Regex: Find strings without character
i build regex expression that matches
2 letters or 2 letters folowed by '/' and next 2 letters for example:
rt bl/ws se gn/wd wk bl/rt
/^(((s+)?[a-zA-Z]{2}(/[a-zA-Z]{2})?)(s+|$))+$/i
and that works without problems.
Next problem what I have is match all "word" not containing '/' character.
and replace all matches by duplicate values separated by '/'. For above example excepted output should be:
rt/rt bl/ws se/se gn/wd wk/wk bl/rt
I tried it some time but without success. Could you help me with that ?
Update 1:
I've started with regex that matches words not containing 'at'
(b((?!(at))w)+b)
Et the and I want to replace matched elements with python like
re.sub(r'(b((?!(at))w)+b)', r'1/1', text)
but first have to find right elements ...
javascript regex regex-negation
edited Nov 22 '18 at 7:57
JohnyL
3,7231924
asked Nov 20 '18 at 22:09
walko1234walko1234
466
|
show 1 more comment
i build regex expression that matches
2 letters or 2 letters folowed by '/' and next 2 letters for example:
rt bl/ws se gn/wd wk bl/rt
/^(((s+)?[a-zA-Z]{2}(/[a-zA-Z]{2})?)(s+|$))+$/i
and that works without problems.
Next problem what I have is match all "word" not containing '/' character.
and replace all matches by duplicate values separated by '/'. For above example excepted output should be:
rt/rt bl/ws se/se gn/wd wk/wk bl/rt
I tried it some time but without success. Could you help me with that ?
Update 1:
I've started with regex that matches words not containing 'at'
(b((?!(at))w)+b)
Et the and I want to replace matched elements with python like
re.sub(r'(b((?!(at))w)+b)', r'1/1', text)
but first have to find right elements ...
javascript regex regex-negation
edited Nov 22 '18 at 7:57
JohnyL
3,7231924
asked Nov 20 '18 at 22:09
walko1234walko1234
466
1
Show us what you tried, it will help identify your problem.
– Wiktor Stribiżew
Nov 20 '18 at 22:11
Also, it would be great if you mentioned the programming language you are using the regex in.
– Wiktor Stribiżew
Nov 20 '18 at 22:15
Your regex is very complicated. You could simplify it. For instance,(s+)?is the same ass*. Also have a look atwandb.
– Socowi
Nov 20 '18 at 22:16
@Wiktor Stribiżew OP never excepts that work will be made by someone else. Is there ask for help not right place on this forum ?
– walko1234
Nov 20 '18 at 22:53
Then you may usere.sub(r'(?<!S)[^Wd_]+(?!S)', r'g<0>/g<0>', s). Orre.sub(r'(?<!S)([^Wd_]+)(?!S)', r'1/1', s). To excludeatword matches, usere.sub(r'(?<!S)(?!at(?!S))([^Wd_]+)(?!S)', r'1/1', s)
– Wiktor Stribiżew
Nov 21 '18 at 0:40
|
show 1 more comment
i build regex expression that matches
2 letters or 2 letters folowed by '/' and next 2 letters for example:
rt bl/ws se gn/wd wk bl/rt
/^(((s+)?[a-zA-Z]{2}(/[a-zA-Z]{2})?)(s+|$))+$/i
and that works without problems.
Next problem what I have is match all "word" not containing '/' character.
and replace all matches by duplicate values separated by '/'. For above example excepted output should be:
rt/rt bl/ws se/se gn/wd wk/wk bl/rt
I tried it some time but without success. Could you help me with that ?
Update 1:
I've started with regex that matches words not containing 'at'
(b((?!(at))w)+b)
Et the and I want to replace matched elements with python like
re.sub(r'(b((?!(at))w)+b)', r'1/1', text)
but first have to find right elements ...
javascript regex regex-negation
edited Nov 22 '18 at 7:57
JohnyL
3,7231924
asked Nov 20 '18 at 22:09
walko1234walko1234
466
i build regex expression that matches
2 letters or 2 letters folowed by '/' and next 2 letters for example:
rt bl/ws se gn/wd wk bl/rt
/^(((s+)?[a-zA-Z]{2}(/[a-zA-Z]{2})?)(s+|$))+$/i
and that works without problems.
Next problem what I have is match all "word" not containing '/' character.
and replace all matches by duplicate values separated by '/'. For above example excepted output should be:
rt/rt bl/ws se/se gn/wd wk/wk bl/rt
I tried it some time but without success. Could you help me with that ?
