to compare these two arrays and the result should only have the single entry instead of duplicates. In the above example result should have the following output.
Also items in the selected should be taken into consideration and should be in the beginning of the result
result = [
{ id: 1, name: "abc" },
{ id: 2, name: "def" }
];
Also when the input is following
const selected = [ {id:5, name: "xyz" }];
const current = [
{ id: 1, name: "abc" },
{ id: 2, name: "def" }
];
result = [[
{ id: 5, name: "xyz" },
{ id: 1, name: "abc" },
{ id: 2, name: "def" }
];
Also when the input is following
const selected = [ {id:1, name: "abc" }, {id:4, name: "lmn" }];
const current = [
{ id: 1, name: "abc" },
{ id: 2, name: "def" }
];
result = [[
{ id: 1, name: "abc" },
{ id: 4, name: "lmn" }
{ id: 2, name: "def" }
];
Note the comparison should be made using name field
Code that I tried
const res = [...(selected || [])].filter((s) =>
current.find((c) => s.name === c.name)
);
Sandbox: https://codesandbox.io/s/nervous-shannon-j1vn5k?file=/src/index.js:115-206
You could get all items and filter the array by checking the name with a Set.
const
filterBy = (key, s = new Set) => o => !s.has(o[key]) && s.add(o[key]),
selected = [{ id: 1, name: "abc" }, { id: 1, name: "lmn" }],
current = [{ id: 1, name: "abc" }, { id: 2, name: "def" }],
result = [...selected, ...current].filter(filterBy('name'));
console.log(result);
.as-console-wrapper { max-height: 100% !important; top: 0; }
Loop through selected, and if there is no object in current with a name that matches the name of the object in the current iteration push it into current.
const selected=[{id:1,name:"abc"},{id:6,name:"def"},{id:4,name:"lmn"}];
const current=[{id:1,name:"abc"},{id:2,name:"def"}];
for (const sel of selected) {
const found = current.find(cur => cur.name === sel.name);
if (!found) current.push(sel);
}
console.log(current);
This is a good use for .reduce, avoids multiple loops/finds and doesn't need filtering with side-effects.
const selected = [ {id:1, name: "abc" }, {id:4, name: "lmn" }];
const current = [
{ id: 1, name: "abc" },
{ id: 2, name: "def" }
];
const result = Object.values(
[...selected, ...current].reduce((obj, item) => {
obj[item.name] = obj[item.name] || item;
return obj;
}, {})
)
console.log(result);
I want to find all objects in items array which is on groupedItems array by 'id' and add to each item isGrouped: true property.
const items = [
{ id: 1, name: "item1" },
{ id: 2, name: "item2" },
{ id: 3, name: "item3" }
];
const groupedItems = [
{ id: 2, name: "item2" },
{ id: 3, name: "item3" }
];
so the result should be:
items = [
{ id: 1, name: "item1" },
{ id: 2, name: "item2", isGrouped:true },
{ id: 3, name: "item3", isGrouped: true }
];
any ideas ?
You can iterate over the items array using map, creating an isGrouped property which is the result of calling findIndex on the item.id property in the groupedItems id values:
const items = [
{ id: 1, name: "item1" },
{ id: 2, name: "item2" },
{ id: 3, name: "item3" }
];
const groupedItems = [
{ id: 2, name: "item2" },
{ id: 3, name: "item3" }
];
const result = items.map(item => ({
...item,
isGrouped : groupedItems.findIndex(g => g.id == item.id) >= 0
}
));
console.log(result);
Note that I've added isGrouped as false for those objects which are not grouped, rather than omitting the property. It seems that it should be easier to just test a boolean value of a property rather than checking whether the property exists. If you really want to omit the property, you could do something like this:
const items = [
{ id: 1, name: "item1" },
{ id: 2, name: "item2" },
{ id: 3, name: "item3" }
];
const groupedItems = [
{ id: 2, name: "item2" },
{ id: 3, name: "item3" }
];
const result = items.map(item =>
groupedItems.find(g => g.id == item.id) ? { ...item, isGrouped : true } : item
);
console.log(result);
I would like to loop through an array of elements and find the index # of the one that matches certain criteria. Take the following array:
services: [
{ _id: <ObjectId>,
name: "initiating"
},
{ _id: <ObjectId>,
name: "evaluating"
},
{ _id: <ObjectId>,
name: "servicing"
},
]
How would I loop through this array and pull out the array index # of the object where the property "name" is equal to "evaluating" (i.e. - array element #1)?
