Sorting dictionary by value by datetime in python-CodePudding

How do I sort a dictionary by value in datetime in shape of this:

{'user_0': [['item_805696', '2021-02-11 13:03:42'],
  ['item_386903', '2021-02-11 13:03:52'],
  ['item_3832', '2021-02-11 13:04:07'],
  ['item_849824', '2021-02-11 13:05:04'],
'user_1': [['item_97057', '2021-02-11 13:03:42'],
  ['item_644971', '2021-02-11 13:09:32'],
  ['item_947129', '2021-02-11 13:15:27'],
  ['item_58840', '2021-02-11 13:16:11'],
  ['item_640213', '2021-02-11 13:17:40'],
...

Im trying to sort values by datetime of second value in values of the dictionary

CodePudding user response：

You can follow this answer to convert the string representing the date to a number of seconds since the Unix epoch (1st January 1970): How to convert python timestamp string to epoch?

Then use can use the key argument of the sorted built-in method to use it to sort the dictionary how you want.

CodePudding user response：

You can give a key to tell list.sort or sorted how to sort.

A key to sort by the first value is lambda x: x[0]; a key to sort by the second value is lambda x: x[1].

d = {'user_0': [['item_805696', '2021-02-11 13:03:42'],
                ['item_849824', '2021-02-11 13:05:04'],
                ['item_386903', '2021-02-11 13:03:52'],
                ['item_3832',   '2021-02-11 13:04:07']],
     'user_1': [['item_58840',  '2021-02-11 13:16:11'],
                ['item_947129', '2021-02-11 13:15:27'],
                ['item_97057',  '2021-02-11 13:03:42'],
                ['item_640213', '2021-02-11 13:17:40'],
                ['item_644971', '2021-02-11 13:09:32']]}

for v in d.values():
    v.sort(key=lambda x: x[1])

print(d)
# {'user_0': [['item_805696', '2021-02-11 13:03:42'],
#             ['item_386903', '2021-02-11 13:03:52'],
#             ['item_3832',   '2021-02-11 13:04:07'],
#             ['item_849824', '2021-02-11 13:05:04']],
#  'user_1': [['item_97057',  '2021-02-11 13:03:42'],
#             ['item_644971', '2021-02-11 13:09:32'],
#             ['item_947129', '2021-02-11 13:15:27'],
#             ['item_58840',  '2021-02-11 13:16:11'],
#             ['item_640213', '2021-02-11 13:17:40']]}

Note that this only works because the timestamps are presented in the very practical format 'yyyy-mm-dd HH:MM:SS', and strings are sorted lexicographically, i.e., from left to right. If the timestamps had been in a less friendly format, such as 'dd-mm-yyyy HH:MM:SS', then you'd need to use the key to parse the timestamps.

CodePudding user response：

You can also use the datetime module in python standard librairy and here is the code

from datetime import datetime

d = {'user_0': [['item_805696', '2021-02-11 13:03:42'],
  ['item_386903', '2021-02-11 13:03:52'],
  ['item_3832', '2021-02-11 13:04:07'],
  ['item_849824', '2021-02-11 13:05:04']],
'user_1': [['item_97057', '2021-02-11 13:03:42'],
  ['item_644971', '2021-02-11 13:09:32'],
  ['item_947129', '2021-02-11 13:15:27'],
  ['item_58840', '2021-02-11 13:16:11'],
  ['item_640213', '2021-02-11 13:17:40']]}

def sort_by_datetime(val):
    return datetime.strptime(val[1], '%Y-%m-%d %H:%M:%S')

sorted_dict = {k: sorted(v, key=sort_by_datetime) for k, v in d.items()}