WebAug 29, 2012 · Random rand = new Random (); var models = garage.OrderBy (c => rand.Next ()).Select (c => c.Model).ToList (); //Model is assuming that's the name of your property Note : Random (), ironically, isn't actually very random but fine for a quick simple solution. There are better algorithms out there to do this, here's one to look at; WebMar 15, 2024 · As opposed to the OrderBy(random) which is biased due to the fact it exploits the underlying QuickSort algorithm and done repeatedly results in a pattern of permutations that are favored over others. So not only is it more time complex, but it results in less statistical variation... which in most games doesn't matter.
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WebMay 29, 2011 · If you want to randomly re-order in place you should shuffle the list, usage with an extension method is then a simple one-liner. This assumes you already have an IList based collection. Usage: myList.Shuffle (); Web假設您不想重復任何獎品,則解決方案會稍微復雜一些,但是您可以帶上Linq和Random的一些技巧: var prizes = new string[] { "vacation to Hawaii with all expenses covered", "used glue stick", // etc "dime" }; var rand = new Random(); var result = (from prize in prizes orderby rand.NextDouble() select prize).Take(3).ToArray(); is speedtest accurate
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WebDec 27, 2016 · Then just order by ctx.Random (); this will do a random ordering at the SQL-Server courtesy of NEWID (). i.e. var cust = (from row in ctx.Customers where row.IsActive // your filter orderby ctx.Random () select row).FirstOrDefault (); Note that this is only suitable for small-to-mid-size tables; for huge tables, it will have a performance ... WebJul 4, 2014 · The first variant, where I am using random.Next(), is working fine. But the variant, where I have call new Random().Next(), does NOT return random numbers; instead it returns a sequence of numbers from 0 to 20. Now my questions are: What is term to denote the second type of initialization of object new Random().Next() in C#? WebOct 31, 2008 · The Developer Fusion VB.Net to C# converter says that the equivalent C# code is: System.Random rnd = new System.Random (); IEnumerable numbers = Enumerable.Range (1, 100).OrderBy (r => rnd.Next ()); For future reference, they also have a C# to VB.Net converter. There are several other tools available for this as well. Share … if it4u