I'm trying to get the list of all digits preceding a hyphen in a given string (let's say in cell A1), using a Google Sheets regex formula :
=REGEXEXTRACT(A1, "\d-")
My problem is that it only returns the first match... how can I get all matches?
Example text:
"A1-Nutrition;A2-ActPhysiq;A2-BioMeta;A2-Patho-jour;A2-StgMrktg2;H2-Bioth2/EtudeCas;H2-Bioth2/Gemmo;H2-Bioth2/Oligo;H2-Bioth2/Opo;H2-Bioth2/Organo;H3-Endocrino;H3-Génétiq"
My formula returns 1-, whereas I want to get 1-2-2-2-2-2-2-2-2-2-3-3- (either as an array or concatenated text).
I know I could use a script or another function (like SPLIT) to achieve the desired result, but what I really want to know is how I could get a re2 regular expression to return such multiple matches in a "REGEX.*" Google Sheets formula.
Something like the "global - Don't return after first match" option on regex101.com
I've also tried removing the undesired text with REGEXREPLACE, with no success either (I couldn't get rid of other digits not preceding a hyphen).
Any help appreciated!
Thanks :)
You can actually do this in a single formula using regexreplace to surround all the values with a capture group instead of replacing the text:
=join("",REGEXEXTRACT(A1,REGEXREPLACE(A1,"(\d-)","($1)")))
basically what it does is surround all instances of the \d- with a "capture group" then using regex extract, it neatly returns all the captures. if you want to join it back into a single string you can just use join to pack it back into a single cell:
You may create your own custom function in the Script Editor:
function ExtractAllRegex(input, pattern,groupId) {
return [Array.from(input.matchAll(new RegExp(pattern,'g')), x=>x[groupId])];
}
Or, if you need to return all matches in a single cell joined with some separator:
function ExtractAllRegex(input, pattern,groupId,separator) {
return Array.from(input.matchAll(new RegExp(pattern,'g')), x=>x[groupId]).join(separator);
}
Then, just call it like =ExtractAllRegex(A1, "\d-", 0, ", ").
Description:
input - current cell value
pattern - regex pattern
groupId - Capturing group ID you want to extract
separator - text used to join the matched results.
Edit
I came up with more general solution:
=regexreplace(A1,"(.)?(\d-)|(.)","$2")
It replaces any text except the second group match (\d-) with just the second group $2.
"(.)?(\d-)|(.)"
1 2 3
Groups are in ()
---------------------------------------
"$2" -- means return the group number 2
Learn regular expressions: https://regexone.com
Try this formula:
=regexreplace(regexreplace(A1,"[^\-0-9]",""),"(\d-)|(.)","$1")
It will handle string like this:
"A1-Nutrition;A2-ActPhysiq;A2-BioM---eta;A2-PH3-Généti***566*9q"
with output:
1-2-2-2-3-
I wasn't able to get the accepted answer to work for my case. I'd like to do it that way, but needed a quick solution and went with the following:
Input:
1111 days, 123 hours 1234 minutes and 121 seconds
Expected output:
1111 123 1234 121
Formula:
=split(REGEXREPLACE(C26,"[a-z,]"," ")," ")
The shortest possible regex:
=regexreplace(A1,".?(\d-)|.", "$1")
Which returns 1-2-2-2-2-2-2-2-2-2-3-3- for "A1-Nutrition;A2-ActPhysiq;A2-BioMeta;A2-Patho-jour;A2-StgMrktg2;H2-Bioth2/EtudeCas;H2-Bioth2/Gemmo;H2-Bioth2/Oligo;H2-Bioth2/Opo;H2-Bioth2/Organo;H3-Endocrino;H3-Génétiq".
Explanation of regex:
.? -- optional character
(\d-) -- capture group 1 with a digit followed by a dash (specify (\d+-) multiple digits)
| -- logical or
. -- any character
the replacement "$1" uses just the capture group 1, and discards anything else
Learn more about regex: https://twiki.org/cgi-bin/view/Codev/TWikiPresentation2018x10x14Regex
This seems to work and I have tried to verify it.
The logic is
(1) Replace letter followed by hyphen with nothing
(2) Replace any digit not followed by a hyphen with nothing
(3) Replace everything which is not a digit or hyphen with nothing
=regexreplace(A1,"[a-zA-Z]-|[0-9][^-]|[a-zA-Z;/é]","")
Result
1-2-2-2-2-2-2-2-2-2-3-3-
Analysis
I had to step through these procedurally to convince myself that this was correct. According to this reference when there are alternatives separated by the pipe symbol, regex should match them in order left-to-right. The above formula doesn't work properly unless rule 1 comes first (otherwise it reduces all characters except a digit or hyphen to null before rule (1) can come into play and you get an extra hyphen from "Patho-jour").
Here are some examples of how I think it must deal with the text
The solution to capture groups with RegexReplace and then do the RegexExctract works here too, but there is a catch.
=join("",REGEXEXTRACT(A1,REGEXREPLACE(A1,"(\d-)","($1)")))
If the cell that you are trying to get the values has Special Characters like parentheses "(" or question mark "?" the solution provided won´t work.
In my case, I was trying to list all “variables text” contained in the cell. Those “variables text “ was wrote inside like that: “{example_name}”. But the full content of the cell had special characters making the regex formula do break. When I removed theses specials characters, then I could list all captured groups like the solution did.
There are two general ('Excel' / 'native' / non-Apps Script) solutions to return an array of regex matches in the style of REGEXEXTRACT:
Method 1)
insert a delimiter around matches, remove junk, and call SPLIT
Regexes work by iterating over the string from left to right, and 'consuming'. If we are careful to consume junk values, we can throw them away.
(This gets around the problem faced by the currently accepted solution, which is that as Carlos Eduardo Oliveira mentions, it will obviously fail if the corpus text contains special regex characters.)
First we pick a delimiter, which must not already exist in the text. The proper way to do this is to parse the text to temporarily replace our delimiter with a "temporary delimiter", like if we were going to use commas "," we'd first replace all existing commas with something like "<<QUOTED-COMMA>>" then un-replace them later. BUT, for simplicity's sake, we'll just grab a random character such as from the private-use unicode blocks and use it as our special delimiter (note that it is 2 bytes... google spreadsheets might not count bytes in graphemes in a consistent way, but we'll be careful later).
=SPLIT(
LAMBDA(temp,
MID(temp, 1, LEN(temp)-LEN(""))
)(
REGEXREPLACE(
"xyzSixSpaces:[ ]123ThreeSpaces:[ ]aaaa 12345",".*?( |$)",
"$1"
)
),
""
)
We just use a lambda to define temp="match1match2match3", then use that to remove the last delimiter into "match1match2match3", then SPLIT it.
Taking COLUMNS of the result will prove that the correct result is returned, i.e. {" ", " ", " "}.
This is a particularly good function to turn into a Named Function, and call it something like REGEXGLOBALEXTRACT(text,regex) or REGEXALLEXTRACT(text,regex), e.g.:
=SPLIT(
LAMBDA(temp,
MID(temp, 1, LEN(temp)-LEN(""))
)(
REGEXREPLACE(
text,
".*?("®ex&"|$)",
"$1"
)
),
""
)
Method 2)
use recursion
With LAMBDA (i.e. lets you define a function like any other programming language), you can use some tricks from the well-studied lambda calculus and function programming: you have access to recursion. Defining a recursive function is confusing because there's no easy way for it to refer to itself, so you have to use a trick/convention:
trick for recursive functions: to actually define a function f which needs to refer to itself, instead define a function that takes a parameter of itself and returns the function you actually want; pass in this 'convention' to the Y-combinator to turn it into an actual recursive function
The plumbing which takes such a function work is called the Y-combinator. Here is a good article to understand it if you have some programming background.
For example to get the result of 5! (5 factorial, i.e. implement our own FACT(5)), we could define:
Named Function Y(f)=LAMBDA(f, (LAMBDA(x,x(x)))( LAMBDA(x, f(LAMBDA(y, x(x)(y)))) ) ) (this is the Y-combinator and is magic; you don't have to understand it to use it)
Named Function MY_FACTORIAL(n)=
Y(LAMBDA(self,
LAMBDA(n,
IF(n=0, 1, n*self(n-1))
)
))
result of MY_FACTORIAL(5): 120
The Y-combinator makes writing recursive functions look relatively easy, like an introduction to programming class. I'm using Named Functions for clarity, but you could just dump it all together at the expense of sanity...
