Random date in C#

Question

I'm looking for some succinct, modern C# code to generate a random date between Jan 1 1995 and the current date.

I'm thinking some solution that utilizes Enumerable.Range somehow may make this more succinct.

Answer 1

private Random gen = new Random();
DateTime RandomDay()
{
    DateTime start = new DateTime(1995, 1, 1);
    int range = (DateTime.Today - start).Days;           
    return start.AddDays(gen.Next(range));
}

For better performance if this will be called repeatedly, create the start and gen (and maybe even range) variables outside of the function.

Answer 2

This is in slight response to Joel's comment about making a slighly more optimized version. Instead of returning a random date directly, why not return a generator function which can be called repeatedly to create a random date.

Func<DateTime> RandomDayFunc()
{
    DateTime start = new DateTime(1995, 1, 1); 
    Random gen = new Random(); 
    int range = ((TimeSpan)(DateTime.Today - start)).Days; 
    return () => start.AddDays(gen.Next(range));
}

Answer 3

I have taken @Joel Coehoorn answer and made the changes he adviced - put the variable out of the method and put all in class. Plus now the time is random too. Here is the result.

class RandomDateTime
{
    DateTime start;
    Random gen;
    int range;

    public RandomDateTime()
    {
        start = new DateTime(1995, 1, 1);
        gen = new Random();
        range = (DateTime.Today - start).Days;
    }

    public DateTime Next()
    {
        return start.AddDays(gen.Next(range)).AddHours(gen.Next(0,24)).AddMinutes(gen.Next(0,60)).AddSeconds(gen.Next(0,60));
    }
}

And example how to use to write 100 random DateTimes to console:

RandomDateTime date = new RandomDateTime();
for (int i = 0; i < 100; i++)
{
    Console.WriteLine(date.Next());
}

Answer 4

Well, if you gonna present alternate optimization, we can also go for an iterator:

 static IEnumerable<DateTime> RandomDay()
 {
    DateTime start = new DateTime(1995, 1, 1);
    Random gen = new Random();
    int range = ((TimeSpan)(DateTime.Today - start)).Days;
    while (true)
        yield return  start.AddDays(gen.Next(range));        
}

you could use it like this:

int i=0;
foreach(DateTime dt in RandomDay())
{
    Console.WriteLine(dt);
    if (++i == 10)
        break;
}

Answer 5

Start with a fixed date object (Jan 1, 1995), and add a random number of days with AddDays (obviusly, pay attention not surpassing the current date).

Answer 6

Random rnd = new Random();
DateTime datetoday = DateTime.Now;

int rndYear = rnd.Next(1995, datetoday.Year);
int rndMonth = rnd.Next(1, 12);
int rndDay = rnd.Next(1, 31);

DateTime generateDate = new DateTime(rndYear, rndMonth, rndDay);
Console.WriteLine(generateDate);

//this maybe is not the best method but is fast and easy to understand

Answer 7

One more solution to the problem, this time a class to which you provide a range you want the dates in. Its down to random minutes in the results.

/// <summary>
/// A random date/time class that provides random dates within a given range
/// </summary>
public class RandomDateTime
{
    private readonly Random rng = new Random();
    private readonly int totalMinutes;
    private readonly DateTime startDateTime;

    /// <summary>
    /// Initializes a new instance of the <see cref="RandomDateTime"/> class.
    /// </summary>
    /// <param name="startDate">The start date.</param>
    /// <param name="endDate">The end date.</param>
    public RandomDateTime(DateTime startDate, DateTime endDate)
    {
        this.startDateTime = startDate;
        TimeSpan timeSpan = endDate - startDate;
        this.totalMinutes = (int)timeSpan.TotalMinutes;
    }

    /// <summary>
    /// Gets the next random datetime object within the range of startDate and endDate provided in the ctor
    /// </summary>
    /// <returns>A DateTime.</returns>
    public DateTime NextDateTime
    {
        get
        {
            TimeSpan newSpan = new TimeSpan(0, rng.Next(0, this.totalMinutes), 0);
            return this.startDateTime + newSpan;
        }
    }
}

Use it like this to spit out 5 random dates between january 1st 2020 and december 31 2022:

RandomDateTime rdt = new RandomDateTime(DateTime.Parse("01/01/2020"), DateTime.Parse("31/12/2022"));

for (int i = 0; i < 5; i++)
    Debug.WriteLine(rdt.NextDateTime);

Answer 8

I am a bit late in to the game, but here is one solution which works fine:

    void Main()
    {
        var dateResult = GetRandomDates(new DateTime(1995, 1, 1), DateTime.UtcNow, 100);
        foreach (var r in dateResult)
            Console.WriteLine(r);
    }

    public static IList<DateTime> GetRandomDates(DateTime startDate, DateTime maxDate, int range)
    {
        var randomResult = GetRandomNumbers(range).ToArray();

        var calculationValue = maxDate.Subtract(startDate).TotalMinutes / int.MaxValue;
        var dateResults = randomResult.Select(s => startDate.AddMinutes(s * calculationValue)).ToList();
        return dateResults;
    }

    public static IEnumerable<int> GetRandomNumbers(int size)
    {
        var data = new byte[4];
        using (var rng = new System.Security.Cryptography.RNGCryptoServiceProvider(data))
        {
            for (int i = 0; i < size; i++)
            {
                rng.GetBytes(data);

                var value = BitConverter.ToInt32(data, 0);
                yield return value < 0 ? value * -1 : value;
            }
        }
    }

Answer 9

Small method that returns a random date as string, based on some simple input parameters. Built based on variations from the above answers:

public string RandomDate(int startYear = 1960, string outputDateFormat = "yyyy-MM-dd")
{
   DateTime start = new DateTime(startYear, 1, 1);
   Random gen = new Random(Guid.NewGuid().GetHashCode());
   int range = (DateTime.Today - start).Days;
   return start.AddDays(gen.Next(range)).ToString(outputDateFormat);
}

Answer 10

Useful extension based of @Jeremy Thompson's solution

public static class RandomExtensions
{
    public static DateTime Next(this Random random, DateTime start, DateTime? end = null)
    {
        end ??= new DateTime();
        int range = (end.Value - start).Days;
        return start.AddDays(random.Next(range));
    }
}