I have just started using visual basic and wanted to create a program that counted the number of times a word appeared. My plan was develop a program that analyses a sentence that contains several words without punctuation. When
a word in that sentence is input, the program identifies all of the positions where the word occurs in the sentence.
I started by making a code that counted the amount of spaces in a sentence but am now stuck.
Module Module1
Sub Main()
Dim Sentence As String
Dim SentenceLength As Integer
Dim Text As String
Console.WriteLine("ASK NOT WHAT YOUR COUNTRY CAN DO FOR YOU ASK WHAT YOU CAN DO FOR YOUR COUNTRY")
Console.WriteLine("Enter your word ") : Sentence = Console.ReadLine
Dim TextCounter As Integer = 0
Dim MainWord As String = Sentence
Dim CountChar As String = " "
Do While InStr(MainWord, CountChar) > 0
MainWord = Mid(MainWord, 1 + InStr(MainWord, CountChar), Len(MainWord))
TextCounter = TextCounter + 1
Text = TextCounter + 2
Console.WriteLine(Text)
Loop
Console.WriteLine(TextCounter)
Console.Write("Press Enter to Exit")
Console.ReadLine()
End Sub
End Module
A quick & dirty method is to split the string into an array of strings, then count how many times a word appears in it:
Dim words() As String = Sentence.Split(new char() {" ", ",", ".", ";"} ' add other punctuation as appropriate
Dim count = words.Count(Function(word) word = MainWord)
This uses the String.Split method to split the string each time a space is encountered. Then it uses the Enumerable.Count extension method to count the words that match a certain condition, that the word is equal to MainWord
To count substrings:
Dim count = UBound(Split("catty cat", "cat")) ' 2
To count words:
Dim countWords = Regex.Matches("catty cat", "\bcat\b").Count ' 1
Related
I'm a programing student, so I've started with vb.net as my first language and I need some help.
I need to know how I delete excess white spaces between words in a sentence, only using these string functions: Trim, instr, char, mid, val and len.
I made a part of the code but it doesn't work, Thanks.
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Knocked up a quick routine for you.
Public Function RemoveMyExcessSpaces(str As String) As String
Dim r As String = ""
If str IsNot Nothing AndAlso Len(str) > 0 Then
Dim spacefound As Boolean = False
For i As Integer = 1 To Len(str)
If Mid(str, i, 1) = " " Then
If Not spacefound Then
spacefound = True
End If
Else
If spacefound Then
spacefound = False
r += " "
End If
r += Mid(str, i, 1)
End If
Next
End If
Return r
End Function
I think it meets your criteria.
Hope that helps.
Unless using those VB6 methods is a requirement, here's a one-line solution:
TextBox2.Text = String.Join(" ", TextBox1.Text.Split(New Char() {" "c}, StringSplitOptions.RemoveEmptyEntries))
Online test: http://ideone.com/gBbi55
String.Split() splits a string on a specific character or substring (in this case a space) and creates an array of the string parts in-between. I.e: "Hello There" -> {"Hello", "There"}
StringSplitOptions.RemoveEmptyEntries removes any empty strings from the resulting split array. Double spaces will create empty strings when split, thus you'll get rid of them using this option.
String.Join() will create a string from an array and separate each array entry with the specified string (in this case a single space).
There is a very simple answer to this question, there is a string method that allows you to remove those "White Spaces" within a string.
Dim text_with_white_spaces as string = "Hey There!"
Dim text_without_white_spaces as string = text_with_white_spaces.Replace(" ", "")
'text_without_white_spaces should be equal to "HeyThere!"
Hope it helped!
I have a string (for example: "Hello there. My name is John. I work very hard. Hello there!") and I am trying to find the number of occurrences of the string "hello there". So far, this is the code I have:
Dim input as String = "Hello there. My name is John. I work very hard. Hello there!"
Dim phrase as String = "hello there"
Dim Occurrences As Integer = 0
If input.toLower.Contains(phrase) = True Then
Occurrences = input.Split(phrase).Length
'REM: Do stuff
End If
Unfortunately, what this line of code seems to do is split the string every time it sees the first letter of phrase, in this case, h. So instead of the result Occurrences = 2 that I would hope for, I actually get a much larger number. I know that counting the number of splits in a string is a horrible way to go about doing this, even if I did get the correct answer, so could someone please help me out and provide some assistance?
Yet another idea:
Dim input As String = "Hello there. My name is John. I work very hard. Hello there!"
Dim phrase As String = "Hello there"
Dim Occurrences As Integer = (input.Length - input.Replace(phrase, String.Empty).Length) / phrase.Length
You just need to make sure that phrase.Length > 0.
the best way to do it is this:
Public Function countString(ByVal inputString As String, ByVal stringToBeSearchedInsideTheInputString as String) As Integer
Return System.Text.RegularExpressions.Regex.Split(inputString, stringToBeSearchedInsideTheInputString).Length -1
End Function
str="Thisissumlivinginsumgjhvgsum in the sum bcoz sum ot ih sum"
b= LCase(str)
array1=Split(b,"sum")
l=Ubound(array1)
msgbox l
the output gives u the no. of occurences of a string within another one.
You can create a Do Until loop that stops once an integer variable equals the length of the string you're checking. If the phrase exists, increment your occurences and add the length of the phrase plus the position in which it is found to the cursor variable. If the phrase can not be found, you are done searching (no more results), so set it to the length of the target string. To not count the same occurance more than once, check only from the cursor to the length of the target string in the Loop (strCheckThisString).
Dim input As String = "hello there. this is a test. hello there hello there!"
Dim phrase As String = "hello there"
Dim Occurrences As Integer = 0
Dim intCursor As Integer = 0
Do Until intCursor >= input.Length
Dim strCheckThisString As String = Mid(LCase(input), intCursor + 1, (Len(input) - intCursor))
Dim intPlaceOfPhrase As Integer = InStr(strCheckThisString, phrase)
If intPlaceOfPhrase > 0 Then
Occurrences += 1
intCursor += (intPlaceOfPhrase + Len(phrase) - 1)
Else
intCursor = input.Length
End If
Loop
You just have to change the input of the split function into a string array and then delare the StringSplitOptions.
Try out this line of code:
Occurrences = input.Split({phrase}, StringSplitOptions.None).Length
I haven't checked this, but I'm thinking you'll also have to account for the fact that occurrences would be too high due to the fact that you're splitting using your string and not actually counting how many times it is in the string, so I think Occurrences = Occurrences - 1
Hope this helps
You could create a recursive function using IndexOf. Passing the string to be searched and the string to locate, each recursion increments a Counter and sets the StartIndex to +1 the last found index, until the search string is no longer found. Function will require optional parameters Starting Position and Counter passed by reference:
Function InStrCount(ByVal SourceString As String, _
ByVal SearchString As String, _
Optional ByRef StartPos As Integer = 0, _
Optional ByRef Count As Integer = 0) As Integer
If SourceString.IndexOf(SearchString, StartPos) > -1 Then
Count += 1
InStrCount(SourceString, _
SearchString, _
SourceString.IndexOf(SearchString, StartPos) + 1, _
Count)
End If
Return Count
End Function
Call function by passing string to search and string to locate and, optionally, start position:
Dim input As String = "Hello there. My name is John. I work very hard. Hello there!"
Dim phrase As String = "hello there"
Dim Occurrences As Integer
Occurrances = InStrCount(input.ToLower, phrase.ToLower)
Note the use of .ToLower, which is used to ignore case in your comparison. Do not include this directive if you do wish comparison to be case specific.
One more solution based on InStr(i, str, substr) function (searching substr in str starting from i position, more info about InStr()):
Function findOccurancesCount(baseString, subString)
occurancesCount = 0
i = 1
Do
foundPosition = InStr(i, baseString, subString) 'searching from i position
If foundPosition > 0 Then 'substring is found at foundPosition index
occurancesCount = occurancesCount + 1 'count this occurance
i = foundPosition + 1 'searching from i+1 on the next cycle
End If
Loop While foundPosition <> 0
findOccurancesCount = occurancesCount
End Function
As soon as there is no substring found (InStr returns 0, instead of found substring position in base string), searching is over and occurances count is returned.
Looking at your original attempt, I have found that this should do the trick as "Split" creates an array.
Occurrences = input.split(phrase).ubound
This is CaSe sensitive, so in your case the phrase should equal "Hello there", as there is no "hello there" in the input
Expanding on Sumit Kumar's simple solution, here it is as a one-line working function:
Public Function fnStrCnt(ByVal str As String, ByVal substr As String) As Integer
fnStrCnt = UBound(Split(LCase(str), substr))
End Function
Demo:
Sub testit()
Dim thePhrase
thePhrase = "Once upon a midnight dreary while a man was in a house in the usa."
If fnStrCnt(thePhrase, " a ") > 1 Then
MsgBox "Found " & fnStrCnt(thePhrase, " a ") & " occurrences."
End If
End Sub 'testit()
I don't know if this is more obvious?
Starting from the beginning of longString check the next characters up to the number characters in phrase, if phrase is not found start looking from the second character etc. If it is found start agin from the current position plus the number of characters in phrase and increment the value of occurences
Module Module1
Sub Main()
Dim longString As String = "Hello there. My name is John. I work very hard. Hello there! Hello therehello there"
Dim phrase As String = "hello There"
Dim occurences As Integer = 0
Dim n As Integer = 0
Do Until n >= longString.Length - (phrase.Length - 1)
If longString.ToLower.Substring(n, phrase.Length).Contains(phrase.ToLower) Then
occurences += 1
n = n + (phrase.Length - 1)
End If
n += 1
Loop
Console.WriteLine(occurences)
End Sub
End Module
I used this in Vbscript, You can convert the same to VB.net as well
Dim str, strToFind
str = "sdfsdf:sdsdgs::"
strToFind = ":"
MsgBox GetNoOfOccurranceOf( strToFind, str)
Function GetNoOfOccurranceOf(ByVal subStringToFind As String, ByVal strReference As String)
Dim iTotalLength, newString, iTotalOccCount
iTotalLength = Len(strReference)
newString = Replace(strReference, subStringToFind, "")
iTotalOccCount = iTotalLength - Len(newString)
GetNoOfOccurranceOf = iTotalOccCount
End Function
I know this thread is really old, but I got another solution too:
Function countOccurencesOf(needle As String, s As String)
Dim count As Integer = 0
For i As Integer = 0 to s.Length - 1
If s.Substring(i).Startswith(needle) Then
count = count + 1
End If
Next
Return count
End Function
I am writing a program in Visual Basic 2010 that lists how many times a word of each length occurs in a user-inputted string. Although most of the program is working, I have one problem:
When looping through all of the characters in the string, the program checks whether there is a next character (such that the program does not attempt to loop through characters that do not exist). For example, I use the condition:
If letter = Microsoft.VisualBasic.Right(input, 1) Then
Where letter is the character, input is the string, and Microsoft.VisualBasic.Right(input, 1) extracts the rightmost character from the string. Thus, if letter is the rightmost character, the program will cease to loop through the string.
This is where the problems comes in. Let us say the string is This sentence has five words. The rightmost character is an s, but an s is also the fourth and sixth character. That means that the first and second s will break the loop just as the others will.
My questions is whether there is a way to ensure that only the last s, or whatever character is the last one in the string can break the loop.
