I want to add (int) value to a backreference.
For this I created a function and pass the appropriate backreference. The backreference if returned without any modification works fine, however when I try to modify or use any other functions on the backreference that was passed it assumes \3 as the argument value and not the backreference value itself.
For eg-
CREATE OR REPLACE FUNCTION add10(text) returns text as $$
DECLARE
t int;
BEGIN
t := to_number($1, '999999') + 10;
return trim(to_char(t, '999999'), ' ');
END;
$$ LANGUAGE plpgsql;
then:
select regexp_replace('890808', '80(\d+)', add10('\1'), 'g');
should give result as
test
-------
89018
(1 row)
However it gives --
test
-------
89011
(1 row)
taking the value of $1 as 1(the backreference number) instead of value 8.
Any ideas why does this happen?
Problem: order of evaluation
My guess (and only a guess, given that the question isn't super clear) is that you're confused by the order of evaluation of arguments within function calls, and are trying to call a function on a backref value, but order of evaluation means that it's called on the backref string before regexp evaluation.
Assuming that add10 and t are the same thing, then:
select regexp_replace('890808', '80(\d+)', add10('\1'), 'g');
is evaluated by first calling add10('\1'). That will in turn run:
select to_number('\1', '999999') + 10 into t;
Since select to_number('\1', '999999') produces the value 1, you'll get 11 in t. You then convert that back to a string (via a rather weird approach, why didn't you just cast it).
So you've replaced '\1' with '11', so your regexp_replace call looks like:
select regexp_replace('890808', '80(\d+)', '11`, 'g');
... from which you can see where your unexpected result came from.
Solution: Split the value, modify it, then reassemble it
I don't think your desired result makes any sense, so I can't really figure out how to produce it. You seem to be trying to retain all digits before "80", discard "80", convert all digits after the "80" to a number and add 10, then substitute it back in. Which is pretty WTFy, why?
Regular expressions are one way to split numbers up, but the best way is usually modulus and remainder:
craig=> SELECT 890808 / 10000, 890808 % 10000;
?column? | ?column?
----------+----------
89 | 808
(1 row)
If you must use regexps (say, if it's mixed alphanumeric or if your criteria are not easily expressed by place values), you probably want to use regexp_split_to_array.
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-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|$)',' ',1,0) from dual;
and the results are;
SNERD JR
Use occurrence parameter of REGEXP_REPLACE function. The docs says:
occurrence is a nonnegative integer indicating the occurrence of the replace operation:
If you specify 0, then Oracle replaces all occurrences of the match.
If you specify a positive integer n, then Oracle replaces the nth occurrenc
https://docs.oracle.com/cd/B28359_01/server.111/b28286/functions137.htm#SQLRF06302
It should look like:
...
REGEXP_REPLACE(IN_STRING,REGEX_SALUTATIONS,' ', 1,0 )
...
I have a table with a column po_number of type varchar in Postgres 8.4. It stores alphanumeric values with some special characters. I want to ignore the characters [/alpha/?/$/encoding/.] and check if the column contains a number or not. If its a number then it needs to typecast as number or else pass null, as my output field po_number_new is a number field.
Below is the example:
SQL Fiddle.
I tired this statement:
select
(case when regexp_replace(po_number,'[^\w],.-+\?/','') then po_number::numeric
else null
end) as po_number_new from test
But I got an error for explicit cast:
Simply:
SELECT NULLIF(regexp_replace(po_number, '\D','','g'), '')::numeric AS result
FROM tbl;
\D being the class shorthand for "not a digit".
And you need the 4th parameter 'g' (for "globally") to replace all occurrences.
Details in the manual.
For a known, limited set of characters to replace, plain string manipulation functions like replace() or translate() are substantially cheaper. Regular expressions are just more versatile, and we want to eliminate everything but digits in this case. Related:
Regex remove all occurrences of multiple characters in a string
PostgreSQL SELECT only alpha characters on a row
Is there a regexp_replace equivalent for postgresql 7.4?
But why Postgres 8.4? Consider upgrading to a modern version.
Consider pitfalls for outdated versions:
Order varchar string as numeric
WARNING: nonstandard use of escape in a string literal
I think you want something like this:
select (case when regexp_replace(po_number, '[^\w],.-+\?/', '') ~ '^[0-9]+$'
then regexp_replace(po_number, '[^\w],.-+\?/', '')::numeric
end) as po_number_new
from test;
That is, you need to do the conversion on the string after replacement.
Note: This assumes that the "number" is just a string of digits.
The logic I would use to determine if the po_number field contains numeric digits is that its length should decrease when attempting to remove numeric digits.
If so, then all non numeric digits ([^\d]) should be removed from the po_number column. Otherwise, NULL should be returned.
select case when char_length(regexp_replace(po_number, '\d', '', 'g')) < char_length(po_number)
then regexp_replace(po_number, '[^0-9]', '', 'g')
else null
end as po_number_new
from test
If you want to extract floating numbers try to use this:
SELECT NULLIF(regexp_replace(po_number, '[^\.\d]','','g'), '')::numeric AS result FROM tbl;
It's the same as Erwin Brandstetter answer but with different expression:
[^...] - match any character except a list of excluded characters, put the excluded charaters instead of ...
\. - point character (also you can change it to , char)
\d - digit character
Since version 12 - that's 2 years + 4 months ago at the time of writing (but after the last edit that I can see on the accepted answer), you could use a GENERATED FIELD to do this quite easily on a one-time basis rather than having to calculate it each time you wish to SELECT a new po_number.
Furthermore, you can use the TRANSLATE function to extract your digits which is less expensive than the REGEXP_REPLACE solution proposed by #ErwinBrandstetter!
I would do this as follows (all of the code below is available on the fiddle here):
CREATE TABLE s
(
num TEXT,
new_num INTEGER GENERATED ALWAYS AS
(NULLIF(TRANSLATE(num, 'ABCDEFGHIJKLMNOPQRSTUVWXYZ. ', ''), '')::INTEGER) STORED
);
You can add to the 'ABCDEFG... string in the TRANSLATE function as appropriate - I have decimal point (.) and a space ( ) at the end - you may wish to have more characters there depending on your input!
And checking:
INSERT INTO s VALUES ('2'), (''), (NULL), (' ');
INSERT INTO t VALUES ('2'), (''), (NULL), (' ');
SELECT * FROM s;
SELECT * FROM t;
Result (same for both):
num new_num
2 2
NULL
NULL
NULL
So, I wanted to check how efficient my solution was, so I ran the following test inserting 10,000 records into both tables s and t as follows (from here):
EXPLAIN (ANALYZE, BUFFERS, VERBOSE)
INSERT INTO t
with symbols(characters) as
(
VALUES ('ABCDEFGHIJKLMNOPQRSTUVWXYZ0123456789')
)
select string_agg(substr(characters, (random() * length(characters) + 1) :: INTEGER, 1), '')
from symbols
join generate_series(1,10) as word(chr_idx) on 1 = 1 -- word length
join generate_series(1,10000) as words(idx) on 1 = 1 -- # of words
group by idx;
The differences weren't that huge but the regex solution was consistently slower by about 25% - even changing the order of the tables undergoing the INSERTs.
However, where the TRANSLATE solution really shines is when doing a "raw" SELECT as follows:
EXPLAIN (ANALYZE, BUFFERS, VERBOSE)
SELECT
NULLIF(TRANSLATE(num, 'ABCDEFGHIJKLMNOPQRSTUVWXYZ. ', ''), '')::INTEGER
FROM s;
and the same for the REGEXP_REPLACE solution.
The differences were very marked, the TRANSLATE taking approx. 25% of the time of the other function. Finally, in the interests of fairness, I also did this for both tables:
EXPLAIN (ANALYZE, BUFFERS, VERBOSE)
SELECT
num, new_num
FROM t;
Both extremely quick and identical!
i have a issue where the there is a amount field which has data like
(- 98765.00),minus{spaces]{numbers} ?, i need to remove the space between the minus and the number and get is as (-98765.00), how do i do it in expression transformation.
field datatype is decimal (8,2).
Thanks,
Kiran
output_port: TO_DECIMAL(REPLACECHR(FALSE,input_port,' ',''))
REPLACECHR replaces the blanks with empty character, essentially removing them. The first argument can be TRUE/FALSE to specify case sensitive or not, but it is not important in this case.
You can use REG_REPLACE function to replace space
To achieve this you need to follow below steps,
* Create two variable ports
* REG_REPLACE - function requires string column, so you need to convert the decimal column to string column using TO_CHAR function
First variable port(string) - TO_CHAR(column_name)
* In previous port data is converted to string, now convert it again to decimal and apply REG_REPLACE function
Second variable port(decimal) - to_decimal(reg_replace(first_variable_port,'s+',''))
s - determines the white spaces in informatica regular expression
See the below image,
same number which you provided is used. Use the same data type and function
Debugger gives the exact result by removing white space in the below image,
May be you have the issue with other transformations which you are passing through. Debug and verify the data once.
Hope you got it, any issues feel free to ask
To have enjoy informatica, have a fun on https://etlinfromatica.wordpress.com/
If my understanding is correct, you need to replace both the spaces and the brackets. Here's the expression:
TO_DECIMAL(
REPLACECHR(0,
REPLACECHR(0, '(- 98765.00)', ' ', '') -- this part does the space replacement
, '()', '') -- this part replaces the brackets
)
I'm trying to extract both ints and chars from names such as 123A America, 234B Britania.
I only want the the number and the attached letter (i.e. 123A) .
I'm using regexp_matches(name, '(\d+)(\D)') and it results as:
{123,A},
{456,B}
I thought using concatenation, getting the first element of an array and the second element using two different functions
(regexp_matches(name, '(\d+)(\D)' )) [1] || (regexp_matches(name, '(\d+)(\D)' )) [2]
But it generates an error:
ERROR: functions and operators can take at most one set argument
How can I get the two element as one string?
You don't have to get the two items you're searching for as different sets, just get them as a single set. Remove the )( between \d+ and \D and that will return a set containing the entire string you're looking for.
Results in this -
regexp_matches('123A America, 234B Britania', '(\d+\D)' )
This will only find the first match. To get all matching substrings, use the g flag -
regexp_matches('123A America, 234B Britania', '(\d+\D)', 'g')
good answer by #Scott S however if you can't achieve what you need within one capture group the solution is to write a function, assign the regexp result to a variable and then use it.
CREATE OR REPLACE FUNCTION do_something(_input character varying)
RETURNS character varying AS
$BODY$
DECLARE
matches text[];
BEGIN
matches := regexp_matches(_input, '^([0-9]{1,}_[^_]{1,})_[a-z]{1,}(.*)$','i');
return substring(matches[1], 0, 24)||matches[2];
END
$BODY$
LANGUAGE plpgsql;