I have this content in my cell in Excel 'TS_Bud03A'
I have other cell content with content that should not be touched by bullet no 1 below. For instance cells with P_Har02b
I would like one nested formula for 2 operations
Substitute 'TS_' with 'TVS_'
Remove the last letter after the 2 digit number if one exists
For the first operation I have made this formula
=SUBSTITUTE(K12;"TS_";"TVS_")
It works. But how do I nest another substitute, that will remove the last letter (any case - if it exists) after the 2 digit number?
By combining the 2 answers below I think I almost got it. Like so:
=SUBSTITUTE(REPLACE(LEFT(K15;LEN(K15)--ISNUMBER(RIGHT(K15;1)*1)-1);2;0;"V");"TS_";"TVS_")
But it still changes this P_Bud04a to this PV_Bud04 and this is wrong. The P cells must be untouched.
Any ideas how I can remove the replace that adds V part of the formula I have here in the question?
See if the following works for you:
Formula in B1:
=REPLACE(LEFT(A1,LEN(A1)--ISNUMBER(RIGHT(A1,1)*1)-1),2,0,"V")
REPLACE(<string>,2,0,"V") - Acts like pasting a substring into a string.
LEFT(A1,LEN(A1)-<number> - Our string in the first parameter of REPLACE() has to be either the full length or minus one. That number is still to be determined.
-ISNUMBER(RIGHT(A1,1)*1)-1 - Will check if the last character in A1 is numeric and therfor will return a boolean value TRUE or FALSE. Their equivalent numeric values are 1 and 0. So if TRUE it will return 0, if FALSE, it will return 1.
EDIT: After your comment I understand your data does not always start with TS_, therefor you need to simply swap REPLACE() with SUBSTITUTE(), e.g:
=SUBSTITUTE(LEFT(A1,LEN(A1)--ISNUMBER(RIGHT(A1,1)*1)-1),"TS_","TVS_")
=IF(ISNUMBER(--RIGHT(K12;2));SUBSTITUTE(LEFT(K12;LEN(K12)-2);"TS_";"TVS_");SUBSTITUTE(K12;"TS_";"TVS_"))
I think this is what you need. It first checks if the right last 2 characters are a number (added -- to convert number stored as text to number). Then, if true, remove the last 2 characters and substitute like you mentioned, or, if false substitute according to your formula without removing the last 2 characters.
OK I got it. I just removed the V between quotes, so as to not put any new content into the cell. And it worked.
=SUBSTITUTE(REPLACE(LEFT(K12;LEN(K12)--ISNUMBER(RIGHT(K12;1)*1)-1);2;0;"");"TS_";"TVS_")
Thanks guys
Related
I'm trying to use the IF function in Excel so that if the first character of a cell is 9, then the value shown should be the eight rightmost characters of that cell, otherwise the value shown should be the four rightmost characters. This formula however does not work:
=IF(LEFT(A2,1)=9,RIGHT(A2,8),RIGHT(A2,4))
It keeps returning the rightmost four numbers even though the number in cell A2 starts with 9.
Could you please point out what I'm doing wrong here?
LEFTreturns text, so the comparison needs to also be against a string:
=IF(LEFT(A2,1)="9",RIGHT(A2,8),RIGHT(A2,4))
or you need to convert the result of LEFT to a number again:
=IF(NUMBERVALUE(LEFT(A2,1))=9,RIGHT(A2,8),RIGHT(A2,4))
try =IF(INT(LEFT(A2,1))=9,RIGHT(A2,8),RIGHT(A2,4))
A shorter version if the condition is used to set the number of characters:
=RIGHT(A2,4+4*(LEFT(A2)="9"))
I've got a fairly simple question and I'm sure that I'm missing something obvious.
I've got say 40 cells and all of them contain a formula in them. Only 38 of those cells actually have string or text in them the remaining two do NOT. They're blank with the exception of the formula.
However when I do a COUNTIF or a COUNTA to try and not count the cells that are filled automatically it is giving me the result of 40.
Ways I've tried this and all go the result of 40:
=COUNTIF(B60:B99,"*")
=COUNTIF(B60:B99)
=(COUNTA(B60-B99)
Does anyone know what I'm doing wrong?
Example of the formula in a blank cell that is being counted:
=IF(ISBLANK('Dodgeball'!B48),"",'Dodgeball'!B48)
Use:
=SUMPRODUCT(--(B60:B99<>""))
as this ignores null strings.
You are attempting to count the cells that are not "" but considering them as blank. Try using
=countif(B60:B99,"<>""""")
To explain that final string of quotes, the first one is an escape character so that the second one is read as quotes within the string, similarly the third and fourth are an escape character and a quote for within the string, and finally we end the string with a quote.
One option would be to insert an additional column and simply use the istext() function. Then you can sum that column to get your text count, because false counts and 0 and true counts as 1.
I have personal ID's in reports I have to find in one cell. Too bad the string in the cell which hides this ID can be anything, the ID can be at the beginning, the end, anywhere, but it is there.
The only thing I know is the pattern "space,letter,letter,number,number,number,number,number,number,space". Jike DB544345
I was looking for the correct word for this "mask", but couldn't find an answer. Thank you for your help.
As the comments are numerous I have created a minimal example that might represent what the OP is dealing with:
A1: 123456789 DB544345 asdfg asdfghjk
A2: creating dummy data is a DB544345 pain
A3: DB5443456 and soething else
parsed a copy of that in ColumnB with Text To Columns (with space as the delimiter) then applied:
=IFERROR(IF(AND(LEN(B1)=8,CODE(LEFT(B1))>64,CODE(LEFT(B1))<91,CODE(MID(B1,2,1))>64,CODE(MID(B1,2,1))<91,ISNUMBER(RIGHT(B1,6)*1),RIGHT(B1,6)*1>99999),B1,""),"")
to K1, copied this across to P1 and then K1:P1 down.
A concise "built-in function only" solution to a problem such as this requires a bit of tinkering as many attempts will dead-end or need workarounds due to deficiencies and quirks in the built-in Excel formulas. I much prefer single cell formulas because they minimally affect the general spreadsheet structure. However, due to the limitations listed above, complex single cell solutions often come at the cost of being rather long and cumbersome (this answer is somehow still only two lines on my formula bar in Excel). I came back to your question and cobbled together a formula that can (as far as I have tested) extract the first occurrence of this pattern with a single cell formula. This is an array formula (Ctrl+Shift+Enter instead of Enter) that assumes your data is in A2. This rough formula returns the first 8 characters if no match is found and throws #REF if the string is shorter than 10 characters.
=MID(A2,MIN(IF(MID(A2,ROW(INDIRECT("A1:A"&(LEN(A2)-9))),1)=" ",IF(MID(A2,ROW(INDIRECT("A1:A"&(LEN(A2)-9)))+9,1)=" ",IF(CODE(MID(A2,ROW(INDIRECT("A1:A"&(LEN(A2)-9)))+1,1))>64,IF(CODE(MID(A2,ROW(INDIRECT("A1:A"&(LEN(A2)-9)))+1,1))<91,IF(CODE(MID(A2,ROW(INDIRECT("A1:A"&(LEN(A2)-9)))+2,1))>64,IF(CODE(MID(A2,ROW(INDIRECT("A1:A"&(LEN(A2)-9)))+2,1))<91,IF(IFERROR(MID(A2,ROW(INDIRECT("A1:A"&(LEN(A2)-9)))+3,6)*1>99999,FALSE),ROW(INDIRECT("A1:A"&(LEN(A2)-9)))))))))))+1,8)
Let me try to break this down at least on a high level. We are splitting the main text into every possible ten character chunk so that we can test each one using the suggestion of #pnuts to verify the Unicode values of the first two characters and run an ISNUMBER check on the rest of the string. This first block recurs throughout my formula. It generates a list of numbers from 1 to n-9 where n is the length of our main text string.
ROW(INDIRECT("A1:A"&(LEN(A2)-9)))
Let's assume our string is 40 characters long and replace the above formula with {1...31}. Using this number sequence generation we can check if characters 1 to 31 are spaces:
IF(MID(A2,{1...31},1)=" "
Then we can check if characters 10 to 40 are spaces:
IF(MID(A2,{1...31}+9,1)=" "
Then we can check if characters 2 to 32 are capital letters:
IF(CODE(MID(A2,ROW(INDIRECT("A1:A"&(LEN(A2)-9)))+1,1))>64,
IF(CODE(MID(A2,ROW(INDIRECT("A1:A"&(LEN(A2)-9)))+1,1))<91
Then we can check if characters 3 to 33 are capital letters:
IF(CODE(MID(A2,ROW(INDIRECT("A1:A"&(LEN(A2)-9)))+2,1))>64,
IF(CODE(MID(A2,ROW(INDIRECT("A1:A"&(LEN(A2)-9)))+2,1))<91
Then we can check if the strings of characters 4 to 9, 5 to 10, ..., 33 to 38, 34 to 39 are six-digit numbers:
IF(IFERROR(MID(A2,ROW(INDIRECT("A1:A"&(LEN(A2)-9)))+3,6)*1>99999,FALSE)
If all conditions are TRUE, that 10 digit chunk test will return the index of its first character in the string via another instance of the original array {1...31}. Otherwise it returns nothing. We take the Min of all return indexes and then use the Mid function to grab the 8 digit string determined by the aforementioned minimum index:
=MID(A2,MIN(matching index list)+1,8)
I think this will work, if we assume that the SPACE at the beginning and end are merely to differentiate the ID from the rest of the string; hence would not be present if the ID were at the beginning or end of the string. This formula is case insensitive. If case sensitivity is required, we could do character code comparisons.
=LOOKUP(2,1/((LEFT(myArr,2)>="AA")*(LEFT(myArr,2)<="ZZ")*(LEN(myArr)=8)*ISNUMBER(-RIGHT(myArr,6))),myArr)
Where myArr refers to:
=TRIM(MID(SUBSTITUTE(TRIM(Sheet2!A1)," ",REPT(" ",99)),(ROW(INDIRECT("1:10"))-1)*99+1,99))
If myArr is initially defined with the cursor in B1, referring to A1 as shown, it will adjust to refer to the cell one column to the left of the cell in which the Name appears.
