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Wanted to check if CONCATENATE is the one to use (not sure if my excel has TEXTJOIN), and how to show just the text that has empty value in the cells.
For example in my attachment below, I want the intended result shown like in B2 and B3, where the texts shown with delimiter, when the values are false (empty).
If I were to use CONCATENATE like in Row 10 and Row 11, it's rather manual and it only capture "positive values" as in non-blank cells.
Purpose: To show pending tasks (empty/blank status cells)
Use MID with CONCATENATED IFS:
=MID(IF(C2="","/"&$C$1,"")&IF(D2="","/"&$D$1,"")&IF(E2="","/"&$E$1,"")&IF(F2="","/"&$F$1,"")&IF(GC2="","/"&$G$1,"")&IF(H2="","/"&$H$1,""),2,999)
I would use TEXJOIN and FILTER if you have the newest version of Excel.
For example: =TEXTJOIN("/",1,FILTER($E$2:$I$2, ISBLANK(E3:I3)))
EDIT: For older versions, a temporary workaround is as follows:
make a temporary array the same size as your original dataframe where each value is determined by a formula such as =IF(ISBLANK(E3), E$2&"/","")
Use something like =LEFT(CONCAT(E15:J15), LEN(CONCAT(E15:J15))-1) to get the desired result (where E15:J15 is where I elected to store the first row of the temporary array created in step 1).
I am not sure of your Excel version, but I think this would work in older versions (formatted for readability - will work if you paste it directly into cell B2 and copy down):
=LEFT(CONCAT( INDEX( CHOOSE({1;2;3},$C$1:$H$1,{"/","/","/","/","/","/"},{"","","","","",""}),
INDEX( IF(ISBLANK(C2:H2),{1;2},{3;3}),
MOD(COLUMN(A1:INDEX(1:1,,12))-1,2)+1,
(COLUMN(A1:INDEX(1:1,,12))-1)/2+1 ),
(COLUMN(A1:INDEX(1:1,,12))-1)/2+1 ) ),
SUM(7*ISBLANK(C2:H2))-1 )
Notes
As this is an array formula, you may have to enter it with CTRL + SHIFT + ENTER with an older version of Excel.
The stat labels must all have a length of 6 characters as shown in your post. If not, then they must at least have the same length and the last line SUM(7*ISBLANK(C2:H2))-1 must be changed to replace the 7 with the string length + 1, e.g. a length of 9 would be SUM(10*ISBLANK(C2:H2))-1.
If they don't have the same length, the LEFT( can be removed along with the SUM(10*ISBLANK(C2:H2))-1) at the end. You will end up having a trailing / delimiter at the end. You could fix that for the case of stat F being the last part by changing {"/","/","/","/","/","/"} to {"/","/","/","/","/",""}, but the other cases would still have a trailing /. Another approach is much more complex, but the component SUM(10*ISBLANK(C2:H2))-1) could be shaped to identify what to cut off or maybe a helper column could be built - in any case, let's hope your situation is that the stat labels all have the same length.
The delimiter "/" can be changed, but must always be a single character. If not, then then last line must be changed to SUM( [label length + delimiter length] *ISBLANK(C2:H2))-1.
This formula is fixed to 6 stat columns. If you need for it to accommodate more, it is possible by extending the {"/","/","/","/","/","/"} and {"","","","","",""} (one element for each new column) and replacing every 12 with 2 times the number of columns. Also, obviously, the references $C$1:$H$1 and C1:H2 must be changed to read in your new columns.
Say I have columns
/670 - White | /650 - black | /680 - Red | /800 - Whitest
These have data in their rows. Basically, I want to SUM their values together if their headers contain my desired string.
For modularity's sake, I wanted to merely specify to sum /670, /650, and /680 without having to mention the rest of the header text.
So, something like =SUMIF(a1:c1; "/NUM & /NUM & /NUM"; a2:c2)
That doesn't work, and honestly I don't know what i should be looking for.
