从给定的两个列表中，如何从一个列表中提取 Dart 中另一个列表中不可用的所有元素？

Question

我在 Dart 中有两个列表，如下所示，

  final List availableIssueComponents = [
    {'id': 1, 'componentName': 'Cash Acceptor'},
    {'id': 2, 'componentName': 'Printer'},
    {'id': 3, 'componentName': 'PIN Pad'},
    {'id': 4, 'componentName': 'Key Board'},
    {'id': 5, 'componentName': 'Touch Screen'},
    {'id': 6, 'componentName': 'Computer'},
    {'id': 7, 'componentName': 'Application'},
    {'id': 8, 'componentName': 'Network'},
    {'id': 9, 'componentName': 'Power'},
    {'id': 10, 'componentName': 'Camera'},
    {'id': 11, 'componentName': 'Safe'},
    {'id': 13, 'componentName': 'Screen'},
    {'id': 14, 'componentName': 'Battery'},
    {'id': 15, 'componentName': 'Ports'},
    {'id': 16, 'componentName': 'Application'},
    {'id': 17, 'componentName': 'Safe'},
    {'id': 18, 'componentName': 'Camera'},
    {'id': 19, 'componentName': 'Power'},
    {'id': 20, 'componentName': 'Key Board'},
    {'id': 21, 'componentName': 'PIN Pad'},
    {'id': 22, 'componentName': 'Printer'},
    {'id': 23, 'componentName': 'Computer'},
    {'id': 24, 'componentName': 'Touch Screen'},
    {'id': 25, 'componentName': 'Application'},
    {'id': 26, 'componentName': 'Network'}
  ];

  final List selectedIssueComponents = [
    {'id': 3, 'componentName': 'PIN Pad'},
    {'id': 6, 'componentName': 'Computer'},
    {'id': 19, 'componentName': 'Power'},
  ];

从以上两个列表中，我试图 select availableIssueComponents 中的所有元素，不包括 selectedIssueComponents 中已经可用的元素。

Ex：由于 ID 为 3、6、19 的组件在两个列表中都很常见，因此我想要第三个列表，其中包含除组件之外的所有组件使用 3、6、19.

的 ID

第三个列表应该如下所示，

  final List availableIssueComponents = [
    {'id': 1, 'componentName': 'Cash Acceptor'},
    {'id': 2, 'componentName': 'Printer'},
    {'id': 4, 'componentName': 'Key Board'},
    {'id': 5, 'componentName': 'Touch Screen'},
    {'id': 7, 'componentName': 'Application'},
    {'id': 8, 'componentName': 'Network'},
    {'id': 9, 'componentName': 'Power'},
    {'id': 10, 'componentName': 'Camera'},
    {'id': 11, 'componentName': 'Safe'},
    {'id': 13, 'componentName': 'Screen'},
    {'id': 14, 'componentName': 'Battery'},
    {'id': 15, 'componentName': 'Ports'},
    {'id': 16, 'componentName': 'Application'},
    {'id': 17, 'componentName': 'Safe'},
    {'id': 18, 'componentName': 'Camera'},
    {'id': 20, 'componentName': 'Key Board'},
    {'id': 21, 'componentName': 'PIN Pad'},
    {'id': 22, 'componentName': 'Printer'},
    {'id': 23, 'componentName': 'Computer'},
    {'id': 24, 'componentName': 'Touch Screen'},
    {'id': 25, 'componentName': 'Application'},
    {'id': 26, 'componentName': 'Network'}
  ];

我尝试使用 Sets 来做到这一点，以下是我的方法，

Set availableComponentsSet = Set.from(availableIssueComponents);
Set issueComponentsSet = Set.from(selectedIssueComponents);

Set resultComponents = availableComponentsSet.difference(issueComponentsSet);

但是当登录到控制台时，它 resultComponents 包含所有组件。这不是我想要的。我也试过嵌套的 for 循环，它也没有用。

Answer 1

当使用 Set<Map> 时，组件对象不会被 Set 过滤，因为 Map 是一个引用类型并且被认为是唯一的，除非被比较的两个对象指向相同的实例（如@Pat9RB 评论）。

我会将选定的 ID 映射到列表，然后使用 List#where(fn)

