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对象JavaScript数组的前n个最大值

2024-04-30 15:09:03 425

假设我们有一个这样的对象数组-

const arr = [
   {"id":0,"start":0,"duration":117,"slide":4,"view":0},
   {"id":0,"start":0,"duration":12,"slide":1,"view":0},
   {"id":0,"start":0,"duration":41,"slide":2,"view":0},
   {"id":0,"start":0,"duration":29,"slide":3,"view":0},
   {"id":0,"start":0,"duration":123,"slide":3,"view":0},
   {"id":0,"start":0,"duration":417,"slide":2,"view":0},
   {"id":0,"start":0,"duration":12,"slide":1,"view":0},
   {"id":0,"start":0,"duration":67,"slide":2,"view":0}
];

我们必须编写一个函数，该函数接受此数组并在另一个数组中返回该数组的前n个元素（top表示持续时间最大的对象）。

因此，让我们为这个问题编写代码-

示例

const arr = [
   {"id":0,"start":0,"duration":117,"slide":4,"view":0},
   {"id":0,"start":0,"duration":12,"slide":1,"view":0},
   {"id":0,"start":0,"duration":41,"slide":2,"view":0},
   {"id":0,"start":0,"duration":29,"slide":3,"view":0},
   {"id":0,"start":0,"duration":123,"slide":3,"view":0},
   {"id":0,"start":0,"duration":417,"slide":2,"view":0},
   {"id":0,"start":0,"duration":12,"slide":1,"view":0},
   {"id":0,"start":0,"duration":67,"slide":2,"view":0}
];
const topN = (arr, n) => {
   if(n > arr.length){
      return false;
   }
   return arr
   .slice()
   .sort((a, b) => {
      return b.duration - a.duration
   })
   .slice(0, n);
};
console.log(topN(arr, 3));
console.log(topN(arr, 4));
console.log(topN(arr, 5));

输出结果

控制台中的输出将为-

[
   { id: 0, start: 0, duration: 417, slide: 2, view: 0 },
   { id: 0, start: 0, duration: 123, slide: 3, view: 0 },
   { id: 0, start: 0, duration: 117, slide: 4, view: 0 }
]
[
   { id: 0, start: 0, duration: 417, slide: 2, view: 0 },
   { id: 0, start: 0, duration: 123, slide: 3, view: 0 },
   { id: 0, start: 0, duration: 117, slide: 4, view: 0 },
   { id: 0, start: 0, duration: 67, slide: 2, view: 0 }
]
[
   { id: 0, start: 0, duration: 417, slide: 2, view: 0 },
   { id: 0, start: 0, duration: 123, slide: 3, view: 0 },
   { id: 0, start: 0, duration: 117, slide: 4, view: 0 },
   { id: 0, start: 0, duration: 67, slide: 2, view: 0 },
   { id: 0, start: 0, duration: 41, slide: 2, view: 0 }
]

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