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)
)
)
))
)
I want to analyze a string by many different patterns for numbers, dates and other strings. So I have an array of patterns I want to check in that order.
let patterns = [... "\\d{6}", "\\d{4}", "\\d" ] // to be extended :-)
let s = "IMG_123456_2006.10.03-13.52.59 Testfile_2009_5"
Starting with the first item in pattern I need a search in string s. If found, the string should be split in found parts e.g. "2006" and "2009" and the non matching parts. The remaining parts will be searched with the next pattern and so on. Assuming I already had the pattern defined for time/date in the middle which should be placed at the first item, the splitted string should look like:
"IMG_", "123456", "_", "2006.10.03-13.52.59", " Testfile_", "2009", "_", "5"
Can I use a build in functionality of regex.matches, or do I have to write everything by my own?
I already been able to find a match. But then I have to use the ranges to split the string and do it again and again for the remaining parts until no further matches are indicated. This will need a lot more calculations than I would expect using the results in match.numberOfRanges. Any small solutions available?
This is a follow up to A regex to detect periodic strings .
A period p of a string w is any positive integer p such that w[i]=w[i+p]
whenever both sides of this equation are defined. Let per(w) denote
the size of the smallest period of w . We say that a string w is
periodic iff per(w) <= |w|/2.
So informally a periodic string is just a string that is made up from a another string repeated at least once. The only complication is that at the end of the string we don't require a full copy of the repeated string as long as it is repeated in its entirety at least once.
For, example consider the string x = abcab. per(abcab) = 3 as x[1] = x[1+3] = a, x[2]=x[2+3] = b and there is no smaller period. The string abcab is therefore not periodic. However, the string ababa is periodic as per(ababa) = 2.
As more examples, abcabca, ababababa and abcabcabc are also periodic.
#horcruz, amongst others, gave a very nice regex to recognize a periodic string. It is
\b(\w*)(\w+\1)\2+\b
I would like to find all maximal periodic substrings in a longer string. These are sometimes called runs in the literature.
Formally a substring w is a maximal periodic substring if it is periodic and neither w[i-1] = w[i-1+p] nor w[j+1] = w[j+1-p]. Informally, the "run" cannot be contained in a larger "run"
with the same period.
The four maximal periodic substrings (runs) of string T = atattatt are T[4,5] = tt, T[7,8] = tt, T[1,4] = atat, T[2,8] = tattatt.
The string T = aabaabaaaacaacac contains the following 7 maximal periodic substrings (runs):
T[1,2] = aa, T[4,5] = aa, T[7,10] = aaaa, T[12,13] = aa, T[13,16] = acac, T[1,8] = aabaabaa, T[9,15] = aacaaca.
The string T = atatbatatb contains the following three runs. They are:
T[1, 4] = atat, T[6, 9] = atat and T[1, 10] = atatbatatb.
Is there a regex (with backreferences) that will capture all maximal
periodic substrings?
I don't really mind which flavor of regex but if it makes a difference, anything that the Python module re supports. However I would even be happy with PCRE if that makes the problem solvable.
(This question is partly copied from https://codegolf.stackexchange.com/questions/84592/compute-the-maximum-number-of-runs-possible-for-as-large-a-string-as-possible . )
Let's extend the regex version to the very powerful https://pypi.python.org/pypi/regex . This supports variable length lookbehinds for example.
This should do it, using Python's re module:
(?<=(.))(?=((\w*)(\w*(?!\1)\w\3)\4+))
Fiddle: https://regex101.com/r/aA9uJ0/2
Notes:
You must precede the string being scanned by a dummy character; the # in the fiddle. If that is a problem, it should be possible to work around it in the regex.
Get captured group 2 from each match to get the collection of maximal periodic substrings.
Haven't tried it with longer strings; performance may be an issue.
Explanation:
(?<=(.)) - look-behind to the character preceding the maximal periodic substring; captured as group 1
(?=...) - look-ahead, to ensure overlapping patterns are matched; see How to find overlapping matches with a regexp?
(...) - captures the maximal periodic substring (group 2)
(\w*)(\w*...\w\3)\4+ - #horcruz's regex, as proposed by OP
(?!\1) - negative look-ahead to group 1 to ensure the periodic substring is maximal
As pointed out by #ClasG, the result of my regex may be incomplete. This happens when two runs start at the same offset. Examples:
aabaab has 3 runs: aabaab, aa and aa. The first two runs start at the same offset. My regex will fail to return the shortest one.
atatbatatb has 3 runs: atatbatatb, atat, atat. Same problem here; my regex will only return the first and third run.
This may well be impossible to solve within the regex. As far as I know, there is no regex engine that is capable of returning two different matches that start at the same offset.
I see two possible solutions:
Ignore the missing runs. If I am not mistaken, then they are always duplicates; an identical run will follow within the same encapsulating run.
Do some postprocessing on the result. For every run found (let's call this X), scan earlier runs trying to find one that starts with the same character sequence (let's call this Y). When found (and not already 'used'), add an entry with the same character sequence as X, but the offset of Y.
I think it is not possible. Regular expressions cannot do complex nondeterministic jobs, even with backreferences. You need an algorithm for this.
This kind of depends on your input criteria... There is no infinite string of characters.. using back references you will be able to create a suitable representation of the last amount of occurrences of the pattern you wish to match.
\
Personally I would define buckets of length of input and then fill them.
I would then use automata to find patterns in the buckets and then finally coalesce them into larger patterns.
It's not how fast the RegEx is going to be in this case it's how fast you are going to be able to recognize a pattern and eliminate the invalid criterion.
I'd like to find a certain string in a webpage. I decided to use RegEx. (I know my RegExes are quite terrible, however, they work). My two expressions are very fast when used in Notepad++ (probably < 1s) and on Regex101, but they are horribly slow when used in AutoHotKey – about 2-5 minutes. How do I fix this?
sWindowInfo2 = http://www.archiwum.wyborcza.pl/Archiwum/1,0,4583161,20060208LU-DLO,Dzis_bedzie_Piast,.html
whr := ComObjCreate("WinHttp.WinHttpRequest.5.1")
whr.Open("GET", sWindowInfo2, false ), whr.Send()
whr.ResponseText
sPage := ""
sPage := whr.ResponseText
; get city name (if exists) – the following is very slooooow
if RegExMatch(sPage, "[\s\S]+<dzial>Gazeta\s(.+)<\/dzial>[\s\S]+")
{
sCity := RegExReplace(sPage, "[\s\S]+<dzial>Gazeta\s(.+)<\/dzial>[\s\S]+", "$1")
;MsgBox, % sCity
city := 1
}
if RegExMatch(sPage, "[\s\S]+<metryczka>GW\s(.+)\snr[\s\S]+")
{
sCity := RegExReplace(sPage, "[\s\S]+<metryczka>GW\s(.+)\snr[\s\S]+", "$1")
city := 1
}
EDIT:
In the page I provided the match is Lublin. Have a look at: https://regex101.com/r/qJ2pF8/1
You do not need to use RegExReplace to get the captured value. As per reference, you can pass the 3rd var into RegExMatch:
OutputVar
OutputVar is the unquoted name of a variable in which to store a match object, which can be used to retrieve the position, length and value of the overall match and of each captured subpattern, if any are present.
So, use a much simpler pattern:
FoundPos := RegExMatch(sPage, "<metryczka>GW\s(.+)\snr", SubPat) ;
It will return the position of the match, and will store "Lublin" in SubPat[1].
With this pattern, you avoid heavy backtracking you had with [\s\S]+<metryczka>GW\s(.+)\snr[\s\S]+ as the first [\s\S]+ matched up to the end of the string, and then backtracked to accommodate for the subsequent subpatterns. The longer the string, the slower the operation is.
I have to parse various strings and determine a prefix, number, and suffix. The problem is the strings can come in a wide variety of formats. The best way for me to think about how to parse it is to find the longest number in the string, then take everything before that as a prefix and everything after that as a suffix.
Some examples:
0001 - No prefix, Number = 0001, No suffix
1-0001 - Prefix = 1-, Number = 0001, No suffix
AAA001 - Prefix = AAA, Number = 001, No suffix
AAA 001.01 - Prefix = AAA , Number = 001, Suffix = .01
1_00001-01 - Prefix = 1_, Number = 00001, Suffix = -01
123AAA 001_01 - Prefix = 123AAA , Number = 001, Suffix = _01
The strings can come with any mixture of prefixes and suffixes, but the key point is the Number portion is always the longest sequential list of digits.
I've tried a variety of RegEx's that work with most but not all of these examples. I might be missing something, or perhaps a RegEx isn't the right way to go in this case?
(The RegEx should be .NET compatible)
UPDATE: For those that are interested, here's the C# code I came up with:
var regex = new System.Text.RegularExpressions.Regex(#"(\d+)");
if (regex.IsMatch(m_Key)) {
string value = "";
int length;
var matches = regex.Matches(m_Key);
foreach (var match in matches) {
if (match.Length >= length) {
value = match.Value;
length = match.Length;
}
}
var split = m_Key.Split(new String[] {value}, System.StringSplitOptions.RemoveEmptyEntries);
m_KeyCounter = value;
if (split.Length >= 1) m_KeyPrefix = split(0);
if (split.Length >= 2) m_KeySuffix = split(1);
}
You're right, this problem can't be solved purely by regular expressions. You can use regexes to "tokenize" (lexically analyze) the input but after that you'll need further processing (parsing).
So in this case I would tokenize the input with (for example) a simple regular expression search (\d+) and then process the tokens (parse). That would involve seeing if the current token is longer than the tokens seen before it.
To gain more understanding of the class of problems regular expressions "solve" and when parsing is needed, you might want to check out general compiler theory, specifically when regexes are used in the construction of a compiler (e.g. http://en.wikipedia.org/wiki/Book:Compiler_construction).
You're input isn't regular so, a regex won't do. I would iterate over the all groups of digits via (\d+) and find the longest and then build a new regex in the form of (.*)<number>(.*) to find your prefix/suffix.
Or if you're comfortable with string operations you can probably just find the start and end of the target group and use substr to find the pre/suf fix.
I don't think you can do this with one regex. I would find all digit sequences within the string (probably with a regex) and then I would select the longest with .NET code, and call Split().
This depends entirely on your Regexp engine. Check your Regexp environment for capturing, there might be something in it like the automatic variables in Perl.
OK, let's talk about your question:
Keep in mind, that both, NFA and DFA, of almost every Regexp engine are greedy, this means, that a (\d+) will always find the longest match, when it "stumbles" over it.
Now, what I can get from your example, is you always need middle portion of a number, try this:
/^(.*\D)?(\d+)(\D.*)?$/ig
The now look at variables $1, $2, $3. Not all of them will exist: if there are all three of them, $2 will hold your number in question, the other vars, parts of the prefix. when one of the prefixes is missing, only variable $1 and $2 will be set, you have to see for yourself, which one is the integer. If both prefix and suffix are missing, $1 will hold the number.
The idea is to make the engine "stumble" over the first few characters and start matching a long number in the middle.
Since the modifier /gis present, you can loop through all available combinations, that the machine finds, you can then simply take the one you like most or something.
This example is in PCRE, but I'm sure .NET has a compatible mode.