减少每个不同对象的对象数组和sum属性 [英] Reduce an array of objects and sum property for each distinct object

问题描述

我不确定我应该考虑哪些功能来实现我想要做的事情。可能减少是不正确的。

I am not sure for what functions I should look at to achieve what I am trying to do. Probably reduce is not correct.

数据库查询返回一个像这样的对象列表：

A database query returns a list of objects like this one:

result = [{group1: 'A', group2: 'A', SUM: 5},
           {group1: 'A', group2: 'B', SUM: 2},
           {group1: 'C', groupe2: 'B', SUM: 3}
         ]

我想减少或分组这个数组，以获得每个不同的 group1 和 group2 值和它的SUM相对值如下所示：

I want to "reduce" or "group" this array to obtain an object for each distinct group1 and group2 values and the SUM relative of it like this object below:

wanted = [{group1: 'A', group1_SUM: 7, group2: 'A', group2_SUM: 5},
          {group1: 'B', group1_SUM: 0, group2: 'B', group2_SUM: 5},
          {group1: 'C', group1_SUM: 3, group2: 'C', group2_SUM: 0}
         ]

或也可能是：

wanted = [{groupName: 'A', group1_SUM: 7, group2_SUM: 5},
          {groupName: 'B', group1_SUM: 0, group2_SUM: 5},
          {groupName: 'C', group1_SUM: 3, group2_SUM: 0}
         ]

推荐答案

第一个函数减少结果以查找每个组的总和。

First function reduces result to find the sums for each group.

然后我们通过每个组返回结果。

Then we go through each group to return the results.

var result = [{group1: 'A', group2: 'A', SUM: 5},
  {group1: 'A', group2: 'B', SUM: 2},
  {group1: 'C', group2: 'B', SUM: 3}
];

(function groupSum(result) {
  var sums = result.reduce(function(a, e) {
    a.group1_SUM[e.group1] = (a.group1_SUM[e.group1] || 0) + e.SUM;
    a.group2_SUM[e.group2] = (a.group2_SUM[e.group2] || 0) + e.SUM;
    return a;
  }, {group1_SUM: {}, group2_SUM: {}});

  return Array.from(new Set(Object.keys(sums.group1_SUM).concat(Object.keys(sums.group2_SUM)))).map(function(e) {
    return {
      groupName: e, group1_SUM: sums.group1_SUM[e] || 0, group2_SUM: sums.group2_SUM[e] || 0
    }
  });
})(result);

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