'Counting number of words in C#
I'm trying to count the number of words from a rich textbox in C# the code that I have below only works if it is a single line. How do I do this without relying on regex or any other special functions.
string whole_text = richTextBox1.Text;
string trimmed_text = whole_text.Trim();
string[] split_text = trimmed_text.Split(' ');
int space_count = 0;
string new_text = "";
foreach(string av in split_text)
{
if (av == "")
{
space_count++;
}
else
{
new_text = new_text + av + ",";
}
}
new_text = new_text.TrimEnd(',');
split_text = new_text.Split(',');
MessageBox.Show(split_text.Length.ToString ());
Solution 1:[1]
char[] delimiters = new char[] {' ', '\r', '\n' };
whole_text.Split(delimiters,StringSplitOptions.RemoveEmptyEntries).Length;
Solution 2:[2]
There are some better ways to do this, but in keeping with what you've got, try the following:
string whole_text = richTextBox1.Text;
string trimmed_text = whole_text.Trim();
// new line split here
string[] lines = trimmed_text.Split(Environment.NewLine.ToCharArray());
// don't need this here now...
//string[] split_text = trimmed_text.Split(' ');
int space_count = 0;
string new_text = "";
Now make two foreach loops. One for each line and one for counting words within the lines.
foreach (string line in lines)
{
// Modify the inner foreach to do the split on ' ' here
// instead of split_text
foreach (string av in line.Split(' '))
{
if (av == "")
{
space_count++;
}
else
{
new_text = new_text + av + ",";
}
}
}
new_text = new_text.TrimEnd(',');
// use lines here instead of split_text
lines = new_text.Split(',');
MessageBox.Show(lines.Length.ToString());
}
Solution 3:[3]
This was a phone screening interview question that I just took (by a large company located in CA who sells all kinds of devices that starts with a letter "i"), and I think I franked... after I got offline, I wrote this. I wish I were able to do it during interview..
static void Main(string[] args)
{
Debug.Assert(CountWords("Hello world") == 2);
Debug.Assert(CountWords(" Hello world") == 2);
Debug.Assert(CountWords("Hello world ") == 2);
Debug.Assert(CountWords("Hello world") == 2);
}
public static int CountWords(string test)
{
int count = 0;
bool wasInWord = false;
bool inWord = false;
for (int i = 0; i < test.Length; i++)
{
if (inWord)
{
wasInWord = true;
}
if (Char.IsWhiteSpace(test[i]))
{
if (wasInWord)
{
count++;
wasInWord = false;
}
inWord = false;
}
else
{
inWord = true;
}
}
// Check to see if we got out with seeing a word
if (wasInWord)
{
count++;
}
return count;
}
Solution 4:[4]
Have a look at the Lines property mentioned in @Jay Riggs comment, along with this overload of String.Split to make the code much simpler. Then the simplest approach would be to loop over each line in the Lines property, call String.Split on it, and add the length of the array it returns to a running count.
EDIT: Also, is there any reason you're using a RichTextBox instead of a TextBox with Multiline set to True?
Solution 5:[5]
I use an extension method for grabbing word count in a string. Do note, however, that double spaces will mess the count up.
public static int CountWords(this string line)
{
var wordCount = 0;
for (var i = 0; i < line.Length; i++)
if (line[i] == ' ' || i == line.Length - 1)
wordCount++;
return wordCount;
}
}
Solution 6:[6]
Your approach is on the right path. I would do something like, passing the text property of richTextBox1 into the method. This however won't be accurate if your rich textbox is formatting HTML, so you'll need to strip out any HTML tags prior to running the word count:
public static int CountWords(string s)
{
int c = 0;
for (int i = 1; i < s.Length; i++)
{
if (char.IsWhiteSpace(s[i - 1]) == true)
{
if (char.IsLetterOrDigit(s[i]) == true ||
char.IsPunctuation(s[i]))
{
c++;
}
}
}
if (s.Length > 2)
{
c++;
}
return c;
}
Solution 7:[7]
We used an adapted form of Yoshi's answer, where we fixed the bug where it would not count the last word in a string if there was no white-space after it:
public static int CountWords(string test)
{
int count = 0;
bool inWord = false;
foreach (char t in test)
{
if (char.IsWhiteSpace(t))
{
inWord = false;
}
else
{
if (!inWord) count++;
inWord = true;
}
}
return count;
}
Solution 8:[8]
using System.Collections;
using System;
class Program{
public static void Main(string[] args){
//Enter the value of n
int n = Convert.ToInt32(Console.ReadLine());
string[] s = new string[n];
ArrayList arr = new ArrayList();
//enter the elements
for(int i=0;i<n;i++){
s[i] = Console.ReadLine();
}
string str = "";
//Filter out duplicate values and store in arr
foreach(string i in s){
if(str.Contains(i)){
}else{
arr.Add(i);
}
str += i;
}
//Count the string with arr and s variables
foreach(string i in arr){
int count = 0;
foreach(string j in s){
if(i.Equals(j)){
count++;
}
}
Console.WriteLine(i+" - "+count);
}
}
}
Solution 9:[9]
public static int WordCount(string str)
{
int num=0;
bool wasInaWord=true;;
if (string.IsNullOrEmpty(str))
{
return num;
}
for (int i=0;i< str.Length;i++)
{
if (i!=0)
{
if (str[i]==' ' && str[i-1]!=' ')
{
num++;
wasInaWord=false;
}
}
if (str[i]!=' ')
{
wasInaWord=true;
}
}
if (wasInaWord)
{
num++;
}
return num;
}
Solution 10:[10]
class Program
{
static void Main(string[] args)
{
string str;
int i, wrd, l;
StringBuilder sb = new StringBuilder();
Console.Write("\n\nCount the total number of words in a string
:\n");
Console.Write("---------------------------------------------------
---\n");
Console.Write("Input the string : ");
str = Console.ReadLine();
l = 0;
wrd = 1;
foreach (var a in str)
{
sb.Append(a);
if (str[l] == ' ' || str[l] == '\n' || str[l] == '\t')
{
wrd++;
}
l++;
}
Console.WriteLine(sb.Replace(' ', '\n'));
Console.Write("Total number of words in the string is : {0}\n",
wrd);
Console.ReadLine();
}
Solution 11:[11]
This should work
input.Split(' ').ToList().Count;
Solution 12:[12]
This can show you the number of words in a line
string line = Console.ReadLine();
string[] word = line.Split(' ');
Console.WriteLine("Words " + word.Length);
Sources
This article follows the attribution requirements of Stack Overflow and is licensed under CC BY-SA 3.0.
Source: Stack Overflow