JS：按每个类别的最大值过滤对象数组

问题描述 投票：0回答：8

什么是最有效/优雅的方式来实现类似sql的过滤效果。我想过滤它们并只获取某个组中最大值的对象。

这是我的代码，它可以工作，但可能不是最好的方法：

uniqueValues = (arr) => [...new Set(arr)];
getMaxTimeOf = (arr) => Math.max(...arr.map(o => o.timeStamp), 0);
selectorName = (name) => (obj) => obj.name === name;
selectorTime = (time) => (obj) => obj.timeStamp === time;
getGroup = (obj, selector) => obj.filter(selector)

onlyLastChangedFrom = (history) => {
const uniqueNames = uniqueValues(history.map(o => o.name))
let filtered = []
 uniqueNames.forEach(name => {
  const group = getGroup(history, selectorName(name))
  const groupLastTime = getMaxTimeOf(group)
  const lastChange = getGroup(group, selectorTime(groupLastTime))
  filtered.push(lastChange[0])
 });
 return filtered
}   
onlyLastChangedFrom(history)

    // Input:
    [ { name: 'bathroom',
        value: 54,
        timeStamp: 1562318089713 },
      { name: 'bathroom',
        value: 55,
        timeStamp: 1562318090807 },
      { name: 'bedroom',
        value: 48,
        timeStamp: 1562318092084 },
      { name: 'bedroom',
        value: 49,
        timeStamp: 1562318092223 },
      { name: 'room',
        value: 41,
        timeStamp: 1562318093467 } ]

    // Output:
    [ { name: 'bathroom',
        value: 55,
        timeStamp: 1562318090807 },
      { name: 'bedroom',
        value: 49,
        timeStamp: 1562318092223 },
      { name: 'room',
        value: 41,
        timeStamp: 1562318093467 } ]
javascript arrays
8个回答
13
投票
使用

name 属性作为键，将数组 

Reduce
为对象。对于每个项目，检查累加器中存在的项目是否具有比当前项目更高的值，如果没有，则将其替换为当前项目。使用 
Object.values()
:

转换回数组 

const arr = [{"name":"bathroom","value":54,"timeStamp":1562318089713},{"name":"bathroom","value":55,"timeStamp":1562318090807},{"name":"bedroom","value":48,"timeStamp":1562318092084},{"name":"bedroom","value":49,"timeStamp":1562318092223},{"name":"room","value":41,"timeStamp":1562318093467}]

const result = Object.values(arr.reduce((r, o) => {
  r[o.name] = (r[o.name] && r[o.name].value > o.value) ? r[o.name] : o

  return r
}, {}))

console.log(result)

2
投票

我喜欢使用 lodash 来做这样的事情。它非常实用，因此非常清晰和直接。

看一下下面的代码：

const DATA = [
  {
    name: "bathroom",
    value: 54,
    timeStamp: 1562318089713
  },
  {
    name: "bathroom",
    value: 55,
    timeStamp: 1562318090807
  },
  {
    name: "bedroom",
    value: 48,
    timeStamp: 1562318092084
  },
  {
    name: "bedroom",
    value: 49,
    timeStamp: 1562318092223
  },
  {
    name: "room",
    value: 41,
    timeStamp: 1562318093467
  }
];

let max = _
  .chain(DATA)
  .groupBy('name')
  .sortBy('value')
  .map(o => _(o).reverse().first())
  .flatten()
  .value();

console.log(max); // returns [{"name":"bathroom","value":55,"timeStamp":1562318090807},{"name":"bedroom","value":49,"timeStamp":1562318092223},{"name":"room","value":41,"timeStamp":1562318093467}]
2
投票

什么是最有效/优雅的方式来实现类似sql的过滤效果。

您可以为每个步骤采用函数，并通过管道传输所有函数以获得单个结果。

例如在 SQL 中，您将有以下查询：

SELECT name, value, MAX(timeStamp) 
FROM data 
GROUP BY name;

使用类似 SQL 的方法，您可以首先进行分组并从结果集中取出最大对象。

result = pipe(
    groupBy('name'),
    select(max('timeStamp'))
)(data);

