Replace() already exists, but that function takes strings as parameters. I need range.
In my string there are two "strings" that are 10 characters long.
Greger with 6 chars and 4 spaces and the other string with 10 characters.
"Greger AASSDDFFGG"
I want to replace "Greger " with "googlioa "
What i'm looking for is basically this:
Replace(MyString,1,10) = "googlioa "
Is there any way to achieve this?
If they're always going to be 10 chars, just pad the names.
strNameFind = "Greger"
strNameReplace = "googlioa"
' Pad the names...
strNameFind = Left(strNameFind & Space(10), 10)
strNameReplace = Left(strNameReplace & Space(10), 10)
MyString = Replace(MyString, strNameFind, strNameReplace)
Alternatively, if you don't want to determine the existing name, just pad your new name appropriately and add the remainder of your string:
' Pad the new name to fit in a 10-char column...
strNameReplace = "googlioa"
strNameReplace = Left(strNameReplace & Space(10), 10)
' Update the record...
MyString = strNameReplace & Mid(MyString, 11)
If you want to replace strictly by position, use concatenation of Left(), new, and Mid(). To get you started:
>> Function replByPos(s, f, l, n)
>> replByPos = Left(s, f-1) & n & Mid(s, f + l - 1)
>> End Function
>> s = "Greger AASSDDFFGG"
>> r = replByPos(s, 1, 10, "googlioa ")
>> WScript.Echo s
>> WScript.Echo r
>>
Greger AASSDDFFGG
googlioa AASSDDFFGG
>>
Further enhancements:
safety: f(rom) - 1 is risky - should be checked
padding of new string wrt l(ength)
perhaps you want to search (Instr()) for old ("Greger ") before the concatenation
On second thought (and stealing Bond's padding):
Maybe I should have interpeted the 10 as a to/till/upto value instead of a length/width specification. So see whether
Option Explicit
Function replByPos(src, from, till, ns)
Dim w : w = till - from
replByPos = Left(src, from - 1) & Left(ns & Space(w), w) & Mid(src, till)
End Function
Dim s : s = "Greger AASSDDFFGG"
Dim ns : ns = "googlioa"
WScript.Echo s
WScript.Echo replByPos(s, 1, 10, ns)
s = "Whatever Greger AASSDDFFGG"
ns = "googlioa"
Dim p : p = Instr(s, "Greger")
WScript.Echo s
WScript.Echo replByPos(s, p, p + 10, ns)
output:
cscript 22811896.vbs
Greger AASSDDFFGG
googlioa AASSDDFFGG
Whatever Greger AASSDDFFGG
Whatever googlioa AASSDDFFGG
matches your specs better.
Related
I need to convert my time to Number. i need it because letter i'm going to divide this dates by yourself to calculate % (agent productive time ).
I tried something like this
'cells(2,3) = 22:12:2
cells(2,3) / (60*60*1000)
Thanks in advance
You can do like this:
a = "78:19:41"
b = "74:23:58"
ta = (Split(a, ":")(0) / 24) + TimeValue("00:" & Split(a, ":", 2)(1))
tb = (Split(b, ":")(0) / 24) + TimeValue("00:" & Split(b, ":", 2)(1))
p = tb / ta * 100
p -> 94.9844138434859
I need to remove the numeric part at the end of a string. Here are some examples:
"abcd1234" -> "abcd"
"a3bc45" -> "a3bc"
"kj3ih5" -> "kj3ih"
You get the idea.
I implemented a function which works well for this purpose.
Function VarStamm(name As String) As String
Dim i, a As Integer
a = 0
For i = Len(name) To 1 Step -1
If IsNumeric(Mid(name, i, 1)) = False Then
i = i + 1
Exit For
End If
Next i
If i <= Len(name) Then
VarStamm = name.Substring(0, i - 1)
Else
VarStamm = name
End If
End Function
The question is: is there any faster (more efficient in speed) way to do this? The problem is, I call this function within a loop with 3 million iterations and it would be nice to have it be more efficient.
I know about the String.LastIndexOf method, but I don't know how to use it when I need the index of the last connected number within a string.
You can use Array.FindLastIndex and then Substring:
Dim lastNonDigitIndex = Array.FindLastIndex(text.ToCharArray(), Function(c) Not char.IsDigit(c))
If lastNonDigitIndex >= 0
lastNonDigitIndex += 1
Dim part1 = text.Substring(0, lastNonDigitIndex)
Dim part2 = text.Substring(lastNonDigitIndex)
End If
I was skeptical that the Array.FindLastIndex method was actually faster, so I tested it myself. I borrowed the testing code posted by Amessihel, but added a third method:
Function VarStamm3(name As String) As String
Dim i As Integer
For i = name.Length - 1 To 0 Step -1
If Not Char.IsDigit(name(i)) Then
Exit For
End If
Next i
Return name.Substring(0, i + 1)
End Function
It uses your original algorithm, but just swaps out the old VB6-style string methods for newer .NET equivalent ones. Here's the results on my machine:
RunTime :
- VarStamm : 00:00:07.92
- VarStamm2 : 00:00:00.60
- VarStamm3 : 00:00:00.23
As you can see, your original algorithm was already quite well tuned. The problem wasn't the loop. The problem was Mid, IsNumeric, and Len. Since Tim's method didn't use those, it was much faster. But, if you stick with a manual for loop, it's twice as fast as using Array.FindLastIndex, all things being equal
Given your function VarStamm and Tim Schmelter's one named VarStamm2, here is a small test performance I wrote. I typed an arbitrary long String with a huge right part, and ran the functions one million times.
Module StackOverlow
Sub Main()
Dim testStr = "azekzoerjezoriezltjreoitueriou7657678678797897898997897978897898797989797"
Console.WriteLine("RunTime :" + vbNewLine +
" - VarStamm : " + getTimeSpent(AddressOf VarStamm, testStr) + vbNewLine +
" - VarStamm2 : " + getTimeSpent(AddressOf VarStamm2, testStr))
End Sub
Function getTimeSpent(f As Action(Of String), str As String) As String
Dim sw As Stopwatch = New Stopwatch()
Dim ts As TimeSpan
sw.Start()
For i = 1 To 1000000
f(str)
Next
sw.Stop()
ts = sw.Elapsed
Return String.Format("{0:00}:{1:00}:{2:00}.{3:00}",
ts.Hours, ts.Minutes, ts.Seconds,
ts.Milliseconds / 10)
End Function
Function VarStamm(name As String) As String
Dim i, a As Integer
a = 0
For i = Len(name) To 1 Step -1
If IsNumeric(Mid(name, i, 1)) = False Then
i = i + 1
Exit For
End If
Next i
If i <= Len(name) Then
VarStamm = name.Substring(0, i - 1)
Else
VarStamm = name
End If
End Function
Function VarStamm2(name As String) As String
Dim lastNonDigitIndex = Array.FindLastIndex(name.ToCharArray(), Function(c) Not Char.IsDigit(c))
If lastNonDigitIndex >= 0 Then
lastNonDigitIndex += 1
Return name.Substring(0, lastNonDigitIndex)
End If
Return name
End Function
End Module
Here is the output I got:
RunTime :
- VarStamm : 00:00:38.33
- VarStamm2 : 00:00:02.72
So yes, you should choose his answer, his code is both pretty and efficient.
Hello everyone.
Dim txt1 As Double = Convert.ToDouble(TextBox1.Text) / 100
Dim txt2 As Double = Convert.ToDouble(TextBox2.Text)
Dim txt3 As Double = Convert.ToDouble(TextBox3.Text)
Dim txtResult As Double = Convert.ToDouble(TextBox4.Text)
Dim result As Double = txt1 * txt2 * txt3
TextBox4.Text = result
As you can see I get my result depending on what the user types in. So I have to add a space after a certain character. Textbox14.text(0) <--- after this do I want my space. It's so that after the value is higher than 999 it should type out 1 000 and not 1000. Thank you very much for any useful help, I've truly looked everywhere, I just can't find anything.
You talking about group separator. Custom Numeric Format Strings
You can use .ToString() method and define group separator in the format.
TextBox4.Text = result.ToString("0,0.000")
Different separators will be used based on the local system's language/region settings.
You can define your custome separator manually
var cultureInfo = new System.Globalization.CultureInfo("en-US");
var numberInfo = cultureInfo.NumberFormat;
numberInfo.NumberGroupSeparator = " ";
TextBox4.Text = result.ToString("0,0.000", numberInfo)
If I get it right, you want every 3 chars a space, right ?
Like 1 000 000 ?
try this :
Dim result As String, str As String, ret As String
Dim i As Integer
Dim arr As Char()
'your text to space
result = "10000000"
'reverte so we start with the end
result = StrReverse(result)
i = 0
ret = ""
' make a char array which each char is an own array element
arr = result.Take(result.Length).ToArray
'iterate through all elements
For Each str In arr
' skip the first element .
' only add a space every 3 elements
If (i <> 0) And (i Mod 3 = 0) Then
ret = ret + " "
End If
ret = ret + str
i = i + 1
Next
' revers again the output
ret = StrReverse(ret)
MsgBox(ret)
So I have the following code to split a string between whitespaces:
text = "I am 'the text'"
for string in text:gmatch("%S+") do
print(string)
end
The result:
I
am
'the
text'
But I need to do this:
I
am
the text --[[yep, without the quotes]]
How can I do this?
Edit: just to complement the question, the idea is to pass parameters from a program to another program. Here is the pull request that I am working, currently in review: https://github.com/mpv-player/mpv/pull/1619
There may be ways to do this with clever parsing, but an alternative way may be to keep track of a simple state and merge fragments based on detection of quoted fragments. Something like this may work:
local text = [[I "am" 'the text' and "some more text with '" and "escaped \" text"]]
local spat, epat, buf, quoted = [=[^(['"])]=], [=[(['"])$]=]
for str in text:gmatch("%S+") do
local squoted = str:match(spat)
local equoted = str:match(epat)
local escaped = str:match([=[(\*)['"]$]=])
if squoted and not quoted and not equoted then
buf, quoted = str, squoted
elseif buf and equoted == quoted and #escaped % 2 == 0 then
str, buf, quoted = buf .. ' ' .. str, nil, nil
elseif buf then
buf = buf .. ' ' .. str
end
if not buf then print((str:gsub(spat,""):gsub(epat,""))) end
end
if buf then print("Missing matching quote for "..buf) end
This will print:
I
am
the text
and
some more text with '
and
escaped \" text
Updated to handle mixed and escaped quotes. Updated to remove quotes. Updated to handle quoted words.
Try this:
text = [[I am 'the text' and '' here is "another text in quotes" and this is the end]]
local e = 0
while true do
local b = e+1
b = text:find("%S",b)
if b==nil then break end
if text:sub(b,b)=="'" then
e = text:find("'",b+1)
b = b+1
elseif text:sub(b,b)=='"' then
e = text:find('"',b+1)
b = b+1
else
e = text:find("%s",b+1)
end
if e==nil then e=#text+1 end
print("["..text:sub(b,e-1).."]")
end
Lua Patterns aren't powerful to handle this task properly. Here is an LPeg solution adapted from the Lua Lexer. It handles both single and double quotes.
local lpeg = require 'lpeg'
local P, S, C, Cc, Ct = lpeg.P, lpeg.S, lpeg.C, lpeg.Cc, lpeg.Ct
local function token(id, patt) return Ct(Cc(id) * C(patt)) end
local singleq = P "'" * ((1 - S "'\r\n\f\\") + (P '\\' * 1)) ^ 0 * "'"
local doubleq = P '"' * ((1 - S '"\r\n\f\\') + (P '\\' * 1)) ^ 0 * '"'
local white = token('whitespace', S('\r\n\f\t ')^1)
local word = token('word', (1 - S("' \r\n\f\t\""))^1)
local string = token('string', singleq + doubleq)
local tokens = Ct((string + white + word) ^ 0)
input = [["This is a string" 'another string' these are words]]
for _, tok in ipairs(lpeg.match(tokens, input)) do
if tok[1] ~= "whitespace" then
if tok[1] == "string" then
print(tok[2]:sub(2,-2)) -- cut off quotes
else
print(tok[2])
end
end
end
Output:
This is a string
another string
these
are
words
This question already has answers here:
How to convert a column number (e.g. 127) into an Excel column (e.g. AA)
(60 answers)
Closed 8 years ago.
I need some help in doing a logic that would convert a numeric value to corresponding MS Excel header value.
For example:
1 = "A"
2 = "B"
3 = "C"
4 = "D"
5 = "E"
.........
25 = "Y"
26 = "Z"
27 = "AA"
28 = "AB"
29 = "AC"
30 = "AD"
.........
Would appreciate some .NET codes (C# or VB) for this. Thanks.
Here's some VBA (with test code) I strung together in Excel which does the trick. Unless VB.NET has changed drastically, it should work okay. Even if it has, you should be able to translate the idea into workable code.
' num2col - translate Excel column number (1-n) into column string ("A"-"ZZ"). '
Function num2col(num As Integer) As String
' Subtract one to make modulo/divide cleaner. '
num = num - 1
' Select return value based on invalid/one-char/two-char input. '
If num < 0 Or num >= 27 * 26 Then
' Return special sentinel value if out of range. '
num2col = "-"
Else
' Single char, just get the letter. '
If num < 26 Then
num2col = Chr(num + 65)
Else
' Double char, get letters based on integer divide and modulus. '
num2col = Chr(num \ 26 + 64) + Chr(num Mod 26 + 65)
End If
End If
End Function
' Test code in Excel VBA. '
Sub main()
MsgBox ("- should be " & num2col(0))
MsgBox ("A should be " & num2col(1))
MsgBox ("B should be " & num2col(2))
MsgBox ("Z should be " & num2col(26))
MsgBox ("AA should be " & num2col(27))
MsgBox ("AB should be " & num2col(28))
MsgBox ("AY should be " & num2col(51))
MsgBox ("AZ should be " & num2col(52))
MsgBox ("BA should be " & num2col(53))
MsgBox ("ZY should be " & num2col(27 * 26 - 1))
MsgBox ("ZZ should be " & num2col(27 * 26))
MsgBox ("- should be " & num2col(27 * 26 + 1))
End Sub
public string ColumnNumberToLetter(int ColumnNumber)
{
if (ColumnNumber > 26)
{
return ((char) (Math.Floor(((double)ColumnNumber - 1) / 26) + 64)).ToString()
+ ((char) (((ColumnNumber - 1) % 26) + 65)).ToString();
}
return ((char)(ColumnNumber+64)).ToString();
}
Try this:
public static string ToExcelString(int number)
{
if (number > 25)
{
int secondaryCounter = 0;
while (number > 25)
{
secondaryCounter = secondaryCounter + 1;
number = number - 25;
}
return ToExcelChar(number) + ToExcelChar(secondaryCounter);
}
else
{
return ToExcelChar(number)
}
}
private const string alphabet = "ABCDEFGHIJKLMNOPQRSTUVWXYZ";
private static string ToExcelChar(int number)
{
if (number > 25 || number < 0)
{
throw new InvalidArgumentException("the number passed in (" + number + ") must be between the range 0-25");
}
return alphabet[number];
}
Use a number base conversion routine. You want to convert from base 10 to base 26. Add each digit to 'A'
As in: http://www.vbforums.com/showthread.php?t=271359
Just Use the activecell.address then manuipulate the string based on where the $ is to what you need.