CODEWITHSUNDEEP
pythonregex
Aiden
Aiden

Reputation: 9

Using regex to filter the timestamp out of message logs?

im new to learning python just so you guys know. Im trying to make a function that uses regex to filter out the timestamp from a message log. Im trying to get my test cases to work but im getting an assertion error.

def get_hr_min(row):
    time_match = re.search(r'(\d{2}):(\d{2})', row)
    hour = time_match.group(1)
    minute = time_match.group(2)
    dhour = {}
    dhour['hour'] = hour
    dmin = {}
    dmin['minute'] = minute
    return dhour, dmin

here is what I have so far. as far as I can tell the function should return the minutes and hours in dictionary format(but I may have formatted it wrong as my professor did not tell me what a dictionary actually is)

log_open_row = '--- Log opened Tue Sep 20 00:01:49 2016'
join_quit_row = '00:01 -!- Guest40341 [[email protected]] has quit [Quit: Bye]'
message_row = '00:25 < ice231> anyone good with exploiting cisco asa with extrabacon?'
        
#assert get_hr_min(log_open_row) == {}
assert get_hr_min(join_quit_row) == {'hour': 0, 'minute': 1}
assert get_hr_min(message_row) == {'hour': 0, 'minute': 25}

here are my test cases. I get an assertion error when I run these test cases. I need to make an if statement to filter out the first assertion but I wanted to get the dictionary to work first(im assuming that is where my mistake is?). Any help or tips are appreciated.

Upvotes: 0

Views: 147

Answers (1)

Agent Biscutt
Agent Biscutt

Reputation: 737

A few problems with your code.

First of all, the functino you have there returns a set of dictionaries, rather than one dictionary. This means the value given is effectively a dictionary of dictionaries rather than a single dictionary. Second problem is that you are trying to compare and integer to a string in your assertion. Despite the fact that regex tells you that '01' is a number, its not. '01' is a string. To fix this, turn it into a number using int().

Here is the code fixed up:

def get_hr_min(row):
    time_match = re.search(r'(\d{2}):(\d{2})', row)
    hour = time_match.group(1)
    minute = time_match.group(2)
    rvalue = {} # return value
    rvalue['hour'] = int(hour) #give the hour as an integer to the "hour" value of rvalue
    rvalue['minute'] = int(minute) #give the minute as an integer to the "minute" value of rvalue
    return rvalue # return the dictionary, as one complete dictinary rather than a set.

Upvotes: 1

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