sort a list using lambda and regex in python

Question

list = ['xxxx ResultDatetime:2017-05-31 09:38:00.000:ResultDatetime', 'xxxx ResultDatetime:2017-05-26 15:36:00.000:ResultDatetime', 'yyyyy' ResultDatetime:2017-10-23 16:16:00.000:ResultDatetime]

datet = re.compile(r'ResultDatetime:(\d{4}-\d{2}-\d{2} \d{2}:\d{2})')

list.sort(key = lambda x: ........)

I want to sort the lists in an order starting with the earliest date. How should I go about it using lambda and regex?

Answer 1

With the code you have there it is sufficient to do:

list.sort(key=lambda x: datet.search(x).group(1))

(but please, don't use list as a variable name).

There is no need to convert the extracted string to a datetime as it is already in a format that will sort naturally.

Note however that if any string does not match the regex this will generate an error, so you may be better to split the key out into a named multi-line function and test for a successful match before returning the matched group.

def sort_key(line):                                                                                                                                               
    match = datet.search(line)                                                                                                                                               
    if match:                                                                                                                                                     
        return match.group(1)                                                                                                                                                    
    return ''        

data = [
    'xxxx ResultDatetime:2017-05-31 09:38:00.000:ResultDatetime',
    'xxxx ResultDatetime:2017-05-26 15:36:00.000:ResultDatetime',
    'yyyyy ResultDatetime:2017-10-23 16:16:00.000:ResultDatetime'
]
data.sort(key=sort_key)