How to get the earliest date from list of dictionaries?

Question

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.

So your question is how to remove the time from the date/time string? (I assume you know how to access the `'taps'` key and how to use the `min` function on the resulting list) — mkrieger1, Oct 19 '22 at 21:10
You have a pair of subproblems: (A.) how to extract y-m-d's from datastructure, and (B.) how to compute the min(ymd). But you didn't show us any code. Which subproblem does your current code fail to properly address? https://stackoverflow.com/help/minimal-reproducible-example — J_H, Oct 19 '22 at 21:13
There's a typo in your dictionary - what have you tried yourself? — Grismar, Oct 19 '22 at 21:19

ErnestBidouille · Answer 1 · 2022-10-19T21:38:43.210

1

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

edited Oct 19 '22 at 21:38

answered Oct 19 '22 at 21:14

ErnestBidouille

1,071
4
16

Note that this works because the dates happen to be formatted in ISO standard format, which has the date written so that it sorts alphanumerically. For other date formats, a conversion to a datetime format would be required as part of the key function. – Grismar Oct 19 '22 at 21:17
Answer edited fir `datetime` formats – ErnestBidouille Oct 19 '22 at 21:24
Are you sure ? @RahulKP because `min('2022-06-05', '2022-10-05') == '2022-06-05'` – ErnestBidouille Oct 19 '22 at 21:27
@ErnestBidouille Yeah you are right, since all digit prefixes with `0` it seems not to make any issue. I am not sure though. – Rahul K P Oct 19 '22 at 21:31
Yeah, only working with strings in ISO format, I agree with you on this point. I really followed his example for my answer – ErnestBidouille Oct 19 '22 at 21:32
@ErnestBidouille If the format is different it will make issues, Refer this, https://stackoverflow.com/questions/20365854/comparing-two-date-strings-in-python – Rahul K P Oct 19 '22 at 21:34
1

@RahulKP Of course, I refer for my answer only to his not very detailed example... – ErnestBidouille Oct 19 '22 at 21:36

score 1 · Answer 2 · answered Oct 19 '22 at 21:18

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.

How to get the earliest date from list of dictionaries?

2 Answers2