Sentence : "I went to hospital and admitted. Date of admission: 12/08/2019 and surgery of Date of surgery: 15/09/2015. Date of admission: 12/05/2018 is admitted Raju"
keyword: "Date of admission:"
Required solution: 12/08/2019,12/05/2018
Is there any solution to get the dates near "Date of admission:" only. Is there any solution
I was unable to reproduce the result in the answer by #Ders. Plus I think .findall() is more appropriate here anyway, so:
import re
pattern = re.compile(r"Date of admission: (\d{2}/\d{2}/\d{4})")
print(pattern.findall(s))
# ['12/08/2019', '12/05/2018']
Use a capturing group. If the re matches, then you can get the contents of the group.
import re
p = re.compile("Date of admission: (\d{2}/\d{2}/\d{4})")
m = p.match(s)
date = m.group(1)
# 12/08/2019
Related
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.
In this assignment, the input wanted is in this format:
Regular: 16Mar2009(mon), 17Mar2009(tues), 18Mar2009(wed) ...
Reward: 26Mar2009(thur), 27Mar2009(fri), 28Mar2009(sat)
Regular or Reward is the name of customer type. I separated this string like this.
entry_list = input.split(":") #input is a variable
client = entry_list[0] # only Regular or Reward
dates = entry_list[1] # only dates
days = dates.split(",")
But now I need to count weekdays or weekend days inside the days list:
days = [' 16Mar2009(mon)', ' 17Mar2009(tues)', ' 18Mar2009(wed)']
When it is mon tues wed thur fri, all count as weekday, and I need to know how many weekdays the input have.
When it is sat sun must be counted as weekend days, and I need to know how many weekends the input have.
How to write a regex in python to search for all weekdays and weekend days inside this list and count them, putting the number of weekdays and weekend days in two different counters?
EDIT
I wrote this function to check if the dates are in the write format but it's not working:
def is_date_valid(date):
date_regex = re.compile(r'(?:\d{1,2}[A-Za-z]{3}\d{4}\([A-Za-z]{3}\),\s+){2}\d{1,2}[A-Za-z]{3}\d{4}\([A-Za-z]{3}\)$')
m = date_regex.search(date)
m is only returning None
You don't really need a regex for this. You can just look for "sat" and "sun" tags directly, since your days are formatted the same way (i.e. no capitals, no "tue" instead of "tues", etc.) you shouldn't need to generalize to a pattern. Just loop through the list and look for "sat" and "sun":
import re #if you are using the re
days = [' 16Mar2009(mon)', ' 17Mar2009(tues)', ' 18Mar2009(wed)', ' 18Mar2009(sat)', ' 18Mar2009(sun)']
weekends = 0
weekdays = 0
for day in days:
if "sat" in day or "sun" in day: #if re.search( '(sat|sun)', day ): also works
weekends = weekends+1
else:
weekdays = weekdays+1
print(weekends)
print(weekdays)
>>>2
>>>3
if you need to use a regex, because this is part of an assignment for example, then this variation of the if statement will do it: if re.search( '(sat|sun)', day ): This isn't too much more useful than just using the strings since the strings are the regex in this case, but seeing how to put multiple patterns together into one regex with or style logic is useful so I'm still including it here.
I'm trying to replace a word (e.g. on) if it falls between two substrings (e.g. <temp> & </temp>) however other words are present which need to be kept.
string = "<temp>The sale happened on February 22nd</temp>"
The desired string after the replace would be:
Result = <temp>The sale happened {replace} February 22nd</temp>
I've tried using regex, I've only been able to figure out how to replace everything lying between the two <temp> tags. (Because of the .*?)
result = re.sub('<temp>.*?</temp>', '{replace}', string, flags=re.DOTALL)
However on may appear later in the string not between <temp></temp> and I wouldn't want to replace this.
re.sub('(<temp>.*?) on (.*?</temp>)', lambda x: x.group(1)+" <replace> "+x.group(2), string, flags=re.DOTALL)
Output:
<temp>The sale happened <replace> February 22nd</temp>
Edit:
Changed the regex based on suggestions by Wiktor and HolyDanna.
P.S: Wiktor's comment on the question provides a better solution.
Try lxml:
from lxml import etree
root = etree.fromstring("<temp>The sale happened on February 22nd</temp>")
root.text = root.text.replace(" on ", " {replace} ")
print(etree.tostring(root, pretty_print=True))
Output:
<temp>The sale happened {replace} February 22nd</temp>
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 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