Sort list of long dictionaries

Question

I have a list of two dictionaries in Python

list = [{'id' : 'ewsfsdf', 
           '_source' : 
                      {'timestamp':'2018-08-08T20:56:01', 
                       'bytes':'43534534'}},
         {'id' : 'ewsfsdz', 
           '_source' : 
                      {'timestamp':'2018-08-08T20:57:01', 
                       'bytes':'4354343534'}}]

I would like to sort the list element by the timestamp in the 2nd depth dictionary. I looked up many other examples of python but could not find a suitable solution to order list element by a dictionary element of a dictionary element in a list.

Can you give me a clue?

You should try to not use `list` as a variable name, as it is one of the built-in functions. — James, Jul 10 '18 at 12:20
@James Very true, although `list` isn't actually a function, it's a type (aka class). — PM 2Ring, Jul 10 '18 at 13:25

Rakesh · Answer 1 · 2018-07-10T12:44:32.417

import datetime
l = [{'id' : 'ewsfsdf', 
           '_source' : 
                      {'timestamp':'2018-08-08T20:56:01', 
                       'bytes':'43534534'}},
         {'id' : 'ewsfsdz', 
           '_source' : 
                      {'timestamp':'2018-08-08T20:57:01', 
                       'bytes':'4354343534'}}]

print( sorted(l, key=lambda k: datetime.datetime.strptime(k['_source']["timestamp"], "%Y-%m-%dT%H:%M:%S")) )

or

print(sorted(l, key=lambda k: k['_source']['timestamp']))

Output:

[{'id': 'ewsfsdf', '_source': {'timestamp': '2018-08-08T20:56:01', 'bytes': '43534534'}}, {'id': 'ewsfsdz', '_source': {'timestamp': '2018-08-08T20:57:01', 'bytes': '4354343534'}}]

You can use lambda in key
Use datetime.datetime.strptime(k['_source']["timestamp"], "%Y-%m-%dT%H:%M:%S") to convert string to datetime object

This is a good solution, but it might be cleaner to factor the lambda out into a named function — Edward Minnix, Jul 10 '18 at 12:22
Do we need to convert a generic ISO timestamp to a datetime object? String ordering is enough, isn't it? — FHTMitchell, Jul 10 '18 at 12:23

score 1 · Answer 2 · answered Jul 10 '18 at 12:21

1

You can do:

print(sorted(list, key=lambda d: d['_source']['timestamp']))

Depending on how you want to sort by timestamp, you may need to parse the timestamp to another format that's comparable time-wise and not just lexicographical order.

answered Jul 10 '18 at 12:21

Kent Munthe Caspersen

5,918
1
35
34

Simple string comparison is sufficient on the ISO 8601 timestamps in the OP's data, no conversion is necessary, assuming that they all use the same representation. – PM 2Ring Jul 10 '18 at 12:33
@PM2Ring That is a good observation – Kent Munthe Caspersen Jul 10 '18 at 13:20

score 1 · Answer 3 · answered Jul 10 '18 at 12:23

1

Another solution is to use list method sort and dparser

from dateutil.parser import dparser
l.sort(key=lambda x: dparser.parse(x['_source']['timestamp']))

Notice that this will return nothing and sort your list inplace

answered Jul 10 '18 at 12:23

rafaelc

57,686
15
58
82

1

alternatively `arrow.Arrow` – FHTMitchell Jul 10 '18 at 12:24

Sunitha · Accepted Answer · 2018-07-15T09:30:52.707

Use dateutil package to parse datetime and sort it based on that. You can install dateutil package by doing pip install python-dateutil

from dateutil import parser
sorted(lst, key=lambda x: parser.parse(x['_source']["timestamp"]))
# [{'id': 'ewsfsdf', '_source': {'timestamp': '2018-08-08T20:56:01', 'bytes': '43534534'}}, {'id': 'ewsfsdz', '_source': {'timestamp': '2018-08-08T20:57:01', 'bytes': '4354343534'}}]

score 0 · Answer 5 · answered Jul 10 '18 at 12:27

I'm not sure how you want to sort the timestamp (from earliest to latest or latest to earliest) but here's how you could get into the second depth:

list[0]['_source']['timestamp']

If you want to cycle through the whole list, simply get the length of the list and cycle through a for loop (make sure not to overflow the list by going out of range with the index).

I hope this helps!

Sort list of long dictionaries

5 Answers5