通过过滤基于另一个数组的数据创建对象数组

Question

我有如下所示的对象数组

const history = [
  {
    name: "Lance",
    baseline: { risk: "112", age: "45", insurance: "Star" },
  },
  {
    name: "Sam",
    baseline: { risk: "9", age: "21", insurance: "Sigma" },
  },
  {
    name: "Jill",
    baseline: { risk: "15", age: "21", insurance: "Sigma" },
  },
  {
    name: "Bill",
    baseline: { risk: "15", age: "21", insurance: "Mercy" },
  },
];

然后是第二个，如下所示

const current = [
  {
    name: "Lance",
    item: "PS",
    "3-Aug-21": "117",
    "4-Aug-21": "120",
    "5-Aug-21": "112",
  },
  {
    name: "Sam",
    item: "PS",
    "3-Aug-21": "10.14",
    "4-Aug-21": "10.2",
    "5-Aug-21": "10",
  },

  {
    name: "Jill",
    item: "PS",
    "3-Aug-21": "10.14",
    "4-Aug-21": "10.2",
    "5-Aug-21": "10",
  },
  {
    name: "Bill",
    item: "PS",
    "3-Aug-21": "10.14",
    "4-Aug-21": "10.2",
    "5-Aug-21": "10",
  },
];

有人能得到如下的输出吗？

const filteredData= [
{
  insurance: 'Star',
  details:
  {
    name: "Lance",
    values: [
      { date: "3-Aug-21", riskLevel: "104.4" }, //117/112(112 is the baseline risk for Lance)
      { date: "4-Aug-21", riskLevel: "107.1" }, //120/112(112 is the baseline risk for Lance)
      { date: "5-Aug-21", riskLevel: "100" }, //112/112(112 is the baseline risk for Lance)
    ]
  }},
  {insurance: 'Sigma'
  ,
  details:
  {
    name: "Sam",
    values: [
      { date: "3-Aug-21", riskLevel: "112.6" }, //10.14/9(9 is the baseline risk for Sam)
      { date: "4-Aug-21", riskLevel: "113.3" }, //10.2/9(9 is the baseline risk for Sam)
      { date: "5-Aug-21", riskLevel: "111.1" }, //10/9(9 is the baseline risk for Sam)
    ],
name: "Jill",
    values: [
      { date: "3-Aug-21", riskLevel: "112.6" }, //10.14/9(9 is the baseline risk for Sam)
      { date: "4-Aug-21", riskLevel: "113.3" }, //10.2/9(9 is the baseline risk for Sam)
      { date: "5-Aug-21", riskLevel: "111.1" }, //10/9(9 is the baseline risk for Sam)
    ],
  },
},
{insurance: 'Mercy',
details:
  {
    name: "Bill",
    values: [
      { date: "3-Aug-21", riskLevel: "104.4" }, //117/112(112 is the baseline risk for Lance)
      { date: "4-Aug-21", riskLevel: "107.1" }, //120/112(112 is the baseline risk for Lance)
      { date: "5-Aug-21", riskLevel: "100" }, //112/112(112 is the baseline risk for Lance)
    ],
  }
},
];

我也回答了类似的问题。链接-> Using an external object in array map() method

这个问题的不同之处在于，在这里，我试图从历史数组中读取保险，然后一起获得相同保险的患者的值。

基于前面的问题。我有名称和保险之间的地图，代码如下。

const historyMap = new Map(history.map(o => [o.name, o.baseline.insurance]));

然而，我不知道如何进一步使用它。

我得到了下面的代码，他们根据“硬币”对物品进行分类(在我们的例子中，保险&数据也是不同的结构)，但我无法弄清楚。

filteredData = {};
  Object.keys(data).forEach((coin) => {
    filteredData[coin] = data[coin]
      .filter((d) => {
        return !(d["price_usd"] == null);
      })
      .map((d) => {
        d["price_usd"] = Number(d["price_usd"]);
        d["24h_vol"] = Number(d["24h_vol"]);
        d["market_cap"] = Number(d["market_cap"]);
        d["date"] = parseTime(d["date"]);
        return d;
      });
  });

Answer 1

您可以采用两个循环方法，对象作为哈希表。

​

const
    history = [{ name: "Lance", baseline: { risk: "112", age: "45", insurance: "Star" } }, { name: "Sam", baseline: { risk: "9", age: "21", insurance: "Sigma" } }, { name: "Jill", baseline: { risk: "15", age: "21", insurance: "Sigma" } }, { name: "Bill", baseline: { risk: "15", age: "21", insurance: "Mercy" } }],
    current = [{ name: "Lance", item: "PS", "3-Aug-21": "117", "4-Aug-21": "120", "5-Aug-21": "112" }, { name: "Sam", item: "PS", "3-Aug-21": "10.14", "4-Aug-21": "10.2", "5-Aug-21": "10" }, { name: "Jill", item: "PS", "3-Aug-21": "10.14", "4-Aug-21": "10.2", "5-Aug-21": "10" }, { name: "Bill", item: "PS", "3-Aug-21": "10.14", "4-Aug-21": "10.2", "5-Aug-21": "10" }],
    temp = current.reduce((r, { name, item, ...data }) => {
        r[name] = Object
            .entries(data)
            .map(([date, riskLevel]) => ({ date, riskLevel }));
        return r;
    }, {}),
    result = Object.values(history.reduce((r, { name, baseline: { risk, insurance } }) => {
        r[insurance] ??= { insurance, details: [] };
        r[insurance].details.push({
            name,
            values: temp[name].map(({ date, riskLevel }) =>
                ({ date, riskLevel: (100 * riskLevel / risk).toFixed(1) }))
        });
        return r;
    }, {}));
  
console.log(result);

