I am trying to write a regular expression which returns a string which is between two other strings. For example: I want to get the string along with spaces which resides between the strings "15/08/2017" and "$610,000"
a='172 211 342 15/08/2017 TRANSFER OF LAND $610,000 CASH & MTGE'
should return
"TRANSFER OF LAND"
Here is the expression I have pieced together so far:
re.search(r'15/08/2017(.*?)$610,000', a).group(1)
It doesn't return any matches. I think it is because we also need to consider spaces in the expression. Is there a way to find strings between two strings ignoring the spaces?
Use Regex Lookbehind & Lookahead
Ex:
import re
a='172 211 342 15/08/2017 TRANSFER OF LAND $610,000 CASH & MTGE'
print(re.search(r"(?<=15/08/2017).*?(?=\$610,000)", a).group())
Output:
TRANSFER OF LAND
>>> re.search(r'15/08/2017(.*)\$610,000',a).group(1)
' TRANSFER OF LAND '
Since $ is a regex metacharacter (standing for the end of a logical line), you need to escape it to use as a literal '$'.
Might be easier to use find:
a = '172 211 342 15/08/2017 TRANSFER OF LAND $610,000 CASH & MTGE'
b = '15/08/2017'
c = '$610,000'
a[a.find(b) + len(b):a.find(c)].strip()
'TRANSFER OF LAND'
Related
I'm using pandas to analyze data from 3 different sources, which are imported into dataframes and require modification to account for human error, as this data was all entered by humans and contains errors.
Specifically, I'm working with street names. Until now, I have been using .str.replace() to remove street types (st., street, blvd., ave., etc.), as shown below. This isn't working well enough, and I decided I would like to use regex to match a pattern, and transform that entire column from the original street name, to the pattern matched by regex.
df['street'] = df['street'].str.replace(r' avenue+', '', regex=True)
I've decided I would like to use regex to identify (and remove all other characters from the address column's fields): any number of integers, followed by a space, and then the first 3 number of alphabetic characters.
For example, "3762 pearl street" might become "3762 pea" if x is 3 with the following regex:
(\d+ )+\w{0,3}
How can I use panda's .str.replace to do this? I don't want to specify WHAT I want to replace with the second argument. I want to replace the original string with the pattern matched from regex.
Something that, in my mind, might work like this:
df['street'] = df['street'].str.replace(ORIGINAL STRING, r' (\d+ )+\w{0,3}, regex=True)
which might make 43 milford st. into "43 mil".
Thank you, please let me know if I'm being unclear.
you could use the extract method to overwrite the column with its own content
pat = r'(\d+\s[a-zA-Z]{3})'
df['street'] = df['street'].str.extract(pat)
Just an observation: The regex you shared (\d+ )+\w{0,3} matches the following patterns and returns some funky stuff as well
1131 1313 street
121 avenue
1 1 1 1 1 1 avenue
42
I've changed it up a bit based on what you described, but i'm not sure if that works for all your datapoints.
I am trying to find words and print using below code. Everything is working perfect but only issue is i am unable to print the last word(which is number).
words = ['Town of','Block No.','Lot No.','Premium (if any) Paid ']
import re
for i in words:
y = re.findall('{} ([^ ]*)'.format(i), textfile)
print(y)
Text file i working with:
textfile = """1, REBECCA M. ROTH , COLLECTOR OF TAXES of the taxing district of the
township of MORRIS for Six Hundred Sixty Seven dollars andFifty Two cents, the land
in said taxing district described as Block No. 10303 Lot No. 10 :
and known as 239 E HANOVER AVE , on the tax Taxes For: 2012
Sewer
Assessments For Improvements
Total Cost of Sale 35.00
Total
Premium (if any) Paid 1,400.00 """
Would like to know where am i making mistake.
Any suggestion is appreciated.
A couple of issues:
As others have mentioned, you need to escape special characters like parentheses ( ) and dots .. Very simply, you can use re.escape
Another issue is the trailing space in Premium \(if any\) Paid (it's trying to match two spaces instead of one as you're also checking for a space in your regex {} ([^ ]*))
You should instead change your code to the following:
See working code here
words = ['Town of','Block No.','Lot No.','Premium (if any) Paid']
import re
for i in words:
y = re.findall('{} ([^ ]*)'.format(re.escape(i)), textfile)
print(y)
Two problems:
Your current 'Premium (if any) Paid ' string ends on a space, and '{} ([^ ]*)' also has a space after {}, which adds them together. Delete the trailing space in 'Premium (if any) Paid '.
You need to escape parenthesis, so if you want to keep your regular expression unchanged, the string in the list should be ['Premium \(if any\) Paid']. You can also use re.escape instead.
For your particular cases, this seems to be an optimal solution:
words = ['Town of','Block No.','Lot No.','Premium (if any) Paid']
import re
for i in words:
y = re.findall('{}\s+([\S]*)'.format(re.escape(i)), text, re.I)
print(y)
I am trying to write a regular expression which returns a part of substring which is after a string. For example: I want to get part of substring along with spaces which resides after "15/08/2017".
a='''S
LINC SHORT LEGAL TITLE NUMBER
0037 471 661 1720278;16;21 172 211 342
LEGAL DESCRIPTION
PLAN 1720278
BLOCK 16
LOT 21
EXCEPTING THEREOUT ALL MINES AND MINERALS
ESTATE: FEE SIMPLE
ATS REFERENCE: 4;24;54;2;SW
MUNICIPALITY: CITY OF EDMONTON
REFERENCE NUMBER: 172 023 641 +71
----------------------------------------------------------------------------
----
REGISTERED OWNER(S)
REGISTRATION DATE(DMY) DOCUMENT TYPE VALUE CONSIDERATION
---------------------------------------------------------------------------
--
---
172 211 342 15/08/2017 AFFIDAVIT OF CASH & MTGE'''
Is there a way to get 'AFFIDAVIT OF' and 'CASH & MTGE' as separate strings?
Here is the expression I have pieced together so far:
doc = (a.split('15/08/2017', 1)[1]).strip()
'AFFIDAVIT OF CASH & MTGE'
Not a regex based solution. But does the trick.
a='''S
LINC SHORT LEGAL TITLE NUMBER
0037 471 661 1720278;16;21 172 211 342
LEGAL DESCRIPTION
PLAN 1720278
BLOCK 16
LOT 21
EXCEPTING THEREOUT ALL MINES AND MINERALS
ESTATE: FEE SIMPLE
ATS REFERENCE: 4;24;54;2;SW
MUNICIPALITY: CITY OF EDMONTON
REFERENCE NUMBER: 172 023 641 +71
----------------------------------------------------------------------------
----
REGISTERED OWNER(S)
REGISTRATION DATE(DMY) DOCUMENT TYPE VALUE CONSIDERATION
---------------------------------------------------------------------------
--
---
172 211 342 15/08/2017 AFFIDAVIT OF CASH & MTGE'''
doc = (a.split('15/08/2017', 1)[1]).strip()
# used split with two white spaces instead of one to get the desired result
print(doc.split(" ")[0].strip()) # outputs AFFIDAVIT OF
print(doc.split(" ")[-1].strip()) # outputs CASH & MTGE
Hope it helps.
