Question

在工作中，我被赋予了以下任务：将数组中具有相似属性的元素分组。

通常，问题如下：

var order = [
    {
        "tot":1,
        "ru":"R401",
        "area":"RFCC",
        "unit":"OFFSITE",
        "equipment":"37 P 552 A",
        "parameter":"Speed"
    }, {
        "tot":1,
        "ru":"R401",
        "area":"RFCC",
        "unit":"OFFSITE",
        "equipment":"37 P 552 A",
        "parameter":"Discharge pressure"
    }, {
        "tot":1,
        "ru":"R401",
        "area":"RFCC",
        "unit":"OFFSITE",
        "equipment":"37 P 552 A",
        "parameter":"Speed"
    }, {
        "tot":1,
        "ru":"R401",
        "area":"RFCC",
        "unit":"RCU",
        "equipment":"37 P 552 B",
        "parameter":"Discharge pressure"
    }
]

如果我按“ ru”，“ area”，“ unit”，“ tot”，“ equipment”和“ parameter”对这些元素进行分组，则会得到以下结果：

var result = [
    {
        "ru":"R401",
        "area":"RFCC",
        "unit":"OFFSITE",
        "tot":2,
        "equipment":"37 P 552 A",
        "parameter":"Speed"
    }, {
        "ru":"R401",
        "area":"RFCC",
        "unit":"OFFSITE",
        "tot":1,
        "equipment":"37 P 552 A",
        "parameter":"Discharge Pressure"
    }, {
        "ru":"R401",
        "area":"RFCC",
        "unit":"RCU",
        "tot":1,
        "equipment":"37 P 552 B",
        "parameter":"Discharge Pressure"
    }];

经过一些实验，我来到了以下代码：

  var groups = _.groupBy(order, function(value) {
    return value.ru + "#" + value.area + "#" + value.unit + "#" + value.equipment + "#" + value.parameter + "#";
  });

  groups = _.map(groups, function(group) {
    return _.extend(group[0], {tot: group.length});
  });

现在我在如何对相似结果元素进行分组方面遇到问题。如果周围有人，请帮助我。预先感谢。

Answer 1

如何？

var order = [
    {
        "tot":1,
        "ru":"R401",
        "area":"RFCC",
        "unit":"OFFSITE",
        "equipment":"37 P 552 A",
        "parameter":"Speed"
    }, {
        "tot":1,
        "ru":"R401",
        "area":"RFCC",
        "unit":"OFFSITE",
        "equipment":"37 P 552 A",
        "parameter":"Discharge pressure"
    }, {
        "tot":1,
        "ru":"R401",
        "area":"RFCC",
        "unit":"OFFSITE",
        "equipment":"37 P 552 A",
        "parameter":"Speed"
    }, {
        "tot":1,
        "ru":"R401",
        "area":"RFCC",
        "unit":"RCU",
        "equipment":"37 P 552 B",
        "parameter":"Discharge pressure"
    }
];
var result = [];
order.forEach(current=>{
    let index = -1;
    result.forEach((c,i)=>{
        if (c.ru==current.ru && c.area==current.area && c.unit==current.unit && c.tot==current.tot && c.equipment==current.equipment && c.parameter==current.parameter){
            index = i;
        }
    });
    // console.log(index);
    if(index==-1){
        result.push(current);
    }else{
        result[index]["tot"] = result[index]["tot"]+current["tot"];
    }
});
console.log(result);

Answer 2

如果遇到类似的元素，也许可以避免分组并直接更新“索引”地图/对象中合并对象的tot属性：

这是一个使用reduce来实现的解决方案：

const order = [
    {
        "tot":1,
        "ru":"R401",
        "area":"RFCC",
        "unit":"OFFSITE",
        "equipment":"37 P 552 A",
        "parameter":"Speed"
    }, {
        "tot":1,
        "ru":"R401",
        "area":"RFCC",
        "unit":"OFFSITE",
        "equipment":"37 P 552 A",
        "parameter":"Discharge pressure"
    }, {
        "tot":1,
        "ru":"R401",
        "area":"RFCC",
        "unit":"OFFSITE",
        "equipment":"37 P 552 A",
        "parameter":"Speed"
    }, {
        "tot":1,
        "ru":"R401",
        "area":"RFCC",
        "unit":"RCU",
        "equipment":"37 P 552 B",
        "parameter":"Discharge pressure"
    }
]

const merged = Object.values(order.reduce((acc, curr) => {
  const key = `${curr.ru}#${curr.area}#${curr.unit}#${curr.equipment}#${curr.parameter}`;
  if (key in acc) acc[key].tot += 1;
  else acc[key] = Object.assign({}, curr);
  return acc;
}, {}));

console.log(merged);
console.log(merged.map(d => d.ru));

通过多个属性下划线对数组中的元素进行分组

2 个答案: