I need to generate random numbers from 1 to 100 and I know how to do that part...
I need to ask user how many numbers he wants to generate(if he says 5 the program needs to generate 5 numbers from 1 to 100). I only now how to make a fixable amount by adding new int's in a list.
I did achieve that before, but then i couldn't make it work, so it would write average of those numbers and min+max value.
Here is my code below:
Random k = new Random();
//here i added in the same way other variables and put them in a list
int j = k.Next(100);
Console.WriteLine("");
double[] list1 = {j};
double povp = list1.Average();
Console.WriteLine(povp);
Console.WriteLine("");
Console.WriteLine(list1.Max());
Console.WriteLine("");
Console.WriteLine(list1.Min());
Console.ReadKey();
You could use the following code to generate N numbers:
IEnumerable<int> numbers = Enumerable.Repeat(1,N).Select(_ => random.Next(100));
// ask user for input
string input = Console.Readline();
int parsed;
// parse to int, needs error checking (will throw exception when input is not a valid int)
int.TryParse(input, out parsed);
Random random = new Random();
List<double> list = new List<double>();
for(int i = 0; i < parsed; parsed++)
{
list.Add(random.Next(100));
}
public void Main()
{
const int NUMBERS_FROM = 1;
const int NUMBERS_TO = 100;
int n = int.Parse(Console.ReadLine());
Random rnd = new Random();
List<int> numbers = new List<int>();
for (int i = 0; i < n; i++)
{
int rndNumber = rnd.Next(NUMBERS_FROM, NUMBERS_TO + 1);
numbers.Add(rndNumber);
}
Console.WriteLine("Numbers : {0}",string.Join(", ",numbers));
}
this will generate N numbers and add them to a list and then print them to the console. I think this is what you were looking for
Related
Hi guy i try to generate 50 number with 5 digit depend on user total summary. For example, User give 500000 and then i need to random number with 5 digit by 50 number equal 500000
i try this but it isn't 5 digit number
int balane = 500000;
int nums = 50;
int max = balane / nums;
Random rand = new Random();
int newNum = 0;
int[] ar = new int[nums];
for (int i = 0; i < nums - 1; i++)
{
newNum = rand.Next(0, max);
ar[i] = newNum;
balane -= newNum;
max = balane / (nums - i - 1);
ar[nums - 1] = balane;
}
int check = 0;
foreach (int x in ar)
{
check += x;
}
i tried already but value inside my array have negative value i want to get only
positive value
Please help me, how to solve this and thank for advance.
I once asked a similar question on codereview.stackexchange.com. I have modified my answer to produce a five digit sequence for you.
Furthermore, this code is fast enough to be used to create tens of thousands of codes in a single request. If you look at the initial question and answer (linked to below) you will find that it checks to see whether the code has been used or not prior to inserting it. Thus, the codes are unique.
void Main()
{
Console.WriteLine(GenerateCode(CodeLength));
}
private const int CodeLength = 10;
// since Random does not make any guarantees of thread-safety, use different Random instances per thread
private static readonly ThreadLocal<Random> _random = new ThreadLocal<Random>(() => new Random());
// Define other methods and classes here
private static string GenerateCode(int numberOfCharsToGenerate)
{
char[] chars = "0123456789".ToCharArray();
var sb = new StringBuilder();
for (int i = 0; i < numberOfCharsToGenerate; i++)
{
int num = _random.Value.Next(0, chars.Length);
sb.Append(chars[num]);
}
return sb.ToString();
}
Original question and answer: https://codereview.stackexchange.com/questions/142049/creating-a-random-code-and-saving-if-it-does-not-exist/142056#142056
Perhaps try this:
var rnd = new Random();
var numbers = Enumerable.Range(0, 50).Select(x => rnd.Next(500_000)).OrderBy(x => x).ToArray();
numbers = numbers.Skip(1).Zip(numbers, (x1, x0) => x1 - x0).ToArray();
numbers = numbers.Append(500_000 - numbers.Sum()).ToArray();
Console.WriteLine(numbers.Count());
Console.WriteLine(numbers.Sum());
This outputs:
50
500000
This works by generating 50 random numbers between 0 and 499,999 inclusively. It then sorts them ascendingly and then gets the difference between each successive pair. This by definition produces a set of 49 values that almost adds up to 500,000. It's then just a matter of adding the one missing number by doing 500_000 - numbers.Sum().
I've searched for a while and been struggling to find this, I'm trying to generate several random, unique numbers is C#. I'm using System.Random, and I'm using a DateTime.Now.Ticks seed:
public Random a = new Random(DateTime.Now.Ticks.GetHashCode());
private void NewNumber()
{
MyNumber = a.Next(0, 10);
}
I'm calling NewNumber() regularly, but the problem is I often get repeated numbers. Some people suggested because I was declaring the random every time I did it, it would not produce a random number, so I put the declaration outside my function. Any suggestions or better ways than using System.Random ? Thank you
I'm calling NewNumber() regularly, but the problem is I often get
repeated numbers.
Random.Next doesn't guarantee the number to be unique. Also your range is from 0 to 10 and chances are you will get duplicate values. May be you can setup a list of int and insert random numbers in the list after checking if it doesn't contain the duplicate. Something like:
public Random a = new Random(); // replace from new Random(DateTime.Now.Ticks.GetHashCode());
// Since similar code is done in default constructor internally
public List<int> randomList = new List<int>();
int MyNumber = 0;
private void NewNumber()
{
MyNumber = a.Next(0, 10);
if (!randomList.Contains(MyNumber))
randomList.Add(MyNumber);
}
You might try shuffling an array of possible ints if your range is only 0 through 9. This adds the benefit of avoiding any conflicts in the number generation.
var nums = Enumerable.Range(0, 10).ToArray();
var rnd = new Random();
// Shuffle the array
for (int i = 0;i < nums.Length;++i)
{
int randomIndex = rnd.Next(nums.Length);
int temp = nums[randomIndex];
nums[randomIndex] = nums[i];
nums[i] = temp;
}
// Now your array is randomized and you can simply print them in order
for (int i = 0;i < nums.Length;++i)
Console.WriteLine(nums[i]);
NOTE, I dont recommend this :).
Here's a "oneliner" as well:
var result = Enumerable.Range(0,9).OrderBy(g => Guid.NewGuid()).ToArray();
I'm posting a correct implementation of a shuffle algorithm, since the other one posted here doesn't produce a uniform shuffle.
As the other answer states, for small numbers of values to be randomized, you can simply fill an array with those values, shuffle the array, and then use however many of the values that you want.
The following is an implementation of the Fisher-Yates Shuffle (aka the Knuth Shuffle). (Read the "implementation errors" section of that link (search for "always selecting j from the entire range of valid array indices on every iteration") to see some discussion about what is wrong with the other implementation posted here.)
using System;
using System.Collections.Generic;
namespace ConsoleApplication2
{
static class Program
{
static void Main(string[] args)
{
Shuffler shuffler = new Shuffler();
List<int> list = new List<int>{ 1, 2, 3, 4, 5, 6, 7, 8, 9 };
shuffler.Shuffle(list);
foreach (int value in list)
{
Console.WriteLine(value);
}
}
}
/// <summary>Used to shuffle collections.</summary>
public class Shuffler
{
public Shuffler()
{
_rng = new Random();
}
/// <summary>Shuffles the specified array.</summary>
/// <typeparam name="T">The type of the array elements.</typeparam>
/// <param name="array">The array to shuffle.</param>
public void Shuffle<T>(IList<T> array)
{
for (int n = array.Count; n > 1; )
{
int k = _rng.Next(n);
--n;
T temp = array[n];
array[n] = array[k];
array[k] = temp;
}
}
private System.Random _rng;
}
}
This is a unity only answer:
Check this ready-to-use method: Give in a range & count of number you want to get.
public static int[] getUniqueRandomArray(int min, int max, int count) {
int[] result = new int[count];
List<int> numbersInOrder = new List<int>();
for (var x = min; x < max; x++) {
numbersInOrder.Add(x);
}
for (var x = 0; x < count; x++) {
var randomIndex = UnityEngine.Random.Range(0, numbersInOrder.Count);
result[x] = numbersInOrder[randomIndex];
numbersInOrder.RemoveAt(randomIndex);
}
return result;
}
Same as #Habib's answer, but as a function:
List<int> randomList = new List<int>();
int UniqueRandomInt(int min, int max)
{
var rand = new Random();
int myNumber;
do
{
myNumber = rand.Next(min, max);
} while (randomList.Contains(myNumber));
return myNumber;
}
If randomList is a class property, UniqueRandomInt will return unique integers in the context of the same instance of that class. If you want it to be unique globally, you will need to make randomList static.
