My code has an array of elements as follows:
element: { fromX: { id: ... } , toX: { id: ... } }
Requirement is to pull all the fromX ids into one array, and all toX ids into other.
There are a couple of different ways,
such as using foreach, reduce, iterating for each respectively, but I'm searching for an optimal functional way to return two arrays with one mapping?
Using Array#reduce and destructuring
const data=[{fromX:{id:1},toX:{id:2}},{fromX:{id:3},toX:{id:4}},{fromX:{id:5},toX:{id:6}},{fromX:{id:7},toX:{id:8}}]
const [fromX,toX] = data.reduce(([a,b], {fromX,toX})=>{
a.push(fromX.id);
b.push(toX.id);
return [a,b];
}, [[],[]]);
console.log(fromX);
console.log(toX);
You could take an array for the wanted keys and map the value. Later take a destructuring assignment for getting single id.
const
transpose = array => array.reduce((r, a) => a.map((v, i) => [...(r[i] || []), v]), []),
array = [{ fromX: { id: 1 }, toX: { id: 2 } }, { fromX: { id: 3 }, toX: { id: 4 } }],
keys = ['fromX', 'toX'],
[fromX, toX] = transpose(array.map(o => keys.map(k => o[k].id)));
console.log(fromX);
console.log(toX);
fromX and toX.fromX = [1,2] and toX = [3,4]. I believe the expected result is fromX = [1, 3] and toX = [2, 4].Try this:
const arr = [{
fromX: {
id: 1
},
toX: {
id: 2
}
}, {
fromX: {
id: 3
},
toX: {
id: 4
}
}]
let {
arr1,
arr2
} = arr.reduce((acc, {
fromX,
toX
}) => ({ ...acc,
arr1: [...acc.arr1, fromX],
arr2: [...acc.arr2, toX]
}), {
arr1: [],
arr2: []
})
console.log(arr1, arr2);You can achieve this by using below solution
var array = [{ fromX: { id: 1 }, toX: { id: 2 } }, { fromX: { id: 3 }, toX: { id: 4 } }],
arrayX = array.map(x => x.fromX), arraytoX = array.map(toX => toX.toX)
console.log(arrayX);
console.log(arraytoX );Returning multiple keys from the list requires using maps in jQuery.
var list = [{ name:"h1",emailId:"[email protected]",password:"h1"},{ name:"h2",emailId:"[email protected]",password:"h2"},name:"h3",emailId:"[email protected]",password:"h3"}]
Obtain only the name and emailId from this list. Subsequently, use the map keyword.
var filter = list.map(x => ({name : x.name, email : x.emailId }));
fromXandfromYvalues are shown?[froms, tos] = transpose(elements.map(e => [e.fromX, e.toX]))is the functional way - unfortunately there is no nativetransposefunction in JS