We Keep Coding

Given a string, extract all the necessary information about the person

In my homework, I need to extract the first name, last name, ID code, phone number, date of birth and address of a person from a given string using Regex. The order of the parameters always remains the same. Each parameter requires a separate pattern.
Requirements are as follows:
Both first and last names always begin with a capital letter followed by at least one lowercase letter.
ID code is always 11 characters long and consists only of numbers.
The phone number itself is a combination of 7-8 numbers. The phone number might be separated from the area code with a whitespace, but not necessarily. It is also possible that there is no area code at all.
Date of birth is formatted as dd-MM-YYYY
Address is everything else that remains.
I got the following patterns for each parameter:
str1 = "HeinoPlekk69712047623+3725688736412-09-2020Tartu mnt 183,Tallinn,16881,Eesti"
first_name_pattern = r"^[A-Z][a-z]+"
last_name_pattern = r"[A-z][a-z]+(?=[0-9])"
id_code_pattern = r"\d{11}(?=\+)"
phone_number_pattern = r"\+\d{3}?\s*\d{7,8}"
date_pattern = r"\d{1,2}\-\d{1,2}\-\d{1,4}"
address_pattern = r"[A-Z][a-z]*\s.*$"
first_name_match = re.findall(first_name_pattern, str1)
last_name_match = re.findall(last_name_pattern, str1)
id_code_match = re.findall(id_code_pattern, str1)
phone_number_match = re.findall(phone_number_pattern, str1)
date_match = re.findall(date_pattern, str1)
address_match = re.findall(address_pattern, str1)
So, given "HeinoPlekk69712047623+3725688736412-09-2020Tartu mnt 183,Tallinn,16881,Eesti", I get ['Heino'] ['Plekk'] ['69712047623'] ['+372 56887364' ] ['12-09-2020'] ['Tartu mnt 183,Tallinn,16881,Eesti'], which suits me perfectly.
The problem starts when the area code is missing, because now id_code_pattern can't find the id code because of (?=\+), and if one tries to use |\d{11} (or) there is another problem because now it finds both id code and phone number (69712047623 and 37256887364). And how to improve phone_number_pattern so that it finds only 7 or 8 digits of the phone number, I do not understand.

A single expression with some well-crafted capture groups will help you immensely:
import re
str1 = "HeinoPlekk69712047623+3725688736412-09-2020Tartu mnt 183,Tallinn,16881,Eesti"
pattern = r"^(?P<first_name>[A-Z][a-z]+)(?P<last_name>[A-Z][a-z]+)(?P<id_code>\d{11})(?P<phone>(?:\+\d{3})?\s*\d{7,8})(?P<dob>\d{1,2}\-\d{1,2}\-\d{1,4})(?P<address>.*)$"
print(re.match(pattern, str1).groupdict())
Repl.it | regex101
Result:
{'first_name': 'Heino', 'last_name': 'Plekk', 'id_code': '69712047623', 'phone': '+37256887364', 'dob': '12-09-2020', 'address': 'Tartu mnt 183,Tallinn,16881,Eesti'}

How to add a missing closing parenthesis to a string in Python?

I have multiple strings to postprocess, where a lot of the acronyms have a missing closing bracket. Assume the string text below, but also assume that this type of missing bracket happens often.
My code below only works by adding the closing bracket to the missing acronym independently, but not to the full string/sentence. Any tips on how to do this efficiently, and preferably without needing to iterate ?
import re
#original string
text = "The dog walked (ABC in the park"
#Desired output:
desired_output = "The dog walked (ABC) in the park"
#My code:
acronyms = re.findall(r'\([A-Z]*\)?', text)
for acronym in acronyms:
if ')' not in acronym: #find those without a closing bracket ')'.
print(acronym + ')') #add the closing bracket ')'.
#current output:
>>'(ABC)'

You may use
text = re.sub(r'(\([A-Z]+(?!\))\b)', r"\1)", text)
With this approach, you can also get rid of the check if the text has ) in it before, see a demo on regex101.com.
In full:
import re
#original string
text = "The dog walked (ABC in the park"
text = re.sub(r'(\([A-Z]+(?!\))\b)', r"\1)", text)
print(text)
This yields
The dog walked (ABC) in the park
See a working demo on ideone.com.

For the typical example you have provided, I don't see the need of using regex
You can just use some string methods:
text = "The dog walked (ABC in the park"
withoutClosing = [word for word in text.split() if word.startswith('(') and not word.endswith(')') ]
withoutClosing
Out[45]: ['(ABC']
Now you have the words without closing parenthesis, you can just replace them:
for eachWord in withoutClosing:
text = text.replace(eachWord, eachWord+')')
text
Out[46]: 'The dog walked (ABC) in the park'

Parsing file name with RegEx - Python

I'm trying to get the "real" name of a movie from its name when you download it.
So for instance, I have
Star.Wars.Episode.4.A.New.Hope.1977.1080p.BrRip.x264.BOKUTOX.YIFY
and would like to get
Star Wars Episode 4 A New Hope
So I'm using this regex:
.*?\d{1}?[ .a-zA-Z]*
which works fine, but only for a movie with a number, as in 'Iron Man 3' for example.
I'd like to be able to get movies like 'Interstellar' from
Interstellar.2014.1080p.BluRay.H264.AAC-RARBG
and I currently get
Interstellar 2
I tried several ways, and spent quite a lot of time on it already, but figured it wouldn't hurt asking you guys if you had any suggestion/idea/tip on how to do it...
Thanks a lot!

Given your examples and assuming you always download in 1080p (or know that field's value):
x = 'Interstellar.2014.1080p.BluRay.H264.AAC-RARBG'
y = x.split('.')
print " ".join(y[:y.index('1080p')-1])
Forget the regex (for now anyway!) and work with the fixed field layout. Find a field you know (1080p) and remove the information you don't want (the year). Recombine the results and you get "Interstellar" and "Star Wars Episode 4 A New Hope".

The following regex would work (assuming the format is something like moviename.year.1080p.anything or moviename.year.720p.anything:
.*(?=.\d{4}.*\d{3,}p)
Regex example (try the unit tests to see the regex in action)
Explanation:

\.(?=.*?(?:19|20)\d{2}\b)|(?:19|20)\d{2}\b.*$
Try this with re.sub.See demo.
https://regex101.com/r/hR7tH4/10
import re
p = re.compile(r'\.(?=.*?(?:19|20)\d{2}\b)|(?:19|20)\d{2}\b.*$', re.MULTILINE)
test_str = "Star.Wars.Episode.4.A.New.Hope.1977.1080p.BrRip.x264.BOKUTOX.YIFY\nInterstellar.2014.1080p.BluRay.H264.AAC-RARBG\nIron Man 3"
subst = " "
result = re.sub(p, subst, test_str)

Assuming, there is always a four-digit-year, or a four-digit-resolution notation within the movie's file name, a simple solution replaces the not-wanted parts as this:
"(?:\.|\d{4,4}.+$)"
by a blank, strip()'ing them afterwards ...
For example:
test1 = "Star.Wars.Episode.4.A.New.Hope.1977.1080p.BrRip.x264.BOKUTOX.YIFY"
test2 = "Interstellar.2014.1080p.BluRay.H264.AAC-RARBG"
res1 = re.sub(r"(?:\.|\d{4,4}.+$)",' ',test1).strip()
res2 = re.sub(r"(?:\.|\d{4,4}.+$)",' ',test2).strip()
print(res1, res2, sep='\n')
>>> Star Wars Episode 4 A New Hope
>>> Interstellar

Python regex to print all sentences that contain two identified classes of markup

I wish to read in an XML file, find all sentences that contain both the markup <emotion> and the markup <LOCATION>, then print those entire sentences to a unique line. Here is a sample of the code:
import re
text = "Cello is a <emotion> wonderful </emotion> parakeet who lives in <LOCATION> Omaha </LOCATION>. He is the <emotion> best </emotion> singer <pronoun> I </pronoun> have ever heard."
out = open('out.txt', 'w')
for match in re.findall(r'(?:(?<=\.)\s+|^)((?=(?:(?!\.(?:\s|$)).)*?\bwonderful(?=\s|\.|$))(?=(?:(?!\.(?:\s|$)).)*?\bomaha(?=\s|\.|$)).*?\.(?=\s|$))', text, flags=re.I):
line = ''.join(str(x) for x in match)
out.write(line + '\n')
out.close()
The regex here grabs all sentences with "wonderful" and "omaha" in them, and returns:
Cello is a <emotion> wonderful </emotion> parakeet who lives in <LOCATION> Omaha </LOCATION>.
Which is perfect, but I really want to print all sentences that contain both <emotion> and <LOCATION>. For some reason, though, when I replace "wonderful" in the regex above with "emotion," the regex fails to return any output. So the following code yields no result:
import re
text = "Cello is a <emotion> wonderful </emotion> parakeet who lives in <LOCATION> Omaha </LOCATION>. He is the <emotion> best </emotion> singer I have ever heard."
out = open('out.txt', 'w')
for match in re.findall(r'(?:(?<=\.)\s+|^)((?=(?:(?!\.(?:\s|$)).)*?\bemotion(?=\s|\.|$))(?=(?:(?!\.(?:\s|$)).)*?\bomaha(?=\s|\.|$)).*?\.(?=\s|$))', text, flags=re.I):
line = ''.join(str(x) for x in match)
out.write(line + '\n')
out.close()
My question is: How can I modify my regular expression in order to grab only those sentences that contain both <emotion> and <LOCATION>? I would be most grateful for any help others can offer on this question.
(For what it's worth, I'm working on parsing my text in BeautifulSoup as well, but wanted to give regular expressions one last shot before throwing in the towel.)

Your problem appears to be that your regex is expecting a space (\s) to follow the matching word, as seen with:
emotion(?=\s|\.|$)
Since when it's part of a tag, it's followed by a >, rather than a space, no match is found since that lookahead fails. To fix it, you can just add the > after emotion, like:
for match in re.findall(r'(?:(?<=\.)\s+|^)((?=(?:(?!\.(?:\s|$)).)*?\bemotion>(?=\s|\.|$))(?=(?:(?!\.(?:\s|$)).)*?\bomaha(?=\s|\.|$)).*?\.(?=\s|$))', text, flags=re.I):
line = ''.join(str(x) for x in match)
Upon testing, this seems to solve your problem. Make sure and treat "LOCATION" similarly:
for match in re.findall(r'(?:(?<=\.)\s+|^)((?=(?:(?!\.(?:\s|$)).)*?\bemotion>(?=\s|\.|$))(?=(?:(?!\.(?:\s|$)).)*?\bLOCATION>(?=\s|\.|$)).*?\.(?=\s|$))', text, flags=re.I):
line = ''.join(str(x) for x in match)

If I do not understand bad what you are trying to do is remove <emotion> </emotion> <LOCATION></LOCATION> ??
Well if is that what you want to do you can do this
import re
text = "Cello is a <emotion> wonderful </emotion> parakeet who lives in <LOCATION> Omaha </LOCATION>. He is the <emotion> best </emotion> singer I have ever heard."
out = open('out.txt', 'w')
def remove_xml_tags(xml):
content = re.compile(r'<.*?>')
return content.sub('', xml)
data = remove_xml_tags(text)
out.write(data + '\n')
out.close()

I have just discovered that the regex may be bypassed altogether. To find (and print) all sentences that contain two identified classes of markup, you can use a simple for loop. In case it might help others who find themselves where I found myself, I'll post my code:
# read in your file
f = open('sampleinput.txt', 'r')
# use read method to convert the read data object into string
readfile = f.read()
#########################
# now use the replace() method to clean data
#########################
# replace all \n with " "
nolinebreaks = readfile.replace('\n', ' ')
# replace all commas with ""
nocommas = nolinebreaks.replace(',', '')
# replace all ? with .
noquestions = nocommas.replace('?', '.')
# replace all ! with .
noexclamations = noquestions.replace('!', '.')
# replace all ; with .
nosemicolons = noexclamations.replace(';', '.')
######################
# now use replace() to get rid of periods that don't end sentences
######################
# replace all Mr. with Mr
nomisters = nosemicolons.replace('Mr.', 'Mr')
#replace 'Mrs.' with 'Mrs' etc.
cleantext = nomisters
#now, having cleaned the input, find all sentences that contain your two target words. To find markup, just replace "Toby" and "pipe" with <markupclassone> and <markupclasstwo>
periodsplit = cleantext.split('.')
for x in periodsplit:
if 'Toby' in x and 'pipe' in x:
print x

Python, Regular Expression Postcode search

I am trying to use regular expressions to find a UK postcode within a string.
I have got the regular expression working inside RegexBuddy, see below:
\b[A-Z]{1,2}[0-9][A-Z0-9]? [0-9][ABD-HJLNP-UW-Z]{2}\b
I have a bunch of addresses and want to grab the postcode from them, example below:
123 Some Road Name Town, City County PA23 6NH
How would I go about this in Python? I am aware of the re module for Python but I am struggling to get it working.
Cheers
Eef

repeating your address 3 times with postcode PA23 6NH, PA2 6NH and PA2Q 6NH as test for you pattern and using the regex from wikipedia against yours, the code is..
import re
s="123 Some Road Name\nTown, City\nCounty\nPA23 6NH\n123 Some Road Name\nTown, City"\
"County\nPA2 6NH\n123 Some Road Name\nTown, City\nCounty\nPA2Q 6NH"
#custom
print re.findall(r'\b[A-Z]{1,2}[0-9][A-Z0-9]? [0-9][ABD-HJLNP-UW-Z]{2}\b', s)
#regex from #http://en.wikipedia.orgwikiUK_postcodes#Validation
print re.findall(r'[A-Z]{1,2}[0-9R][0-9A-Z]? [0-9][A-Z]{2}', s)
the result is
['PA23 6NH', 'PA2 6NH', 'PA2Q 6NH']
['PA23 6NH', 'PA2 6NH', 'PA2Q 6NH']
both the regex's give the same result.

Try
import re
re.findall("[A-Z]{1,2}[0-9][A-Z0-9]? [0-9][ABD-HJLNP-UW-Z]{2}", x)
You don't need the \b.

#!/usr/bin/env python
import re
ADDRESS="""123 Some Road Name
Town, City
County
PA23 6NH"""
reobj = re.compile(r'(\b[A-Z]{1,2}[0-9][A-Z0-9]? [0-9][ABD-HJLNP-UW-Z]{2}\b)')
matchobj = reobj.search(ADDRESS)
if matchobj:
print matchobj.group(1)
Example output:
[user#host]$ python uk_postcode.py
PA23 6NH