.as-console-wrapper { max-height: 100% !important; top: 0; }

​

Answer 2

我使用reduce()对保险进行分类，并使用Object.values()提取数据。

注:精度我不知道你的要求。我要让你调整一下。我现在使用toFixed(1)。

​

const history = [ { name: "Lance", baseline: { risk: "112", age: "45", insurance: "Star" }, }, { name: "Sam", baseline: { risk: "9", age: "21", insurance: "Sigma" }, }, { name: "Jill", baseline: { risk: "15", age: "21", insurance: "Sigma" }, }, { name: "Bill", baseline: { risk: "15", age: "21", insurance: "Mercy" }, }, ]; 

const current = [ { name: "Lance", item: "PS", "3-Aug-21": "117", "4-Aug-21": "120", "5-Aug-21": "112", }, { name: "Sam", item: "PS", "3-Aug-21": "10.14", "4-Aug-21": "10.2", "5-Aug-21": "10", }, { name: "Jill", item: "PS", "3-Aug-21": "10.14", "4-Aug-21": "10.2", "5-Aug-21": "10", }, { name: "Bill", item: "PS", "3-Aug-21": "10.14", "4-Aug-21": "10.2", "5-Aug-21": "10", }, ];

const categorizedInsurance = history.reduce((acc, h) => {
  const cur = current.find(c => c.name === h.name);
  const insurance = h.baseline.insurance;
  const keys = Object.keys(cur).filter(key => key !== "name" && key !== "item");
  const values = keys.map(key => ({
    date: key,
    riskLevel: ((cur[key] / h.baseline.risk) * 100).toFixed(1),
  }));

  if (acc[insurance]) {
    acc[insurance].details.push({
      name: h.name,
      values: values,
    });
  } else {
    acc[insurance] = {
      insurance: insurance,
      details: [
        {
          name: h.name,
          values: values,
        },
      ],
    };
  }

  return acc;
}, {});

const output = Object.values(categorizedInsurance);

console.log(output);

​

Answer 3

使用filteredData的语法检查。但是，似乎每个details数组对象中的filteredData节点必须是一个数组，因为有多个以键作为名称的节点。

逻辑

• 循环通过current数组。
• 为current数组中的每个对象找到一个来自具有相同名称的history数组的匹配节点。
• 用于current数组中的每个节点，循环遍历键。除了dates.
• Create和item以外的键应该是name，一个detail对象，它从历史数组中以name作为匹配的名称，并为每个日期的值加上risk.
• Push的计算值。详细信息推到accumulator array.
• Before，验证在accumulator数组中是否存在一个具有相同保险名称的节点。如果已经存在节点，则将新创建的详细信息对象推送到该节点的详细信息数组，或者将一个新节点推送到accumulator数组。

工作示例

​

const history = [
  {
    name: "Lance",
    baseline: { risk: "112", age: "45", insurance: "Star" },
  },
  {
    name: "Sam",
    baseline: { risk: "9", age: "21", insurance: "Sigma" },
  },
  {
    name: "Jill",
    baseline: { risk: "15", age: "21", insurance: "Sigma" },
  },
  {
    name: "Bill",
    baseline: { risk: "15", age: "21", insurance: "Mercy" },
  },
];
const current = [
  {
    name: "Lance",
    item: "PS",
    "3-Aug-21": "117",
    "4-Aug-21": "120",
    "5-Aug-21": "112",
  },
  {
    name: "Sam",
    item: "PS",
    "3-Aug-21": "10.14",
    "4-Aug-21": "10.2",
    "5-Aug-21": "10",
  },

  {
    name: "Jill",
    item: "PS",
    "3-Aug-21": "10.14",
    "4-Aug-21": "10.2",
    "5-Aug-21": "10",
  },
  {
    name: "Bill",
    item: "PS",
    "3-Aug-21": "10.14",
    "4-Aug-21": "10.2",
    "5-Aug-21": "10",
  },
];

const output = current.reduce((acc, curr) => {
  // Find the node from the array `history` having the same name from array `current`
  const historyNode = history.find(node => node.name === curr.name);
  const details = [];
  const detail = {
    name: historyNode.name,
    values: [],
  }
  // Loop through each keys in the object `curr`
  // If the node is not `name` and `item` then defenitely its a date
  Object.keys(curr).forEach((key) => {
    if (key !== 'name' && key !== 'item') {
      // Push the date values against the name
      detail.values.push({
        date: key,
        risk: (+curr[key] / +historyNode.baseline.risk * 100).toFixed(1)
      })
    }
  })
  details.push(detail)

  // Find for a node an insurance same as that we found match from the history array.
  // If there is a match found, then this details must be pushed to the `details` array of that particular node
  const accumulatorNode = acc.find((node) => node.insurance === historyNode.baseline.insurance);
  if (accumulatorNode) {
    accumulatorNode.details.push(detail);
  } else {
    // If there is no node existing with same insurance name, then we need to insert a new object to the accumulator
    acc.push({
      insurance: historyNode.baseline.insurance,
      details,
    })
  }
  return acc;
}, []);
console.log(output)

​