I've tried lots of searches for this but I'm still not coming up with anything that works.
I have a range of strings in Column A
Amend.Clause_1.1.AddMCQ
Amend.Clause_1.1.AddNo
Amend.Clause_1.1.AddRepeat
Amend.Clause_1.13.AddRepeat
Amend.Clause_1.13.AddTitle
Amend.Clause_1.13.AddUTQ
Amend.Clause_2.8.Heading_Edit
Amend.Clause_2.8.MCQ
Amend.Clause_2.8.Remove
Amend.Clause_4.26.AddUTQ
Amend.Clause_4.26.Heading_Edit
Amend.Clause_4.26.MCQ
Amend.Clause_5.15.AddMCQ
Amend.Clause_5.15.AddNo
Amend.Clause_5.15.AddRepeat
As you can see, the numbers always start in the same place, after the underscore "_" at position 13.
I need to extract the decimal numbers from these strings into a new column so I'm left with 1.1, 1.13, 1.14, 4.26 etc.
I've tried all sorts of combos of MID, LEFT, LEN, RIGHT but to no avail, trying to find the position of the last period.
Could anyone explain how to accomplish this? Ideally I'd like to do this without VBA.
Thanks
Here you are:
=VALUE(MID(A1,SEARCH("_",A1)+1,SEARCH(".",A1,SEARCH(".",A1,SEARCH("_",A1)+1)+1)-(SEARCH("_",A1)+1)))
Here's what inside =VALUE(MID(...)):
A1 - the whole string itself
SEARCH("_",A1)+1 - find the number starting position - right after "_".
SEARCH(".",A1,SEARCH(".",A1,SEARCH("_",A1)+1)+1)-(SEARCH("_",A1)+1) - find number length - position of second "." after first "." minus number starting position.
Try with three functions:
=MID(A1,14,FIND("#",SUBSTITUTE(A1,".","#",3))-14)
Try this - If the position of _ is not necessarily 13.
=MID(A1,FIND("_",A1,1)+1,FIND("¬¬",SUBSTITUTE(A1,".","¬¬",LEN(A1)-LEN(SUBSTITUTE(A1,".",""))))-FIND("_",A1,1)-1)
Or this if the _ is always 13
=MID(A1,14,FIND("¬¬",SUBSTITUTE(A1,".","¬¬",LEN(A1)-LEN(SUBSTITUTE(A1,".",""))))-14)
Use This:
=VALUE(TRIM(LEFT(SUBSTITUTE(RIGHT(A1;LEN(A1)-FIND("_";A1));".";REPT(" ";LEN(A1));2);LEN(A1))))
assuming value is in A1
Far from ideal, but with a shorter formula than the solutions offered so far:
=SUBSTITUTE(A1,".","_",3)
Catch is that formulae would then need to be converted to values, parsed with delimiter _ (being careful to ensure Column data format is Text) and surplus columns deleted.
When the string Amend.Clause_1.1.AddMCQ is in A1
=Find(".",A1,Find(".",A1)+1)
will give the position of the second decimal point, then you should be able to extract the decimal number.
The syntax is
FIND(find_text, within_text, [start_num])
Related
I am not sure where to begin with the formula as I have gotten myself so confused with everything. I have a cell the contains "PON " or "PON: " or "PON = " then the actual PON (Example: PON 123467) I want to formula to return 123467 in the cell.
