I have 3 rows. All of them have a (number)&(letter) format. For example, Say 5&b is present on 3 rows. My question is:
How do I do sum just the numbers? I have tried:
=SUM(REGEXEXTRACT(B3:B, "(.)&."))
I am trying to add the first value of all the rows.
Try:
=ARRAYFORMULA(SUM(--REGEXEXTRACT(B3:B27, "(\d+)&\w+")))
\d+ stands for one or more digits. \w+ will stand for one or more words.
You must use ARRAYFORMULA to extract data from the whole array B3:B27.
-- will convert text to number
Related
I'd need to split or extract only numbers made of 8 digits from a string in Google Sheets.
I've tried with SPLIT or REGEXREPLACE but I can't find a way to get only the numbers of that length, I only get all the numbers in the string!
For example I'm using
=SPLIT(lower(N2),"qwertyuiopasdfghjklzxcvbnm`-=[]\;' ,./!:##$%^&*()")
but I get all the numbers while I only need 8 digits numbers.
This may be a test value:
00150412632BBHBBLD 12458 32354 1312548896 ACT inv 62345471
I only need to extract "62345471" and nothing else!
Could you please help me out?
Many thanks!
Please use the following formula for a single cell.
Drag it down for more cells.
=INDEX(TRANSPOSE(QUERY(TRANSPOSE(IF(LEN(SPLIT(REGEXREPLACE(A2&" ","\D+"," ")," "))=8,
SPLIT(REGEXREPLACE(A2&" ","\D+"," ")," "),"")),"where Col1 is not null ",0)))
Functions used:
QUERY
INDEX
TRANSPOSE
IF
LEN
SPLIT
REGEXREPLACE
If you only need to do this for one cell (or you have your heart set on dragging the formula down into individual cells), use the following formula:
=REGEXEXTRACT(" "&N2&" ","\s(\d{8})\s")
However, I suspect you want to process the eight-digit number out of all cells running N2:N. If that is the case, clear whatever will be your results column (including any headers) and place the following in the top cell of that otherwise cleared results column:
=ArrayFormula({"Your Header"; IF(N2:N="",,IFERROR(REGEXEXTRACT(" "&N2:N&" ","\s(\d{8})\s")))})
Replace the header text Your Header with whatever you want your actual header text to be. The formula will show that header text and will return all results for all rows where N2:N is not null. Where no eight-digit number is found, null will be returned.
By prepending and appending a space to the N2:N raw strings before processing, spaces before and after string components can be used to determine where only eight digits exist together (as opposed to eight digits within a longer string of digits).
The only assumption here is that there are, in fact, spaces between string components. I did not assume that the eight-digit number will always be in a certain position (e.g., first, last) within the string.
Try this, take a look at Example sheet
=FILTER(TRANSPOSE(SPLIT(B2," ")),LEN(TRANSPOSE(SPLIT(B2," ")))=8)
Or this to get them all.
=JOIN(" ,",FILTER(TRANSPOSE(SPLIT(B2," ")),LEN(TRANSPOSE(SPLIT(B2," ")))=8))
Explanation
SPLIT with the dilimiter set to " " space TRANSPOSE and FILTER TRANSPOSE(SPLIT(B2," ") with the condition1 set to LEN(TRANSPOSE(SPLIT(B2," "))) is = 8
JOIN the outputed column whith " ," to gat all occurrences of number with a length of 8
Note: to get the numbers with the length of N just replace 8 in the FILTER function with a cell refrence.
Using this on a cell worked just fine for me:
(cell_with_data)=REGEXEXTRACT(A1,"[0-9]{8}$")
I need to validate with regex a date in format yyyy-mm-dd (2019-12-31) that should be within the range 2019-12-20 - 2020-01-10.
What would be the regex for this?
Thanks
Regex only deal with characters. so we have to work out at each position in the date what are the valid characters.
The first part is easy. The first two characters have to be 20
Now it gets complicated the next character can be a 1 or a 2 but what follows depends on the value of that character so we split the rest of the regex into two sections the first if the third character matches 1 and the second if it matches 2
We know that if the third character is a 1 then what must follow is the characters 9-12- as the range starts at 2019-12-20 now for the day part. The 9th character is the tens for the day this can only be 2 or 3 as we are already in the last month and the minimum date is 20. The last character can be any digit 0-9. This gives us a day match of [23][0-9]. Putting this together we now have a pattern for years starting 2019 as 19-12-[23][0-9]
It the third character is a 2 then we can match up to the day part of the date a gain as the range ends in January. This gives us a partial match of 20-01- leaving us to work on the day part. Hear we know that the first character of the day can either be a 1 or 0 however if it's a 1 then the last character must be a 0 and if it's a 0 then the last character can only be in the range 1 to 9. This give us another alteration (?:0[1-9]|10) Putting the second part together we get 20-01-(?:0[1-9]|10).
