Python-如何格式化字典但保留相对时间戳？

Question

I have a dictionary like below with the key as a date and the items in the dictionary numerical characters with times. I'd like to consolidate the lines, and not lose the timestamp (it can be rounded a little, but should still represent hour and minutes is possible). I'd like to capture from decimal to decimal but I have no idea how to go about this. The dictionary has been parsed from a text file and is contained in a dict offset_dict.

Here is the dictionary:

offset_dict = {'2019/07/15': ['. 11:47:24',
                             '0 11:47:24 ',
                             '0 11:47:24 ',
                             '0 11:47:24',
                             '1 11:47:24 ', 
                             '. 11:47:47',
                             '0 11:47:47',
                             '0 11:47:47',
                             '0 11:47:47',
                             '4 11:47:48',
                             '. 11:51:46',
                             '0 11:51:47',
                             '0 11:51:47',
                             '0 11:51:48',
                             '3 11:51:48'],
              '2019/07/16': ['. 06:24:52',
                             '1 06:24:53',
                             '0 06:24:53',
                             '2 06:24:56', 
                             '8 06:24:57', 
                             '0 06:24:57',
                             '0 06:24:59',
                             '8 06:24:59',
                             '0 06:24:59',
                             '8 06:25:00',
                             '0 06:25:03',
                             '. 06:25:04',
                             '5 06:25:04',
                             '0 06:25:05',
                             '. 06:34:19',
                             '0 06:34:19',
                             '0 06:34:19',
                             '5 06:34:35']}

Running this:

for key, line in offset_dict.items():
    print(key)
    print(line)

Will output the dict as

2019/07/15
{'. 11:47:24', '3 11:51:48', '1 11:47:24 ', '0 11:47:47', '0 11:47:24 ', '0 11:51:47', '. 11:47:47', '0 11:51:48', '4 11:47:48', '. 11:51:46', '0 11:47:24'}
2019/07/16
{'1 06:24:53', '0 06:25:03', '0 06:24:53', '8 06:24:57', '. 06:34:19', '0 06:25:05', '5 06:34:35', '8 06:24:59', '. 06:24:52', '8 06:25:00', '5 06:25:04', '2 06:24:56', '. 06:25:04', '0 06:34:19', '0 06:24:57', '0 06:24:59'}

The format I'd like to see:

2019/07/15
{'.0001 11:47:00', '.0004 11:47:00', '.0003 11:51:00'}

2019/07/16
{'.1028008080 06:24:00' '.50 06:25:00', '.005 06:34:00'}

Answer 1

Not the cleanest solution but it gets close to the desired output as long as order isn't important and using the timestamps don't need to end with 00 . If order of the output is important I will need to change this slightly.

for date in offset_dict:
    times_for_date = []
    decimal_time = []
    start_time = None

    for entry in offset_dict[date]:
        start, time = entry.split()
        if start == '.':
            if start_time:
                times_for_date.append((''.join(decimal_time), start_time))
                decimal_time = []
            start_time = time
        decimal_time.append(start)
    if decimal_time:
        times_for_date.append((''.join(decimal_time), start_time))

    output_set = set(f"{decimal_time} {start_time}" for decimal_time, start_time in times_for_date)
    print(f"{date} {output_set}")

Ouput:

2019/07/15 {'.0004 11:47:47', '.0001 11:47:24', '.0003 11:51:46'}
2019/07/16 {'.50 06:25:04', '.005 06:34:19', '.1028008080 06:24:52'}

Answer 2

Using itertools.groupby and statistics.mean to compute the average time:

offset_dict = {'2019/07/15': ['. 11:47:24',
                             '0 11:47:24 ',
                             '0 11:47:24 ',
                             '0 11:47:24',
                             '1 11:47:24 ',
                             '. 11:47:47',
                             '0 11:47:47',
                             '0 11:47:47',
                             '0 11:47:47',
                             '4 11:47:48',
                             '. 11:51:46',
                             '0 11:51:47',
                             '0 11:51:47',
                             '0 11:51:48',
                             '3 11:51:48'],
              '2019/07/16': ['. 06:24:52',
                             '1 06:24:53',
                             '0 06:24:53',
                             '2 06:24:56',
                             '8 06:24:57',
                             '0 06:24:57',
                             '0 06:24:59',
                             '8 06:24:59',
                             '0 06:24:59',
                             '8 06:25:00',
                             '0 06:25:03',
                             '. 06:25:04',
                             '5 06:25:04',
                             '0 06:25:05',
                             '. 06:34:19',
                             '0 06:34:19',
                             '0 06:34:19',
                             '5 06:34:35']}

from itertools import groupby
from datetime import datetime
from statistics import mean

out = {}
for k, v in offset_dict.items():
    out[k] = []
    for vv, gg in groupby(v, lambda k, d={'g':0}: (d.update(g=d['g']+1), d['g']) if k.startswith('.') else (None, d['g'])):
        l = [*gg]
        d = [datetime.strptime(i.split()[1], '%H:%M:%S').timestamp() for i in l]
        out[k].append(''.join(i.split()[0] for i in l) + ' ' + datetime.strftime(datetime.fromtimestamp(mean(d)), '%H:%M:00'))

from pprint import pprint
pprint(out)

Prints:

{'2019/07/15': ['.0001 11:47:00', '.0004 11:47:00', '.0003 11:51:00'],
 '2019/07/16': ['.1028008080 06:24:00', '.50 06:25:00', '.005 06:34:00']}