CODEWITHSUNDEEP
pythonsortingdictionary
sagar
sagar

Reputation: 1435

How to sort a list of dictionaries in python?

Input data:

results= [
        {
      "timestamp_datetime": "2014-03-31 18:10:00 UTC",
      "job_id": 5,
      "processor_utilization_percentage": 72
    },
        {
      "timestamp_datetime": "2014-03-31 18:20:00 UTC",
      "job_id": 2,
      "processor_utilization_percentage": 60
    },
        {
      "timestamp_datetime": "2014-03-30 18:20:00 UTC",
      "job_id": 2,
      "processor_utilization_percentage": 0
    }]

Output has to be sorted like below, grouping by job_id in ascending order:

newresult = {
    '2':[{ "timestamp_datetime": "2014-03-31 18:20:00 UTC",
            "processor_utilization_percentage": 60},

          {"timestamp_datetime": "2014-03-30 18:20:00 UTC",
          "processor_utilization_percentage": 0},]

    '5':[{
          "timestamp_datetime": "2014-03-31 18:10:00 UTC",
          "processor_utilization_percentage": 72},
        ],
    }

What is pythonic way to do this?

Upvotes: 1

Views: 86

Answers (3)

tobias_k
tobias_k

Reputation: 82949

You can use itertools.groupby to group the results by their job_id:

from itertools import groupby
new_results = {k: list(g) for k, g in groupby(results, key=lambda d: d["job_id"])}

The result is a dictionary, i.e. it has no particular order. If you want to iterate the values in ascending order, you can just do something like this:

for key in sorted(new_results):
    entries = new_results[key]
    # do something with entries

Update: as Martijn points out, this requires the results list to be sorted by the job_ids (as it is in your example), otherwise entries might be lost.

Upvotes: 3

davef
davef

Reputation: 91

Assuming you really didn't want the the job_id in the newresult:

from collections import defaultdict
newresult = defaultdict(list)
for result in results:
    job_id = result['job_id']
    newresult[job_id].append( 
        {'timestamp_datetime':result['timestamp_datetime'],
         'processor_utilization_percentage':result['processor_utilization_percentage']}
        )
#print newresult

I don't really see a way to do this with a dictionary comprehension, but I'm sure there's someone out there with more experience in doing that sort of thing who could pull it off. This is pretty straightforward, though.

Upvotes: 0

Martijn Pieters
Martijn Pieters

Reputation: 1124858

You are grouping; this is easiest with a collections.defaultdict() object:

from collections import defaultdict

newresult = defaultdict(list)

for entry in result:
    job_id = entry.pop('job_id')
    newresult[job_id].append(entry)

newresult is a dictionary and these are not ordered; if you need to access job ids in ascending order, sort the keys as you list them:

for job_id in sorted(newresult):
    # loops over the job ids in ascending order.
    for job in newresult[job_id]:
        # entries per job id

Upvotes: 5

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