JS（ES6）：基于对象属性减少数组答案

【问题标题】：JS (ES6): Reduce array based on object attributeJS（ES6）：基于对象属性减少数组
【发布时间】：2018-09-29 19:05:27
【问题描述】：

我有一个这样的数组：

const fruits = [ 
    { fruit: 'apple',  year: 2018 },
    { fruit: 'apple',  year: 2018 },
    { fruit: 'banana', year: 2018 },
    { fruit: 'orange', year: 2018 },
    { fruit: 'apple',  year: 2017 },
    { fruit: 'apple',  year: 2016 }
];

现在我想减少这个数组，看看每年有多少水果，总数是多少。所以结果看起来像这样：

const result = [ 
    { year: 2018, apple: 2, banana: 1, orange: 1 total: 4 },
    { year: 2017, apple: 1, total: 1 },
    { year: 2016, apple: 1, total: 1 }
];

我尝试过使用 reduce，但我不知道如何根据年份对其进行分组。也许还有一个 lodash 助手？

【问题讨论】：

标签： javascript arrays ecmascript-6 reduce

【解决方案1】：

使用Object.values 和reduce

var output = Object.values(fruits.reduce( (a,c) => {
  a[c.year] = a[c.year] || { year : c.year };  
  a[c.year]["total"] = (a[c.year]["total"] || 0) + 1;
  a[c.year][c.fruit] = (a[c.year][c.fruit] || 0) + 1;
  return a;
},{}));

演示

var fruits = [{
    fruit: 'apple',
    year: 2018
  },
  {
    fruit: 'apple',
    year: 2018
  },
  {
    fruit: 'banana',
    year: 2018
  },
  {
    fruit: 'orange',
    year: 2018
  },
  {
    fruit: 'apple',
    year: 2017
  },
  {
    fruit: 'apple',
    year: 2016
  }
];

var output = Object.values(fruits.reduce((a, c) => {
  a[c.year] = a[c.year] || {
    year: c.year
  };
  a[c.year]["total"] = (a[c.year]["total"] || 0) + 1;
  a[c.year][c.fruit] = (a[c.year][c.fruit] || 0) + 1;
  return a;
}, {}));

console.log(output);

【讨论】：

【解决方案2】：

您可以使用array#reduce 根据年份对数据进行分组，并计算对象累加器中的水果出现次数。然后使用Object.values()从这个对象中获取所有值

const fruits = [ { fruit: 'apple', year: 2018 }, { fruit: 'apple', year: 2018 }, { fruit: 'banana', year: 2018 }, { fruit: 'orange', year: 2018 }, { fruit: 'apple', year: 2017 }, { fruit: 'apple', year: 2016 } ],
    result = Object.values(fruits.reduce((r,{fruit, year}) => {
      r[year] = r[year] || {year};
      r[year][fruit] = (r[year][fruit] || 0) + 1;
      r[year]['total'] = (r[year]['total'] || 0) + 1;
      return r;
    },{}));
console.log(result);

【讨论】：

【解决方案3】：

您可以使用数组作为结果集，它保持给定的年份顺序。

var fruits = [{ fruit: 'apple',  year: 2018 }, { fruit: 'apple',  year: 2018 }, { fruit: 'banana', year: 2018 }, { fruit: 'orange', year: 2018 }, { fruit: 'apple',  year: 2017 }, { fruit: 'apple',  year: 2016 }],
    result = fruits.reduce((r, { year, fruit }) => {
        var item = r.find(o => o.year === year);
        if (!item) {
            r.push(item = { year });
        }
        item[fruit] = (item[fruit] || 0) + 1;
        return r;
    }, []);
    
console.log(result);

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【讨论】：