There are a few methods you can use for this, one as Neolisk shows; here are a couple of others:
Dim breakChar As Char = "s"
Dim str As String = "This sentence has five words"
str = str.Replace(".", " ")
str = str.Replace(",", " ")
str = str.Replace(vbTab, " ")
' other chars to replace
Dim words() As String = str.ToLower.Split(New Char() {" "}, StringSplitOptions.RemoveEmptyEntries)
For Each word In words
If word.StartsWith(breakChar) Then Exit For
Console.WriteLine("M1 Word: ""{0}"" Length: {1:N0}", word, word.Length)
Next
If you need to loop though chars for whatever reason, you can use something like this:
Dim breakChar As Char = "s"
Dim str As String = "This sentence has five words"
str = str.Replace(".", " ")
str = str.Replace(",", " ")
str = str.Replace(vbTab, " ")
' other chars to replace
'method 2
Dim word As New StringBuilder
Dim words As New List(Of String)
For Each c As Char In str.ToLower.Trim
If c = " "c Then
If word.Length > 0 'support multiple white-spaces (double-space etc.)
Console.WriteLine("M2 Word: ""{0}"" Length: {1:N0}", word.ToString, word.ToString.Length)
words.Add(word.ToString)
word.Clear()
End If
Else
If word.Length = 0 And c = breakChar Then Exit For
word.Append(c)
End If
Next
If word.Length > 0 Then
words.Add(word.ToString)
Console.WriteLine("M2 Word: ""{0}"" Length: {1:N0}", word.ToString, word.ToString.Length)
End If
I wrote these specifically to break on the first letter in a word as you ask, adjust as needed.
VB.NET code to calculate how many times a word of each length occurs in a user-inputted string:
Dim sentence As String = "This sentence has five words"
Dim words() As String = sentence.Split(" ")
Dim v = From word As String In words Group By L = word.Length Into Group Order By L
Line 2 may need to be adjusted to remove punctuation characters, trim extra spaces etc.
In the above example, v(i) contains word length, and v(i).Group.Count contains how many words of this length were encountered. For debugging purposes, you also have v(i).Group, which is an array of String, containing all words belonging to this group.
I have a text file on my website and I download the whole string via webclient.downloadstring.
The text file contains this :
cookies,dishes,candy,(new line)
back,forward,refresh,(new line)
mail,media,mute,
This is just an example it's not the actual string , but it will do for help purposes.
What I want is I want to download the whole string , find the line that contains the word that was entered by the user in a textbox, get that line into a string, then I want to use the string.split with as delimiter the "," and output each word that is in the string into an richtextbox.
Now here is the code that I have used (some fields are removed for privacy reasons).
If TextBox1.TextLength > 0 Then
words = web.DownloadString("webadress here")
If words.Contains(TextBox1.Text) Then
'retrieval code here
Dim length As Integer = TextBox1.TextLength
Dim word As String
word = words.Substring(length + 1) // the plus 1 is for the ","
Dim cred() As String
cred = word.Split(",")
RichTextBox1.Text = "Your word: " + cred(0) + vbCr + "Your other word: " + cred(1)
Else
MsgBox("Sorry, but we could not find the word you have entered", MsgBoxStyle.Critical)
End If
Else
MsgBox("Please fill in an word", MsgBoxStyle.Critical)
End If
Now it works and no errors , but it only works for line 1 and not on line 2 or 3
what am I doing wrong ?
It's because the string words also contains the new line characters that you seem to be omitting in your code. You should first split words with the delimiter \n (or \r\n, depending on the platform), like this:
Dim lines() As String = words.Split("\n")
After that, you have an array of strings, each element representing a single line. Loop it through like this:
For Each line As String In lines
If line.Contains(TextBox1.Text) Then
'retrieval code here
End If
Next
Smi's answer is correct, but since you're using VB you need to split on vbNewLine. \n and \r are for use in C#. I get tripped up by that a lot.
Another way to do this is to use regular expressions. A regular expression match can both find the word you want and return the line that contains it in a single step.
Barely tested sample below. I couldn't quite figure out if your code was doing what you said it should be doing so I improvised based on your description.
