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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)
)
)
))
)
I need to replace multiple words such as (dog|cat|bird) with nothing in a string where there may be multiple consecutive occurrences of a word. The actual code is to remove salutations and suffixes from a name. Unfortunately the garbage data I get sometimes contains "SNERD JR JR."
I was able to create a regular expression pattern that accomplishes my goal but only for the first occurrence. I implemented a stupid hack to get rid of the second occurrence, but I believe there has to be a better way. I just can't figure it out.
Here is my "hacked" code;
FUNCTION REMOVE_SALUTATIONS(IN_STRING VARCHAR2) RETURN VARCHAR2 DETERMINISTIC
AS
REGEX_SALUTATIONS VARCHAR2(4000) := '(^|\s)(MR|MS|MISS|MRS|DR|MD|M D|SR|SIR|PHD|P H D|II|III|IV|JR)(\.?)(\s|$)';
BEGIN
RETURN TRIM(REGEXP_REPLACE(REGEXP_REPLACE(IN_STRING,REGEX_SALUTATIONS,' '),REGEX_SALUTATIONS,''));
END REMOVE_SALUTATIONS;
I was actually proud that I was able to get this far, as regular expression are not very regular to me. All help is appreciated.
EDIT:
The default for regexp_replace based on my understanding is to do a global replace. But on the outside chance my DB is configured different I did try;
select REGEXP_REPLACE('SNERD JR JR','(^|\s)(MR|MS|MISS|MRS|DR|MD|M D|SR|SIR|PHD|P H D|II|III|IV|JR)(\.?)(\s|$)',' ',1,0) from dual;
and the results are;
SNERD JR
Use occurrence parameter of REGEXP_REPLACE function. The docs says:
occurrence is a nonnegative integer indicating the occurrence of the replace operation:
If you specify 0, then Oracle replaces all occurrences of the match.
If you specify a positive integer n, then Oracle replaces the nth occurrenc
https://docs.oracle.com/cd/B28359_01/server.111/b28286/functions137.htm#SQLRF06302
It should look like:
...
REGEXP_REPLACE(IN_STRING,REGEX_SALUTATIONS,' ', 1,0 )
...
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).
I have a column in Oracle which can contain up to 5 separate values, each separated by a '|'. Any of the values can be present or missing. Here are come examples of how the data might look:
100-1
10-3|25-1|120/240
15-1|15-3|15-2|120/208
15-1|15-3|15-2|120/208|STA-2
112-123|120/208|STA-3
The values are arbitrary except for the order. The numerical values separated by dashes always come first. There can be 1 to 3 of these values present. The numerical values separated by a slash (if it is present) is next. The string, 'STA', and a numerical value separated by a dash is always last, if it is present.
What I would like to do is reformat this column to only ever include the first three possible values, those being the three numerical values separated by dashes. Afterwards, I want to replace 2nd numeric in each value (the numeric after the dash) using the following pattern:
1 = A
2 = B
3 = C
I would also like to remove the dash afterwards, but not the '|' that separates the values unless there is a trailing '|'.
To give you an idea, here's how the values at the beginning of the post would look after the reformatting:
100A
10C|25A
15A|15C|15B
15A|15C|15B
112ABC
I'm thinking this can be done with regex expressions but it's got me a little confused. Does anyone have a solution?
If I have to solve this problem I will solve it in following ways.
SELECT
REGEXP_REPLACE(column,'\|\d+\/\d+(\|STA-\d+)?',''),
REGEXP_REPLACE(column,'(\d+)-(1)([^\d])','\1A\3'),
REGEXP_REPLACE(column,'(\d+)-(2)([^\d])','\1B\3'),
REGEXP_REPLACE(column,'(\d+)-(3)([^\d])','\1C\3'),
REGEXP_REPLACE(column,'(\d+)-(123)([^\d])','\1ABC')
FROM table;
Explanation: Let us break down each REGEXP_REPLACE statement one by one.
