Flutter/Dart how to groupBy list of maps

Question

I have this list of maps.

[
    {title: \'Avengers\', release_date: \'10/01/2019\'},
    {title: \'Creed\', release_date: \'10/01/2019\'}
    {title: \'Juman         


        

Answer 1

If you have Dart 2.7, you can extend Iterable to add a useful groupBy method:

extension Iterables<E> on Iterable<E> {
  Map<K, List<E>> groupBy<K>(K Function(E) keyFunction) => fold(
      <K, List<E>>{},
      (Map<K, List<E>> map, E element) =>
          map..putIfAbsent(keyFunction(element), () => <E>[]).add(element));
}

Now, you're List of Maps, could be grouped using something like:

final releaseDateMap = listOfMaps.groupBy((m) => m['release_date'])

Data like this:

[
    {title: 'Avengers', release_date: '10/01/2019'},
    {title: 'Creed', release_date: '10/01/2019'}
    {title: 'Jumanji', release_date: '30/10/2019'},
]

would turn into:

{
  '10/01/2019': [
    {title: 'Avengers', release_date: '10/01/2019'},
    {title: 'Creed', release_date: '10/01/2019'}
  ],
  '30/10/2019': [
    {title: 'Jumanji', release_date: '30/10/2019'},
  ]
}

Answer 2

Using the supercharged package, you'd write it like this:

List list = [
  { title: 'Avengers', release_date: '10/01/2019' },
  { title: 'Creed', release_date: '10/01/2019' }
  { title: 'Jumanji', release_date: '30/10/2019' },
];

final map = list.groupBy<String, Map>((item) => 
  item['release_date'],
  valueTransform: (item) => item..remove('release_date'),
);

Answer 3

The package collection implements the groupBy function.

For grouping by date:

import "package:collection/collection.dart";

main(List<String> args) {
  var data = [
    {"title": 'Avengers', "release_date": '10/01/2019'},
    {"title": 'Creed', "release_date": '10/01/2019'},
    {"title": 'Jumanji', "release_date": '30/10/2019'},
  ];


  var newMap = groupBy(data, (obj) => obj['release_date']);

  print(newMap);
}

For removing the release_date key from each map entry:

var newMap = groupBy(data, (obj) => obj['release_date']).map(
    (k, v) => MapEntry(k, v.map((item) { item.remove('release_date'); return item;}).toList()));

For changing a key:

var newMap = groupBy(data, (obj) => obj['release_date']).map(
    (k, v) => MapEntry(k, v.map((item) => {'name': item['title']}).toList()));

Answer 4

It may not be the best solution. But it can give you an idea

List arrayData = [
  {"name": 'John', "gender": 'male'},
  {"name": 'James', "gender": 'male'},
  {"name": 'Mary', "gender": 'female'}
];

Retrieve list ​​by gender:

List males = arrayData.where((o) => o['gender'] == "male").toList();
List females = arrayData.where((o) => o['gender'] == "female").toList();

Make new map with desired format:

List result = [
  {
    "male": males.map((f) => {"name": f['name']}).toList()
  },
  {
    "female": females.map((f) => {"name": f['name']}).toList()
  }
];

print:

debugPrint('${result}');

result:

[{male: [{name: John}, {name: James}]}, {female: [{name: Mary}]}]

Answer 5

This is a method naively implemented (in case you don't want to use the groupBy function from the collections package):

List<Map<String, List<Map<String, String>>>> MapByKey(String keyName, String newKeyName, String keyForNewName, List<Map<String,String>> input) {
  Map<String, Map<String, List<Map<String, String>>>> returnValue = Map<String, Map<String, List<Map<String, String>>>>();
  for (var currMap in input) {
    if (currMap.containsKey(keyName)) {
      var currKeyValue = currMap[keyName];
      var currKeyValueForNewName = currMap[keyForNewName];
      if (!returnValue.containsKey(currKeyValue)){
        returnValue[currKeyValue] = {currKeyValue : List<Map<String, String>>()};  
      }
      returnValue[currKeyValue][currKeyValue].add({newKeyName : currKeyValueForNewName});
    }
  }
  return returnValue.values.toList();
}

void main() {
    var test = [
    {"title": 'Avengers', "release_date": '10/01/2019'},
    {"title": 'Creed', "release_date": '10/01/2019'},
    {"title": 'Jumanji', "release_date": '30/10/2019'},
  ];

  var testMapped = MapByKey("release_date", "name", "title", test);

  print("$testMapped");
}

The output is:

[
    {
        10/01/2019: [
            {name: Avengers
            },
            {name: Creed
            }
        ]
    },
    {
        30/10/2019: [
            {name: Jumanji
            }
        ]
    }
]