Pythonic way to get the difference in timestamps in the same dict from one value to another-CodePudding

Is there any pythonic way of getting the values minus from the same dictionary.

I have a dictionary

{'ZLER3B8I': {1: datetime.datetime(2022, 6, 30, 7, 32, 34, 586000, tzinfo=datetime.timezone.utc),
     3: datetime.datetime(2022, 7, 1, 6, 4, 10, 446000, tzinfo=datetime.timezone.utc),
     5: datetime.datetime(2022, 7, 1, 6, 4, 18, 596000, tzinfo=datetime.timezone.utc),
     6: datetime.datetime(2022, 7, 1, 11, 8, 7, 151000, tzinfo=datetime.timezone.utc),
     7: datetime.datetime(2022, 7, 1, 13, 16, 44, 104000, tzinfo=datetime.timezone.utc),
    11: datetime.datetime(2022, 7, 1, 6, 41, 45, 749000, tzinfo=datetime.timezone.utc),
    12: datetime.datetime(2022, 7, 1, 10, 19, 55, 470000, tzinfo=datetime.timezone.utc)}}

Is an efficient pythonic way of calculating the timestamp from the 1st key, value to the all the other key values ? Without many nested for loops.

like from 1 key to 3 key time taken is 28 hrs, 30 mins

{(1-3): 28 Hrs 30 mins, (1-5): 30 hrs 2 mins,.. so on  }

CodePudding user response：

data = {'ZLER3B8I': {1: datetime.datetime(2022, 6, 30, 7, 32, 34, 586000, tzinfo=datetime.timezone.utc),
     3: datetime.datetime(2022, 7, 1, 6, 4, 10, 446000, tzinfo=datetime.timezone.utc),
     5: datetime.datetime(2022, 7, 1, 6, 4, 18, 596000, tzinfo=datetime.timezone.utc),
     6: datetime.datetime(2022, 7, 1, 11, 8, 7, 151000, tzinfo=datetime.timezone.utc),
     7: datetime.datetime(2022, 7, 1, 13, 16, 44, 104000, tzinfo=datetime.timezone.utc),
    11: datetime.datetime(2022, 7, 1, 6, 41, 45, 749000, tzinfo=datetime.timezone.utc),
    12: datetime.datetime(2022, 7, 1, 10, 19, 55, 470000, tzinfo=datetime.timezone.utc)}}

out = {i:{f'1-{j}':y-x[1] for j, y in x.items() if j != 1} for i, x in data.items()}
print(out)

Output:

{'ZLER3B8I': {'1-3': datetime.timedelta(seconds=81095, microseconds=860000),
              '1-5': datetime.timedelta(seconds=81104, microseconds=10000),
              '1-6': datetime.timedelta(days=1, seconds=12932, microseconds=565000),
              '1-7': datetime.timedelta(days=1, seconds=20649, microseconds=518000),
              '1-11': datetime.timedelta(seconds=83351, microseconds=163000),
              '1-12': datetime.timedelta(days=1, seconds=10040, microseconds=884000)}}

{i:{f'1-{j}':f'{y-x[1]}' for j, y in x.items() if j != 1} for i, x in data.items()}
{'ZLER3B8I': {'1-3': '22:31:35.860000',
              '1-5': '22:31:44.010000',
              '1-6': '1 day, 3:35:32.565000',
              '1-7': '1 day, 5:44:09.518000',
              '1-11': '23:09:11.163000',
              '1-12': '1 day, 2:47:20.884000'}}

CodePudding user response：

Sounds like you're looking for itertools.combinations:

d = {
    1: datetime.datetime(2022, 6, 30, 7, 32, 34, 586000, tzinfo=datetime.timezone.utc),
    3: datetime.datetime(2022, 7, 1, 6, 4, 10, 446000, tzinfo=datetime.timezone.utc),
    5: datetime.datetime(2022, 7, 1, 6, 4, 18, 596000, tzinfo=datetime.timezone.utc),
    6: datetime.datetime(2022, 7, 1, 11, 8, 7, 151000, tzinfo=datetime.timezone.utc),
    7: datetime.datetime(2022, 7, 1, 13, 16, 44, 104000, tzinfo=datetime.timezone.utc),
    11: datetime.datetime(2022, 7, 1, 6, 41, 45, 749000, tzinfo=datetime.timezone.utc),
    12: datetime.datetime(2022, 7, 1, 10, 19, 55, 470000, tzinfo=datetime.timezone.utc),
}

for i, j in itertools.combinations(d.keys(), 2):
    print(f'{i=} - {j=} = {d[i] - d[j]}')

Sample execution:

i=1 - j=3 = -1 day, 1:28:24.140000
i=1 - j=5 = -1 day, 1:28:15.990000
i=1 - j=6 = -2 days, 20:24:27.435000
i=1 - j=7 = -2 days, 18:15:50.482000
i=1 - j=11 = -1 day, 0:50:48.837000
i=1 - j=12 = -2 days, 21:12:39.116000
i=3 - j=5 = -1 day, 23:59:51.850000
i=3 - j=6 = -1 day, 18:56:03.295000
i=3 - j=7 = -1 day, 16:47:26.342000
i=3 - j=11 = -1 day, 23:22:24.697000
i=3 - j=12 = -1 day, 19:44:14.976000
i=5 - j=6 = -1 day, 18:56:11.445000
i=5 - j=7 = -1 day, 16:47:34.492000
i=5 - j=11 = -1 day, 23:22:32.847000
i=5 - j=12 = -1 day, 19:44:23.126000
i=6 - j=7 = -1 day, 21:51:23.047000
i=6 - j=11 = 4:26:21.402000
i=6 - j=12 = 0:48:11.681000
i=7 - j=11 = 6:34:58.355000
i=7 - j=12 = 2:56:48.634000
i=11 - j=12 = -1 day, 20:21:50.279000

There's also permutations if you want to compare all possible pairings (i.e., 1 - 3 and 3 - 1 are distinct, not just 1 - 3).