Question

有人可以给我一个提示吗？我想按列表的长度对地图的值进行排序。

var chordtypes = {
  "maj": [0, 4, 7],
  "M7": [0, 4, 7, 11],
  "m7": [0, 3, 7, 10],
  "6": [0, 4, 7, 9],
  "9": [0, 4, 7, 10, 14],
  "sus2": [0, 2, 7],
  "sus4": [0, 5, 7],
  "omit3": [0, 7],
  "#5": [0, 4, 8],
  "+7b9#11": [0, 4, 8, 10, 13, 18],
  "+9": [0, 4, 8, 10, 14]
};

Answer 1

一个按照长度对List of Map进行排序的函数。

import 'dart:collection';

/// sorts the ListMap (== A Map of List<V>) on the length
/// of the List values.
LinkedHashMap sortListMap(LinkedHashMap map) {
    List mapKeys = map.keys.toList(growable : false);
    mapKeys.sort((k1, k2) => map[k1].length - map[k2].length);
    LinkedHashMap resMap = new LinkedHashMap();
    mapKeys.forEach((k1) { resMap[k1] = map[k1] ; }) ;        
    return resMap;
}

结果：

var res = sortListMap(chordtypes);
print(res);

==＆GT;

{ omit3: [0, 7], 
  maj: [0, 4, 7], 
  sus2: [0, 2, 7], 
  sus4: [0, 5, 7], 
  #5: [0, 4, 8], 
  M7: [0, 4, 7, 11], 
  m7: [0, 3, 7, 10], 
  6: [0, 4, 7, 9], 
  9: [0, 4, 7, 10, 14], 
  +9: [0, 4, 8, 10, 14], 
  +7b9#11: [0, 4, 8, 10, 13, 18] }

Answer 2

这样的事可能适合你：

Map chordtypes = {
  "maj": [0, 4, 7],
  "M7": [0, 4, 7, 11],
  "m7": [0, 3, 7, 10],
  "6": [0, 4, 7, 9],
  "9": [0, 4, 7, 10, 14],
  "sus2": [0, 2, 7],
  "sus4": [0, 5, 7],
  "omit3": [0, 7],
  "#5": [0, 4, 8],
  "+7b9#11": [0, 4, 8, 10, 13, 18],
  "+9": [0, 4, 8, 10, 14]
};

List keys = chordtypes.keys.toList();
keys.sort((k1, k2) {
  if(chordtypes[k1].length > chordtypes[k2].length)
    return -1;
  if(chordtypes[k1].length < chordtypes[k2].length)
    return 1;
  return 0;
});
keys.forEach((String k) {
  print('$k ${chordtypes[k]}');
});

Answer 3

以@Leonardo Rignanese的答案为基础。扩展功能，可提供更实用的方法：

extension MapExt<T, U> on Map<T, U> {
  Map<T, U> sortedBy(Comparable value(U u)) {
    final entries = this.entries.toList();
    entries.sort((a, b) => value(a.value).compareTo(value(b.value)));
    return Map<T, U>.fromEntries(entries);
  }
}

一般用法：

foos.sortedBy((it) => it.bar);

OP的用法：

final sortedChordtypes = chordtypes.sortedBy((it) => it.length);

要点在这里：https://gist.github.com/nmwilk/68ae0424e848b9f05a8239db6b708390

Answer 4

import 'package:queries/collections.dart';

void main() {
  var query = Dictionary.fromMap(chordtypes)
      .orderBy((e) => e.value.length)
      .toDictionary$1((kv) => kv.key, (kv) => kv.value);
  print(query.toMap());
}

var chordtypes = {
  "maj": [0, 4, 7],
  "M7": [0, 4, 7, 11],
  "m7": [0, 3, 7, 10],
  "6": [0, 4, 7, 9],
  "9": [0, 4, 7, 10, 14],
  "sus2": [0, 2, 7],
  "sus4": [0, 5, 7],
  "omit3": [0, 7],
  "#5": [0, 4, 8],
  "+7b9#11": [0, 4, 8, 10, 13, 18],
  "+9": [0, 4, 8, 10, 14]
};

结果：

{
omit3: [0, 7],
maj: [0, 4, 7],
sus2: [0, 2, 7],
sus4: [0, 5, 7],
#5: [0, 4, 8],
M7: [0, 4, 7, 11],
m7: [0, 3, 7, 10],
6: [0, 4, 7, 9],
9: [0, 4, 7, 10, 14],
+9: [0, 4, 8, 10, 14],
+7b9#11: [0, 4, 8, 10, 13, 18]}
}

Answer 5

使用DART语言：

比方说，您要使用整数键和值类型为Foo的Map进行排序：

class Foo {
  int x; //it can be any type
}

因此，您可以获得所有条目的列表，将它们像普通列表一样进行排序，然后重建地图：

Map<int, Foo> map = //fill map

var entries = map.entries.toList();
entries.sort((MapEntry<int, Foo> a, MapEntry<int, Foo> b) => a.value.x.compareTo(b.value.x));
map = Map<int, Foo>.fromEntries(entries);

如何对地图的值进行排序？

5 个答案: