Sort two lists the same way - sorting

Question

I need to sort a list of DateTime from earliest to latest.
List<DateTime> list = [2021-01-15 12:26:40.709246, 2021-02-25 13:26:40.709246, 2021-02-20 19:26:40.709246];
datetimeList.sort();
I have another list of Strings.
List<String> list = ["one", "two", "three"];
The indexes of stringList have to match the indexes of datetimeList. So the index of "one" always has to be the same as the index of 2021-01-15 12:26:40.709246 and so on.
If I sort the lists individually, the DateTime is sorted by DateTime and the Strings are sorted alphabetically. This way, the String does not go with its initial date anymore.
How can I sort one list (datetimeList) with the other list (stringList) sorting exactly the same way?

Answer 1

The easiest solution would be to create a struct/class to combine both variables so you don't have to worry about keeping the objects in the arrays aligned. The last thing you need to do is to sort the array ob new objects by the date. For that, I cannot help you due to missing knowledge about Dart.

Answer 2

You could us a SplayTreeMap as well.https://api.dart.dev/stable/2.8.4/dart-collection/SplayTreeMap-class.html.
SplayTreeMap ensures that its keys are in sorted order.You could use your datetime as key and the its contents of other list as value.
main() {
final SplayTreeMap<DateTime, String> map =
new SplayTreeMap<DateTime, String>();
map[DateTime.parse("2021-01-15 12:26:40.709246")] = "one";
map[DateTime.parse("2021-02-25 13:26:40.709246")] = "three";
map[DateTime.parse("2021-02-20 19:26:40.709246")] = "two";
for (final DateTime key in map.keys) {
print("$key : ${map[key]}");
}
}

Answer 3

I recommend the simpler suggestions given here.
For completeness, I'll provide one more approach: Compute the permutation by sorting a list of indices:
List<int> sortedPermutation<T>(List<T> elements, int compare(T a, T b)) =>
[for (var i = 0; i < elements.length; i++) i]
..sort((i, j) => compare(elements[i], elements[j]));
Then you can reorder the existing lists to match:
List<T> reorder<T>(List<T> elements, List<int> permutation) =>
[for (var i = 0; i < permutation.length; i++) elements[permutation[i]]];
If you do:
var sorted = reorder(original, sortedPermutation(original, compare));
it should give you a sorted list.
It's less efficient than sorting in-place because you create a new list,
but you can apply the same reordering to multiple lists afterwards.

Answer 4

Fast and very effective way.
void main() {
final l1 = [3, 1, 2];
final l2 = ['three', 'one', 'two'];
final l3 = ['drei', 'ein', 'zwei'];
print(l1);
print(l2);
print(l3);
myCompare(int x, int y) => x.compareTo(y);
l1.sortLists([l2, l3], myCompare);
print('============');
print(l1);
print(l2);
print(l3);
}
extension SortListByList<E> on List<E> {
sortLists(Iterable<List> lists, int Function(E, E) compare) {
for (final list in lists) {
if (list.length != length) {
throw StateError('The length of lists must be equal');
}
}
final rules = <int>[];
sort((x, y) {
final rule = compare(x, y);
rules.add(rule);
return rule;
});
for (final list in lists) {
var rule = 0;
list.sort((x, y) => rules[rule++]);
}
}
}
Output:
[3, 1, 2]
[three, one, two]
[drei, ein, zwei]
============
[1, 2, 3]
[one, two, three]
[ein, zwei, drei]