Question 109723 in Json Question posted in
Our archive of expertly curated questions and answers provides insights and solutions to common problems related to this popular data interchange format. From parsing and manipulating JSON data to integrating it with various programming languages and web services, our archive has got you covered. Start exploring today and take your JSON skills to the next level

Json – how can match two list and remove duplicate based on condition on dictionary in list

I want to merge two lists and get data matched without duplicates and to alias them to new structure
I have two lists

here is a given two list try to merge

cats = [
    'orange', 'apple', 'banana'
]

and second list

types = [
    {
        "id": 1,
        "type": "orange"
    },
    {
        "id": 2,
        "type": "apple"
    },
    {
        "id": 3,
        "type": "apple"
    },
    {
        "id": 4,
        "type": "orange"
    },
    {
        "id": 5,
        "type": "banana"
    }
]

and I want to combine them to get this result:

[
    {'orange': {
        'UNIT': [1, 4]
    }
    },
    {'apple': {
        'UNIT': [2, 3]
    }
    },
    {'banana': {
        'UNIT': [5]
    }
    }
]

and my code, this after my tries i get this result :

for item in types:
    for cat in cats:
        if item['type'] == cat:
     
            matched.append(
                {
                    cat: {
                        "UNIT": [i['id'] for i in types if
                                 'id' in i]
                    }
                }
            )

and my result is like this


[{'orange': {'UNIT': [1, 2, 3, 4, 5]}},
 {'apple': {'UNIT': [1, 2, 3, 4, 5]}},
 {'apple': {'UNIT': [1, 2, 3, 4, 5]}},
 {'orange': {'UNIT': [1, 2, 3, 4, 5]}},
 {'banana': {'UNIT': [1, 2, 3, 4, 5]}}]
Tags:

3

Answers


  1. 0

    With list comprehension:

    [{cat:{'UNIT':[type_['id'] for type_ in types if type_['type'] == cat]}} for cat in cats]
    Login or Signup to reply.
  2. 0

    Your problem is the in inside your list comprehension – beside that your code is complex. You get multiples due to your for loops and never checking if that fruit was already added to matched.

    To reduce the 2 lists to the needed values you can use

    cats = [    'orange', 'apple', 'banana']

types = [    { "id": 1, "type": "orange" },
    { "id": 2, "type": "apple" },
    { "id": 3, "type": "apple" },
    { "id": 4, "type": "orange" },
    { "id": 5, "type": "banana" }]

rv = {}
for inner in types:
    t = inner["type"]
    if t not in cats: # for current data not needed, only needed
        continue      # if some listelements don't occure in dict
    rv.setdefault(t, [])
    rv[t].append(inner["id"])
        
print(rv)

    which leads to an easier dictionary with all the data you need:

    {'orange': [1, 4], 'apple': [2, 3], 'banana': [5]}

    From there you can build up your overly complex list of dicts with 1 key each:

    
lv = [{k:{"UNIT":v}} for k,v in rv.items()]
print (lv)

    to get

    [{'orange': {'UNIT': [1, 4]}}, 
 {'apple': {'UNIT': [2, 3]}}, 
 {'banana': {'UNIT': [5]}}]

    Answer to extended problem from comment:
    If you need to add more things you need to capture you can leverage the fact that lists and dicts store by ref:

    cats = [    'orange', 'apple', 'banana']
types = [    { "id": 1, "type": "orange" , "bouncyness": 42 },
             { "id": 2, "type": "apple"  , "bouncyness": 21 },    
             {"id": 3, "type": "apple"  , "bouncyness": 63},    
             { "id": 4, "type": "orange"  , "bouncyness": 84},    
             { "id": 5, "type": "banana"  , "bouncyness": 99}]

rv = []   # list of dicts - single key is "fruitname"
pil = {}  # dict that keeps track which fruit is on what position in rv
          # to avoid iterating over list to find correct fruit-dict

di = None # the current fruits dictionary 

for inner in types:
    t = inner["type"]
    if t not in cats: # for current data not needed, only needed
        continue      # if some listelements don't occure in dict

    # step1: find correct fruit dict in rv or create new one and add it
    di = None
    if t in pil:
        # get cached dict by position of fruit in rv
        di = rv[pil[t]]
    else:
        # create fruit dict, cache position in rv in pil
        di = {}
        rv.append(di)
        pil[t] = len(rv)-1

    # step2: create all the needed inner lists         
    #   you can speed this up using defaultdict(list) if speed gets
    #   problematic - until then dict.setdefault should be fine
    di.setdefault(t, [])
    di.setdefault("bouncyness", [])

    # step3: fill with values
    di[t].append(inner["id"])
    di["bouncyness"].append(inner["bouncyness"])
        
print(rv)

    to get

    [{'orange': [1, 4], 'bouncyness': [42, 84]}, 
 {'apple': [2, 3], 'bouncyness': [21, 63]}, 
 {'banana': [5], 'bouncyness': [99]}]
    Login or Signup to reply.
  3. 0

    Here is an alternative approach using filter() method and lambda.

    dict_lst = []
for cat in cats:
    cat_items = filter(lambda x: x['type'] == cat, types)
    cat_dict = {cat: {'UNIT': [x['id'] for x in cat_items]}} 
    dict_lst.append(cat_dict)
print(dict_lst)

    [{'orange': {'UNIT': [1, 4]}},
 {'apple': {'UNIT': [2, 3]}},
 {'banana': {'UNIT': [5]}}]
    Login or Signup to reply.
Please signup or login to give your own answer.

Back To Top
Search