I have a string such as K68272CAA6A1
And need to do that, formula will pass the first character (I mean string will be 68272CAA6A1 in mind) and formula will find the first text character. And cell value will be 7. Because first text character is "C" and it's the 7th character of my string (include "K" character).
And after that I'll split rest of them. But I'm confused about this issue.
If I understand you correctly, you are looking for the position of the 2nd letter in your string. That number is given by the following array-entered formula.
To enter an array formula, hold down ctrl+shift while hitting Enter. If you do this correctly, in the Formula Bar you will see braces {...} around the formula:
=MATCH(FALSE,ISNUMBER(MID(A1,ROW(INDIRECT("2:99")),1)/1),0)+1
The 99 just needs to be some number larger than the length of your longest string.
If I understood you correctly, a formula that implements this functionality (assuming cell A1 = K68272CAA6A1 and B1 = K) would be:
=FIND(RIGHT(SUBSTITUTE(SUBSTITUTE(SUBSTITUTE(SUBSTITUTE(SUBSTITUTE(SUBSTITUTE(SUBSTITUTE(SUBSTITUTE(SUBSTITUTE(RIGHT(A1,LEN(A1)-FIND(B1,A1)),"1",""),"2",""),"3",""),"4",""),"5",""),"6",""),"7",""),"8",""),"9",""),1),RIGHT(A1,LEN(A1)-FIND(B1,A1)))-1
The long sequence of substitute is there to remove the numbers (I couldn't find a specific formula to remove them).
This gigantic formula for your example would simply give the answer 6.
To get the strings separated as you want all you need to do is =LEFT(A1,D1) supposing the long formula is on D1 and =RIGHT(A1,D1), which in your example would yield respectively K68272 and CAA6A1
Related
I have a comma separated value in A2 and same numbers in different cells B1, C1, D1.... I want to match them from comma separated value and find out the count in B2, C2, D2. Please see the image attached you will get the context.
Can we achieve this by formula or macro in excel?
Tried formula:
=LEN(TRIM($A$2))-LEN(SUBSTITUTE(TRIM($A$2),C1,","""))
Also, I have two data sets where I will be using this formula to find out the count of number from comma-separated value and based on count I want the repeated ones to come in a different cell please refer the image for better understanding.
Probably not the best solution but get the job done. Please note it is case-sensitive and please make sure to press Ctrl+Shift+Enter upon finishing this formula.
{=SUM(--(EXACT("!"&TRIM(MID(SUBSTITUTE($A2,",",REPT(" ",100)),(ROW(INDIRECT("1:"&LEN($A2)))-1)*100+1,100)),"!"&B1)))}
You can replace ! in the above formula with a unique symbol that will never appear in the text string to be safer.
The logic is to SUBSTITUTE the comma , with and long string of blanks, then use MID to find each value in the text string and return the result as an array, then use EXACT to match each value in the array with the look up value and return a new array of TRUE and FALSE, then SUM up all TRUE which will give the count of the look up value.
UPDATE #2
As requested by OP, here is one way of solving the second query which is to match the same value with the same occurrence from two text strings separated by comma ,.
The formula in Cell C2 is from the original solution which is used to find the occurrence of a given value in a text string;
The formula for Range C6:K6 is an array formula as shown below. I used a helper row to layout the matching values, and excluding the one that has 0 count for both data set;
{=IFERROR(INDEX($C$1:$K$1,,AGGREGATE(15,7,COLUMN(INDIRECT("1:"&COLUMNS($C$1:$K$1)))/($C$2:$K$2=$C$3:$K$3)/($C$2:$K$2>0),COLUMN()-2))&",","")}
The formula in Cell L8 is concatenating all values from Range C6:K6 and remove the last comma , from the final text string:
=LEFT(CONCATENATE(C6,D6,E6,F6,G6,H6,I6,J6,K6),LEN(CONCATENATE(C6,D6,E6,F6,G6,H6,I6,J6,K6))-1)
The following worked for me, give it a try:
Formula in B2:
=(LEN(","&SUBSTITUTE($A$2,",",",,")&",")-LEN(SUBSTITUTE(","&SUBSTITUTE($A$2,",",",,")&",",","&B$1&",","")))/LEN(","&B$1&",")
Drag right...
A simpler way of doing it is to simply calculate the difference in the length of the string minus the length of the string when replacing the value searched by nothing and dividing by the length of the string searched
The formula would be:
=(LEN($A$1)+1-LEN(SUBSTITUTE($A$1&",",B1&",","")))/LEN(B1&",")
There is a much simpler solution:
=COUNTIF(SPLIT($A$2, ","), B1)
I have a scenario where I want my Microsoft excel field to have the same length of the longest word in the column.
Basically lets say if I have:
ACBBASDBBADSAD
BADFDFDDF
So here I want to have the second word with less characters to have white spaces at its end to match the length of the first word.
=&" " this definitely helps but I am unable to achieve the above scenario
Consider this screenshot:
In column B the length of each cell of column A is established with the formula =len(A1) copied down.
Cell D2 has the range name MaximumLength and the formula =max(B:B).
With that in place, you can create the padded values with this formula in cell G1, copied down:
=A1&REPT("*",MaximumLength-LEN(A1))
If you don't want to use the helper column and helper cell, you can use this array formula instead:
=A2&REPT("*",MAX(LEN(A1:A15))-LEN(A2))
This formula must be confirmed with Ctrl-Shift-Enter. It is advisable to use defined ranges, not whole columns in array formulas, hence the range in LEN(A1:A15). Adjust as desired.
I've used the "*" character so it is visible. Replace it with a space " " in your scenario.
You can add this formula to count maximum characters and use on some cell, because you will need to press a command for it to work, so every cell can't contain this formula, let's say it is on Z1:
=MAX(LEN($A:$A))
Certify to press ctrl+shift+enter on the formula
Then you use this formula on your cells:=REPT(" ";Z1-LEN(A2))&A2
Edit: Sorry, anwsered late, teylyn is more complete.
I have a column of survey responses to a question in which the data is an integer 1 through 5 inclusive followed by text (e.g., "5 stars - I love it"). I want to sum the integers in the 1st characters. The dirty way would be to create a new column in which all but the first character is stripped, but I thought I'd be slick and avoid this.
However, my CSE array formula seems to be summing COUNTA(...) instead of summing over the LEFT(...,1) in the cells
{=SUM(NUMBERVALUE(LEFT(AC$62:AC$9999,1)))}
I'm wondering why there's an implicit COUNTA included here and whether CSE formula is not a good way to approach this.
with data in A1 through A5, use:
=SUMPRODUCT(--LEFT(A1:A5,1))
As you see, the formula discards everything after the leading digit.
NOTE:
If you replace blanks with zeros, the formula will work.
EDIT#2:
You can avoid the helper column if you use the array formula:
=SUMPRODUCT(--(LEFT(IF(A1:A6="",0,A1:A6),1)))
Array formulas must be entered with Ctrl + Shift + Enter rather than just the Enter key.
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)
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.