Combining these together gives the final regex 20(?:19-12-[23][0-9]|20-01-(?:0[1-9]|10))
Note that I'm assuming that the date you are testing against is a validly formatted date.
Try this:
(2019|2020)\-(12|01)\-([0-3][0-9]|[0-9])
But be aware that this will allow number up to where the first digit is between zero and three and the second digit between zero and nine for the dd value. You could specify all numbers you want to allow (from 20 to 10) like this (20|21|22|23|24|25|26|27|28|29|30|31|01|1|02|2|03|3|04|4|05|5|06|6|07|7|08|8|09|9|10).
(2019|2020)\-(12|01)\-(20|21|22|23|24|25|26|27|28|29|30|31|01|1|02|2|03|3|04|4|05|5|06|6|07|7|08|8|09|9|10)
But honestly... Regular-Expressions are not the right tool for this. RegExp gives a mask to something, not a logical context. Use regex to extract the data/value from a string and validate those values using another language.
The above 2nd Regex will, f.e. match your dates, but also values outside of this range since there is no context between 2019|2020 and the second group 12|01 so they match values like 2019-12-11 but also 2020-12-11.
To only match the values you want this will be a really large regex like this (inner brackets only if you need them) ((2019)-(12)-(20)|(2019)-(12)-(21)|(2019)-(12)-(22)|...) and continue with all possible dates - and ask yourself: what would you do if you find such a regex in a project you have to work with ;)
Better solution (quick and dirty, there might be better solutions):
(?<yyyy>20[0-9]{2})\-(?<mm>[01][0-9]|[0-9])\-(?<dd>[0-3][0-9]|[0-9])
This way you have three named groups (yyyy, mm, dd) you can access and validate the matched values... The regex is smaller, you have a better association between code and regex and both are easier to maintain.
I have a dataframe and in 2 columns i have to change values removing all that is not a number or ".". The final result should be only dotted numbers. thank you very much for those of you can help me. I attach how columns areenter image description here
The regex you're after is:
/[0-9.]+/
Explanation:
[0-9.] Looks for one character that is a digit or .
+ repeats last character unlimited times
let regex = /[0-9.]+/;
console.log("1.0.0".match(regex)); //["1.0.0"]
console.log("1.2.4".match(regex)); //["1.2.4"]
console.log("NaN".match(regex)); //[]
console.log("1.1".match(regex)); //["1.1"]
console.log("6.1.61.1".match(regex)); //["6.1.61.1"]
console.log("4.0.3 and up".match(regex)); //["4.0.3"]
This will match any combination of numbers and . of any character length.
For example:
Sample sheet.
As the title says, given a column of arbitrary number of words of arbitrary length, Want a single ArrayFormula to get the first letters of all words in the said column.
I have tried two methods, seen in sample sheet.
1) Using SPLIT and ARRAYFORMULA, can get it one cell but cannot extend down column.
2) Using 2 REGEXEXTRACT, can get for first 2 initials and extend down
But is it possible to get for arbitrary number of words for whole column using ArrayFormula.
Is it possible to use REGEXEXTRACT to return the first letters of many words?
This replaces every word with the captured first letter
=ARRAYFORMULA(UPPER(REGEXREPLACE(A1:A6,"(\w)\S*\s?","$1")))
I have a cell with some text content.
For example: "Red, shirt, size,"
I need to count how many times the comma is used in this cell.
The result should be "3"
Any ideas?
You can use the following formula: LEN(Cell)-LEN(SUBSTITUTE(Cell;"YourCharacter";""))
In your case, the formula would be: LEN(Cell)-LEN(SUBSTITUTE(Cell;",";"").
LEN(Cell) does the following: Counts the number of characters in your cell.
LEN(SUBSTITUTE(Cell;"YourCharacter";"")) counts the number of characters in your cell without the character ",". By subtracting the second formula, you get the number of occurrences of your character.
You can use LEN() function as below
=LEN(cell)-LEN(SUBSTITUTE(cell;",";""))