How to obtain the n-th letter of the alphabet - excel

Is there a short formula to get the n-th letter of the alphabet?
For example, if I give parameter 5 to the function, I would get the letter e.


There is a function CHAR which gives a character with the specified code:
CHAR(96 + 5)
will yield your "e".
But there is no direct way to get a character of the alphabet.


An alternate, although not as short as the CHAR function, is the CHOOSE function
=CHOOSE(5,"a","b","c","d","e","f","g","h","I","j","k","l","m",
"n","o","p","q","r","s","t","u","v","w","x","y","z")
The index number '5' returns the fifth value in the list. The list could be an Excel range of data e.g. (A1:A26).
If the index number is outside the range, #VALUE! is returned


you could use an ascii function since every letter has a numeric value in ascii
Not sure what language your using... in T-SQL you can use an ASCII and CHAR functions:
PRINT CHAR(ASCII('A') + #i) -- where #i is your numeric value


There is also another simpler way: CHAR(CODE("A")+TRUNC(RAND()*26)).


This gives you the position of the letter in question (C3, for example) if it is capital or not.
=IF(AND(CODE(C3)>=65,CODE(C3)<=90),CODE(C3)-64,IF(AND(CODE(C3)>=97,CODE(C3)<=122),CODE(C3)-96,"Error"))

Related

Excel get substring from nth position up to the end of string

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.

Remove all text and characters except some

I have here some text strings
"16cg-301 -request","16cg-3368 - for review","16cg-3684 - for process"
what i would like to do is to remove all the text and characters except the number and the letters "cg" and - which is within the reference code.

If the string you want to extract is always before the first space in the full string then you can use SEARCH and LEFT to extract your reference code:
=LEFT(A1,SEARCH(" ",A1)-1)
This formula would take 16cg-3368 from 16cg-3368 - for review.

I suggest using something like suggested here
How to use Regular Expressions (Regex) in Microsoft Excel both in-cell and loops
With a replace regex similar to this
[^\dcg]*
or a match regex like this
^([0-9cg- ]+).*
else you could also work with a strange formule similar to this
=CONCATENATE(IF(NOT(ISERROR(SEARCH(MID(A2;1;1);"01234567890cg-")>0));MID(A2;1;1);"");IF(NOT(ISERROR(SEARCH(MID(A2;2;1);"01234567890cg-")>0));MID(A2;2;1);"");IF(NOT(ISERROR(SEARCH(MID(A2;3;1);"01234567890cg-")>0));MID(A2;3;1);"");IF(NOT(ISERROR(SEARCH(MID(A2;4;1);"01234567890cg-")>0));MID(A2;4;1);"");IF(NOT(ISERROR(SEARCH(MID(A2;5;1);"01234567890cg-")>0));MID(A2;5;1);"");IF(NOT(ISERROR(SEARCH(MID(A2;6;1);"01234567890cg-")>0));MID(A2;6;1);"");IF(NOT(ISERROR(SEARCH(MID(A2;7;1);"01234567890cg-")>0));MID(A2;7;1);"");IF(NOT(ISERROR(SEARCH(MID(A2;8;1);"01234567890cg-")>0));MID(A2;8;1);"");IF(NOT(ISERROR(SEARCH(MID(A2;9;1);"01234567890cg-")>0));MID(A2;9;1);"");IF(NOT(ISERROR(SEARCH(MID(A2;10;1);"01234567890cg-")>0));MID(A2;10;1);"");IF(NOT(ISERROR(SEARCH(MID(A2;11;1);"01234567890cg-")>0));MID(A2;11;1);"");IF(NOT(ISERROR(SEARCH(MID(A2;12;1);"01234567890cg-")>0));MID(A2;12;1);"");IF(NOT(ISERROR(SEARCH(MID(A2;13;1);"01234567890cg-")>0));MID(A2;13;1);"");IF(NOT(ISERROR(SEARCH(MID(A2;14;1);"01234567890cg-")>0));MID(A2;14;1);"");IF(NOT(ISERROR(SEARCH(MID(A2;15;1);"01234567890cg-")>0));MID(A2;15;1);"");IF(NOT(ISERROR(SEARCH(MID(A2;16;1);"01234567890cg-")>0));MID(A2;16;1);"");IF(NOT(ISERROR(SEARCH(MID(A2;17;1);"01234567890cg-")>0));MID(A2;17;1);"");IF(NOT(ISERROR(SEARCH(MID(A2;18;1);"01234567890cg-")>0));MID(A2;18;1);"");IF(NOT(ISERROR(SEARCH(MID(A2;19;1);"01234567890cg-")>0));MID(A2;19;1);"");IF(NOT(ISERROR(SEARCH(MID(A2;20;1);"01234567890cg-")>0));MID(A2;20;1);"");IF(NOT(ISERROR(SEARCH(MID(A2;21;1);"01234567890cg-")>0));MID(A2;21;1);"");IF(NOT(ISERROR(SEARCH(MID(A2;22;1);"01234567890cg-")>0));MID(A2;22;1);"");IF(NOT(ISERROR(SEARCH(MID(A2;23;1);"01234567890cg-")>0));MID(A2;23;1);"");IF(NOT(ISERROR(SEARCH(MID(A2;24;1);"01234567890cg-")>0));MID(A2;24;1);"");IF(NOT(ISERROR(SEARCH(MID(A2;25;1);"01234567890cg-")>0));MID(A2;25;1);"");IF(NOT(ISERROR(SEARCH(MID(A2;26;1);"01234567890cg-")>0));MID(A2;26;1);"");IF(NOT(ISERROR(SEARCH(MID(A2;27;1);"01234567890cg-")>0));MID(A2;27;1);"");IF(NOT(ISERROR(SEARCH(MID(A2;28;1);"01234567890cg-")>0));MID(A2;28;1);"");IF(NOT(ISERROR(SEARCH(MID(A2;29;1);"01234567890cg-")>0));MID(A2;29;1);"");IF(NOT(ISERROR(SEARCH(MID(A2;30;1);"01234567890cg-")>0));MID(A2;30;1);"");IF(NOT(ISERROR(SEARCH(MID(A2;31;1);"01234567890cg-")>0));MID(A2;31;1);"");IF(NOT(ISERROR(SEARCH(MID(A2;32;1);"01234567890cg-")>0));MID(A2;32;1);""))
only works by now for less than 33 signs.
problem here will be that you will get unexpected behavior like this:
123cg-123 - Process => 123cg-123-c
after rereading , I think you should try an other approach than described in the question ;-)

If you want to return everything up to and including the last digit, then try:
=LEFT(A1,LOOKUP(2,1/ISNUMBER(-MID(A1,seq,1)),seq))
seq is a named formula: Formula ► Define Name
Name: seq
Refers to: =ROW(INDEX($1:$65535,1,1):INDEX($1:$65535,255,1))
seq returns an array of sequential numbers from 1 to 255.
mid(a1,seq,1)
returns an array consisting of the individual characters in the string in A1. The leading minus sign converts the digits from strings to numbers.
The lookup function will then return the position of the last digit

Remove text from excel cell before first occurance of special character [duplicate]

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 /

How to print a number within a string in matlab

I would like to use the command text to type numbers within 57 hexagons. I want to use a loop:
for mm=1:57
text(x(m),y(m),'m')
end
where x(m) and y(m) are the coordinates of the text .
The script above types the string "m" and not the value of m. What am I doing wrong?

Jubobs pretty much told you how to do it. Use the num2str function. BTW, small typo in your for loop. You mean to use mm:
for mm=1:57
text(x(mm),y(mm),num2str(mm));
end
The reason why I've even decided to post an answer is because you can do this vectorized without a loop, which I'd also like to write an answer for. What you can do place each number into a character array where each row denotes a unique number, and you can use text to print out all numbers simultaneously.
m = sprintfc('%2d', 1:57);
d = reshape([m{:}], 2, 57).';
text(x, y, d);
The (undocumented!) function sprintfc takes a formatting specifier and an array and creates a cell array of strings where each cell is the string version of each element in the array you supply. In order to ensure that the character array has the same number of columns per row, I ensure that each string takes up 2 characters, and so any number less than 10 will have a blank space at the beginning. I then convert the cell array of strings into a character array by converting the cell array into a comma-separated list of strings and I reshape the matrix into an acceptable form, and then I call text with all of the pairs of x and y, with the corresponding labels in m together on the screen.

Return the characters after Nth character in a string

I need help! Can someone please let me know how to return the characters after the nth character?
For example, the strings I have is "001 baseball" and "002 golf", I want my code to return baseball and golf, not the number part. Since the word after the number is not always the same length, I cannot use = Right(String, n)
Any help will be greatly appreciated

If your numbers are always 4 digits long:
=RIGHT(A1,LEN(A1)-5) //'0001 Baseball' returns Baseball
If the numbers are variable (i.e. could be more or less than 4 digits) then:
=RIGHT(A1,LEN(A1)-FIND(" ",A1,1)) //'123456 Baseball’ returns Baseball

Mid(strYourString, 4) (i.e. without the optional length argument) will return the substring starting from the 4th character and going to the end of the string.

Alternately, you could do a Text to Columns with space as the delimiter.

Since there is the [vba] tag, split is also easy:
str1 = "001 baseball"
str2 = Split(str1)
Then use str2(1).

Another formula option is to use REPLACE function to replace the first n characters with nothing, e.g. if n = 4
=REPLACE(A1,1,4,"")

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