My text string is in cell D2:
Decision, ERC Case No. 2009-094 MC, In the Matter of the Application for Authority to Secure Loan from the National Electrification Administration (NEA), with Prayer for Issuance of Provisional Authority, Dinagat Island Electric Cooperative, Inc. (DIELCO) applicant(12/29/2011)
This function:
=regexextract(D2,"\([A-Z]*\)")
will grab (NEA) but not (DIELCO)
I would like it to extract both (NEA) and (DIELCO)
You can use capture groups, which will cause regexextract() to return an array. You can use this as the cell result, in which case you will get a range of results, or you can feed the array to another function to reformat it to your purpose. For example:
regexextract( "abracadabra" ; "(bra).*(bra)" )
will return the array:
{bra,bra}
Another approach would be to use regexreplace(). This has the advantage that the replace is global (like s/pattern/replacement/g), so you do not need to know the number of results in advance. For example:
regexreplace( "aBRAcadaBRA" ; "[a-z]+" ; "..." )
will return the string:
...BRA...BRA
Here are two solutions, one using the specific terms in the author's example, the other one expanding on the author's sample regex pattern which appears to match all ALLCAPS terms. I'm not sure which is wanted, so I gave both.
(Put the block of text in A1)
Generic solution for all words in ALLCAPS
=regexreplace(regexreplace(REGEXREPLACE(A1,"\b\w[^A-Z]*\b","|"),"\W+","|"),"^\||\|$","")
Result:
ERC|MC|NEA|DIELCO
NB: The brunt of the work is in the CAPITALIZED formula, the lowercase functions are just for cleanup.
If you want space separation, the formula is a little simpler:
=trim(regexreplace(REGEXREPLACE(A1,"\b\w[^A-Z]*\b"," "),"\W+"," "))
Result:
ERC MC NEA DIELCO
(One way I like playing with regex in google spreadsheets is to read the regex pattern from another cell so I can change it without having to edit or re-paste into all the cells using that pattern. This looks so:
Cell A1:
Block of text
Cell B1 (no quote marks):
\b\w[^A-Z]*\b
Formula, in any cell:
=trim(regexreplace(REGEXREPLACE(A1,B$1," "),"\W+"," "))
By anchoring it to B$1, I can fill all my rows at once and the reference won't increment.)
Previous answer:
Specific solution for selected terms (ERC, DIELCO)
=regexreplace(join("|",IF(REGEXMATCH(A1,"ERC"),"ERC",""),IF(REGEXMATCH(A1,"DIELCO"),"DIELCO","")),"(^\||\|$)","")
Result:
ERC|DIELCO
As before, the brunt of the work is in the CAPITALIZED formula, the lowercase functions are just for cleanup.
This formula will find any ERC or DIELCO, or both in the block of text. The initial order doesn't matter, but the output will always be ERC followed by DIELCO (the order of appearance is lost). This fixes the shortcoming with the previous answer using "(bra).*(bra)" in that isolated ERC or DIELCO can still be matched.
This also has a simpler form with space separation:
=trim(join(" ",IF(REGEXMATCH(A1,"ERC"),"ERC",""),IF(REGEXMATCH(A1,"DIELCO"),"DIELCO","")))
Result:
ERC DIELCO
Please try:
=SPLIT(regexreplace(A1 ; "(?s)(.)?\(([A-Z]+)\)|(.)" ; "๐งธ$2");"๐งธ")
or
=REGEXEXTRACT(A1;"\Q"®EXREPLACE(A1;"\([A-Z]+\)";"\\E(.*)\\Q")&"\E")
Related
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-Gรฉnรฉtiq"
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-Gรฉnรฉti***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-Gรฉnรฉtiq".
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;/รฉ]","")
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-)","($1)")))
If the cell that you are trying to get the values has Special Characters like parentheses "(" or question mark "?" the solution provided wonยดt work.
