将三个列表合并为一个字典

Question

我需要将三个列表合并为一个字典.这些列表来自读取我格式化的txt文件,这里是该文件的一个片段:

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

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

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

进入以下:

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

这是我做的:

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)

当我输入这个时,我得到:

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

并非显示列表中的所有数据,我不希望defaultdict显示.

当我尝试以下内容时:

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   ####

只有部分数据显示:

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

我该怎么办？

Answer 1

你走在正确的轨道上zip,但要注意:

返回的列表的长度被截断为最短参数序列的长度.

如果您对此感到满意,可以将数据压缩到元组列表中,压缩键,然后将所有内容都移到dict().

如果你想处理缺失值,结账itertools izip_longest(Python的2)或zip_longest(Python 3中),其中

如果迭代的长度不均匀,则使用fillvalue填充缺失值.

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)

输出:

{'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

这个怎么样:

>>> 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')}