將三個列表合併爲一個字典

Question

我需要將三個列表合併到一個字典中。這些列表是從閱讀我格式化的txt文件，並在這裏是從該文件中的一個片段：

maker =['Horsey', 'Ford', 'Overland', 'Scripps-Booth'] 

year = ['1899', '1909', '1911', '1913'] 

model = ['Horseless', 'Model T', 'OctoAuto', 'Bi-Autogo'] 

爲以下：

car_dict = {'Horsey':1899,'Horseless','Ford':1909,'Model T','Overland' : 1911, 'OctoAuto', 'Scripps-Booth' : 1913, 'Bi-Autogo'} 

這裏是我做過什麼：

def car_data_merge(car_maker,car_model,car_year): 
    car_dict = {} 
    car_merge = [] 

    car_dict = defaultdict(partial(defaultdict,list)) 

    for (car_maker,car_model,car_year) in zip(car_maker,car_model,car_year): 
     car_dict[car_year][car_model].append(car_maker) 
    print(car_dict) 

當我輸入這個時，我得到：

{'Horsey': defaultdict(<class 'list'>, {'1899': ['Horseless']}) 

並非來自列表的所有數據都顯示，我不希望顯示defaultdict。

當我嘗試了以下內容：數據

def car_data_merge(car_maker,car_data): 
    car_dict = {} 
    car_merge = [] 
    car_merge = zip(car_maker,car_data) 
    car_dict = dict(car_merge) 
    print(car_dict) 

    ### car_data holds both year and model #### 

只有部分顯示出來：

'Horsey':'Horseless',':1909,'Model T 

我該怎麼辦？

Answer 1

你在正確的軌道上與zip，bu牛逼提防：

The returned list is truncated in length to the length of the shortest argument sequence.

如果你罰款，你可以壓縮你的數據到元組的列表，拉鍊的鑰匙，和手都斷到dict()。

如果你想處理缺失值，結賬itertoolsizip_longest (Python 2)或zip_longest (Python 3)其中

If the iterables are of uneven length, missing values are filled-in with fillvalue.

try: 
    # Python 2 
    from itertools import izip_longest 
    zip_longest = izip_longest 
except ImportError: 
    # Python 3 
    from itertools import zip_longest 

from pprint import pprint 


def main(): 
    maker =['Horsey', 'Ford', 'Overland', 'Scripps-Booth', 'FutureX', 'FutureY'] 
    year = ['1899', '1909', '1911', '1913', '20xx'] 
    model = ['Horseless', 'Model T', 'OctoAuto', 'Bi-Autogo'] 

    car_data = dict(zip(maker, zip(year, model))) 
    car_data_longest = {mk: (yr, md) for mk, yr, md in zip_longest(maker, year, model)} 

    pprint(car_data) 
    pprint(car_data_longest) 

輸出：

{'Ford': ('1909', 'Model T'), 
'Horsey': ('1899', 'Horseless'), 
'Overland': ('1911', 'OctoAuto'), 
'Scripps-Booth': ('1913', 'Bi-Autogo')} 
{'Ford': ('1909', 'Model T'), 
'FutureX': ('20xx', None), 
'FutureY': (None, None), 
'Horsey': ('1899', 'Horseless'), 
'Overland': ('1911', 'OctoAuto'), 
'Scripps-Booth': ('1913', 'Bi-Autogo')} 

Answer 2

這是你尋找的東西嗎？

>>> car_dict = {maker[i]:[year[i], model[i]] for i in xrange(len(maker))} 
>>> car_dict 
{'Overland': ['1911', 'OctoAuto'], 'Horsey': ['1899', 'Horseless'], 'Scripps-Booth': ['1913', 'Bi-Autogo'], 'Ford': ['1909', 'Model T']} 

Answer 3

這聽起來像你想要的是：

{maker: (year, model) for maker, model, year in zip(car_maker,car_model,car_year)} 

，這將給你：

{'Horsey':(1899,'Horseless'),'Ford':(1909,'Model T'),'Overland':(1911,'OctoAuto')...} 

Answer 4

這個怎麼樣：

>>> maker =['Horsey', 'Ford', 'Overland', 'Scripps-Booth'] 
>>> year = ['1899', '1909', '1911', '1913'] 
>>> model = ['Horseless', 'Model T', 'OctoAuto', 'Bi-Autogo'] 
>>> d = dict(zip(maker,zip(year,model))) 
{'Overland': ('1911', 'OctoAuto'), 'Horsey': ('1899', 'Horseless'), 'Scripps-Booth': ('1913', 'Bi-Autogo'), 'Ford': ('1909', 'Model T')}