Imports System.Text.RegularExpressions
Public Class Form1
Private Sub ButtonFind_Click(sender As System.Object, e As System.EventArgs) Handles ButtonFind.Click
Dim downloadedString As String
downloadedString = "cookies,dishes,candy," _
& vbNewLine & "back,forward,refresh," _
& vbNewLine & "mail,media,mute,"
'Use the regular expression anchor characters (^$) to match a line that contains the given text.
Dim wordToFind As String = TextBox1.Text & "," 'Include the comma that comes after each word to avoid partial matches.
Dim pattern As String = "^.*" & wordToFind & ".*$"
Dim rx As Regex = New Regex(pattern, RegexOptions.Multiline + RegexOptions.IgnoreCase)
Dim M As Match = rx.Match(downloadedString)
'M will either be Match.Empty (no matching word was found),
'or it will be the matching line.
If M IsNot Match.Empty Then
Dim words() As String = M.Value.Split(","c)
RichTextBox1.Clear()
For Each word As String In words
If Not String.IsNullOrEmpty(word) Then
RichTextBox1.AppendText(word & vbNewLine)
End If
Next
Else
RichTextBox1.Text = "No match found."
End If
End Sub
End Class
I'm making an application that will change position of two characters in Word.
Imports System.IO
Module Module1
Sub Main()
Dim str As String = File.ReadAllText("File.txt")
Dim str2 As String() = Split(str, " ")
For i As Integer = 0 To str2.Length - 1
Dim arr As Char() = CType(str2(i), Char())
For ia As Integer = 0 To arr.Length() - 1 Step 2
Dim pa As String
pa = arr(ia + 1)
arr(ia + 1) = arr(ia)
arr(ia) = pa
Next ia
For ib As Integer = 0 To arr.Length - 1
Console.Write(arr(ib))
File.WriteAllText("File2.txt", arr(ib))
Next ib
File.WriteAllText("File2.txt", " ")
Console.Write(" ")
Next i
Console.Read()
End Sub
End Module
For example:
Input: ab
Output: ba
Input: asdasd asdasd
Output: saadds saadds
Program works good, it is mixing characters good, but it doesn't write text to the file. It will write text in console, but not in file.
Note: Program is working only with words that are divisible by 2, but it's not a problem.
Also, it does not return any error message.
Your code is overwriting the file that you have already written with a single space (" ") each time round.
You should only open the file once, and append to it using a stream writer:
Using output = File.CreateText("file2.txt")
' Put the for loop here.
End Using
There are some other things wrong with your code. Firstly, use For Each instead of For, this makes your code much more simple and readable. Secondly, try to avoid For loops altogether where possible. For instance, instead of iterating over the characters to output them one at a time, just create a new string from the char array, and write that:
Dim shuffledWord As New String(arr)
output.Write(shuffledWord)
Some of your types are plain wrong, i.e. you are using String in places instead of Char. You should always use Option Strict On. Then the compiler will not tolerate such code.
You should also prefer to use framework methods over VB-specific methods. This makes it easier to understand for C# programmers, and also makes it easier to translate and change (that is, use the Split method of strings instead of a free function, use ToCharArray instead of a cast to Char() …).
Finally, use meaningful variable names. str, str2 and arr are particularly cryptic because they don’t tell the reader of the code anything of interest about the variables.
Sub Main()
Dim text As String = File.ReadAllText("File.txt")
Dim words As String() = str.Split(" "c)
Using output = File.CreateText("file2.txt")
For Each word In words
dim wordChars = word.ToCharArray()
For i As Integer = 0 To wordChars.Length - 1 Step 2
Dim tmp As Char = wordChars(i + 1)
wordChars(i + 1) = wordChars(i)
arr(i) = tmp
Next
Dim shuffledWord As New String(wordChars)
output.Write(shuffledWord + " ")
Console.Write(huffledWord + " ")
Next
End Using
Console.Read()
End Sub