对多个属性的数组进行排序，然后根据前一项更新属性答案

【问题标题】：Sort array on multiple properties and then update a property based on previous item对多个属性的数组进行排序，然后根据前一项更新属性
【发布时间】：2019-01-22 17:26:59
【问题描述】：

我想创建电视节目列表应用。我需要调整 tvProgram 单元格大小的函数。

输入

[
  {channel: "x", start: "2019-01-08T06:00:00", distanceFromTop: 6},
  {channel: "y", start: "2019-01-08T06:30:00", distanceFromTop: 9},
  {channel: "x", start: "2019-01-08T07:00:00", distanceFromTop: 9}, 
  {channel: "x", start: "2019-01-08T07:30:00", distanceFromTop: 18},
  {channel: "y", start: "2019-01-08T08:00:00", distanceFromTop: 13},
  {channel: "x", start: "2019-01-08T08:00:00", distanceFromTop: 15}
]

输出：

[
  {channel: "x", start: "2019-01-08T06:00:00", distanceFromTop: 6},
  {channel: "y", start: "2019-01-08T06:30:00", distanceFromTop: 9},
  {channel: "x", start: "2019-01-08T07:00:00", distanceFromTop: 10},
  {channel: "x", start: "2019-01-08T07:30:00", distanceFromTop: 18},
  {channel: "y", start: "2019-01-08T08:00:00", distanceFromTop: 19},
  {channel: "y", start: "2019-01-08T08:00:00", distanceFromTop: 19}
]

我想根据以下条件将 distanceFromTop 加 1。

“开始”必须排序。
必须对“distanceFromTop”进行排序。
如果有相同的“start”对象，distanceFromTop 也相同。
不要减少距离从顶部，只需加 1。

  // wrong
  {channel: "y", start: "2019-01-08T08:00:00", distanceFromTop: 13}, // 13 !== 15
  {channel: "y", start: "2019-01-08T08:00:00", distanceFromTop: 15}
  distanceFromTop must be also same.
  // correct
  {channel: "y", start: "2019-01-08T08:00:00", distanceFromTop: 15}, // 15 === 15
  {channel: "y", start: "2019-01-08T08:00:00", distanceFromTop: 15}

ideal data
const ideal = [
  {channel: "x", start: "2019-01-08T06:00:00", distanceFromTop: 6},
  {channel: "y", start: "2019-01-08T06:30:00", distanceFromTop: 9},
  {channel: "x", start: "2019-01-08T07:00:00", distanceFromTop: 10},
  {channel: "x", start: "2019-01-08T07:30:00", distanceFromTop: 18},
  {channel: "y", start: "2019-01-08T08:00:00", distanceFromTop: 19},
  {channel: "y", start: "2019-01-08T08:00:00", distanceFromTop: 19}
]

如何创建这个函数

ideal = func(ideal)

我试过下面的代码。

const isSorted = (arr) => {
    var len = arr.length - 1;
    for(var i = 0; i < len; ++i) {
        if(arr[i].distanceFromTop > arr[i+1].distanceFromTop) {
            return false;
        }
    }
    return true;
}

const func = (arr) => {
  output = arr.slice()
  while (isSort(ideal)) {
    addOne(ideal)
  }

  return output
}

请告诉我如何编写 addOne 方法。

【问题讨论】：

不清楚。 I want to add 1 是什么意思？如果两个项目具有相同的start 值，并且具有不同的distanceFromTop，则将distanceFromTop 更新为相同。应该取哪个对象的distanceFromTop？如果两个项目具有相同的start 但具有不同的channel，那么它们是否被视为匹配数据？您应该提供一个涵盖您的场景的输入和输出。

标签： javascript arrays node.js

【解决方案1】：

首先是sort 数组，然后是start，然后是distanceFromTop。然后，您可以使用reduce 执行类似的操作。

获取新distanceFromTop的逻辑：

获取累加器中的last项
如果当前项的start与last.start相同，则使用last.distanceFromTop
否则，检查当前distanceFromTop > last.distanceFromTop
如果是，请保持原样。将其添加到累加器中
否则，请使用last.distanceFromTop + 1

const input = [
  {channel: "x", start: "2019-01-08T06:00:00", distanceFromTop: 6},
  {channel: "y", start: "2019-01-08T06:30:00", distanceFromTop: 9},
  {channel: "x", start: "2019-01-08T07:00:00", distanceFromTop: 9}, 
  {channel: "x", start: "2019-01-08T07:30:00", distanceFromTop: 18},
  {channel: "y", start: "2019-01-08T08:00:00", distanceFromTop: 13},
  {channel: "x", start: "2019-01-08T08:00:00", distanceFromTop: 15}
]

const sorted = input.sort((a, b) => a.start.localeCompare(b.start) 
                                || a.distanceFromTop - b.distanceFromTop);
                                
const output = sorted.reduce((acc, { channel, start, distanceFromTop }) => {
     if (acc.length === 0)
       return [{channel,start,distanceFromTop}];

     const last = acc[acc.length - 1];
     const newDistanceFromTop = last.start === start 
                                 ? last.distanceFromTop 
                                 : distanceFromTop > last.distanceFromTop 
                                 ? distanceFromTop 
                                 : last.distanceFromTop + 1;

     return [...acc, { channel, start, distanceFromTop: newDistanceFromTop}];
  }, [])

console.log(output)

【讨论】：