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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)
)
)
))
)
This is a cross-post from TeX, but it did not get any answers there. And since I assume the problem has more to do with my understanding of regular expressions (or better, lack thereof) than with LaTeX itself, StackOverflow may have been the better place to ask to begin with.
I would like to use BibTool (which was written in C, if this is of any consequence here) to enclose some strings in a bib-file in curly braces. The test bib entry looks like this:
#Article{Cite1,
author = {Adelbert, A.},
date = {2020},
journaltitle = {A Journal},
title = {A title with just \textit{Test} structure and some chemistry \ce{CO2}},
number = {2},
pages = {1--4},
volume = {1},
}
I have created the following BibTool resource file:
resource {biblatex}
preserve.keys = on
preserve.key.case = on
rewrite.rule = {"\\\(.*{.*}\)" "{{\1}}"}
The rewrite.rule is supposed to be the following:
Find all strings within any field that start with \, like \ce{}, \textit{}, etc. This is done by the \\ at the beginning of the regular expression.
When this string is found save the following in a group, denoted by \(\): A random string at the beginning, followed by {, a random string, followed by }; i.e. the string textit{Test}.
Write this string back into the same position, but enclose it in a double-set of curly braces "{{\1}}".
What it manages so far:
It apparently finds all commands starting with \.
It saves the strings and writes them back into the file.
So far, the code returns the following
#Article{Cite1,
Author = {Adelbert, A.},
Date = {2020},
JournalTitle = {A Journal},
Title = {A title with just {{textit{Test} structure and some chemistry {{ce{CO2}}}}}},
Number = {2},
Pages = {1--4},
Volume = {1},
}
You see it finds the strings and puts {{ at the beginning of each string. Unfortunately, it puts }} at the end of the field, not the string, so I now have 6 curly braces at the end of the title field. The braces do match, just two of them should be after {{textit{Test} not at the very end. I tried various constructions like rewrite.rule = {"\\\(.*{.*}\)$" "{{\1}}"}, rewrite.rule = {"\\\(.*{.*}\) ?$" "{{\1}}"}, rewrite.rule = {"\\\(.*{.*}\)*$" "{{\1}}"} but this all did not work.
When trying to get the \ back at the beginning of the string, using rewrite.rule = {"\\\(.*{.*}\)" "{{\\\1}}"} I get the \ back, but also thousands of {} until I get a Rewrite limit exceeded error.
I am not very good with regular expressions and would be happy for any comments.
My approach would use two phases. In the first phase I would process the macro with one argument and replace in the result the \ by a replacement representation (here ##). In the second pahe I simply replace ## by \.
In BibTool this looks as follows:
rewrite.rule {"\\\(\([a-zA-Z]+\|.\){[^{}]*}\)" "{##\1}"}
rewrite.rule {"##" "\\"}
Note, that in general the task depicted can not be solved with regular expressions...
The behavior of .* by default is to match as many characters as possible. This is called 'greedy matching' in regex terms.
Your pattern is likely matching the following on hitting the first \:
\textit{Test} structure and some chemistry \ce{CO2}}
Replacing the text to:
{{textit{Test} structure and some chemistry \ce{CO2}}}}
And then finding the next \ and replacing:
\ce{CO2}}}} becomes {{ce{CO2}}}}}}
Total effect:
{A title with just \textit{Test} structure and some chemistry \ce{CO2}}
{A title with just {{textit{Test} structure and some chemistry {{ce{CO2}}}}}}
To change the behaviour in most regex flavors you can put a ? after the quantifier: .*? to make it 'lazy', that is match the least amount of characters.
I have a string, and I want to extract, using regular expressions, groups of characters that are between the character : and the other character /.
typically, here is a string example I'm getting:
'abcd:45.72643,4.91203/Rou:hereanotherdata/defgh'
and so, I want to retrieved, 45.72643,4.91203 and also hereanotherdata
As they are both between characters : and /.
I tried with this syntax in a easier string where there is only 1 time the pattern,
[tt]=regexp(str,':(\w.*)/','match')
tt = ':45.72643,4.91203/'
but it works only if the pattern happens once. If I use it in string containing multiples times the pattern, I get all the string between the first : and the last /.
How can I mention that the pattern will occur multiple time, and how can I retrieve it?
Use lookaround and a lazy quantifier:
regexp(str, '(?<=:).+?(?=/)', 'match')
Example (Matlab R2016b):
>> str = 'abcd:45.72643,4.91203/Rou:hereanotherdata/defgh';
>> result = regexp(str, '(?<=:).+?(?=/)', 'match')
result =
1×2 cell array
'45.72643,4.91203' 'hereanotherdata'
In most languages this is hard to do with a single regexp. Ultimately you'll only ever get back the one string, and you want to get back multiple strings.
I've never used Matlab, so it may be possible in that language, but based on other languages, this is how I'd approach it...
I can't give you the exact code, but a search indicates that in Matlab there is a function called strsplit, example...
C = strsplit(data,':')
That should will break your original string up into an array of strings, using the ":" as the break point. You can then ignore the first array index (as it contains text before a ":"), loop the rest of the array and regexp to extract everything that comes before a "/".
So for instance...
'abcd:45.72643,4.91203/Rou:hereanotherdata/defgh'
Breaks down into an array with parts...
1 - 'abcd'
2 - '45.72643,4.91203/Rou'
3 - 'hereanotherdata/defgh'
Then Ignore 1, and extract everything before the "/" in 2 and 3.
As John Mawer and Adriaan mentioned, strsplit is a good place to start with. You can use it for both ':' and '/', but then you will not be able to determine where each of them started. If you do it with strsplit twice, you can know where the ':' starts :
A='abcd:45.72643,4.91203/Rou:hereanotherdata/defgh';
B=cellfun(#(x) strsplit(x,'/'),strsplit(A,':'),'uniformoutput',0);
Now B has cells that start with ':', and has two cells in each cell that contain '/' also. You can extract it with checking where B has more than one cell, and take the first of each of them:
C=cellfun(#(x) x{1},B(cellfun('length',B)>1),'uniformoutput',0)
C =
1×2 cell array
'45.72643,4.91203' 'hereanotherdata'
Starting in 16b you can use extractBetween:
>> str = 'abcd:45.72643,4.91203/Rou:hereanotherdata/defgh';
>> result = extractBetween(str,':','/')
result =
2×1 cell array
{'45.72643,4.91203'}
{'hereanotherdata' }
If all your text elements have the same number of delimiters this can be vectorized too.
I've got a file that uses an outdated macro to read 32 bit integers,
READ32(dest, src)
I need to replace all calls with
dest = readUint32(&src);
I'm trying to write a SED style Vim search & replace command, but not having luck.
I can match the 1st part using READ32([a-z]\+, cmd) using the / search prompt, but it does not seem to match in the :s syntax.
Here's what I finally figured out to work:
:%s/READ32(\(\a\+\),\(\a\+\)/\1 = readUint32(\&\2);
The trick is wrapping the values you want to store in \1 & \2 in \( and \) The other trick was you have to escape the & operator as & in vim replacement is "the whole match".
EDIT: improved further as I refined it:
:%s/READ32(\(\w\+\),\s*\(\w\+\)/\1 = readUint32(\&\2);
Changed \a to \w as I had variables with _ in them.
Added \s* to take care of white space issues between the , and second variable.
Now just trying to deal with c++ style variables of style class.variable.subvariable
EDIT 2:
replaced \w with [a-zA-Z0-9_.] to catch all of the ways my variables were named.
This should do what you want or at least get you started:
%s-READ32(\s*\(\i\+\)\s*,\s*\(\i\+\)\s*)-\1 = readUint32(\&\2);-g
I'd do the macro style again: hit * to 'highlight' search for READ32.
Now, we are going to record a macro (q..qq):
n (move to next match)
cwreadUint32Esc (change the function name)
wwdt, (delete the first argument)
"_dw (remove the redundant ,)
bbPa=Esc (insert the result variable appending = before readUint32)
A; (append ; to the end of the line)
Now you can just repeat the macro (1000#q).
