使用lambda查找数组中最频繁的数字，并使用JavaScript查找正则表达式

Question

I know this question has been asked several times, however I didn't find the desired solution.

My task is to find the most frequent number in an array. First I have to convert to number because our input is given as strings. After that I sort the array with the lambda expression. (I don't understand how it works, I just know it does). And now I want to filter the array [ 1, 1, 1, 2, 2, 3, 3, 4, 4, 4, 4, 4, 9, 13 ] so only [4, 4, 4, 4, 4] is left. I would like to use either .filter or .reduce .

function z(input) 
{
let numbers = input.map(n => +n)
                   .sort((a, b) => a - b)
                   .filter();

console.log(numbers);
}
z(['13', '4', '1', '1', '4', '2', '3', '4', '4', '1', '2', '4', '9', '3']);

Also if someone can explain:

function mode(arr){
    return arr.sort((a,b) => 
          arr.filter(v => v===a).length 
        - arr.filter(v => v===b).length
    ).pop();
}

mode(['pear', 'apple', 'orange', 'apple']); // apple

The callback on arr.sort takes two parameters a and b however after that I cannot understand what is going on.

In fact any guide on how to use lambda will be appreciated.

Answer 1

You could use a Map for collecting the elements, with the same value and take a single loop approach by checking the lenght of the array and check the actual length against the previously store result.

This approach take the possibility in account if more than one value has the same length, then both all arrays are returned with the maximum length.

 const z = a => { var m = new Map; return a.reduce((r, v, i) => { // create map var array = m.get(v) || m.set(v, []).get(v); // get/create array array.push(v); // push value if (!i || r[0].length < array.length) { // take first or check return [array]; // return array } if (r[0] !== array && r[0].length === array.length) { // prevent push same a r.push(array); // push with same len } return r; }, undefined); }; console.log(z(['13', '4', '1', '1', '4', '2', '3', '4', '4', '1', '2', '4', '9', '3'])); console.log(z(['pear', 'apple', 'orange', 'apple'])); 

 .as-console-wrapper { max-height: 100% !important; top: 0; } 

Answer 2

First of all I don't think you need to sort the array if you traverse it until its end?

function z(input) {
    let map = new Map;
    let numbers = input.map(i => parseInt(i))
                   //.sort((a, b) => a - b)
                   .forEach(e => {
                        map.set(e, (map.get(e) || 0) + 1);
                   });

    return Array.from(map).reduce((a, b) => a[1] > b[1] ? a : b)[0];
}

Explanation:

1. Create a Map which can you use to set how many times your number appears in the array.
2. Use parseInt to transform each element of the array to a Number
3. For each item in the array (which now has Numbers instead of Strings) you need to get the number of occurrences from the Map (using get function) and then set it back to the Map using the set function with the increamented value. (map.get(e) || 0) is used in case this is the first time you encounter this number.
4. In the end you can extract the content of the map using Array.from which will give you a jagged array (like [[1, 2], [3, 4], [4, 5]]). Each inner array will hold a pair of (number, occurrences). You need to take the the pair which has the highest value in the second item of the pair which means 5 in the jagged array example and extract the first item of this pair. You do it using the reduce function of the array. The reduce function gets two items in the array and expects to return a value that has the same shape of the items that entered this function something like this reduce(a: A, b: A) => A which means a function that accepts two arguments of type A and return an instance of type A. In your example it is: reduce(a: [Int, Int], b: [Int, Int]) => [Int, Int] which means gets two parameters of type array and returns an array.

So the reduce function checks whether the first array has a greater number in the second index then the second array in the second index. If yes it returns the first array and not the value in the second index, otherwise the second array. Because the reduce function output must be the same type of each one of its arguments than the type of the return value is also an array which has the number in its first index and the occurences in the second so you need to get the value from the first index.

Answer 3

You can find the most frequent element in an array like this:

 function z(arr) { return Object.entries(arr.reduce((res, v) => { if(v in res) res[v]++; else res[v] = 1; return res; }, {})).sort((a, b) => a[1] < b[1])[0][0]; } let a = z(['13', '4', '1', '1', '4', '2', '3', '4', '4', '1', '2', '4', '9', '3']); console.log(a); 

Explanation:

1. The call to arr.reduce serves to count the occurrences of every element in the arr .
2. We then convert the object from arr.reduce to a 2D array like [[key, value]] using Object.entries .
3. We then sort the Object.entries array in descending order based on the second element of each sub array. (This element is the value part)
4. We then pick the first element in the sorted array, and return the key part of that element [0][0] .