I need help grabbing just K334-76A9 from this string:
b'\x0cWelcome, Pepo \r\nToday is Mon 04/29/2019 \r\n\r\n Volume in drive C has no label.\r\n Volume Serial Number is K334-76A9\r\n
Please help, I have tried so many things but none have worked.
Sorry if my question is bad :/
If you want to find the format xxxx-xxxx, no matter what string you have you can do it like this:
import re
b = '\x0cWelcome, Pepo \r\nToday is Mon 04/29/2019 \r\n\r\n Volume in drive C has no label.\r\n Volume Serial Number is K334-76A9\r\n'
splitString = []
splitString = b.split()
r = re.compile('.{4}-.{4}')
for string in splitString:
if r.match(string):
print(string)
Output:
K334-76A9
Here's code that grabs everything after "Serial Number is " up to the next whitespace character.
import re
data = b'\x0cWelcome, Pepo \r\nToday is Mon 04/29/2019 \r\n\r\n Volume in drive C has no label.\r\n Volume Serial Number is K334-76A9\r\n'
pat = re.compile(r"Serial Number is ([^\s]+)")
match = pat.search(data.decode("ASCII"))
if match:
print(match.group(1))
Result:
K334-76A9
You can adjust the regular expression per your needs. Regular expressions are Da Bomb! This one's really simple, but you can do amazingly complex things with them.
Related
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'}
string = "At Donald Trump<U+2019>s Properties, a Showcase for a Brand and a President-Elect"
I want to get rid of the <U + 2019> and replace it with '. Is there a way to do this in python?
Edit : I also have instances of <U + 2014>, <U + 201C> etc. Looking for something which can replace all of this with appropriate values
Replace them all at once with re.sub:
import re
string = "testing<U+2019> <U+2014> <U+201C>testing<U+1F603>"
result = re.sub(r'<U\+([0-9a-fA-F]{4,6})>', lambda x: chr(int(x.group(1),16)), string)
print(result)
Output:
testing’ — “testing😃
The regular expression matches <U+hhhh> where hhhh can be 4-6 hexadecimal characters. Note that Unicode defines code points from U+0000 to U+10FFFF so this accounts for that. The lambda replacement function converts the string hhhh to an integer using base 16 and then converts that number to a Unicode character.
Here's my solution for all code points denoted as U+0000 through U+10FFFF ("U+" followed by the code point value in hexadecimal, which is prepended with leading zeros to a minimum of four digits):
import re
def UniToChar(unicode_notation):
return chr(int(re.findall(r'<U\+([a-hA-H0-9]{4,5})>',unicode_notation)[0],16))
xx= '''
At Donald<U+2019>s <U+2016>Elect<U+2016> in <U+2017>2019<U+2017>
<U+00C0> la Donald<U+2019>s friend <U+1F986>. <U+1F929><U+1F92A><U+1F601>
'''
for x in xx.split('\n'):
abc = re.findall(r'<U\+[a-hA-H0-9]{4,5}>',x)
if len(abc) > 0:
for uniid in set(abc): x=x.replace(uniid, UniToChar(uniid))
print(repr(x).strip("'"))
Output: 71307293.py
At Donald’s ‖Elect‖ in ‗2019‗
À la Donald’s friend 🦆. 🤩🤪😁
In fact, private range from U+100000 to U+10FFFD (Plane 16) isn't detected using above simplified regex… Improved code follows:
import re
def UniToChar(unicode_notation):
aux = int(re.findall(r'<U\+([a-hA-H0-9]{4,6})>',unicode_notation)[0],16)
# circumvent the "ValueError: chr() arg not in range(0x110000)"
if aux <= 0x10FFFD:
return chr(aux)
else:
return chr(0xFFFD) # Replacement Character
xx= '''
At Donald<U+2019>s <U+2016>Elect<U+2016> in <U+2017>2019<U+2017>
<U+00C0> la Donald<U+2019>s friend <U+1F986>. <U+1F929><U+1F92A><U+1F601>
Unassigned: <U+05ff>; out of Unicode range: <U+110000>.
'''
for x in xx.split('\n'):
abc = re.findall(r'<U\+[a-hA-H0-9]{4,6}>',x)
if len(abc) > 0:
for uniid in set(abc): x=x.replace(uniid, UniToChar(uniid))
print(repr(x).strip("'"))
Output: 71307293.py
At Donald’s ‖Elect‖ in ‗2019‗
À la Donald’s friend 🦆. 🤩🤪😁
Unassigned: \u05ff; out of Unicode range: �.
I guess this solves the problem if its just one or two of these characters.
>>> string = "At Donald Trump<U+2019>s Properties, a Showcase for a Brand and a President-Elect"
>>> string.replace("<U+2019>","'")
"At Donald Trump's Properties, a Showcase for a Brand and a President-Elect"
If there are many if these substitutions to be done, consider using 'map()' method.
Source: Removing \u2018 and \u2019 character
You can replace using .replace()
print(string.replace('<U+2019>', "'"))
Or if your year changes, you can use re. But make it more attractive than mine.
import re
string = "At Donald Trump<U+2019>s Properties, a Showcase for a Brand and a President-Elect"
rep = re.search('[<][U][+]\d{4}[>]', string).group()
print(string.replace(rep, "'"))
what version of python are u using?
I edited my answer so it can bee used with multiple code point in the same string
well u need to convert the unicode's code point that is between < >, to unicode char
I used regex to get the unicode's code point and then convert it to the corresponding uniode char
import re
string = "At Donald Trump<U+2019>s Properties, a Showcase for a Brand and a President<U+2014>Elect"
repbool = re.search('[<][U][+]\d{4}[>]', string)
while repbool:
rep = re.search('[<][U][+]\d{4}[>]', string).group()
string=string.replace(rep, chr(int(rep[1:-1][2:], 16)))
repbool = re.search('[<][U][+]\d{4}[>]', string)
print(string)
My software outputs these two types of output:
-rwx------ Administrators/Domain Users 456220672 0% 2018-04-16 16:04:40 E:\\_WiE10-18.0.100-77.iso
-rwxrwx--- Administrators/unknown 6677 0% 2018-04-17 01:33:23 E:\\program files\\cluster groups\\sql server (mssqlserver)\\logs\\progress-MOD-1523883344023-3001-Windows.log
I would like to get the file names from both outputs:
E:\\_WiE10-18.0.100-77.iso, for the first one
E:\\program files\\cluster groups\\sql server (mssqlserver)\\logs\\progress-MOD-1523883344023-3001-Windows.log, for the second one
If i use something like the code below, it won't work if the second parameter has spaces in it. It works if there aren't any spaces in the Domain Username.
for item in outputs:
outputs.extend(item.split())
for item2 in [' '.join(outputs[6:])]:
new_list.append(item2)
How can I get all the parameters individually, including the filenames?
If regex is an option:
text = """-rwx------ Administrators/Domain Users 456220672 0% 2018-04-16 16:04:40 E:\\_WiE10-18.0.100-77.iso
-rwxrwx--- Administrators/unknown 6677 0% 2018-04-17 01:33:23 E:\\program files\\cluster groups\\sql server (mssqlserver)\\logs\\progress-MOD-1523883344023-3001-Windows.log"""
import re
for h in re.findall(r"^.*?\d\d:\d\d:\d\d (.*)",text,flags=re.MULTILINE):
print(h)
Output:
E:\_WiE10-18.0.100-77.iso
E:\program files\cluster groups\sql server (mssqlserver)\logs\progress-MOD-1523883344023-3001-Windows.log
Pattern explained:
The pattern r"^.*?\d\d:\d\d:\d\d (.*)" looks for linestart '^' + as less anythings as possible '.*?' + the time-stamp '\d\d:\d\d:\d\d ' followed by a space and captures all behind it till end of line into a group.
It uses the re.MULTILINE flag for that.
Edit:
Capturing the individual things needs some more capturing groups:
import re
for h in re.findall(r"^([rwexXst-]+) ([^0-9]+) +\d+.+? +(\d{4}-\d{2}-\d{2} \d{2}:\d{2}:\d{2}) (.*)",text,flags=re.MULTILINE):
# ^^^^^^^^^^^^ ^^^^^^^^ ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ ^^^^
# flags grpName datetime filename
for k in h:
print(k)
print("")
Output:
-rwx------
Administrators/Domain Users
2018-04-16 16:04:40
E:\_WiE10-18.0.100-77.iso
-rwxrwx---
Administrators/unknown
2018-04-17 01:33:23
E:\program files\cluster groups\sql server (mssqlserver)\logs\progress-MOD-1523883344023-3001-Windows.log
You could use a regular expression like
\b[A-Z]:\\\\.+
Aside from using regex, you can try something similar to this.
output = '-rwx------ ... 2018-04-16 16:04:40 E:\\\\_WiE10-18.0.100-77.iso'
drive_letter_start = output.find(':\\\\')
filename = output[drive_letter_start - 1:]
It looks for the first occurrence of ':\\'and gets the drive letter before the substring (i.e. ':\\') and the full file path after the substring.
EDIT
Patrick Artner's answer is better and completely answers OP's question compared to this answer. This only encompasses capturing the file path. I am leaving this answer here should anyone find it useful.
import re
col4="""May god bless our families studied. CiteSeerX 2009-05-24 2007-11-19 2004"""
b=re.findall(r'\sCiteSeerX',col4)
print b
I have to print "May god bless our families studied". I'm using pythton regular expressions to extract the file name but i'm only getting CiteSeerX as output.I'm doing this on a very large dataset so i only want to use regular expression if there is any other efficient and faster way please point out.
Also I want the last year 2004 as a output.
I'm new to regular expressions and I now that my above implementation is wrong but I can't find a correct one. This is a very naive question. I'm sorry and Thank you in advance.
Here is an answer that doesn't use regex.
>>> s = "now is the time for all good men"
>>> s.find("all")
20
>>> s[:20]
'now is the time for '
>>>
If the structure of all your data is similar to the sample you provided, this should get you going:
import re
data = re.findall("(.*?) CiteSeerX.*(\d{4})$", col4)
if data:
# we have a match extract the first capturing group
title, year = data[0]
print(title, year)
else:
print("Unable to parse the string")
# Output: May god bless our families studied. 2004
This snippet extracts everything before CiteSeerX as the title and the last 4 digits as the year (again, assuming that the structure is similar for all the data you have). The brackets mark the capturing groups for the parts that we are interested in.
Update:
For the case, where there is metadata following the year of publishing, use the following regular expression:
import re
YEAR = "\d{4}"
DATE = "\d\d\d\d-\d\d-\d\d"
def parse_citation(s):
regex = "(.*?) CiteSeerX\s+{date} {date} ({year}).*$".format(date=DATE, year=YEAR)
data = re.findall(regex, s)
if data:
# we have a match extract the first group
return data[0]
else:
return None
c1 = """May god bless our families studied. CiteSeerX 2009-05-24 2007-11-19 2004"""
c2 = """May god bless our families studied. CiteSeerX 2009-05-24 2007-11-19 2004 application/pdf text http //citeseerx.ist.psu.edu/viewdoc/summary?doi=10.1.1.1.1483 http //www.biomedcentral.com/content/pdf/1471-2350-5-20.pdf en Metadata may be used without restrictions as long as the oai identifier remains attached to it."""
print(parse_citation(c1))
print(parse_citation(c2))
# Output:
# ('May god bless our families studied.', '2004')
# ('May god bless our families studied.', '2004')
I want to get an HTML page with python and then print out all the IPs from it.
I will define an IP as the following:
x.x.x.x:y
Where:
x = a number between 0 and 256.
y = a number with < 7 digits.
Thanks.
Right. The only part I cant do is the regular expression one. – das 9 mins ago If someone shows me that, I will be fine. – das 8 mins ago
import re
ip = re.compile(r"\b(?:(?:25[0-5]|2[0-4]\d|[01]?\d\d?)\.){3}(?:25[0-5]|2[0-4]\d|[01]?\d\d?):\d{1,6}\b")
junk = " 1.1.1.1:123 2.2.2.2:321 312.123.1.12:123 "
print ip.findall(junk)
# outputs ['1.1.1.1:123', '2.2.2.2:321']
Here is a complete example:
import re, urllib2
f = urllib2.urlopen("http://www.samair.ru/proxy/ip-address-01.htm")
junk = f.read()
ip = re.compile(r"\b(?:(?:25[0-5]|2[0-4]\d|[01]?\d\d?)\.){3}(?:25[0-5]|2[0-4]\d|[01]?\d\d?):\d{1,6}\b")
print ip.findall(junk)
# ['114.30.47.10:80', '118.228.148.83:80', '119.70.40.101:8080', '12.47.164.114:8888', '121.
# 17.161.114:3128', '122.152.183.103:80', '122.224.171.91:3128', '123.234.32.27:8080', '124.
# 107.85.115:80', '124.247.222.66:6588', '125.76.228.201:808', '128.112.139.75:3128', '128.2
# 08.004.197:3128', '128.233.252.11:3124', '128.233.252.12:3124']
The basic approach would be:
Use urllib2 to download the contents of the page
Use a regular expression to extract IPv4-like addresses
Validate each match according to the numeric constraints on each octet
Print out the list of matches
Please provide a clearer indication of what specific part you are having trouble with, along with evidence to show what it is you've tried thus far.
Not to turn this into a who's-a-better-regex-author-war but...
(\d{1,3}\.){3}\d{1,3}\:\d{1,6}
Try:
re.compile("\d?\d?\d.\d?\d?\d.\d?\d?\d.\d?\d?\d:\d+").findall(urllib2.urlopen(url).read())
In action:
\b(?: # A.B.C in A.B.C.D:port
(?:
25[0-5]
| 2[0-4][0-9]
| 1[0-9][0-9]
| [1-9]?[0-9]
)\.
){3}
(?: # D in A.B.C.D:port
25[0-5]
| 2[0-4][0-9]
| 1[0-9][0-9]
| [1-9]?[0-9]
)
:[1-9]\d{0,5} # port number any number in (0,999999]
\b