First of all I do not know if these types of questions are appropriate in stackoverflow.
I have the following dict:
file_dict = {"asda_20221003:ada":1, "scdfws_20221104eaf2we":5, "iasndcfiopsadn":9}
The key values of the dict always contains a date with the format %Y%m%d, what i want i to obtain the value for the key that have the highest date.
What I have done is the following:
OrderedDict(sorted(file_dict.items(), key=lambda t: datetime.strptime(re.findall("\d{8}",t[0])[0], "%Y%m%d") if len(re.findall("\d{8}",t[0])) > 0 else datetime.min, reverse=True))
This expression works but it is unreadable. Is there any way in order to improve it?
what i would like is to asigne at some point re.findall("\d{8}",t[0])
to a variable (for example date) and use this one for all the expression.
Something like this:
OrderedDict(sorted(file_dict.items(), key=lambda t: datetime.strptime(x[0], "%Y%m%d") if len(re.findall("\d{8}",t[0]) as x) > 0 else None, reverse=True))
I am also open for other ways to perform this operation
You can simply compare list of strings and not compare datetimes.
As @Steven Rumbalski said: Note that as of Python 3.7 dictionaries are guaranteed to retain the insertion order of the keys, so OrderedDict()
can be replaced by dict()
in this example.
OrderedDict(
sorted(file_dict.items(),
key=lambda t: re.findall(r"\d{8}", t[0]),
reverse=True))
# OR
dict(
sorted(file_dict.items(),
key=lambda t: re.findall(r"\d{8}", t[0]),
reverse=True))
Results:
OrderedDict([('scdfws_20221104eaf2we', 5),
('asda_20221003:ada', 1),
('iasndcfiopsadn', 9)])
# OR
{'scdfws_20221104eaf2we': 5, 'asda_20221003:ada': 1, 'iasndcfiopsadn': 9}
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