How do I delete duplicates from a list without fooling around with a set? Is there something like list.distinct()? or list.unique()?
void main() {
print("Hello, World!");
List<String> list = ['abc',"abc",'def'];
list.forEach((f) => print("this is list $f"));
Set<String> set = new Set<String>.from(list);
print("this is #0 ${list[0]}");
set.forEach((f) => print("set: $f"));
List<String> l2= new List<String>.from(set);
l2.forEach((f) => print("This is new $f"));
}
Hello, World!
this is list abc
this is list abc
this is list def
this is #0 abc
set: abc
set: def
This is new abc
This is new def
Set seems to be way faster!! But it loses the order of the items :/
Use toSet and then toList
var ids = [1, 4, 4, 4, 5, 6, 6];
var distinctIds = ids.toSet().toList();
Result: [1, 4, 5, 6]
Or with spread operators:
var distinctIds = [...{...ids}];
I didn't find any of the provided answers very helpful.
Here is what I generally do:
final ids = Set();
myList.retainWhere((x) => ids.add(x.id));
Of course you can use any attribute which uniquely identifies your objects. It doesn't have to be an id field.
Benefits over other approaches:
Preserves the original order of the list
Works for rich objects not just primitives/hashable types
Doesn't have to copy the entire list to a set and back to a list
Update 09/12/21
You can also declare an extension method once for lists:
extension Unique<E, Id> on List<E> {
List<E> unique([Id Function(E element)? id, bool inplace = true]) {
final ids = Set();
var list = inplace ? this : List<E>.from(this);
list.retainWhere((x) => ids.add(id != null ? id(x) : x as Id));
return list;
}
}
This extension method does the same as my original answer. Usage:
// Use a lambda to map an object to its unique identifier.
myRichObjectList.unique((x) => x.id);
// Don't use a lambda for primitive/hashable types.
hashableValueList.unique();
Set works okay, but it doesn't preserve the order. Here's another way using LinkedHashSet:
import "dart:collection";
void main() {
List<String> arr = ["a", "a", "b", "c", "b", "d"];
List<String> result = LinkedHashSet<String>.from(arr).toList();
print(result); // => ["a", "b", "c", "d"]
}
https://api.dart.dev/stable/2.4.0/dart-collection/LinkedHashSet/LinkedHashSet.from.html
Try the following:
List<String> duplicates = ["a", "c", "a"];
duplicates = duplicates.toSet().toList();
Check this code on Dartpad.
If you want to keep ordering or are dealing with more complex objects than primitive types, store seen ids to the Set and filter away those ones that are already in the set.
final list = ['a', 'a', 'b'];
final seen = Set<String>();
final unique = list.where((str) => seen.add(str)).toList();
print(unique); // => ['a', 'b']
//This easy way works fine
List<String> myArray = [];
myArray = ['x', 'w', 'x', 'y', 'o', 'x', 'y', 'y', 'r', 'a'];
myArray = myArray.toSet().toList();
print(myArray);
// result => myArray =['x','w','y','o','r', 'a']
I am adding this to atreeon's answer. For anyone that want use this with Object:
class MyObject{
int id;
MyObject(this.id);
#override
bool operator ==(Object other) {
return other != null && other is MyObject && hashCode == other.hashCode;
}
#override
int get hashCode => id;
}
main(){
List<MyObject> list = [MyObject(1),MyObject(2),MyObject(1)];
// The new list will be [MyObject(1),MyObject(2)]
List<MyObject> newList = list.toSet().toList();
}
Remove duplicates from a list of objects:
class Stock {
String? documentID; //key
Make? make;
Model? model;
String? year;
Stock({
this.documentID,
this.make,
this.model,
this.year,
});
}
List of stock, from where we want to remove duplicate stocks
List<Stock> stockList = [stock1, stock2, stock3];
Remove duplicates
final ids = stockList.map((e) => e.documentID).toSet();
stockList.retainWhere((x) => ids.remove(x.documentID));
Using Dart 2.3+, you can use the spread operators to do this:
final ids = [1, 4, 4, 4, 5, 6, 6];
final distinctIds = [...{...ids}];
Whether this is more or less readable than ids.toSet().toList() I'll let the reader decide :)
For distinct list of objects you can use Equatable package.
Example:
// ignore: must_be_immutable
class User extends Equatable {
int id;
String name;
User({this.id, this.name});
#override
List<Object> get props => [id];
}
List<User> items = [
User(
id: 1,
name: "Omid",
),
User(
id: 2,
name: "Raha",
),
User(
id: 1,
name: "Omid",
),
User(
id: 2,
name: "Raha",
),
];
print(items.toSet().toList());
Output:
[User(1), User(2)]
Here it is, a working solution:
var sampleList = ['1', '2', '3', '3', '4', '4'];
//print('original: $sampleList');
sampleList = Set.of(sampleList).toList();
//print('processed: $sampleList');
Output:
original: [1, 2, 3, 3, 4, 4]
processed: [1, 2, 3, 4]
Using the fast_immutable_collections package:
[1, 2, 3, 2].distinct();
Or
[1, 2, 3, 2].removeDuplicates().toList();
Note: While distinct() returns a new list, removeDuplicates() does it lazily by returning an Iterable. This means it is much more efficient when you are doing some extra processing. For example, suppose you have a list with a million items, and you want to remove duplicates and get the first five:
// This will process five items:
List<String> newList = list.removeDuplicates().take(5).toList();
// This will process a million items:
List<String> newList = list.distinct().sublist(0, 5);
// This will also process a million items:
List<String> newList = [...{...list}].sublist(0, 5);
Both methods also accept a by parameter. For example:
// Returns ["a", "yk", "xyz"]
["a", "yk", "xyz", "b", "xm"].removeDuplicates(by: (item) => item.length);
If you don't want to include a package into your project but needs the lazy code, here it is a simplified removeDuplicates():
Iterable<T> removeDuplicates<T>(Iterable<T> iterable) sync* {
Set<T> items = {};
for (T item in iterable) {
if (!items.contains(item)) yield item;
items.add(item);
}
}
Note: I am one of the authors of the fast_immutable_collections package.
void uniqifyList(List<Dynamic> list) {
for (int i = 0; i < list.length; i++) {
Dynamic o = list[i];
int index;
// Remove duplicates
do {
index = list.indexOf(o, i+1);
if (index != -1) {
list.removeRange(index, 1);
}
} while (index != -1);
}
}
void main() {
List<String> list = ['abc', "abc", 'def'];
print('$list');
uniqifyList(list);
print('$list');
}
Gives output:
[abc, abc, def]
[abc, def]
As for me, one of the best practices is sort the array, and then deduplicate it. The idea is stolen from low-level languages. So, first make the sort by your own, and then deduplicate equal values that are going after each other.
// Easy example
void dedup<T>(List<T> list, {removeLast: true}) {
int shift = removeLast ? 1 : 0;
T compareItem;
for (int i = list.length - 1; i >= 0; i--) {
if (compareItem == (compareItem = list[i])) {
list.removeAt(i + shift);
}
}
}
// Harder example
void dedupBy<T, I>(List<T> list, I Function(T) compare, {removeLast: true}) {
int shift = removeLast ? 1 : 0;
I compareItem;
for (int i = list.length - 1; i >= 0; i--) {
if (compareItem == (compareItem = compare(list[i]))) {
list.removeAt(i + shift);
}
}
}
void main() {
List<List<int>> list = [[1], [1], [2, 1], [2, 2]];
print('$list');
dedupBy(list, (innerList) => innerList[0]);
print('$list');
print('\n removeLast: false');
List<List<int>> list2 = [[1], [1], [2, 1], [2, 2]];
print('$list2');
dedupBy(list2, (innerList) => innerList[0], removeLast: false);
print('$list2');
}
Output:
[[1], [1], [2, 1], [2, 2]]
[[1], [2, 1]]
removeLast: false
[[1], [1], [2, 1], [2, 2]]
[[1], [2, 2]]
This is another way...
final reducedList = [];
list.reduce((value, element) {
if (value != element)
reducedList.add(value);
return element;
});
reducedList.add(list.last);
print(reducedList);
It works for me.
var list = [
{"id": 1, "name": "Joshua"},
{"id": 2, "name": "Joshua"},
{"id": 3, "name": "Shinta"},
{"id": 4, "name": "Shinta"},
{"id": 5, "name": "Zaidan"}
];
list.removeWhere((element) => element.name == element.name.codeUnitAt(1));
list.sort((a, b) => a.name.compareTo(b.name));
Output:
[{"id": 1, "name": "Joshua"},
{"id": 3, "name": "Shinta"},
{"id": 5, "name": "Zaidan"}]
List<Model> bigList = [];
List<ModelNew> newList = [];
for (var element in bigList) {
var list = newList.where((i) => i.type == element.type).toList();
if(list.isEmpty){
newList.add(element);
}
}
Create method to remove duplicates from Array and return Array of unique elements.
