基于偶数/奇数索引将数组转换为对象

Question

I am trying to convert an array to an object based on whether the index of array is odd or even.

For example,,

Input: ["name", "Tom", "age", 20]

output: { "name": "tom","age": 20 }

It can be implemented using basic functions of JavaScript such as forEach , map , and filter . But I want more simple code.

So I checked docs of underscore.js, but I couldn't find a good way. Is there any way to solve this simply?

Answer 1

Interesting question, my two cents:

Simple and performant for loop:

const simpleArray = ["name", "Tom", "age", 20];

// Modifying the Array object
Array.prototype.toObject = function() {
    let r = {};

    for(let i = 0; i < this.length; i += 2) {
        let key = this[i], value = this[i + 1];
        r[key] = value;
    }

    return r;
}

// Or as a function
const toObject = arr => {
    let r = {};

    for(let i = 0; i < arr.length; i += 2) {
        let key = arr[i], value = arr[i + 1];
        r[key] = value;
    }

    return r;
}

const simpleObjectOne = simpleArray.toObject(); // First method
const simpleObjectTwo = toObject(simpleArray); // Second method

Answer 2

You could use Array#forEach and a check for the index, if uneven, then assign the element to the key from the last item.

 var array = ["name", "Tom", "age", 20], object = {}; array.forEach(function (a, i, aa) { if (i & 1) { object[aa[i - 1]] = a; } }); console.log(object); 

The same with Array#reduce

 var array = ["name", "Tom", "age", 20], object = array.reduce(function (r, a, i, aa) { if (i & 1) { r[aa[i - 1]] = a; } return r; }, {}); console.log(object); 

Answer 3

var input = ["name", "Tom", "age", 20] ;
var output = {}
input.forEach((x, i, arr) => {
  if (i % 2 === 0) output[x] = arr[i+1];
});
console.log(output); //{ name: 'Tom', age: 20 }

Answer 4

You may also use reduce

 var arr = ["name", "Tom", "age", 20], obj = arr.reduce((p,c,i,a) => (i%2 ? p[a[i-1]] = c : p[c],p),{}); console.log(obj); 

for this operation as follows;

Answer 5

Underscore solution:

output = _.object(..._.partition(input, (_, i) => !(i % 2)))

_.partition will partition the array into [['name', 'age'], ["tom", "20"]] . In other words, it returns an array containing two subarrays--in this case, one array of keys and one array of values. _.object takes an array of keys and an array of values as parameters, so we use ... to pass the subarrays in the value returned by _.partition to it as two parameters.

If you're into very functional, semantic code:

const even = i => !(i % 2);
const index = fn => (_, i) => fn(i);

output = _.object(..._.partition(input, index(even)))

Recursive solution:

function arrayToObject([prop, value, ...rest], result = {}) {
  return prop ? arrayToObject(rest, Object.assign(result, {[prop]: value})) : result;
}

Iterative solution:

function arrayToObject(arr) {
  const result = {};
  while (arr.length) result[arr.shift()] = arr.shift();
  return result;
}

Answer 6

A simple for loop

var arr = ["name", "Tom", "age", 20, "address"];
var obj = {};
for (var i = 0; i < arr.length; i+=2){
    //First iteration: assign obj["name"] to "Tom"
    //Second iteration: assign obj["age"] to 20
    //Third iteration: assign obj["address"] to "" because arr[5] does not exist
    obj[arr[i]] = arr[i+1] || "";
}