Depending on what you are really after you can do something like this:
using System;
using System.Collections.Generic;
using System.Linq;
namespace SO14473321
{
class Program
{
static void Main()
{
UniqueRandom u = new UniqueRandom(Enumerable.Range(1,10));
for (int i = 0; i < 10; i++)
{
Console.Write("{0} ",u.Next());
}
}
}
class UniqueRandom
{
private readonly List<int> _currentList;
private readonly Random _random = new Random();
public UniqueRandom(IEnumerable<int> seed)
{
_currentList = new List<int>(seed);
}
public int Next()
{
if (_currentList.Count == 0)
{
throw new ApplicationException("No more numbers");
}
int i = _random.Next(_currentList.Count);
int result = _currentList[i];
_currentList.RemoveAt(i);
return result;
}
}
}
And here my version of finding N random unique numbers using HashSet.
Looks pretty simple, since HashSet can contain only different items.
It's interesting - would it be faster then using List or Shuffler?
using System;
using System.Collections.Generic;
namespace ConsoleApplication1
{
class RnDHash
{
static void Main()
{
HashSet<int> rndIndexes = new HashSet<int>();
Random rng = new Random();
int maxNumber;
Console.Write("Please input Max number: ");
maxNumber = int.Parse(Console.ReadLine());
int iter = 0;
while (rndIndexes.Count != maxNumber)
{
int index = rng.Next(maxNumber);
rndIndexes.Add(index);
iter++;
}
Console.WriteLine("Random numbers were found in {0} iterations: ", iter);
foreach (int num in rndIndexes)
{
Console.WriteLine(num);
}
Console.ReadKey();
}
}
}
I noted that the accepted answer keeps adding int to the list and keeps checking them with if (!randomList.Contains(MyNumber)) and I think this doesn't scale well, especially if you keep asking for new numbers.
I would do the opposite.
Generate the list at startup, linearly
Get a random index from the list
Remove the found int from the list
This would require a slightly bit more time at startup, but will scale much much better.
public class RandomIntGenerator
{
public Random a = new Random();
private List<int> _validNumbers;
private RandomIntGenerator(int desiredAmount, int start = 0)
{
_validNumbers = new List<int>();
for (int i = 0; i < desiredAmount; i++)
_validNumbers.Add(i + start);
}
private int GetRandomInt()
{
if (_validNumbers.Count == 0)
{
//you could throw an exception here
return -1;
}
else
{
var nextIndex = a.Next(0, _validNumbers.Count - 1);
var number = _validNumbers[nextIndex];
_validNumbers.RemoveAt(nextIndex);
return number;
}
}
}
It's may be a little bit late, but here is more suitable code, for example when you need to use loops:
List<int> genered = new List<int>();
Random rnd = new Random();
for(int x = 0; x < files.Length; x++)
{
int value = rnd.Next(0, files.Length - 1);
while (genered.Contains(value))
{
value = rnd.Next(0, files.Length - 1);
}
genered.Add(value);
returnFiles[x] = files[value];
}
with Functional way*
static Func<int> GetNextUniqueIntegerFunc(int min, int max)
{
var list = new List<int>();
var random = new Random();
int getNextValue()
{
while (true)
{
var random_number = random.Next(min, max);
if (!list.Contains(random_number))
{
list.Add(random_number);
return random_number;
}
}
}
return getNextValue;
}
unique random number from 0 to 9
int sum = 0;
int[] hue = new int[10];
for (int i = 0; i < 10; i++)
{
int m;
do
{
m = rand.Next(0, 10);
} while (hue.Contains(m) && sum != 45);
if (!hue.Contains(m))
{
hue[i] = m;
sum = sum + m;
}
}
You could also use a dataTable storing each random value, then simply perform the random method while != values in the dataColumn
randomNumber function return unqiue integer value between 0 to 100000
bool check[] = new bool[100001];
Random r = new Random();
public int randomNumber() {
int num = r.Next(0,100000);
while(check[num] == true) {
num = r.Next(0,100000);
}
check[num] = true;
return num;
}
hi here i posted one video ,and it explains how to generate unique random number
public List<int> random_generator(){
Random random = new Random();
List<int> random_container = new List<int>;
do{
int random_number = random.next(10);
if(!random_container.contains(random_number){
random_container.add(random_number)
}
}
while(random_container.count!=10);
return random_container;
}
here ,,, in random container you will get non repeated 10 numbers starts from 0 to 9(10 numbers) as random.. thank you........