In my case, I was trying to list all โvariables textโ contained in the cell. Those โvariables text โ was wrote inside like that: โ{example_name}โ. But the full content of the cell had special characters making the regex formula do break. When I removed theses specials characters, then I could list all captured groups like the solution did.
There are two general ('Excel' / 'native' / non-Apps Script) solutions to return an array of regex matches in the style of REGEXEXTRACT:
Method 1)
insert a delimiter around matches, remove junk, and call SPLIT
Regexes work by iterating over the string from left to right, and 'consuming'. If we are careful to consume junk values, we can throw them away.
(This gets around the problem faced by the currently accepted solution, which is that as Carlos Eduardo Oliveira mentions, it will obviously fail if the corpus text contains special regex characters.)
First we pick a delimiter, which must not already exist in the text. The proper way to do this is to parse the text to temporarily replace our delimiter with a "temporary delimiter", like if we were going to use commas "," we'd first replace all existing commas with something like "<<QUOTED-COMMA>>" then un-replace them later. BUT, for simplicity's sake, we'll just grab a random character such as ๏ท from the private-use unicode blocks and use it as our special delimiter (note that it is 2 bytes... google spreadsheets might not count bytes in graphemes in a consistent way, but we'll be careful later).
=SPLIT(
LAMBDA(temp,
MID(temp, 1, LEN(temp)-LEN("๏ท"))
)(
REGEXREPLACE(
"xyzSixSpaces:[ ]123ThreeSpaces:[ ]aaaa 12345",".*?( |$)",
"$1๏ท"
)
),
"๏ท"
)
We just use a lambda to define temp="match1๏ทmatch2๏ทmatch3๏ท", then use that to remove the last delimiter into "match1๏ทmatch2๏ทmatch3", then SPLIT it.
Taking COLUMNS of the result will prove that the correct result is returned, i.e. {" ", " ", " "}.
This is a particularly good function to turn into a Named Function, and call it something like REGEXGLOBALEXTRACT(text,regex) or REGEXALLEXTRACT(text,regex), e.g.:
=SPLIT(
LAMBDA(temp,
MID(temp, 1, LEN(temp)-LEN("๏ท"))
)(
REGEXREPLACE(
text,
".*?("®ex&"|$)",
"$1๏ท"
)
),
"๏ท"
)
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 have a text in cell (A1) like this:
โ๐๐
๐
โ๏ธ๐๐๐ช๐ง๐ง
I want to extract the unique emojis from this cell into separate cells:
โ๐๐
โ๏ธ๐๐๐ช๐ง
Is this possible?
You want to put each character of โ๐๐
๐
โ๏ธ๐๐๐ช๐ง๐ง to each cell by splitting using the built-in function of Google Spreadsheet.
Sample formula:
=SPLIT(REGEXREPLACE(A1,"(.)","$1#"),"#")
โ๐๐
๐
โ๏ธ๐๐๐ช๐ง๐ง is put in a cell "A1".
Using REGEXREPLACE, # is put to between each character like โ#๐#๐
#๐
#โ#๏ธ#๐#๐#๐ช#๐ง#๐ง#.
Using SPLIT, the value is splitted with #.
Result:
Note:
In your question, the value of ๏ธ which cannot be displayed is included. It's \ufe0f. So "G1" can be seen like no value. But the value is existing. So please be careful this. If you want to remove the value, you can use โ๐๐
๐
โ๐๐๐ช๐ง๐ง.
References:
REGEXREPLACE
SPLIT
Added:
From marikamitsos's comment, I could notice that my understanding was not correct. So the final result is as follows. This is from marikamitsos.
