Creating Random Names and Dates

Often, website demos are more compelling when using a copy of production data. However, my current client has no production data. Nonetheless, I want to give them a realistic view of my work. So I needed to create a great deal of realistic looking data.

I began by creating random Customer objects. For the first and last names I could have created random strings. But this particular client would be distracted by the random character names. Silly I know, but that’s my reality sometimes. First names are often either female or male, so the firstnames collection is an IDictionary. The key is the first name and the char is the gender, either ‘M’ or ‘F’. This links a name with its typical gender. I wrote extension methods to return a random element from the firstname and lastname collections. The last extension method generates a random date for the DOB. This extension method accepts another date, figures out the number of days between the two dates, let’s call this value x, and adds a random number of days between 0 and x to the lesser date to create a new random date, which should always be something between the two date values.

using System;
using System.Collections.Generic;
using System.Configuration;
using System.Linq;
using JCS.Model;
 
namespace LoadTestData
{
  internal class Program
  {
    private static readonly string connection = ConfigurationManager.ConnectionStrings["XXXAAA"].ConnectionString;
 
    // class variables so they can be used for other names, e.g. parent names
    private static readonly Dictionary<string, char> firstnames = new Dictionary<string, char>
      {{"Troy", 'M'},{"Jamal", 'M'},{"Alicia", 'F'},{"Gabriel", 'M'},{"Christian", 'M'},
      {"Benjamin", 'M'},{"Gavin", 'M'},{"Jose", 'M'},{"Brandon", 'M'},{"Dylan", 'M'},
      {"Victoria", 'F'},{"Savannah", 'F'},{"Riley", 'F'},{"Hailey", 'F'},{"Nevaeh", 'F'},
      {"Lily", 'F'},{"Kandis", 'F'},{"Dominique", 'F'}};
 
 
    // class variables so they can be used for other names, e.g. parent names
    private static readonly List<string> lastnames = new List<string>
      {"Scott","Stewart","Sanchez","Morris","Rogers","Reed","Cook",
        "Morgan","Bell","Murphy","Rivera","Ward","Peterson",
        "Watson","Brooks","Bennett","Wood","Henderson","Jenkins","Sanders","Cox"};
 
 
    static void Main()
    {
      Random random = new Random(DateTime.Now.Millisecond);
      DateTime minCustomerDOB = new DateTime(1940, 1, 1);
      DateTime maxCustomerDOB = new DateTime(1989, 1, 1);
 
      string firstname = firstnames.RandomKey(random);
      var customer = new Customer
                       {
                         FirstName = firstname,
                         LastName = lastnames.RandomElement(random),
                         Gender = firstnames[firstname],
                         DOB = minCustomerDOB.RandomDate(maxCustomerDOB, random)
                       };
 
      using (var context = new JCSDBDataContext(connection))
      {
        context.Customers.InsertOnSubmit(customer);
        context.SubmitChanges();
      }
    }
  }
 
  public static class Extensions
  {
    static public TKey RandomKey<TKey, TValue>(this Dictionary<TKey, TValue> dictionary, Random random)
    {
      int i = random.Next(0, dictionary.Count - 1);
      return dictionary.ElementAt(i).Key;
    }
 
    static public T RandomElement<T>(this List<T> list, Random random)
    {
      int i = random.Next(0, list.Count - 1);
      return list.ElementAt(i);
    }
 
    static public DateTime RandomDate(this DateTime min, DateTime max, Random random)
    {
      int days = (max - min).Days;
 
      // use absolute value in case max is lesser than min
      int i = random.Next(Math.Abs(days));
      DateTime ranDate = min.AddDays(i);
 
      //maybe the max date is actually lesser than the min date.
      if (max < min)
        ranDate = max.AddDays(i);
 
      return ranDate;
    }
  }
}

This entry was posted on Friday, June 19th, 2009 at 2:02 pm and is filed under .NET, C#. You can follow any responses to this entry through the RSS 2.0 feed. You can leave a response, or trackback from your own site.

Leave a Reply