The regex in string.format of LUA - regex

I use string.format(str, regex) of LUA to fetch some key word.
local RICH_TAGS = {
"texture",
"img",
}
--\[((img)|(texture))=
local START_OF_PATTER = "\\[("
for index = 1, #RICH_TAGS - 1 do
START_OF_PATTER = START_OF_PATTER .. "(" .. RICH_TAGS[index]..")|"
end
START_OF_PATTER = START_OF_PATTER .. "("..RICH_TAGS[#RICH_TAGS].."))"
function RichTextDecoder.decodeRich(str)
local result = {}
print(str, START_OF_PATTER)
dump({string.find(str, START_OF_PATTER)})
end
output
hello[img=123] \[((texture)|(img))
dump from: [string "utils/RichTextDecoder.lua"]:21: in function 'decodeRich'
"<var>" = {
}
The output means:
str = hello[img=123]
START_OF_PATTER = \[((texture)|(img))
This regex works well with some online regex tools. But it find nothing in LUA.
Is there any wrong using in my code?

You cannot use regular expressions in Lua. Use Lua's string patterns to match strings.
See How to write this regular expression in Lua?
Try dump({str:find("\\%[%("))})
Also note that this loop:
for index = 1, #RICH_TAGS - 1 do
START_OF_PATTER = START_OF_PATTER .. "(" .. RICH_TAGS[index]..")|"
end
will leave out the last element of RICH_TAGS, I assume that was not your intention.
Edit:
But what I want is to fetch several specific word. For example, the
pattern can fetch "[img=" "[texture=" "[font=" any one of them. With
the regex string I wrote in my question, regex can do the work. But
with Lua, the way to do the job is write code like string.find(str,
"[img=") and string.find(str, "[texture=") and string.find(str,
"[font="). I wonder there should be a way to do the job with a single
pattern string. I tryed pattern string like "%[%a*=", but obviously it
will fetch a lot more string I need.
You cannot match several specific words with a single pattern unless they are in that string in a specific order. The only thing you could do is to put all the characters that make up those words into a class, but then you risk to find any word you can build from those letters.
Usually you would match each word with a separate pattern or you match any word and check if the match is one of your words using a look up table for example.
So basically you do what a regex library would do in a few lines of Lua.

Related

Regex Multiple rows [duplicate]

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-)","($1)")))
If the cell that you are trying to get the values has Special Characters like parentheses "(" or question mark "?" the solution provided won´t work.
In my case, I was trying to list all “variables text” contained in the cell. Those “variables text “ was wrote inside like that: “{example_name}”. But the full content of the cell had special characters making the regex formula do break. When I removed theses specials characters, then I could list all captured groups like the solution did.
There are two general ('Excel' / 'native' / non-Apps Script) solutions to return an array of regex matches in the style of REGEXEXTRACT:
Method 1)
insert a delimiter around matches, remove junk, and call SPLIT
Regexes work by iterating over the string from left to right, and 'consuming'. If we are careful to consume junk values, we can throw them away.
(This gets around the problem faced by the currently accepted solution, which is that as Carlos Eduardo Oliveira mentions, it will obviously fail if the corpus text contains special regex characters.)
First we pick a delimiter, which must not already exist in the text. The proper way to do this is to parse the text to temporarily replace our delimiter with a "temporary delimiter", like if we were going to use commas "," we'd first replace all existing commas with something like "<<QUOTED-COMMA>>" then un-replace them later. BUT, for simplicity's sake, we'll just grab a random character such as  from the private-use unicode blocks and use it as our special delimiter (note that it is 2 bytes... google spreadsheets might not count bytes in graphemes in a consistent way, but we'll be careful later).
=SPLIT(
LAMBDA(temp,
MID(temp, 1, LEN(temp)-LEN(""))
)(
REGEXREPLACE(
"xyzSixSpaces:[ ]123ThreeSpaces:[ ]aaaa 12345",".*?( |$)",
"$1"
)
),
""
)
We just use a lambda to define temp="match1match2match3", then use that to remove the last delimiter into "match1match2match3", then SPLIT it.
Taking COLUMNS of the result will prove that the correct result is returned, i.e. {" ", " ", " "}.
This is a particularly good function to turn into a Named Function, and call it something like REGEXGLOBALEXTRACT(text,regex) or REGEXALLEXTRACT(text,regex), e.g.:
=SPLIT(
LAMBDA(temp,
MID(temp, 1, LEN(temp)-LEN(""))
)(
REGEXREPLACE(
text,
".*?("&regex&"|$)",
"$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)
)
)
))
)

Replace multiple words in pig

I am new to Pig. In the script that I am writing I want to perform an operation similar to this:
foreach X GENERATE REPLACE(word,'.*abc.*','abc') OR REPLACE(word,'.*def.*','def').
If the first pattern matches then abc is replaced else if second pattern is matched then def is replaced. But I suppose the syntax is incorrect. Can someone help me with the syntax?
There are a few ways to do this, but since if the regex doesn't match the string, you'll just get your string back, this is pretty compact:
Y = FOREACH X GENERATE REPLACE(REPLACE(word, '.*abc.*', 'abc'), '.*def.*', 'def');

regular expressions and vba

Does anyone know how to extract matches as strings from a RegExp.Execute() function?
Let me show you what I've gotten to so far:
Regex.Pattern = "^[^*]*[*]+"
Set myMatches = Regex.Execute(temp)
I want the object "myMatches" which is holding the matches, to be converted to a string. I know that there is only going to be one match per execution.
Does anyone know how to extract the matches from the object as Strings to be displayed lets say via a MsgBox?
Try this:
Dim sResult As String
'// Your expression code here...
sResult = myMatches.Item(0)
'// or
sResult = myMatches(0)
Msgbox("The matching text was: " & sResult)
The Execute method returns a match collection and you can use the item property to retrieve the text using an index.
As you stated you only ever have one match then the index is zero. If you have more than one match you can return the index of the match you require or loop over the entire collection.
This page has a lot of information on regex and seems to have what you want.
http://www.regular-expressions.info/vbscript.html

Is there a RegEx that can parse out the longest list of digits from a string?

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.

Regex to replace string with another string in MS Word?

Can anyone help me with a regex to turn:
filename_author
to
author_filename
I am using MS Word 2003 and am trying to do this with Word's Find-and-Replace. I've tried the use wildcards feature but haven't had any luck.
Am I only going to be able to do it programmatically?
Here is the regex:
([^_]*)_(.*)
And here is a C# example:
using System;
using System.Text.RegularExpressions;
class Program
{
static void Main()
{
String test = "filename_author";
String result = Regex.Replace(test, #"([^_]*)_(.*)", "$2_$1");
}
}
Here is a Python example:
from re import sub
test = "filename_author";
result = sub('([^_]*)_(.*)', r'\2_\1', test)
Edit: In order to do this in Microsoft Word using wildcards use this as a search string:
(<*>)_(<*>)
and replace with this:
\2_\1
Also, please see Add power to Word searches with regular expressions for an explanation of the syntax I have used above:
The asterisk (*) returns all the text in the word.
The less than and greater than symbols (< >) mark the start and end
of each word, respectively. They
ensure that the search returns a
single word.
The parentheses and the space between them divide the words into
distinct groups: (first word) (second
word). The parentheses also indicate
the order in which you want search to
evaluate each expression.
Here you go:
s/^([a-zA-Z]+)_([a-zA-Z]+)$/\2_\1/
Depending on the context, that might be a little greedy.
Search pattern:
([^_]+)_(.+)
Replacement pattern:
$2_$1
In .NET you could use ([^_]+)_([^_]+) as the regex and then $2_$1 as the substitution pattern, for this very specific type of case. If you need more than 2 parts it gets a lot more complicated.
Since you're in MS Word, you might try a non-programming approach. Highlight all of the text, select Table -> Convert -> Text to Table. Set the number of columns at 2. Choose Separate Text At, select the Other radio, and enter an _. That will give you a table. Switch the two columns. Then convert the table back to text using the _ again.
Or you could copy the whole thing to Excel, construct a formula to split and rejoin the text and then copy and paste that back to Word. Either would work.
In C# you could also do something like this.
string[] parts = "filename_author".Split('_');
return parts[1] + "_" + parts[0];
You asked about regex of course, but this might be a good alternative.