re based code snippet
import re
foo = '''S
LINC SHORT LEGAL TITLE NUMBER
0037 471 661 1720278;16;21 172 211 342
LEGAL DESCRIPTION
PLAN 1720278
BLOCK 16
LOT 21
EXCEPTING THEREOUT ALL MINES AND MINERALS
ESTATE: FEE SIMPLE
ATS REFERENCE: 4;24;54;2;SW
MUNICIPALITY: CITY OF EDMONTON
REFERENCE NUMBER: 172 023 641 +71
----------------------------------------------------------------------------
----
REGISTERED OWNER(S)
REGISTRATION DATE(DMY) DOCUMENT TYPE VALUE CONSIDERATION
---------------------------------------------------------------------------
--
---
172 211 342 15/08/2017 AFFIDAVIT OF CASH & MTGE'''
pattern = '.*\d{2}/\d{2}/\d{4}\s+(\w+\s+\w+)\s+(\w+\s+.*\s+\w+)'
result = re.findall(pattern, foo, re.MULTILINE)
print "1st match: ", result[0][0]
print "2nd match: ", result[0][1]
Output
1st match: AFFIDAVIT OF
2nd match: CASH & MTGE
We can try using re.findall with the following pattern:
PHASED OF ((?!\bCONDOMINIUM PLAN).)*)(?=CONDOMINIUM PLAN)
Searching in multiline and DOTALL mode, the above pattern will match everything occurring between PHASED OF until, but not including, CONDOMINIUM PLAN.
input = "182 246 612 01/10/2018 PHASED OF CASH & MTGE\n CONDOMINIUM PLAN"
result = re.findall(r'PHASED OF (((?!\bCONDOMINIUM PLAN).)*)(?=CONDOMINIUM PLAN)', input, re.DOTALL|re.MULTILINE)
output = result[0][0].strip()
print(output)
CASH & MTGE
Note that I also strip off whitespace from the match. We might be able to modify the regex pattern to do this, but in a general solution, maybe you want to keep some of the whitespace, in certain cases.
Why regular expressions?
It looks like you know the exact delimiting string, just str.split() by it and get the first part:
In [1]: a='172 211 342 15/08/2017 TRANSFER OF LAND $610,000 CASH & MTGE'
In [2]: a.split("15/08/2017", 1)[0]
Out[2]: '172 211 342 '
I would avoid using regex here, because the only meaningful separation between the logical terms appears to be 2 or more spaces. Individual terms, including the one you want to match, may also have spaces. So, I recommend doing a regex split on the input using \s{2,} as the pattern. These will yield a list containing all the terms. Then, we can just walk down the list once, and when we find the forward looking term, we can return the previous term in the list.
import re
a = "172 211 342 15/08/2017 TRANSFER OF LAND $610,000 CASH & MTGE"
parts = re.compile("\s{2,}").split(a)
print(parts)
for i in range(1, len(parts)):
if (parts[i] == "15/08/2017"):
print(parts[i-1])
['172 211 342', '15/08/2017', 'TRANSFER OF LAND', '$610,000', 'CASH & MTGE']
172 211 342
positive lookbehind assertion**
m=re.search('(?<=15/08/2017).*', a)
m.group(0)
You have to return the right group:
re.match("(.*?)15/08/2017",a).group(1)
You nede to use group(1)
import re
re.match("(.*?)15/08/2017",a).group(1)
Output
'172 211 342 '
Building on your expression, this is what I believe you need:
import re
a='172 211 342 15/08/2017 TRANSFER OF LAND $610,000 CASH & MTGE'
re.match("(.*?)(\w+/)",a).group(1)
Output:
'172 211 342 '
You can do this by using group(1)
re.match("(.*?)15/08/2017",a).group(1)
UPDATE
For updated string you can use .search instead of .match
re.search("(.*?)15\/08\/2017",a).group(1)
Your problem is that your string is formatted the way it is.
The line you are looking for is
182 246 612 01/10/2018 PHASED OF CASH & MTGE
And then you are looking for what ever comes after 'PHASED OF' and some spaces.
You want to search for
(?<=PHASED OF)\s*(?P.*?)\n
in your string. This will return a match object containing the value you are looking for in the group value.
m = re.search(r'(?<=PHASED OF)\s*(?P<your_text>.*?)\n', a)
your_desired_text = m.group('your_text')
Also: There are many good online regex testers to fiddle around with your regexes.
And only after finishing up the regex just copy and paste it into python.
