Question

我下面有对象。

car = [{
    Acc: "X1",
    primary: true,
    Name: "Park Street",
    Date: "2018/01/01"
},{
    Acc: "X1",
    primary: false,
    Name: "Park Street 2",
    Date: "2018/03/01"
},{
    Acc: "X2",
    primary: false,
    Name: "Park Street 3",
    Date: "2018/01/01"
},{
    Acc: "X2",
    primary: false,
    Name: "Park Street 4",
    Date: "2018/05/01"
}];

我只需要从“ Acc”的每个组中获取一个元素。 Acc：“ X1”和Acc：“ X2”中的每一个元素。逻辑上找到应该为primary：true的逻辑，然后从组中选择该元素，否则选择具有最新Date：值的元素。

我确实知道使用下划线js可以实现此要求，但我是新手，无法弄清楚相关方法。

有人可以在这里帮助我吗？

Answer 1

执行此操作的一种方法是，首先按Acc进行分组，然后在组中按primary和Date降序排序，然后仅选择每个组中的第一个。 / p>

例如

const car = [{
    Acc: "X1",
    primary: true,
    Name: "Park Street",
    Date: "2018/01/01"
},{
    Acc: "X1",
    primary: false,
    Name: "Park Street 2",
    Date: "2018/03/01"
},{
    Acc: "X2",
    primary: false,
    Name: "Park Street 3",
    Date: "2018/01/01"
},{
    Acc: "X2",
    primary: false,
    Name: "Park Street 4",
    Date: "2018/05/01"
}];


//lets group
const grouped = car.reduce((a, v) => {
  if (!a[v.Acc]) a[v.Acc] = [];
  a[v.Acc].push(v);
  return a;
}, {});

//now for each group sort by primary, and then Date
const result = Object.values(grouped).map((g) => {
  g.sort((a, b) => b.primary - a.primary || b.Date.localeCompare(a.Date));
  return g[0]; //now sorted, pick first.
});

console.log(result);

另一种更有效的方法可能只是跟踪对象文字中的最佳内容。

例如。

const car = [{
    Acc: "X1",
    primary: true,
    Name: "Park Street",
    Date: "2018/01/01"
},{
    Acc: "X1",
    primary: false,
    Name: "Park Street 2",
    Date: "2018/03/01"
},{
    Acc: "X2",
    primary: false,
    Name: "Park Street 3",
    Date: "2018/01/01"
},{
    Acc: "X2",
    primary: false,
    Name: "Park Street 4",
    Date: "2018/05/01"
}];

const best = {};

car.forEach((c) => {
  const b = best[c.Acc] = best[c.Acc] || c;
  if (b.primary) return;
  if (c.Date.localeCompare(best.Date) < 0) best[c.Acc] = c;
});

console.log(Object.values(best));

Answer 2

由于您提到了Underscore，所以我了解您可以使用外部库。此方法与@Keith相同，只是使用Lodash。

此外，这应该适用于所有浏览器。

var car = [{
    Acc: "X1",
    primary: true,
    Name: "Park Street",
    Date: "2018/01/01"
},{
    Acc: "X1",
    primary: false,
    Name: "Park Street 2",
    Date: "2018/03/01"
},{
    Acc: "X2",
    primary: false,
    Name: "Park Street 3",
    Date: "2018/01/01"
},{
    Acc: "X2",
    primary: false,
    Name: "Park Street 4",
    Date: "2018/05/01"
}];

var sorted = _.orderBy(car, ['primary', 'Date'], ['desc', 'desc']);
var groupedSorted = _.groupBy(sorted, 'Acc');
var result = _.flatMap(groupedSorted, function(o) { return o[0]; });
console.log(result);

<script src="https://cdnjs.cloudflare.com/ajax/libs/lodash.js/4.17.10/lodash.min.js"></script>

Answer 3

如果primary：true，则从组中选择该元素，否则选择具有最新Date：值的元素

根据上述说明，每个组（primary:true或X1）中只有一个X2。如果我的理解是正确的，则可以尝试使用以下代码：

// Input object
var car = [{
	Acc: "X1",
	primary: true,
	Name: "Park Street",
	Date: "2018/01/01"
}, {
	Acc: "X1",
	primary: false,
	Name: "Park Street 2",
	Date: "2018/03/01"
}, {
	Acc: "X2",
	primary: false,
	Name: "Park Street 3",
	Date: "2018/01/01"
}, {
	Acc: "X2",
	primary: false,
	Name: "Park Street 4",
	Date: "2018/05/01"
}];

// fetch the primary:true objects from the Input array.
var primaryTrueArray = car.filter(obj => {
  if (obj.primary === true) {
    return obj;
  }
});

// filtered out the primary:true objects from the Input array based on the primaryTrueArray.Hence, we will get all primary:false.
var filteredOutPrimaryTrue = car.filter(obj => primaryTrueArray.filter(item => item.Acc != obj.Acc));

// Fetch the latest date from the group who has primary:false from the filteredOutPrimaryTrue.
var latestDateFromGroup = filteredOutPrimaryTrue.sort(function(a,b){
                                   if (a.Date < b.Date)
                                      return 1;
                                   else if (a.Date == b.Date)
                                      return 0;
                                   else
                                      return -1;
                               });
                               
// Concatenate both the arrays "primaryTrueArray & latestDateFromGroup" into single one.                           
var finalResult = primaryTrueArray.concat([latestDateFromGroup[0]]); 

// Desired output.
console.log(finalResult);

分组，然后过滤

3 个答案: