I want to find words starting with a single non-alphanumerical character, say '$', in a string with re.findall
Example of matching words
$Python
$foo
$any_word123
Example of non-matching words
$$Python
foo
foo$bar
Why \b does not work
If the first character were to be alphanumerical, I could do this.
re.findall(r'\bA\w+', s)
But this does not work for a pattern like \b\$\w+ because \b matches the empty string only between a \w and a \W.
# The line below matches only the last '$baz' which is the one that should not be matched
re.findall(r'\b\$\w+', '$foo $bar x$baz').
The above outputs ['$baz'], but the desired pattern should output ['$foo', '$bar'].
I tried replacing \b by a positive lookbehind with pattern ^|\s, but this does not work because lookarounds must be fixed in length.
What is the correct way to handle this pattern?
The following will match a word starting with a single non-alphanumerical character.
re.findall(r'''
(?: # start non-capturing group
^ # start of string
| # or
\s # space character
) # end non-capturing group
( # start capturing group
[^\w\s] # character that is not a word or space character
\w+ # one or more word characters
) # end capturing group
''', s, re.X)
or just:
re.findall(r'(?:^|\s)([^\w\s]\w+)', s, re.X)
results in:
'$a $b a$c $$d' -> ['$a', '$b']
One way is to use a negative lookbehind with the non-whitespace metacharacter \S.
s = '$Python $foo foo$bar baz'
re.findall(r'(?<!\S)\$\w+', s) # output: ['$Python', '$foo']
Related
I want to match any string that starts with . and word and then optionally any character after a space.
r"^\.(\w+)(?:\s+(.+)\b)?"
eg:
should match
.just one two
.just
.blah one#nine
.blah
.jargon blah
should not match
.jargon
I want this second group mandatory if first group is jargon
Using Python you can exclude matching only jargon using a negative lookahead, and then match 1 or more word characters
Then optionally match 1 or more whitespace characters excluding newlines followed by at least 1 or more characters without newlines.
^\.(?!jargon$)\w+(?:[^\S\n]+.+)?$
The pattern matches:
^ Start of string
\. Match a dot
(?!jargon$) Exlude matching jargon as the only word on the line
\w+ Match 1+ word characters
(?: Non capture group
[^\S\n]+.+ match 1+ whitespace chars excluding newline and then 1+ chars except newlines
)? Close non capture group and make it optional
$ End of string
See a regex demo and a Python demo.
Example
import re
strings = [
".just one two",
".just",
".blah one#nine",
".blah",
".jargon blah",
".jargon"
]
for s in strings:
m = re.match(r"\.(?!jargon$)\w+(?:[^\S\n]+.+)?$", s)
if m:
print(m.group())
Output
.just one two
.just
.blah one#nine
.blah
.jargon blah
One approach would be to phrase your requirement using an alternation:
^\.(?:(?!jargon\b)\w+(?: \S+)*|jargon(?: \S+)+)$
This pattern says to match:
^ from the start of the input
\. match dot
(?:
(?!jargon\b)\w+ match a first term which is NOT "jargon"
(?: \S+)* then match optional following terms zero or more times
| OR
jargon match "jargon" as the first term
(?: \S+)+ then match mandatory one or more terms
)
$ end of the input
Here is a sample Python script:
inp = [".just one two", ".just", ".blah one#nine", ".blah", ".jargon blah", "jargon"]
matches = [x for x in inp if re.search(r'^\.(?:(?!jargon\b)\w+(?: \S+)*|jargon(?: \S+)+)$', x)]
print(matches) # ['.just one two', '.just', '.blah one#nine', '.blah', '.jargon blah']
You could attempt to match the following regular expression:
^\.(?!jargon$)\w+(?= .|$).*
Demo
If successful, this will match the entire string. If one simply wants to know if the string conforms to the requirements .* can be dropped.
(?!jargon$) is a negative lookahead that asserts that the period is not immediately followed by 'jargon' at the end of the string.
(?= .|$) is a positive lookahead that asserts that the string of word characters is followed by a space followed by any character or they terminate the string.
I want to split strings like:
(so) what (are you trying to say)
what (do you mean)
Into lists like:
[(so), what, (are you trying to say)]
[what, (do you mean)]
The code that I tried is below. In the site regexr, the regex expression match the parts that I want but gives a warning, so... I'm not a expert in regex, I don't know what I'm doing wrong.
import re
string = "(so) what (are you trying to say)?"
rx = re.compile(r"((\([\w \w]*\)|[\w]*))")
print(re.split(rx, string ))
Using [\w \w]* is the same as [\w ]* and also matches an empty string.
Instead of using split, you can use re.findall without any capture groups and write the pattern like:
\(\w+(?:[^\S\n]+\w+)*\)|\w+
\( Match (
\w+ Match 1+ word chars
(?:[^\S\n]+\w+)* Optionally repeat matching spaces and 1+ word chars
\) Match )
| Or
\w+ Match 1+ word chars
Regex demo
import re
string = "(so) what (are you trying to say)? what (do you mean)"
rx = re.compile(r"\(\w+(?:[^\S\n]+\w+)*\)|\w+")
print(re.findall(rx, string))
Output
['(so)', 'what', '(are you trying to say)', 'what', '(do you mean)']
For your two examples you can write:
re.split(r'(?<=\)) +| +(?=\()', str)
Python regex<¯\(ツ)/¯>Python code
This does not work, however, for string defined in the OP's code, which contains a question mark, which is contrary to the statement of the question in terms of the two examples.
The regular expression can be broken down as follows.
(?<=\)) # positive lookbehind asserts that location in the
# string is preceded by ')'
[ ]+ # match one or more spaces
| # or
[ ]+ # match one or more spaces
(?=\() # positive lookahead asserts that location in the
# string is followed by '('
In the above I've put each of two space characters in a character class merely to make it visible.
String1: {{word1|word2|word3 (word4 word5)|word6}}
String2: {{word1|word2|word3|word6}}
With this regex sentence:
(?<=\{\{)(\w+(?:\s+\w+)*)\|(\w+(?:\s+\w+)*)\|(\w+(?:\s+\w+)*)\|(\w+(?:\s+\w+)*)(?=\}\})
I capture String2 as groups. How can I change the regex sentence to capture (word4 word5) also as a group?
You can add a (?:\s*(\([^()]*\)))? subpattern:
(?<=\{\{)(\w+(?:\s+\w+)*)\|(\w+(?:\s+\w+)*)\|(\w+(?:\s+\w+)*)(?:\s*(\([^()]*\)))?\|(\w+(?:\s+\w+)*)(?=\}\})
See the regex demo.
The (?:\s*(\([^()]*\)))? part is an optional non-capturing group that matches one or zero occurrences of
\s* - zero or more whitespaces
( - start of a capturing group:
\( - a ( char
[^()]* - zero or more chars other than ( and )
\) - a ) char
) - end of the group.
If you need to make sure only whitespace separated words are allowed inside parentheses, replace [^()]* with \w+(?:\s+\w+)* and insert (?:\s*(\(\w+(?:\s+\w+)*\)))?:
(?<=\{\{)(\w+(?:\s+\w+)*)\|(\w+(?:\s+\w+)*)\|(\w+(?:\s+\w+)*)(?:\s*(\(\w+(?:\s+\w+)*\)))?\|(\w+(?:\s+\w+)*)(?=\}\})
See this regex demo.
You could simplify the expression by matching the desired substrings rather than capturing them. For that you could use the following regular expression.
(?<=[{| ])\w+(?=[}| ])|\([\w ]+\)
Regex demo <¯\(ツ)/¯> Python demo
The elements of the expression are as follows.
(?<= # begin a positive lookbehind
[{| ] # match one of the indicated characters
) # end the positive lookbehind
\w+ # match one or more word characters
(?= # begin a positive lookahead
[}| ] # match one of the indicated characters
) # end positive lookahead
| # or
\( # match character
[\w ]+ # match one or more of the indicated characters
\) # match character
Note that this does not validate the format of the string.
I am trying to extract the name and profession as a list of tuples from the below string using regex.
Input string
text = "Mr John,Carpenter,Mrs Liza,amazing painter"
As you can see the first word is the name followed by the profession which repeats in a comma seperated fashion. The problem is that, I want to get rid of the adjectives that comes along with the profession. For e.g "amazing" in the below example.
Expected output
[('Mr John', 'Carpenter'), ('Mrs Liza', 'painter')]
I stripped out the adjective from the text using "replace" and used the below code using "regex" to get the output. But I am looking for a single regex function to avoid running the string replace. I figured that this has something to do with look ahead in regex but couldn't make it work. Any help would be appreciated.
text.replace("amazing ", "")
txt_new = re.findall("([\w\s]+),([\w\s]+)",text)
If you only want to use word and whitespace characters, this could be another option:
(\w+(?:\s+\w+)*)\s*,\s*(?:\w+\s+)*(\w+)
Explanation
( Capture group 1
\w+(?:\s+\w+)* Match 1+ word chars and optionally repeat 1+ whitespace chars and 1+ word chars
) Close group 1
\s*,\s* Match a comma between optional whitespace chars
(?:\w+\s+)* Optionally repeat 1+ word and 1+ whitespace chars
(\w+) Capture group 2, match 1+ word chars
Regex demo | Python demo
import re
regex = r"(\w+(?:\s+\w+)*)\s*,\s*(?:\w+\s+)*(\w+)"
s = ("Mr John,Carpenter,Mrs Liza,amazing painter")
print(re.findall(regex, s))
Output
[('Mr John', 'Carpenter'), ('Mrs Liza', 'painter')]
Here is one regex approach using re.findall:
text = "Mr John,Carpenter,Mrs Liza,amazing painter"
matches = re.findall(r'\s*([^,]+?)\s*,\s*.*?(\S+)\s*(?![^,])', text)
print(matches)
This prints:
[('Mr John', 'Carpenter'), ('Mrs Liza', 'painter')]
Here is an explanation of the regex pattern:
\s* match optional whitespace
([^,]+?) match the name
\s* optional whitespace
, first comma
\s* optional whitespace
.*? consume all content up until
(\S+) the last profession word
\s* optional whitespace
(?![^,]) assert that what follows is either comma or the end of the input
I want to get patterns involving complete words, not pieces of words.
E.g. 12345 [some word] 1234567 [some word] 123 1679. Random text and the pattern appears again 1111 123 [word] 555.
This should return
[[12345, 1234567, 123, 1679],[1111, 123, 555]]
I am only tolerating one word between the numbers otherwise the whole string would match.
Also note that it is important to capture that 2 matches were found and so a two-element list was returned.
I am running this in python3.
I have tried:
\b(\d+)\b\s\b(\w+)?\b\s\b(\d+)\b
but I am not sure how to scale this to an unrestricted number of matches.
re.findall('\b(\d+)\b\s\b(\w+)?\b\s\b(\d+)\b', string)
This matches [number] [word] [number] but not any number that might follow with or without a word in between.
Are you expecting re.findall() to return a list of lists? It will only return a list - no matter what regex you use.
One approach is to split your input string into sentences and then loop through them
import re
inputArray = re.split('<pattern>',inputText)
outputArray = []
for item in inputArray:
outputArray.append(re.findall('\b(\d+)\b\s\b(\w+)?\b\s\b(\d+)\b', item))
the trick is to find a <pattern> to split your input.
You can't do this in one operation with the Python re engine.
But you could match the sequence with one match, then extract the
digits with another.
This matches the sequence
r"(?<!\w)\d+(?:(?:[^\S\r\n]+[a-zA-Z](?:\w*[a-zA-Z])*)?[^\S\r\n]+\d+)*(?!\w)"
https://regex101.com/r/73AYLU/1
Explained
(?<! \w ) # Not a word behind
\d+ # Many digits
(?: # Optional word block
(?: # Optional words
[^\S\r\n]+ # Horizontal whitespace
[a-zA-Z] # Starts with a letter
(?: \w* [a-zA-Z] )* # Can be digits in middle, ends with a letter
)? # End words, do once
[^\S\r\n]+ # Horizontal whitespace
\d+ # Many digits
)* # End word block, do many times
(?! \w ) # Not a word ahead
This gets the array of digits from the sequence matched above (use findall)
r"(?<!\S)(\d+)(?!\S)"
https://regex101.com/r/BHov38/1
Explained
(?<! \S ) # Whitespace boundary
( \d+ ) # (1)
(?! \S ) # Whitespace boundary
This is a bit complicated, maybe this expression would be just something to look into:
(((\d+)\s*)*(?:\s*\[.*?\]\s*)((\d+)\s*)*)|([A-za-z\s]+)
and script the rest of the problem for a valid solution.
Demo