使用两个条件在 JavaScript 中聚合数据

问题描述 投票：0回答：1

我一直在尝试汇总数据以满足两个标准。

假设我的初始数组 (

rawData
) 如下：

rawData = [
  { date: '2023-11-01', fruit: 'oranges', state: 'NY', salePct: 0.10 },
  { date: '2023-11-01', fruit: 'apples', state: 'NY', salePct: 0.25 },
  { date: '2023-11-01', fruit: 'pears', state: 'NY', salePct: 0.10 },
  { date: '2023-11-01', fruit: 'figs', state: 'NY', salePct: 0.01 },
  { date: '2023-11-01', fruit: 'strawberries', state: 'NY', salePct: 0.15 },
  { date: '2023-11-01', fruit: 'raspberries', state: 'NY', salePct: 0.15 },
  { date: '2023-11-01', fruit: 'blueberries', state: 'NY', salePct: 0.15 },
  { date: '2023-11-01', fruit: 'cranberries', state: 'NY', salePct: 0.02 },
  { date: '2023-11-01', fruit: 'pineapples', state: 'NY', salePct: 0.02 },
  { date: '2023-11-01', fruit: 'peaches', state: 'NY', salePct: 0.05 },
  { date: '2023-11-02', fruit: 'oranges', state: 'NY', salePct: 0.20 },
  { date: '2023-11-02', fruit: 'apples', state: 'NY', salePct: 0.20 },
  { date: '2023-11-02', fruit: 'pears', state: 'NY', salePct: 0.10 },
  { date: '2023-11-02', fruit: 'figs', state: 'NY', salePct: 0.10 },
  { date: '2023-11-02', fruit: 'strawberries', state: 'NY', salePct: 0.20 },
  { date: '2023-11-02', fruit: 'raspberries', state: 'NY', salePct: 0.10 },
  { date: '2023-11-02', fruit: 'blueberries', state: 'NY', salePct: 0.03 },
  { date: '2023-11-02', fruit: 'cranberries', state: 'NY', salePct: 0.03 },
  { date: '2023-11-02', fruit: 'pineapples', state: 'NY', salePct: 0.04},
  { date: '2023-11-03', fruit: 'oranges', state: 'NY', salePct: 0.20 },
  { date: '2023-11-03', fruit: 'apples', state: 'NY', salePct: 0.20 },
  { date: '2023-11-03', fruit: 'pears', state: 'NY', salePct: 0.10 },
  { date: '2023-11-03', fruit: 'figs', state: 'NY', salePct: 0.10 },
  { date: '2023-11-03', fruit: 'strawberries', state: 'NY', salePct: 0.30 },
  { date: '2023-11-03', fruit: 'raspberries', state: 'NY', salePct: 0.05 },
  { date: '2023-11-03', fruit: 'blueberries', state: 'NY', salePct: 0.05 },
  { date: '2023-11-05', fruit: 'oranges', state: 'NY', salePct: 0.10 },
  { date: '2023-11-05', fruit: 'apples', state: 'NY', salePct: 0.10 },
  { date: '2023-11-05', fruit: 'pears', state: 'NY', salePct: 0.10 },
  { date: '2023-11-05', fruit: 'figs', state: 'NY', salePct: 0.10 },
  { date: '2023-11-05', fruit: 'strawberries', state: 'NY', salePct: 0.10 },
  { date: '2023-11-05', fruit: 'raspberries', state: 'NY', salePct: 0.25 },
  { date: '2023-11-05', fruit: 'blueberries', state: 'NY', salePct: 0.25 },
];

我希望将可用日期的销售百分比低于所有可用日期的最小值的水果聚合为 

others
。因此，对于 
minSalePct = 0.10
，
aggregatedByThreshold
数组如下：

aggregatedByThreshold = [
  { date: '2023-11-01', fruit: 'oranges', state: 'NY', salePct: 0.10 },
  { date: '2023-11-01', fruit: 'apples', state: 'NY', salePct: 0.25 },
  { date: '2023-11-01', fruit: 'pears', state: 'NY', salePct: 0.10 },
  { date: '2023-11-01', fruit: 'strawberries', state: 'NY', salePct: 0.15 },
  { dat3: '2023-11-01', fruit: 'others', state: 'NY', salePct: 0.40},
  { date: '2023-11-02', fruit: 'oranges', state: 'NY', salePct: 0.20 },
  { date: '2023-11-02', fruit: 'apples', state: 'NY', salePct: 0.20 },
  { date: '2023-11-02', fruit: 'pears', state: 'NY', salePct: 0.10 },
  { date: '2023-11-02', fruit: 'strawberries', state: 'NY', salePct: 0.20 },
  { date: '2023-11-02', fruit: 'others', state: 'NY', salePct: 0.30 },
  { date: '2023-11-03', fruit: 'oranges', state: 'NY', salePct: 0.20 },
  { date: '2023-11-03', fruit: 'apples', state: 'NY', salePct: 0.20 },
  { date: '2023-11-03', fruit: 'pears', state: 'NY', salePct: 0.10 },
  { date: '2023-11-03', fruit: 'strawberries', state: 'NY', salePct: 0.30 },
  { date: '2023-11-03', fruit: 'others', state: 'NY', salePct: 0.10 },
  { date: '2023-11-05', fruit: 'oranges', state: 'NY', salePct: 0.10 },
  { date: '2023-11-05', fruit: 'apples', state: 'NY', salePct: 0.10 },
  { date: '2023-11-05', fruit: 'pears', state: 'NY', salePct: 0.10 },
  { date: '2023-11-05', fruit: 'strawberries', state: 'NY', salePct: 0.10 },
  { date: '2023-11-05', fruit: 'others', state: 'NY', salePct: 0.50 },
];

