I use following Groovy code to select random elements from a list.
Random rnd = new Random()
log.info("Random element:" + list[rnd.nextInt(list.size)])
I get previous or next element as a "random" value in most cases.
Example with seeding.
In the past, I found that when generating random numbers, they were not quite 'random enough' and I have tried a few different way to make it more random.
E.g.
Date date = new Date();
def now = date.getTime();
log.info(now); // This value is the number of millis since January 1, 1970
Random rnd = new Random();
rnd.setSeed(now); // Seed the Random with the millis value
def play = randomValue.nextInt(10); // Value in brackets is the max number of interest
log.info(play);
When not working in ranges and just need a x-digit random value, I have done something like...
randomValue = new Random();
val1 = randomValue.nextInt(10);
val2 = randomValue.nextInt(10);
val3 = randomValue.nextInt(10);
val4 = randomValue.nextInt(10);
val5 = randomValue.nextInt(10);
orderRef = val1.toString() + val2.toString() + val3.toString()+ val4.toString()+ val5.toString();
Try this
def <T> T getRandomElement(List<T> list) {
if (list) {
Collections.shuffle(list)
list.first()
}
}
Be aware that Collections.shuffle mutates the list. This might matter if there are references to the list elsewhere.
Related
I am programming in Kotlin and have a MutableList from which I would like to remove the first n elements from that specific list instance. This means that functions like MutableList.drop(n) are out of the question.
One solution would of course be to loop and call MutableList.removeFirst() n times, but this feels inefficient, being O(n). Another way would be to choose another data type, but I would prefer not to clutter my project by implementing my own data type for this, if I can avoid it.
Is there a faster way to do this with a MutableList? If not, is there another built-in data type that can achieve this in less than O(n)?
In my opinion the best way to achieve this is
abstract fun subList(fromIndex: Int, toIndex: Int): List<E>.
https://kotlinlang.org/api/latest/jvm/stdlib/kotlin.collections/-list/sub-list.html
Under the hood it creates a new instance of list(SubList class for AbstractClass) with elements between the selected indexes.
Using:
val yourList = listOf<YourType>(...)
val yourNewList = yourList.subList(5, yourList.size)
// return list from 6th elem to last
One method which seems to be faster if n is sufficiently large seems to be the following:
Store the last listSize - n bytes to keep in a temporary list,
Clear original list instance
Add temporary list to original list
Here is a quick benchmark for some example values that happen to fit my use case:
val numRepetitions = 15_000
val listSize = 1_000
val maxRemove = listSize
val rnd0 = Random(0)
val rnd1 = Random(0)
// 1. Store the last `listSize - n` bytes to keep in a temporary list,
// 2. Clear original list
// 3. Add temporary list to original list
var accumulatedMsClearAddAll = 0L
for (i in 0 until numRepetitions) {
val l = Random.nextBytes(listSize).toMutableList()
val numRemove = rnd0.nextInt(maxRemove)
val numKeep = listSize - numRemove
val startTime = System.currentTimeMillis()
val expectedOutput = l.takeLast(numKeep)
l.clear()
l.addAll(expectedOutput)
val endTime = System.currentTimeMillis()
assert(l == expectedOutput)
accumulatedMsClearAddAll += endTime - startTime
}
// Iteratively remove the first byte `n` times.
var accumulatedMsIterative = 0L
for (i in 0 until numRepetitions) {
val numRemove = rnd1.nextInt(maxRemove)
val l = Random.nextBytes(listSize).toMutableList()
val expectedOutput = l.takeLast(listSize - numRemove)
val startTime = System.currentTimeMillis()
for (ii in 0 until numRemove) {
l.removeFirst()
}
val endTime = System.currentTimeMillis()
assert(l == expectedOutput)
accumulatedMsIterative += endTime - startTime
}
println("clear+addAll removal: $accumulatedMsClearAddAll ms")
println("Iterative removal: $accumulatedMsIterative ms")
Output:
Clear+addAll removal: 478 ms
Iterative removal: 12683 ms
I am creating a list of random numbers using the following approach
val randomList = List(4) { Random.nextInt(0, 100) }
However, this approach doesn't work as I want to avoid repetitions
One way is to shuffle a Range and take as many items as you want:
val randomList = (0..99).shuffled().take(4)
This is not so efficient if the range is big and you only need just a few numbers.In this case it's better to use a Set like this:
val s: MutableSet<Int> = mutableSetOf()
while (s.size < 4) { s.add((0..99).random()) }
val randomList = s.toList()
Create:
val list = (0 until 100).toMutableList()
val randList = mutableListOf<Int>()
for (i in 0 until 4) {
val uniqueRand = list.random()
randList.add(uniqueRand)
list.remove(uniqueRand)
}
One line approach to get a list of n distinct random elements. Random is not limited in any way.
val list = mutableSetOf<Int>().let { while (it.size() < n) it += Random.nextInt(0, 100) }.toList()
Edit: Here is the code I have so far for generating the Patient Oct-Tuples.