Update 1:
I've started with regex that matches words not containing 'at'
(b((?!(at))w)+b)
Et the and I want to replace matched elements with python like
re.sub(r'(b((?!(at))w)+b)', r'1/1', text)
but first have to find right elements ...
javascript regex regex-negation
javascript regex regex-negation
edited Nov 22 '18 at 7:57
JohnyL
3,7231924
asked Nov 20 '18 at 22:09
walko1234walko1234
466
edited Nov 22 '18 at 7:57
JohnyL
3,7231924
asked Nov 20 '18 at 22:09
walko1234walko1234
466
edited Nov 22 '18 at 7:57
JohnyL
3,7231924
edited Nov 22 '18 at 7:57
JohnyL
3,7231924
edited Nov 22 '18 at 7:57
JohnyL
3,7231924
3,7231924
asked Nov 20 '18 at 22:09
walko1234walko1234
466
asked Nov 20 '18 at 22:09
walko1234walko1234
466
asked Nov 20 '18 at 22:09
walko1234walko1234
466
466
1
Show us what you tried, it will help identify your problem.
– Wiktor Stribiżew
Nov 20 '18 at 22:11
Also, it would be great if you mentioned the programming language you are using the regex in.
– Wiktor Stribiżew
Nov 20 '18 at 22:15
Your regex is very complicated. You could simplify it. For instance,(s+)?is the same ass*. Also have a look atwandb.
– Socowi
Nov 20 '18 at 22:16
@Wiktor Stribiżew OP never excepts that work will be made by someone else. Is there ask for help not right place on this forum ?
– walko1234
Nov 20 '18 at 22:53
Then you may usere.sub(r'(?<!S)[^Wd_]+(?!S)', r'g<0>/g<0>', s). Orre.sub(r'(?<!S)([^Wd_]+)(?!S)', r'1/1', s). To excludeatword matches, usere.sub(r'(?<!S)(?!at(?!S))([^Wd_]+)(?!S)', r'1/1', s)
– Wiktor Stribiżew
Nov 21 '18 at 0:40
|
show 1 more comment
1
Show us what you tried, it will help identify your problem.
– Wiktor Stribiżew
Nov 20 '18 at 22:11
Also, it would be great if you mentioned the programming language you are using the regex in.
– Wiktor Stribiżew
Nov 20 '18 at 22:15
Your regex is very complicated. You could simplify it. For instance,(s+)?is the same ass*. Also have a look atwandb.
– Socowi
Nov 20 '18 at 22:16
@Wiktor Stribiżew OP never excepts that work will be made by someone else. Is there ask for help not right place on this forum ?
– walko1234
Nov 20 '18 at 22:53
Then you may usere.sub(r'(?<!S)[^Wd_]+(?!S)', r'g<0>/g<0>', s). Orre.sub(r'(?<!S)([^Wd_]+)(?!S)', r'1/1', s). To excludeatword matches, usere.sub(r'(?<!S)(?!at(?!S))([^Wd_]+)(?!S)', r'1/1', s)
– Wiktor Stribiżew
Nov 21 '18 at 0:40
1
1
Show us what you tried, it will help identify your problem.
– Wiktor Stribiżew
Nov 20 '18 at 22:11
Show us what you tried, it will help identify your problem.
– Wiktor Stribiżew
Nov 20 '18 at 22:11
Also, it would be great if you mentioned the programming language you are using the regex in.
– Wiktor Stribiżew
Nov 20 '18 at 22:15
Also, it would be great if you mentioned the programming language you are using the regex in.
– Wiktor Stribiżew
Nov 20 '18 at 22:15
Your regex is very complicated. You could simplify it. For instance,
(s+)? is the same as s*. Also have a look at w and b.– Socowi
Nov 20 '18 at 22:16
Your regex is very complicated. You could simplify it. For instance,
(s+)? is the same as s*. Also have a look at w and b.– Socowi
Nov 20 '18 at 22:16
@Wiktor Stribiżew OP never excepts that work will be made by someone else. Is there ask for help not right place on this forum ?
– walko1234
Nov 20 '18 at 22:53
@Wiktor Stribiżew OP never excepts that work will be made by someone else. Is there ask for help not right place on this forum ?
– walko1234
Nov 20 '18 at 22:53
Then you may use
re.sub(r'(?<!S)[^Wd_]+(?!S)', r'g<0>/g<0>', s). Or re.sub(r'(?<!S)([^Wd_]+)(?!S)', r'1/1', s). To exclude at word matches, use re.sub(r'(?<!S)(?!at(?!S))([^Wd_]+)(?!S)', r'1/1', s)– Wiktor Stribiżew
Nov 21 '18 at 0:40
Then you may use
re.sub(r'(?<!S)[^Wd_]+(?!S)', r'g<0>/g<0>', s). Or re.sub(r'(?<!S)([^Wd_]+)(?!S)', r'1/1', s). To exclude at word matches, use re.sub(r'(?<!S)(?!at(?!S))([^Wd_]+)(?!S)', r'1/1', s)– Wiktor Stribiżew
Nov 21 '18 at 0:40
|
show 1 more comment
2 Answers
2
active
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votes
in python you can do something like:
re.sub(r'(?:(?<=^)|(?<= ))(w+)(?= )',r'1/1','rt bl/ws se gn/wd wk bl/rt')
'rt/rt bl/ws se/se gn/wd wk/wk bl/rt'
answered Nov 20 '18 at 23:17
OnyambuOnyambu
16k1522
add a comment |
If you're using a flavour of regex that supports negative lookbehind this regex should find the strings you want to replace:
(?<!/)([A-Za-z]{2}(?!/))
(?<!/) - makes sure there isn't a / behind the match([A-Za-z]{2} looks for two alphabetic characters(?!/)) makes sure there isn't a trailing /
In Python you would use it like this:
print(re.sub(r'(?<!/)([A-Za-z]{2}(?!/))',r'1/1','rt bl/ws se gn/wd wk bl/rt'))
Output:
rt/rt bl/ws se/se gn/wd wk/wk bl/rt
Demo on rextester
In PHP you would use it like so:
$str = 'rt bl/ws se gn/wd wk bl/rt';
echo preg_replace('/(?<!/)([A-Za-z]{2}(?!/))/', '$1/$1', $str);
Output:
rt/rt bl/ws se/se gn/wd wk/wk bl/rt
Demo on 3v4l.org
answered Nov 20 '18 at 23:03
NickNick
33.3k132042
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
in python you can do something like:
re.sub(r'(?:(?<=^)|(?<= ))(w+)(?= )',r'1/1','rt bl/ws se gn/wd wk bl/rt')
'rt/rt bl/ws se/se gn/wd wk/wk bl/rt'
answered Nov 20 '18 at 23:17
OnyambuOnyambu
16k1522
add a comment |
in python you can do something like:
re.sub(r'(?:(?<=^)|(?<= ))(w+)(?= )',r'1/1','rt bl/ws se gn/wd wk bl/rt')
'rt/rt bl/ws se/se gn/wd wk/wk bl/rt'
answered Nov 20 '18 at 23:17
OnyambuOnyambu
16k1522
add a comment |
in python you can do something like:
re.sub(r'(?:(?<=^)|(?<= ))(w+)(?= )',r'1/1','rt bl/ws se gn/wd wk bl/rt')
'rt/rt bl/ws se/se gn/wd wk/wk bl/rt'
answered Nov 20 '18 at 23:17
OnyambuOnyambu
16k1522
in python you can do something like:
re.sub(r'(?:(?<=^)|(?<= ))(w+)(?= )',r'1/1','rt bl/ws se gn/wd wk bl/rt')
'rt/rt bl/ws se/se gn/wd wk/wk bl/rt'
answered Nov 20 '18 at 23:17
OnyambuOnyambu
16k1522
answered Nov 20 '18 at 23:17
OnyambuOnyambu
16k1522
answered Nov 20 '18 at 23:17
OnyambuOnyambu
16k1522
answered Nov 20 '18 at 23:17
OnyambuOnyambu
16k1522
16k1522
add a comment |
add a comment |
If you're using a flavour of regex that supports negative lookbehind this regex should find the strings you want to replace:
(?<!/)([A-Za-z]{2}(?!/))
(?<!/) - makes sure there isn't a / behind the match([A-Za-z]{2} looks for two alphabetic characters(?!/)) makes sure there isn't a trailing /
In Python you would use it like this:
print(re.sub(r'(?<!/)([A-Za-z]{2}(?!/))',r'1/1','rt bl/ws se gn/wd wk bl/rt'))
Output:
rt/rt bl/ws se/se gn/wd wk/wk bl/rt
Demo on rextester
In PHP you would use it like so:
$str = 'rt bl/ws se gn/wd wk bl/rt';
echo preg_replace('/(?<!/)([A-Za-z]{2}(?!/))/', '$1/$1', $str);
Output:
rt/rt bl/ws se/se gn/wd wk/wk bl/rt
Demo on 3v4l.org
answered Nov 20 '18 at 23:03
NickNick
33.3k132042
add a comment |
If you're using a flavour of regex that supports negative lookbehind this regex should find the strings you want to replace:
(?<!/)([A-Za-z]{2}(?!/))
(?<!/) - makes sure there isn't a / behind the match([A-Za-z]{2} looks for two alphabetic characters(?!/)) makes sure there isn't a trailing /
In Python you would use it like this:
print(re.sub(r'(?<!/)([A-Za-z]{2}(?!/))',r'1/1','rt bl/ws se gn/wd wk bl/rt'))
Output:
rt/rt bl/ws se/se gn/wd wk/wk bl/rt
Demo on rextester
In PHP you would use it like so:
$str = 'rt bl/ws se gn/wd wk bl/rt';
echo preg_replace('/(?<!/)([A-Za-z]{2}(?!/))/', '$1/$1', $str);
Output:
rt/rt bl/ws se/se gn/wd wk/wk bl/rt
Demo on 3v4l.org
answered Nov 20 '18 at 23:03
NickNick
33.3k132042
add a comment |
If you're using a flavour of regex that supports negative lookbehind this regex should find the strings you want to replace:
(?<!/)([A-Za-z]{2}(?!/))
(?<!/) - makes sure there isn't a / behind the match([A-Za-z]{2} looks for two alphabetic characters(?!/)) makes sure there isn't a trailing /
In Python you would use it like this:
print(re.sub(r'(?<!/)([A-Za-z]{2}(?!/))',r'1/1','rt bl/ws se gn/wd wk bl/rt'))
Output:
rt/rt bl/ws se/se gn/wd wk/wk bl/rt
Demo on rextester
In PHP you would use it like so:
$str = 'rt bl/ws se gn/wd wk bl/rt';
echo preg_replace('/(?<!/)([A-Za-z]{2}(?!/))/', '$1/$1', $str);
Output:
rt/rt bl/ws se/se gn/wd wk/wk bl/rt
Demo on 3v4l.org
answered Nov 20 '18 at 23:03
NickNick
33.3k132042
If you're using a flavour of regex that supports negative lookbehind this regex should find the strings you want to replace:
(?<!/)([A-Za-z]{2}(?!/))
(?<!/) - makes sure there isn't a / behind the match([A-Za-z]{2} looks for two alphabetic characters(?!/)) makes sure there isn't a trailing /
In Python you would use it like this:
print(re.sub(r'(?<!/)([A-Za-z]{2}(?!/))',r'1/1','rt bl/ws se gn/wd wk bl/rt'))
Output:
rt/rt bl/ws se/se gn/wd wk/wk bl/rt
Demo on rextester
In PHP you would use it like so:
$str = 'rt bl/ws se gn/wd wk bl/rt';
echo preg_replace('/(?<!/)([A-Za-z]{2}(?!/))/', '$1/$1', $str);
Output:
rt/rt bl/ws se/se gn/wd wk/wk bl/rt
Demo on 3v4l.org
answered Nov 20 '18 at 23:03
NickNick
33.3k132042
edited Nov 22 '18 at 7:09
answered Nov 20 '18 at 23:03
NickNick
33.3k132042
answered Nov 20 '18 at 23:03
NickNick
33.3k132042
answered Nov 20 '18 at 23:03
NickNick
33.3k132042
33.3k132042
add a comment |
add a comment |
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1
Show us what you tried, it will help identify your problem.
– Wiktor Stribiżew
Nov 20 '18 at 22:11
Also, it would be great if you mentioned the programming language you are using the regex in.
– Wiktor Stribiżew
Nov 20 '18 at 22:15
Your regex is very complicated. You could simplify it. For instance,
(s+)?is the same ass*. Also have a look atwandb.– Socowi
Nov 20 '18 at 22:16
@Wiktor Stribiżew OP never excepts that work will be made by someone else. Is there ask for help not right place on this forum ?
– walko1234
Nov 20 '18 at 22:53
Then you may use
re.sub(r'(?<!S)[^Wd_]+(?!S)', r'g<0>/g<0>', s). Orre.sub(r'(?<!S)([^Wd_]+)(?!S)', r'1/1', s). To excludeatword matches, usere.sub(r'(?<!S)(?!at(?!S))([^Wd_]+)(?!S)', r'1/1', s)– Wiktor Stribiżew
Nov 21 '18 at 0:40