I need to hide part of the string. Hide all before some ending part.
It easy to implement by regexp like this:
replace("123-134-04", ".(?=.*-)", " ")
replace any symbol if future part of string contains "-".
So result is: " -04"
It is important to keep spaces.
But, I can't use lookahead or lookbehind.
I can catch the group before ending part, but how to replace this for right number of spaces?
Or maybe some other ways to resolve this with regex?
Tnanks in advance!
If the number of to be replaced characters does not differ too much, and you have a means to match the part to be preserved, you could run through a series of search and replace:
replace("12-14-04", "^.{5}(-[^-]+)$", " \1")
replace("123-134-04", "^.{7}(-[^-]+)$", " \1")
replace("adfasd-adf-da7474-04", "^.{17}(-[^-]+)$", " \1")
Or you do:
split the string at the position, where the to be preserved part begins,
run the replace("ALL OF THIS SHOULD BECOME BLANKS", ".", " ") on the first part, and
join them up again.
Related
Ok,
it looks simple but when the words start or finish with an accents it is the mess.
I've looked on Stack Overflow and others and haven't really found a way to solve this problem.
I would like, to be able with a Google sheet formula, to extract from a cell, words only built with the ASCII characters that follow:
A,B,C,D,E,F,G,H,I,J,K,L,M,N,O,P,Q,R,S,T,U,V,W,X,Y,Z,À,Á,Â,Ã,Ä,Å,Æ,Ç,È,É,Ê,Ë,Ì,Í,Î,Ï,Ð,Ñ,Ò,Ó,Ô,Õ,Ö,Ø,Ù,Ú,Û,Ü,Ý
For example with "Éléonorä-Camilliâ ÀLËMMNIÖ DE SANTORINÕ" or "ÀLËMMNIÖ DE SANTORINÕ Éléonorä Camilliâ" the result has to be the same "ÀLËMMNIÖ DE SANTORINÕ"
This formula works when no accent all:
=REGEXEXTRACT(A2;"\b[A-Z]+(?:\s+[A-Z]+)*\b")
These formula work sometimes when the names are easy.
=REGEXEXTRACT(A2;"\b[A-Ý]+(?:\s+[A-Ý]+)*\b")
=REGEXEXTRACT(A2;"\B[A-Ý]+(?:\S+[A-Ý]+)*\B")
Can anybody help me or give me some hint?
It seems your expected matches are simply between whitespace or start/end of string. If you add a space before and after the cell value, you may simply extract all the chunks of whitespace-separated uppercase letter words between whitespaces, and the formula will boil down to
=REGEXEXTRACT(" " & A2 & " "; "\s([A-ZÀ-ÖØ-Ý]+(?:\s+[A-ZÀ-ÖØ-Ý]+)*)\s")
See the Google sheets demo:
Regex details:
\s - a whitespace
([A-ZÀ-ÖØ-Ý]+(?:\s+[A-ZÀ-ÖØ-Ý]+)*) - Group 1 (the actual value returned by REGEXEXTRACT): one or more uppercase letters from the specified ranges followed with zero or more repetitions of one or more whitespace and then one or more uppercase letters
\s - a whitespace.
You may use an ARRAYFORMULA, as well:
=ARRAYFORMULA(IFERROR(REGEXEXTRACT(" " & A:A & " ", "\s([A-ZÀ-ÖØ-Ý]+(?:\s+[A-ZÀ-ÖØ-Ý]+)*)\s"),""))
Supposing your sample name were in A2, this should work:
=TRIM(REGEXEXTRACT(A2&" ","([A-ZÀ-Ý ]+)\s"))
By appending a space to the end of the string first, we can then look for the [uppercase letter set or space] in any number up ending with a space. This rules out strings like "Éléonorä" and "Camilliâ" because those uppercase letters are not followed by a space.
Put a different way, the rule here says, "Grab as many uppercase letters or spaces in this set as possible, as long as you still have a space left over at the end." And since we appended a space to the end of the entire string, we can catch such groupings anywhere in the modified string.
Try this- backslashing the non A-Z characters.
[A-Z\À\Á\Â\Ã\Ä\Å\Æ\Ç\È\É\Ê\Ë\Ì\Í\Î\Ï\Ð\Ñ\Ò\Ó\Ô\Õ\Ö\Ø\Ù\Ú\Û\Ü\Ý]
If that fails you can encode each one of those letters like below:
Look up for characters: https://www.w3schools.com/charsets/ref_utf_latin1_supplement.asp
[A-Z\u00C0\u00C1... and so on...]
use:
=ARRAYFORMULA(TRIM(TRANSPOSE(QUERY(TRANSPOSE(IF(""<>
IFERROR(REGEXEXTRACT(SPLIT(A1:A, " "), "["&TEXTJOIN("", 1,
UNIQUE(QUERY({UPPER(CHAR(ROW(65:1500))), LOWER(CHAR(ROW(65:1500)))},
"select Col2 where Col1<>Col2")))&"]+")),,IFERROR(SPLIT(A1:A, " ")))),,9^9))))
or 10 characters shorter:
=INDEX(TRIM(TRANSPOSE(QUERY(TRANSPOSE(IF(""<>
IFERROR(REGEXEXTRACT(SPLIT(A:A; " "); "["&JOIN(;
UNIQUE(LOWER(QUERY(CHAR(ROUNDUP(SEQUENCE(1500; 2; 65)/2));
"select Col1 where lower(Col1)<>upper(Col2)"))))&"]+"));;
IFERROR(SPLIT(A:A; " "))));;9^9))))
works with all Europe-based alphabets and captures all diacritics out there. it can differentiate between:
LOWER
and
UPPER
I need to extract all matches from a huge text that start with [" and end with "]. These special characters separate each record from database. I need to extract all records.