过滤掉这些 ID

final availableIssueComponents = [
  {'id': 1, 'componentName': 'Cash Acceptor'},
  {'id': 2, 'componentName': 'Printer'},
  {'id': 3, 'componentName': 'PIN Pad'},
  {'id': 4, 'componentName': 'Key Board'},
  {'id': 5, 'componentName': 'Touch Screen'},
  {'id': 6, 'componentName': 'Computer'},
  {'id': 7, 'componentName': 'Application'},
  {'id': 8, 'componentName': 'Network'},
  {'id': 9, 'componentName': 'Power'},
  {'id': 10, 'componentName': 'Camera'},
  {'id': 11, 'componentName': 'Safe'},
  {'id': 13, 'componentName': 'Screen'},
  {'id': 14, 'componentName': 'Battery'},
  {'id': 15, 'componentName': 'Ports'},
  {'id': 16, 'componentName': 'Application'},
  {'id': 17, 'componentName': 'Safe'},
  {'id': 18, 'componentName': 'Camera'},
  {'id': 19, 'componentName': 'Power'},
  {'id': 20, 'componentName': 'Key Board'},
  {'id': 21, 'componentName': 'PIN Pad'},
  {'id': 22, 'componentName': 'Printer'},
  {'id': 23, 'componentName': 'Computer'},
  {'id': 24, 'componentName': 'Touch Screen'},
  {'id': 25, 'componentName': 'Application'},
  {'id': 26, 'componentName': 'Network'}
];

final selectedIssueComponents = [
  {'id': 3, 'componentName': 'PIN Pad'},
  {'id': 6, 'componentName': 'Computer'},
  {'id': 19, 'componentName': 'Power'},
];

final selectedIds = selectedIssueComponents.map((component) => component['id']).toList();
final filtered = availableIssueComponents.where((element) => !selectedIds.contains(element["id"])).toList();

print(filtered);

如果您更喜欢使用 Set 和 difference，您可以创建一组 ID。这将创建一组 int (Set<int>)，这是原始类型并允许预期的过滤类型：

final availableIssueComponents = [
  {'id': 1, 'componentName': 'Cash Acceptor'},
  {'id': 2, 'componentName': 'Printer'},
  {'id': 3, 'componentName': 'PIN Pad'},
  {'id': 4, 'componentName': 'Key Board'},
  {'id': 5, 'componentName': 'Touch Screen'},
  {'id': 6, 'componentName': 'Computer'},
  {'id': 7, 'componentName': 'Application'},
  {'id': 8, 'componentName': 'Network'},
  {'id': 9, 'componentName': 'Power'},
  {'id': 10, 'componentName': 'Camera'},
  {'id': 11, 'componentName': 'Safe'},
  {'id': 13, 'componentName': 'Screen'},
  {'id': 14, 'componentName': 'Battery'},
  {'id': 15, 'componentName': 'Ports'},
  {'id': 16, 'componentName': 'Application'},
  {'id': 17, 'componentName': 'Safe'},
  {'id': 18, 'componentName': 'Camera'},
  {'id': 19, 'componentName': 'Power'},
  {'id': 20, 'componentName': 'Key Board'},
  {'id': 21, 'componentName': 'PIN Pad'},
  {'id': 22, 'componentName': 'Printer'},
  {'id': 23, 'componentName': 'Computer'},
  {'id': 24, 'componentName': 'Touch Screen'},
  {'id': 25, 'componentName': 'Application'},
  {'id': 26, 'componentName': 'Network'}
];

final selectedIssueComponents = [
  {'id': 3, 'componentName': 'PIN Pad'},
  {'id': 6, 'componentName': 'Computer'},
  {'id': 19, 'componentName': 'Power'},
];
final availableIds = availableIssueComponents.map((component) => component['id']).toSet();
final selectedIds = selectedIssueComponents.map((component) => component['id']).toSet();
final filteredIds = availableIds.difference(selectedIds);
final filteredComponents = availableIssueComponents.where((element) => filteredIds.contains(element["id"])).toList();

print(filteredComponents);

从给定的两个列表中，如何从一个列表中提取 Dart 中另一个列表中不可用的所有元素？

From given two lists, how to extract all the elements from one list that are not available in the other list in Dart?

flutter

dart

dart-editor

dart-2

dartdoc