test_string: '**Amount** : $25k **Name** : James **Excess** : None Returned \n **In Suit?** Y **Venue** : SF **Insurance** : N/A \n **FTSA** : None listed'
import re
regex = r"(?:^|[^.?*,!-]*(?<=[.?\s*,!-]))(n/a)(?=[\s.?*!,-])[^.?*,!-]*[.?*,!-]"
subst = ""
result = re.sub(regex, subst, test_str, 0, re.IGNORECASE | re.MULTILINE)
I tried to extract '**Insurance** : N/A' from the string. But my above code doesn't work. How can I make it?
Thanks in advance!
I would treat the content like a (semi-structured) key-value file format.
You can match the key-value pairs with a regex like this:
(\*\*[a-zA-Y ?]+\*\*) : ((?:(?!\*\*).)*)(?= |$)
Demo
Explanation:
(\*\*[a-zA-Y ?]+\*\*) the key: you may have to adjust the character range
: the kv separator with surrounded by spaces
((?:(?!\*\*).)*) the value is captured with a tempered greedy token: everything but literal ** followed by (?= |$) the end of string $ or a separating space.
(?= |$)
Sample Code:
import re
regex = r"(\*\*[a-zA-Z ?]+\*\*) : ((?:(?!\*\*).)*)(?= |$)"
test_str = "**Amount** : $25k **Name** : James **Excess** : None Returned \\n **In Suit?** : Y **Venue** : SF **Insurance** : N/A \\n **FTSA** : None listed"
matches = re.finditer(regex, test_str, re.MULTILINE)
for match in matches:
if match.group(1) == "**Insurance**":
print (match.group(2))
Related
Let's start with the following input.
Input = 'blue, blueblue, b l u e'
I want to match everything that is not the string 'blue'. Note that blueblue should not match, but single characters should (even if present in match string).
From this, If I replace the matches with an empty string, it should return:
Result = 'blueblueblue'
I have tried with [^\bblue\b]+
but this matches the last four single characters 'b', 'l','u','e'
Another solution:
(?<=blue)(?:(?!blue).)+(?=blue|$)|^(?:(?!blue).)+(?=blue|$)
Regex demo
If you regex engine support the \K flag, then we can try:
/blue\K|.*?(?=blue|$)/gm
Demo
This pattern says to match:
blue match "blue"
\K but then forget that match
| OR
.*? match anything else until reaching
(?=blue|$) the next "blue" or the end of the string
Edit:
On JavaScript, we can try the following replacement:
var input = "blue, blueblue, b l u e";
var output = input.replace(/blue|.*?(?=blue|$)/g, (x) => x != "blue" ? "" : "blue");
console.log(output);
I have a list of string i.e.
slist = ["-args", "-111111", "20-args", "20 - 20", "20-10", "args-deep"]
I want to remove the '-' from string where it is the first character and is followed by strings but not numbers or if before the '-' there is number/alphabet but after it is alphabets, then it should replace the '-' with space
So for the list slist I want the output as
["args", "-111111", "20 args", "20 - 20", "20-10", "args deep"]
I have tried
slist = ["-args", "-111111", "20-args", "20 - 20", "20-10", "args-deep"]
nlist = list()
for estr in slist:
nlist.append(re.sub("((^-[a-zA-Z])|([0-9]*-[a-zA-Z]))", "", estr))
print (nlist)
and i get the output
['rgs', '-111111', 'rgs', '20 - 20', '20-10', 'argseep']
You may use
nlist.append(re.sub(r"-(?=[a-zA-Z])", " ", estr).lstrip())
or
nlist.append(re.sub(r"-(?=[^\W\d_])", " ", estr).lstrip())
Result: ['args', '-111111', '20 args', '20 - 20', '20-10', 'args deep']
See the Python demo.
The -(?=[a-zA-Z]) pattern matches a hyphen before an ASCII letter (-(?=[^\W\d_]) matches a hyphen before any letter), and replaces the match with a space. Since - may be matched at the start of a string, the space may appear at that position, so .lstrip() is used to remove the space(s) there.
Here, we might just want to capture the first letter after a starting -, then replace it with that letter only, maybe with an i flag expression similar to:
^-([a-z])
DEMO
Test
# coding=utf8
# the above tag defines encoding for this document and is for Python 2.x compatibility
import re
regex = r"^-([a-z])"
test_str = ("-args\n"
"-111111\n"
"20-args\n"
"20 - 20\n"
"20-10\n"
"args-deep")
subst = "\\1"
# You can manually specify the number of replacements by changing the 4th argument
result = re.sub(regex, subst, test_str, 0, re.MULTILINE | re.IGNORECASE)
if result:
print (result)
# Note: for Python 2.7 compatibility, use ur"" to prefix the regex and u"" to prefix the test string and substitution.
Demo
const regex = /^-([a-z])/gmi;
const str = `-args
-111111
20-args
20 - 20
20-10
args-deep`;
const subst = `$1`;
// The substituted value will be contained in the result variable
const result = str.replace(regex, subst);
console.log('Substitution result: ', result);
RegEx
If this expression wasn't desired, it can be modified or changed in regex101.com.
RegEx Circuit
jex.im visualizes regular expressions:
One option could be to do 2 times a replacement. First match the hyphen at the start when there are only alphabets following:
^-(?=[a-zA-Z]+$)
Regex demo
In the replacement use an empty string.
Then capture 1 or more times an alphabet or digit in group 1, match - followed by capturing 1+ times an alphabet in group 2.
^([a-zA-Z0-9]+)-([a-zA-Z]+)$
Regex demo
In the replacement use r"\1 \2"
For example
import re
regex1 = r"^-(?=[a-zA-Z]+$)"
regex2 = r"^([a-zA-Z0-9]+)-([a-zA-Z]+)$"
slist = ["-args", "-111111", "20-args", "20 - 20", "20-10", "args-deep"]
slist = list(map(lambda s: re.sub(regex2, r"\1 \2", re.sub(regex1, "", s)), slist))
print(slist)
Result
['args', '-111111', '20 args', '20 - 20', '20-10', 'args deep']
Python demo
I would like to select all long words from a string: re.findall("[a-z]{3,}")
However, for a reason I can use substitute only. Hence I need to substitute everything but words of 3 and more letters by space. (e.g. abc de1 fgh ij -> abc fgh)
How would such a regex look like?
The result should be all "[a-z]{3,}" concatenated by spaces. However, you can use substitution only.
Or in Python: Find a regex such that
re.sub(regex, " ", text) == " ".join(re.findall("[a-z]{3,}", text))
Here is some test cases
import re
solution_regex="..."
for test_str in ["aaa aa aaa aa",
"aaa aa11",
"11aaa11 11aa11",
"aa aa1aa aaaa"
]:
expected_str = " ".join(re.findall("[a-z]{3,}", test_str))
print(test_str, "->", expected_str)
if re.sub(solution_regex, " ", test_str)!=expected_str:
print("ERROR")
->
aaa aa aaa aa -> aaa aaa
aaa aa11 -> aaa
11aaa11 11aa11 -> aaa
aa aa1aa aaaa -> aaaa
Note that space is no different than any other symbol.
\b(?:[a-z,A-Z,_]{1,2}|\w*\d+\w*)\b
Explanation:
\b means that the substring you are looking for start and end by border of word
(?: ) - non captured group
\w*\d+\w* Any word that contains at least one digit and consists of digits, '_' and letters
Here you can see the test.
You can use the regex
(\s\b(\d*[a-z]\d*){1,2}\b)|(\s\b\d+\b)
and replace with an empty string, here is a python code for the same
import re
regex = r"(\s\b(\d*[a-z]\d*){1,2}\b)|(\s\b\d+\b)"
test_str = "abcd abc ad1r ab a11b a1 11a 1111 1111abcd a1b2c3d"
subst = ""
# You can manually specify the number of replacements by changing the 4th argument
result = re.sub(regex, subst, test_str, 0)
if result:
print (result)
here is a demo
In Autoit this works for me
#include <Array.au3>
$a = StringRegExp('abc de1 fgh ij 234234324 sdfsdfsdf wfwfwe', '(?i)[a-z]{3,}', 3)
ConsoleWrite(_ArrayToString($a, ' ') & #CRLF)
Result ==> abc fgh sdfsdfsdf wfwfwe
import re
regex = r"(?:^|\s)[^a-z\s]*[a-z]{0,2}[^a-z\s]*(?:\s|$)"
str = "abc de1 fgh ij"
subst = " "
result = re.sub(regex, subst, str)
print (result)
Output:
abc fgh
Explanation:
(?:^|\s) : non capture group, start of string or space
[^a-z\s]* : 0 or more any character that is not letter or space
[a-z]{0,2} : 0, 1 or 2 letters
[^a-z\s]* : 0 or more any character that is not letter or space
(?:\s|$) : non capture group, space or end of string
With the other ideas posted here, I came up with an answer. I can't believe I missed that:
([^a-z]+|(?<![a-z])[a-z]{1,2}(?![a-z]))+
https://regex101.com/r/IIxkki/2
Match either non-letters, or up to two letters bounded by non-letters.
I need a regex that extract pattern after specific word (her like Limits::)
i have teststring ,So let's say the text is always between delimiter !Limits::****! :
*ksjfl kfj sdfasdfaf dfasf asd sdf a dfasd fdaf ad f afdfaf dfad bla bla ksfajs ldsfskj !Limits::WLo1/WHi1/WHi1/WHi1,WLo2/WHi2/WHi/WHi2,.hier repeated pattern..,WLon/WHin/CLon/CHin!
fasdfakl skdfkas sflas fasf sdf afasf
i just want only words :
WLo1
WHi1
WHi1
WHi1
WLo2
WHi2
WHi
WHi2
.
.
.
WLon
WHin
CLon
CHin
i have tested like (?:!\w+::(?:(\w+)/(\w+)/(\w+)/(\w+)))|(?:,(\w+)/(\w+)/(\w+)/(\w+))+.*!, with fail
Regular expressions:
/(W.*|C.*)(?=\/|!|,)/g : match words beginning with W or C followed by / , !, or ,
/\/|,.*(?=,)|,/ : remove / or , or any characters followed by , or , from string returned from first RegExp
var str = "*ksjfl kfj sdfasdfaf dfasf asd sdf a dfasd fdaf ad f afdfaf dfad bla bla ksfajs ldsfskj !Limits::WLo1/WHi1/WHi1/WHi1,WLo2/WHi2/WHi/WHi2,.hier repeated pattern..,WLon/WHin/CLon/CHin! fasdfakl skdfkas sflas fasf sdf afasf";
var res = str.match(/(W.*|C.*)(?=\/|!|,)/g)[0].split(/\/|,.*(?=,)|,/);
document.body.textContent = res.join(" ")
I don't know what the ending delimiter is, so if it matters, update your question and I'll amend this expression:
/(?<=Limits::)(?:(.+?)\/)+/i
Searches for Limits::, then repeating strings ending with /, your words will be in group 1.
I've been trying to do this Regex for a while now. I'd like to create one that matches all the spaces of a text, except those in literal string.
Exemple:
123 Foo "String with spaces"
Space between 123 and Foo would match, as well as the one between Foo and "String with spaces", but only those two.
Thanks
A common, simple strategy for this is to count the number of quotes leading up to your location in the string. If the count is odd, you are inside a quoted string; if the amount is even, you are outside a quoted string. I can't think of a way to do this in regular expressions, but you could use this strategy to filter the results.
You could use re.findall to match either a string or a space and then afterwards inspect the matches:
import re
hits = re.findall("\"(?:\\\\.|[^\\\"])*\"|[ ]", 'foo bar baz "another\\" test\" and done')
for h in hits:
print "found: [%s]" % h
yields:
found: [ ]
found: [ ]
found: [ ]
found: ["another\" test"]
found: [ ]
found: [ ]
A short explanation:
" # match a double quote
(?: # start non-capture group 1
\\\\. # match a backslash followed by any character (except line breaks)
| # OR
[^\\\"] # match any character except a '\' and '"'
)* # end non-capture group 1 and repeat it zero or more times
" # match a double quote
| # OR
[ ] # match a single space
If this ->123 Foo "String with spaces" <- is your structure for a line that is to say text followed by a quoted text you could create 2 groups the quoted and the unquoted text and an tackle them separately.
ex.regex -> (.*)(".*") where $1 should contain ->123 Foo <- and $2 ->"String with spaces"<-
java example.
String aux = "123 Foo \"String with spaces\"";
String regex = "(.*)(\".*\")";
String unquoted = aux.replaceAll(regex, "$1").replace(" ", "");
String quoted = aux.replaceAll(regex, "$2");
System.out.println(unquoted+quoted);
javascript example.
<SCRIPT LANGUAGE="JavaScript">
<!--
str='1 23 Foo \"String with spaces\"';
re = new RegExp('(.*)(".*")') ;
var quoted = str.replace(re, "$1");
var unquoted = str.replace(re, "$2");
document.write (quoted.split(' ').join('')+unquoted);
// -->
</SCRIPT>