=LAMBDA(Y,
Y(LAMBDA(self, LAMBDA(n, IF(n=0,1,n*self(n-1))) ))(5)
)(
LAMBDA(f, (LAMBDA(x,x(x)))( LAMBDA(x, f(LAMBDA(y, x(x)(y)))) ) )
)
How does this apply to the problem at hand? Well a recursive solution is as follows:
in pseudocode below, I use 'function' instead of LAMBDA, but it's the same thing:
// code to get around the fact that you can't have 0-length arrays
function emptyList() {
return {"ignore this value"}
}
function listToArray(myList) {
return OFFSET(myList,0,1)
}
function allMatches(text, regex) {
allMatchesHelper(emptyList(), text, regex)
}
function allMatchesHelper(resultsToReturn, text, regex) {
currentMatch = REGEXEXTRACT(...)
if (currentMatch succeeds) {
textWithoutMatch = SUBSTITUTE(text, currentMatch, "", 1)
return allMatches(
{resultsToReturn,currentMatch},
textWithoutMatch,
regex
)
} else {
return listToArray(resultsToReturn)
}
}
Unfortunately, the recursive approach is quadratic order of growth (because it's appending the results over and over to itself, while recreating the giant search string with smaller and smaller bites taken out of it, so 1+2+3+4+5+... = big^2, which can add up to a lot of time), so may be slow if you have many many matches. It's better to stay inside the regex engine for speed, since it's probably highly optimized.
You could of course avoid using Named Functions by doing temporary bindings with LAMBDA(varName, expr)(varValue) if you want to use varName in an expression. (You can define this pattern as a Named Function =cont(varValue) to invert the order of the parameters to keep code cleaner, or not.)
Whenever I use varName = varValue, write that instead.
to see if a match succeeds, use ISNA(...)
It would look something like:
Named Function allMatches(resultsToReturn, text, regex):
UNTESTED:
LAMBDA(helper,
OFFSET(
helper({"ignore"}, text, regex),
0,1)
)(
Y(LAMBDA(helperItself,
LAMBDA(results, partialText,
LAMBDA(currentMatch,
IF(ISNA(currentMatch),
results,
LAMBDA(textWithoutMatch,
helperItself({results,currentMatch}, textWithoutMatch)
)(
SUBSTITUTE(partialText, currentMatch, "", 1)
)
)
)(
REGEXEXTRACT(partialText, regex)
)
)
))
)
Here is a sample string.
"BLAH, blah, going to the store &^5, light Version 12.7(2)L6, anyway
plus other stuff Version 3.3.4.6. Then goes on an on for several lines..."
I want to capture only the first version number without including the word version if possible but not include the periods and parenthesis. The result would stop when it encounters a comma. The result would be:
"1272L6"
I don't want it to include other instances of version in the text. Can this be done?
I've tried (?<=version)[^,]* I know it does not address removing the periods and parens and does not address the subsequent versions.
This exact RegEx, maybe not the best solution, but it might help you to get 1272L6:
([0-9]{2})\.([0-9]{1})\(([0-9]{1})\)([A-Z]{1}[0-9]{1})
It creates four groups (where $1$2$3$4 is your target 1272L6) and passes ., ) and (.
You might change {1} to other numbers of repetitions, such as {1,2}.
Assuming the version number is fixed on format but not on the specific digits or letters, you could do this.
String s = "this is a test 12.7(2)L6, 13.7(2)L6, 14.7(2)L6";
String reg = "(\\d\\d\\.\\d\\(\\d\\)[A-Z]\\d),";
Matcher m = Pattern.compile(reg).matcher(s);
if (m.find()) { // should only find first one
System.out.println(m.group(1).replaceAll("[.()]", ""));
}
I use the following regex in SSRS to test for a particular column name in a parameter:
=IIf(InStr(Join(Parameters!ColumnNames.Value, ","), "x"), False, True)
This will hide a column on a report if it is not one of the chosen columns. This works just fine if there is not another column called "xy". The string being tested may be "z,x,w", in which case the test works fine; but it may also be "z,xy,w", in which case it will find "x" and display both "x" and "xy".
I tried checking for "x," which only works if "x" is not the last character of the string. I need to know the syntax to check for both "x," OR "x as the last piece of the string". Unfortunately "x" can have any length. The basic problem is I do not know how to use an OR in the IIF statement.
I tried the most obvious ways and kept getting errors. Using "\b" also does not work because there are no spaces in the string (so word boundaries are not applicable).
What you can do is add the delimiter to your check, so that way you're checking the exact string only and not any that just include it:
=IIf
(InStr("," & Join(Parameters!ColumnNames.Value, ",") & ",", ",x,") > 0
, False
, True)
So this will catch x but not xy.
One thing to note:
I have added a check to see of InStr > 0, as this returns an integer and not a boolean.
You want to match a specific column name in an array of column names but do this on a single line to include in the IIF statement.
Based on the last technique suggested in How can I quickly determine if a string exists within an array? your code would need to be.
=IIf((UBound(Filter(Parameters!ColumnNames.Value, "x", True, compare)) > -1), False, True)
It doesn't look like there is an actual Regex anywhere?
I'm trying to find any occurrences of a character repeating more than 2 times in a user entered string. I have this, but it doesn't go into the if statement.
password = asDFwe23df333
s = re.compile('((\w)\2{2,})')
m = s.search(password)
if m:
print ("Password cannot contain 3 or more of the same characters in a row\n")
sys.exit(0)
You need to prefix your regex with the letter 'r', like so:
s = re.compile(r'((\w)\2{2,})')
If you don't do that, then you'll have to double up on all your backslashes since Python normally treats backlashes like an escape character in its normal strings. Since that makes regexes even harder to read then they normally are, most regexes in Python include that prefix.
Also, in your included code your password isn't in quotes, but I'm assuming it has quotes in your code.
Can't you simply go through the whole string and everytime you found a character equal to the previous, you incremented a counter, till it reached the value of 3? If the character was different from the previous, it would only be a matter of setting the counter back to 0.
EDIT:
Or, you can use:
s = 'aaabbb'
re.findall(r'((\w)\2{2,})', s)
And check if the list returned by the second line has any elements.
well i am currently writing a script that is meant to check the logs of another script i wrote to see if it has had three or more unsuccessful pings in a row before a successful one, this is just barebones at the moment but it should look something like this
fileread,x,C:\Users\Michael\Desktop\ping.txt
result:=RegExMatch(%x% ,failure success)
msgbox,,, The file is = %x% `n the result is = %result%
now the file that is trying to read is
success failure success
and for some reason, when it reads the file it says that the variable %x% 'contains illegal characters
when i copy and paste the contents of ping.txt into the script and save it as a variable it works
i have made sure that the file has windows line endings CR +LF
i have assigned the variable generated in file read as another variable thus stripping any trailing or leading whitespace characters
the file is encoded in ANSI and still has the problem with UTF8
Function parameters take variable names without the % symbol, simply remove them.
I also want to point out that if the second parameter is meant to be a regular expression,
instead of a variable containing a regular expression, you will need quotes around it.
As is your script passes an empty string as the pattern which will always return 1
(failure is interpreted as a variable with an empty string associated with it.).
To quote Lexikos:
"An empty string, when compiled as a regex pattern, will match exactly
zero characters at whatever position you attempt to match it. Think of
it this way: For any position n in any string, the next 0 characters
are always the same."
Because you are simply truth testing,
or finding the index I want to point out that Autohotkey has a useful shorthand operator for this.
string := "this is a test"
f1::
result := RegExMatch(string, "\sis")
traytip,, %result%
Return
f2::
result := string ~= "\sis"
traytip,, % result
Return
These hotkeys both do the same thing; the second uses the shorthand operator ~=
and notice how the traytip parameter in the second example has only one %
When you start a command parameter with a % that starts an expression,
and within an expression variables are not enclosed with %.
The ternary operator ?: is also very useful:
string := "this is a test"
f3::traytip,, % (result := string ~= "\sis") ? (result) : ("nothing")
It might look complicated but it's very simple.
Think of
% as if
? as then
: as else
If (true) then (a) else (b)
% (true) ? (a) : (b)
A variable will be evaluated as False if 0 (or nothing) is assigned to it.
But in this example "\sis" is matched and the index of the space is returned (5),
so it is evaluated as True.
You can read more about variables and operators here:
http://l.autohotkey.net/docs/Variables.htm