python dictionary datetime as key, keyError

问题:

I'm trying to run a Python script using cron in Linux, which should construct a dictionary of data. I'm attempting to use datetime().now().time()as keys in the dictionary, but it seems to raise an error.

Can't the datetime type be used as a dictionary key in Python? If that is the case, what are my alternatives?

Code:

time_now = dt.datetime.now().time()
date_today = dt.datetime.now().date()
usage_dict_hourly = {}
date_wise_dict = {}

def constructing_dict(data_int):
    date_wise_dict[usage_dict_hourly[time_now]] = data_int
    print date_wise_dict

Error:

<ipython-input-9-ef6a500cc71b> in constructing_dict(data_int)
     36 
     37 def constructing_dict(data_int):
---> 38     date_wise_dict[usage_dict_hourly[time_now]] = data_int
     39     print date_wise_dict
     40 

KeyError: datetime.time(22, 40, 33, 746509)

回答1:

Answering your question about datetime as a dictionary key:

Yes, time object of datetime can be used as dictionary key.

Conditions to use an object as a dictionary key:

To be used as a dictionary key, an object must support the hash function (e.g. through __hash__), equality comparison (e.g. through __eq__ or __cmp__)(...)

(Source: DictionaryKeys)

datetime support for cited conditions:

Supported operations:

...

• hash, use as dict key

(Source: datetime - time Objects)

However, you are getting this exception because dict usage_dict_hourly is empty.

usage_dict_hourly is initiated with {}. So, when you try to look for the element with key time_now in your function constructing_dict, it raises KeyError exception, because that key has not been found.

回答2:

You aren't actually setting useage_dict_hourly to any value that way, that's your error. You should ideally do something like:

useage_dict_hourly[time_now] = None
date_wise_dict[useage_dict_hourly[time_now]] = data_int

Or, better yet, just use datetime.datetime.now() as the key:

datetime_wise_date[datetime.datetime.now()] = data_int.

This way, there is no possibility of clashes in the value (which is possible - though unlikely - for useage_dict_hourly)