2024 年更新

现在与

Object.groupBy

const
    // functions
    pipe = (...functions) => input => functions.reduce((acc, fn) => fn(acc), input),
    groupBy = key => array => Object.values(Object.groupBy(
        array, 
        object => object[key]
    )),
    max = key => array => array.reduce((a, b) => a[key] > b[key] ? a : b),
    select = fn => array => array.map(fn),

    // data and result
    data = [{ name: 'bathroom', value: 54, timeStamp: 1562318089713 }, { name: 'bathroom', value: 55, timeStamp: 1562318090807 }, { name: 'bedroom', value: 48, timeStamp: 1562318092084 }, { name: 'bedroom', value: 49, timeStamp: 1562318092223 }, { name: 'room', value: 41, timeStamp: 1562318093467 }],
    result = pipe(
        groupBy('name'),
        select(max('timeStamp'))
    )(data);

console.log(result);
.as-console-wrapper { max-height: 100% !important; top: 0; }

1
投票

这是另一种减少替代方案：

var arr = [{"name":"bathroom","value":54,"timeStamp":1562318089713},{"name":"bathroom","value":55,"timeStamp":1562318090807},{"name":"bedroom","value":48,"timeStamp":1562318092084},{"name":"bedroom","value":49,"timeStamp":1562318092223},{"name":"room","value":41,"timeStamp":1562318093467}];

var obj = arr.reduce((r, o) => (o.value < (r[o.name] || {}).value || (r[o.name] = o), r), {});

console.log( Object.values(obj) );

0
投票

您可以通过保留累积对象来使用 

.reduce()
，该对象保留当前找到的最大组，然后使用 
Object.values()
 来获取这些对象的数组（而不是键值对对象关系）。

参见下面的示例：

const arr=[{name:"bathroom",value:54,timeStamp:1562318089713},{name:"bathroom",value:55,timeStamp:1562318090807},{name:"bedroom",value:48,timeStamp:1562318092084},{name:"bedroom",value:49,timeStamp:1562318092223},{name:"room",value:41,timeStamp:1562318093467}];

const res = Object.values(arr.reduce((acc, o) => {
  acc[o.name] = acc[o.name] || o;
  if (o.value > acc[o.name].value)
    acc[o.name] = o;
  return acc;
}, {}));

console.log(res);

0
投票

分阶段进行。

  1. 获取一组名称
  2. 创建一个按降序排列的数组
  3. 然后使用 find 进行映射以获得第一个

下面是一个例子。

const input = [{"name":"bathroom","value":54,"timeStamp":1562318089713},{"name":"bathroom","value":55,"timeStamp":1562318090807},{"name":"bedroom","value":48,"timeStamp":1562318092084},{"name":"bedroom","value":49,"timeStamp":1562318092223},{"name":"room","value":41,"timeStamp":1562318093467}];

const output = [...new Set(input.map(m => m.name))].
  map(m => [...input].sort(
    (a,b) => b.value - a.value).
    find(x => m === x.name));
  
console.log(output);

0
投票

使用 

map()
和 
foreach()
获得所需的输出

const arr=[{name:"bathroom",value:54,timeStamp:1562318089713}, 
    {name:"bathroom",value:55,timeStamp:1562318090807}, 
    {name:"bedroom",value:48,timeStamp:1562318092084}, 
    {name:"bedroom",value:49,timeStamp:1562318092223}, 
    {name:"room",value:41,timeStamp:1562318093467}];

let res = new Map();
arr.forEach((obj) => {
    let values = res.get(obj.name);
    if(!(values && values.value > obj.value)){ 
        res.set(obj.name, obj) 
    }
})
console.log(res);
console.log([...res])
0
投票

您可以使用排序和过滤方法

const arr = [{"name":"bathroom","value":54,"timeStamp":1562318089713},{"name":"bathroom","value":55,"timeStamp":1562318090807},{"name":"bedroom","value":48,"timeStamp":1562318092084},{"name":"bedroom","value":49,"timeStamp":1562318092223},{"name":"room","value":41,"timeStamp":1562318093467}]
arr.sort(function (a, b) {
    return b.value - a.value;
}).filter((v, i, a) => a.findIndex((v2) => v2.name === v.name) === i);

console.log(arr);

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