Question

我有一个表单

中不同属性的词典列表

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

这些是我的清单

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}]

我需要做的是为每个用户创建一个列表，例如：

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]

编辑：为了解释我做了什么，我有一个调查django应用程序，我使用Chart.JS来呈现结果，必须采用以下形式： a = [number1，number2，number3]

现在这些数字是来自作为团队一部分的多个用户值的计算，所以我创建了一个方法，对于特定团队的每个成员，我提取一些值进行计算以给我一个分数然后我追加使用{id：score}

形式的member_id在列表中得分

每个a，b，c，d都有一个计算方法来创建一个分数

这是方法之一的一个例子：

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：就像你可以看到我在编码领域的几个月，所以任何提示都欢迎进步;）

我该怎么做呢？非常感谢你

Answer 1

您可能希望将列表存储在新的词典中：

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

首先将您的dicts列表转换为更好的数据结构，如字典：

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

现在获取所有可能的密钥：

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

最后形成每个键的列表：

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

Answer 3

您可以尝试两种方法：

数据是：

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]

第一种方法：

一个是使用默认字典：

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)

输出：

{'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]}

第二种方法：

您可以在不导入任何逻辑的情况下实现自己的逻辑：

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)

输出：

{'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]}

将字典列表转换为值的字典

3 个答案: