Question

I am trying to filter a List of strings based on the number of words in each string. I am assuming that you would trim any white-space at the ends of the string, and then count the number of spaces left in the string, so that WordCount = NumberOfSpaces + 1. Is that the most efficient way to do this? I know that for filtering based on character count the following is working fine…just cant figure out how to write it succinctly using C#/LINQ.

if (checkBox_MinMaxChars.Checked)
{
    int minChar = int.Parse(numeric_MinChars.Text);
    int maxChar = int.Parse(numeric_MaxChars.Text);

    myList = myList.Where(x => 
                              x.Length >= minChar && 
                              x.Length <= maxChar).ToList();
}

Any ideas of for counting words?

UPDATE: This Worked like a charm…Thanks Mathew:

int minWords = int.Parse(numeric_MinWords.Text);
int maxWords = int.Parse(numeric_MaxWords.Text);

sortBox1 = sortBox1.Where(x => x.Trim().Split(new char[] { ' ' }, StringSplitOptions.RemoveEmptyEntries).Count() >= minWords &&
                               x.Trim().Split(new char[] { ' ' }, StringSplitOptions.RemoveEmptyEntries).Count() <= maxWords).ToList();

Answer 1

I would approach it in a more simplified manner since you have indicated that a space can be used reliably as a delimiter like so:

var str = "     the string to split and count        ";
var wordCount = str.Trim().Split(new char[] { ' ' }, StringSplitOptions.RemoveEmptyEntries).Count();

EDIT:

If optimal perforamnce is necessary and memory usage is a concern you could write your own method and leverage IndexOf() (although there are many avenues for implementation on a problem like this, I just prefer reuse rather than from-scratch code design):

    public int WordCount(string s) {
        const int DONE = -1;
        var wordCount = 0;
        var index = 0;
        var str = s.Trim();
        while (index != DONE) {
            wordCount++;
            index = str.IndexOf(" ", index + 1);
        }
        return wordCount;
    }