Try Array.prototype.findIndex:
const services = [
{ _id: 1,
name: "initiating"
},
{ _id: 2,
name: "evaluating"
},
{ _id: 3,
name: "servicing"
},
];
console.log(
services.findIndex(({ name }) => name === 'evaluating')
);
let services = [{
_id: 0,
name: "initiating"
},
{
_id: 1,
name: "evaluating"
},
{
_id: 2,
name: "servicing"
},
];
let index = services.findIndex(item => item.name === "evaluating");
document.write(index);
This will return the index after searching through the each object's name property and comparing it to "evaluating".
let services = [{
_id: 0,
name: "initiating"
},
{
_id: 1,
name: "evaluating"
},
{
_id: 2,
name: "servicing"
},
]
let selected = services.findIndex(service => service.name == "evaluating")
console.log(selected)
You may use .reduce():
var services = [
{
_id: '<ObjectId>',
name: "initiating"
},
{
_id: '<ObjectId>',
name: "evaluating"
},
{
_id: '<ObjectId>',
name: "servicing"
},
];
var idx = services.reduce((x, ele, idx) => (ele.name=='evaluating') ? idx : x, -1);
console.log(idx);
This question already has answers here:
How to get the difference between two arrays of objects in JavaScript
(22 answers)
Closed 1 year ago.
I need some help. How can I get the array of the difference on this scenario:
var b1 = [
{ id: 0, name: 'john' },
{ id: 1, name: 'mary' },
{ id: 2, name: 'pablo' },
{ id: 3, name: 'escobar' }
];
var b2 = [
{ id: 0, name: 'john' },
{ id: 1, name: 'mary' }
];
I want the array of difference:
// [{ id: 2, name: 'pablo' }, { id: 3, name: 'escobar' }]
How is the most optimized approach?
I´m trying to filter a reduced array.. something on this line:
var Bfiltered = b1.filter(function (x) {
return x.name !== b2.reduce(function (acc, document, index) {
return (document.name === x.name) ? document.name : false
},0)
});
console.log("Bfiltered", Bfiltered);
// returns { id: 0, name: 'john' }, { id: 2, name: 'pablo' }, { id: 3, name: 'escobar' } ]
Thanks,
Robot
.Filter() and .some() functions will do the trick
var b1 = [
{ id: 0, name: 'john' },
{ id: 1, name: 'mary' },
{ id: 2, name: 'pablo' },
{ id: 3, name: 'escobar' }
];
var b2 = [
{ id: 0, name: 'john' },
{ id: 1, name: 'mary' }
];
var res = b1.filter(item1 =>
!b2.some(item2 => (item2.id === item1.id && item2.name === item1.name)))
console.log(res);
You can use filter to filter/loop thru the array and some to check if id exist on array 2
var b1 = [{ id: 0, name: 'john' }, { id: 1, name: 'mary' }, { id: 2, name: 'pablo' }, { id: 3, name: 'escobar' } ];
var b2 = [{ id: 0, name: 'john' }, { id: 1, name: 'mary' }];
var result = b1.filter(o => !b2.some(v => v.id === o.id));
console.log(result);
Above example will work if array 1 is longer. If you dont know which one is longer you can use sort to arrange the array and use reduce and filter.