I'm trying to retrieve a filename without the extension in ColdFusion. I am using the following function:
REMatchNoCase( "(.+?)(\.[^.]*$|$)" , "Doe, John 8.15.2012.docx" );
I would like this to return an array like: ["Doe, John 8.15.2012","docx"]
but instead I always get an array with one element - the entire filename:["Doe, John 8.15.2012.docx"]
I tried the regex string above on rexv.org and it works as expected, but not on ColdFusion. I got the string from this SO question: Regex: Get Filename Without Extension in One Shot?
Does ColdFusion use a different syntax? Or am I doing something wrong?
Thanks.
Why you're not getting expected results...
The reason you are getting a one-item array with the whole filename is because your pattern matches the entire filename, and matches once.
It is capturing the two groups, but rematch returns arrays of matches, not arrays of the captured groups, so you don't see those groups.
How to solve the problem...
If you are dealing with simple files (i.e. no .htaccess or similar), then the simplest solution is to just use...
ListLast( filename , '.' )
....to get only the file extension and to get the name without extension you can do...
rematch( '.+(?=\.[^.]+$)' , filename )
This uses a lookahead to ensure there is a . followed by at least one non-. at the end of the string, but (since it's a lookahead) it is excluded from the match (so you only get the pre-extension part in your match).
To deal with non-extensioned files (e.g. .htaccess or README) you can modify the above regex to .+(?=(?:\.[^.]+)?$) which basically does the same thing except making the extension optional. However, there isn't a trivial way to get update the ListLast method for these (guess you'd need to check len(extension) LT len(filename)-1 or similar).
(optional) Accessing captured groups...
If you want to get at the actual captured groups, the closest native way to do this in CF is using the refind function, with the fourth argument set to true - however, this only gives you positions and lengths - requiring that you use mid to extract them yourself.
For this reason (amongst many others), I've created an improved regex implementation for CF, called cfRegex, which lets you return the group text directly (i.e. no messing around with mid).
If you wanted to use cfRegex, you can do so with your original pattern like so:
RegexMatch( '(.+?)(\.[^.]*$|$)' , filename , 1 , 0 , 'groups' )
Or with named arguments:
RegexMatch( pattern='(.+?)(\.[^.]*$|$)' , text=filename , returntype='groups' )
And you get returned an array of matches, within each element being an array of the captured groups for that match.
If you're doing lots of regex work dealing with captured groups, cfRegex is definitely better than doing it with CF's re methods.
If all you care about is getting the extension and/or the filename with extension excluded then the previous examples above are sufficient.
#Peter's response is great, however the approach is perhaps a bit longer-winded than necessary. One can do this with reMatch() with a slight tweak to the regex.
<cfscript>
param name="URL.filename";
sRegex = "^.+?(?=(?:\.[^.]+?)?$)";
aMatch = reMatch(sRegex, URL.filename);
writeDump(aMatch);
</cfscript>
This works on the following filename patterns:
foo.bar
foo
.htaccess
John 8.15.2012.docx
Explanation of the regex:
^ From the beginning of the string
.+? One or more (+) characters (.), but the fewest (?) that will work with the rest of the regex. This is the file name.
(?=) Look ahead. Make sure the stuff in here appears in the string, but don't actually match it. This is the key bit to NOT return any file extension that might be present.
(?: Group this stuff together, but don't remember it for a back reference.
. A dot. This is the separator between file name and file extension.
[^.]+? One or more (+) single ([]) non-dot characters (^.), again matching the fewest possible (?) that will allow the regex as a whole to work.
? (This is the one after the (?:) group). Zero or one of those groups: ie: zero or one file extensions.
$ To the end of the string
I've only tested with those four file name patterns, but it seems to work OK. Other people might be able to finetune it.
A few more ways of achieving the same result. They all execute in roughly the same amount of time.
<cfscript>
str = 'Doe, John 8.15.2012.docx';
// sans regex
arr1 = [
reverse( listRest( reverse( str ), '.' ) ),
listLast( str, '.' )
];
// using Java String lastIndexOf()
arr2 = [
str.substring( 0, str.lastIndexOf( '.' ) ),
str.substring( str.lastIndexOf( '.' ) + 1 )
];
// using listToArray with non-filename safe character replace
arr3 = listToArray( str.replaceAll( '\.([^\.]+)$', '|$1' ), '|' );
</cfscript>