I use this one: https://regex101.com/
I have following requirements in date which can be any of the following format.
mm/dd/yyyy or dd Mon YYYY
Few examples are shown below
04/20/2009 and 24 Jan 2001
To handle this I have written regular expression as below
Few text scenarios are metnioned below
txt1 = 'Lithium 0.25 (7/11/77). LFTS wnl. Urine tox neg. Serum tox
+ fluoxetine 500; otherwise neg. TSH 3.28. BUN/Cr: 16/0.83. Lipids unremarkable. B12 363, Folate >20. CBC: 4.9/36/308 Pertinent Medical
Review of Systems Constitutional:'
txt2 = "s The patient is a 44 year old married Caucasian woman,
unemployed Decorator, living with husband and caring for two young
children, who is referred by Capitol Hill Hospital PCP, Dr. Heather
Zubia, for urgent evaluation/treatment till first visit with Dr. Toney
Winkler IN EIGHT WEEKS on 24 Jan 2001."
date = re.findall(r'(?:\b(?<!\.)[\d{0,2}]+)'
'(?:[/-]\d{0,}[/-]\d{2,4}) | (?:\b(?<!\.)[\d{1,2}]+)[th|st|nd]*'
' (?:[Jan|Feb|Mar|Apr|May|Jun|Jul|Aug|Sep|Oct|Nov|Dec][a-z]*) \d{2,4}', txtData)
I am not getting 24 Jan 2001 where as if I run individually (?:\b(?<!\.)[\d{1,2}]+)[th|st|nd]* (?:[Jan|Feb|Mar|Apr|May|Jun|Jul|Aug|Sep|Oct|Nov|Dec][a-z]*) \d{2,4}' I am able to get output.
Question 1: What is bug in above expression?
Question 2: I want to combine both to make more readable as I have to parse any other formats so I used join as shown below
RE1 = '(?:\b(?<!\.)[\d{0,2}]+) (?:[/-]\d{0,}[/-]\d{2,4})'
RE2 = '(?:\b(?<!\.)[\d{1,2}]+)[th|st|nd]* (?:[Jan|Feb|Mar|Apr|May|Jun|Jul|Aug|Sep|Oct|Nov|Dec][a-z]*) \d{2,4}'
regex_all = '|'.join([RE1, RE2])
regex_all = re.compile(regex_all)
date = regex_all.findall(txtData) // notice here txtData can be any one of the above string.
I am getting output as NaN in case of above for date.
Please suggest what is the mistake if I join.
Thanks for your help.
Note that it is a very bad idea to join such long patterns that also match at the same location within the string. That would cause the regex engine to backtrack too much, and possibly lead to crashes and slowdown. If there is a way to re-write the alternations so that they could only match at different locations, or even get rid of them completely, do it.
Besides, you should use grouping constructs (...) to groups sequences of patterns, and only use [...] character classes when you need to matches specific chars.
Also, your alternatives are overlapping, you may combine them easily. See the fixed regex:
\b(?<!\.)\d{1,2}(?:[/-]\d+[/-]|(?:th|st|[nr]d)?\s*(?:(?:Jan|Feb|Mar|Apr|May|Jun|Jul|Aug|Sep|Oct|Nov|Dec)[a-z]*))\s*(?:\d{4}|\d{2})\b
See the regex demo.
Details
\b - a word boundary
(?<!\.) - no . immediately to the left of the current location
\d{1,2} - 1 or 2 digits
(?: - start of a non-capturing alternation group:
[/-]\d+[/-] - / or -, 1+ digits, - or /
| - or
(?:th|st|[nr]d)?\s*(?:
(?:Jan|Feb|Mar|Apr|May|Jun|Jul|Aug|Sep|Oct|Nov|Dec)[a-z]*)) - th, st, nd or rd (optionally), followed with 0+ whitespaces, and then month names
\s* - 0+ whitespaces
(?:\d{4}|\d{2}) - 2 or 4 digits
\b - trailing word boundary.
Another note: if you want to match the date-like strings with two matching delimiters, you will need to capture the first one, and use a backreference to match the second one, see this regex demo. In Python, you would need a re.finditer to get those matches.
See this Python demo:
import re
rx = r"\b(?<!\.)\d{1,2}(?:([/-])\d+\1|(?:th|st|[nr]d)?\s*(?:(?:Jan|Feb|Mar|Apr|May|Jun|Jul|Aug|Sep|Oct|Nov|Dec)[a-z]*))\s*(?:\d{4}|\d{4})\b"
s = "Lithium 0.25 (7/11/77). LFTS wnl. Urine tox neg. Serum tox\nfluoxetine 500; otherwise neg. TSH 3.28. BUN/Cr: 16/0.83. Lipids unremarkable. B12 363, Folate >20. CBC: 4.9/36/308 Pertinent Medical\nReview of Systems Constitutional:\n\nThe patient is a 44 year old married Caucasian woman, unemployed Decorator, living with husband and caring for two young children, who is referred by Capitol Hill Hospital PCP, Dr. Heather Zubia, for urgent evaluation/treatment till first visit with Dr. Toney Winkler IN EIGHT WEEKS on 24 Jan 2001"
print([x.group(0) for x in re.finditer(rx, s, re.I)])
# => ['7/11/77', '24 Jan 2001']
I think your approach is too complicated. I suggest using a combination of a simple regex and strptime().
import re
from datetime import datetime
date_formats = ['%m/%d/%Y', '%d %b %Y']
pattern = re.compile(r'\b(\d\d?/\d\d?/\d{4}|\d\d? \w{3} \d{4})\b')
data = "... your string ..."
for match in re.findall(pattern, data):
print("Trying to parse '%s'" % match)
for fmt in date_formats:
try:
date = datetime.strptime(match, fmt)
print(" OK:", date)
break
except:
pass
The advantage of this approach is, besides a much more manageable regex, that it won't pick dates that look plausible but do not exist, like 2/29/2000 (whereas 2/29/2004 works).
r'(?:\b(?<!\.)[\d{0,2}]+)'
'(?:[/-]\d{0,}[/-]\d{2,4}) | (?:\b(?<!\.)[\d{1,2}]+)[th|st|nd]*'
' (?:[Jan|Feb|Mar|Apr|May|Jun|Jul|Aug|Sep|Oct|Nov|Dec][a-z]*) \d{2,4}'
you should use raw strings (r'foo') for each string, not only the first one. This way backslashes (\) will be considered as normal character and usable by the re library.
[abc|def] matches any character between the [], while (one|two|three) matches any expression (one, two, or three)
I have a string of a couple thousand addresses in python, like such:
'123 Chestnut Way 4567 Oak Lane 890 South Pine Court'
What is the easiest way to split this long string into separate addresses? I'm trying to write a program that splits based on 3 or 4 characters in a row where 47 < ord(i) < 58, but I'm having trouble.
Assuming all of the addresses are like those given, you can use re.findall:
>>> from re import findall
>>> string = '123 Chestnut Way 4567 Oak Lane 890 South Pine Court'
>>> findall("\d+\D+(?=\s\d|$)", string)
['123 Chestnut Way', '4567 Oak Lane', '890 South Pine Court']
>>>
All of the Regex syntax used above is explained here, but below is a quick breakdown:
\d+ # One or more digits
\D+ # One or more non-digits
(?= # The start of a lookahead assertion
\s # A space
\d|$ # A digit or the end of the string
) # The end of the lookahead assertion
You can do this with regular expressions fairly easily,
import re
txt = '123 Chestnut Way 4567 Oak Lane 890 South Pine Court'
re.findall( r'\d+', txt )
the last will return all runs of digits,
['123', '4567', '890']
you can then use that information to parse the string. there are lots of ways, but you could just find the index of the numbers in the original string and get the text in between. you could also make the regeular expression a little more advanced. The following will match any number of digits followed by a space followed by any number of non-digits (including spaces)
re.findall( r'\d+ \D+', txt )
and will return,
['123 Chestnut Way ', '4567 Oak Lane ', '890 South Pine Court']