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up vote 3 down vote favorite

I am writing a python library to parse different working hours string and produce the standard format of hours. I stuck in the following case:

My regex should return groups for Mon - Fri 7am - 5pm Sat 9am - 3pm as ['Mon - Fri 7am - 5pm ', 'Sat 9am - 3pm'] but if there is a comma between first and second then it should return [].

Also the comma can be in anywhere but should not between the two weekdays & duration. eg: Mon - Fri 7am - 5pm Sat 9am - 3pm and available upon email, phone call should return ['Mon - Fri 7am - 5pm ', 'Sat 9am - 3pm'].

This is what I have tried,

import re
pattern = """(
    (?:mon|tue|wed|thu|fri|sat|sun|mo|tu|we|th|fr|sa|su|m|w|f|thurs) # Start weekday
\s*[-|to]+\s* # Seperator
(?:mon|tue|wed|thu|fri|sat|sun|mo|tu|we|th|fr|sa|su|^(?![ap])m|w|f|thurs)?  # End weekday
\s*[from]*\s* # Seperator
(?:\d{1,2}(?:[:]\d{1,2})?)\s*(?:[ap][.]?m.?) # Start hour
\s*[-|to]+\s* # Seperator
(?:\d{1,2}(?:[:]\d{1,2})?)\s*(?:[ap][.]?m.?) # Close hour
)"""

regEx = re.compile(pattern, re.IGNORECASE|re.VERBOSE)

print re.findall(regEx, "Mon - Fri 7am - 5pm Sat 9am - 3pm")
# output ['Mon - Fri 7am - 5pm ', 'Sat 9am - 3pm']
print re.findall(regEx, "Mon - Fri 7am - 5pm Sat - Sun 9am - 3pm")
# output ['Mon - Fri 7am - 5pm ', 'Sat - Sun 9am - 3pm']
print re.findall(regEx, "Mon - Fri 7am - 5pm, Sat 9am - 3pm")
# expected output []
# but I get ['Mon - Fri 7am - 5pm,', 'Sat 9am - 3pm']
print re.findall(regEx, "Mon - Fri 7am - 5pm , Sat 9am - 3pm")
# expected output []
# but I get ['Mon - Fri 7am - 5pm ', 'Sat 9am - 3pm']

Also I tried negative look ahead pattern in my regex

pattern = """(
(?:mon|tue|wed|thu|fri|sat|sun|mo|tu|we|th|fr|sa|su|m|w|f|thurs)
\s*[-|to]+\s*
(?:mon|tue|wed|thu|fri|sat|sun|mo|tu|we|th|fr|sa|su|^(?![ap])m|w|f|thurs)?
\s*[from]*\s*
(?:\d{1,2}(?:[:]\d{1,2})?)\s*(?:[ap][.]?m.?)
\s*[-|to]+\s*
(?:\d{1,2}(?:[:]\d{1,2})?)\s*(?:[ap][.]?m.?)
(?![^,])
)"""

But I didnt get expected one. Should I explicitly write code for checking condition? Is there any way to just changing my regex instead of writing explicit condition checking?

Another way I like to implement is infix the comma between two weekday- duration if comma doesn't exist and change my regex to group by/split by comma. "Mon - Fri 7am - 5pm Sat 9am - 3pm" => "Mon - Fri 7am - 5pm, Sat 9am - 3pm"

share|improve this question
    
How about you remove the comma? Wouldn't that be easier? Filter out the comma before sending it to re. – CppLearner Feb 7 '13 at 9:24
    
I require comma for further processing, so I would like to add comma between two week-duration if "Mon - Fri 7am - 5pm Sat 9am - 3pm". I will edit my question now. "further processsing"- I already have a parser to standardize the string if comma exists. – underscore Feb 7 '13 at 9:40

3 Answers 3

up vote 1 down vote

I think that you can doing it simply by matching the whole expression so that comma (and other characters are not allowed :

pattern = """^(
(
    (?:mon|tue|wed|thu|fri|sat|sun|mo|tu|we|th|fr|sa|su|m|w|f|thurs) # Start weekday
\s*[-|to]+\s* # Seperator
(?:mon|tue|wed|thu|fri|sat|sun|mo|tu|we|th|fr|sa|su|^(?![ap])m|w|f|thurs)?  # End weekday
\s*[from]*\s* # Seperator
(?:\d{1,2}(?:[:]\d{1,2})?)\s*(?:[ap][.]?m.?) # Start hour
\s*[-|to]+\s* # Seperator
(?:\d{1,2}(?:[:]\d{1,2})?)\s*(?:[ap][.]?m.?) # Close hour
)
)+$""

This will output :

[('Sat 9am - 3pm', 'Sat 9am - 3pm')]
[('Sat - Sun 9am - 3pm', 'Sat - Sun 9am - 3pm')]
[]
[]

Hope it helps,

share|improve this answer
    
Did you notice it both the items in the array? – underscore Feb 7 '13 at 10:17
    
Yes I'm trying to find a better solution :) (actually I can't figure out why it doesn't work when you remove the first enclosing parenthesis...) – Y__ Feb 7 '13 at 10:19
up vote 0 down vote

I wrote few lines of code to check and insert comma after every if the comma doesn't exists between two weekday duration. So I could able to get a same format "Mon - Fri 7am - 5pm, Sat 9am - 3pm" and I can proceed further.

share|improve this answer
up vote 0 down vote

Couldn't figure how to do that in a single regex, its hard you got a nice question. I could do what you need, but be aware that im not proud of it.

Supossing you have a function to do that...

def sample_funct(unparsed_schedule)
    result = []

    # Day Pattern
    pattern = """
    (?:mon|tue|wed|thu|fri|sat|sun|mo|tu|we|th|fr|sa|su|m|w|f|thurs) # Start weekday
    \s*[-|to]+\s* # Seperator
    (?:mon|tue|wed|thu|fri|sat|sun|mo|tu|we|th|fr|sa|su|^(?![ap])m|w|f|thurs)?  # End weekday
    \s*[from]*\s* # Seperator
    (?:\d{1,2}(?:[:]\d{1,2})?)\s*(?:[ap][\.]?m\.?) # Start hour
    \s*[-|to]+\s* # Seperator
    (?:\d{1,2}(?:[:]\d{1,2})?)\s*(?:[ap][\.]?m\.?) # Close hour
    """

    # No commas pattern
    pattern2 = "%s\s*[^,]\s*%s" % (pattern, pattern)

    # Actual Regex Pattern Items
    schedule     = re.compile(pattern, re.IGNORECASE|re.VERBOSE)
    remove_comma = re.compile(pattern2, re.IGNORECASE|re.VERBOSE)

    # Check we have no commas in the middle
    valid_result = re.search(remove_comma, unparsed_schedule)
    if valid_result:
        # Positive result, return the list with schedules
        result = re.findall(schedule, validresult.group(0))

    # If no valid results will return empty list
    return result
share|improve this answer
    
Thanks! I made the string to unique format i.e, inserted commas for which has no comma. – underscore Feb 14 '13 at 8:06

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