如何獲得2個日期范圍數組之間的差異？

Question

我有2個日期范圍數組，我試圖找出兩者之間的區別。 我們以數字為例：

我有2個范圍[1-7, 9-16] ，我想減去[2-3, 7-9, 14-20]並得到結果范圍[1-1, 4-6, 10-13]

我被弄得一頭霧水，試圖弄清楚。 當然，我不知道有一個通用的解決方案嗎？

diffDateRangesArray(rangesArray1, rangesArray2) {
    //rangesArray = [{startDate, endDate}]
    let diffedRanges = [];
    rangesArray1.forEach(function(range1){
      //loop through rangesArray2 removing from range1
      rangesArray2.forEach(function(range2){
        // breaks if array returned
        // perhaps should always return array and flatten?
        range1 = diffDateRanges(range1, range2);
      });
      diffedRanges.push(range1);
    });
    //probably should do some sort of union here
    return diffedRanges;
  }

  diffDateRanges(range1, range2) {
    //range = {startDate, endDate}
    let diffedRange = {};
    // if not in range
    if(range2.endDate <= range1.startDate || range2.startDate >= range1.endDate){
      return range1;
    //if envelops range
    } else if(range2.endDate >= range1.endDate && range2.startDate <= range1.startDate){
      return null;
    //if cuts off end of range
    } else if(range2.startDate <= range1.endDate && range2.endDate >= range1.endDate){
      return {startDate:range1.startDate, endDate: range2.startDate};
    // if cuts off start of range
    } else if(range2.endDate >= range1.startDate && range2.startDate <= range1.startDate){
      return {startDate:range2.endDate, endDate: range1.endDate};
    // if inside of range - should better handle arrays
    } else if(range2.startDate >= range1.startDate && range2.endDate <= range1.endDate){
      return [
        {startDate:range1.startDate, endDate: range2.startDate},
        {startDate:range2.endDate, endDate: range1.endDate},
      ];
    }
  }

Answer 1

如果我正確地理解了您的問題，則可以通過以下步驟完成您想要的工作：

讓我們首先創建一些實用函數：

function range(start, end) {
  return [...Array(end - start + 1)].map((_, i) => start + i)
}

function unique(a) {
  return Array.from(new Set(a))
}

function immutableSort(arr) {
  return arr.concat().sort((a, b) => a - b)
}

Array.prototype.has = function(e) {
  return this.indexOf(e) >= 0
}

Object.prototype.isEmpty = function() {
  return Object.keys(this).length === 0 && this.constructor === Object
}

function arrayDifference(A, B) {
  return A.filter((e) => B.indexOf(e) < 0)
}

現在，讓我們執行一些功能來解決您的特定問題：

function arrayToRangeObjects(A) {
  const preparedA = immutableSort(unique(A))
  const minA = preparedA[0]
  const maxA = preparedA[preparedA.length - 1]

  const result = []
  let rangeObject = {}
  range(minA, maxA).forEach((v) => {
    if (!preparedA.has(v)) {
      if (rangeObject.hasOwnProperty('start')) {
          if (!rangeObject.hasOwnProperty('end')) {
            rangeObject.end = rangeObject.start
          }
        result.push(rangeObject)
      }
      rangeObject = {}
    } else {
      if (rangeObject.hasOwnProperty('start')) {
        rangeObject.end = v
      } else {
        rangeObject.start = v
      }
    }
  })
  if (!rangeObject.isEmpty()) {
    result.push(rangeObject)
  }
  return result
}

function rangeObjectToRange(rangeObject) {
  return range(rangeObject.start, rangeObject.end)
}

function rangeObjectsToRange(A) {
  return immutableSort(
    unique(
      A
      .map((rangeObject) => {
        return rangeObjectToRange(rangeObject)
      })
      .reduce((a, b) => {
        return a.concat(b)
      }, [])
    )
  )
}

這樣，您的問題的答案是：

function yourAnswer(A, B) {
  return arrayToRangeObjects(
    arrayDifference(rangeObjectsToRange(A), rangeObjectsToRange(B))
  )
}

讓我們測試一下：

const A = [
  {
    start: 1,
    end: 7
  },
  {
    start: 9,
    end: 16
  }
]

const B = [
  {
    start: 2,
    end: 3
  },
  {
    start: 7,
    end: 9
  },
  {
    start: 14,
    end: 20
  }
]

> yourAnswer(A, B)
[
  {
    start: 1,
    end: 1
  },
  {
    start: 4,
    end: 6
  },
  {
    start: 10,
    end: 13
  }
]

就像個人觀點一樣，我認為您的“范圍對象”數據結構有點難以操作，而且有點不靈活（所有這些麻煩都是為了獲得與范圍集合不重疊的范圍）：可能想看看用於存儲范圍的更有效的數據結構 。

Answer 2

因此事實證明這被稱為間隔代數，並且有用於此https://www.npmjs.com/package/qintervals的庫