Examples What I want returned
I have PON 123467 for shoes 123467
I have PON: 234567-AB for food 234567-AB
I have PON - 569874-Weird for accessories 569874-Weird
I have PON = DOG-564-987 for dog food DOG-564-987
I am currently using Excel 365
Filterxml() will give you best companion here in this case. Try-
=FILTERXML("<t><s>"&SUBSTITUTE(FILTERXML("<t><s>"&SUBSTITUTE(A1," for","</s><s>")&"</s></t>","//s[1]")," ","</s><s>")&"</s></t>","//s[last()]")
Using FILTERXML, and testing for a substring following PON, you can try:
=FILTERXML("<t><s>"&SUBSTITUTE(TRIM(A1)," ","</s><s>") & "</s></t>","//s[contains(.,'PON')]/following-sibling::*[string-length(.)>2][1]")
Note that FILTERXML solution will cause a PON that is solely numeric, but with a leading zero, to drop the leading zero. Unfortunately, the xPath implementation in that function does not include the string() function
If dropping the leading zero might be a problem, you can add a character to the node that will force the number to be seen as a string. In the modified formula below, I use the unicode zero-width space, but there are others you can use. Note that this will count as a character for the string=length function, so be sure to maintain the >2 parameter:
=FILTERXML("<t><s>"&SUBSTITUTE(TRIM(A1)," ","</s><s>"&UNICHAR(8203)) & "</s></t>","//s[contains(.,'PON')]/following-sibling::*[string-length(.)>2][1]")
Because of the variablity in your data, that sometimes there are extraneous space-separated substrings between PON and your desired extract, the xpath:
locates the substring PON
returns all subsequent siblings that have a string-length of more than two (adjust if necessary)
returns the first sibling that meets that criterion.
You might try this formula.
=TRIM(LEFT(MID(A2,FIND(#{1,2,3,4,5,6,7,8,9},A2),100),FIND(" ",MID(A2,FIND(#{1,2,3,4,5,6,7,8,9},A2),100))))
It extracts the text between the first number and the first space following that number. The size of that extract is limited to 100 characters.
So How to get substring from nth position up to the end of string?
Input at cell A1 Name: Thomas B.
Expected output: Thomas B.
I know some way to do it but I wonder if there are other elegant ways than them? (some kind of =RIGHT(A1, -6)....)
=MID(A1, 6, 999999) //999999 looks not so good
=MID(A1, 6, LEN(A1) - 5) //must calculate 2 times, first get len, then get substring, seems too much works?
REPLACE
As Dominique already wrote:
'Why don't you just replace the first six characters by an empty string?'
=REPLACE(A1,1,6,"")
I've done some time measuring, but the difference is less than a second at 50000 records (for LEFT, MID, REPLACE & SUSTITUTE). So I'm afraid ELEGANCE is all you're going to get.
A Small Study
I created this study due to the fact that when you say from the n-th character, your n-th character is 7 (your MID-s are wrong), but you want to remove the first n-1 (6) characters. So depending on how you formulate your question, you might have a different approach in RIGHT or MID, and you will remember REPLACE and SUBSTITUTE or you may not.
Small Study Formulas for A1 (*) and B1 (#, ?, *)
Get String From N-th Character to the End, e.g. 7
=RIGHT(A1,LEN(A1)-(B1-1))
=RIGHT(A1,LEN(A1)-B1+1)
=RIGHT(A1,LEN(A1)-6)
=MID(A1,B1,LEN(A1)-(B1-1))
=MID(A1,B1,LEN(A1)-B1+1)
=MID(A1,B1,LEN(A1))
=MID(A1,7,LEN(A1)-6)
=MID(A1,7,LEN(A1))
Remove N First Characters of a String, e.g. 6
=RIGHT(A1,LEN(A1)-B1)
=RIGHT(A1,LEN(A1)-6)
=MID(A1,B1+1,LEN(A1)-B1)
=MID(A1,B1+1,LEN(A1))
=MID(A1,7,LEN(A1)-6)
=MID(A1,7,LEN(A1))
Get String After a Character e.g. " "
=RIGHT(A1,LEN(A1)-(FIND(B1,A1)))
=RIGHT(A1,LEN(A1)-(FIND(" ",A1)))
=MID(A1,FIND(B1,A1)+1,LEN(A1)-FIND(B1,A1))
=MID(A1,FIND(B1,A1)+1,LEN(A1))
=MID(A1,FIND(" ",A1)+1,LEN(A1)-FIND(" ",A1))
=MID(A1,FIND(" ",A1)+1,LEN(A1))
Get String After a String e.g. ": "
=RIGHT(A1,LEN(A1)-(FIND(B1,A1)+LEN(B1))+1)
=RIGHT(A1,LEN(A1)-FIND(B1,A1)-LEN(B1)+1)
=RIGHT(A1,LEN(A1)-FIND(": ",A1)-LEN(": ")+1)
=MID(A1,FIND(B1,A1)+LEN(B1),LEN(A1)-(FIND(B1,A1)+LEN(B1))+1)
=MID(A1,FIND(B1,A1)+LEN(B1),LEN(A1)-FIND(B1,A1)-LEN(B1)+1)
=MID(A1,FIND(B1,A1)+LEN(B1),LEN(A1))
=MID(A1,FIND(": ",A1)+LEN(": "),LEN(A1)-FIND(": ",A1)-LEN(": ")+1)
=MID(A1,FIND(": ",A1)+LEN(": "),LEN(A1))
Back to Remove N First Characters of a String, e.g. 6
=SUBSTITUTE(A1,LEFT(A1,6),"",1)
=REPLACE(A1,1,6,"")
Well, both of your methods already work, but you could also use this one:
=RIGHT(A1,LEN(A1)-6)
(you nearly had this one in your own question)
or this one:
=TRIM(MID(A1,FIND(":",A1)+1,100))
(the FIND() function returns the numeric position of a search string, so is great for doing dynamic substrings)
Why don't you just replace the first six characters by an empty string?
=SUBSTITUTE(A1;LEFT(A1;6);"";1)
Another possibility is that you create a constant with the value 2^31-1 (=2147483647), which is the maximum signed integer value on 32-bit systems, and you give it a nice name, like MaxInt, then your first formula will be efficient and nice looking, too:
=MID(A1, 6, MaxInt)
You can add the Name with Ctrl+F3. If you are interested in fast calculations, giving it as 2147483647 rather than 2^31-1 may have some (very little) advantage.
I know how to add leading 0's, but I was wondering if it is posible to add them in a middle of a cell to = 12 characters. The first two characthers are always going to be letters. and then nummbers.
Example
AB45686 = AB0000045686
DS456 = DS0000000456
Parse the string with LEFT and MID, then use text to add the needed 0s:
=LEFT(A1,2) & TEXT(--MID(A1,3,LEN(A1)),"0000000000")
Here is the other way to get the result.
Is there an efficient way to identify the last character/string match in a string using base functions? I.e. not the last character/string of the string, but the position of a character/string's last occurrence in a string. Search and find both work left-to-right so I can't think how to apply without lengthy recursive algorithm. And this solution now seems obsolete.
I think I get what you mean. Let's say for example you want the right-most \ in the following string (which is stored in cell A1):
Drive:\Folder\SubFolder\Filename.ext
To get the position of the last \, you would use this formula:
=FIND("#",SUBSTITUTE(A1,"\","#",(LEN(A1)-LEN(SUBSTITUTE(A1,"\","")))/LEN("\")))
That tells us the right-most \ is at character 24. It does this by looking for "#" and substituting the very last "\" with an "#". It determines the last one by using
(len(string)-len(substitute(string, substring, "")))\len(substring)
In this scenario, the substring is simply "\" which has a length of 1, so you could leave off the division at the end and just use:
=FIND("#",SUBSTITUTE(A1,"\","#",LEN(A1)-LEN(SUBSTITUTE(A1,"\",""))))
Now we can use that to get the folder path:
=LEFT(A1,FIND("#",SUBSTITUTE(A1,"\","#",LEN(A1)-LEN(SUBSTITUTE(A1,"\","")))))
Here's the folder path without the trailing \
=LEFT(A1,FIND("#",SUBSTITUTE(A1,"\","#",LEN(A1)-LEN(SUBSTITUTE(A1,"\",""))))-1)
And to get just the filename:
=MID(A1,FIND("#",SUBSTITUTE(A1,"\","#",LEN(A1)-LEN(SUBSTITUTE(A1,"\",""))))+1,LEN(A1))
However, here is an alternate version of getting everything to the right of the last instance of a specific character. So using our same example, this would also return the file name:
=TRIM(RIGHT(SUBSTITUTE(A1,"\",REPT(" ",LEN(A1))),LEN(A1)))
How about creating a custom function and using that in your formula? VBA has a built-in function, InStrRev, that does exactly what you're looking for.
Put this in a new module:
Function RSearch(str As String, find As String)
RSearch = InStrRev(str, find)
End Function
And your function will look like this (assuming the original string is in B1):
=LEFT(B1,RSearch(B1,"\"))
New Answer | 31-3-2022:
With even newer functions come even shorter answers. At time of writing in BETA, but probably widely available in the near future, we can use TEXTBEFORE():
=LEN(TEXTBEFORE(A2,B2,-1))+1
The trick here is that the 3rd parameter tells the function to retrieve the last occurence of the substring we give in the 2nd parameter. At time of writing this function is still case-sensitive by default which could be handeld by the optional 4th parameter.
Original Answer | 17-6-2020:
With newer versions of excel come new functions and thus new methods. Though it's replicable in older versions (yet I have not seen it before), when one has Excel O365 one can use:
=MATCH(2,1/(MID(A1,SEQUENCE(LEN(A1)),1)="Y"))
This can also be used to retrieve the last position of (overlapping) substrings:
=MATCH(2,1/(MID(A1,SEQUENCE(LEN(A1)),2)="YY"))
| Value | Pattern | Formula | Position |
|--------|---------|------------------------------------------------|----------|
| XYYZ | Y | =MATCH(2,1/(MID(A2,SEQUENCE(LEN(A2)),1)="Y")) | 3 |
| XYYYZ | YY | =MATCH(2,1/(MID(A3,SEQUENCE(LEN(A3)),2)="YY")) | 3 |
| XYYYYZ | YY | =MATCH(2,1/(MID(A4,SEQUENCE(LEN(A4)),2)="YY")) | 4 |
Whilst this both allows us to no longer use an arbitrary replacement character and it allows overlapping patterns, the "downside" is the useage of an array.
Note: You can force the same behaviour in older Excel versions through either
=MATCH(2,1/(MID(A2,ROW(A1:INDEX(A:A,LEN(A2))),1)="Y"))
Entered through CtrlShiftEnter, or using an inline INDEX to get rid of implicit intersection:
=MATCH(2,INDEX(1/(MID(A2,ROW(A1:INDEX(A:A,LEN(A2))),1)="Y"),))
tigeravatar and Jean-François Corbett suggested to use this formula to generate the string right of the last occurrence of the "\" character
=TRIM(RIGHT(SUBSTITUTE(A1,"\",REPT(" ",LEN(A1))),LEN(A1)))
If the character used as separator is space, " ", then the formula has to be changed to:
=SUBSTITUTE(RIGHT(SUBSTITUTE(A1," ",REPT("{",LEN(A1))),LEN(A1)),"{","")
No need to mention, the "{" character can be replaced with any character that would not "normally" occur in the text to process.
Just came up with this solution, no VBA needed;
Find the last occurance of "_" in my example;
=IFERROR(FIND(CHAR(1);SUBSTITUTE(A1;"_";CHAR(1);LEN(A1)-LEN(SUBSTITUTE(A1;"_";"")));0)
Explained inside out;
SUBSTITUTE(A1;"_";"") => replace "_" by spaces
LEN( *above* ) => count the chars
LEN(A1)- *above* => indicates amount of chars replaced (= occurrences of "_")
SUBSTITUTE(A1;"_";CHAR(1); *above* ) => replace the Nth occurence of "_" by CHAR(1) (Nth = amount of chars replaced = the last one)
FIND(CHAR(1); *above* ) => Find the CHAR(1), being the last (replaced) occurance of "_" in our case
IFERROR( *above* ;"0") => in case no chars were found, return "0"
Hope this was helpful.
You could use this function I created to find the last instance of a string within a string.
Sure the accepted Excel formula works, but it's much too difficult to read and use. At some point you have to break out into smaller chunks so it's maintainable. My function below is readable, but that's irrelevant because you call it in a formula using named parameters. This makes using it simple.
Public Function FindLastCharOccurence(fromText As String, searchChar As String) As Integer
Dim lastOccur As Integer
lastOccur = -1
Dim i As Integer
i = 0
For i = Len(fromText) To 1 Step -1
If Mid(fromText, i, 1) = searchChar Then
lastOccur = i
Exit For
End If
Next i
FindLastCharOccurence = lastOccur
End Function
I use it like this:
=RIGHT(A2, LEN(A2) - FindLastCharOccurence(A2, "\"))
Considering a part of a Comment made by #SSilk my end goal has really been to get everything to the right of that last occurence an alternative approach with a very simple formula is to copy a column (say A) of strings and on the copy (say ColumnB) apply Find and Replace. For instance taking the example: Drive:\Folder\SubFolder\Filename.ext
This returns what remains (here Filename.ext) after the last instance of whatever character is chosen (here \) which is sometimes the objective anyway and facilitates finding the position of the last such character with a short formula such as:
=FIND(B1,A1)-1
I'm a little late to the party, but maybe this could help. The link in the question had a similar formula, but mine uses the IF() statement to get rid of errors.
If you're not afraid of Ctrl+Shift+Enter, you can do pretty well with an array formula.
String (in cell A1):
"one.two.three.four"
Formula:
{=MAX(IF(MID(A1,ROW($1:$99),1)=".",ROW($1:$99)))} use Ctrl+Shift+Enter
Result: 14
First,
ROW($1:$99)
returns an array of integers from 1 to 99: {1,2,3,4,...,98,99}.
Next,
MID(A1,ROW($1:$99),1)
returns an array of 1-length strings found in the target string, then returns blank strings after the length of the target string is reached: {"o","n","e",".",..."u","r","","",""...}
Next,
IF(MID(I16,ROW($1:$99),1)=".",ROW($1:$99))
compares each item in the array to the string "." and returns either the index of the character in the string or FALSE: {FALSE,FALSE,FALSE,4,FALSE,FALSE,FALSE,8,FALSE,FALSE,FALSE,FALSE,FALSE,14,FALSE,FALSE.....}
Last,
=MAX(IF(MID(I16,ROW($1:$99),1)=".",ROW($1:$99)))
returns the maximum value of the array: 14
Advantages of this formula is that it is short, relatively easy to understand, and doesn't require any unique characters.
Disadvantages are the required use of Ctrl+Shift+Enter and the limitation on string length. This can be worked around with a variation shown below, but that variation uses the OFFSET() function which is a volatile (read: slow) function.
Not sure what the speed of this formula is vs. others.
Variations:
=MAX((MID(A1,ROW(OFFSET($A$1,,,LEN(A1))),1)=".")*ROW(OFFSET($A$1,,,LEN(A1)))) works the same way, but you don't have to worry about the length of the string
=SMALL(IF(MID(A1,ROW($1:$99),1)=".",ROW($1:$99)),2) determines the 2nd occurrence of the match
=LARGE(IF(MID(A1,ROW($1:$99),1)=".",ROW($1:$99)),2) determines the 2nd-to-last occurrence of the match
=MAX(IF(MID(I16,ROW($1:$99),2)=".t",ROW($1:$99))) matches a 2-character string **Make sure you change the last argument of the MID() function to the number of characters in the string you wish to match!
In newer versions of Excel (2013 and up) flash fill might be a simple and quick solution see: Using Flash Fill in Excel
.
For a string in A1 and substring in B1, use:
=XMATCH(B1,MID(A1,SEQUENCE(LEN(A1)),LEN(B1)),,-1)
Working from inside out, MID(A1,SEQUENCE(LEN(A1)),LEN(B1)) splits string A1 into a dynamic array of substrings, each the length of B1. To find the position of the last occurrence of substring B1, we use XMATCH with its Search_mode argument set to -1.
A simple way to do that in VBA is:
YourText = "c:\excel\text.txt"
xString = Mid(YourText, 2 + Len(YourText) - InStr(StrReverse(YourText), "\" ))
Very late to the party, but A simple solution is using VBA to create a custom function.
Add the function to VBA in the WorkBook, Worksheet, or a VBA Module
Function LastSegment(S, C)
LastSegment = Right(S, Len(S) - InStrRev(S, C))
End Function
Then the cell formula
=lastsegment(B1,"/")
in a cell and the string to be searched in cell B1 will populate the cell with the text trailing the last "/" from cell B1.
No length limit, no obscure formulas.
Only downside I can think is the need for a macro-enabled workbook.
Any user VBA Function can be called this way to return a value to a cell formula, including as a parameter to a builtin Excel function.
If you are going to use the function heavily you'll want to check for the case when the character is not in the string, then string is blank, etc.
If you're only looking for the position of the last instance of character "~" then
=len(substitute(String,"~",""))+1
I'm sure there is version that will work with the last instance of a string but I have to get back to work.
Cell A1 = find/the/position/of/the last slash
simple way to do it is reverse the text and then find the first slash as normal. Now you can get the length of the full text minus this number.
Like so:
=LEN(A1)-FIND("/",REVERSETEXT(A1),1)+1
This returns 21, the position of the last /
I have a set of data that shown below on excel.
R/V(208,0,32) YR/V(255,156,0) Y/V(255,217,0)
R/S(184,28,16) YR/S(216,128,0) Y/S(209,171,0)
R/B(255,88,80) YR/B(255,168,40) Y/B(255,216,40)
And I want to separate the data in each cell look like this.
R/V 208 0 32
R/S 184 28 16
R/B 255 88 80
what is the function in excel that I can use for this case.
Thank you in advance.
kennytm doesn't provide an example so here's how you do substrings:
=MID(text, start_num, char_num)
Let's say cell A1 is Hello.
=MID(A1, 2, 3)
Would return
ell
Because it says to start at character 2, e, and to return 3 characters.
In Excel, the substring function is called MID function, and indexOf is called FIND for case-sensitive location and SEARCH function for non-case-sensitive location. For the first portion of your text parsing the LEFT function may also be useful.
See all the text functions here: Text Functions (reference).
Full worksheet function reference lists available at:
Excel functions (by category)
Excel functions (alphabetical)
Another way you can do this is by using the substitute function. Substitute "(", ")" and "," with spaces.
e.g.
=SUBSTITUTE(SUBSTITUTE(SUBSTITUTE(A1, "(", " "), ")", " "), ",", " ")
I believe we can start from basic to achieve desired result.
For example, I had a situation to extract data after "/". The given excel field had a value of 2rko6xyda14gdl7/VEERABABU%20MATCHA%20IN131621.jpg . I simply wanted to extract the text from "I5" cell after slash symbol. So firstly I want to find where "/" symbol is (FIND("/",I5). This gives me the position of "/". Then I should know the length of text, which i can get by LEN(I5).so total length minus the position of "/" . which is LEN(I5)-(FIND("/",I5)) . This will first find the "/" position and then get me the total text that needs to be extracted.
The RIGHT function is RIGHT(I5,12) will simply extract all the values of last 12 digits starting from right most character. So I will replace the above function "LEN(I5)-(FIND("/",I5))" for 12 number in the RIGHT function to get me dynamically the number of characters I need to extract in any given cell and my solution is presented as given below
The approach was
=RIGHT(I5,LEN(I5)-(FIND("/",I5))) will give me out as VEERABABU%20MATCHA%20IN131621.jpg . I think I am clear.
Update on 11/30/2022
With new excel functions, you can use the following in cell C1 for the input in A1:
=TEXTJOIN(" ",,TEXTSPLIT(A1,{"(",",",")"}))
Here is the output:
What about using Replace all?
Just replace All on bracket to space.
And comma to space. And I think you can achieve it.