如何從兩個部分串列和基于一個部分串列創建一個新串列？

這需要一些我在 Dart 中還沒有的天才、經驗和知識，所以我正在尋求幫助。

我正在嘗試填充折線圖，當用戶第一次開始使用軟體時，圖表資料中會有“空”條目。也可能有“零”條目或在某些周或條目中根本沒有條目。我想確保基于始終顯示 13 個條目的圖表，我的折線圖顯示任何缺失條目的零（對于銷售收入 - yAxis）。

例如，mockupGraphData 將是一種現有情況，其中幾個銷售日期缺少條目（基于每周條目）并且沒有足夠的條目來滿足圖表需求。

我想自動輸入“歸零”日期。日期是缺失日期的索引，GraphData.yAxis（收入）為零。

下面無法編譯，當然，考慮的是什么，我認為最后的回圈偽代碼，如果我能得到周圍的mockupGraphData不存在該指數將作業...

void main() {
  
  // Actual "mock" input data
  List<GraphData> mockupGraphData = [
    GraphData(1586761200000, 145333),
    GraphData(1587366000000, 433343),
    GraphData(1587970800000, 223432),
    GraphData(1589180400000, 122221),
    GraphData(1589785200000, 982347),
    GraphData(1590994800000, 234322),
  ];
  
  // Required graph dates as per below calculations (13 entries):
    // 1583737200000
    // 1584342000000
    // 1584946800000
    // 1585551600000
    // 1586156400000
    // 1586761200000
    // 1587366000000
    // 1587970800000
    // 1588575600000
    // 1589180400000
    // 1589785200000
    // 1590390000000
    // 1590994800000

  int entries = 13;
  int latestDateReported = DateTime(2020, 6, 1).millisecondsSinceEpoch;
  int oneWeek = Duration(days: 7).inMilliseconds;
  int earliestEntryDate = latestDateReported - (oneWeek * entries);
  List<int> graphRequiredDates = [];
  List<GraphData> finalData = [];
  
  for (int i = 0; i < entries; i  ) {
    earliestEntryDate  = oneWeek;
    graphRequiredDates.add(earliestEntryDate);
  }
  // What I would like to do
  for (int i = 0; i < entries; i  ) {
    if (!mockupGraphData.contains(GraphData(graphRequiredDates[i]))) {
      finalData.add(GraphData(graphRequiredDates[i], 0));
    }
  }
  
  print(finalData);
}

class GraphData {
    int xAxis; // Sale date
    int yAxis; // Sale income
  
  GraphData(this.xAxis, this.yAxis);
  
  @override
  String toString() {
    return '$xAxis $yAxis';
  }
  }

// IDEAL OUTPUT:
  List<GraphData> mockupGraphData = [
    GraphData(1583737200000, 0),
    GraphData(1584342000000, 0),
    GraphData(1584946800000, 0),
    GraphData(1585551600000, 0),
    GraphData(1586156400000, 0),
    GraphData(1586761200000, 145333),
    GraphData(1587366000000, 433343),
    GraphData(1587970800000, 223432),
    GraphData(1588575600000, 0),
    GraphData(1589180400000, 122221),
    GraphData(1589785200000, 982347),
    GraphData(1590390000000, 0),
    GraphData(1590994800000, 234322),
  ];

uj5u.com熱心網友回復：

我將使用將時間戳映射到值mockupGraphData的List<GraphData>，而不是存盤為Map<int, int>。從那里，很容易迭代您的預期時間戳串列，檢查時間戳是否已存在于 中mockupGraphData，然后GraphData適當地構造物件。

您還可以使用 collection-for使您的一些代碼更整潔：

void main() {
  // Actual "mock" input data

  var mockupGraphData = <int, int>{
    1586761200000: 145333,
    1587366000000: 433343,
    1587970800000: 223432,
    1589180400000: 122221,
    1589785200000: 982347,
    1590994800000: 234322,
  };

  int entries = 13;
  int latestDateReported = DateTime(2020, 6, 1).millisecondsSinceEpoch;
  int oneWeek = Duration(days: 7).inMilliseconds;
  int earliestEntryDate = latestDateReported - (oneWeek * (entries - 1));
  List<int> graphRequiredDates = [
    for (var i = 0; i < entries; i  )
      earliestEntryDate   i * oneWeek,
  ];
  List<GraphData> finalData = [
    for (var requiredDate in graphRequiredDates)
      GraphData(requiredDate, mockupGraphData[requiredDate] ?? 0),
  ];

  finalData.forEach(print);
}

class GraphData {
  int xAxis; // Sale date
  int yAxis; // Sale income

  GraphData(this.xAxis, this.yAxis);

  @override
  String toString() {
    return 'GraphData($xAxis, $yAxis)';
  }
}

uj5u.com熱心網友回復：

你可以嘗試這樣的事情：

void main() {
  List<GraphData> mockupGraphData = [
    GraphData(1586761200000, 145333),
    GraphData(1587366000000, 433343),
    GraphData(1587970800000, 223432),
    GraphData(1589180400000, 122221),
    GraphData(1589785200000, 982347),
    GraphData(1590994800000, 234322),
  ];

  int entries = 13;
  int latestDateReported = DateTime(2020, 6, 1).millisecondsSinceEpoch;
  int oneWeek = Duration(days: 7).inMilliseconds;
  int earliestEntryDate = 1583737200000;

  final result = [
    for (int i = 0, v = 0; i < entries; i  , earliestEntryDate  = oneWeek)
      if (mockupGraphData[v].xAxis == earliestEntryDate)
        mockupGraphData[v  ]
      else
        GraphData(earliestEntryDate, 0),
  ];

  result.forEach(print);
}

class GraphData {
  int xAxis; // Sale date
  int yAxis; // Sale income

  GraphData(this.xAxis, this.yAxis);

  @override
  String toString() {
    return '$xAxis $yAxis';
  }
}

我硬編碼了 的值，earliestEntryDate因為計算似乎與預期的輸出不匹配。上述解決方案假定mockupGraphData已排序，并且值至少相隔一周。

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