You can use basic Random Functions of C#
Random ran = new Random();
int randomno = ran.Next(0,100);
you can now use the value in the randomno in anything you want but keep in mind that this will generate a random number between 0 and 100 Only and you can extend that to any figure.
Try this:
private void NewNumber()
{
Random a = new Random(Guid.newGuid().GetHashCode());
MyNumber = a.Next(0, 10);
}
Some Explnations:
Guid : base on here : Represents a globally unique identifier (GUID)
Guid.newGuid() produces a unique identifier like "936DA01F-9ABD-4d9d-80C7-02AF85C822A8"
and it will be unique in all over the universe base on here
Hash code here produce a unique integer from our unique identifier
so Guid.newGuid().GetHashCode() gives us a unique number and the random class will produce real random numbers throw this
Sample:
https://rextester.com/ODOXS63244
generated ten random numbers with this approach with result of:
-1541116401
7
-1936409663
3
-804754459
8
1403945863
3
1287118327
1
2112146189
1
1461188435
9
-752742620
4
-175247185
4
1666734552
7
we got two 1s next to each other, but the hash codes do not same.
I want to generate an unique 4 digit random number. This is the below code what I have tried:
Code for generating random number
//Generate RandomNo
public int GenerateRandomNo()
{
int _min = 0000;
int _max = 9999;
Random _rdm = new Random();
return _rdm.Next(_min, _max);
}
The problem is I have received a random no with value 241 which is not a 4 digit number. Is there any problems with the code?
//Generate RandomNo
public int GenerateRandomNo()
{
int _min = 1000;
int _max = 9999;
Random _rdm = new Random();
return _rdm.Next(_min, _max);
}
you need a 4 digit code, start with 1000
private Random _random = new Random();
public string GenerateRandomNo()
{
return _random.Next(0, 9999).ToString("D4");
}
241 is a four digit number, if you use leading zeros: 0241.
Display the returned number with a format string like this:
String.Format("{0:0000}", n);
Just one line code
int num = new Random().Next(1000, 9999);
0 is the same as 0000.
241 is the same as 0241.
You could format the integer to a string with a leading zero.
Random generator = new Random();
string number = generator.Next(1, 10000).ToString("D4");
use: int _min = 1000;
or use leading 0 in case if you want 0241
I suggest to create new list and check if this list contains any of number
var IdList = new List<int>();
do
{
billId = random.Next(1, 9000);
} while (IdList.Contains(billId));
IdList.Add(billId);
int NoDigits = 4;
Random rnd = new Random();
textBox2.Text = rnd.Next((int)Math.Pow(10, (NoDigits - 1)), (int)Math.Pow(10, NoDigits) -1).ToString();
Expanding on the answer from brij but with 0000 to 9999 rather than 1000 to 9999
string formatting = "0000"; //Will pad out to four digits if under 1000
int _min = 0;
int _max = 9999;
Random randomNumber = new Random();
var randomNumberString = randomNumber.Next(_min, _max).ToString(formatting);
or if you want to minimalize lines:
Random randomNumber = new Random();
var randomNumberString = randomNumber.Next(0, 9999).ToString("0000");
Using this you will avoid starting numbers with 00[...] and you can also specify the length.
string RandomNumbers(int Length)
{
Random Rand = new Random();
StringBuilder SB = new StringBuilder();
for (int i = 0; i < Length; i++)
SB.Append(Rand.Next(0, 9));
return SB.ToString();
}
RandomNumbers(4) // OUTPUT: 9301, 4936, 0692, etc ...
Here is Method to generate any digits number. The loop inside will regenerate the number if it contains duplicate digits, so the random number will consist from unique digits only.
using System.Linq;
public static int GenerateRandomNum()
{
// Number of digits for random number to generate
int randomDigits = 4;
int _max = (int)Math.Pow(10, randomDigits);
Random _rdm = new Random();
int _out = _rdm.Next(0, _max);
while (randomDigits != _out.ToString().ToArray().Distinct().Count())
{
_out = _rdm.Next(0, _max);
}
return _out;
}
You can consider something like this.
int length = 4;
int number = 50;
string asString = number.ToString("D" + length);
The above code gives the result 0050.
