How to get the earliest date from list of dictionaries?-CodePudding

I have a list of dictionaries and would like to get the earliest date value (and date only!) from the list. See example below:

{'taps': [{'duration': 0,
   'datetime': '2022-06-05T09:35:56.131498'},
  {'duratin': 518,
   'datetime': '2022-06-05T09:35:56.649846',
  {'duration': 500,
   'datetime': '2022-06-06T09:35:57.150820'}]}

From the example above, I want to get 2022-06-05.

CodePudding user response：

You can do with dateutil.parser

import dateutil.parser

date_list = sorted([dateutil.parser.isoparse(item['datetime']) for item in d['taps']])
print(date_list[0].strftime("%Y-%m-%d")) # Convert back to string
'2022-06-05'

Conver the into string format datetime to datetime format. Since it's in ISO standard you can use dateutil.parser and sort the list and take the first element.

CodePudding user response：

In your example, datetimes are strings only, so this will print the earliest date: (working if strings are all in the format : YYYY-MM-DD...)

items = {
    'taps': [{
        'duration': 0,
        'datetime': '2022-06-05T09:35:56.131498'
    }, {
        'duratin': 518,
        'datetime': '2022-06-05T09:35:56.649846'
    }, {
        'duration': 500,
        'datetime': '2022-06-06T09:35:57.150820'
    }]
}

min(items['taps'],key=lambda x:x['datetime'])['datetime'][:10] # 2022-06-05

If you are working with datetime objects, use this:

items = {
    'taps': [{
        'duration': 0,
        'datetime': datetime(2022, 6, 5, 9, 35, 56, 131498)
    }, {
        'duratin': 518,
        'datetime': datetime(2022, 6, 5, 9, 35, 56, 649846)
    }, {
        'duration': 500,
        'datetime': datetime(2022, 6, 6, 9, 35, 57, 150820)
    }]
}

min(items['taps'], key=lambda x: x['datetime'])['datetime'].strftime('%Y-%m-%d') # 2022-06-05