Trying to find multiple word match in given text.For example :
text = "oracle sql"
regex = "(oracle\\ sql|sql)"
re.findall(regex,text,re.I)
Output actual
oracle sql
Expected output
oracle sql,sql
Can anyone tell me, where is problem with regex expression ?
Updated:
#jim it won't work ,if multiple overlapping comes, for example :
re.findall("(?=(spark|spark sql|sql))","spark sql",re.I)
Actual Output
['spark','sql']
Expected Output :
['spark','sql','spark sql']]
Note : In the above case if both are matched then it won't match combination of words.
Updated :
Check link : repl.it/repls/NewFaithfulMath
You don't need to escape whitespace.
import re
text = "oracle sql"
regex = "(oracle sql|sql)"
print re.findall(regex, text, re.I)
From the documentation:
Return all non-overlapping matches of pattern in string, as a list of
strings.
This counts as an overlapping match.
Returning overlapping matches
You can use a lookahead to capture the string you're looking for, but because it's technically matching on the lookahead they are not overlapping.
import re
text = "oracle sql"
regex = "(?=(oracle sql|sql))"
print re.findall(regex, text, re.I)
Output:
['oracle sql', 'sql']
See it in action.
The downside of this implementation is that it will only find 1 match for each word at a particular position in a string. This is due to overlapping matches.
For example (my test|my|test) will only find ['my test', 'test'].
You could always use a regex replacement that will find overlapping matches too, such as regex, but this will still only find ['my test', 'test'] with the pattern (my test|my|test):
import regex as re
text = "oracle sql"
regex = "(oracle sql|sql)"
print re.findall(regex, text, re.I, overlapped=True)
Recursion
Regex will only find one match per character. It has already found the match for the first character based on "oracle sql" so you can't get a match on just oracle. You can't find every single one.
However... You could use a recursive function to try to match the same string with all of the items - what has already been matched.
I am not sure how performant this code will be as you could execute a lot of regex searches.
import re
def find_all_matches(text, items):
regex_items = '|'.join(items)
regex = "(?=({}))".format(regex_items)
matches = re.findall(regex, text, re.I)
new_items = [i for i in items if i not in matches]
if new_items:
new_matches = find_all_matches(text, new_items)
return matches + new_matches
return matches
print find_all_matches("oracle sql", ['oracle sql', 'oracle', 'sql'])
Output:
['oracle sql', 'sql', 'oracle']
No regex
Lastly you could implement this without regex. Again I haven't looked at the performance of this.
def find_all_matches(text, items):
return [i for i in items if i in text]
print find_all_matches("oracle sql", ['oracle sql', 'oracle', 'sql'])
Output:
['oracle sql', 'oracle', 'sql']
Related
In regex alternation, is there a way to retrieve which alternation was matched? I just need the first alternation match, not all the alternations that match.
For example, I have a regex like this
pattern = r'(abc.*def|mno.*pqr|mno.*pqrt|.....)'
string = 'mnoxxxpqrt'
I want the output to be 'mno.*pqr'
How should I write the regex statement? Python language is preferred.
To do this efficiently without any iterations, you can put your desired sub-patterns in a list and join them into one alternation pattern with each sub-pattern enclosed in a capture group (so the resulting pattern looks like (abc.*def)|(mno.*pqr) instead of (abc.*def|mno.*pqr)). You can then obtain the group number of the sub-pattern with the Match object's lastindex attribute and in turn obtain the matching sub-pattern from the original list of sub-patterns:
import re
patterns = [r'abc.*def', r'mno.*pqr', r'mno.*pqrt']
pattern = '|'.join(map('({})'.format, patterns))
string = 'mno_foobar_pqrt'
print(pattern)
print(patterns[re.search(pattern, string).lastindex - 1])
This outputs:
(abc.*def)|(mno.*pqr)|(mno.*pqrt)
mno.*pqr
Demo: https://replit.com/#blhsing/JointBruisedMention
You can use capture groups:
import re
string = 'abcxxxdef'
patterns = ['abc.*def', 'mno.*pqr']
match = re.match(r'((abc.*def)|(mno.*pqr))',string)
groups = match.groups()
alternations = []
for i in range(1, len(groups)):
if (groups[i] != None):
pattern = patterns[i-1]
break
print(pattern)
Result: mno.*pqr
Expressions inside round brackets are capture groups, they correspond to the 1st to last index of the response. The 0th index is the whole match.
Then you would need to find the index which matched. Except your patterns would need to be fined before hand.
Well you could iterate the terms in the regex alternation:
string = 'abcxxxdef'
pattern = r'(abc.*def|mno.*pqr)'
terms = pattern[1:-1].split("|")
for term in terms:
if re.search(term, string):
print("MATCH => " + term)
This prints:
MATCH => abc.*def
The right answer to the question How should I write the regex statement? should actually be:
There is no known way to write the regex statement using the provided regex pattern which will allow to extract from the regex search result the information which of the alternatives have triggered the match.