The 10 in 1:10 is the maximum number of words in the string -- can be adjusted if required.
I have a column with values like
WI60P-14E64F5167E6-01138
or
WI60-34D03E185267-01051
etc....
i need to find the first occurrence of - till the second occurrence of - and remove the resulting character.
I am using this function =MID(A1,FIND("-",A1),13) which returns me -34D03E185267 or -14E64F5167E6 from the above strings.
But I want the output like WI60P-01138 or WI60-01051
Can anyone help me?
Assuming that the length of text between the two hyphens never varies, you can use your existing function, along with SUBSTITUTE to get the desired string, like so:
=SUBSTITUTE(A1,MID(A1,FIND("-",A1),13),"")
Alternately, you can use FIND to get the hyphen positions and again use SUBSTITUTE to get the replaced string, like so:
=SUBSTITUTE(A1,MID(A1,FIND("-",A1),FIND("-",A1,FIND("-",A1)+1)-FIND("-",A1)),"")
Notice that one of the FIND expressions takes 3 parameters - by passing FIND("-",A1) + 1 as the third parameter, we get the second occurence of '-' within the cell.
This will work for any position of the two hyphens:
=CONCATENATE(MID(A1;FIND("-";A1)+1;FIND("-";A1;FIND("-";A1)+1)-FIND("-";A1)-1))
Solution using multiple cells as support (with dynamic length of the string):
Put this in cell B1 to obtain the first part of the code:
=MID(A1,1,FIND("-",A1)-1)
Put this in cell C1 to obtain the last part of the code:
=MID(A1,FIND("-",A1)+1,LEN(A1))
Put this in cell D1 to concatenate both and get your result:
=concat(B1,C1)
Solution in a single cell (with dynamic length of the string)
Put this in cell B1 to get the same result as above :
=concat(MID(A1,1,FIND("-",A1)-1),MID(MID(A1,FIND("-",A1)+1,LEN(A1)),FIND("-",MID(A1,FIND("-",A1)+1,LEN(A1))),LEN(MID(A1,FIND("-",A1)+1,LEN(A1)))))
A program that exports to Excel creates a file with an indented list in a single column like this:
Column A
First Text
Second Text
Third Text
Fourth Text
Fifth Text
How can I create a function in excel that counts the number of white spaces before the string of text?
So as to return: 1 for the first text row and 3 for the for the thirst row, etc in this example.
Preferably seeking a non-VBA solution.
TRIM doesn't help here because it removes double spaces also between words.
The main idea is to find the FIRST letter in the trimmed string and find its position in the original string:
=FIND(LEFT(TRIM(A1),1),A1)-1
You can try this function in Ms Excel itself:
=LEN(A1)-LEN(SUBSTITUTE(A1," ",""))
This would apply if the results are in a single cell. If it is for a whole row/column, just drag the formula accordingly.
Try below:
=FIND(" ",A1,1)-1
It calculates the position of the first found whitespace character in a cell and reduces it by 1 to reflect number of characters before that position.
As per http://www.mrexcel.com/forum/excel-questions/61485-counting-spaces.html, you may try:
=LEN(Cell)-LEN(SUBSTITUTE(Cell," ",""))
where Cell is your target cell (i.e. A1, B1, D3, etc.).
My example:
B8: =LEN(F8)-LEN(SUBSTITUTE(F8," ",""))
F8: [ this is a test ]
produces 4 in B8.
The above method will count spaces before the string if any were inserted, between individual words and after the string, if any were inserted. It won't count available space that does not have an actual white space character. So, if I inserted two spaces after test in the above example, the total count would be raised to 6.
As has been pointed out in the other answers, you can't really use TRIM or SUBSTITUTE as potential spaces in between words or at the end will give you the wrong result.
However, this formula will work:
=MATCH(TRUE,MID(A1,COLUMN($A$1:$J$1),1)<>" ",0)-1
You need to enter it as an array formula, i.e. press Ctrl-Shift-Enter instead of Enter.
In case you expect more than 10 spaces, replace the $J with a column letter further down in the alphabet!
Here's my solution. If the left 5 characters equals "_____" (5 blank spaces), then return 5, else look for 4 spaces, and so on.
=IF(LEFT(B1,5)=" ",5,IF(LEFT(B1,4)=" ",4,IF(LEFT(B1,3)=" ",3,IF(LEFT(B1,2)=" ",2,1))))
You almost got it with LEN + TRIM in answers before, you only need to combine both:
=LEN(Cell)-LEN(TRIM(Cell))
If it is Indented you could create a Personal Function like this:
Function IndentLevel(Cell As Range)
'This function returns the indentation of a cell content
Application.Volatile
'With "Application.Volatile" you can make sure, that the function will be
recalculated once the worksheet is recalculated
'for example, when you press F9 (Windows) or press enter in a cell
IndentLevel = Cell.IndentLevel
'Return the IndentLevel
End Function
This will work only if it is Indented, you can see this property in the Cell Format -> Alignment.
After This you could see the Indentation Level.