如何迭代和比较 JS Object 中的值？

Question

The result should be a 2 element array with the 2 number that are most frequent; I can return the top 2 most amount of times a number has been spotted (my //comments). But I can not figure out how to return the correct keys from the pairs object that appear the most times. Please help.

 /* majorityElementTopTwo * Write a function which accept an array of integers and returns in a new two item array * the two integers in the input that appear most frequently. * majorityElementTopTwo([3,3,1,2,1,1,4,4,4,4]); // [4,1] */ function majorityElementTopTwo(array) { let result = []; let pairs = {}; for (let i = 0; i < array.length; i++) { if (array[i] in pairs) { pairs[array[i]] += 1; } else { pairs[array[i]] = 1; } } return pairs; // obj.sort(); // for(let i = obj.length; i > 0; i--){ // let num = obj.pop(); // console.log(num); // result.push(num); // } // return result.slice(0,2); } console.log(majorityElementTopTwo([3, 3, 1, 2, 1, 1, 4, 4, 4, 4])); // [4,1]

Answer 1

I hope this is the answer you are looking for what I have done here is

1. first map all the entries as an array ie: [["1",3],["2",1],["3",2],["4",4]]
2. Then sort it based upon the second element which is number of times an element has occured
3. Then slice the first two elements that are most occurred [["4",4],["1",3]]

 /* majorityElementTopTwo * Write a function which accept an array of integers and returns in a new two item array * the two integers in the input that appear most frequently. * majorityElementTopTwo([3,3,1,2,1,1,4,4,4,4]); // [4,1] */ function majorityElementTopTwo(array) { let pairs = {}; for (let i = 0; i < array.length; i++) { if (array[i] in pairs) { pairs[array[i]] += 1; } else { pairs[array[i]] = 1; } } return Object.entries(pairs).map((elements) => elements).sort((a,b) => b[1]-a[1]).slice(0,2) } console.log(majorityElementTopTwo([3, 3, 1, 2, 1, 1, 4, 4, 4, 4])); // [4,1]

Answer 2

As of JavaScript you can use Object.keys(obj) to get an Array of properties defined on the object itself (the ones that return true for obj.hasOwnProperty(key)).

Object.keys(obj).forEach(function(key,index) {
    // key: the name of the object key
    // index: the ordinal position of the key within the object 
});

This is better (and more readable) than using a for-in loop.

Its supported on these browsers:

Firefox (Gecko): 4 (2.0) Chrome: 5 Internet Explorer: 9

ref: Iterate through object properties

Answer 3

Using Array.prototype.map(),Array.prototype.filter() and Array.prototype.sort().

 /* majorityElementTopTwo * Write a function which accept an array of integers and returns in a new two item array * the two integers in the input that appear most frequently. * majorityElementTopTwo([3,3,1,2,1,1,4,4,4,4]); // [4,1] */ function majorityElementTopTwo(array) { const [first, second] = [...new Set(array)].map((num) => [num, array.filter((n) => n === num).length]).sort((a, b) => b[1] - a[1]).slice(0, 2); return [first[0], second[0]]; } console.log(majorityElementTopTwo([3, 3, 1, 2, 1, 1, 4, 4, 4, 4]));