C# 检查字符串是否包含字符串数组中的任何匹配项

Question

What would be the fastest way to check if a string contains any matches in a string array in C#? I can do it using a loop, but I think that would be too slow.

Answer 1

Using LINQ:

 return array.Any(s => s.Equals(myString))

Granted, you might want to take culture and case into account, but that's the general idea. Also, if equality is not what you meant by "matches", you can always you the function you need to use for "match".

Answer 2

I really couldn't tell you if this is absolutely the fastest way, but one of the ways I have commonly done this is:

This will check if the string contains any of the strings from the array:

string[] myStrings = { "a", "b", "c" };
string checkThis = "abc";

if (myStrings.Any(checkThis.Contains))
{
    MessageBox.Show("checkThis contains a string from string array myStrings.");
}

To check if the string contains all the strings (elements) of the array, simply change myStrings.Any in the if statement to myStrings.All .

I don't know what kind of application this is, but I often need to use:

if (myStrings.Any(checkThis.ToLowerInvariant().Contains))

So if you are checking to see user input, it won't matter, whether the user enters the string in CAPITAL letters, this could easily be reversed using ToLowerInvariant().

Hope this helped!

Answer 3

That works fine for me:

string[] characters = new string[] { ".", ",", "'" };
bool contains = characters.Any(c => word.Contains(c));

Answer 4

You could combine the strings with regex or statements, and then "do it in one pass," but technically the regex would still performing a loop internally. Ultimately, looping is necessary.

Answer 5

If the "array" will never change (or change only infrequently), and you'll have many input strings that you're testing against it, then you could build a HashSet<string> from the array. HashSet<T>.Contains is an O(1) operation, as opposed to a loop which is O(N).

But it would take some (small) amount of time to build the HashSet. If the array will change frequently, then a loop is the only realistic way to do it.