如何从另一个字典值制作字典

Question

I have a list of dictionaries, something like this

input = [
    {"0_question": "how are you"}, 
    {"0_answer": "good"},
    {"4_question": "what time is it"},
    {"4_answer": "morning"},
]

and I want this:

expected_result = {
    0: {"how are you": "good"},
    4: {"what time is it" : "morning"},
}

this is what I tried:

x = {}
l= []
for i, d in enumerate(a):
    for k, v in d.items():
        l.append(v)
    x[l[i]] = l[i+1]

Answer 1

Split the original dict key into numeric and type part:

a = [
    {
        "0_question": "how are you"
    }, {
        "0_answer": "good"
    }, {
        "4_question": "what time is it"
    }, {
        "4_answer": "morning"
    }
]

dic = {}

for item in a:
    value = list(item.values())[0]
    num, itemType = list(item.keys())[0].split('_')
    num = int(num)

    if itemType == 'question':
        dic[num] = {value: ''}
    else:
        key = list(dic[num].keys())[0]
        dic[num][key] = value
print(dic)

Out:

{0: {'how are you': 'good'}, 4: {'what time is it': 'morning'}}

Answer 2

You can use a nested loop and the modulo operator ( % ). Use regex to find the question/answer number:

import re

l = [
    {"0_question": "how are you"},
    {"0_answer": "good"},
    {"4_question": "what time is it"},
    {"4_answer": "morning"},
    {"157_question": "which fruit do you want"},
    {"157_answer": "apple"}
]
new_d = {}
for i, d in enumerate(l):
    for k, v in d.items():
        result = re.search("\d+", k).group(0)
        num = int(result)

        if ((i + 1) % 2) == 0:
            for new_k in new_d[num].keys():
                new_d[num][new_k] = v
        else:
            new_d[num] = {v: ""}

print(new_d)

Output

{0: {'how are you': 'good'}, 4: {'what time is it': 'morning'}, 157: {'which fruit do you want': 'apple'}}

Answer 3

Use This Code

a = [{"0_question": "how are you"},
     {"0_answer": "good"},
     {"4_question": "what time is it"},
     {"4_answer": "morning"}]


dic = {}

for i in range(len(a) - 1):
    if i % 2 == 0:
        dic[list(a[i].keys())[0]] = {list(a[i].values())[0]: list(a[i + 1].values())[0]}

print(dic)

Answer 4

This has fewer lines and hopefully increases readability. The downside is it uses more resources by restructuring the initial data.

sample_data = [{"0_question" : "how are you"},
{"0_answer": "good"},
{"4_question" : "what time is it"},
{"4_answer": "morning"}]

restructured_data = {k: v for data in sample_data for k, v in data.items()}
nums = list(set([int(k.split("_")[0]) for k, v in restructured_data.items()]))
new_dict = {num: {restructured_data[f'{num}_question']: restructured_data[f'{num}_answer']} for num in nums}

print(new_dict)

output:

{0: {'how are you': 'good'}, 4: {'what time is it': 'morning'}}

Answer 5

data = [{"0_question" : "how are you"}, 
        {"0_answer": "good"},
        {"4_question" : "what time is it"},
        {"4_answer": "morning"}]
        
# unpack dict values into any iterable data structure
# list in my case
unpacked_data = list(map(lambda dict_: list(dict_.items()), data))

# dict_items is not subscriptable, so unpack it again
# also exctracting number from key.
processed_data = list(
                        map(
                            lambda values_tuple: (values_tuple[0][0][0], values_tuple[0][1]),
                            unpacked_data
                            )
                    )

result_dict = {}

# according to input data we assuame that there is a pairs of (question: answer).
# using zip and slicing for pair iteration
for question, answer in zip(processed_data[0::2], processed_data[1::2]):
    result_dict[question[0]] = {question[1] : answer[1]}
    
print(result_dict)

UPD: this may be not the most optimal solution, but i hope that is pretty clear for understanding

Answer 6

Check Below Code:

{ int(list(i1)[0][0]) : {list(i1.values())[0] : list(i2.values())[0] } for i1, i2 in zip(input[::2], input[1::2])}

Output: