Welcome to OGeek Q&A Community for programmer and developer-Open, Learning and Share
Welcome To Ask or Share your Answers For Others

Categories

0 votes
161 views
in Technique[技术] by (71.8m points)

c# - Checking if a list is empty with LINQ

What's the "best" (taking both speed and readability into account) way to determine if a list is empty? Even if the list is of type IEnumerable<T> and doesn't have a Count property.

Right now I'm tossing up between this:

if (myList.Count() == 0) { ... }

and this:

if (!myList.Any()) { ... }

My guess is that the second option is faster, since it'll come back with a result as soon as it sees the first item, whereas the second option (for an IEnumerable) will need to visit every item to return the count.

That being said, does the second option look as readable to you? Which would you prefer? Or can you think of a better way to test for an empty list?

Edit @lassevk's response seems to be the most logical, coupled with a bit of runtime checking to use a cached count if possible, like this:

public static bool IsEmpty<T>(this IEnumerable<T> list)
{
    if (list is ICollection<T>) return ((ICollection<T>)list).Count == 0;

    return !list.Any();
}
See Question&Answers more detail:os

与恶龙缠斗过久,自身亦成为恶龙;凝视深渊过久,深渊将回以凝视…
Welcome To Ask or Share your Answers For Others

1 Reply

0 votes
by (71.8m points)

You could do this:

public static Boolean IsEmpty<T>(this IEnumerable<T> source)
{
    if (source == null)
        return true; // or throw an exception
    return !source.Any();
}

Edit: Note that simply using the .Count method will be fast if the underlying source actually has a fast Count property. A valid optimization above would be to detect a few base types and simply use the .Count property of those, instead of the .Any() approach, but then fall back to .Any() if no guarantee can be made.


与恶龙缠斗过久,自身亦成为恶龙;凝视深渊过久,深渊将回以凝视…
OGeek|极客中国-欢迎来到极客的世界,一个免费开放的程序员编程交流平台!开放,进步,分享!让技术改变生活,让极客改变未来! Welcome to OGeek Q&A Community for programmer and developer-Open, Learning and Share
Click Here to Ask a Question

...