I'm trying to figure out a way to extract the last alpha+numeric sequence in a string made up of similar patterns. The sequence is an alpha+numeric pair: an alpha string (one or more letters) plus a numeric string (one or more numbers). For instance:
G98Y8RT9 -- I need to isolate "RT9"
H8L77 -- I need to isolate "L77"
D64RL19HT7899 -- I need to isolate "HT7899"
As shown above, there are a variable number of characters in each part of the pair and also in the number of pairs preceding the last one. I've tried Excel formulas using FIND, ISNUMBER, etc., but I couldn't figure out the logic to make it work for these variables.
Is there a formula that would help? Or is some kind of regex VBA function the way to go?
I think this should work, as a user-defined function you can place it in a standard module, and call it like:
=GetLastPair($A$1), etc.
Here is the function:
Function GetLastPair(str As String)
Dim numPart As Integer
Dim strPart As Integer
Do Until Not IsNumeric(Mid(str, Len(str) - numPart, 1))
numPart = numPart + 1
Loop
Do Until IsNumeric(Mid(str, Len(str) - numPart - strPart, 1))
strPart = strPart + 1
Loop
GetLastPair = Right(str, numPart + strPart)
End Function
Results:
A bit long formula, but seems to work:
=RIGHT(A1,MATCH(TRUE,ISNUMBER(1*MID(A1,LEN(A1)-MATCH(FALSE,ISNUMBER(1*MID(A1,LEN(A1)-{0,1,2,3,4,5,6,7,8},1)),0)-{0,1,2,3,4,5,6,7,8},1)),0)+MATCH(FALSE,ISNUMBER(1*MID(A1,LEN(A1)-{0,1,2,3,4,5,6,7,8},1)),0)-1)
Related
Good day All,
I would like some input on the following.
I have a function that split parts of a string and converts single digits to double digits.
The function uses '.' as a delimiter then joins the parts again by a different delimiter.
When I call the function I simply add "=Outliner(C3)" to the cell I want the formatting to happen.
Everything works beautiful. If I put 14.1.1.1 the function returns 14-01-01-01 which is exactly what I need it to do.
Now a new scenario have come up where I am getting the number as 14.1.1-1 and now the function returns 14-01-01-43831 which is not what I want. I still want to have 14-01-01-01.
Is there an easy fix to the function. I have added the code below.
Function Outliner(S As String) As String
Dim x As Long, Parts() As String
Parts = Split(S, ".")
For x = 0 To UBound(Parts)
Parts(x) = Format(Parts(x), "00")
Next
Outliner = Join(Parts, "-")
End Function
Replace
Parts = Split(S, ".")
with
Parts = Split(Replace(S, "-", "."), ".")
I have the following Text sample:
Ins-Si_079_GM_SOC_US_VI SI_SOC_FY1920_US_FY19/20_A2554_Si Resp_2_May
I want to get the number 079, So what I need is the first instance of digits of length 3. There are certain times the 3 digits are at the end, but they usually found with the first 2 underscores. I only want the digits with length three (079) and not 19, 1920, or 2554 which are different lengths.
Sometimes it can look like this with no underscore:
1920 O-B CLI 353 Tar Traf
Or like this with the 3 digit number at the end:
Ins-Si_GM_SOC_US_VI SI_SOC_FY1920_US_FY19/20_A2554_Si Resp_2_079
There are also times where what I need is 2 digits but when it's 2 digits its always at the end like this:
FY1920-Or-OLV-B-45
How would I get what I need in all cases?
You can split the listed items and check for 3 digits via Like:
Function Get3Digits(s As String) As String
Dim tmp, elem
tmp = Split(Replace(Replace(s, "-", " "), "_", " "), " ")
For Each elem In tmp
If elem Like "###" Then Get3Digits = elem: Exit Function
Next
If Get3Digits = vbNullString Then Get3Digits = IIf(Right(s, 2) Like "##", Right(s, 2), "")
End Function
Edited due to comment:
I would execute a 2 digit search when there are no 3 didget numbers before the end part and the last 2 digits are 2. if 3 digits are fount at end then get 3 but if not then get 2. there are times when last is a number but only one number. I would only want to get last if there are 2 or 3 numbers. The - would not be relevant to the 2 digets. if nothing is found that is desired then would return " ".
If VBA is not a must you could try:
=TEXT(INDEX(FILTERXML("<t><s>"&SUBSTITUTE(SUBSTITUTE(SUBSTITUTE(A1,"_"," "),"-"," ")," ","</s><s>")&"</s></t>","//s[.*0=0][string-length()=3 or (position()=last() and string-length()=2)]"),1),"000")
It worked for your sample data.
Edit: Some explaination.
SUBSTITUTE(SUBSTITUTE(SUBSTITUTE(A1,"_"," "),"-"," ")," ","</s><s>") - The key part to transform all three potential delimiters (hyphen, underscore and space) to valid XML node end- and startconstruct.
The above concatenated using ampersand into a valid XML construct (adding a parent node <t>).
FILTERXML can be used to now 'split' the string into an array.
//s[.*0=0][string-length()=3 or last() and string-length()=2] - The 2nd parameter of FILTERXML which should be valid XPATH syntax. It reads:
//s 'Select all <s> nodes with
following conditions:
[.*0=0] 'Check if an <s> node times zero
returns zero (to check if a node
is numeric. '
[string-length()=3 or (position()=last() and string-length()=2)] 'Check if a node is 3 characters
long OR if it's the last node and
only 2 characters long.
INDEX(.....,1) - I mentioned in the comments that usually this is not needed, but since ExcelO365 might spill the returned array, we may as well implemented to prevent spilling errors for those who use the newest Excel version. Now we just retrieving the very first element of whatever array FILTERXML returns.
TEXT(....,"000") - Excel will try delete leading zeros of a numeric value so we use TEXT() to turn it into a string value of three digits.
Now, if no element can be found, this will return an error however a simple IFERROR could fix this.
Try this function, please:
Function ExtractThreeDigitsNumber(x As String) As String
Dim El As Variant, arr As Variant, strFound As String
If InStr(x, "_") > 0 Then
arr = Split(x, "_")
Elseif InStr(x, "-") > 0 Then
arr = Split(x, "-")
Else
arr = Split(x, " ")
End If
For Each El In arr
If IsNumeric(El) And Len(El) = 3 Then strFound = El: Exit For
Next
If strFound = "" Then
If IsNumeric(Right(x, 2)) Then ExtractThreeDigitsNumber = Right(x, 2)
Else
ExtractThreeDigitsNumber = strFound
End If
End Function
It can be called in this way:
Sub testExtractThreDig()
Dim x As String
x = "Ins-Si_079_GM_SOC_US_VI SI_SOC_FY1920_US_FY19/20_A2554_Si Resp_2_May"
Debug.Print ExtractThreeDigitsNumber(x)
End Sub
I have two string arrays and I want to find where each string from the first array is in the second array, so i tried this:
for i = 1:length(array1);
cmp(i) = strfind(array2,array1(i,:));
end
This doesn't seem to work and I get an error: "must be one row".
Just for the sake of completeness, an array of strings is nothing but a char matrix. This can be quite restrictive because all of your strings must have the same number of elements. And that's what #neerad29 solution is all about.
However, instead of an array of strings you might want to consider a cell array of strings, in which every string can be arbitrarily long. I will report the very same #neerad29 solution, but with cell arrays. The code will also look a little bit smarter:
a = {'abcd'; 'efgh'; 'ijkl'};
b = {'efgh'; 'abcd'; 'ijkl'};
pos=[];
for i=1:size(a,1)
AreStringFound=cellfun(#(x) strcmp(x,a(i,:)),b);
pos=[pos find(AreStringFound)];
end
But some additional words might be needed:
pos will contain the indices, 2 1 3 in our case, just like #neerad29 's solution
cellfun() is a function which applies a given function, the strcmp() in our case, to every cell of a given cell array. x will be the generic cell from array b which will be compared with a(i,:)
the cellfun() returns a boolean array (AreStringFound) with true in position j if a(i,:) is found in the j-th cell of b and the find() will indeed return the value of j, our proper index. This code is more robust and works also if a given string is found in more than one position in b.
strfind won't work, because it is used to find a string within another string, not within an array of strings. So, how about this:
a = ['abcd'; 'efgh'; 'ijkl'];
b = ['efgh'; 'abcd'; 'ijkl'];
cmp = zeros(1, size(a, 1));
for i = 1:size(a, 1)
for j = 1:size(b, 1)
if strcmp(a(i, :), b(j, :))
cmp(i) = j;
break;
end
end
end
cmp =
2 1 3
I wanted to perform a XOR calculation of two Binary numbers for example: on Sheet 1
Range A1 = 10101010
Range A2 = 11100010
Now I need to perform XOR of A1, A2 result in A3. I tried different formula's two perform XOR calculations like: A1^A2, (BITXOR (A1, A2)) but unfortunately it didn't worked I think because I am using excel 2007 "XOR" doesn't support.
I'm expecting a result of 1001000.
First, you should note that Excel pre-Excel2013 has no bitwise operators or functions built-in (Excel's OR() function is logical even if the operands are numeric). Excel 2013 finally adds this glaringly missing functionality.
Using VBA
The simplest way is to create a User Defined Function that does it. Formulae can work if you are prepared for either a decimal output, or helper columns, or a very repetitive Concatenate formula but VBA gets around these limitations - I recommend it if you are able to have code in the workbook.
Decimal Input, Decimal Output
The below examples just expose the built-in bitwise operators to use as functions in Excel formulae. I assume an integral type, although you could change it to accept decimals etc.
You can convert your string binary numbers (e.g. "1010") to decimals (10, for the previous example) using the BIN2DEC() function built-in to Excel, although this only handles 9 bits + sign bit, alternatively you can use an array formula to convert it for you (see my section on "Using Formulas" below).
Public Function BITWISE_OR(operand1, operand2)
BITWISE_OR = CLng(operand1) Or CLng(operand2)
End Function
Public Function BITWISE_AND(operand1, operand2)
BITWISE_AND = CLng(operand1) And CLng(operand2)
End Function
Public Function BITWISE_XOR(operand1, operand2)
BITWISE_XOR = CLng(operand1) Xor CLng(operand2)
End Function
Converting the numeric results back to binary strings is pretty annoying with formulas - if you need more than the range covered by DEC2BIN() (a paltry -512 to +511) function built in to Excel then I would suggest either using VBA (see below), or building up your binary string bit by bit using columns or rows (see my Using Formulas section below).
Binary string input, Binary string output
The below essentially iterates through a string setting each bit in turn based on the corresponding bits in the input strings. It performs the bit changes on the string in-place using Mid$ statement. Bit strings can be arbitrary length.
The below looks complicated but really it is the same basic stuff repeated 3 times for each of And, Or and XOr.
'str1, str2: the two bit strings. They can be different lengths.
'significantDigitsAreLeft: optional parameter to dictate how different length strings should be padded. Default = True.
Public Function Bitstr_AND(str1 As String, str2 As String, Optional significantDigitsAreLeft As Boolean = True)
Dim maxLen As Long, resStr As String, i As Long
If Len(str1) > Len(str2) Then maxLen = Len(str1) Else maxLen = Len(str2) 'get max length of the two strings
str1 = getPaddedString(str1, maxLen, significantDigitsAreLeft) 'pad left or right to the desired length
str2 = getPaddedString(str2, maxLen, significantDigitsAreLeft) 'pad left or right to the desired length
resStr = String$(maxLen, "0") 'prepare the result string into memory (Mid$ can operate without creating a new string, for performance)
For i = 1 To maxLen
If Mid$(str1, i, 1) = "1" And Mid$(str2, i, 1) = "1" Then
Mid$(resStr, i, 1) = "1" 'in-place overwrite of the existing "0" with "1"
End If
Next i
Bitstr_AND = resStr
End Function
'For explanatory comments, see Bitstr_AND
Public Function Bitstr_OR(str1 As String, str2 As String, Optional significantDigitsAreLeft As Boolean = True)
Dim maxLen As Long
Dim resStr As String
Dim i As Long
If Len(str1) > Len(str2) Then maxLen = Len(str1) Else maxLen = Len(str2)
str1 = getPaddedString(str1, maxLen, significantDigitsAreLeft)
str2 = getPaddedString(str2, maxLen, significantDigitsAreLeft)
resStr = String$(maxLen, "0")
For i = 1 To maxLen
If Mid$(str1, i, 1) = "1" Or Mid$(str2, i, 1) = "1" Then
Mid$(resStr, i, 1) = "1"
End If
Next i
Bitstr_OR = resStr
End Function
'For explanatory comments, see Bitstr_AND
Public Function Bitstr_XOR(str1 As String, str2 As String, Optional significantDigitsAreLeft As Boolean = True)
Dim maxLen As Long
Dim resStr As String
Dim i As Long
If Len(str1) > Len(str2) Then maxLen = Len(str1) Else maxLen = Len(str2)
str1 = getPaddedString(str1, maxLen, significantDigitsAreLeft)
str2 = getPaddedString(str2, maxLen, significantDigitsAreLeft)
resStr = String$(maxLen, "0")
For i = 1 To maxLen
If Mid$(str1, i, 1) = "1" Then
If Not Mid$(str2, i, 1) = "1" Then
Mid$(resStr, i, 1) = "1"
End If
ElseIf Mid$(str2, i, 1) = "1" Then 'Save an If check by assuming input string contains only "0" or "1"
Mid$(resStr, i, 1) = "1"
End If
Next i
Bitstr_XOR = resStr
End Function
'Helper to pad string
Private Function getPaddedString(str As String, length As Long, padLeft As Boolean) As String
If Len(str) < length Then
If padLeft Then
getPaddedString = String$(length - Len(str), "0") & str
Else
getPaddedString = str & String$(length - Len(str), "0")
End If
Else
getPaddedString = str
End If
End Function
Using Formulas
You can do an XOR operation using Text functions or Sumproduct. This may be more appropriate if you do not want to use VBA but formulas are painful to ensure they covers all situations, like negatives or different length binary strings. I refer you to the superb blog post http://www.excelhero.com/blog/2010/01/5-and-3-is-1.html for examples using Sumproduct, and http://chandoo.org/wp/2011/07/29/bitwise-operations-in-excel/ for examples using Text functions.
I cooked up my own formulae that handles certain cases and I explain them below to guide you.
Binary string Input, Decimal Output
In the below, A2 and B2 refer to the two binary numbers in up to 32-bits string form. The strings can be variable length, as the formula will pad with 0's to the necessary length. It should be obvious how to increase it to more bits. They must be entered using Ctrl+Shift+Enter.
The most significant bit is on the left. To make it least significant bit on the left, you can remove the little subtraction in the powers of 2 part, and make it pad to the right.
Bitwise And:
=SUM((((MID(REPT("0",32-LEN($A$2))&$A$2,ROW($1:$32),1)="1")+(MID(REPT("0",32-LEN($B$2))&$B$2,ROW($1:$32),1)="1"))=2)*(2^(32-ROW($1:$32))))
Bitwise Or:
=SUM((((MID(REPT("0",32-LEN($A$2))&$A$2,ROW($1:$32),1)="1")+(MID(REPT("0",32-LEN($B$2))&$B$2,ROW($1:$32),1)="1"))>0)*(2^(32-ROW($1:$32))))
Bitwise Xor:
=SUM((((MID(REPT("0",32-LEN($A$2))&$A$2,ROW($1:$32),1)="1")+(MID(REPT("0",32-LEN($B$2))&$B$2,ROW($1:$32),1)="1"))=1)*(2^(32-ROW($1:$32))))
Binary string input, Binary string Output
A single cell solution would be arduous because there is no array concatenation formula in Excel. You could do it using the CONCATENATE function glueing together each bits, with each bit being the result of an If comparing each binary string returning 1 or 0 as appropriate. As I said, though easy (just build it up like =IF(Mid(A1,1,1) = "1",...), this would be boring so I personally won't do it for you ;)
Alternatively, you could do it more simply using columns or rows to build up the string, like:
If A1 and B1 have your binary strings, then in C1 put (for AND, or for OR change the =2 at the end to >0 and for XOR change it to =1):
=IF((MID($A1,1,1)="1")+(MID($B1,1,1)="1"))=2,"1","0")
Then in D1 put:
=C1 & IF((MID($A1,COLUMN()-COLUMN($C1),1)="1")+(MID($B1,COLUMN()-COLUMN($C1),1)="1"))=2,"1","0")
Then drag this across as many columns as bits
I have this formula in a table which basically collects data from two columns and combines them. Now, I'm looking to combine this formula with a REPLACE formula that basically takes these characters æ,ø,å and replaces them with a,o,a.
Here's the formula:
=LOWER(LEFT(tableFaste[[#This Row];[Fornavn:]])&tableFaste[[#This Row];[Etternavn:]])
Sorry, don't know of a Formula way to remove any of a list of characters from a string. You might have to revert to vba for this. Here's a user defined function to do it. Your formula will become
=DeleteChars([#UserName],{"æ","ø","å";"a","o","a"})
To replace the characters use {"æ","ø","å";"a","o","a"} where the list up to the ; is the old characters, after the ; the new. You can make the list as long as you need, just make sure the lists are the same length.
To Delete the characters replace use {"æ","ø","å"} an array list of characters you want to remove
UDF code:
Function DeleteChars(r1 As Range, ParamArray c() As Variant) As Variant
Dim i As Long
Dim s As String
s = r1
If UBound(c(0), 1) = 1 Then
For i = LBound(c(0), 2) To UBound(c(0), 2)
s = Replace(s, c(0)(1, i), "")
Next
Else
For i = LBound(c(0), 2) To UBound(c(0), 2)
s = Replace(s, c(0)(1, i), c(0)(2, i))
Next
End If
DeleteChars = s
End Function
You can use SUBSTITUTE
=SUBSTITUTE(SUBSTITUTE(SUBSTITUTE(LOWER(LEFT(tableFaste[[#This Row];[Fornavn:]])&tableFaste[[#This Row];[Etternavn:]]),"æ","a"),"ø","o"),"å","a")