REGEXP_REPLACE(column,'\|\d+\/\d+(\|STA-\d+)?','')
This will replace the end part like 120/208|STA-2 with empty string so that further processing is easy.
Finding match was easy but replacing A for 1, B for 2 and C for 3 was not possible ( as per my knowledge ) So I did those matching and replacements separately.
In each regex from second statement (\d+)-(yourNumber)([^\d]) first group is number before - then yourNumber is either 1,2,3 or 123 followed by |.
So the replacement will be according to yourNumber.
All demos here from version 1 to 5.
Note:- I have just done replacement for combination of yourNUmber for those present in question. You can do likewise for other combinations too.
you can do this in one line, but you can write simple function to do that
SELECT str, REGEXP_REPLACE(str,'(\|\d+\/\d+)?(\|STA-\d+)?','') cut
, REGEXP_REPLACE(REGEXP_REPLACE(str,'(\|\d+\/\d+)?(\|STA-\d+)?',''), '(\-)([1,2]*)(3)([1,2]*)', '\1\2C\4') rep3toC
, REGEXP_REPLACE(REGEXP_REPLACE(REGEXP_REPLACE(str,'(\|\d+\/\d+)?(\|STA-\d+)?',''), '(\-)([1,2]*)(3)([1,2]*)', '\1\2C\4'), '(\-)([1,C]*)(2)([1,C]*)', '\1\2B\4') rep2toB
, REGEXP_REPLACE(REGEXP_REPLACE(REGEXP_REPLACE(REGEXP_REPLACE(str,'(\|\d+\/\d+)?(\|STA-\d+)?',''), '(\-)([1,2]*)(3)([1,2]*)', '\1\2C\4'), '(\-)([1,C]*)(2)([1,C]*)', '\1\2B\4'), '(\-)([B,C]*)(1)([B,C]*)', '\1\2A\4') rep1toA
, REGEXP_REPLACE(REGEXP_REPLACE(REGEXP_REPLACE(REGEXP_REPLACE(REGEXP_REPLACE(str,'(\|\d+\/\d+)?(\|STA-\d+)?',''), '(\-)([1,2]*)(3)([1,2]*)', '\1\2C\4'), '(\-)([1,C]*)(2)([1,C]*)', '\1\2B\4'), '(\-)([B,C]*)(1)([B,C]*)', '\1\2A\4'), '-', '') "rep-"
FROM (
SELECT '100-1' str FROM dual UNION
SELECT '10-3|25-1|120/240' str FROM dual UNION
SELECT '15-1|15-3|15-2|120/208' str FROM dual UNION
SELECT '15-1|15-3|15-2|120/208|STA-2' str FROM dual UNION
SELECT '112-123|120/208|STA-3' FROM dual
) tab
I am trying to add a new calculated column that counts the number of semi colons in a string and adds one to it. So the column i have contains a bunch of aliases and I need to know how many for each row.
For example,
A; B; C; D
So basically this means there are 4 aliases (3 semi colons + 1)
Need to do this for over 2 million rows. Help please!
Basic idea is to subtract length of your string without ; characters from it's original length:
len([columnName])-len(Substitute([columnName],";",""))+1
Here it is with a regular expression:
Len(RXReplace([Column 1], "(?!;).", "", "gis"))+1
RXReplace takes as arguments:
The string you are wanting to work on (in this case it is on Column 1)
The regular expression you want to use (here it is (?!;). )
What you want to replace matches with (blank in this situation so
that everything that matches the regex is removed)
Finally a parameter saying how you want it to work (we are passing
in gis which means replace all matches not just the first, ignore case, replace newlines)
We wrap this in a Len which gives us the amount of semicolons since that is all that is left and finally we add 1 to it to get the final result.
You can read more about the regular expression here: https://msdn.microsoft.com/en-us/library/az24scfc(v=vs.110).aspx but in a nutshell it says match everything that isn't a semi colon.
You can read more about RXReplace and Len here: https://docs.tibco.com/pub/spotfire/6.0.0-november-2013/userguide-webhelp/ncfe/ncfe_text_functions.htm