I am trying to get a random object within linq. Here is how I did.
//get all the answers
var Answers = q.Skip(1).Take(int.MaxValue);
//get the random number by the number of answers
int intRandomAnswer = r.Next(1, Answers.Count());
int count = 0;
//locate the answer
foreach(var Answer in Answers)
{
if (count == intRandomAnswer)
{
SelectedPost = Answer;
break;
}
count++;
}
Is this the best way to do this?
What about:
SelectedPost = q.ElementAt(r.Next(1, Answers.Count()));
Further reading:
The comments below make good contributions to closely related questions, and I'll include them here, since as @Rouby points out, people searching for an answer to these may find this answer and it won't be correct in those cases.
Random Element Across Entire Input
To make all elements a candidate in the random selection, you need to change the input to r.Next
:
SelectedPost = Answers.ElementAt(r.Next(0, Answers.Count()));
@Zidad adds a helpful extension method to get random element over all elements in the sequence:
public static T Random<T>(this IEnumerable<T> enumerable)
{
if (enumerable == null)
{
throw new ArgumentNullException(nameof(enumerable));
}
// note: creating a Random instance each call may not be correct for you,
// consider a thread-safe static instance
var r = new Random();
var list = enumerable as IList<T> ?? enumerable.ToList();
return list.Count == 0 ? default(T) : list[r.Next(0, list.Count)];
}
Use a Fisher-Yates-Durstenfeld shuffle.
(You could use a helper/extension method to shuffle your IEnumerable<T>
sequence. Alternatively, if you were using an IList<T>
you could perform an in-place shuffle, if you prefer.)
Another wacky approach (not the most efficient for larger data sets):
SelectedPost = q.OrderBy(qu => Guid.NewGuid()).First();
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