Count Instances Of String Within String

[Derek Hart]

Is there an efficient line of code to count the number of instances of one
string within another.

If I have the sentence:
"I want to go to the park, and then go home."

It would give me a count of 2 for the word "go"

Derek

[Michel Posseth [MCP]]

Re: Count Instances Of String Within String

use the string compare method

Dim s As String = "I want to go to the park, and then go home."
MsgBox((String. Compare(s, "go") + 1).ToString & " Occurances of go")

regards

Michel Posseth [MCP]


"Derek Hart" <derekmhart@yah oo.com> schreef in bericht
news:%23x99Nifc GHA.3952@TK2MSF TNGP04.phx.gbl. ..[color=blue]
> Is there an efficient line of code to count the number of instances of one
> string within another.
>
> If I have the sentence:
> "I want to go to the park, and then go home."
>
> It would give me a count of 2 for the word "go"
>
> Derek
>[/color]

[Nathan Sokalski]

Re: Count Instances Of String Within String

I don't believe there is a way to put it in one line (at least not in
VB.NET, other languages might have a built-in method that does this or
possibly give you the ability to put multiple commands on one line), but
here is a simple loop that does what you want in VB.NET:


Dim teststring As String = "I want to go to the park, and then go home."
Dim i As Integer = -1
Dim stringcount As Integer = 0
While teststring.Inde xOf("go", i + 1) <> -1
stringcount += 1
i = teststring.Inde xOf("go", i + 1)
End While


You can also write it as a function, which I would strongly suggest if you
plan on doing this more than once:


Public Function CountInstances( ByVal lookfor As String, ByVal lookin As
String) As Integer
Dim stringcount As Integer = 0
Dim i As Integer = -1
While lookin.IndexOf( lookfor, i + 1) <> -1
stringcount += 1
i = lookin.IndexOf( lookfor, i + 1)
End While
Return stringcount
End Function


Writing it as a function will require the few lines of code to write the
function, but after that you can call it from just one line, making your
code simpler to write and debug:


stringcount = CountInstances( "go", teststring)


If you have any questions, feel free to ask. Good Luck!
--
Nathan Sokalski
njsokalski@hotm ail.com

http://www.nathansokalski.com/

"Derek Hart" <derekmhart@yah oo.com> wrote in message
news:%23x99Nifc GHA.3952@TK2MSF TNGP04.phx.gbl. ..[color=blue]
> Is there an efficient line of code to count the number of instances of one
> string within another.
>
> If I have the sentence:
> "I want to go to the park, and then go home."
>
> It would give me a count of 2 for the word "go"
>
> Derek
>[/color]

[jayeldee]

Re: Count Instances Of String Within String

Derek,

You could also use regular expressions to match the instances of a
string inside another string. I *believe* the pattern I use returns
all instances of 'go' surrounded by spaces, but then again I'm not very
good with RegEx yet.


Imports System.Text.Reg ularExpressions

Module main
Sub main()

Dim strtoSearch As String = "I want to go to the park, and then
go home."
Console.WriteLi ne(GetStringOcc urences(strtoSe arch,
"go").ToStr ing)
Console.ReadLin e()

End Sub

Private Function GetStringOccure nces(ByVal searchString, ByVal
searchWord) As Integer

Dim r As New Regex(String.Fo rmat("\s{0}\s", searchWord))

Return r.Matches(searc hString).Count

End Function

End Module

[Michel Posseth [MCP]]

Re: Count Instances Of String Within String


Ahum :-(

Embarased mode :

Private Sub Button1_Click(B yVal sender As System.Object, ByVal e As
System.EventArg s) Handles Button1.Click
Dim s As String = "I want to go to the park, and then go home."
MsgBox(countsub strings(s, "go"))

End Sub
Function countsubstrings (ByVal source As String, ByVal search As String)
As Integer
Dim count As Integer = -1
Dim index As Integer = -1
Do
count += 1
index = source.IndexOf( search, index + 1)
Loop Until index < 0
Return count
End Function

this wil work




"Michel Posseth [MCP]" <michel@posseth .com> schreef in bericht
news:OmwGe9fcGH A.4108@TK2MSFTN GP03.phx.gbl...[color=blue]
> use the string compare method
>
> Dim s As String = "I want to go to the park, and then go home."
> MsgBox((String. Compare(s, "go") + 1).ToString & " Occurances of
> go")
>
> regards
>
> Michel Posseth [MCP]
>
>
> "Derek Hart" <derekmhart@yah oo.com> schreef in bericht
> news:%23x99Nifc GHA.3952@TK2MSF TNGP04.phx.gbl. ..[color=green]
>> Is there an efficient line of code to count the number of instances of
>> one string within another.
>>
>> If I have the sentence:
>> "I want to go to the park, and then go home."
>>
>> It would give me a count of 2 for the word "go"
>>
>> Derek
>>[/color]
>
>[/color]

[+Vice]

Re: Count Instances Of String Within String

Dim str$ = "I want to go to the park, and then go home."
Dim findstr$ = "go"
Dim wordcount% = (Len(str) - Len(Replace(str , findstr, ""))) / Len(findstr)
MsgBox(wordcoun t)

"Derek Hart" <derekmhart@yah oo.com> wrote in message
news:%23x99Nifc GHA.3952@TK2MSF TNGP04.phx.gbl. ..[color=blue]
> Is there an efficient line of code to count the number of instances of one
> string within another.
>
> If I have the sentence:
> "I want to go to the park, and then go home."
>
> It would give me a count of 2 for the word "go"
>
> Derek
>[/color]

[Cor Ligthert [MVP]]

Re: Count Instances Of String Within String

Derek,

As once tested in this newsgroup is this the fastest method for that (I have
changed the fieldnames now so watch that).
\\\
Public Function CountString(ByV al SearchItem As String, ByVal ToCountString
_
As String) As Integer
Dim Start as Integer = 1
Dim Count as Interger = 0
Dim Result as Interger
Do
Result = InStr(Start, SearchItem, ToCountString)
If Result = 0 Then Exit Do
Count += 1
Start = Result + 1
Loop
Return Count
End Function
///

If the ToCountString becomes a ToCountChar than there are better methods.

I hope this helps,

Cor

"Derek Hart" <derekmhart@yah oo.com> schreef in bericht
news:%23x99Nifc GHA.3952@TK2MSF TNGP04.phx.gbl. ..[color=blue]
> Is there an efficient line of code to count the number of instances of one
> string within another.
>
> If I have the sentence:
> "I want to go to the park, and then go home."
>
> It would give me a count of 2 for the word "go"
>
> Derek
>[/color]

[Greg Young]

Re: Count Instances Of String Within String

Cor ...

surely the following will be much faster.
string tofind = "test"
string foo = "testtest footestfoo";
string changed = foo.Replace(tof ind, "");
return (foo.length - changed.length) / tofind.length

Cheers,

Greg Young
MVP - C#
"Cor Ligthert [MVP]" <notmyfirstname @planet.nl> wrote in message
news:u9riclmcGH A.3908@TK2MSFTN GP04.phx.gbl...[color=blue]
> Derek,
>
> As once tested in this newsgroup is this the fastest method for that (I
> have changed the fieldnames now so watch that).
> \\\
> Public Function CountString(ByV al SearchItem As String, ByVal
> ToCountString _
> As String) As Integer
> Dim Start as Integer = 1
> Dim Count as Interger = 0
> Dim Result as Interger
> Do
> Result = InStr(Start, SearchItem, ToCountString)
> If Result = 0 Then Exit Do
> Count += 1
> Start = Result + 1
> Loop
> Return Count
> End Function
> ///
>
> If the ToCountString becomes a ToCountChar than there are better methods.
>
> I hope this helps,
>
> Cor
>
> "Derek Hart" <derekmhart@yah oo.com> schreef in bericht
> news:%23x99Nifc GHA.3952@TK2MSF TNGP04.phx.gbl. ..[color=green]
>> Is there an efficient line of code to count the number of instances of
>> one string within another.
>>
>> If I have the sentence:
>> "I want to go to the park, and then go home."
>>
>> It would give me a count of 2 for the word "go"
>>
>> Derek
>>[/color]
>
>[/color]

[Cor Ligthert [MVP]]

Re: Count Instances Of String Within String

Greg,

I am sure it will not.
[color=blue]
>
> surely the following will be much faster.
> string tofind = "test"
> string foo = "testtest footestfoo";
> string changed = foo.Replace(tof ind, "");
> return (foo.length - changed.length) / tofind.length
>[/color]
Although some little changes can make that it probably does.

:-)

I said I thought yesterday already that I found it a nice idea.

I forgot that with some changes you can use it for this as well.

Cor

[Cor Ligthert [MVP]]

Re: Count Instances Of String Within String

It does probably, I misreaded something, I am testing it, what it real means
because I am in doubt about some side effects.

Cor

"Cor Ligthert [MVP]" <notmyfirstname @planet.nl> schreef in bericht
news:OIiJw3ncGH A.1208@TK2MSFTN GP02.phx.gbl...[color=blue]
> Greg,
>
> I am sure it will not.
>[color=green]
>>
>> surely the following will be much faster.
>> string tofind = "test"
>> string foo = "testtest footestfoo";
>> string changed = foo.Replace(tof ind, "");
>> return (foo.length - changed.length) / tofind.length
>>[/color]
> Although some little changes can make that it probably does.
>
> :-)
>
> I said I thought yesterday already that I found it a nice idea.
>
> I forgot that with some changes you can use it for this as well.
>
> Cor
>[/color]

[Greg Young]

Re: Count Instances Of String Within String

I am a strong believer in the pragmatic programmer idea of ... if in doubt
test it ...

added a loop to run 100 tests, countstring2 beat countstring 100/100 times
... even with varying data counts (including no data)... perhaps you have
some code shoing the opposite?

Cheers,

Greg Young
MVP - C#

Module Module1

Public Function CountString(ByV al SearchItem As String, ByVal
ToCountString As String) As Integer
Dim Start As Integer = 1
Dim Count As Integer = 0
Dim Result As Integer
Do
Result = InStr(Start, SearchItem, ToCountString)
If Result = 0 Then Exit Do
Count += 1
Start = Result + 1
Loop
Return Count
End Function

Public Function CountString2(By Val SearchItem As String, ByVal
ToCountString As String) As Integer
Dim tmp As String = SearchItem.Repl ace(ToCountStri ng, "")
Return (SearchItem.Len gth - tmp.Length) / ToCountString.L ength
End Function

Sub Main()
Dim ToCountString As String = "test"
Dim SearchItem As String = "testtesttestfo otesttesttest"
Dim i As Integer
Dim starttime As DateTime = DateTime.Now
Dim endtime As DateTime
For i = 0 To 1000000
CountString(Sea rchItem, ToCountString)
Next
endtime = DateTime.Now
Console.WriteLi ne("CountStrin g - " & (endtime -
starttime).ToSt ring())
starttime = DateTime.Now
For i = 0 To 1000000
CountString2(Se archItem, ToCountString)
Next
endtime = DateTime.Now
Console.WriteLi ne("CountString 2 - " & (endtime -
starttime).ToSt ring())

End Sub

End Module
"Cor Ligthert [MVP]" <notmyfirstname @planet.nl> wrote in message
news:OIiJw3ncGH A.1208@TK2MSFTN GP02.phx.gbl...[color=blue]
> Greg,
>
> I am sure it will not.
>[color=green]
>>
>> surely the following will be much faster.
>> string tofind = "test"
>> string foo = "testtest footestfoo";
>> string changed = foo.Replace(tof ind, "");
>> return (foo.length - changed.length) / tofind.length
>>[/color]
> Although some little changes can make that it probably does.
>
> :-)
>
> I said I thought yesterday already that I found it a nice idea.
>
> I forgot that with some changes you can use it for this as well.
>
> Cor
>[/color]

[Greg Young]

Re: Count Instances Of String Within String

The problem exists in both methods in use ..

If I want to look for the word "GO" and I also have worgs like gong or pogo,
they will be detected as being instances of the word, an easy way to work
around this is to pass in spaces i.e. " go " but then I will not detect
patterns such as "lets go!" because there is no traling space or "go to the
beach" because there is no leading space.

It is these items that make the implementation of an algorithm like this
tricky.

Cheers,

Greg Young
MVP - C#
"Cor Ligthert [MVP]" <notmyfirstname @planet.nl> wrote in message
news:%23mJl6Boc GHA.1208@TK2MSF TNGP02.phx.gbl. ..[color=blue]
> It does probably, I misreaded something, I am testing it, what it real
> means because I am in doubt about some side effects.
>
> Cor
>
> "Cor Ligthert [MVP]" <notmyfirstname @planet.nl> schreef in bericht
> news:OIiJw3ncGH A.1208@TK2MSFTN GP02.phx.gbl...[color=green]
>> Greg,
>>
>> I am sure it will not.
>>[color=darkred]
>>>
>>> surely the following will be much faster.
>>> string tofind = "test"
>>> string foo = "testtest footestfoo";
>>> string changed = foo.Replace(tof ind, "");
>>> return (foo.length - changed.length) / tofind.length
>>>[/color]
>> Although some little changes can make that it probably does.
>>
>> :-)
>>
>> I said I thought yesterday already that I found it a nice idea.
>>
>> I forgot that with some changes you can use it for this as well.
>>
>> Cor
>>[/color]
>
>[/color]

[Cor Ligthert [MVP]]

Re: Count Instances Of String Within String

Greg,

The problem as you wrote was my idea too, however that side effect was not
there in my simple test. 100* a string "Cor Greg GregCor CorGreg ", that I
tested 100.000 times each time both methods and counting the time.

The method with the replace beats the methode with the moving instr at least
about 5:7.

Both methods are therefore quicker than any I have seen until now, while
that replace method is for me now the fastest.

Cor

"Greg Young" <DruckDruckGoos e@hotmail.com> schreef in bericht
news:uVRYdEocGH A.3352@TK2MSFTN GP03.phx.gbl...[color=blue]
>I am a strong believer in the pragmatic programmer idea of ... if in doubt
>test it ...
>
> added a loop to run 100 tests, countstring2 beat countstring 100/100 times
> .. even with varying data counts (including no data)... perhaps you have
> some code shoing the opposite?
>
> Cheers,
>
> Greg Young
> MVP - C#
>
> Module Module1
>
> Public Function CountString(ByV al SearchItem As String, ByVal
> ToCountString As String) As Integer
> Dim Start As Integer = 1
> Dim Count As Integer = 0
> Dim Result As Integer
> Do
> Result = InStr(Start, SearchItem, ToCountString)
> If Result = 0 Then Exit Do
> Count += 1
> Start = Result + 1
> Loop
> Return Count
> End Function
>
> Public Function CountString2(By Val SearchItem As String, ByVal
> ToCountString As String) As Integer
> Dim tmp As String = SearchItem.Repl ace(ToCountStri ng, "")
> Return (SearchItem.Len gth - tmp.Length) / ToCountString.L ength
> End Function
>
> Sub Main()
> Dim ToCountString As String = "test"
> Dim SearchItem As String = "testtesttestfo otesttesttest"
> Dim i As Integer
> Dim starttime As DateTime = DateTime.Now
> Dim endtime As DateTime
> For i = 0 To 1000000
> CountString(Sea rchItem, ToCountString)
> Next
> endtime = DateTime.Now
> Console.WriteLi ne("CountStrin g - " & (endtime -
> starttime).ToSt ring())
> starttime = DateTime.Now
> For i = 0 To 1000000
> CountString2(Se archItem, ToCountString)
> Next
> endtime = DateTime.Now
> Console.WriteLi ne("CountString 2 - " & (endtime -
> starttime).ToSt ring())
>
> End Sub
>
> End Module
> "Cor Ligthert [MVP]" <notmyfirstname @planet.nl> wrote in message
> news:OIiJw3ncGH A.1208@TK2MSFTN GP02.phx.gbl...[color=green]
>> Greg,
>>
>> I am sure it will not.
>>[color=darkred]
>>>
>>> surely the following will be much faster.
>>> string tofind = "test"
>>> string foo = "testtest footestfoo";
>>> string changed = foo.Replace(tof ind, "");
>>> return (foo.length - changed.length) / tofind.length
>>>[/color]
>> Although some little changes can make that it probably does.
>>
>> :-)
>>
>> I said I thought yesterday already that I found it a nice idea.
>>
>> I forgot that with some changes you can use it for this as well.
>>
>> Cor
>>[/color]
>
>[/color]

[Göran Andersson]

Re: Count Instances Of String Within String

The replace method might be more efficient for short strings, but the
method using InStr (or IndexOf) scales better, as it doesn't create
another string that is almost as big as the string being searched.

Cor Ligthert [MVP] wrote:[color=blue]
> It does probably, I misreaded something, I am testing it, what it real means
> because I am in doubt about some side effects.
>
> Cor
>
> "Cor Ligthert [MVP]" <notmyfirstname @planet.nl> schreef in bericht
> news:OIiJw3ncGH A.1208@TK2MSFTN GP02.phx.gbl...[color=green]
>> Greg,
>>
>> I am sure it will not.
>>[color=darkred]
>>> surely the following will be much faster.
>>> string tofind = "test"
>>> string foo = "testtest footestfoo";
>>> string changed = foo.Replace(tof ind, "");
>>> return (foo.length - changed.length) / tofind.length
>>>[/color]
>> Although some little changes can make that it probably does.
>>
>> :-)
>>
>> I said I thought yesterday already that I found it a nice idea.
>>
>> I forgot that with some changes you can use it for this as well.
>>
>> Cor
>>[/color]
>
>[/color]

[Cor Ligthert [MVP]]

Re: Count Instances Of String Within String

Goran,

Indexof with strings is twice as slow as InStr.

With char it beats InStr that it is not to mention, but it is than of course
comparing apples with pears, because InStr(char) does not exist.

Cor

"Göran Andersson" <guffa@guffa.co m> schreef in bericht
news:eHs4RZocGH A.3952@TK2MSFTN GP04.phx.gbl...[color=blue]
> The replace method might be more efficient for short strings, but the
> method using InStr (or IndexOf) scales better, as it doesn't create
> another string that is almost as big as the string being searched.
>
> Cor Ligthert [MVP] wrote:[color=green]
>> It does probably, I misreaded something, I am testing it, what it real
>> means because I am in doubt about some side effects.
>>
>> Cor
>>
>> "Cor Ligthert [MVP]" <notmyfirstname @planet.nl> schreef in bericht
>> news:OIiJw3ncGH A.1208@TK2MSFTN GP02.phx.gbl...[color=darkred]
>>> Greg,
>>>
>>> I am sure it will not.
>>>
>>>> surely the following will be much faster.
>>>> string tofind = "test"
>>>> string foo = "testtest footestfoo";
>>>> string changed = foo.Replace(tof ind, "");
>>>> return (foo.length - changed.length) / tofind.length
>>>>
>>> Although some little changes can make that it probably does.
>>>
>>> :-)
>>>
>>> I said I thought yesterday already that I found it a nice idea.
>>>
>>> I forgot that with some changes you can use it for this as well.
>>>
>>> Cor
>>>[/color]
>>[/color][/color]