Inside this record there are letters, numbers and special characters like -, ., &, (), /, {space} or so.
I'm writing this in Office VBA.
The pattern I have come so far looks like this: .Pattern = "[[][""][a-z|A-Z|w|W]*".
With this pattern, I am able to extract the first word from each record, with the starting characters [". The count of found matches is correct.
Example of one record:
["blabla","blabla","blabla","\u00e1no","nie","\u00e1no","\u00e1no","\u00e1no","\u003Ca class=\u0022btn btn-default\u0022 href=\u0022\u0026#x2F;siea\u0026#x2F;suppliers\u0026#x2F;42\u0022\u003E\u003Ci class=\u0022fa fa-pencil\u0022\u003E\u003C\/i\u003E Upravi\u0165\u003C\/a\u003E \u003Ca class=\u0022btn btn-default\u0022 href=\u0022\u0026#x2F;siea\u0026#x2F;suppliers\u0026#x2F;form\u0026#x2F;42\u0022\u003E\u003Ci class=\u0022fa fa-file-pdf-o\u0022\u003E\u003C\/i\u003E Zmluva\u003C\/a\u003E \u003Ca class=\u0022btn btn-default\u0022 href=\u0022\u0026#x2F;siea\u0026#x2F;suppliers\u0026#x2F;crz-form\u0026#x2F;42\u0022\u003E\u003Ci class=\u0022fa fa-file-pdf-o\u0022\u003E\u003C\/i\u003E Zmluva CRZ\u003C\/a\u003E"]
The question is : How can I extract the all records starting with [" and ending with "]?
I don't necessary need the starting and ending characters, but I can clean that up later.
Thanks for help.
The easiest way is to get rid of the initial and trailing [" and "] with either Replace or Left/Right/Mid functions, and then Split with "," (in VBA, """,""").
E.g.
input = "YOUR_STRING"
input = Replace(Replace(input, """]", ""), "[""", "")
result = Split(input, """,""")
If you plan to use Regex, you can use \["[\s\S]*?"] pattern, but it is not that efficient with long inputs and may even freeze the macro if timeout issue occurs. You can unroll it as
\["[^"]*(?:"(?!])[^"]*)*"]
See the regex demo. In VBA, Pattern = "\[""[^""]*(?:""(?!])[^""]*)*""]"
Note that with this unrolled pattern, you do not even need to use the workarounds for dot matching newline issue (negated character class [^"] matches any char but ", including a newline).
Pattern details:
\[" - [" literally
[^"]* - zero or more characters other than "
(?:"(?!])[^"]*)* - zero or more sequences of
"(?!]) - " not followed with ]
[^"]* - zero or more characters other than "
"] - literal character sequence "]
I have a list of strings like this:
/soccer/poland/ekstraklasa-2008-2009/results/
/soccer/poland/orange-ekstraklasa-2007-2008/results/
/soccer/poland/orange-ekstraklasa-youth-2010-2011/results/
From each string I want to take a middle part resulting in respectively:
ekstraklasa
orange ekstraklasa
orange ekstraklasa youth
My code here does the job but it feels like it can be done in fewer steps and probably with regex alone.
name = re.search('/([-a-z\d]+)/results/', string).group(1) # take the middle part
name = re.search('[-a-z]+', name).group() # trim numbers
if name.endswith('-'):
name = name[:-1] # trim tailing `-` if needed
name = name.replace('-', ' ')
Can anyone see how make it better?
This regex should do the work:
/(?:\/\w+){2}\/([\w\-]+)(?:-\d+){2}/
Explanation:
(?:\/\w+){2} - eat the first two words delimited by /
\/ - eat the next /
([\w\-]+)- match the word characters of hyphens (this is what we're looking for)
(?:-\d+){2} - eat the hyphens and the numbers after the part we're looking for
The result is in the first match group
I cant test it because i am not using python, but i would use an Expression like
^(/soccer/poland/)([a-z\-]*)(.*)$
or
^(/[a-z]*/[a-z]*/)([a-z\-]*)(.*)$
This Expressen works like "/soccer/poland/" at the beginning, than "everything with a to z (small) or -" and the rest of the string.
And than taking 2nd Group!
The Groups should hold this Strings:
/soccer/poland/
orange-ekstraklasa-youth-
2010-2011/results/
And then simply replacing "-" with " " and after that TRIM Spaces.
PS: If ur Using regex101.com e.g., u need to escape / AND just use one Row of String!
Expression
^(\/soccer\/poland\/)([a-z\-]*)(.*)$
And one Row of ur String.
/soccer/poland/orange-ekstraklasa-youth-2010-2011/results/
If u prefere to use the Expression not just for soccer and poland, use
^(\/[a-z]*\/[a-z]*\/)([a-z\-]*)(.*)$
I've been working on this for a while, but it's beyond my understanding of regex.
I'm using Yahoo Pipes on an RSS, and I want to create hashtags from titles; so, I'd like to remove space from everything between quotes, but, if there's a colon within the quotes, I only want the space removed between the words before the colon.
And, it would be great if I could also capture the unspaced words as a group, to be able to use: #$1 to output the hashtag in one step.
So, something like:
"The New Apple: Worlds Within Worlds" Before We Begin...
Could be substituted like #$1 - with this result:
"#TheNewApple: Worlds Within Worlds" Before We Begin...
After some work, I was able to come up with, this regex:
\s(?=\s)?|(‘|’|(Review)|:.*)
("Review" was a word that often came before colons and wouldn't be stripped, if it were later in the title; that's what that's for, but I would like to not require that, to be more universal)
But, it has two problems:
I have to use multiple steps. The result of that regex would be:
"TheNewApple: Worlds Within Worlds" Before We Begin...
And I could then add another regex step, to put the hash # in front
But, it only works if the quotes are first, and I don't know how to fix that...
You can do this all in one step with regex, with a caveat. You run into problems with a repeated capturing group because only the last iteration is available in the replacement string. Searching for ( (\w+))+ and replacing with $2 will replace all the words with just the last match - not what we want.
The way around this is to repeat the pattern an arbitrary number of times that will suffice for your use. Each separate group can be referenced.
Search: "(\w+)(?: (\w+))?(?: (\w+))?(?: (\w+))?(?: (\w+))?(?: (\w+))?
Replace: "#$1$2$3$4$5$6
This will replace up to 6-word titles, exactly as you need them. First, "(\w+) matches any word following a quote. In the replacement string, it is put back as "#$1, adding the hashtag. The rest is a repeated list of (?: (\w+))? matches, each matching a possible space and word. Notice the space is part of a non-capturing group; only the word is part of the inner capture group. In the replacement string, I have $1$2$3$4$5$6, which puts back the words, without the spaces. Notice that a colon will not match any part of this, so it will stop once it hits a colon.
Examples:
"The New Apple: Worlds Within Worlds" Before We Begin...
"The New Apple" Before We Begin...
"One: Two"
only "One" word
this has "Two Words"
"The Great Big Apple Dumpling"
"The Great Big Apple Dumpling Again: Part 2"
Results:
"#TheNewApple: Worlds Within Worlds" Before We Begin...
"#TheNewApple" Before We Begin...
"#One: Two"
only "#One" word
this has "#TwoWords"
"#TheGreatBigAppleDumpling"
"#TheGreatBigAppleDumplingAgain: Part 2"
You can match the text with
"([^:]*)(.*?)"(.*)
then use some programming language to output the result like this:
'"#' + removeSpace($1) + $2 + '"' + $3
I have no idea what language you're using, but this seems like a poor choice for regex. In Python I'd do this:
# Python 3
import re
titles = ['''"The New Apple: Worlds Within Worlds" Before We Begin...''',
'''"Made Up Title: For Example Only" So We Can Continue...''']
hashtagged_titles = list()
for title in titles:
hashtagme, *restofstring = title.split(":")
hashtag = '"#'+hashtagme[1:].translate(str.maketrans('', '', " "))
result = "{}:{}".format(hashtag, restofstring)
hashtagged_titles.append(result)
Do a global search for
\ (?=.*:)
Replaced with nothing. Example
You'll need a second search on the results of that if you want to capture "TheNewApple" as a single word.
I want to extract excerpt from a long string using Regular expression
Example string: "" Is it possible that Germany, which beat Argentina 1-0 today to win the World Cup, that will end up as a loser in terms of economic growth? ""
String to search: " that "
Expected result from regex
" possible that Germany "
" rd Cup, that will end "
I want to search the desired text from the string with -9 and +9 characters from the forward and the backward of the occurence of the searched string. Search string can occur multiple times within the given string.
I am working on an iOS app
using iOS 7.
I have so far created this expression with my little knowledge about reguler expressions but not able to get desired result from that
" (.){0,9} (that) {0,9} "
Remove the spaces in your regex. If you want to capture the matched ones. Then enclose the pattern within capturing groups (ie, ()),
.{9}that.{9}
OR
(?:.{9}|.{0,9})that(?:.{9}|.{0,9})
DEMO
Make the preceding and following characters as optional to match the line which looks like that will change history
Well, in your expression you were just missing the second "." and maybe the "?" for spaces.
.{0,9} ?that ?.{0,9}
Try that.
You can add ( ) for making groups if you want. I added the "?" to make it comply with your other example:
" that will change history"