Question

我有这样的JSON：

[
  {
     platformId: 1,
     payout: 15,
     numOfPeople: 4
  },
  {
     platformId: 1,
     payout: 12,
     numOfPeople: 3

  },
  {
     platformId: 2,
     payout: 6,
     numOfPeople: 5

  },
  {
     platformId: 2,
     payout: 10,
     numOfPeople: 1
  },

]

我希望按platformId与payout和numOfPeople之和进行分组。

即。结果我想要这样的JSON：

[
  "1": {
     payout: 27,
     numOfPeople: 7
   },

  "2": {
     payout: 16,
     numOfPeople: 6
  }
]

我尝试使用underscore.js的{{1}}方法，并且它组合得很好，但我如何才能获得上面演示的对象属性值的SUM？

Answer 1

这是解决此类问题的Lodash解决方案。它与Underscore类似，但具有一些更高级的功能。

const data = [{
    platformId: 1,
    payout: 15,
    numOfPeople: 4
  },
  {
    platformId: 1,
    payout: 12,
    numOfPeople: 3

  },
  {
    platformId: 2,
    payout: 6,
    numOfPeople: 5

  },
  {
    platformId: 2,
    payout: 10,
    numOfPeople: 1
  },
];

const ans = _(data)
  .groupBy('platformId')
  .map((platform, id) => ({
    platformId: id,
    payout: _.sumBy(platform, 'payout'),
    numOfPeople: _.sumBy(platform, 'numOfPeople')
  }))
  .value()

console.log(ans);

<script src="https://cdn.jsdelivr.net/lodash/4.17.4/lodash.min.js"></script>

Answer 2

我尝试以功能方式执行此操作，结果就像这样

console.log(_.chain(data)
    .groupBy("platformId")
    .map(function(value, key) {
        return [key, _.reduce(value, function(result, currentObject) {
            return {
                payout: result.payout + currentObject.payout,
                numOfPeople: result.numOfPeople + currentObject.numOfPeople
            }
        }, {
            payout: 0,
            numOfPeople: 0
        })];
    })
    .object()
    .value());

Answer 3

您可以在没有下划线的情况下执行此操作：

var result = data.reduce(function(acc, x) {
  var id = acc[x.platformId]
  if (id) {
    id.payout += x.payout
    id.numOfPeople += x.numOfPeople
  } else {
    acc[x.platformId] = x
    delete x.platformId
  }
  return acc
},{})

但是为什么你想要一个带数字键的对象？您可以将其转换回集合：

var toCollection = function(obj) {
  return Object.keys(obj)
    .sort(function(x, y){return +x - +y})
    .map(function(k){return obj[k]})
}

toCollection(result)

请注意，对象是变异的，因此如果要维护原始数据，可以先将它们克隆。

Answer 4

A（在我看来）使用underscore.js的很好的功能方式：

var sum = function(t, n) { return t + n; };
_.mapObject(
    _.groupBy(data, 'platformId'),
    function(values, platformId) {
        return {
            payout: _.reduce(_.pluck(values, 'payout'), sum, 0),
            numOfPeople: _.reduce(_.pluck(values, 'numOfPeople'), sum, 0)
        };
    }
);

Answer 5

本着lodash的眼光

_.map(_.groupBy(your_list, 'group_by'), (o,idx) => { return { id: idx, summed: _.sumBy(o,'sum_me') }})

Answer 6

//Input 
[{
  "platformId": 1,
  "payout": 15,
  "numOfPeople": 4
}, {
  "platformId": 1,
  "payout": 12,
  "numOfPeople": 3
}, {
  "platformId": 2,
  "payout": 6,
  "numOfPeople": 5
}, {
  "platformId": 2,
  "payout": 10,
  "numOfPeople": 1
}]


//Output
{
 "1": {
    "payout": 43,
    "numOfPeople": 13
 },
 "2": {
    "payout": 43,
    "numOfPeople": 13
 }
}


// Code
output = _.reduce(input, function(acc, val, key) {
  acc[val.platformId] = {
     payout: _.sum(a, 'payout'),
     numOfPeople: _.sum(a, 'numOfPeople')
   };
  return acc;
  }, {});

使用Underscore / Lodash分组By和Sum

6 个答案: