转换字典列表，根据键组合列表项

Question

Given a list of dictionaries like:

history = [
  {
    "actions": [{"action": "baz", "people": ["a"]}, {"action": "qux", "people": ["d", "e"]}],
    "events": ["foo"]
  },
  {
    "actions": [{"action": "baz", "people": ["a", "b", "c"]}],
    "events": ["foo", "bar"]
  },
]

What is the most efficient (whilst still readable) way to get a list of dicts, where each dict is an unique event and the list of actions for that event have been merged based on the action key. For example, for the above list, the desired output is:

output = [
    {
      "event": "foo", 
      "actions": [
        {"action": "baz", "people": ["a", "b", "c"]}, 
        {"action": "qux", "people": ["d", "e"]}
      ]
    },
    {
      "event": "bar", 
      "actions": [
        {"action": "baz", "people": ["a", "b", "c"]}
      ]
    },
]

I can't change the output structure as it's being consumed by something external. I've wrote the following code which works but is very verbose and has poor readability.

from collections import defaultdict

def transform(history):
    d = defaultdict(list)
    for item in history:
        for event in item["events"]:
            d[event] = d[event] + item["actions"]
    transformed = []
    for event, actions in d.items():
        merged_actions = {}
        for action in actions:
            name = action["action"]
            if merged_actions.get(name):
                merged_actions[name]["people"] = list(set(action["people"]) | set(merged_actions[name]["people"]))
            else:
                merged_actions[name] = {
                    "action": action["action"],
                    "people": action["people"]
                }
        transformed.append({
            "event": event,
            "actions": list(merged_actions.values())
        })
    return transformed

I'm only targeting python3.6+

Answer 1

You can use collections.defaultdict with itertools.groupby :

from collections import defaultdict
from itertools import groupby as gb
d = defaultdict(list)
for i in history:
  for b in i['events']:
    d[b].extend(i['actions'])

new_d = {a:[(j, list(k)) for j, k in gb(sorted(b, key=lambda x:x['action']), key=lambda x:x['action'])] for a, b in d.items()}
result = [{'event':a, 'actions':[{'action':c, 'people':list(set([i for k in b for i in k['people']]))} for c, b in d]} for a, d in new_d.items()]

Output:

[
 {'event': 'foo', 
  'actions': [
     {'action': 'baz', 'people': ['b', 'a', 'c']}, 
     {'action': 'qux', 'people': ['d', 'e']}
    ]
  }, 
 {'event': 'bar', 
   'actions': [{'action': 'baz', 'people': ['b', 'a', 'c']}]
  }
 ]

Answer 2

It is not a less verbose answer, but maybe a bit better readable. Also it does not depend on anything else and is just standard python.

tmp_dict = {}
for d in history:
    for event in d["events"]:
        if event not in tmp_dict:
            tmp_dict[event] = {}
            for actions in d["actions"]:
                tmp_dict[event][actions["action"]] = actions["people"]
        else:
            for actions in d["actions"]:
                if actions["action"] in tmp_dict[event]:
                    tmp_dict[event][actions["action"]].extend(actions["people"])
                else:
                    tmp_dict[event][actions["action"]] = actions["people"]

output = [{"event": event, "actions": [{"action": ac, "people": list(set(peop))} for ac, peop in tmp_dict[event].items()]} for event in tmp_dict]

print (output)

Output:

[
   {'event': 'foo',
    'actions': [
                {'action': 'qux', 'people': ['e', 'd']},
                {'action': 'baz', 'people': ['a', 'c', 'b']}
               ]
   },
   {'event': 'bar',
    'actions': [{'action': 'baz', 'people': ['a', 'c', 'b']}]
   }
]