按数组中的对象分组最有效的方法是什么?

例如,给定此对象数组:

[ 
    { Phase: "Phase 1", Step: "Step 1", Task: "Task 1", Value: "5" },
    { Phase: "Phase 1", Step: "Step 1", Task: "Task 2", Value: "10" },
    { Phase: "Phase 1", Step: "Step 2", Task: "Task 1", Value: "15" },
    { Phase: "Phase 1", Step: "Step 2", Task: "Task 2", Value: "20" },
    { Phase: "Phase 2", Step: "Step 1", Task: "Task 1", Value: "25" },
    { Phase: "Phase 2", Step: "Step 1", Task: "Task 2", Value: "30" },
    { Phase: "Phase 2", Step: "Step 2", Task: "Task 1", Value: "35" },
    { Phase: "Phase 2", Step: "Step 2", Task: "Task 2", Value: "40" }
]

我正在表格中显示这些信息。我想通过不同的方法进行分组,但我想对值求和。

我将Undercore.js用于其groupby函数,这很有用,但并不能完成全部任务,因为我不希望它们“拆分”,而是“合并”,更像SQL groupby方法。

我要找的是能够合计特定值(如果需要)。

因此,如果我按阶段分组,我希望收到:

[
    { Phase: "Phase 1", Value: 50 },
    { Phase: "Phase 2", Value: 130 }
]

如果我组了阶段/步骤,我会收到:

[
    { Phase: "Phase 1", Step: "Step 1", Value: 15 },
    { Phase: "Phase 1", Step: "Step 2", Value: 35 },
    { Phase: "Phase 2", Step: "Step 1", Value: 55 },
    { Phase: "Phase 2", Step: "Step 2", Value: 75 }
]

是否有一个有用的脚本,或者我应该坚持使用Undercore.js,然后遍历生成的对象,自己计算总数?


当前回答

为了补充Scott Sauyet的答案,一些人在评论中询问如何使用他的函数按值1、值2等分组,而不是仅对一个值分组。

只需编辑他的求和函数:

DataGrouper.register("sum", function(item) {
    return _.extend({}, item.key,
        {VALUE1: _.reduce(item.vals, function(memo, node) {
        return memo + Number(node.VALUE1);}, 0)},
        {VALUE2: _.reduce(item.vals, function(memo, node) {
        return memo + Number(node.VALUE2);}, 0)}
    );
});

保持主组(DataGrouper)不变:

var DataGrouper = (function() {
    var has = function(obj, target) {
        return _.any(obj, function(value) {
            return _.isEqual(value, target);
        });
    };

    var keys = function(data, names) {
        return _.reduce(data, function(memo, item) {
            var key = _.pick(item, names);
            if (!has(memo, key)) {
                memo.push(key);
            }
            return memo;
        }, []);
    };

    var group = function(data, names) {
        var stems = keys(data, names);
        return _.map(stems, function(stem) {
            return {
                key: stem,
                vals:_.map(_.where(data, stem), function(item) {
                    return _.omit(item, names);
                })
            };
        });
    };

    group.register = function(name, converter) {
        return group[name] = function(data, names) {
            return _.map(group(data, names), converter);
        };
    };

    return group;
}());

其他回答

var arr = [ 
    { Phase: "Phase 1", `enter code here`Step: "Step 1", Task: "Task 1", Value: "5" },
    { Phase: "Phase 1", Step: "Step 1", Task: "Task 2", Value: "10" },
    { Phase: "Phase 1", Step: "Step 2", Task: "Task 1", Value: "15" },
    { Phase: "Phase 1", Step: "Step 2", Task: "Task 2", Value: "20" },
    { Phase: "Phase 2", Step: "Step 1", Task: "Task 1", Value: "25" },
    { Phase: "Phase 2", Step: "Step 1", Task: "Task 2", Value: "30" },
    { Phase: "Phase 2", Step: "Step 2", Task: "Task 1", Value: "35" },
    { Phase: "Phase 2", Step: "Step 2", Task: "Task 2", Value: "40" }
];

创建并清空对象。循环遍历arr并添加使用Phase作为obj的唯一键。在循环遍历arr时,保持更新obj中的键总数。

const obj = {};
arr.forEach((item) => {
  obj[item.Phase] = obj[item.Phase] ? obj[item.Phase] + 
  parseInt(item.Value) : parseInt(item.Value);
});

结果如下:

{ "Phase 1": 50, "Phase 2": 130 }

循环通过obj形成表单和resultArr。

const resultArr = [];
for (item in obj) {
  resultArr.push({ Phase: item, Value: obj[item] });
}
console.log(resultArr);

我不认为给出的答案是对问题的回应,我认为以下内容应回答第一部分:

常量arr=[{阶段:“阶段1”,步骤:“步骤1”,任务:“任务1”,值:“5”},{阶段:“阶段1”,步骤:“步骤1”,任务:“任务2”,值:“10”},{阶段:“阶段1”,步骤:“步骤2”,任务:“任务1”,值:“15”},{阶段:“阶段1”,步骤:“步骤2”,任务:“任务2”,值:“20”},{阶段:“阶段2”,步骤:“步骤1”,任务:“任务1”,值:“25”},{阶段:“阶段2”,步骤:“步骤1”,任务:“任务2”,值:“30”},{阶段:“阶段2”,步骤:“步骤2”,任务:“任务1”,值:“35”},{阶段:“阶段2”,步骤:“步骤2”,任务:“任务2”,值:“40”}]const groupBy=(key)=>arr.sort((a,b)=>a[key].localeCompare(b[key])).reduce((total,currentValue)=>{const newTotal=总计;如果(总长度&&总计[total.length-1][key]==当前值[key])新总计[total.length-1]={…总计[总长度-1],…当前值,值:parseInt(total[total.length-1].Value)+parseInt(currentValue.Value,};else newTotal[total.length]=当前值;return newTotal;}, []);console.log(groupBy(“阶段”));//=>[{阶段:“阶段1”,值:50},{阶段“阶段2”,值130}]console.log(groupBy(“步骤”));//=>[{步骤:“步骤1”,值:70},{步骤“步骤2”,值为110}]

具有排序功能

export const groupBy = function groupByArray(xs, key, sortKey) {
      return xs.reduce(function(rv, x) {
        let v = key instanceof Function ? key(x) : x[key];
        let el = rv.find(r => r && r.key === v);

        if (el) {
          el.values.push(x);
          el.values.sort(function(a, b) {
            return a[sortKey].toLowerCase().localeCompare(b[sortKey].toLowerCase());
          });
        } else {
          rv.push({ key: v, values: [x] });
        }

        return rv;
      }, []);
    };

示例:

var state = [
    {
      name: "Arkansas",
      population: "2.978M",
      flag:
  "https://upload.wikimedia.org/wikipedia/commons/9/9d/Flag_of_Arkansas.svg",
      category: "city"
    },{
      name: "Crkansas",
      population: "2.978M",
      flag:
        "https://upload.wikimedia.org/wikipedia/commons/9/9d/Flag_of_Arkansas.svg",
      category: "city"
    },
    {
      name: "Balifornia",
      population: "39.14M",
      flag:
        "https://upload.wikimedia.org/wikipedia/commons/0/01/Flag_of_California.svg",
      category: "city"
    },
    {
      name: "Florida",
      population: "20.27M",
      flag:
        "https://upload.wikimedia.org/wikipedia/commons/f/f7/Flag_of_Florida.svg",
      category: "airport"
    },
    {
      name: "Texas",
      population: "27.47M",
      flag:
        "https://upload.wikimedia.org/wikipedia/commons/f/f7/Flag_of_Texas.svg",
      category: "landmark"
    }
  ];
console.log(JSON.stringify(groupBy(state,'category','name')));
groupByArray(xs, key) {
    return xs.reduce(function (rv, x) {
        let v = key instanceof Function ? key(x) : x[key];
        let el = rv.find((r) => r && r.key === v);
        if (el) {
            el.values.push(x);
        }
        else {
            rv.push({
                key: v,
                values: [x]
            });
        }
        return rv;
    }, []);
}

这一个输出数组。

Ceasar的答案很好,但只适用于数组中元素的内部财产(字符串的长度)。

这个实现的工作方式更像:这个链接

const groupBy = function (arr, f) {
    return arr.reduce((out, val) => {
        let by = typeof f === 'function' ? '' + f(val) : val[f];
        (out[by] = out[by] || []).push(val);
        return out;
    }, {});
};

希望这有帮助。。。