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;
}
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));
}
}
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;
}
}
}
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));
}
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);
}
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).
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;
}
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;
}
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());
}
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).
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));
}
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).
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));
}
}
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());
}
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).
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));
}
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);
}
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;
}
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;
}
}
}
Source: Stackoverflow.com