但是，非聚合水果的最大数量 (

maxFruit
) 也需要强制执行。各日期销售百分比总和较低的水果将逐步添加到 
others
，直到满足 
maxFruit
。如果 
maxFruit = 3
，则 
aggregatedByThresholdAndQuantity
数组将如下所示：

aggregatedByThresholdAndQuantity = [
  { date: '2023-11-01', fruit: 'oranges', state: 'NY', salePct: 0.10 },
  { date: '2023-11-01', fruit: 'apples', state: 'NY', salePct: 0.25 },
  { date: '2023-11-01', fruit: 'strawberries', state: 'NY', salePct: 0.15 },
  { dat3: '2023-11-01', fruit: 'others', state: 'NY', salePct: 0.50},
  { date: '2023-11-02', fruit: 'oranges', state: 'NY', salePct: 0.20 },
  { date: '2023-11-02', fruit: 'apples', state: 'NY', salePct: 0.20 },
  { date: '2023-11-02', fruit: 'strawberries', state: 'NY', salePct: 0.20 },
  { date: '2023-11-02', fruit: 'others', state: 'NY', salePct: 0.40 },
  { date: '2023-11-03', fruit: 'oranges', state: 'NY', salePct: 0.20 },
  { date: '2023-11-03', fruit: 'apples', state: 'NY', salePct: 0.20 },
  { date: '2023-11-03', fruit: 'strawberries', state: 'NY', salePct: 0.30 },
  { date: '2023-11-03', fruit: 'others', state: 'NY', salePct: 0.20 },
  { date: '2023-11-05', fruit: 'oranges', state: 'NY', salePct: 0.10 },
  { date: '2023-11-05', fruit: 'apples', state: 'NY', salePct: 0.10 },
  { date: '2023-11-05', fruit: 'strawberries', state: 'NY', salePct: 0.10 },
  { date: '2023-11-05', fruit: 'others', state: 'NY', salePct: 0.60 },
];

此外，我想要一个数组，其中包含与其他水果一起聚合的水果名称：

otherFruit = ['pears', 'figs', 'raspberries', 'cranberries', 'blueberries', 'pineapples', 'peaches'];

这些解决方案解决了部分问题，但我无法将它们结合起来 生成所需的输出：

尽管我使用了两个数组来解释场景（

aggregatedByThreshold
和
aggregatedByThresholdAndQuantity
），但解决方案不需要中间步骤。 
aggregatedByThresholdAndQuantity
和
otherFruit
，并且设置两个标准（
minSalePct
和
maxFruit
）的可能性就足够了。我也同意使用其他库（例如 lodash）的解决方案。

谢谢！

丹妮拉

javascript aggregate data-wrangling
1个回答
0
投票

如果效率不重要，您可以简单地循环列表两次，首先确定顶部

maxFruit
水果列表中的水果，然后在下一个循环中进行实际聚合：

const maxFruit = 3;
const minSalePct = 0.10;

//find top N SalePct fruits
totalSalePct = {}
for (const item of rawData) {
  if (item["fruit"] in totalSalePct === false)
    totalSalePct[item["fruit"]] = 0;
  totalSalePct[item["fruit"]] += item["salePct"];
}
const topNFruites = Object.keys(totalSalePct).sort((a,b) => totalSalePct[b]- totalSalePct[a]).slice(0,maxFruit);

//aggregate
let otherFruit = new Set();
let result = {}
for (const {date, fruit, state, salePct} of rawData) {
  if(date in result === false){
      result[date] = {}
  }
  const isOther = salePct < minSalePct || !topNFruites.includes(fruit);
  const name = isOther? "others" : fruit;
  if(name in result[date] === false){
    result[date][name] = {"state":state, "salePct":0.0}
  }
  result[date][name]["salePct"] += salePct;
  if(isOther)
    otherFruit.add(fruit);
}

//convert result to original list format
let resultList = []
for (const [date, fruits] of Object.entries(result)) {
  for (const [fruit, item] of Object.entries(fruits)) {
    resultList.push({
      "date": date,
      "fruit": fruit,
      "state": item["state"],
      "salePct": item["salePct"].toFixed(2)
    })
  }
}

console.log(resultList)
console.log(otherFruit)
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