преобразовать 2-й список в словарь

Question

Я хочу создать словарь из трех 2-х списков. Например:

list1 = [['Brand', 'Release Date', 'Related'], ['Structure', 'Screen', 'Photos', 'Videos']]

list2 = [['Brand', 'Aliases'], ['Release date', 'State'], ['Predecessor', 'Successors'], ['Size', 'Aspect Ratio', 'Weight', 'Usable surface', 'Materials', 'Colors', 'Origami']]

list3 = [['OppoSmartphones by Oppo', 'PAFM00, Oppo Find X Standard Edition'], ['June 2018, 10 months ago', 'On Sale'], ['Oppo Find 7', 'Oppo Reno'], ['74.2 mm x 156.7 mm x 9.4 mm', '19:9', '186 g', '87 %', 'Glass', 'Turquoise Violet', 'Print 3D Model']]

и я хочу получить результат как

{'Brand':{'Brand': 'OppoSmartphones by Oppo', 'Aliases': 'PAFM00, Oppo Find X Standard Edition'}, 'Release Date':{ 'Release date': 'June 2018, 10 months ago', 'State': 'On Sale'}, 'Related':{'Predecessor': 'Oppo Find 7', 'Successors': 'Oppo Reno'}}

Answer 1

Ну, вы можете использовать циклы, dict() и zip() для достижения этого, как показано ниже:

list1 = [['Brand', 'Release Date', 'Related'], 
         ['Structure', 'Screen', 'Photos', 'Videos']]

list2 = [['Brand', 'Aliases'], ['Release date', 'State'],
         ['Predecessor', 'Successors'],
         ['Size', 'Aspect Ratio', 'Weight', 'Usable surface', 'Materials', 'Colors', 'Origami']]

list3 = [['OppoSmartphones by Oppo', 'PAFM00, Oppo Find X Standard Edition'],
         ['June 2018, 10 months ago', 'On Sale'],
         ['Oppo Find 7', 'Oppo Reno'],
         ['74.2 mm x 156.7 mm x 9.4 mm', '19:9', '186 g', '87 %', 'Glass', 'Turquoise Violet', 'Print 3D Model']]

d1 = {}
for i in range(len(list1)):
    for j in range(len(list1[i])):
        #print(i,j)
        d1[list1[i][j]] = dict(zip(list2[j], list3[j]))


# to get this patter: dict = {list1[0][0]:{list2[0][0]:list3[0][0], list2[0][1]:list3[0][1]},
#                             list1[0][1]:{list2[1][0]:list3[1][0], list2[1][1]:list3[1][1]},
#                             list1[0][2]:{list2[2][0]:list3[2][0], list2[2][1]:list3[2][1]}}
d2 = {}
for i in range(len(list1[0])):
    if list1[0][i] not in d2.keys(): d2[list1[0][i]] = {}
    if list1[0][i] not in d2.keys(): d2[list1[0][i]] = {}

    d2[list1[0][i]][list2[i][0]] = list3[i][0] 
    d2[list1[0][i]][list2[i][1]] = list3[i][1] 


# if you want to apply the behaviour on all elements:
d3 = {}
for k in range(len(list1)): 
    for i in range(len(list1[k])):
        if list1[k][i] not in d3.keys(): d3[list1[k][i]] = {}
        if list1[k][i] not in d3.keys(): d3[list1[k][i]] = {}

        d3[list1[k][i]][list2[i][0]] = list3[i][0] 
        d3[list1[k][i]][list2[i][1]] = list3[i][1] 

Выход (d2):

{      'Brand': {'Aliases': 'PAFM00, Oppo Find X Standard Edition',
                 'Brand': 'OppoSmartphones by Oppo'},

     'Related': {'Predecessor': 'Oppo Find 7', 
                  'Successors': 'Oppo Reno'},

'Release Date': {'Release date': 'June 2018, 10 months ago',
                        'State': 'On Sale'}}

Answer 2

Попробуйте:

res = {}

for i in range(3):
    res[list1[0][i]] = {list2[i][0]:list3[i][0], list2[i][1]:list3[i][1]}