And as there is no way to do it using the given pattern it is necessary to change the regex pattern which then makes it possible to extract from the match the requested information.
A possible way around this regex engine limitation is proposed below, but it requires an additional regex pattern search and has the disadvantage that there is a chance that it fails for some special search pattern alternatives.
The below provided code allows usage of simpler regex patterns without defining groups and works the "other way around" by checking which of the alternate patterns triggers a match in the found match for the entire regex:
import re
pattern = r'abc.*def|mno.*pqr|mno.*pqrt'
text = 'mnoxxxpqrt'
match = re.match(pattern,text)[0]
print(next(p for p in pattern.split('|') if re.match(p, match)))
It might fail in case when in the text found match string fails to be also a match for the single regex pattern what can happen for example if a non-word boundary \B requirement is used in the search pattern ( as mentioned in the comments by Kelly Bundy ).
A not failing alternative solution is to perform the regex search using a modified regex pattern. Below an approach using a dictionary for defining the alternatives and a function returning the matched group:
import re
dct_alts = {1:r'(abc.*def)',2:r'(mno.*pqr)',3:r'(mno.*pqrt)'}
# ^-- the dictionary index is the index of the matching group in the found match.
text = 'mnoxxxpqrt'
def get_matched_group(dct_alts, text):
pattern = '|'.join(dct_alts.values())
re_match = re.match(pattern, text)
return(dct_alts[re_match.lastindex])
print(get_matched_group(dct_alts, text))
prints
(mno.*pqr)
For the sake of completeness a function returning a list of all of the alternatives which give a match (not only the first one which matches):
import re
lst_alts = [r'abc.*def', r'mno.*pqr', r'mno.*pqrt']
text = 'mnoxxxpqrt'
def get_all_matched_groups(lst_alts, text):
matches = []
for pattern in lst_alts:
re_match = re.match(pattern, text)
if re_match:
matches.append(pattern)
return matches
print(get_all_matched_groups(lst_alts, text))
prints
['mno.*pqr', 'mno.*pqrt']
The problem is simple, I'm given a random string and a random pattern and I'm told to get all the posible combinations of that pattern that occur in the string and mark then with [target] and [endtarget] at the beggining and end.
For example:
given the following text: "XuyZB8we4"
and the following pattern: "XYZAB"
The expected output would be: "[target]X[endtarget]uy[target]ZB[endtarget]8we4".
I already got the part that identifies all the words, but I can't find a way of placing the [target] and [endtarget] strings after and before the pattern (called in the code match).
import re
def tagger(text, search):
place_s = "[target]"
place_f = "[endtarget]"
pattern = re.compile(rf"[{search}]+")
matches = pattern.finditer(text)
for match in matches:
print(match)
return test_string
test_string = "alsikjuyZB8we4 aBBe8XAZ piarBq8 Bq84Z "
pattern = "XYZAB"
print(tagger(test_string, pattern))
I also tried the for with the sub method, but I couldn't get it to work.
for match in matches:
re.sub(match.group(0), place_s + match.group(0) + place_f, text)
return text
re.sub allows you to pass backreferences to matched groups within your pattern. so you do need to enclose your pattern in parentheses, or create a named group, and then it will replace all matches in the entire string at once with your desired replacements:
In [10]: re.sub(r'([XYZAB]+)', r'[target]\1[endtarget]', test_string)
Out[10]: 'alsikjuy[target]ZB[endtarget]8we4 a[target]BB[endtarget]e8[target]XAZ[endtarget] piar[target]B[endtarget]q8 [target]B[endtarget]q84[target]Z[endtarget] '
With this approach, re.finditer is not not needed at all.
describe aws_security_group({:group_id=>"sg-ezsrzerzer", :vpc_id=>"vpc-zfds54zef4s"}) do
I try to filter the sg-ezsrzerzer out of it (so I want to filter on start sg- till double quote). I'm using python
I currently have:
import re
a = 'describe aws_security_group({:group_id=>"sg-ezsrzerzer", :vpc_id=>"vpc-zfds54zef4s"}) do'
test = re.findall(r'\bsg-.*\b', a)
print(test)
output is
['sg-ezsrzerzer", :vpc_id=>"vpc-zfds54zef4s"}) do']
How do I only get ['sg-ezsrzerzer']?
The pattern (?<=group_id=\>").+?(?=\") would work nicely if the goal is to extract the group_id value within a given string formatted as in your example.
(?<=group_id=\>") Looks behind for the sub-string group_id=>" before the string to be matched.
.+? Matches one or more of any character lazily.
(?=\") Looks ahead for the character " following the match (effectively making the expression .+ match any character except a closing ").
If you only want to extract sub-strings where the group_id starts with sg- then you can simply add this to the matching part of the pattern as follows (?<=group_id=\>")sg\-.+?(?=\")
import re
s = 'describe aws_security_group({:group_id=>"sg-ezsrzerzer", :vpc_id=>"vpc-zfds54zef4s"}) do'
results = re.findall('(?<=group_id=\>").+?(?=\")', s)
print(results)
Output
['sg-ezsrzerzer']
Of course you could alternatively use re.search instead of re.findall to find the first instance of a sub-string matching the above pattern in a given string - depends on your use case I suppose.
import re
s = 'describe aws_security_group({:group_id=>"sg-ezsrzerzer", :vpc_id=>"vpc-zfds54zef4s"}) do'
result = re.search('(?<=group_id=\>").+?(?=\")', s)
if result:
result = result.group()
print(result)
Output
'sg-ezsrzerzer'
If you decide to use re.search you will find that it returns None if there is no match found in your input string and an re.Match object if there is - hence the if statement and call to s.group() to extract the matching string if present in the above example.
The pattern \bsg-.*\b matches too much as the .* will match until the end of the string, and will then backtrack to the first word boundary, which is after the o and the end of string.
If you are using re.findall you can also use a capture group instead of lookarounds and the group value will be in the result.
:group_id=>"(sg-[^"\r\n]+)"
The pattern matches:
:group_id=>" Match literally
(sg-[^"\r\n]+) Capture group 1 match sg- and 1+ times any char except " or a newline
" Match the double quote
See a regex demo or a Python demo
For example
import re
pattern = r':group_id=>"(sg-[^"\r\n]+)"'
s = "describe aws_security_group({:group_id=>\"sg-ezsrzerzer\", :vpc_id=>\"vpc-zfds54zef4s\"}) do"
print(re.findall(pattern, s))
Output
['sg-ezsrzerzer']
Match until the first word boundary with \w+:
import re
a = 'describe aws_security_group({:group_id=>"sg-ezsrzerzer", :vpc_id=>"vpc-zfds54zef4s"}) do'
test = re.findall(r'\bsg-\w+', a)
print(test[0])
See Python proof.
EXPLANATION
--------------------------------------------------------------------------------
\b the boundary between a word char (\w) and
something that is not a word char
--------------------------------------------------------------------------------
sg- 'sg-'
--------------------------------------------------------------------------------
\w+ word characters (a-z, A-Z, 0-9, _) (1 or
more times (matching the most amount
possible))
Results: g-ezsrzerzer
I need to extract people's names from a really long string.
Their names are in this format: LAST, FIRST.
Some of these people have hyphenated names. Some don't.
My attempt with a smaller string:
Input:
import re
text = 'Smith-Jones, Robert&Epson, Robert'
pattern = r'[A-Za-z]+(-[A-Za-z]+)?,\sRobert'
print re.findall(pattern, text)
Expected output:
['Smith-Jones, Robert', 'Epson, Robert']
Actual output:
['-Jones', '']
What am I doing wrong?
Use
import re
text = 'Smith-Jones, Robert&Epson, Robert'
pattern = r'[A-Za-z]+(?:-[A-Za-z]+)?,\sRobert'
print re.findall(pattern, text)
# => ['Smith-Jones, Robert', 'Epson, Robert']
Just make the capturing group non-capturing. The thing is that findall returns capture group values if they are specified in the regex pattern. So, the best way to solve this in this pattern is just replace (...)? with (?:...)?.
See IDEONE demo
I've come up with a regex expression that works well enough for my purposes for finding phone numbers.
I would like to take it a step further and use it in large text blocks to identify matching strings that follow the words 'cell' or 'mobile' by at most 10 characters. I would like it to return the number in Cell Phone: (954) 555-4444 as well as Mobile 555-777-9999 but not Fax: (555) 444-6666
something like (in pseudocode)
regex = re.compile(r'(\+?[2-9]\d{2}\)?[ -]?\d{3}[ -]?\d{4})')
bigstring = # Some giant string added together from many globbed files
matches = regex.search(bigstring)
for match in matches:
if match follows 'cell' or match follows 'mobile':
print match.group(0)
You can do:
txt='''\
Call me on my mobile anytime: 555-666-1212
The office is best at 555-222-3333
Dont ever call me at 555-666-2345 '''
import re
print re.findall(r'(?:(mobile|office).{0,15}(\+?[2-9]\d{2}\)?[ -]?\d{3}[ -]?\d{4}))', txt)
Prints:
[('mobile', '555-666-1212'), ('office', '555-222-3333')]
You can do that with your regular expression. In the re documentation, you will find that the pattern r'(?<=abc)def' matches 'def' only if it is preceded by 'abc'.
Similarly r'Hello (?=World)' matches 'Hello ' if followed by 'World'