Question

我知道，关于对象数组中的一个属性进行分组的问题不计其数。但是我想做的是更具体一些：

const lists = [
{ groupKey: 'ABC', key: 'r8', timestamp: '2014', index: 2 },
{ groupKey: 'ABC', key: 'q8', timestamp: '2012', index: 0 }, 
{ groupKey: 'ABC', key: 'w8', timestamp: '2013', index: 1 },
{ groupKey: 'CDE', key: 'r7', timestamp: '2019', index: 0 } 
]

结果应按groupKey分组，并按索引排序

（这里的索引是迭代器0、1、2、3 ...，因此无需实际排序，而是按数组的正确顺序放置。例如：array [index] = ...）。

这应该看起来像：

{ 
  ABC: [
    {  key: 'q8', timestamp: '2012', index: 0 },
    {  key: 'w8', timestamp: '2013', index: 1 },
    {  key: 'r8', timestamp: '2014', index: 2 }
  ],
  CDE: [
    { key: 'r7', timestamp: '2019', index: 0 } 
  ]
}

我尝试不进行排序就进行分组：

const result = lists.reduce((r, item) => {
      let { groupKey, ...rest } = item
      r[item.groupKey] = [...(r[item.groupKey] || []), rest]
      return r
    }, {})

通过排序，并不成功，但是您知道我的意思：

const result = lists.reduce((r, item) => {
          let { groupKey, ...rest } = item
          r[item.groupKey][item.index] = rest //err: can not set property 2 of undefined
          return r
        }, {})

任何建议都值得赞赏

Answer 1

您可以直接获取索引，而无需稍后排序。

const
    lists = [{ groupKey: 'ABC', key: 'r8', timestamp: '2014', index: 2 }, { groupKey: 'ABC', key: 'q8', timestamp: '2012', index: 0 }, { groupKey: 'ABC', key: 'w8', timestamp: '2013', index: 1 }, { groupKey: 'CDE', key: 'r7', timestamp: '2019', index: 0 }],
    result = lists.reduce((r, { groupKey, ...rest }) => {
        r[groupKey] = r[groupKey] || [];
        r[groupKey][rest.index] = rest;
        return r
    }, {});

console.log(result);

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Answer 2

尝试这样

const lists = [
{ groupKey: 'ABC', key: 'r8', timestamp: '2014', index: 2 },
{ groupKey: 'ABC', key: 'q8', timestamp: '2012', index: 0 }, 
{ groupKey: 'ABC', key: 'w8', timestamp: '2013', index: 1 },
{ groupKey: 'CDE', key: 'r7', timestamp: '2019', index: 0 } 
]

var groupBy = function(datas, key) {
  return datas.reduce(function(data, x) {
    var dKey = x[key];
    delete x[key];
    (data[dKey] = data[dKey] || []).push(x);
    
    return data;
  }, {});
};

console.log(groupBy(lists, 'groupKey'));

Answer 3

您快到了，您只需要在reduce完成后对进行排序：

const lists = [{
    groupKey: 'ABC',
    key: 'r8',
    timestamp: '2014',
    index: 2
  },
  {
    groupKey: 'ABC',
    key: 'q8',
    timestamp: '2012',
    index: 0
  },
  {
    groupKey: 'ABC',
    key: 'w8',
    timestamp: '2013',
    index: 1
  },
  {
    groupKey: 'CDE',
    key: 'r7',
    timestamp: '2019',
    index: 0
  }
]
const result = lists.reduce((r, item) => {
  const { groupKey, ...rest } = item
  r[groupKey] = [...(r[groupKey] || []), rest];
  return r;
}, {})
Object.values(result).forEach((arr) => {
  arr.sort((a, b) => a.index - b.index);
});
console.log(result);

Answer 4

首先可以按键分组，然后需要根据索引进行排序

const lists = [{ groupKey: 'ABC', key: 'r8', timestamp: '2014', index: 2 },{ groupKey: 'ABC', key: 'q8', timestamp: '2012', index: 0 }, { groupKey: 'ABC', key: 'w8', timestamp: '2013', index: 1 },{ groupKey: 'CDE', key: 'r7', timestamp: '2019', index: 0 } ]

// group by groupkey
let grouped = lists.reduce((op, inp) => {
  let groupKey = inp.groupKey
  op[groupKey] = op[groupKey] || []
  op[groupKey].push(inp)
  return op
},{})

// sort by index 
let output = Object.entries(grouped).reduce((op,[key,value]) => {
  value = value.sort((a,b)=> a.index-b.index)
  op[key] = value
  return op
},{})

console.log(output)

Answer 5

尝试一下：

const result  = lists.reduce((r, item) => {
      let { groupKey, ...rest } = item
      r[item.groupKey] = [...(r[item.groupKey] || []), rest];
      (r[item.groupKey]).sort((a, b) => a.index - b.index);
      return r
    }, {});

Javascript-按对象数组中的一个对象属性分组

5 个答案: