With the following code:
public class Main {
public static void main(String[] args) {
final List<Integer> items =
IntStream.rangeClosed(0, 23).boxed().collect(Collectors.toList());
final String s = items
.stream()
.map(Object::toString)
.collect(Collectors.joining(","))
.toString()
.concat(".");
System.out.println(s);
}
}
I get:
0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23.
What I would like to do, is to break the line every 10 items, in order to get:
0,1,2,3,4,5,6,7,8,9,
10,11,12,13,14,15,16,17,18,19,
20,21,22,23.
I have try a lot of things after googling without any success !
Can you help me ?
Thanks,
Olivier.
If you're open to using a third-party library, the following will work using Eclipse Collections Collectors2.chunk(int).
String s = IntStream.rangeClosed(0, 23)
.boxed()
.collect(Collectors2.chunk(10))
.collectWith(MutableList::makeString, ",")
.makeString("", ",\n", ".");
The result of Collectors2.chunk(10) will be a MutableList<MutableList<Integer>>. At this point I switch from the Streams APIs to using native Eclipse Collections APIs which are available directly on the collections. The method makeString is similar to Collectors.joining(). The method collectWith is like Stream.map() with the difference that a Function2 and an extra parameter are passed to the method. This allows a method reference to be used here instead of a lambda. The equivalent lambda would be list -> list.makeString(",").
If you use just Eclipse Collections APIs, this problem can be simplified as follows:
String s = Interval.zeroTo(23)
.chunk(10)
.collectWith(RichIterable::makeString, ",")
.makeString("", ",\n", ".");
Note: I am a committer for Eclipse Collections.
If all you want to do is process these ascending numbers, you can do it like
String s = IntStream.rangeClosed(0, 23).boxed()
.collect(Collectors.groupingBy(i -> i/10, LinkedHashMap::new,
Collectors.mapping(Object::toString, Collectors.joining(","))))
.values().stream()
.collect(Collectors.joining(",\n", "", "."));
This solution can be adapted to work on an arbitrary random access list as well, e.g.
List<Integer> items = IntStream.rangeClosed(0, 23).boxed().collect(Collectors.toList());
String s = IntStream.range(0, items.size()).boxed()
.collect(Collectors.groupingBy(i -> i/10, LinkedHashMap::new,
Collectors.mapping(ix -> items.get(ix).toString(), Collectors.joining(","))))
.values().stream()
.collect(Collectors.joining(",\n", "", "."));
However, there is no simple and elegant solution for arbitrary streams, a limitation which applies to all kind of tasks having a dependency to the element’s position.
Here is an adaptation of the already linked in the comments Collector:
private static Collector<String, ?, String> partitioning(int size) {
class Acc {
int count = 0;
List<List<String>> list = new ArrayList<>();
void add(String elem) {
int index = count++ / size;
if (index == list.size()) {
list.add(new ArrayList<>());
}
list.get(index).add(elem);
}
Acc merge(Acc right) {
List<String> lastLeftList = list.get(list.size() - 1);
List<String> firstRightList = right.list.get(0);
int lastLeftSize = lastLeftList.size();
int firstRightSize = firstRightList.size();
// they are both size, simply addAll will work
if (lastLeftSize + firstRightSize == 2 * size) {
System.out.println("Perfect!");
list.addAll(right.list);
return this;
}
// last and first from each chunk are merged "perfectly"
if (lastLeftSize + firstRightSize == size) {
System.out.println("Almost perfect");
int x = 0;
while (x < firstRightSize) {
lastLeftList.add(firstRightList.remove(x));
--firstRightSize;
}
right.list.remove(0);
list.addAll(right.list);
return this;
}
right.list.stream().flatMap(List::stream).forEach(this::add);
return this;
}
public String finisher() {
return list.stream().map(x -> x.stream().collect(Collectors.joining(",")))
.collect(Collectors.collectingAndThen(Collectors.joining(",\n"), x -> x + "."));
}
}
return Collector.of(Acc::new, Acc::add, Acc::merge, Acc::finisher);
}
And usage would be:
String result = IntStream.rangeClosed(0, 24)
.mapToObj(String::valueOf)
.collect(partitioning(10));
Related
How can I type something like "print(list[1,4]);" in Dart?
For example:
int main() {
var products = new List(5);
products[0] = "Laptop";
products[1] = "Mouse";
products[2] = "Keyboard";
products[3] = "Monitor";
products[4] = "Microphone";
print(products[1]); // Mouse
print(products[1,3]); // I want to see 'Mouse,Monitor'
}
This is not directly supported in the SDK but you can easily make a extension on e.g. List to add this feature:
void main() {
final products = List<String>(5);
products[0] = "Laptop";
products[1] = "Mouse";
products[2] = "Keyboard";
products[3] = "Monitor";
products[4] = "Microphone";
print(products[1]); // Mouse
print(products.selectMultiple([1,3]).join(',')); // Mouse,Monitor
}
extension MultiSelectListExtension<E> on List<E> {
Iterable<E> selectMultiple(Iterable<int> indexes) sync* {
for (final index in indexes) {
yield this[index];
}
}
}
You can't make it so [1,3] (as in you own example) would be valid since the [] operator does only allow one argument. So instead, we need to make a method which takes our requested indexes as argument.
Starting to take in hand Flutter for a study project, I'm wondering about sorting a list of files.
Indeed, my program has a list of 4 files initialized like this :
List<File> imageFiles = List(4);
This initialization actually implies that my list is like this : [null,null,null,null].
When the user performs actions, this list can fill up. However, the user can delete a file at any time, which can give us the following situation: [file A, null, null, file d].
My question is, how to sort the list when a deletion arrives in order to have a list where null objects are always last ([file A, file D, null, null]).
I've looked at a lot of topics already, but they never concern the DART.
Thank you in advance for your help.
You can sort the list with list.sort((a, b) => a == null ? 1 : 0);
Here's a full example, with String instead of File, that you can run on DartPad
void main() {
List<String> list = List(4);
list[0] = "file1";
list[3] = "file4";
print("list before sort: $list");
// list before sort: [file1, null, null, file4]
list.sort((a, b) => a == null ? 1 : 0);
print("list after sort: $list");
// list after sort: [file1, file4, null, null]
}
If it's a business requirement to have a max of 4 files, I would suggest creating a value object that can handle with that.
For example:
class ImageList {
final _images = List<String>();
void add(String image) {
if(_images.length < 4) {
_images.add(image);
}
}
void removeAt(int index) {
_images.removeAt(index);
}
String get(int index) {
return _images[index];
}
List getAll() {
return _images;
}
}
And you could run it like this:
void main() {
ImageList imageList = ImageList();
imageList.add("file1");
imageList.add("file2");
imageList.add("file3");
imageList.add("file4");
imageList.add("file5"); // won't be add
print("imagelist: ${imageList.getAll()}");
// imagelist: [file1, file2, file3, file4]
imageList.removeAt(2); // remove file3
print("imagelist: ${imageList.getAll()}");
// imagelist: [file1, file2, file4]
}
This will make it easier to have control. (This example was again with String instead of File)
You can try this:
This place all null at end.
sortedList.sort((a, b) {
int result;
if (a == null) {
result = 1;
} else if (b == null) {
result = -1;
} else {
// Ascending Order
result = a.compareTo(b);
}
return result;
})
How to sort a list of objects by the alphabetical order of one of its properties (Not the name but the actual value the property holds)?
You can pass a comparison function to List.sort.
someObjects.sort((a, b) => a.someProperty.compareTo(b.someProperty));
In general, you can provide a custom comparison function to List.sort.
/// Desired relation | Result
/// -------------------------------------------
/// a < b | Returns a negative value.
/// a == b | Returns 0.
/// a > b | Returns a positive value.
///
int mySortComparison(SomeClass a, SomeClass b) {
final propertyA = someProperty(a);
final propertyB = someProperty(b);
if (propertyA < propertyB) {
return -1;
} else if (propertyA > propertyB) {
return 1;
} else {
return 0;
}
}
list.sort(mySortComparison);
If you're sorting some custom class you own, you alternatively could make your class implement the Comparable interface:
class MyCustomClass implements Comparable<MyCustomClass> {
...
#override
int compareTo(MyCustomClass other) {
if (someProperty < other.someProperty) {
return -1;
} else if (someProperty > other.someProperty) {
return 1;
} else {
return 0;
}
}
}
and then you can use list.sort() directly without supplying a callback.
Note that if you're sorting by a single property that already implements the Comparable interface, implementing the comparison functions is much simpler. For example:
class MyCustomClass implements Comparable<MyCustomClass> {
...
#override
int compareTo(MyCustomClass other) =>
someProperty.compareTo(other.someProperty);
}
Reversing
If you want to reverse the sort order, you can:
Make your comparison function return a value with the opposite sign.
Alternatively just explicitly reverse the list after sorting:
list = (list..sort()).reversed.toList();
Sorting by multiple properties (a.k.a. subsorting)
There are a variety of ways to sort by multiple properties.
A general way is to perform a stable sort for each property in reverse order of importance. For example, if you want to sort names primarily by surname and then subsort within surnames by given name, then you would first sort by given names, and then perform a stable sort by surname. See below for how to perform a stable sort.
Alternatively, you could sort with a comparison function that itself checks multiple properties. For example:
class Name {
Name({String? surname, String? givenName})
: surname = surname ?? "",
givenName = givenName ?? "";
final String surname;
final String givenName;
}
int compareNames(Name name1, Name name2) {
var comparisonResult = name1.surname.compareTo(name2.surname);
if (comparisonResult != 0) {
return comparisonResult;
}
// Surnames are the same, so subsort by given name.
return name1.givenName.compareTo(name2.givenName);
}
package:collection provides an extension to chain comparison functions so that combining them is a bit more straightforward and less error-prone:
import 'package:collection/collection.dart';
int compareSurnames(Name name1, Name name2) =>
name1.surname.compareTo(name2.surname);
int compareGivenNames(Name name1, Name name2) =>
name1.givenName.compareTo(name2.givenName);
final compareNames = compareSurnames.then(compareGivenNames);
My dartbag package also provides a compareIterables function that allows comparing Lists of property values in order of importance:
import 'package:dartbag/collection.dart';
int compareNames(Name name1, Name name2) =>
compareIterables(
[name1.surname, name1.givenName],
[name2.surname, name2.givenName],
);
Okay, I want a stable sort
List.sort is not guaranteed to be a stable sort. If you need a stable sort, package:collection provides insertionSort and mergeSort implementations that are stable.
But comparing might be expensive
Suppose you have a custom comparison function that looks something like:
int compareMyCustomClass(MyCustomClass a, MyCustomClass b) {
var a0 = computeValue(a);
var b0 = computeValue(b);
return a0.compareTo(b0);
}
The sorting process might call computeValue multiple times for the same object, which is particularly wasteful if computeValue() is expensive. In such cases, a Schwartzian transform could be faster (at the expense of using more memory). This approach maps your objects to directly sortable keys, sorts those keys, and extracts the original objects. (This is how Python's sort and sorted functions work.)
Here's one possible implementation:
class _SortableKeyPair<T, K extends Comparable<Object>>
implements Comparable<_SortableKeyPair<T, K>> {
_SortableKeyPair(this.original, this.key);
final T original;
final K key;
#override
int compareTo(_SortableKeyPair<T, K> other) => key.compareTo(other.key);
}
/// Returns a sorted *copy* of [items] according to the computed sort key.
List<E> sortedWithKey<E, K extends Comparable<Object>>(
Iterable<E> items,
K Function(E) toKey,
) {
final keyPairs = [
for (var element in items) _SortableKeyPair(element, toKey(element)),
]..sort();
return [
for (var keyPair in keyPairs) keyPair.original,
];
}
void main() {
final list = <MyCustomClass>[ ... ];
final sorted = sortedWithKeys(list, computeValue);
}
My dartbag package provides such a sortWithKey function (and also a sortWithAsyncKey function if the key needs to be generated asynchronously).
If you want to sort the object "objects" by the property "name" do something like this
objects.sort((a, b) {
return a.value['name'].toString().toLowerCase().compareTo(b.value['name'].toString().toLowerCase());
});
Immutable extension sortedBy for List.
extension MyIterable<E> on Iterable<E> {
Iterable<E> sortedBy(Comparable key(E e)) =>
toList()..sort((a, b) => key(a).compareTo(key(b)));
}
And use
list.sortedBy((it) => it.name);
Here is my contribution to this good question. If someone is facing difficulty to understand how the #Nate Bosch answer is working & you want to sort your custom model class list then you can do this way.
1. You have to implement Comparable abstract class in your model class.
It has the method compareTo which you have to override.
For example, I have this StudentMarks model class which has marks property in it.
class StudentMarks implements Comparable {
int marks;
StudentMarks({
this.marks,
});
#override
int compareTo(other) {
if (this.marks == null || other == null) {
return null;
}
if (this.marks < other.marks) {
return 1;
}
if (this.marks > other.marks) {
return -1;
}
if (this.marks == other.marks) {
return 0;
}
return null;
}
}
2. Now you can call compareTo method inside the sort method.
void _sortStudents({bool reversed: false}) {
_students.sort((a, b) {
return a.compareTo(b);
});
if (reversed) {
_students = _students.reversed.toList();
}
setState(() {});
}
Refer to this link you want to know more about the Comparable class
https://api.dart.dev/stable/2.1.0/dart-core/Comparable-class.html
Its worked for me:
myList..sort((a, b) => a.name.toLowerCase().compareTo(b.name.toLowerCase()));
Using Comparatorfunction, sort Users by id.
Comparator<UserModel> sortById = (a, b) => a.id.compareTo(b.id);
users.sort(sortById);
Now we can sort it in reversed/descending order.
users = users.reversed.toList();
To sort it in reverse order :
list.sort((a, b) {
return b.status.toLowerCase().compareTo(a.status.toLowerCase());
});
What's more, you can use Comparable.compare for more clear, for example:
class _Person {
final int age;
final String name;
_Person({required this.age, required this.name});
}
void _test() {
final array = [
_Person(age: 10, name: 'Dean'),
_Person(age: 20, name: 'Jack'),
_Person(age: 30, name: 'Ben'),
];
// ascend with age
// Dean Jack Ben
array.sort((p1, p2) {
return Comparable.compare(p1.age, p2.age);
});
// decend with age
// Ben Jack Dean
array.sort((p1, p2) {
return Comparable.compare(p2.age, p1.age);
});
// ascend with name
// Ben Dean Jack
array.sort((p1, p2) {
return Comparable.compare(p1.name, p2.name);
});
}
Similar to #pavel-shorokhovs answer, but strongly typed:
extension IterableExtensions<T> on Iterable<T> {
Iterable<T> sortBy<TSelected extends Comparable<TSelected>>(
TSelected Function(T) selector) =>
toList()..sort((a, b) => selector(a).compareTo(selector(b)));
Iterable<T> sortByDescending<TSelected extends Comparable<TSelected>>(
TSelected Function(T) selector) =>
sortBy(selector).toList().reversed;
}
i had fpgrowth machine learning output/result with each element of list contains another list and frequency field i was to sort by frequency in descending order so i used a bit of recursion for that try it might work i know i am late but i am posting maybe someone else could benefit.
sort(List<FrequentItem> fqItems) {
int len = fqItems.length;
if(len==2){
if(fqItems[0].frequency>fqItems[1].frequency){
sortedItems.add(fqItems[0]);
sortedItems.add(fqItems[1]);
}else{
sortedItems.add(fqItems[1]);
sortedItems.add(fqItems[0]);
}
return;
}else{
FrequentItem max = fqItems[0];
int index =0;
for(int i=0;i<len-2;i++){
if(max.frequency<fqItems[i+1].frequency){
max = fqItems[i+1];
index = i+1;
}
}
sortedItems.add(max);
fqItems.removeAt(index);
sort(fqItems);
}
}
Step 1: Add compareTo method to class:
class Student {
String? name;
int? age;
Student({this.name, this.age});
int getAge() {
if (age == null) return 0;
return age!;
}
#override
int compareTo(Student other) {
var a = getAge();
var b = other.getAge();
if (a < b) {
return -1;
} else if (a > b) {
return 1;
} else {
return 0;
}
}
}
Step 2: Sorting your list:
By ascending:
studentList.sort((a, b) {
return a.compareTo(b);
});
By descending:
studentList.sort((a, b) {
return b.compareTo(a);
});
How to sort a list of objects by the alphabetical order of one of its properties (Not the name but the actual value the property holds)?
You can pass a comparison function to List.sort.
someObjects.sort((a, b) => a.someProperty.compareTo(b.someProperty));
In general, you can provide a custom comparison function to List.sort.
/// Desired relation | Result
/// -------------------------------------------
/// a < b | Returns a negative value.
/// a == b | Returns 0.
/// a > b | Returns a positive value.
///
int mySortComparison(SomeClass a, SomeClass b) {
final propertyA = someProperty(a);
final propertyB = someProperty(b);
if (propertyA < propertyB) {
return -1;
} else if (propertyA > propertyB) {
return 1;
} else {
return 0;
}
}
list.sort(mySortComparison);
If you're sorting some custom class you own, you alternatively could make your class implement the Comparable interface:
class MyCustomClass implements Comparable<MyCustomClass> {
...
#override
int compareTo(MyCustomClass other) {
if (someProperty < other.someProperty) {
return -1;
} else if (someProperty > other.someProperty) {
return 1;
} else {
return 0;
}
}
}
and then you can use list.sort() directly without supplying a callback.
Note that if you're sorting by a single property that already implements the Comparable interface, implementing the comparison functions is much simpler. For example:
class MyCustomClass implements Comparable<MyCustomClass> {
...
#override
int compareTo(MyCustomClass other) =>
someProperty.compareTo(other.someProperty);
}
Reversing
If you want to reverse the sort order, you can:
Make your comparison function return a value with the opposite sign.
Alternatively just explicitly reverse the list after sorting:
list = (list..sort()).reversed.toList();
Sorting by multiple properties (a.k.a. subsorting)
There are a variety of ways to sort by multiple properties.
A general way is to perform a stable sort for each property in reverse order of importance. For example, if you want to sort names primarily by surname and then subsort within surnames by given name, then you would first sort by given names, and then perform a stable sort by surname. See below for how to perform a stable sort.
Alternatively, you could sort with a comparison function that itself checks multiple properties. For example:
class Name {
Name({String? surname, String? givenName})
: surname = surname ?? "",
givenName = givenName ?? "";
final String surname;
final String givenName;
}
int compareNames(Name name1, Name name2) {
var comparisonResult = name1.surname.compareTo(name2.surname);
if (comparisonResult != 0) {
return comparisonResult;
}
// Surnames are the same, so subsort by given name.
return name1.givenName.compareTo(name2.givenName);
}
package:collection provides an extension to chain comparison functions so that combining them is a bit more straightforward and less error-prone:
import 'package:collection/collection.dart';
int compareSurnames(Name name1, Name name2) =>
name1.surname.compareTo(name2.surname);
int compareGivenNames(Name name1, Name name2) =>
name1.givenName.compareTo(name2.givenName);
final compareNames = compareSurnames.then(compareGivenNames);
My dartbag package also provides a compareIterables function that allows comparing Lists of property values in order of importance:
import 'package:dartbag/collection.dart';
int compareNames(Name name1, Name name2) =>
compareIterables(
[name1.surname, name1.givenName],
[name2.surname, name2.givenName],
);
Okay, I want a stable sort
List.sort is not guaranteed to be a stable sort. If you need a stable sort, package:collection provides insertionSort and mergeSort implementations that are stable.
But comparing might be expensive
Suppose you have a custom comparison function that looks something like:
int compareMyCustomClass(MyCustomClass a, MyCustomClass b) {
var a0 = computeValue(a);
var b0 = computeValue(b);
return a0.compareTo(b0);
}
The sorting process might call computeValue multiple times for the same object, which is particularly wasteful if computeValue() is expensive. In such cases, a Schwartzian transform could be faster (at the expense of using more memory). This approach maps your objects to directly sortable keys, sorts those keys, and extracts the original objects. (This is how Python's sort and sorted functions work.)
Here's one possible implementation:
class _SortableKeyPair<T, K extends Comparable<Object>>
implements Comparable<_SortableKeyPair<T, K>> {
_SortableKeyPair(this.original, this.key);
final T original;
final K key;
#override
int compareTo(_SortableKeyPair<T, K> other) => key.compareTo(other.key);
}
/// Returns a sorted *copy* of [items] according to the computed sort key.
List<E> sortedWithKey<E, K extends Comparable<Object>>(
Iterable<E> items,
K Function(E) toKey,
) {
final keyPairs = [
for (var element in items) _SortableKeyPair(element, toKey(element)),
]..sort();
return [
for (var keyPair in keyPairs) keyPair.original,
];
}
void main() {
final list = <MyCustomClass>[ ... ];
final sorted = sortedWithKeys(list, computeValue);
}
My dartbag package provides such a sortWithKey function (and also a sortWithAsyncKey function if the key needs to be generated asynchronously).
If you want to sort the object "objects" by the property "name" do something like this
objects.sort((a, b) {
return a.value['name'].toString().toLowerCase().compareTo(b.value['name'].toString().toLowerCase());
});
Immutable extension sortedBy for List.
extension MyIterable<E> on Iterable<E> {
Iterable<E> sortedBy(Comparable key(E e)) =>
toList()..sort((a, b) => key(a).compareTo(key(b)));
}
And use
list.sortedBy((it) => it.name);
Here is my contribution to this good question. If someone is facing difficulty to understand how the #Nate Bosch answer is working & you want to sort your custom model class list then you can do this way.
1. You have to implement Comparable abstract class in your model class.
It has the method compareTo which you have to override.
For example, I have this StudentMarks model class which has marks property in it.
class StudentMarks implements Comparable {
int marks;
StudentMarks({
this.marks,
});
#override
int compareTo(other) {
if (this.marks == null || other == null) {
return null;
}
if (this.marks < other.marks) {
return 1;
}
if (this.marks > other.marks) {
return -1;
}
if (this.marks == other.marks) {
return 0;
}
return null;
}
}
2. Now you can call compareTo method inside the sort method.
void _sortStudents({bool reversed: false}) {
_students.sort((a, b) {
return a.compareTo(b);
});
if (reversed) {
_students = _students.reversed.toList();
}
setState(() {});
}
Refer to this link you want to know more about the Comparable class
https://api.dart.dev/stable/2.1.0/dart-core/Comparable-class.html
Its worked for me:
myList..sort((a, b) => a.name.toLowerCase().compareTo(b.name.toLowerCase()));
Using Comparatorfunction, sort Users by id.
Comparator<UserModel> sortById = (a, b) => a.id.compareTo(b.id);
users.sort(sortById);
Now we can sort it in reversed/descending order.
users = users.reversed.toList();
To sort it in reverse order :
list.sort((a, b) {
return b.status.toLowerCase().compareTo(a.status.toLowerCase());
});
What's more, you can use Comparable.compare for more clear, for example:
class _Person {
final int age;
final String name;
_Person({required this.age, required this.name});
}
void _test() {
final array = [
_Person(age: 10, name: 'Dean'),
_Person(age: 20, name: 'Jack'),
_Person(age: 30, name: 'Ben'),
];
// ascend with age
// Dean Jack Ben
array.sort((p1, p2) {
return Comparable.compare(p1.age, p2.age);
});
// decend with age
// Ben Jack Dean
array.sort((p1, p2) {
return Comparable.compare(p2.age, p1.age);
});
// ascend with name
// Ben Dean Jack
array.sort((p1, p2) {
return Comparable.compare(p1.name, p2.name);
});
}
Similar to #pavel-shorokhovs answer, but strongly typed:
extension IterableExtensions<T> on Iterable<T> {
Iterable<T> sortBy<TSelected extends Comparable<TSelected>>(
TSelected Function(T) selector) =>
toList()..sort((a, b) => selector(a).compareTo(selector(b)));
Iterable<T> sortByDescending<TSelected extends Comparable<TSelected>>(
TSelected Function(T) selector) =>
sortBy(selector).toList().reversed;
}
i had fpgrowth machine learning output/result with each element of list contains another list and frequency field i was to sort by frequency in descending order so i used a bit of recursion for that try it might work i know i am late but i am posting maybe someone else could benefit.
sort(List<FrequentItem> fqItems) {
int len = fqItems.length;
if(len==2){
if(fqItems[0].frequency>fqItems[1].frequency){
sortedItems.add(fqItems[0]);
sortedItems.add(fqItems[1]);
}else{
sortedItems.add(fqItems[1]);
sortedItems.add(fqItems[0]);
}
return;
}else{
FrequentItem max = fqItems[0];
int index =0;
for(int i=0;i<len-2;i++){
if(max.frequency<fqItems[i+1].frequency){
max = fqItems[i+1];
index = i+1;
}
}
sortedItems.add(max);
fqItems.removeAt(index);
sort(fqItems);
}
}
Step 1: Add compareTo method to class:
class Student {
String? name;
int? age;
Student({this.name, this.age});
int getAge() {
if (age == null) return 0;
return age!;
}
#override
int compareTo(Student other) {
var a = getAge();
var b = other.getAge();
if (a < b) {
return -1;
} else if (a > b) {
return 1;
} else {
return 0;
}
}
}
Step 2: Sorting your list:
By ascending:
studentList.sort((a, b) {
return a.compareTo(b);
});
By descending:
studentList.sort((a, b) {
return b.compareTo(a);
});
I'm comparing 2 ways to filter lists, with and without using streams. It turns out that the method without using streams is faster for a list of 10,000 items. I'm interested in understanding why is it so. Can anyone explain the results please?
public static int countLongWordsWithoutUsingStreams(
final List<String> words, final int longWordMinLength) {
words.removeIf(word -> word.length() <= longWordMinLength);
return words.size();
}
public static int countLongWordsUsingStreams(final List<String> words, final int longWordMinLength) {
return (int) words.stream().filter(w -> w.length() > longWordMinLength).count();
}
Microbenchmark using JMH:
#Benchmark
#BenchmarkMode(Throughput)
#OutputTimeUnit(MILLISECONDS)
public void benchmarkCountLongWordsWithoutUsingStreams() {
countLongWordsWithoutUsingStreams(nCopies(10000, "IAmALongWord"), 3);
}
#Benchmark
#BenchmarkMode(Throughput)
#OutputTimeUnit(MILLISECONDS)
public void benchmarkCountLongWordsUsingStreams() {
countLongWordsUsingStreams(nCopies(10000, "IAmALongWord"), 3);
}
public static void main(String[] args) throws RunnerException {
final Options opts = new OptionsBuilder()
.include(PracticeQuestionsCh8Benchmark.class.getSimpleName())
.warmupIterations(5).measurementIterations(5).forks(1).build();
new Runner(opts).run();
}
java -jar target/benchmarks.jar -wi 5 -i 5 -f 1
Benchmark Mode Cnt Score Error Units
PracticeQuestionsCh8Benchmark.benchmarkCountLongWordsUsingStreams thrpt 5 10.219 ± 0.408 ops/ms
PracticeQuestionsCh8Benchmark.benchmarkCountLongWordsWithoutUsingStreams thrpt 5 910.785 ± 21.215 ops/ms
Edit: (as someone deleted the update posted as an answer)
public class PracticeQuestionsCh8Benchmark {
private static final int NUM_WORDS = 10000;
private static final int LONG_WORD_MIN_LEN = 10;
private final List<String> words = makeUpWords();
public List<String> makeUpWords() {
List<String> words = new ArrayList<>();
final Random random = new Random();
for (int i = 0; i < NUM_WORDS; i++) {
if (random.nextBoolean()) {
/*
* Do this to avoid string interning. c.f.
* http://en.wikipedia.org/wiki/String_interning
*/
words.add(String.format("%" + LONG_WORD_MIN_LEN + "s", i));
} else {
words.add(String.valueOf(i));
}
}
return words;
}
#Benchmark
#BenchmarkMode(AverageTime)
#OutputTimeUnit(MILLISECONDS)
public int benchmarkCountLongWordsWithoutUsingStreams() {
return countLongWordsWithoutUsingStreams(words, LONG_WORD_MIN_LEN);
}
#Benchmark
#BenchmarkMode(AverageTime)
#OutputTimeUnit(MILLISECONDS)
public int benchmarkCountLongWordsUsingStreams() {
return countLongWordsUsingStreams(words, LONG_WORD_MIN_LEN);
}
}
public static int countLongWordsWithoutUsingStreams(
final List<String> words, final int longWordMinLength) {
final Predicate<String> p = s -> s.length() >= longWordMinLength;
int count = 0;
for (String aWord : words) {
if (p.test(aWord)) {
++count;
}
}
return count;
}
public static int countLongWordsUsingStreams(final List<String> words,
final int longWordMinLength) {
return (int) words.stream()
.filter(w -> w.length() >= longWordMinLength).count();
}
Whenever your benchmark says that some operation over 10000 elements takes 1ns (edit: 1µs), you probably found a case of clever JVM figuring out that your code doesn't actually do anything.
Collections.nCopies doesn't actually make a list of 10000 elements. It makes a sort of a fake list with 1 element and a count of how many times it's supposedly there. That list is also immutable, so your countLongWordsWithoutUsingStreams would throw an exception if there was something for removeIf to do.
You do not return any values from your benchmark methods, thus, JMH has no chance to escape the computed values and your benchmark suffers dead code elimination. You compute how long it takes to do nothing. See the JMH page for further guidance.
Saying this, streams can be slower in some cases: Java 8: performance of Streams vs Collections