=TRANSPOSE(UNIQUE(TRANSPOSE(SPLIT(REGEXREPLACE(A1,"(.)","$1#"),"#"))))
or try:
=TRANSPOSE(UNIQUE(TRANSPOSE(REGEXEXTRACT(A1, REPT("(.)", LEN(A1))))))
Formula
Appears, one of the best formula solutions would be:
=SPLIT(REGEXREPLACE(A1,"(.)","$1#"),"#")
You may also add some additional checks like skin tones & intermediate chars:
=TRANSPOSE(SPLIT(REGEXREPLACE(A2,"(.[๐ป๐ผ๐ฝ๐พ๐ฟ"&CHAR(8205)&CHAR(65039)&"]*)","#$1"),"#"))
It will help to join some emojis as a single emoji.
Script
More precise way is to use the script:
https://github.com/orling/grapheme-splitter/blob/master/index.js
โ
Add the code to Script editor
Add code for sample usage:
function splitEmojis(string) {
var splitter = new GraphemeSplitter();
// split the string to an array of grapheme clusters (one string each)
var graphemes = splitter.splitGraphemes(string);
return graphemes;
}
Tests
Not 100% precise
1
Please note: some emojis are not correctly shown in sheets
๐ด๓ ง๓ ข๓ ท๓ ฌ๓ ณ๓ ฟ๐ด๓ ง๓ ข๓ ณ๓ ฃ๓ ด๓ ฟ๐ด๓ ง๓ ข๓ ฅ๓ ฎ๓ ง๓ ฟ๐ด
โ emojis:
flag: England
flag: Scotland
flag: Wales
black flag
are the same for Google Sheets.
2
Vlookup function in #GoogleSheets and in #Excel thinks chars
#๏ธโฃ and
*๏ธโฃ
are the same!
In Blue Prism, I need to identify specific elements of a Data Item (text), in order to use the information later in my process.
The text string reads:
REKVISITION_NR: 1234567 Dato: 23-07-2018 Rekvirent: ABC, DEF GHI, JKL 60, 8600 MNO Sted: JKL 60, 8600 MNO, Kl.:14:00:00, Bestilt_tid: 60 min Tolkensnavn: PQR STU Koert_fra: VXY , 8600 Silkeborg Vedr.: Z CPR: 123456-7890 Sprog: Arabisk Type: Personlig fremmรธde Godkendt: 24-07-2018
As you can see, each element has these traits (e.g. Kl.:14:00:00 or Sprog: Arabisk):
A string name (starting with an uppercase letter)
Optionally, a period character (.)
A colon character (:)
Optionally, a space character ( )
The value part of the string
A space character ( ), which is followed by the next element.
I believe I should use the Business Object Utility - Strings' action Extract Regex Values, but have not sucessfully been able to match any data that can be copied into the Named Values-collection.
However, I have found that ([A-Z])\w+\.?: ?(\w(\d\-){0,3})+ brings me some of the way in terms of matching.
I want the solution to copy the field names and values into the Named Values collection generated by the action.
Final notes: I am using Blue Prism 6.2.1, and the action's underlying code is based on VB.net's Regex.Match method.
What you seem to be missing are the actual Named Groups. To capture the values in a Blue Prism collection, you need to make sure that you assign proper group names like this:
(?<YourGroupName>[A-Z])
Here's the regex pattern that you may use, although you need to verify if it really works for your case in all possible scenarios.
(?<Name>\b\S*?):\s(?<Value>.*?)\s*(?=(?:\b\S*?:\s)|$)
You can also check and test it here.
EDIT: But please be aware that Blue Prism's original code for extracting multiple values to a collection is barely usable, you may be better off at modifying it or creating your own. For example, what I would expect from such an action is a collection where each row will be a pattern match, with each column being a named group. Sadly, that's not how the default action works.
EDIT:
How can I extract the text up to the 4th instance of a character in a column?
I'm selecting text out of a column called filter_type up to the fourth > character.
To accomplish this, I've been trying to find the position of the fourth > character, but it's not working:
select substring(filter_type from 1 for position('>' in filter_type))
You can use the pattern matching function in Postgres.
First figure out a pattern to capture everything up to the fourth > character.
