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having this data in excel cell
kerass(xcvbn=2, abcdefg_iD='510863005')
having 100000 of records so, how to extract id number using excel formula from that excel sheet
Thanks in advance
Assuming your value is in cell A2. Here is a formula to be inserted
=SUBSTITUTE(RIGHT(A2,LEN(A2)-SEARCH("_iD='",A2)-4),"')","")
This means:
Search for "_iD='"
Once found, subtract the length of "_iD='" of that location (hence the 4)
Use the total length of the string, subtracted with location and 4, in order to find where the actual identifier begins
Take the right part of the string, only the part behind "_id='...'" remains
In that part, replace "')" by an empty string (otherwise you get "...005')"
Alternative answer:
=MID(A1,FIND("'",A1)+1,9)
The find formula finds the first "'" starting position
Assuming all the ID's are 9 digits, the MID formula uses the starting pos. from find and takes the following 9 digits.
So assuming your text is in cell A2, and that the id is the first 9 characters of the last 11 then this is the simplest I can think of:
=LEFT(RIGHT(A2,11),9)
But if extra characters get added or the ID changes length.
Does have the benefit of being minimalist on functions and low volativity.
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 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.
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.
I have column B that I need to test the length to see if it is longer than 200 characters. If it is longer than 200 characters, I need it to go from right to left and find the occurrence of the semicolon ";" and split the field from the right of the semicolon into column C. Can this be done? Before I was having to do this with 4 columns and have reduced it to one column. Please advise to the best formula to do this.
=IF(LEN(B1)>200,MID(B1,SEARCH("#",SUBSTITUTE(B1,";","#",LEN(B1)-LEN(SUBSTITUTE(B1,";",""))))+1,LEN(B1)),"")
Explanation:
Remove all instances of the delimiter: SUBSTITUTE(B1,";","")
Subtract the length of (1) from the length of the entire string to get the number of occurrences of the delimiter: LEN(B1)-LEN([1])
Substitute the last occurrence of the delimiter with an #: SUBSTITUTE(B1,";","#",[2])
Find the location of the #: SEARCH("#",[3])
Get the substring of everything to the right of the # location: MID(B1, [4] +1,LEN(B1))
Add if condition to only process strings of length > 200: =IF(LEN(B1)>200,[5],"")
I searched the web for formulas for this and concluded that you either need a lot of nesting and a difficult to follow formula or a VBA function. I would suggest using a VBA function such as the one I have written below (FindLast) within a simple formula. Let me know if you need instructions on how to create this VBA formula:
Function FindLast(find_text As String, within_text As Range) As Double
Dim i As Integer
i = Len(within_text.Value) ' start at last character and work back
Do While Mid(within_text.Value, i, 1) <> find_text
i = i - 1
Loop
FindLast = i
End Function
You will then be able to use FindLast within a formula such as the below in C1:
=IF(LEN(B1)>200,MID(B1,FindLast(";",B1)+1,500),"")
UPDATE
The 500 in the above is just a long number I picked to mean the rest of the cell. If there may be more than 500 characters after the final ; then use a larger number. Unfortunately I don't think the MID function allows you to specify that you want to return the rest of the cell. I have put to return nothing if B1 is not >200 characters, let me know if this is not the requirement.