CODEWITHSUNDEEP
pythonsortingdatetime
Ben
Ben

Reputation: 391

Sort tuples by time interval? Python

How can I sort these tuples by time interval, say every hour? 

 [('172.18.74.146', datetime.time(11, 28, 58)), ('10.227.211.244',
 datetime.time(11, 54, 19)), ('10.227.215.68', datetime.time(11, 54, 34)),
 ('10.227.209.139', datetime.time(12, 14, 47)), ('10.227.147.98',
 datetime.time(14, 47, 25))]

The result should be: 

    [["172.18.74.146, 10.227.211.244, 10.227.215.68", "11-12"], etc...]

I tried to use group by, but doesnt get what I want: 

   for dd in data[1:]:
    ips = dd[1].split(",")
    dates = dd[2].split(",")
    i = 0
    while(i < len(dates)):
        ips[i] = ips[i].strip()
        hour, mins, second = dates[i].strip().split(":")
        dates[i] = datetime.time(int(hour), int(mins), int(second))
        i+=1    
    order = [(k, ', '.join(str(s[0]) for s in v)) for k, v in groupby(sorted(zip(ips, dates), key=operator.itemgetter(1)), lambda x: x[1].hour)]

Upvotes: 2

Views: 1048

Answers (3)

Cyphase
Cyphase

Reputation: 12022

This should work for you:

from __future__ import print_function

import datetime
import itertools


def iter_len(iterable):
    return sum(1 for __ in iterable)


def by_hour(item):  # Hour key
    timestamp = item[1]
    return '{}-{}'.format(timestamp.hour, (timestamp.hour+1) % 24)


def by_half_hour(item):  # Half-hour key
    timestamp = item[1]
    half_hour = timestamp.hour + (0.5 * (timestamp.minute // 30))
    return '{:.1f}-{:.1f}'.format(half_hour, (half_hour+0.5) % 24)


def get_results(data, key):  # Name this more appropriately
    data = sorted(data, key=key)
    for key, grouper in itertools.groupby(data, key):
        yield (key, iter_len(grouper))


data = [
    ('172.18.74.146', datetime.time(11, 28, 58)),
    ('10.227.211.244', datetime.time(11, 54, 19)),
    ('10.227.215.68', datetime.time(11, 54, 34)),
    ('10.227.209.139', datetime.time(12, 14, 47)),
    ('10.227.147.98', datetime.time(14, 47, 25)),
    ]

print('By Hour')
print(list(get_results(data, by_hour)))
print()
print("By Half Hour")
print(list(get_results(data, by_half_hour)))

Output:

$ ./SO_32081251.py 
By Hour
[('11-12', 3), ('12-13', 1), ('14-15', 1)]

By Half Hour
[('11.0-11.5', 1), ('11.5-12.0', 2), ('12.0-12.5', 1), ('14.5-15.0', 1)]

Upvotes: 1

Randy
Randy

Reputation: 14847

In [17]: a = [('172.18.74.146', datetime.time(11, 28, 58)), ('10.227.211.244',
 datetime.time(11, 54, 19)), ('10.227.215.68', datetime.time(11, 54, 34)),
 ('10.227.209.139', datetime.time(12, 14, 47)), ('10.227.147.98',
 datetime.time(14, 47, 25))]

In [18]: [(k, ', '.join(str(s[0]) for s in v)) for k, v in groupby(a, lambda x: x[1].hour)]
Out[18]: 
[(11, '172.18.74.146, 10.227.211.244, 10.227.215.68'),
 (12, '10.227.209.139'),
 (14, '10.227.147.98')]

Upvotes: 3

AlG
AlG

Reputation: 15157

This is almost what you want. Use the hour to group by:

for k,g in itertools.groupby(order, lambda x: x[1].hour):
    print k,list(g)

Results in:

11 [('172.18.74.146', datetime.time(11, 28, 58)), ('10.227.211.244', datetime.time(11, 54, 19)), ('10.227.215.68', datetime.time(11, 54, 34))]
12 [('10.227.209.139', datetime.time(12, 14, 47))]
14 [('10.227.147.98', datetime.time(14, 47, 25))]

Upvotes: 1

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