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计算JSON数组中的值的总和

ecmascript-6 javascript

AbreQueVoy · 技术社区 · 5 年前

有一个这样的数组:

const data = [
    {
        "name": "Dave",
        "coins": 14,
        "weapons": 2,
        "otherItems": 3,
        "color": "red"
    },
    {
        "name": "Vanessa",
        "coins": 18,
        "weapons": 1,
        "otherItems": 5,
        "color": "blue"
    },
    {
        "name": "Sharon",
        "coins": 9,
        "weapons": 5,
        "otherItems": 1,
        "color": "pink"
    },
    {
        "name": "Walter",
        "coins": 9,
        "weapons": 2,
        "otherItems": 4,
        "color": "white"
    }
]

如何计算总和 coins , weapons 和 otherItems 使用ES6功能?(我不喜欢这个:任何简单的方法都是好的。)

data.reduce((first, last) => first + last) 生成一个链 [object Object][object Object] s

4 回复 | 直到 5 年前

mickl 5 年前

您必须单独处理每个字段(请注意,当您没有为指定第二个参数时 reduce 它将把第一个数组对象作为种子,并从第二个数组对象开始处理):

const data = [
    {
        "name": "Dave",
        "coins": 14,
        "weapons": 2,
        "otherItems": 3,
        "color": "red"
    },
    {
        "name": "Vanessa",
        "coins": 18,
        "weapons": 1,
        "otherItems": 5,
        "color": "blue"
    },
    {
        "name": "Sharon",
        "coins": 9,
        "weapons": 5,
        "otherItems": 1,
        "color": "pink"
    },
    {
        "name": "Walter",
        "coins": 9,
        "weapons": 2,
        "otherItems": 4,
        "color": "white"
    }
]

let result = data.reduce((a,c)=> ({ 
     coins: a.coins + c.coins, 
     weapons: a.weapons + c.weapons, 
     otherItems: a.otherItems + c.otherItems })
)

console.log(result);

Nina Scholz 5 年前

您可以为总和获取一组所需的键,并为总和创建一个对象,然后添加所需的值。

const
    data = [{ name: "Dave", coins: 14, weapons: 2, otherItems: 3, color: "red" }, { name: "Vanessa", coins: 18, weapons: 1, otherItems: 5, color: "blue" }, { name: "Sharon", coins: 9, weapons: 5, otherItems: 1, color: "pink" }, { name: "Walter", coins: 9, weapons: 2, otherItems: 4, color: "white" }],
    keys = ['coins', 'weapons', 'otherItems'],
    sums = data.reduce(
        (r, o) => (keys.forEach(k => r[k] += o[k]), r), 
        Object.fromEntries(keys.map(k => [k, 0]))
    );

console.log(sums);

Fullstack Guy 5 年前

您可以使用 Array.prototype.reduce 为了这个。

为了使其更加灵活和动态,制作一个 Set 你想要清点的钥匙数量。

然后逐一检查钥匙在 设置 如果该密钥在 obj ,将其总结为 accumulator reduce回调中的对象:

const data = [{"name":"Dave","coins":14,"weapons":2,"otherItems":3,"color":"red"},{"name":"Vanessa","coins":18,"weapons":1,"otherItems":5,"color":"blue"},{"name":"Sharon","coins":9,"weapons":5,"otherItems":1,"color":"pink"},{"name":"Walter","coins":9,"weapons":2,"otherItems":4,"color":"white"}]

//Keys to count
const keys = new Set(["coins", "weapons", "otherItems"]);

const count = data.reduce((acc, obj) => {
  const objKeys = keys.forEach(key => {
    if (obj.hasOwnProperty(key)) {
      acc[key] = (acc[key] || 0) + obj[key];
    }
  });
  return acc;
}, {});
console.log(count);

Antoni SilvestroviÄ 5 年前

你的想法是对的,你需要使用 reduce 方法。问题是,你是在求两个对象的和,而不是它们的属性。你所需要做的就是将代码更改为以下内容(对硬币求和):

data.reduce((first, last) => first.coins + last.coins, 0)

以下是武器:

data.reduce((first, last) => first.weapons + last.weapons, 0)