I have some sentence like
"Apartment at Chennai has 4 rooms with a swimming pool"
"this apartment has 2/3 room with a coridor"
How to extract the only number before the word "room|rooms".
looking for an answer which must be 4 and 2/3.
The code i tried,
room_found =re.findall(r"\d\s?\/?\d?\s?(?=(rooms)|(room))", str_arg)
print(room_found)
This prints [('', '4 room')] and [('', '2/3 room')], but i am expecting only 4 and 2/3 to be printed.
You can use
\d+(?:/\d+)?(?=\s?rooms?\b)
Explanation
\d+ Match 1+ digits (Or \d for a single digit)
(?:/\d+)? Optionally match / and 1+ digits
(?= Positive lookahead to assert what is directly at the right is
\s?rooms?\b Match an optional whitspace char followed by room or rooms
) Close the lookahead
Regex demo
import re
str_arg = "\"Apartment at Chennai has 4 rooms with a swimming pool\" \"this apartment has 2/3 room with a coridor\""
room_found =re.findall(r"\d+(?:/\d+)?(?=\s?rooms?\b)", str_arg)
print(room_found)
Output
['4', '2/3']
You can add a condition to find all the numbers before the word room and rooms.
And You can enclose all the valid values inside the square brackets like you are allowing forward slash apart from digits as well.
[\d\/]+ => It selects the digits and forward slash.
(?=\srooms?) => It selects the numbers before the word room and rooms.
[\d\/]+(?=\srooms?)
Please find demo here
You can use ([0-9]+) rooms. So your final code would look like:
import re
str_arg = "\"Apartment at Chennai has 4 rooms with a swimming pool\" \"this apartment has 2/3 room with a coridor\""
room_found = re.findall(r"([0-9]+) rooms", str_arg)
print(room_found)
Related
I want to get bold parts in sentences below.
Examples:
SmellNice Coffee 450 gr
Clean 2 k Rice
LukaLuka 1,5lt cold drink
Jumbo 7 gutgut eggs 12'li
Espresso 5 Klasik 10 Ad
Expression below works well until to the last two.
\d+[.,]?\d*\s*[’']?\s*(gr|g|kg|k|adet|ad|lı|li|lu|lü|cc|cl|ml|lt|l|mm|cm|mt|m)
I have added \s|$ end of the expression. Thinking that If the unit is not the last word then there should be a space after it. But it didn't work. Briefly, how can I capture all bold expressions?
It works with brackets:
\d+[.,]?\d*\s*[’']?\s*(gr|g|kg|k|adet|ad|lı|li|lu|lü|cc|cl|ml|lt|l|mm|cm|mt|m)(\s+|$)
x2 = (
"\d+" #digit
"[,'\s]" #space comma apostrophe
"[\d*\s*]?" #opt digit or space
"((gr)|g|(kg)|k|(adet)|([Aa]d)|(lı)|(li)|(lu)|(lü)|(cc)|(cl)|(ml)|(lt)|l|(mm)|(cm)|(mt)|m)" #all the weights to look for
"(\s+|$)" #it's gotta be followed with a space, or with end of line.
)
Basically, I want to remove the certain phrase patterns embedded in my text data:
Starts with an upper case letter and ends with an Em Dash "—"
Starts with an Em Dash "—" and ends with a "Read Next"
Say, I've got the following data:
CEBU CITY—The widow of slain human rights lawyer .... citing figures from the NUPL that showed that 34 lawyers had been killed in the past two years. —WITH REPORTS FROM JULIE M. AURELIO AND DJ YAPRead Next
and
Manila, Philippines—President .... but justice will eventually push its way through their walls of impunity, ... —REPORTS FROM MELVIN GASCON, JULIE M. AURELIO, DELFIN T. MALLARI JR., JEROME ANING, JOVIC YEE, GABRIEL PABICO LALU, PATHRICIA ANN V. ROXAS, DJ YAP, AFP, APRead Next
I want to remove the following phrases:
"CEBU CITY—"
"—WITH REPORTS FROM JULIE M. AURELIO AND DJ YAPRead Next"
"Manila, Philippines—"
"—REPORTS FROM MELVIN GASCON, JULIE M. AURELIO, DELFIN T. MALLARI JR., JEROME ANING, JOVIC YEE, GABRIEL PABICO LALU, PATHRICIA ANN V. ROXAS, DJ YAP, AFP, APRead Next"
I am assuming this would be needing two regex for each patterns enumerated above.
The regex: —[A-Z].*Read Next\s*$ may work on the pattern # 2 but only when there are no other em dashes in the text data. It will not work when pattern # 1 occurs as it will remove the chunk from the first em dash it has seen until the "Read Next" string.
I have tried the following regex for pattern # 1:
^[A-Z]([A-Za-z]).+(—)$
But how come it does not work. That regex was supposed to look for a phrase that starts with any upper case letter, followed by any length of string as long as it ends with an "—".
What you are considering a hyphen - is not indeed a hyphen instead called Em Dash, hence you need to use this regex which has em dash instead of hyphen in start,
^—[A-Z].*Read Next\s*$
Here is the explanation for this regex,
^ --> Start of input
— --> Matches a literal Em Dash whose Unicode Decimal Code is 8212
[A-Z] --> Matches an upper case letter
.* --> Matches any character zero or more times
Read Next --> Matches these literal words
\s* --> This is for matching any optional white space that might be present at the end of line
$ --> End of input
Online demo
The regex that should take care of this -
^—[A-Z]+(.)*(Read Next)$
You can try implementing this regex on your data and see if it works out.
I have some sentence like
1:
"RLB shows Oubre Jr. (WAS) legally ties up Nurkic (POR), and a held
ball is correctly called."
2:
"Nurkic (POR) maintains legal
guarding position and makes incidental contact with Wall (WAS) that
does not affect his driving shot attempt."
I need to use Python regex to find the name "Oubre Jr." ,"Nurkic" and "Nurkic", "Wall".
p = r'\s*(\w+?)\s[(]'
use this pattern,
I can find "['Nurkic', 'Wall']", but in sentence 1, I just can find ['Nurkic'], missed "Oubre Jr."
Who can help me?
You can use the following regex:
(?:[A-Z][a-z][\s\.a-z]*)+(?=\s\()
|-----Main Pattern-----|
Details:
(?:) - Creates a non-capturing group
[A-Z] - Captures 1 uppercase letter
[a-z] - Captures 1 lowercase letter
[\s\.a-z]* - Captures spaces (' '), periods ('.') or lowercase letters 0+ times
(?=\s\() - Captures the main pattern if it is only followed by ' (' string
str = '''RLB shows Oubre Jr. (WAS) legally ties up Nurkic (POR), and a held ball is correctly called.
Nurkic (POR) maintains legal guarding position and makes incidental contact with Wall (WAS) that does not affect his driving shot attempt.'''
res = re.findall( r'(?:[A-Z][a-z][\s\.a-z]*)+(?=\s\()', str )
print(res)
Demo: https://repl.it/#RahulVerma8/OvalRequiredAdvance?language=python3
Match: https://regex101.com/r/OsLTrY/1
Here is one approach:
line = "RLB shows Oubre Jr (WAS) legally ties up Nurkic (POR), and a held ball is correctly called."
results = re.findall( r'([A-Z][\w+'](?: [JS][r][.]?)?)(?= \([A-Z]+\))', line, re.M|re.I)
print(results)
['Oubre Jr', 'Nurkic']
The above logic will attempt to match one name, beginning with a capital letter, which is possibly followed by either the suffix Jr. or Sr., which in turn is followed by a ([A-Z]+) term.
You need a pattern that you can match - for your sentence you cou try to match things before (XXX) and include a list of possible "suffixes" to include as well - you would need to extract them from your sources
import re
suffs = ["Jr."] # append more to list
rsu = r"(?:"+"|".join(suffs)+")? ?"
# combine with suffixes
regex = r"(\w+ "+rsu+")\(\w{3}\)"
test_str = "RLB shows Oubre Jr. (WAS) legally ties up Nurkic (POR), and a held ball is correctly called. Nurkic (POR) maintains legal guarding position and makes incidental contact with Wall (WAS) that does not affect his driving shot attempt."
matches = re.finditer(regex, test_str, re.MULTILINE)
names = []
for matchNum, match in enumerate(matches,1):
for groupNum in range(0, len(match.groups())):
names.extend(match.groups(groupNum))
print(names)
Output:
['Oubre Jr.', 'Nurkic ', 'Nurkic ', 'Wall ']
This should work as long as you do not have Names with non-\w in them. If you need to adapt the regex, use https://regex101.com/r/pRr9ZU/1 as starting point.
Explanation:
r"(?:"+"|".join(suffs)+")? ?" --> all items in the list suffs are strung together via | (OR) as non grouping (?:...) and made optional followed by optional space.
r"(\w+ "+rsu+")\(\w{3}\)" --> the regex looks for any word characters followed by optional suffs group we just build, followed by literal ( then three word characters followed by another literal )
For this regular expression:
(?<!Mr|Dr|Ms|Jr|Sr)[.?!]+(\s)[A-Z0-9]
I want the input string to be split by the captured matching \s character - the green matches as seen over here.
However, when I run this:
import re
p = re.compile(ur'(?<!Mr|Dr|Ms|Jr|Sr)[.?!]+(\s)[A-Z0-9]')
test_str = u"Mr. Smith bought cheapsite.com for 1.5 million dollars i.e. he paid a lot for it. Did he mind? Adam Jones Jr. thinks he didn't. In any case, this isn't true... Well, with a probability of .9 it isn't.23 is the ish. My name is! Why wouldn't you... this is.\nAndrew"
re.split(p, test_str)
It seems to split the string at the regions given by [.?!]+ and [A-Z0-9] (thus incorrectly omitting them) and leaves \s in the results.
To clarify:
Input: he paid a lot for it. Did he mind
Received Output: ['he paid a lot for it','\s','id he mind']
Expected Output: ['he paid a lot for it.','Did he mind']
You need to remove the capturing group from around (\s) and put the last character class into a look-ahead to exclude it from the match:
p = re.compile(ur'(?<!Mr|Dr|Ms|Jr|Sr)[.?!]+\s(?=[A-Z0-9])')
# ^^^^^ ^
test_str = u"Mr. Smith bought cheapsite.com for 1.5 million dollars i.e. he paid a lot for it. Did he mind? Adam Jones Jr. thinks he didn't. In any case, this isn't true... Well, with a probability of .9 it isn't.23 is the ish. My name is! Why wouldn't you... this is.\nAndrew"
print(p.split(test_str))
See IDEONE demo and the regex demo.
Any capturing group in a regex pattern will create an additional element in the resulting array during re.split.
To force the punctuation to appear inside the "sentences", you can use this matching regex with re.findall:
import re
p = re.compile(r'\s*((?:(?:Mr|Dr|Ms|Jr|Sr)\.|\.(?!\s+[A-Z0-9])|[^.!?])*[.?!]|[^.!?]+)')
test_str = "Mr. Smith bought cheapsite.com for 1.5 million dollars i.e. he paid a lot for it. Did he mind? Adam Jones Jr. thinks he didn't. In any case, this isn't true... Well, with a probability of .9 it isn't.23 is the ish. My name is! Why wouldn't you... this is.\nAndrew"
print(p.findall(test_str))
See IDEONE demo
Results:
['Mr. Smith bought cheapsite.com for 1.5 million dollars i.e. he paid a lot for it.', 'Did he mind?', "Adam Jones Jr. thinks he didn't.", "In any case, this isn't true...", "Well, with a probability of .9 it isn't.23 is the ish.", 'My name is!', "Why wouldn't you... this is.", 'Andrew']
The regex demo
The regex follows the rules in your original pattern:
\s* - matches 0 or more whitespace to omit from the result
(?:(?:Mr|Dr|Ms|Jr|Sr)\.|\.(?!\s+[A-Z0-9])|[^.!?])*[.?!]|[^.!?]+) - 2 aternatives that are captured and returned by re.findall:
(?:(?:Mr|Dr|Ms|Jr|Sr)\.|\.(?!\s+[A-Z0-9])|[^.!?])* - 0 or more sequences of...
(?:Mr|Dr|Ms|Jr|Sr)\. - abbreviated titles
\.(?!\s+[A-Z0-9]) - matches a dot not followed by 1 or more whitespace and then uppercase letters or digits
[^.!?] - any character but a ., !, and ?
or...
[^.!?]+ - any one or more characters but a ., !, and ?
I have a text file containing entries like this:
#markwarner VIRGINIA - Mark Warner
#senatorleahy VERMONT - Patrick Leahy NO
#senatorsanders VERMONT - Bernie Sanders
#orrinhatch UTAH - Orrin Hatch NO
#jimdemint SOUTH CAROLINA - Jim DeMint NO
#senmikelee UTAH -- Mike Lee
#kaybaileyhutch TEXAS - Kay Hutchison
#johncornyn TEXAS - John Cornyn
#senalexander TENNESSEE - Lamar Alexander
I have written the following to remove the 'NO' and the dashes using regular expressions:
import re
politicians = open('testfile.txt')
text = politicians.read()
# Grab the 'no' votes
# Should be 11 entries
regex = re.compile(r'(no\s#[\w+\d+\.]*\s\w+\s?\w+?\s?\W+\s\w+\s?\w+)', re.I)
no = regex.findall(text)
## Make the list a string
newlist = ' '.join(no)
## Replace the dashes in the string with a space
deldash = re.compile('\s-*\s')
a = deldash.sub(' ', newlist)
# Delete 'NO' in the string
delno = re.compile('NO\s')
b = delno.sub('', a)
# make the string into a list
# problem with #jimdemint SOUTH CAROLINA Jim DeMint
regex2 = re.compile(r'(#[\w\d\.]*\s[\w\d\.]*\s?[\w\d\.]\s?[\w\d\.]*?\s+?\w+)', re.I)
lst1 = regex2.findall(b)
for i in lst1:
print i
When I run the code, it captures the twitter handle, state and full names other than the surname of Jim DeMint. I have stated that I want to ignore case for the regex.
Any ideas? Why is the expression not capturing this surname?
It's missing it because his state name contains two words: SOUTH CAROLINA
Have your second regex be this, it should help
(#[\w\d\.]*\s[\w\d\.]*\s?[\w\d\.]\s?[\w\d\.]*?\s+?\w+(?:\s\w+)?)
I added
(?:\s\w+)?
Which is a optional, non capturing group matching a space followed by one or more alphanumeric underscore characters
http://regexr.com?31fv5 shows that it properly matches the input with the NOs and dashes stripped
EDIT:
If you want one master regex to capture and split everything properly, after you remove the Nos and dashes, use
((#[\w]+?\s)((?:(?:[\w]+?)\s){1,2})((?:[\w]+?\s){2}))
Which you can play with here: http://regexr.com?31fvk
The full match is available in $1, the Twitter handle in $2, the State in $3 And the name in $4
Each capturing group works as follows:
(#[\w]+?\s)
This matches an # sign followed by at least one but as few characters as possible until a space.
((?:(?:[\w]+?)\s){1,2})
This matches and captures 1 or two words, which should be the state. This only works because of the next piece, which MUST have two words
((?:[\w]+?\s){2})
Matches and captures exactly two words, which is defined as few characters as possible followed by a space
text=re.sub(' (NO|-+)(?= |$)','',text)
And to capture everything:
re.findall('(#\w+) ([A-Z ]+[A-Z]) (.+?(?= #|$))',text)
Or all at once:
re.findall('(#\w+) ([A-Z ]+[A-Z])(?: NO| -+)? (.+?(?= #|$))',text)