(thanks Anon for giving me the bost on how to calculate weighted probability/setting the seed)
fun genPatients(x:int) =
let
val seed=let
val m=Date.minute(Date.fromTimeLocal(Time.now()))
val s=Date.second(Date.fromTimeLocal(Time.now()))
in Random.rand(m,s)
end;
val survivalrate = ref(1)
val numl = ref(1)
val td = ref(1)
val xray = ref(false)
val count= ref(0)
val emnum= ref(1000)
val ageList = [1, 2, 3, 3];
val xrayList=[false,true];
val age = Random.randRange (0, 3) seed;(* random age*)
val nextInt1 = Random.randRange(0, 1)(* random xray*)
val r1 = Random.rand(1,1)
val nextInt2 = Random.randRange(1, 10000000)(* random td*)
val r2 = Random.rand(1,1)
val r1hold= ref(1);
in
while !count < x do
(
count:= !count + 1;
List.nth(ageList, age);
r1hold:= nextInt1 r1;
td:= nextInt2 r2;
(!emnum,age,survivalrate,numl,[],[],xray,td);
emnum:= !emnum + 1
)
end;
My question now is now how to go about indexing a boolean list?
So I was looking for some help defining my Oct-tuple to finish up my project and lo and behold I find someone posting the entirety of my project hoping for a handout answer. Not only that, but I'm almost certain we're in the same class, and you think posting this the night before the morning the project is due is what a responsible student does? Pretty sure nobody on SO is gonna do your homework for you anyway, in fact I'm not even sure it's allowed.
Maybe do some work and then ask for help when you've actually done anything. Or maybe in the next phase try a little harder.
EDIT: I'll give you something to get you started.
To calculate weighted probability you need a seed.
val seed=let
val m=Date.minute(Date.fromTimeLocal(Time.now()))
val s=Date.second(Date.fromTimeLocal(Time.now()))
in Random.rand(m,s)
end;
Here's one. Then you can calculate probability, at least for the age, like this:
val ageList = [1, 2, 3, 3];
val ageInt = Random.randRange (0, 3) seed;
List.nth(ageList, ageInt)
This was how I decided to do the weighted probability portion, you can equate this to the other weighted sections if you're creative. Good luck.
i m using Zendframework2 and using count function to count the number of values, it fails when i try to count the integer values
$q = $dm->createQueryBuilder('Admin\Document\Institution')
->field('id')->notEqual($id)
->field('coreid')->equals($post['coreid']);
$coreid = $q->getQuery()->execute()->count();
does count function works for integer valeues in Doctrine2 ODM?
If you change your query slightly then possibly get your count value.
$q = $dm->createQueryBuilder('Admin\Document\Institution')
->select('count(id) as id')
->where('id !=:id AND coreid != :coreid')
->setParameters(array('id'=> $id,'coreid'=>$coreid);
$coreid = $q->getQuery()->execute();
I'm looking every where on the web (dart website, stackoverflow, forums, etc), and I can't find my answer.
So there is my problem: I need to write a function, that print a random sort of a list, witch is provided as an argument. : In dart as well.
I try with maps, with Sets, with list ... I try the method with assert, with sort, I look at random method with Math on dart librabry ... nothing can do what I wana do.
Can some one help me with this?
Here some draft:
var element03 = query('#exercice03');
var uneliste03 = {'01':'Jean', '02':'Maximilien', '03':'Brigitte', '04':'Sonia', '05':'Jean-Pierre', '06':'Sandra'};
var alluneliste03 = new Map.from(uneliste03);
assert(uneliste03 != alluneliste03);
print(alluneliste03);
var ingredients = new Set();
ingredients.addAll(['Jean', 'Maximilien', 'Brigitte', 'Sonia', 'Jean-Pierre', 'Sandra']);
var alluneliste03 = new Map.from(ingredients);
assert(ingredients != alluneliste03);
//assert(ingredients.length == 4);
print(ingredients);
var fruits = <String>['bananas', 'apples', 'oranges'];
fruits.sort();
print(fruits);
There is a shuffle method in the List class. The methods shuffles the list in place. You can call it without an argument or provide a random number generator instance:
var list = ['a', 'b', 'c', 'd'];
list.shuffle();
print('$list');
The collection package comes with a shuffle function/extension that also supports specifying a sub range to shuffle:
void shuffle (
List list,
[int start = 0,
int end]
)
Here is a basic shuffle function. Note that the resulting shuffle is not cryptographically strong. It uses Dart's Random class, which produces pseudorandom data not suitable for cryptographic use.
import 'dart:math';
List shuffle(List items) {
var random = new Random();
// Go through all elements.
for (var i = items.length - 1; i > 0; i--) {
// Pick a pseudorandom number according to the list length
var n = random.nextInt(i + 1);
var temp = items[i];
items[i] = items[n];
items[n] = temp;
}
return items;
}
main() {
var items = ['foo', 'bar', 'baz', 'qux'];
print(shuffle(items));
}
You can use shuffle() with 2 dots like Vinoth Vino said.
List cities = ["Ankara","London","Paris"];
List mixed = cities..shuffle();
print(mixed);
// [London, Paris, Ankara]