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I am trying to create a list =5 alphanumeric characters.
They cannot contain 1, and i and there cannot be duplicates when dragging / copying the code down.
The characters that are allowed are:
023456789ABCDEFGHJKLMNOPQRSTUWVXYZ (Capital)
I have tried numerous of options but I can't seem to figure this one out.
Cheers
If your allowable character string is in cell A1 then the following formula will result in random codes that are each five characters in length:
=MID(A1,RANDBETWEEN(1,34),1) & MID(A1,RANDBETWEEN(1,34),1) & MID(A1,RANDBETWEEN(1,34),1) & MID(A1,RANDBETWEEN(1,34),1) & MID(A1,RANDBETWEEN(1,34),1)
But note that there is no guarantee that the codes will be unique.
As #ScottCraner pointed out... if you should happen to have Office 365, you can use this much shorter formula that takes advantage of two new functions only available in Excel 365:
=CONCAT(MID(A1,RANDARRAY(5,,1,34,TRUE),1))
But again, there is no guarantee that the resulting codes will be unique.
This formula will generate the codes in order
=SUBSTITUTE(SUBSTITUTE(BASE(K, 34,5),"1","Z"),"I","Y")
Here K can be 0, 1, 2, .... One way to generate the first ~1,048,576 K's is to use ROW()-1. You could get higher values of K by using something like K = 1048576*(COLUMN()-1) + ROW()-1.
The formula works by
(a) calling BASE(K, 34, 5) to get a 5-char long base-34 representation of K
(b) substituting Z for 1 since 1 is not a valid char
(c) substituting Y for I since I is not a valid char
Please take a look at the attached image. I have a long list of items and I've created a common keywords to search in that list. I'm using this formula:
=INDEX(A:A,MATCH((("*"&B2&"*")&("*"&C2&"*")&("*"&D2&"*")&("*"&E2&"*")&("*"&F2&"*")),A:A,0))
The problem that the search is going through the same sequence that I entered.
It gives error if the sequence of the words in the cell is different than the sequence in my formula which make sense.
Is there a way I can search for 3 or more words that are existing in any cell in any sequence?
I am open to using VBA if necessary.
My search results:
Here is the user defined function:
Public Function indexMX(rng As Range, pat1 As Range, pat2 As Range, pat3 As Range, pat4 As Range, pat5 As Range) As Variant
Dim r As Range, rngx As Range, s(1 To 5) As String, Kount As Long, j As Long
s(1) = pat1.Value
s(2) = pat2.Value
s(3) = pat3.Value
s(4) = pat4.Value
s(5) = pat5.Value
Set rngx = Intersect(rng, rng.Parent.UsedRange)
For Each r In rngx
v = r.Value
Kount = 0
For j = 1 To 5
If InStr(1, v, s(j)) > 0 Or s(j) = "" Then Kount = Kount + 1
Next j
If Kount = 5 Then
indexMX = v
Exit Function
End If
Next r
indexMX = "no luck"
End Function
Here is an example of its usage:
As you see, we give the UDF() the address of the column and the addresses of the five keywords and the UDF() finds the first item containing all five words.
If a keyword is blank, it is not used. (so if you want to search for only two keywords, leave the other three blank). If no matches are found the phrase no luck is returned.
User Defined Functions (UDFs) are very easy to install and use:
ALT-F11 brings up the VBE window
ALT-I
ALT-M opens a fresh module
paste the stuff in and close the VBE window
If you save the workbook, the UDF will be saved with it.
If you are using a version of Excel later then 2003, you must save
the file as .xlsm rather than .xlsx
To remove the UDF:
bring up the VBE window as above
clear the code out
close the VBE window
To use the UDF from Excel:
=myfunction(A1)
To learn more about macros in general, see:
http://www.mvps.org/dmcritchie/excel/getstarted.htm
and
http://msdn.microsoft.com/en-us/library/ee814735(v=office.14).aspx
and for specifics on UDFs, see:
http://www.cpearson.com/excel/WritingFunctionsInVBA.aspx
Macros must be enabled for this to work!
EDIT#1:
to remove case sensitivity, replace:
If InStr(1, v, s(j)) > 0 Or s(j) = "" Then Kount = Kount + 1
with:
If InStr(1, LCase(v), LCase(s(j))) > 0 Or s(j) = "" Then Kount = Kount + 1
Yes, it is possible for a single cell to return three matching words from a different cell. The answer in this example uses a formula to return 6 matches. VBA and special array functions are not used.
This is the formula:
=SUBSTITUTE(SUBSTITUTE(SUBSTITUTE(SUBSTITUTE(SUBSTITUTE(SUBSTITUTE(SUBSTITUTE(SUMPRODUCT((IFERROR(SEARCH(FIRST,TARGET),0)>0)*100000+(IFERROR(SEARCH(SECOND,TARGET),0)>0)*20000+(IFERROR(SEARCH(THIRD,TARGET),0)>0)*3000+(IFERROR(SEARCH(FOURTH,TARGET),0)>0)*400+(IFERROR(SEARCH(FIFTH,TARGET),0)>0)*50+(IFERROR(SEARCH(SIXTH,TARGET),0)>0)*6),"1",FIRST&DELIM),"2",SECOND&DELIM),"3",THIRD&DELIM),"4",FOURTH&DELIM),"5",FIFTH&DELIM),"6",SIXTH&DELIM),"0","")
Did you notice the named ranges? FIRST, SECOND, THIRD, etc are individual cells and each one holds a word. We are trying to find those words inside TARGET. If we find the words, then we will write them in this cell holding this formula and each word will be separated by DELIM
The ranges are optional. In the picture below, you'll see cell A2 contains the word "named". This is the first of the six words we're tying to find and it can be expressed as FIRST = "A2" = "named" Inside the formula you'll see that FIRST appears twice. You could replace it with "named" and the cell A2 would become blank but the functionality of the formula would not change.
Even TARGET is optional. It could be written as E1 or typed out word for word.
I don't know why anyone would do that... but it is possible.
DELIM is at cell B2, it is a double space
Now to explain how it works
SEARCH(search for what?, search where?) This is responsible for determining if a match exists or not. If you understand what the named ranges then you've already figured out the syntax is. The location of first letter that of match in TARGET is returned. In this formula it is always 1 If it is not found then the number is 0
IFERROR(value,value) tries to perform the operation. If successful then the result is displayed. If there's an error the second result is displayed. Every IFERROR in this formula is practically the same: IFERROR(SEARCH(FIRST,TARGET),0) It searches inside TARGET trying to find the FIRST word. Result if found is 1 and if not found is 0
It gets a little more complicated from here so lets recap. We're calling SEARCH 6 times. Once for each word we want to find and we're always looking in TARGET. Result will be a 1 if match is found or a 0 if not. Ironically, us humans can put it together and see the match but the formula can't determine which words have been matched without more information
SUMPRODUCT takes the sum (addition) of the product (multiplication) of two or more arrays.
multiply two arrays to get the product
a, b, c * e, f, g = ae, bf, cg
takethe sum of the product to get the SUMPRODUCT
ae + bf + cg`
This is easiest when thought of a price and quantity. If one array is the price of a group of items and the other is quantity of the same group of items, then multiplying the two arrays will create a new array where each element is the cost to buy all items of that type in the group the total, and adding all those numbers give you the total cost you'd pay for all of the items
Here we multiply two arrays:
Qty Price
12.0 0.3
70.0 0.1
20.0 0.4
Multiply them to get the product:
Qty Price Total
12.0 0.3 3.8
70.0 0.1 7.0
20.0 0.4 8.0
Take the sum of the product:
Qty Price Total
12.0 0.3 3.8
70.0 0.1 7.0
20.0 0.4 8.0
18.8 SUMPRODUCT
Lets look at part of the formula:
SUMPRODUCT((IFERROR(SEARCH(FIRST,TARGET),0)>0)*100000+IFERROR(SEARCH(SECOND,TARGET),0)>0)*20000+...
It is easy to see this segment is looking for two words. We know SUMPRODUCT want's to multiply and add arrays. If you're thinking (IFERROR(SEARCH(FIRST,TARGET),0)>0) is an array, you'd be right! It's not an array in the technical sense of the word, but it does evaluate into a single value which can thought of as a 1x1 array, or, a cell. The sharp eyed and quick witted, may have noticed there is something on this array we have mentioned. It's the inequality at the end! Many of you know that you can take numerical values and turn them into boolean by testing them with an inequality. So lets evaluate.... SEARCH for FIRST inside TARGET = 1 because FIRST = "named" which is inside TARGET waaaaaaay in the back and because it wasn't an error we get to keep the 1. Next we do the inequality 1 > 0 = TRUE One is greater than zero and evaluates to TRUE
This is what we have right now
SUMPRODUCT((TRUE*100000+IFERROR(SEARCH(SECOND,TARGET),0)>0)*20000+....
Can you identify the arrays now? We know TRUE is an array, a 1x1. You know the IFERROR bit all the way to the inequality is also an array. Lets evaluate that IFERROR .... Mathematically we should still be working from left to right but trust me, we're ok if we let it slide this once.
IFERROR(SEARCH(SECOND,TARGET),0)>0) SECOND = "array" = 1 = TRUE
Did you follow my short hand? It's ok if you didn't just back up and practice on FIRST until you understand.
Plugging in the value gives us something like this
SUMPRODUCT(TRUE*100000+TRUE*20000+...
SUMPRODUCT is the SUM(addition) of the PRODUCT(multiplication)
So we're adding the stuff we multiply
SUMPRODUCT = (TRUE * 100000) + (TRUE * 20000)
Remember how easy it was to go from 1 > 0 to TRUE. We're "getting all up into that boolean TRUE that" equals 1 Here's a fun fact, -1 is also equal to TRUE. If you've ever seen a formula with a double negative in it like this STUFF(--(MORESTUFF( that's just some Excel wizard person making sure they get a +1 instead of a -1 ... ok, so lets get back on track and evalute
SUMPRODCUT = 1 * 100000 + 1 * 20000+....
SUMPRODCUT = 100000 + 20000+....
SUMPRODCUT = 120000+.....
I know you've been asking about those numbers. One hundred thousand? What's one hundred thousand for? I've been purposefully ignoring until it became convenient to talk about it. And now it's convenient. Go look at the whoooole formula and you'll find a pattern. Those numbers are in a decreasing sequence. Anyone who's ever done bitwise logic can see where this is going.I'm running short on time so I'll cut to the chase. Assume a hypothetical situation where every word was matched. You'd end up with
SUMPRODUCT = 100000 + 20000 + 3000 + 400 + 50 + 6
SUMPRODCUT = 123456
123456 are you pulling my leg? No, I am not. We're almost done so if you're still with me then you're gonna drive it home.
We have a large group of SUBSTITUTE teachers at the front of the line and we gotta get rid of them.
We also have this to contend with :"1",FIRST&DELIM),"2",SECOND&DELIM),"3",THIRD&DELIM),"4",FOURTH&DELIM),"5",FIFTH&DELIM),"6",SIXTH&DELIM),"0","")
Thankfully for us, they are part of the same problem. We worked our way from the middle out.
SUBSTITUTE(SUBSTITUTE(text, old text, new text)
SUBSTITUTE(SUBSTITUTE("123456","1", FIRST & DELIM),"2",SECOND & DELIM)....
Remember at top DELIM was pointing to a cell holding a double space. Each DELIM can be replaced with " " or any other delimiter you want.
SUBSTITUTE(SUBSTITUTE("123456","1", "named" & " "),"2",SECOND & DELIM)...
SUBSTITUTE("named 23456","2",SECOND & DELIM)...
SUBSTITUTE("named 23456","2","array"& " ")...
("named array 3456")... and so on.
Any questions?
Ok, class is dismissed!
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.
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 /