var b1 = [{ id: 0, name: 'john' }, { id: 1, name: 'mary' }, { id: 2, name: 'pablo' }, { id: 3, name: 'escobar' } ];
var b2 = [{ id: 0, name: 'john' }, { id: 1, name: 'mary' }];
var result = [b1, b2].sort((a,b)=> b.length - a.length)
.reduce((a,b)=>a.filter(o => !b.some(v => v.id === o.id)));
console.log(result);
Another possibility is to use a Map, allowing you to bring down the time complexity to O(max(n,m)) if dealing with a Map-result is fine for you:
function findArrayDifferences(arr1, arr2) {
const map = new Map();
const maxLength = Math.max(arr1.length, arr2.length);
for (let i = 0; i < maxLength; i++) {
if (i < arr1.length) {
const entry = arr1[i];
if (map.has(entry.id)) {
map.delete(entry.id);
} else {
map.set(entry.id, entry);
}
}
if (i < arr2.length) {
const entry = arr2[i];
if (map.has(entry.id)) {
map.delete(entry.id);
} else {
map.set(entry.id, entry);
}
}
}
return map;
}
const arr1 = [{id:0,name:'john'},{id:1,name:'mary'},{id:2,name:'pablo'},{id:3,name:'escobar'}];
const arr2 = [{id:0,name:'john'},{id:1,name:'mary'},{id:99,name:'someone else'}];
const resultAsArray = [...findArrayDifferences(arr1,arr2).values()];
console.log(resultAsArray);
I've got follow code:
list1 = {
Items: [
{
ID: 1,
Name: "Zurich"
},
{
ID: 2,
Name: "London"
}, {
ID: 3,
Name: "New York"
}
]
};
list2 = {
Items: [
{
ID: -1,
Name: "Dummy"
},
{
ID: 0,
Name: "Dummy2"
}
]
};
list1.push(list2);
I expect follow result:
list1:
0: Object (Zurich)
1: Object (London)
3: Object (New York)
4: Object (Dummy)
5: Object (Dummy2)
But I get this one:
list1:
0: Object (Zurich)
1: Object (London)
2: Object (New York)
3: Object (Items)
0: Object (Dummy)
1: Object (Dummy2)
How can I get my expectet result?
Thanks and cheers.
Beside Array#concat, you could use Array#push.apply for it
var list1 = { Items: [{ ID: 1, Name: "Zurich" }, { ID: 2, Name: "London" }, { ID: 3, Name: "New York" }] },
list2 = { Items: [{ ID: -1, Name: "Dummy" }, { ID: 0, Name: "Dummy2" }] };
[].push.apply(list1.Items, list2.Items);
console.log(list1);
The question was how to do this with push() not concat():
for (var i = 0; i < list2.Items.length; i++) {
list1.Items.push(list2.Items[i]);
}
Use the spread operator:
list1.Items.push(...list2.Items)
Spread is an ES2015 feature. Your target browsers or runtime may not support it yet, so check the compatibility table (or use a transpiler like babel).
list1 = {
Items: [
{
ID: 1,
Name: "Zurich"
},
{
ID: 2,
Name: "London"
}, {
ID: 3,
Name: "New York"
}
]
};
list2 = {
Items: [
{
ID: -1,
Name: "Dummy"
},
{
ID: 0,
Name: "Dummy2"
}
]
};
list1.Items = list1.Items.concat(list2.Items);
console.log(list1);
try with:
list2.items.forEach(function (item) {
list1.items.push(item)
})
You need to loop through each items in list2 and then fetch them to push into list1.. Below is the snippet using $.each
var list1 = {
Items: [
{
ID: 1,
Name: "Zurich"
},
{
ID: 2,
Name: "London"
}, {
ID: 3,
Name: "New York"
}
]
};
var list2 = {
Items: [
{
ID: -1,
Name: "Dummy"
},
{
ID: 0,
Name: "Dummy2"
}
]
};
$(list2.Items).each(function(k,v){
list1.Items.push(v);
})
console.log(list1);
<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>