javascript - 使用map/reduce/filter优化Javascript脚本

这是我的问题，我收到了这类json数据，但我事先不知道键(unknownProperty＃)，但是此数据数组中的不同对象具有相同的键。我需要做的是计算每个警报级别的数量(这些级别是固定的)。
这是我的最终结果应该是什么样的

var result = { unknownProperty1: { unknown: 0, ok: 2, warning: 0, ko: 1 },
  unknownProperty2: { unknown: 0, ok: 0, warning: 2, ko: 1 },
  unknownProperty3: { unknown: 3, ok: 0, warning: 0, ko: 0 },
  unknownProperty4: { unknown: 0, ok: 0, warning: 3, ko: 0 },
  unknownProperty5: { unknown: 0, ok: 0, warning: 2, ko: 1 } }

我的脚本可以运行，但是我想知道是否有任何方法可以使用map / reduce / filter javascript函数对其进行优化/清理

const data = [
  {
    unknownProperty1: { alert: "ok", unusedItem1: "something" },
    unknownProperty2: { alert: "warning", unusedItem1: "something" },
    unknownProperty3: { alert: "unknown", unusedItem1: "something" },
    unknownProperty4: { alert: "warning", unusedItem1: "something" },
    unknownProperty5: { alert: "ko", unusedItem1: "something" }
  },
  {
    unknownProperty1: { alert: "ok", unusedItem1: "something" },
    unknownProperty2: { alert: "warning", unusedItem1: "something" },
    unknownProperty3: { alert: "unknown", unusedItem1: "something" },
    unknownProperty4: { alert: "warning", unusedItem1: "something" },
    unknownProperty5: { alert: "warning", unusedItem1: "something" }
  },
  {
    unknownProperty1: { alert: "ko", unusedItem1: "something" },
    unknownProperty2: { alert: "ko", unusedItem1: "something" },
    unknownProperty3: { alert: "unknown", unusedItem1: "something" },
    unknownProperty4: { alert: "warning", unusedItem1: "something" },
    unknownProperty5: { alert: "warning", unusedItem1: "something" }
  }
];

var result = new Object();
var dataKeys = data.map(function(element) {
  return Object.keys(element);
});
dataKeys[0].map(function(element) {
  result[element] = { unknown: 0, ok: 0, warning: 0, ko: 0 };
});
//At this point the result is like what we expect but no calculation has been made yet

//We then increment the right values
data.map(function(element) {
  for (var prop in element) {
    if (Object.prototype.hasOwnProperty.call(element, prop)) {
      // element correspond to the current object
      // prop corresponds to the name of the key, for example 'unknownProperty1'
      switch (element[prop].alert) {
        case "ok":
          result[prop].ok++;
          break;
        case "warning":
          result[prop].warning++;
          break;
        case "unknown":
          result[prop].unknown++;
          break;
        case "ko":
          result[prop].ko++;
          break;

        default:
          break;
      }
    }
  }
});
console.log(result);

最佳答案

简化为由unknownProperty_键索引的对象，迭代内部对象，并在累加器上创建适当的键(如果尚不存在)，并将其值设置为{ unknown: 0, ok: 0, warning: 0, ko: 0 }。然后，在对unknownProperty对象的每次迭代中，只需提取alert并增加a[key][alert]:

const data = [
  {
    unknownProperty1: { alert: "ok", unusedItem1: "something" },
    unknownProperty2: { alert: "warning", unusedItem1: "something" },
    unknownProperty3: { alert: "unknown", unusedItem1: "something" },
    unknownProperty4: { alert: "warning", unusedItem1: "something" },
    unknownProperty5: { alert: "ko", unusedItem1: "something" }
  },
  {
    unknownProperty1: { alert: "ok", unusedItem1: "something" },
    unknownProperty2: { alert: "warning", unusedItem1: "something" },
    unknownProperty3: { alert: "unknown", unusedItem1: "something" },
    unknownProperty4: { alert: "warning", unusedItem1: "something" },
    unknownProperty5: { alert: "warning", unusedItem1: "something" }
  },
  {
    unknownProperty1: { alert: "ko", unusedItem1: "something" },
    unknownProperty2: { alert: "ko", unusedItem1: "something" },
    unknownProperty3: { alert: "unknown", unusedItem1: "something" },
    unknownProperty4: { alert: "warning", unusedItem1: "something" },
    unknownProperty5: { alert: "warning", unusedItem1: "something" }
  }
];

const dataByProperty = data.reduce((a, obj) => {
  Object.entries(obj).forEach(([key, { alert }]) => {
    if (!a[key]) {
      a[key] = { unknown: 0, ok: 0, warning: 0, ko: 0 };
    }
    a[key][alert]++;
  });
  return a;
}, {});
console.log(dataByProperty);