将字典列表转换为值字典

Question

I have a list of dictionaries for different attributes in the form

attribute = [{user_id1:Value},{user_id2:Value};{User_id3:value3}]

these are my list

a = [{6: 81}, {7: 79}, {8: 67}]
b = [{6: 68}, {7: 77}, {8: 71}]
c = [{6: 71}, {7: 86}, {8: 68}]
d = [{6: 71}, {7: 86}, {8: 68}]
e = [{6: 67}, {7: 59}, {8: 85}]
f = [{6: -72}, {7: -71}, {8: -66}]

What I would need to do is to create a list a value for each user for example:

list_user_id_6 = [81,68,71,71,67,-72]
list_user_id_7 = [79,77,86,86,59,-71]
list_user_id_8 = [67,71,68,68,85,-66]

edit: To explain what I did, I have a survey django app, and I am using Chart.JS to render the results, which have to be in the form : a = [number1,number2,number3]

Now those numbers are a calculation from multiple users values that are part of a team, so I created a method that for each member of a particular team I extract some value make a calculation to give me a score and then I append that score in a list with the member_id in the form {id:score}

each a,b,c,d has a calculation method to create a score

this is an example of one of the method:

def get_chunk_score2(self, format=None, *args, **kwargs):
    current_response_list = get_current_team(self)
    chunk_list = []
    for resp in current_response_list:
        current_response = list(resp.values())[0]

        answer_question1 = current_response.answers.get(question_id = 2)
        answer_question2 = current_response.answers.get(question_id = 3)
        json_answer_question1 = json.loads(answer_question1.body)
        json_answer_question2 = json.loads(answer_question2.body)
        answer_key_question1 = list(json_answer_question1.keys())[0][0]
        answer_key_question2 = list(json_answer_question2.keys())[0][0]
        if answer_key_question1 == "1" or "3":
            score1 = list(json_answer_question1.values())[0]
        else:
            score1 = -list(json_answer_question1.values())[0]

        if answer_key_question2 == "1" or "3":
            score2 = list(json_answer_question2.values())[0]
        else:
            score2 = -list(json_answer_question2.values())[0]

        chunk_score = math.ceil((score1+score2)/2)
        chunk_list.append({current_response.user_id:chunk_score})

    return chunk_list

ps: like you can see I am a few month in the world of coding so any tips is welcome to progress;)

how can I properly do it ? thx you very much

Answer 1

You may want to store your list in a new dict:

a = [{6: 81}, {7: 79}, {8: 67}]
b = [{6: 68}, {7: 77}, {8: 71}]
c = [{6: 71}, {7: 86}, {8: 68}]
d = [{6: 71}, {7: 86}, {8: 68}]
e = [{6: 67}, {7: 59}, {8: 85}]
f = [{6: -72}, {7: -71}, {8: -66}]

res = {}

group = [a,b,c,d,e,f]

for a in group:
    for d in a:
        # If actual key is already in dict, get his list, else create an empty list
        res[d.keys()[0]] = res.get(d.keys()[0], [])
        # append to list the value
        res[d.keys()[0]].append(d.values()[0])

print res
# {8: [67, 71, 68, 68, 85, -66], 6: [81, 68, 71, 71, 67, -72], 7: [79, 77, 86, 86, 59, -71]}

Answer 2

First convert your lists of dicts into a better data structure like a dictionary:

lists = [a, b, c, d, e, f]
dictionaries = [dict(k.items()[0] for k in x) for x in lists]

Now get all of the possible keys:

keys = {y for y in x.keys() for x in dictionaries}

And finally forms the lists for each of the keys:

result = {k: [d[k] for d in dictionaries] for k in keys}

Answer 3

You can try with two methods :

Data is :

a = [{6: 81}, {7: 79}, {8: 67}]
b = [{6: 68}, {7: 77}, {8: 71}]
c = [{6: 71}, {7: 86}, {8: 68}]
d = [{6: 71}, {7: 86}, {8: 68}]
e = [{6: 67}, {7: 59}, {8: 85}]
f = [{6: -72}, {7: -71}, {8: -66}]

group = [a,b,c,d,e,f]

First method:

one is using default dict:

import collections
default=collections.defaultdict(list)
for i in group:
    for dict_s in i:
        for key,value in dict_s.items():
            default["list_user_id_{}".format(key)].append(value)


print(default)

output:

{'list_user_id_8': [67, 71, 68, 68, 85, -66], 'list_user_id_7': [79, 77, 86, 86, 59, -71], 'list_user_id_6': [81, 68, 71, 71, 67, -72]}

Second method:

You can implement your own logic without importing any logic :

track={}
for i in group:
    for dict_s in i:
        for key,value in dict_s.items():
            if "list_user_id_{}".format(key) not in track:
                track["list_user_id_{}".format(key)]=[value]
            else:
                track["list_user_id_{}".format(key)].append(value)

print(track)

output:

{'list_user_id_8': [67, 71, 68, 68, 85, -66], 'list_user_id_7': [79, 77, 86, 86, 59, -71], 'list_user_id_6': [81, 68, 71, 71, 67, -72]}