Additional stuff:
I'm trying to think of the answer myself, is it possible to mention the header text as condition for ifs? Like: if A2="/650 - Black" then proceed to sum the next header. Is this possible?
Possibility it would not involve VBA, a draggable formula would be preferable!
At this point, I may as well request a version which handles the complete header name rather than just a part of it as I believe it to be difficult for formula code alone.
Thanks for having a look!
Let me know if I need to elaborate.
EDIT: In regards to data samples, any positive number will do actually, damn shame stack overflow doesn't support table markdown. Anyway, for example then..:
+-------------+-------------+-------------+-------------+-------------+
| A | B | C | D | E |
+---+-------------+-------------+-------------+-------------+-------------+
| 1 |/650 - Black |/670 - White |/800 - White |/680 - Red |/650 - Black |
+---+-------------+-------------+-------------+-------------+-------------+
| 2 | 250 | 400 | 100 | 300 | 125 |
+---+-------------+-------------+-------------+-------------+-------------+
I should have clarified:
The number range for these headers would go from /100 - /9999 and no more than that.
EDIT:
Progress so far:
https://docs.google.com/spreadsheets/d/1GiJKFcPWzG5bDsNt93eG7WS_M5uuVk9cvkt2VGSbpxY/edit?usp=sharing
Formula:
=SUMPRODUCT((A2:D2*
(MID($A$1:$D$1,2,4)=IF(LEN($H$1)=4,$H$1&"",$H$1&" ")))+(A2:D2*
(MID($A$1:$D$1,2,4)=IF(LEN($I$1)=4,$I$1&"",$I$1&" ")))+(A2:D2*
(MID($A$1:$D$1,2,4)=IF(LEN($J$1)=4,$J$1&"",$J$1&" "))))
Apparently, each MID function is returning false with each F9 calculation.
EDIT EDIT:
Okay! I found my issue, it's the /being read when you ALSO mentioned that it wasn't required. Man, I should stop skimming!
Final Edit:
=SUMPRODUCT((RETURNSUM*
(MID(HEADER,2,4)=IF(LEN(Match5)=4,Match5&"",Match5&" ")))+(RETURNSUM*
(MID(HEADER,2,4)=IF(LEN(Match6)=4,Match6&"",Match6&" ")))+(RETURNSUM*
(MID(HEADER,2,4)=IF(LEN(Match7)=4,Match7&"",Match7&" ")))
The idea is that Header and RETURNSUM will become match criteria like the matches written above, that way it would be easier to punch new criterion into the search table. As of the moment, it doesn't support multiple rows/dragging.
I have knocked up a couple of formulas that will achieve what you are looking for. For ease I have made the search input require the number only as pressing / does not automatically type into the formula bar. I apologise for the length of the answer, I got a little carried away with the explanation.
I have set this up for 3 criteria located in J1, K1 and L1.
Here is the output I achieved:
Formula 1 - SUMPRODUCT():
=SUMPRODUCT((A4:G4*(MID($A$1:$G$1,2,4)=IF(LEN($J$1)=4,$J$1&"",$J$1&" ")))+(A4:G4*(MID($A$1:$G$1,2,4)=IF(LEN($K$1)=4,$K$1&"",$K$1&" ")))+(A4:G4*(MID($A$1:$G$1,2,4)=IF(LEN($L$1)=4,$L$1&"",$L$1&" "))))
Sumproduct(array1,[array2]) behaves as an array formula without needed to be entered as one. Array formulas break down ranges and calculate them cell by cell (in this example we are using single rows so the formula will assess columns seperately).
(A4:G4*(MID($A$1:$G$1,2,4)=IF(LEN($J$1)=4,$J$1&"",$J$1&" ")))
Essentially I have broken the Sumproduct() formula into 3 identical parts - 1 for each search condition. (A4:G4*: Now, as the formula behaves like an array, we will multiply each individual cell by either 1 or 0 and add the results together.
1 is produced when the next part of the formula is true and 0 for when it is false (default numeric values for TRUE/FALSE).
(MID($A$1:$G$1,2,4)=IF(LEN($J$1)=4,$J$1&"",$J$1&" "))
MID(text,start_num,num_chars) is being used here to assess the 4 digits after the "/" and see whether they match with the number in the 3 cells that we are searching from (in this case the first one: J1). Again, as SUMPRODUCT() works very much like an array formula, each cell in the range will be assessed individually.
I have then used the IF(logical_test,[value_if_true],[value_if_false]) to check the length of the number that I am searching. As we are searching for a 4 digit text string, if the number is 4 digits then add nothing ("") to force it to a text string and if it is not (as it will have to be 3 digits) add 1 space to the end (" ") again forcing it to become a text string.
The formula will then perform the calculation like so:
The MID() formula produces the array: {"650 ","670 ","800 ","680 ","977 ","9999","143 "}. This combined with the first search produces {TRUE,FALSE,FALSE,FALSE,FALSE,FALSE,FALSE} which when multiplied by A4:G4
(remember 0 for false and 1 for true) produces this array: {250,0,0,0,0,0,0} essentially pulling the desired result ready to be summed together.
Formula 2: =SUM(IF(Array)): [This formula does not work for 3 digit numbers as they will exist within the 4 digit numbers! I have included it for educational purposes only]
=SUM(IF(ISNUMBER(SEARCH($J$1,$A$1:$G$1)),A8:G8),IF(ISNUMBER(SEARCH($K$1,$A$1:$G$1)),A8:G8),IF(ISNUMBER(SEARCH($L$1,$A$1:$G$1)),A8:G8))
The formula will need to be entered as an array (once copy and pasted while still in the formula bar hit CTRL+SHIFT+ENTER)
This formula works in a similar way, SUM() will add together the array values produced where IF(ISNUMBER(SEARCH() columns match the result column.
SEARCH() will return a number when it finds the exact characters in a cell which represents it's position in number of characters. By using ISNUMBER() I am avoiding having to do the whole MID() and IF(LEN()=4,""," ") I used in the previous formula as TRUE/FALSE will be produced when a match is found regardless of it's position or cell formatting.
As previously mentioned, this poses a problem as 999 can be found within 9999 etc.
The resulting array for the first part is: {250,FALSE,FALSE,FALSE,FALSE,FALSE,FALSE} (if you would like to see the array you can highlight that part of the formula and calculate with F9 but be sure to highlight the exact brackets for that part of the formula).
I hope I have explained this well, feel free to ask any questions about stuff that you don't understand. It is good to see people keen to learn and not just fishing for a fast answer. I would be more than happy to help and explain in more depth.
I start this solution with the names in an array, you can read the header names into an array with not too much difficulty.
Sub test()
Dim myArray(1 To 4) As String
myArray(1) = "/670 - White"
myArray(2) = "/650 - black"
myArray(3) = "/680 - Red"
myArray(4) = "/800 - Whitest"
For Each ArrayValue In myArray
'Find position of last character
endposition = InStr(1, ArrayValue, " - ", vbTextCompare)
'Grab the number section from the string, based on starting and ending positions
stringvalue = Mid(ArrayValue, 2, endposition - 2)
'Convert to number
NumberValue = CLng(stringvalue)
'Add to total
Total = Total + NumberValue
Next ArrayValue
'Print total
Debug.Print Total
End Sub
This will print the answer to the debug window.
I have multiple items in a column that look like this:
1. (758,01) 1516,01€
2. (380,95) 761,90€
3. (480) 903,90€
4. (350,06) 700,06€
5. (344) 688,75€
6. (681,16) 1361,16€
And I wanted to know how can I do two things:
Extract the number between ( ) and the number next to ) without the € part so that the final result is for example: 1516,01 758,01
Thanks
There are a number of ways. You can use the MID formula to get get the numbers between (), likewise for the second half and just replace the € with a blank. This works if your data is in cell B2,
=MID(B2, FIND("(", B2)+1, FIND(")", B2)-2)
And
=SUBSTITUTE(RIGHT(B2, FIND(" ", B2)-1), "€", "")
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 /
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 /