To start your pattern you should create a sub-group that captures non > characters, and one > character:
([^>]*>)
Then capture that four times to get to the fourth instance of >
([^>]*>){4}
Then, you will need to wrap that in a group so that the match brings back all four instances:
(([^>]*>){4})
and put a start of string symbol for good measure to make sure it only matches from the beginning of the String (not in the middle):
^(([^>]*>){4})
Here's a working regex101 example of that!
Once you have the pattern that will return what you want in the first group element (which you can tell at the online regex on the right side panel), you need to select it back in the SQL.
In Postgres, the substring function has an option to use a regex pattern to extract text out of the input using a 'from' statement in the substring.
To finish, put it all together!
select substring(filter_type from '^(([^>]*>){4})')
from filter_table
See a working sqlfiddle here
If you want to match the entire string whenever there are less than four instances of >, use this regular expression:
^(([^>]*>){4}|.*)
You can also use a simple, non-regex solution:
SELECT array_to_string((string_to_array(filter_type, '>'))[1:4], '>')
The above query:
splits your string into an array, using '>' as delimeter
selects only the first 4 elements
transforms the array back to a string
substring(filter_type from '^(([^>]*>){4})')
This form of substring lets you extract the portion of a string that matches a regex pattern.
You can also split the string, then choose the N'th element inside the result list. For example:
SELECT SPLIT_PART('aa,bb,cc', ',', 2)
will return: bb.
This function is defined as:
SPLIT_PART(string, delimiter, position)
In order to look at this problem, I did the following (all of the code below is available on the fiddle here):
CREATE TABLE s
(
a TEXT
);
I then created a PL/pgSQL function to generate random strings as follows.
CREATE FUNCTION f() RETURNS TEXT LANGUAGE SQL AS
$$
SELECT STRING_AGG(SUBSTR('abcdef>', CEIL(RANDOM() * 7)::INTEGER, 1), '')
FROM GENERATE_SERIES(1, 40)
$$;
I got the code from here and modified it so that it would produce strings with lots of > characters for testing purposes.
I then manually inserted a few strings at the beginning so that a quick look would tell me if the code was working as anticipated.
INSERT INTO s VALUES
('afsad>adfsaf>asfasf>afasdX>asdffs>asfdf>'),
('23433>433453>4>4559>455>3433>'),
('adfd>adafs>afadsf>'), -- only 3 '>'s!
('babedacfab>feaefbf>fedabbcbbcdcfefefcfcd'),
('e>>>>>'), -- edge case - multiple terminal '>'s
('aaaaaaa'); -- edge case - no '>'s whatsoever
The reason I put in the records with fewer than 4 >s is because the accepted answer (see discussion at the end of this answer) puts forward a solution which should return the entire string if this is the case!
On the fiddle, I then added 50,000 records as follows:
INSERT INTO s
SELECT f() FROM GENERATE_SERIES(1, 50000);
I also created a table s on a home laptop (16GB RAM, 500MB NVMe SSD) and populated it with 40,000,000 (50M) records - times also shown.
Now, my reading of the question is that we need to extract the string up to but not including the 4th > character.
The first solution (from treecon) was this one (I also show them running on the fiddle, but to save space here, I've only included the partial output of EXPLAIN (ANALYZE, BUFFERS, VERBOSE)) - the times shown are typical over a few runs:
EXPLAIN (ANALYZE, BUFFERS, VERBOSE)
SELECT
ARRAY_TO_STRING((STRING_TO_ARRAY(a, '>'))[1:4], '>'),
a
FROM s;
Result (only key parts included):
Seq Scan on public.s
Execution Time: 81.807 ms
40M Time: 46 seconds
A regex solution which works (significantly faster):
EXPLAIN (ANALYZE, BUFFERS, VERBOSE)
SELECT
SUBSTRING(a FROM '^(?:[^>]*>){0,3}[^>]*'),
a
FROM s;
Result:
Seq Scan on public.s
Execution Time: 74.757 ms
40M Time: 32 seconds
The accepted answer fails on many levels (see the fiddle). It leaves a > at the end and fails on various strings even when modified. Also, the solution proposed to include strings with fewer than 4 >s (i.e. ^(([^>]*>){4}|.*)) merely returns the original string (see end of fiddle).
At one point in my app, I need to match some strings against a pattern. Let's say that some of the sample strings look as follows:
Hi there, John.
What a lovely day today!
Lovely sunset today, John, isn't it?
Will you be meeting Linda today, John?
Most (not all) of these strings are from pre-defined patterns as follows:
"Hi there, %s."
"What a lovely day today!"
"Lovely sunset today, %s, isn't it?"
"Will you be meeting %s today, %s?"
This library of patterns is ever-expanding (currently at around 1,500), but is manually maintained. The input strings though (the first group) is largely unpredictable. Though most of them will match one of the patterns, some of them will not.
So, here's my question: Given a string (from the first group) as input, I need to know which of the patterns (known second group) it matched. If nothing matched, it needs to tell me that.
I'm guessing the solution involves building a regex out of the patterns, and iteratively checking which one matched. However, I'm unsure what the code to build those regexes looks like.
Note: The strings I've given here are for illustration purposes. In reality, the strings aren't human generated, but are computer-generated human-friendly strings as shown above from systems I don't control. Since they aren't manually typed in, we don't need to worry about things like typos and other human errors. Just need to find which pattern it matches.
Note 2: I could modify the patterns library to be some other format, if that makes it easier to construct the regexes. The current structure, with the printf style %s, isn't set in stone.
I am looking at this as a parsing problem. The idea is that the parser function takes a string and determines if it is valid or not.
The string is valid if you can find it among the given patterns. That means you need an index of all the patterns. The index must be a full text index. Also it must match according to the word position. eg. it should short circuit if the first word of the input is not found among the first word of the patterns. It should take care of the any match ie %s in the pattern.
One solution is to put the patterns in an in memory database (eg. redis) and do a full text index on it. (this will not match according to word position) but you should be able to narrow down to the correct pattern by splitting the input into words and searching. The searches will be very fast because you have a small in memory database. Also note that you are looking for the closest match. One or more words will not match. The highest number of matches is the pattern you want.
An even better solution is to generate your own index in a dictionary format. Here is an example index for the four patterns you gave as a JavaScript object.
{
"Hi": { "there": {"%s": null}},
"What: {"a": {"lovely": {"day": {"today": null}}}},
"Lovely": {"sunset": {"today": {"%s": {"isnt": {"it": null}}}}},
"Will": {"you": {"be": {"meeting": {"%s": {"today": {"%s": null}}}}}}
}
This index is recursive descending according to the word postion. So search for the first word, if found search for the next within the object returned by the first and so on. Same words at a given level will have only one key. You should also match the any case. This should be blinding fast in memory.
My first thought would be to have the regexp engine take all the trouble of handling this. They're usually optimised to handle large amounts of text so it shouldn't be that much of a performance hassle. It's brute force but the performance seems to be okay. And you could split the input into pieces and have multiple processes handle them. Here's my moderately tested solution (in Python).
import random
import string
import re
def create_random_sentence():
nwords = random.randint(4, 10)
sentence = []
for i in range(nwords):
sentence.append("".join(random.choice(string.lowercase) for x in range(random.randint(3,10))))
ret = " ".join(sentence)
print ret
return ret
patterns = [ r"Hi there, [a-zA-Z]+.",
r"What a lovely day today!",
r"Lovely sunset today, [a-zA-Z]+, isn't it?",
r"Will you be meeting [a-zA-Z]+ today, [a-zA-Z]+\?"]
for i in range(95):
patterns.append(create_random_sentence())
monster_pattern = "|".join("(%s)"%x for x in patterns)
print monster_pattern
print "--------------"
monster_regexp = re.compile(monster_pattern)
inputs = ["Hi there, John.",
"What a lovely day today!",
"Lovely sunset today, John, isn't it?",
"Will you be meeting Linda today, John?",
"Goobledigoock"]*2000
for i in inputs:
ret = monster_regexp.search(i)
if ret:
print ".",
else:
print "x",
I've created a hundred patterns. This is the maximum limit of the python regexp library. 4 of them are your actual examples and the rest are random sentences just to stress performance a little.
Then I combined them into a single regexp with 100 groups. (group1)|(group2)|(group3)|.... I'm guessing you'll have to sanitise the inputs for things that can have meanings in regular expressions (like ? etc.). That's the monster_regexp.
Testing one string against this tests it against 100 patterns in a single shot. There are methods that fetch out the exact group which was matched. I test 10000 strings 80% of which should match and 10% which will not. It short cirtcuits so if there's a success, it will be comparatively quick. Failures will have to run through the whole regexp so it will be slower. You can order things based on the frequency of input to get some more performance out of it.
I ran this on my machine and this is my timing.
python /tmp/scratch.py 0.13s user 0.00s system 97% cpu 0.136 total
which is not too bad.
However, to run a pattern against such a large regexp and fail will take longer so I changed the inputs to have lots of randomly generated strings that won't match and then tried. 10000 strings none of which match the monster_regexp and I got this.
python /tmp/scratch.py 3.76s user 0.01s system 99% cpu 3.779 total
Similar to Noufal's solution, but returns the matched pattern or None.
import re
patterns = [
"Hi there, %s.",
"What a lovely day today!",
"Lovely sunset today, %s, isn't it",
"Will you be meeting %s today, %s?"
]
def make_re_pattern(pattern):
# characters like . ? etc. have special meaning in regular expressions.
# Escape the string to avoid interpretting them as differently.
# The re.escape function escapes even %, so replacing that with XXX to avoid that.
p = re.escape(pattern.replace("%s", "XXX"))
return p.replace("XXX", "\w+")
# Join all the pattens into a single regular expression.
# Each pattern is enclosed in () to remember the match.
# This will help us to find the matched pattern.
rx = re.compile("|".join("(" + make_re_pattern(p) + ")" for p in patterns))
def match(s):
"""Given an input strings, returns the matched pattern or None."""
m = rx.match(s)
if m:
# Find the index of the matching group.
index = (i for i, group in enumerate(m.groups()) if group is not None).next()
return patterns[index]
# Testing with couple of patterns
print match("Hi there, John.")
print match("Will you be meeting Linda today, John?")
Python solution. JS should be similar.
>>> re2.compile('^ABC(.*)E$').search('ABCDE') == None
False
>>> re2.compile('^ABC(.*)E$').search('ABCDDDDDDE') == None
False
>>> re2.compile('^ABC(.*)E$').search('ABX') == None
True
>>>
The trick is to use ^ and $ to bound your pattern and making it a "template". Use (.*) or (.+) or whatever it is that you want to "search" for.
The main bottleneck for you, imho, will be iterating through a list of these patterns. Regex searches are computationally expensive.
If you want a "does any pattern match" result, build a massive OR based regex and let your regex engine handle the 'OR'ing for you.
Also, if you have only prefix patterns, check out the TRIE data structure.
This could be a job for sscanf, there is an implementation in js: http://phpjs.org/functions/sscanf/; the function being copied is this: http://php.net/manual/en/function.sscanf.php.
You should be able to use it without changing the prepared strings much, but I have doubts about the performances.
the problem isn't clear to me. Do you want to take the patterns and build regexes out of it?
Most regex engines have a "quoted string" option. (\Q \E). So you could take the string and make it
^\QHi there,\E(?:.*)\Q.\E$
these will be regexes that match exactly the string you want outside your variables.
if you want to use a single regex to match just a single pattern, you can put them in grouped patterns to find out which one matched, but that will not give you EVERY match, just the first one.
if you use a proper parser (I've used PEG.js), it might be more maintainable though. So that's another option if you think you might get stuck in regex hell