JavaScript - 使用 for...of 循环计算数组中的每个唯一元素

Question

大家好，我被困在一个练习题上，真的想不出一个好的解决方案。

问题是：

"loopOverArray() 函数将一个方向数组作为参数，并且有一个空对象保存为一个变量，该变量存储每个方向以及它在数组中的次数。使用 for 循环，此函数需要循环遍历数组中的每个元素并更新对象以显示在数组中找到每个方向的次数。"

预期的输出应如下所示：

loopOverArray(["n", "s", "e", "e"] ==> returns {n: 1, s: 1, e: 2}
loopOverArray(["north", "south", "south", "north"] ==> returns {north: 2, south: 2}
loopOverArray([]) ==> returns {}

这是我迄今为止尝试过的。我是 JS 新手，所以请原谅非常笨拙的尝试。

function loopOverArray(directions) {
        let ncount = 0;
        let ecount = 0;
        let scount = 0;
        let wcount = 0;
 

    for (let direction of directions) {

        if (direction == "n") {
            ncount += 1;
        } else if (direction == "e") {
            ecount += 1;
        } else if (direction == "s") {
            scount += 1;
        } else {
            wcount += 1;
        }
    }
        console.log(`n: ${ncount}, s: ${scount}, e: ${ecount}, w: ${wcount}`)
}

loopOverArray(["n", "n", "s"]);

问题是我认为这不能正确满足问题要求，而且非常难看。肯定有更好的办法。任何帮助将不胜感激

Answer 1

您需要use an object。所以你快到了。只需将您的变量名称、方向视为对象属性键即可。

function loopOverArray(directions) {

  // Define the object
  const obj = {};

  for (let direction of directions) {

    // If the object key (the direction) on the object doesn't exist
    // create it and set the value to zero
    if (!obj[direction]) obj[direction] = 0;

    // Then just increment the value
    obj[direction] += 1;
  }

  // And then return the object from the function
  // once all the array iterations are done
  return obj;

}

console.log(loopOverArray([ 'n', 'n', 's' ]));
console.log(loopOverArray([ 'north', 'north', 'south', 'east' ]));

Answer 2

我们创建一个空对象

let counts = {};

并循环遍历方向数组。

在每次迭代中，我们检查counts 对象中是否已经有方向计数，或者改为考虑0，然后加1。

counts[direction] = (counts[direction] || 0) + 1;

function loopOverArray(directions) {
  let counts = {};
  for (const direction of directions) {
    counts[direction] = (counts[direction] || 0) + 1;
  }
  return counts;
}

console.log(loopOverArray(["n", "s", "e", "e"]));

Answer 3

检查这是否适合您：

var counts = {};
for (var i = 0; i < yourArray.length; i++) {
    counts[yourArray[i]] = 1 + (counts[yourArray[i]] || 0);
}

您可以在这里查看更多答案：Count unique elements in array without sorting

Answer 4

你可以试试这个

function loopOverArray(directions) {
    let result = {}
    for (let direction of directions) {
        direction in result ? result[direction]++ : result[direction] = 1
    }
    console.log(result)
}
loopOverArray(["n", "n", "s"]);

Answer 5

function findOccurences(directions,elementToFind){    
    var result = directions.reduce(function (elementToFind, total, number){
            return total += number==elementToFind;
        }.bind(this, elementToFind), 0);
    return result;
}
var ar = ['n', 'e', 'w', 's', 's', 's','e', 'e'];   
        
console.log(`n: ${findOccurences(ar, 'n')}, s: ${findOccurences(ar, 's')}, e: ${findOccurences(ar, 'e')}, w: ${findOccurences(ar, 'w')}`);