Space delimiter

[simonc]

One of the things that drives me mad about vb.net is
spending time trying to make a program do something then
finding that there's already an inbuild procedure that
does exactly the same thing.

So before I begin on this can anyone tell me if there is
already a routine to do this:

I am dividing up a delimited textstring using
string.split. The delimiter character is a space, but
where there are adjacent spaces I want to treat these as a
single delimiter. What is the smart way to do this?

Grateful for any ideas

[EricJ]

yust a thought
strNew = strOld.replace(" "," ")
you could put it in a for next loop if you want to replace 3,4,... spaces 2

eric

"simonc" <an*******@discussions.microsoft.com> wrote in message
news:0d****************************@phx.gbl...

One of the things that drives me mad about vb.net is
spending time trying to make a program do something then
finding that there's already an inbuild procedure that
does exactly the same thing.

So before I begin on this can anyone tell me if there is
already a routine to do this:

I am dividing up a delimited textstring using
string.split. The delimiter character is a space, but
where there are adjacent spaces I want to treat these as a
single delimiter. What is the smart way to do this?

Grateful for any ideas

[Cor]

Hi Simonc

In addition to Eric did I make it with stringbuilder,
you asked something new and this is a new function who is real much faster
than all old string replace funtions. But I don't know if there is not a
better way.

Dim a As String = "There is a lot of for air here"
Dim sb As New System.Text.StringBuilder(a)
Do While sb.ToString.IndexOf(" ") > -1 '2 spaces
sb.Replace(" ", " ")
Loop
Dim b As String = sb.ToString
Dim c() As String = Split(b)
I hope this helps a little bit?

Cor
"simonc" <an*******@discussions.microsoft.com> schreef in bericht
news:0d****************************@phx.gbl...

One of the things that drives me mad about vb.net is
spending time trying to make a program do something then
finding that there's already an inbuild procedure that
does exactly the same thing.

So before I begin on this can anyone tell me if there is
already a routine to do this:

I am dividing up a delimited textstring using
string.split. The delimiter character is a space, but
where there are adjacent spaces I want to treat these as a
single delimiter. What is the smart way to do this?

Grateful for any ideas

[Chris Dunaway]

On Wed, 5 Nov 2003 04:27:09 -0800, simonc wrote:

I am dividing up a delimited textstring using
string.split. The delimiter character is a space, but
where there are adjacent spaces I want to treat these as a
single delimiter. What is the smart way to do this?

Yet another method is to use regular expressions. This expression replaces
all occurrences of 2 or more spaces with a single space.

Imports System.Text.RegularExpressions

strNew = Regex.Replace(strOld, " {2,}", " ")

--
Chris

To send me an E-mail, remove the underscores and lunchmeat from my E-Mail
address.

[EricJ]

i think we have a winner :)
ill have a look at those 2 ;)

Imports System.Text.RegularExpressions
strNew = Regex.Replace(strOld, " {2,}", " ")

[AlexS]

Possibly more reliable way would be to use

Regex.Replace(yourStr,@"\s+",@"\s");

Please check definition of \s in regular expressions reference

HTH
Alex
"Chris Dunaway" <dunawayc@_lunchmeat_sbcglobal.net> wrote in message
news:hg*****************************@40tude.net...

On Wed, 5 Nov 2003 04:27:09 -0800, simonc wrote:

I am dividing up a delimited textstring using
string.split. The delimiter character is a space, but
where there are adjacent spaces I want to treat these as a
single delimiter. What is the smart way to do this?
Yet another method is to use regular expressions. This expression

replaces all occurrences of 2 or more spaces with a single space.

Imports System.Text.RegularExpressions

strNew = Regex.Replace(strOld, " {2,}", " ")

--
Chris

To send me an E-mail, remove the underscores and lunchmeat from my E-Mail
address.

[Cor]

Hi Eric,

Did you test it?

strNew = Regex.Replace(strOld, " {2,}", " ")

This is 50 to 100 times slower than that loop with the stringbuilder.

(I did test it)

So it depends on the size of the string what to use.

Cor

[Jay B. Harlow [MVP - Outlook]]

Simonc,
But wait there is more! ;-)

I would not bother with the Replace if I was using the Regular Expression.

I would simply do the Split based on the Regular Expression!

Imports System.Text.RegularExpressions
Dim strNew() As String ' an array of strings

Split based on one or more space characters:
strNew = Regex.Split(strOld, " +")

Alternatively you can split based on whitespace characters.
strNew = Regex.Split(strOld, "\s+")

"\s" which is a short cut for white space, includes more than just the space
character.

Hope this helps
Jay
"simonc" <an*******@discussions.microsoft.com> wrote in message
news:10****************************@phx.gbl... Many thanks. I like this one best of all.

Simon C

-----Original Message-----
On Wed, 5 Nov 2003 04:27:09 -0800, simonc wrote:

I am dividing up a delimited textstring using
string.split. The delimiter character is a space, but
where there are adjacent spaces I want to treat these as a single delimiter. What is the smart way to do this?

Yet another method is to use regular expressions. This

expression replaces

all occurrences of 2 or more spaces with a single space.

Imports System.Text.RegularExpressions

strNew = Regex.Replace(strOld, " {2,}", " ")

--
Chris

To send me an E-mail, remove the underscores and

lunchmeat from my E-Mail

address.
.

[Chris Dunaway]

On Wed, 5 Nov 2003 17:51:13 +0100, Cor wrote:

Hi Eric,

Did you test it?

strNew = Regex.Replace(strOld, " {2,}", " ")

This is 50 to 100 times slower than that loop with the stringbuilder.

(I did test it)

So it depends on the size of the string what to use.

Cor

Thanks for pointing that out, but I am not seeing that. Here is the code I
used to time it. Here are the results:

First Run:

Method1: 156225 ticks
Method2: 0 ticks

Second Run (And every run after that):

Method1: 0 ticks
Method2: 0 ticks

Any idea why the first run on the regular expression would be slower, but
each subsequent run is not?

Thanks

'\\\\\\
Private Sub Button1_Click(...) Handles Button1.Click

Dim sNumber As String
Dim sResult As String
Dim ts As TimeSpan
Dim dStart As DateTime

sNumber = "This is a test string with lots of spaces"
dStart = Now
sResult = Method1(sNumber)
ts = Now.Subtract(dStart)
MsgBox(sResult & " Elapsed Time (1): " & ts.Ticks.ToString & " Ticks.")

dStart = Now
sResult = Method2(sNumber)
ts = Now.Subtract(dStart)

MsgBox(sResult & " Elapsed Time (2): " & ts.Ticks.ToString & " Ticks.")

End Sub

Private Function Method1(ByVal sInput As String) As String
Return RegularExpressions.Regex.Replace(sInput, " {2,}", " ")
End Function

Private Function Method2(ByVal sInput As String) As String
Dim sb As New System.Text.StringBuilder(sInput)
Do While sb.ToString.IndexOf(" ") > -1
sb.Replace(" ", " ")
Loop
Return sb.ToString
End Function

'//////
--
Chris

To send me an E-mail, remove the underscores and lunchmeat from my E-Mail
address.

[Chris Dunaway]

On Wed, 5 Nov 2003 13:47:46 -0600, Jay B. Harlow [MVP - Outlook] wrote:

Simonc,
But wait there is more! ;-)

I would not bother with the Replace if I was using the Regular Expression.

I would simply do the Split based on the Regular Expression!

"We're not worthy!, We're not worthy!"

:)

I forgot about the split method in the RegEx!

--
Chris

To send me an E-mail, remove the underscores and lunchmeat from my E-Mail
address.

[Jay B. Harlow [MVP - Outlook]]

Chris,
I have an little "FAQ" (my OE Sent items currently) that I use to answer
questions from, here is the normal snippet I use when people ask about
Split.

---x--- cut here ---x---
There are three Split functions in VB.NET:

Use Microsoft.VisualBasic.Strings.Split if you need to split a string based
on a specific word (string). It is the Split function from VB6.

Use System.String.Split if you need to split a string based on a collection
of specific characters. Each individual character is its own delimiter.

Alternatively use System.Text.RegularExpressions.RegEx.Split to split based
on matching patterns.
---x--- cut here ---x---

Hope this helps
Jay

"Chris Dunaway" <dunawayc@_lunchmeat_sbcglobal.net> wrote in message
news:1v*****************************@40tude.net...

On Wed, 5 Nov 2003 13:47:46 -0600, Jay B. Harlow [MVP - Outlook] wrote:

Simonc,
But wait there is more! ;-)

I would not bother with the Replace if I was using the Regular Expression.
I would simply do the Split based on the Regular Expression!

"We're not worthy!, We're not worthy!"

:)

I forgot about the split method in the RegEx!

--
Chris

To send me an E-mail, remove the underscores and lunchmeat from my E-Mail
address.

[Cor]

Hi Chris
I did test your test.
This is a short string that I did y times
With that the
first regex : Stringbuilder : Jay regex split 1 : Jay regex split 2
9:3:9:18

I changed your code and did it loop I also did a test with a real long
string and then I get differences from 50 till 100 time. That code I did not
suply if you want it tell it.

This was just for fun, your somewhere changed code is below

Cor

\\\
Private Function Method1(ByVal sInput As String) As String()
Return Split(System.Text.RegularExpressions.Regex.Replace (sInput, " {2,}", "
"))
End Function
Private Function Method2(ByVal sInput As String) As String()
Dim sb As New System.Text.StringBuilder(sInput)
Do While sb.ToString.IndexOf(" ") > -1
sb.Replace(" ", " ")
Loop
Return Split(sb.ToString)
End Function
Private Function Method3(ByVal sInput As String) As String()
Return System.Text.RegularExpressions.Regex.Split(sInput, " +")
End Function
Private Function Method4(ByVal sInput As String) As String()
Return System.Text.RegularExpressions.Regex.Split(sInput, "\s+")
End Function
Private Sub Button1_Click(ByVal sender As System.Object, _
ByVal e As System.EventArgs) Handles Button1.Click
Dim sNumber As String
Dim sResult As String()
Dim ts As Integer
Dim res As String
Dim m As Integer
Dim i As Double
Dim y As Double = 100000
sNumber = "This is a test string with lots of spaces"
ts = Environment.TickCount
For i = 0 To y
sResult = Method1(sNumber)
Next
ts = Environment.TickCount - ts
res = ""
For m = 0 To sResult.Length - 1
res = res & sResult(m).ToString & " "
Next
MsgBox(res & " Elapsed Time (1): " & ts.ToString & " Ticks.")
ts = Environment.TickCount
For i = 0 To y
sResult = Method2(sNumber)
Next
ts = Environment.TickCount - ts
res = ""
For m = 0 To sResult.Length - 1
res = res & sResult(m).ToString & " "
Next
MsgBox(res & " Elapsed Time (2): " & ts.ToString & " Ticks.")
ts = Environment.TickCount
For i = 0 To y
sResult = Method3(sNumber)
Next
ts = Environment.TickCount - ts
res = ""
For m = 0 To sResult.Length - 1
res = res & sResult(m).ToString & " "
Next
MsgBox(res & " Elapsed Time (3): " & ts.ToString & " Ticks.")
ts = Environment.TickCount
For i = 0 To y
sResult = Method4(sNumber)
Next
ts = Environment.TickCount - ts
res = ""
For m = 0 To sResult.Length - 1
res = res & sResult(m).ToString & " "
Next
MsgBox(res & " Elapsed Time (4): " & ts.ToString & " Ticks.")
End Sub
///

[Cor]

HI Jay B.

I thought your last metod would be it, but again stringbuilder wins.

I did only wanted to correct the testprogram from Chris so he could do the
test.

Look for it at Chris, I stop for tonight

If I did something wrong in the test, than it was an accident.

And tell it me?

Cor

[Chris Dunaway]

On Wed, 5 Nov 2003 22:19:04 +0100, Cor wrote:

I did test your test.
This is a short string that I did y times
With that the
first regex : Stringbuilder : Jay regex split 1 : Jay regex split 2
9:3:9:18

That would seem to indicate that the stringbuilder is 3 times faster. I
got similar results using the code you posted (unchanged):

Method1 Method2 Method3 Method4
2141 750 2328 5141 ticks

Again the stringbuilder seems to be about 3 times faster, but not 50 to
100, unless I'm missing something.

It is an interesting experiement, to be sure.

Cheers!
--
Chris

To send me an E-mail, remove the underscores and lunchmeat from my E-Mail
address.

[Jay B. Harlow [MVP - Outlook]]

Cor,
I would not expect the Regex to be the fastest, I would however expect it to
be 'easier' to enter.

Which is where:

strNew = Regex.Split(strOld, " +")

Does seem very easy to enter ;-)

Thanks for the testing.
Jay

"Cor" <no*@non.com> wrote in message
news:%2******************@TK2MSFTNGP11.phx.gbl...

HI Jay B.

I thought your last metod would be it, but again stringbuilder wins.

I did only wanted to correct the testprogram from Chris so he could do the
test.

Look for it at Chris, I stop for tonight

If I did something wrong in the test, than it was an accident.

And tell it me?

Cor

[Cor]

Chris,

Now we where testing 100000 times a very short string in a loop.

In another test I did make a string from 1000000 long (I don't know it exact
anymore) and than you get those differences. The efficiency from the
stringbuilder is in building string, so there would be an upgoing line when
the strings become longer.

Cor

[Fergus Cooney]

Hi Chris,

Any timing test that comes back with 0 is highly suspect!! ;-))

For a fair comparison you want some sort of loop that is going to take a
few seconds for each method.

For I = 1 To 100000 'add zeros as required

You might also want to experiment with different sizes of string

N = 100000
S = "h sl sd js da a sj kl sdj sd "
For Test = 1 To 5
DoTheTimingTests (N, S)
N = N / 2
S &= S
Next

Regards,
Fergus