class Utilities {
static List<String> uniqueArray(List<String> arr) {
List<String> newArr = [];
for (var obj in arr) {
if (newArr.contains(obj)) {
continue;
}
newArr.add(obj);
}
return newArr;
}
}
You can use the following way:
void main(List <String> args){
List<int> nums = [1, 2, 2, 2, 3, 4, 5, 5];
List<int> nums2 = nums.toSet().toList();
}
NOTE: This will not work if the items in the list are objects of class and have the same attributes. So, to solve this, you can use the following way:
void main() {
List<Medicine> objets = [Medicine("Paracetamol"),Medicine("Paracetamol"), Medicine("Benylin")];
List <String> atributs = [];
objets.forEach((element){
atributs.add(element.name);
});
List<String> noDuplicates = atributs.toSet().toList();
print(noDuplicates);
}
class Medicine{
final String name;
Medicine(this.name);
}
This is my solution
List<T> removeDuplicates<T>(List<T> list, IsEqual isEqual) {
List<T> output = [];
for(var i = 0; i < list.length; i++) {
bool found = false;
for(var j = 0; j < output.length; j++) {
if (isEqual(list[i], output[j])) {
found = true;
}
}
if (found) {
output.add(list[i]);
}
}
return output;
}
Use it like this:
var theList = removeDuplicates(myOriginalList, (item1, item2) => item1.documentID == item2.documentID);
or...
var theList = removeDuplicates(myOriginalList, (item1, item2) => item1.equals(item2));
or...
I have a library called Reactive-Dart that contains many composable operators for terminating and non-terminating sequences. For your scenario it would look something like this:
final newList = [];
Observable
.fromList(['abc', 'abc', 'def'])
.distinct()
.observe((next) => newList.add(next), () => print(newList));
Yielding:
[abc, def]
I should add that there are other libraries out there with similar features. Check around on GitHub and I'm sure you'll find something suitable.
I am using an unordered_list tag in django.
I have the following list:
foo = ['A', ['B', 'C', 'D'], 'E']
And the following tag:
{{ foo|unordered_list }}
Which produces, as expected the following:
<li>A
<ul>
<li>B</li>
<li>C</li>
<li>D</li>
</ul>
</li>
<li>E</li>
What I would like to do is to add id and class properties to each "li" node.
I understand that I can do it in JavaScript, or implement my own template tag in django. But I wanted to ask first. May be there already exists an easy build-in way to do it in django template?
As you can see in the source, the <li> is hardcoded so you can't do it without creating your own templatefilter.
def unordered_list(value, autoescape=None):
"""
Recursively takes a self-nested list and returns an HTML unordered list --
WITHOUT opening and closing <ul> tags.
The list is assumed to be in the proper format. For example, if ``var``
contains: ``['States', ['Kansas', ['Lawrence', 'Topeka'], 'Illinois']]``,
then ``{{ var|unordered_list }}`` would return::
<li>States
<ul>
<li>Kansas
<ul>
<li>Lawrence</li>
<li>Topeka</li>
</ul>
</li>
<li>Illinois</li>
</ul>
</li>
"""
if autoescape:
from django.utils.html import conditional_escape
escaper = conditional_escape
else:
escaper = lambda x: x
def convert_old_style_list(list_):
"""
Converts old style lists to the new easier to understand format.
The old list format looked like:
['Item 1', [['Item 1.1', []], ['Item 1.2', []]]
And it is converted to:
['Item 1', ['Item 1.1', 'Item 1.2]]
"""
if not isinstance(list_, (tuple, list)) or len(list_) != 2:
return list_, False
first_item, second_item = list_
if second_item == []:
return [first_item], True
old_style_list = True
new_second_item = []
for sublist in second_item:
item, old_style_list = convert_old_style_list(sublist)
if not old_style_list:
break
new_second_item.extend(item)
if old_style_list:
second_item = new_second_item
return [first_item, second_item], old_style_list
def _helper(list_, tabs=1):
indent = u'\t' * tabs
output = []
list_length = len(list_)
i = 0
while i < list_length:
title = list_[i]
sublist = ''
sublist_item = None
if isinstance(title, (list, tuple)):
sublist_item = title
title = ''
elif i < list_length - 1:
next_item = list_[i+1]
if next_item and isinstance(next_item, (list, tuple)):
# The next item is a sub-list.
sublist_item = next_item
# We've processed the next item now too.
i += 1
if sublist_item:
sublist = _helper(sublist_item, tabs+1)
sublist = '\n%s<ul>\n%s\n%s</ul>\n%s' % (indent, sublist,
indent, indent)
output.append('%s<li>%s%s</li>' % (indent,
escaper(force_unicode(title)), sublist))
i += 1
return '\n'.join(output)
value, converted = convert_old_style_list(value)
return mark_safe(_helper(value))
unordered_list.is_safe = True
unordered_list.needs_autoescape = True
It should be fairly easy to rewrite the filter to add id/class support though, depending on how you want it to work.