从给定条件的映射中获取平均值​

Question

鉴于这张地图：

<DateTime, double> {
2014-09-01 01:46:00.000: 1.29, 
2014-09-01 01:47:00.000: 1.27, 
2014-09-01 01:48:00.000: 1.27,
2014-09-01 01:48:00.000: 1.27,
2014-09-01 01:49:00.000: 1.26, 
2014-09-01 01:50:00.000: 1.26,
2014-09-01 01:50:00.000: 1.26,
2014-09-01 01:51:00.000: 1.26, 
2014-09-01 01:52:00.000: 1.26,
2014-09-01 01:52:00.000: 1.26,
2014-09-01 01:53:00.000: 1.26, 
2014-09-01 01:54:00.000: 1.27,
2014-09-01 01:54:00.000: 1.27
}

如何仅在日期重复的情况下计算值​​的平均值，例如:对于值1.26和1.27，日期重复两次，但使用不同的日期

2014-09-01 01:50:00.000: 1.26,
2014-09-01 01:50:00.000: 1.26,

我怎样才能在地图上做到这一点？

Answer 1

当前的映射包含多个键，这些键将被视为相同的对象，因为DateTime覆盖了==和.hashCode，因此指向同一时间点的两个DateTime将被视为相等：https://api.dart.dev/stable/2.17.6/dart-core/DateTime/operator_equals.html。

这意味着，如果我们正常地尝试创建Map，我们将得到以下结果：

void main() {
  final map = <DateTime, double>{
    DateTime.parse('2014-09-01 01:46:00.000'): 1.29,
    DateTime.parse('2014-09-01 01:47:00.000'): 1.27,
    DateTime.parse('2014-09-01 01:48:00.000'): 1.27,
    DateTime.parse('2014-09-01 01:48:00.000'): 1.27,
    DateTime.parse('2014-09-01 01:49:00.000'): 1.26,
    DateTime.parse('2014-09-01 01:50:00.000'): 1.26,
    DateTime.parse('2014-09-01 01:50:00.000'): 1.26,
    DateTime.parse('2014-09-01 01:51:00.000'): 1.26,
    DateTime.parse('2014-09-01 01:52:00.000'): 1.26,
    DateTime.parse('2014-09-01 01:52:00.000'): 1.26,
    DateTime.parse('2014-09-01 01:53:00.000'): 1.26,
    DateTime.parse('2014-09-01 01:54:00.000'): 1.27,
    DateTime.parse('2014-09-01 01:54:00.000'): 1.27,
  };

  print(map.length); // 9
  map.entries.forEach(print);
  // MapEntry(2014-09-01 01:46:00.000: 1.29)
  // MapEntry(2014-09-01 01:47:00.000: 1.27)
  // MapEntry(2014-09-01 01:48:00.000: 1.27)
  // MapEntry(2014-09-01 01:49:00.000: 1.26)
  // MapEntry(2014-09-01 01:50:00.000: 1.26)
  // MapEntry(2014-09-01 01:51:00.000: 1.26)
  // MapEntry(2014-09-01 01:52:00.000: 1.26)
  // MapEntry(2014-09-01 01:53:00.000: 1.26)
  // MapEntry(2014-09-01 01:54:00.000: 1.27)
}

因此，即使我们将9元素放入Map中，长度也是Map。

我们可以使用Map.identity()映射，只需查看每个对象键的标识，就可以确定密钥是否相等(因此不再使用==和.hashCode)：

void main() {
  final identityMap = Map<DateTime, double>.identity()
    ..[DateTime.parse('2014-09-01 01:46:00.000')] = 1.29
    ..[DateTime.parse('2014-09-01 01:47:00.000')] = 1.27
    ..[DateTime.parse('2014-09-01 01:48:00.000')] = 1.27
    ..[DateTime.parse('2014-09-01 01:48:00.000')] = 1.26
    ..[DateTime.parse('2014-09-01 01:49:00.000')] = 1.26
    ..[DateTime.parse('2014-09-01 01:50:00.000')] = 1.26
    ..[DateTime.parse('2014-09-01 01:50:00.000')] = 1.26
    ..[DateTime.parse('2014-09-01 01:51:00.000')] = 1.26
    ..[DateTime.parse('2014-09-01 01:52:00.000')] = 1.26
    ..[DateTime.parse('2014-09-01 01:52:00.000')] = 1.26
    ..[DateTime.parse('2014-09-01 01:53:00.000')] = 1.26
    ..[DateTime.parse('2014-09-01 01:54:00.000')] = 1.27
    ..[DateTime.parse('2014-09-01 01:54:00.000')] = 1.27;

  print(identityMap.length); // 13
  identityMap.entries.forEach(print);
  // MapEntry(2014-09-01 01:46:00.000: 1.29)
  // MapEntry(2014-09-01 01:47:00.000: 1.27)
  // MapEntry(2014-09-01 01:48:00.000: 1.27)
  // MapEntry(2014-09-01 01:48:00.000: 1.26)
  // MapEntry(2014-09-01 01:49:00.000: 1.26)
  // MapEntry(2014-09-01 01:50:00.000: 1.26)
  // MapEntry(2014-09-01 01:50:00.000: 1.26)
  // MapEntry(2014-09-01 01:51:00.000: 1.26)
  // MapEntry(2014-09-01 01:52:00.000: 1.26)
  // MapEntry(2014-09-01 01:52:00.000: 1.26)
  // MapEntry(2014-09-01 01:53:00.000: 1.26)
  // MapEntry(2014-09-01 01:54:00.000: 1.27)
  // MapEntry(2014-09-01 01:54:00.000: 1.27)
}

如果我们想将应该被视为相等的键组合起来，我们可以这样做：

void main() {
  final identityMap = Map<DateTime, double>.identity()
    ..[DateTime.parse('2014-09-01 01:46:00.000')] = 1.29
    ..[DateTime.parse('2014-09-01 01:47:00.000')] = 1.27
    ..[DateTime.parse('2014-09-01 01:48:00.000')] = 1.27
    ..[DateTime.parse('2014-09-01 01:48:00.000')] = 1.26
    ..[DateTime.parse('2014-09-01 01:49:00.000')] = 1.26
    ..[DateTime.parse('2014-09-01 01:50:00.000')] = 1.26
    ..[DateTime.parse('2014-09-01 01:50:00.000')] = 1.26
    ..[DateTime.parse('2014-09-01 01:51:00.000')] = 1.26
    ..[DateTime.parse('2014-09-01 01:52:00.000')] = 1.26
    ..[DateTime.parse('2014-09-01 01:52:00.000')] = 1.26
    ..[DateTime.parse('2014-09-01 01:53:00.000')] = 1.26
    ..[DateTime.parse('2014-09-01 01:54:00.000')] = 1.27
    ..[DateTime.parse('2014-09-01 01:54:00.000')] = 1.27;

  // Step 1 - First create a map where we collect all values with the same key
  //          value (based on `==` and `.hashCode`).
  final tempMap = <DateTime, List<double>>{};
  for (final entry in identityMap.entries) {
    tempMap.update(
      entry.key,
      (value) => value..add(entry.value),
      ifAbsent: () => [entry.value],
    );
  }
  tempMap.entries.forEach(print);
  // MapEntry(2014-09-01 01:46:00.000: [1.29])
  // MapEntry(2014-09-01 01:47:00.000: [1.27])
  // MapEntry(2014-09-01 01:48:00.000: [1.27, 1.26])
  // MapEntry(2014-09-01 01:49:00.000: [1.26])
  // MapEntry(2014-09-01 01:50:00.000: [1.26, 1.26])
  // MapEntry(2014-09-01 01:51:00.000: [1.26])
  // MapEntry(2014-09-01 01:52:00.000: [1.26, 1.26])
  // MapEntry(2014-09-01 01:53:00.000: [1.26])
  // MapEntry(2014-09-01 01:54:00.000: [1.27, 1.27])

  // Step 2 - Go though each entry in our `tempMap` and take the average of the
  //          values collected and create a new `Map` based on this.
  final resultMap = tempMap.map((key, value) =>
      MapEntry(key, value.reduce((a, b) => a + b) / value.length));

  resultMap.entries.forEach(print);
  // MapEntry(2014-09-01 01:46:00.000: 1.29)
  // MapEntry(2014-09-01 01:47:00.000: 1.27)
  // MapEntry(2014-09-01 01:48:00.000: 1.2650000000000001)
  // MapEntry(2014-09-01 01:49:00.000: 1.26)
  // MapEntry(2014-09-01 01:50:00.000: 1.26)
  // MapEntry(2014-09-01 01:51:00.000: 1.26)
  // MapEntry(2014-09-01 01:52:00.000: 1.26)
  // MapEntry(2014-09-01 01:53:00.000: 1.26)
  // MapEntry(2014-09-01 01:54:00.000: 1.27)
}