Similarly you can try converting to string and verify.
I have created a short program that creates 3 random integers between 1-9 and stores them in an array, however, I would not like any of them to repeat, that is, I would like each to be unique. Is there an easier way to generate 3 unique integers other than having to iterate through the array and comparing each integer to each other? That just seems so tedious if I were to increase my array to beyond 3 integers.
This is my code to generate 3 random numbers. I saw other code in Java, but I thought maybe C# has a easier and more efficient way to do it.
var number = new Numbers[3];
Random r = new Random();
for ( int i = 0; i < number.Length; i++)
{
number[i] = new Numbers(r.Next(1,9));
}
Console.WriteLine("The Three Random Numbers Are:");
foreach(Numbers num in number)
{
Console.WriteLine("{0}", num.Number);
}
I would do something like this:
var range = Enumerable.Range(1, 8);
var rnd = new Random();
var listInts = range.OrderBy(i => rnd.Next()).Take(3).ToList();
You could make an array or a list of the numbers that might be generated, e.g. 0, 1, 2, 3. Then you generate a number from 0 to this list's length, e.g. 2 and pick list[2] so for the next time you only have 0, 1, 3 in your list.
It takes longer to generate it, especially for long lists but it doesn't repeat numbers.
using System;
using System.Collections.Generic;
public class Test
{
static Random random = new Random();
public static List<int> GenerateRandom(int count)
{
// generate count random values.
HashSet<int> candidates = new HashSet<int>();
// top will overflow to Int32.MinValue at the end of the loop
for (Int32 top = Int32.MaxValue - count + 1; top > 0; top++)
{
// May strike a duplicate.
if (!candidates.Add(random.Next(top))) {
candidates.Add(top);
}
}
// load them in to a list.
List<int> result = new List<int>();
result.AddRange(candidates);
// shuffle the results:
int i = result.Count;
while (i > 1)
{
i--;
int k = random.Next(i + 1);
int value = result[k];
result[k] = result[i];
result[i] = value;
}
return result;
}
public static void Main()
{
List<int> vals = GenerateRandom(10);
Console.WriteLine("Result: " + vals.Count);
vals.ForEach(Console.WriteLine);
}
}
Grate explanation and answers from here
Source http://ideone.com/Zjpzdh
How do you program c# to make an array of 1000 random integers between 1-100.
And then how do you get when a person enters a number e.g. 68 how can you make the program say 68 appears so and so many times or that it doesn't work at all!
I am not asking for the complete answer I just need a hint where to get started.
Here is what I know:
I have to use the random function and an if but I dont know what to put where!
int[] iArray = new int[1000];
int counter = 0;
Random random = new Random();
for(int i = 0; i < 1000; i++){
iArray[i] = random.Next(1, 101); //1 - 100, including 100
}
int number = Convert.ToInt32(Console.ReadLine());
foreach(int i in iArray){
if(i == number)count++;
}
Console.WriteLine("The number "+ number+" appears "+count+" times!");
Start with a for loop, in each iterative you call the random function and put the result in a public list. After that you make an dialog for the user to type a number. You can serach with lambda expression in the list to see how many matches you get.
Make an array of 1000 random integers between 1-100 and when a person enters a number e.g. 68 how can you make the program say 68 appears so and so many times
I think you're looking for a method like this:
private static Random rnd = new Random();
public static IEnumerable<int> getRandomNumbers(int count, int lowerbound, int upperbound, int specialNumber = int.MinValue, int specialNumberCount = int.MinValue)
{
List<int> list = new List<int>(count);
HashSet<int> specialNumPositions = new HashSet<int>();
if (specialNumberCount > 0)
{
// generate random positions for the number that must be create at least n-times
for (int i = 0; i < specialNumberCount; i++)
{
while (!specialNumPositions.Add(rnd.Next(0, count)))
;
}
}
while (list.Count < count)
{
if (specialNumPositions.Contains(list.Count))
list.Add(specialNumber);
else
list.Add(rnd.Next(lowerbound, upperbound + 1));
}
return list;
}
which you can use in this way:
// ensure that 68 is generated at least 10 times
var list = getRandomNumbers(1000, 1, 100, 68, 10);
Demo
If you instead just want to know how often a number appears in the list, you can ue Linq:
int count = list.Count(i => i == 68);