I am trying to figure out how to split a column that contains alpha numeric characters. For example, I have in Cell A1, ABC1234567,
Cell A2 has AB12345678, Cell A3 has ABCDE12345 and Cell A4 has ABC12 34 567. Is there any way of turning column A into 2 separate columns, one containing all the alpha characters and another containing all the numeric characters?
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Splitting alpha numeric cells (Microsoft 97)
Home » Forums » AskWoody support » Productivity software by function » MS Excel and spreadsheet help » Splitting alpha numeric cells (Microsoft 97)
- This topic has 19 replies, 7 voices, and was last updated 23 years, 4 months ago.
AuthorTopicWSMike Chez
AskWoody LoungerDecember 3, 2001 at 5:18 pm #363677Viewing 5 reply threadsAuthorReplies-
WSWassim
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WSSammyB
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WSWassim
AskWoody LoungerDecember 3, 2001 at 9:00 pm #555792Sammy
NO one can compete with BobU on these kind of formulas. Bob is one of a kind
…
But to be honest I did have his formula as a basis to start from. Maybe something that would test if IsNumber. But then how do you split the cell after you find where to split it.This has to be two column-formula.
Wassim
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WSJohnBF
AskWoody LoungerDecember 3, 2001 at 10:23 pm #555811You guys need to cheat! King Bob Umlas’ answer is right here (two tweaks courtesy of me). This part extracts the numbers, must be array-entered:
{=1*MID(SUBSTITUTE(A1,” “,””),MATCH(FALSE,ISERROR(1*MID(A1,ROW($1:$10),1)),0),255)}
(So who cares if I don’t understand it?)
And, assuming the source data is in Column A, and the above is in column B, extarcting the letters is easy:
=LEFT(A1,LEN(SUBSTITUTE(A1,” “,””))-LEN(B1))
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WSSammyB
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WSJohnBF
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WSWassim
AskWoody LoungerDecember 4, 2001 at 2:07 pm #555887John
I just hope Mike is paying attention to all what we are offering.
BUT I guess you need to go back to the drawing board because your formula needs some work.
If the Number are followed by Text then (1*MID(A1,ROW($1:$10),1)) gets to be Number #Value.
But all in all, its impressive.
Wassim
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WSJohnBF
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WSSammyB
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H. Legare Coleman
AskWoody PlusDecember 3, 2001 at 9:14 pm #555795The following VBA code will split the values in Column A on Sheet1 like you want into columns B and C.
Public Sub SplitA() Dim I As Long, J As Long I = 0 With Worksheets("Sheet1").Range("A1") While .Offset(I, 0).Value "" For J = 1 To Len(.Offset(I, 0).Value) If IsNumeric(Mid(.Offset(I, 0).Value, J, 1)) Then Exit For End If Next J .Offset(I, 1).Value = Left(.Offset(I, 0), J - 1) .Offset(I, 2).Value = Right(.Offset(I, 0), Len(.Offset(I, 0)) - J + 1) I = I + 1 Wend End With End Sub
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WSWassim
AskWoody Lounger -
WSMike Chez
AskWoody LoungerDecember 4, 2001 at 11:17 pm #555977Thank you all for all of your responses. It’s 6:10PM on 12/4. Due to other problems that came up at woork, this is the first time that I had a chance to do anything with your suggestions.
LegareColeman: I just typed your code in and I got a Run time error ’13’: Type Mismatch relating to the line containing
While.Offset(I,0).Value””. I don’t know what I did wrong or could be wrong. -
H. Legare Coleman
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WSMike Chez
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H. Legare Coleman
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WSAndrew Cronnolly
AskWoody LoungerDecember 5, 2001 at 12:57 pm #556159Just anther option for you in the form of a function. The following function will return the left part of a string from th estart up to the last text character. If all your entries start with Text and then switch to numbers it should work fine.
Function ExtractText(rng As Range) Dim i As Long, strTemp As String strTemp = rng.Value For i = 1 To Len(strTemp) If Not IsNumeric(Mid(strTemp, i, 1)) Then ExtractText = Left(strTemp, i) End If Next End Function
If A1 contains ABC123,
then in B1 =ExtractText(A1) will return ABC,
and in C1 =RIGHT(A1,(LEN(A1)-LEN(B1))) will return 123.
Just one more option for you.
Andrew C
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WSSammyB
AskWoody LoungerDecember 5, 2001 at 1:12 pm #556167Mike, Legare’s code works for me. When you want to use anyone’s post, select the text and copy it to Word, then select it again in Word and copy it to VBA.
Since you seem to prefer a VBA solution, here is a general-pupose function, STRIP. It strips off whatever you don’t want, letters, numbers, or punctuation. So, in your case =STRIP(A1,”Numbers”) would give “ABC”. HTH –Sam
Option Explicit Public Function STRIP(TEXT As Variant, OP As Variant) ' TEXT is any string ' OP is what to strip: NUMBERS, LETTERS, or PUNCTUATION ' (only the first letter is needed: N, L, or P) Dim i As Integer Dim letter As String Dim s As String s = "" For i = 1 To Len(TEXT) letter = Mid(TEXT, i, 1) If IsNumeric(letter) Then If Left(UCase(OP), 1) = "L" Then s = s & letter End If Else If Left(UCase(OP), 1) = "N" Then s = s & letter End If End If If Left(UCase(OP), 1) = "P" Then If IsNumeric(letter) Or _ (Asc(UCase(letter)) >= Asc("A") And Asc(UCase(letter)) <= Asc("Z")) Then s = s & letter End If End If Next i STRIP = s End Function
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WSMike Chez
AskWoody Lounger
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WSAladin Akyurek
AskWoody LoungerJanuary 13, 2002 at 9:51 am #563461Given the structure of your examples,
{“ABC1234567”;
“AB12345678”;
“ABC12 34 567″}which, say, A1:A3 houses,
in B1 enter: =SUBSTITUTE(A1,RIGHT(A1,SUMPRODUCT((LEN(A1)-LEN(SUBSTITUTE(A1,{” “,0,1,2,3,4,5,6,7,8,9},””))))),””)
in C1 enter: =SUBSTITUTE(A1,B1,””) or, depending on what you’d prefer, =SUBSTITUTE(SUBSTITUTE(A1,B1,””),” “,””)+0
Select B1:C1 and copy down.
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