I have addresses in a column. I need to split just the last but one name from that column into another column.
The last but one substring is delimited by a space from both the right and left sides.
The content in cell A2 is "via Milano, 25 ROMA RM" and I entered the below formula in cell B2 (I need "ROMA" in cell B2)
I tried using the following formula:
=MID(A2,FIND(CHAR(1),SUBSTITUTE(A2," ",CHAR(1),LEN(A2)-LEN(SUBSTITUTE(A2," ",""))))+1,LEN(A2))
But it returns the last substring from a given string (in this case "RM"). Can't manage to modify it to get the last but one substring. If useful: currently using macOS Catalina 10.15.7.
Edit: the string address is ALWAYS formatted in the same way "via", "street name","house number", "comune", "province".
You could do something like the hideous:
=LEFT(
RIGHT(
SUBSTITUTE(A2," ","!",LEN(A2)-LEN(SUBSTITUTE(A2," ",""))-1),
LEN(SUBSTITUTE(A2," ","!",LEN(A2)-LEN(SUBSTITUTE(A2," ",""))-1))-
SEARCH("!", SUBSTITUTE(A2," ","!",LEN(A2)-LEN(SUBSTITUTE(A2," ",""))-1))),
SEARCH(" ",
RIGHT(SUBSTITUTE(A2," ","!",LEN(A2)-LEN(SUBSTITUTE(A2," ",""))-1),
LEN(SUBSTITUTE(A2," ","!",LEN(A2)-LEN(SUBSTITUTE(A2," ",""))-1))-
SEARCH("!", SUBSTITUTE(A2," ","!",LEN(A2)-LEN(SUBSTITUTE(A2," ",""))-1)))))
though I am pretty sure there are faster/cleaner ways to get what you want. Note that "!" was just chosen as a character that wouldn't otherwise appear in the string. Also note this is why best practice is to store the smallest pieces of data as separate items rather than in one string!
REQUESTED EDIT: If you are just looking to get everything to the right of a comma and to the left of a space (e.g. "25" from "Via Milano, 25 ROMA RM") you can use something like the below
=LEFT(
TRIM(RIGHT(A2,LEN(A2)-SEARCH(",",A2))),
SEARCH(" ", TRIM(RIGHT(A2,LEN(A2)-SEARCH(",",A2)))))
Related
In excel, TRIM() will remove all spaces before and after text, while also removing any duplicate spaces in between words.
Is there a formula or combination thereof that will do the same as TRIM() but leave spaces between words as-is?
In the following example, I'm looking for a formula that will accomplish that of the fictitious formula "WXYZ":
TRIM(" Omicron Persei 8 ") = "Omicron Persei 8"
WXYZ(" Omicron Persei 8 ") = "Omicron Persei 8"
Note that I've read somewhere that TRIM() in VBA will work like that of WXYZ above. However, I'm looking for a formula solution.
I believe this should work (assuming your string is located at A1):
=MID(A1,
FIND(LEFT(TRIM(A1),1),A1),
(LEN(A1)-MATCH(RIGHT(TRIM(A1),1),INDEX(MID(A1,LEN(A1)-ROW(INDIRECT("1:"&LEN(A1)))+1,1),0),0)-FIND(LEFT(TRIM(A1),1),A1)+2)
FIND(LEFT(TRIM(A1),1),A1) returns the location of the first non-space character in the string
MATCH(RIGHT(TRIM(A1),1),INDEX(MID(A1,LEN(A1)-ROW(INDIRECT("1:"&LEN(A1)))+1,1),0),0) returns the location of the last non-space character in the string from right-to-left.
How would this look in Excel 365? A bit easier I think with let, sequence and xmatch but not particularly short:
=IFERROR(LET(seq,SEQUENCE(LEN(A2)),
array,MID(A2,seq,1),
start,XMATCH(TRUE,array<>" "),
finish,XMATCH(TRUE,array<>" ",0,-1),
MID(A2,start,finish-start+1)),"")
Just to add to all the valuable content:
Formula in B1:
=LET(x,TEXTSPLIT(A2," ",,1),TEXTJOIN(DROP(DROP(TEXTSPLIT(" "&A2&" ",x),,1),,-1),,x))
A Formula Array could also be used.
Assuming the string is located at A1 enter this Formula Array in B2. It's highly suggested to ensure this part of the formula ROW(B:B) refers always to the same column were the formula is located (column B in this case), this is in order to avoid the formula returning an error if the column to which it refers is deleted.
=MID($A1,
FIND(LEFT(TRIM($A1),1),$A1),
1+MAX(ROW(B:B)*(ROW(B:B)<=LEN($A1))*(MID($A1,ROW(B:B),1)<>" "))
-FIND(LEFT(TRIM($A1),1),$A1))
FormulaArrays are entered pressing [Ctrl] + [Shift] + [Enter] simultaneously, you shall see { and } around the formula if entered correctly
As regards the formula provided by #Aakash I suggest to replace the INDIRECT function in this part:
-ROW(INDIRECT("1:"&LEN($A7)))
with this:
-ROW(B:B)
So the formula will become Non-Volatile:
=MID($A1,
FIND(LEFT(TRIM($A1),1),$A1),
(LEN($A1)-MATCH(RIGHT(TRIM($A1),1),INDEX(MID($A1,LEN($A1)-ROW(B:B)+1,1),0),0)
-FIND(LEFT(TRIM($A1),1),$A1)+2))
In Microsoft Excel I wish to count the frequency of a specific word in a cell. The cell contains a few sentences. I am using a formula right now that is working, but not the way I want it to.
A1
my uncle ate potatos. potato was his favorite food. Don't mash the potato, just keep it simple.
B1 (word to count the frequency of)
potato
C1 (forumula)
=(LEN(A2)-LEN(SUBSTITUTE(A2;B2;"")))/LEN(B2)
C1 Results:
3
In C1, I am getting a count 3. I want it just to be 2. So, the formula is counting potatos.
How do I make the function only count exact matches?
I've got a solution here but it's not pretty.
The problem, as I indicate in my comment, is that Excel has no internal function to see if a cell contains an 'exact match'. You can check if the total value in a cell is an exact match, but you can't check whether a search term has been conjugated like that. So, we'll need to create a special method which checks for every 'acceptable' ending to a word. In my eyes, this would be anything that ends with space, anything that ends with punctuation, and anything at the end of a cell with nothing after it.
ARRAY FORMULAS
You were on the right track with the LEN - SUBSTITUTE method, but the formula will need to be an array formula to work. Array formulas calculate the same thing multiple times over a given range of cells, instead of just once. They resolve the calculation for each individual cell in a formula and provide an array of results. This array of results must be collapsed together to get a single total result.
Consider as follows:
=LEN(C1:C6)
Confirm this formula with CTRL + SHIFT + ENTER instead of just ENTER. This gives us the LEN of C1, followed by C2, C3... etc., resulting in an array of results that looks like this [assume C1 had "a", C2 had "aa", C3 had "a", C4 had "", C5 had "aaa", and C6 had ""]:
={1;2;1;0;3;0}
To get that as a single number providing the total length of each cell individually, wrap that in a SUM function:
=SUM(LEN(C1:C6))
Confirmed again with CTRL + SHIFT + ENTER instead of just ENTER. This results in the total length of all cells: 7.
DEFINING AN EXACT MATCH
Now to take your question, you are looking to find all 'acceptable' matches of given word B1, within text A1. As I said before, we can define an acceptable answer as one which ends in punctuation, a space, or the end of the cell. Something at the end of the cell is a special case which we will consider later. First, take a look at the formula below. In cells C1:C6, I have manually typed a comma, a period, a semi-colon; a hyphen, a space, and a slash. These will be the 'acceptable' ways to end the word found in B1.
=LEN(SUBSTITUTE(A1,B1&C1:C6,""))
Confirmed with CTRL + SHIFT + ENTER, this takes the length of the substitution for the search term in B1 appended with the acceptable word-end in C1:C6. So it gives the length for 6 new SUBSTITUTED words. But as this is an array of results, we need to add them together to get a single number, like so:
=SUM(LEN(SUBSTITUTE(A1,B1&C1:C6,"")))
FORMULIZING THE RESULT
To work it as you have in your sentence, we will now need to subtract this length from the length of the original word. Note that there is a problem with doing this simply - since we are searching multiple times, we will need to add the length of the original word multiple times. Consider something like this:
=LEN(A1)-SUM(LEN(SUBSTITUTE(A1,B1&C1:C6,"")))
This won't work, because it only adds the length of A1 once, but it subtracts the length of the substituted strings multiple times. How about this?
=LEN(A1)*6-SUM(LEN(SUBSTITUTE(A1,B1&C1:C6,"")))
This works, because there are 6 word-end terms we search for with C1:C6, so the substitution there will occur 6 times. So we have the original length of the word 6 times, and the length of each substituted word 6 times [keep in mind that if there is no match for, say, "potato;", then that term will give the length of the original word, thus negating one of the times we added the length of that word, as expected].
To finalize this, we need to divide by the number of letters in the search term. Keep in mind that where you have "/LEN(B1)", we will need to add a character for the length of each of our word-ends.
=(LEN(A1)*6-SUM(LEN(SUBSTITUTE(A1,B1&C1:C6,""))))/(LEN(B1)+1)
Finally, we need to add the special case where the last portion of A1 is equal to the search term, with no word-end. Alone, this would be:
=IF(RIGHT(A1,LEN(B1))=B1,1,0)
This will give us a 1 if the last part of A1 is equal to B1, otherwise it gives 0. So now simply add this to our previous formula, as follows:
=(LEN(A1)*6-SUM(LEN(SUBSTITUTE(A1,B1&C1:C6,""))))/(LEN(B1)+1)+IF(RIGHT(A1,LEN(B1))=B1,1,0)
Remember to confirm with CTRL + SHIFT + ENTER, instead of just ENTER. That's it, it now gives you the count of all "exact matches" of your search term.
ALTERNATE APPROACH TO ARRAY FORMULAS
Note that instead of using C1:C6, you could instead hardcode your formula to look for specific punctuation as the word-end. This will be harder to maintain but, in my opinion, just as readable. It will look like this:
=(LEN(A1)*6-SUM(LEN(SUBSTITUTE(A1,B1&{",",".",";"," ","/","-"},""))))/(LEN(B1)+1)+IF(RIGHT(A1,LEN(B1))=B1,1,0)
This is still technically an "array formula", and it works on the same principles as I have described above. However, one benefit here is that you can confirm this type of entry with just ENTER. This is good, in case someone accidentally edits your cell and presses ENTER without noticing. Otherwise, this is equivilent to the format above.
Let me know if you would like any portion of this elaborated on.
I do have an alternate solution for you to consider. I takes a bit more space and the formulas are a little more convoluted, but in some senses it will be simpler.
Use column C as a new helper column. Column C will take the text from column A, and will substitute out all instances of punctuation with a " ". Once this has been done, the formula to count the instances of the search term from column B will be a simple formula essentially as you have it in your OP.
=SUBSTITUTE(SUBSTITUTE(SUBSTITUTE(SUBSTITUTE(SUBSTITUTE(A1,","," "),"."," "),";"," "),"-"," "),"/"," ")
This formula first substitutes all slashes for spaces, then with that substituted text it substitutes dashes for spaces, then with that substited text it substitutes semicolons with spaces, etc. As you indicated, if you use semi-colons as delimiters, you will need to replace my commas separating terms with semi-colons.
Then the formula in D1 is simply what you have above in your OP, with two changes: we will be searching for B1 & " ", because we know all of the 'exact matches' now end in spaces, and we will be adding in an extra '1' if the last part of the text in C1 is the same as the search term in B1 - because if a cell ends in that word, it won't have a space, but it is still an 'exact match'. Like so:
=(LEN(C1)-LEN(SUBSTITUTE(C1,B1&" ","")))/(LEN(B1)+1)+IF(RIGHT(C1,LEN(B1))=B1,1,0)
EDIT
My list of punctuation was only a suggestion; I recommend you really go through some sample text and make sure you don't have any weird characters after words. Also, consider changing uncommon ones I have (like "/", or "-") with "?" or "!". If you want to add more, just follow the pattern of the SUBSTITUTE formula.
To make this case-insensitive, you just need to change the formula in column C to make the result all lower case, and then ensure your search terms in column B are lower case. Change column C like so:
=LOWER(SUBSTITUTE(SUBSTITUTE(SUBSTITUTE(SUBSTITUTE(SUBSTITUTE(A1,","," "),"."," "),";"," "),"-"," "),"/"," "))
Sorry for making it "a new answer". You may move it wherever you want.
I have just found a solution for the answer Liu Kang asked on Aug 3 2015 at 12:15. :)
Unfortunately, I do not have "50 reputation" to comment on Grade 'Eh' Bacon's solution above, where the last comment is this:
Discovered a slight problem. Using =IF(B1<>"";(LEN(A1)-LEN(SUBSTITUTE(A1;B1&" ";"")))/(LEN(B1)+1)+IF(RIGHT(A1;LEN(B1))=B1;1;0);"") with shoe in B1 gives the following result: shoe in A1 = 1 (correct), shoes in A1 = 0 (correct), ladyshoe in A1 = 1 (wrong). Guess this have to do with "RIGHT" in the formula. Is it possible to make the formula non-matching for prefix words? E.g if B1 is containing shoe and A1 is containing ladyshoe dogshoe catshoes shoes I want C1 to result in 0. – Liu Kang Aug 3 '15 at 12:15
The solution is to search for a space at the beginning of the word as well (" "&B1&" ") and to add "one" more LEN(B1)+2. So, it becomes =IF(B1<>"";(LEN(A1)-LEN(SUBSTITUTE(A1;" "&B1&" ";"")))/(LEN(B1)+2)+IF(RIGHT(A1;LEN(B1))=B1;1;0);"").
There is one more problem if the word we are looking for is at the beginning. Because there is obviously no space " " at the beginning of the sentence. I use a workaround for it - I have my sentence in A1, but then I have a hidden column B where there is =" "&A1 in B1 and it puts the "space" I need to the beginning of the sentence and everything from the original Grade 'Eh' Bacon's solution is shifted (A1->B1, B1->C1, C1->D1).
I hope it can help and thanks to all who participated in this thread, you helped me A LOT!
Do you need this to be a single formula? I have an idea, but it takes a few (relaitvely simple) steps.
Since you have a long sentence in A1, what about going to Data -> Text to Columns, and send this sentence into a Row, delimited by spaces. Then, remove any punctuation. Then, just do a simple Countif()?
Put the info in A1, then go to Data --> Text to Columns, choose "Delimited", click Next, and choose "Space":
Click Finish, and it'll put the entire thing into Row 1, with a word in each cell. Now just Find/Replace "." and "," with nothing.
Then, Countif to the rescue!
If that works, we can automate into VB, so you don't have to manually find/replace the puncutation. Before I jump into that, does this method work?
Take the length of the string and minus the length of the string with the keyword replaced with nothing then divide the result by the length of the keyword:
=(LEN(A1)-LEN(SUBSTITUTE(A1,B1,"")))/LEN(B1)
I have a field (column) in excel in the format of "LastName, FirstName MiddleInitial" with a space between the comma after the last name and the first name and a second space between the middle initial and the first name (no comma after the first name). Is there a way to identify which cells have a middle initial on the right hand side and then eliminate the middle initial for all cells such that the output will look like "LastName, FirstName"?
Thanks!
What you want to do is to be able to parse the field into multiple fields, and then rejoin then with a simple excel formula. To take a field and put it into multiple columns you can use the following approach for "Text to Columns"
Text to Columns
Once you have 3 columns of text, you can group the first two together by using a formula like =A1&", "&B1 in the column you want to have the value. (In my example, the last name would be in A1, and first name in B1)
You could also do it as a cell formula without changing your original data. I built this up using the models presented on How to split a string based on “:” in MS-Excel?, over at SuperUser.
Assuming you have a name (Public, John Q) in a1:
=LEFT(A1,FIND(" ",A1)) & LEFT(MID(A1,FIND(" ",A1)+1,LEN(A1)),IFERROR(FIND(" ",MID(A1,FIND(" ",A1)+1,LEN(A1)))-1,LEN(MID(A1,FIND(" ",A1)+1,LEN(A1)))))
It looks very complicated, but really isn't:
Using these two components:
Get everything to the left of the first space in a string: LEFT(A1,FIND(" ",A1))
Get everything to the right of the first space in a string: MID(A1,FIND(" ",A1)+1,LEN(A1)))
We can grab these pieces:
Get the last name: LEFT(A1,FIND(",",A1)+1)
Get the first name and middle name: MID(A1,FIND(" ",A1)+1,LEN(A1)))
Then recurse by putting the "left of a space" construction around the "right of a string" construction:
MID( LEFT(A1,FIND(",",A1)+1) ,FIND(" ", LEFT(A1,FIND(",",A1)+1) )+1,LEN( LEFT(A1,FIND(",",A1)+1) )))
Unless there is no second space (when there's no middle name.) That's what the Iferror is for - in that case, you want the entire length of what you got after the first space:
LEFT(MID(A1,FIND(" ",A1)+1,LEN(A1)),IFERROR(FIND(" ",MID(A1,FIND(" ",A1)+1,LEN(A1)))-1,LEN(MID(A1,FIND(" ",A1)+1,LEN(A1)))))
Put them together for lastname, firstname:
=LEFT(A1,FIND(" ",A1)) & LEFT(MID(A1,FIND(" ",A1)+1,LEN(A1)),IFERROR(FIND(" ",MID(A1,FIND(" ",A1)+1,LEN(A1)))-1,LEN(MID(A1,FIND(" ",A1)+1,LEN(A1)))))
I have a column full of names written as:
"lastName, firstName"
I want to have a nother column that has this name list written as:
"firstName LastName"
So, how can I switch a string from "lastName, firstName" to "firstName LastName" ?
If the first name is in A2 try this formula in B2 copied down
=MID(A2&" "&A2,FIND(" ",A2)+1,LEN(A2)-1)
Enter data into cells e.g.
Brown, John
Green, Bob
Smith, Will
(Note that the comma in this case is the delimiter which the system will use to separate the entries)
Highlight these cells.
Click on the "Data" tab,
click on "Text to Column".
Choose options offered.
The worked for me
=RIGHT(C3,LEN(C3)-LEN(LEFT(C3,FIND(",",C3)-1))-1) & " " & LEFT(C3,FIND(",",C3)-1)
Barry's answer is correct, if the single cell (A2) doesn't ALSO contain errors like "firstname lastname" -- as is often the case.
So, to deal with the cell data being either "Last, First" or "First Last", we'll need to compare the cell and take action appropriately. Also note, in my example, I moved the first name to it's own column, and last name to it's own column. You can of course join them into a single column.
First, let's break down what we need. This will return the first name when there is a "," in the cell contents: (assumes the field you are wanting to split is in A2, adjust as needed)
=MID(A2;FIND(",";A2)+2;99)
and this will return the first name when there isn't:
=MID(A2;1;FIND(" ";A2))
And for Last name when there is a comma:
=MID(A2;1;FIND(",";A2)-1)
and last name when there isn't:
=MID(A2;FIND(" ";A2)+1;99)
So, we need to test if there is a "," in the string, and deal with an error condition when there isn't (FIND returns #value when not found, ISERROR() can test for that). Optionally (and my approach) was to check if the return of the FIND yielded a number (an error won't). In this case, an "error condition of not finding it" would yield a non-number #Value.
=IF( ISNUMBER( FIND(",";A2)); MID(A2;FIND(",";A2)+2;99); MID(A2;1;FIND(" ";A2)))
Doing the same thing for the Last name looks like this:
=IF( ISNUMBER( FIND(","; A2)); MID(A2;1;FIND(",";A2)-1); MID(A2;FIND(" ";A2)+1;99))
Please note that my solution was using my spreadsheet application OpenOffice. I think Excel has a different find method (Search?) - so make adjustments as necessary.
The above answer is incorrect.
In fact, for my own name, Jay Jacob Wind, the formula breaks.
The correct formula is:
Assuming Last, First is in column A, separated by one comma and a space bar
For first name:
=mid(A2,find(",",A2)+2,99)
For last name:
=mid(A2,1,find(",",A2)-1)
Put together, assuming Column B is First Name and Column C is Last Name
=B2&" "&C2
or simply (but more complexly)
=mid(A2,find(",",A2)+2,99)&" "&mid(A2,1,find(",",A2)-1)
but I like separating them into two columns,
then concatenating them
I want to delete first characters in a cell and show it in another cell.
I have A1 - 1900: (1873, 'asd#asd.com
I want to show B1 - asd#asd.com
You can use the RIGHT function to 'keep' a number of characters:
=RIGHT(A1, LEN(A1)-B1)
Will chop off B1 characters of the content of cell A1 (in this case, B1 should contain 14, but you can make it a calculation, too)
If you have always a quote before your email, just select you column, go to Data / convert and use "'" as separator
If the apostrophe precede every e-mail address then this formula can be used in column B.
=MID(A1,FIND("'",A1)+1,LEN(A1))
The find function returns the first occurrence of the apostrophe.
The Mid function extracts the text after the apostrophe.