按值加入列表

Question

是否有一种基于公共值在python中合并两个元组列表的有效方法.目前,我正在做以下事情:

name = [
         (9, "John", "Smith"),
         (11, "Bob", "Dobbs"),
         (14, "Joe", "Bloggs")
         ]

occupation = [
              (9, "Builder"),
              (11, "Baker"),
              (14, "Candlestick Maker")
              ]

name_and_job = []

for n in name:
    for o in occupation:
        if n[0] == o[0]:
            name_and_job.append( (n[0], n[1], n[2], o[1]) )


print(name_and_job)

收益:

[(9, 'John', 'Smith', 'Builder'), (11, 'Bob', 'Dobbs', 'Baker'), (14, 'Joe', 'Bloggs', 'Candlestick Maker')]

尽管此代码工作完全正常的小名单,这是令人难以置信的对数以百万计的记录更长的名单慢.有没有更有效的方式来写这个？

编辑第一列中的数字是唯一的.

编辑修改了@John Kugelman的代码.添加了一个get(),以防名称字典在职业字典中没有匹配的密钥:

>>>> names_and_jobs = {id: names[id] + (jobs.get(id),) for id in names}
>>>> print(names_and_jobs)
{9: ('John', 'Smith', None), 11: ('Bob', 'Dobbs', 'Baker'), 14: ('Joe', 'Bloggs', 'Candlestick Maker')}

Answer 1

使用词典而不是平面列表.

names = {
    9:  ("John", "Smith"),
    11: ("Bob", "Dobbs"),
    14: ("Joe", "Bloggs")
} 

jobs = {
    9:  "Builder",
    11: "Baker",
    14: "Candlestick Maker"
}

如果您需要将它们转换为此格式,您可以执行以下操作:

>>> {id: (first, last) for id, first, last in name}
{9: ('John', 'Smith'), 11: ('Bob', 'Dobbs'), 14: ('Joe', 'Bloggs')}
>>> {id: job for id, job in occupation}
{9: 'Builder', 11: 'Baker', 14: 'Candlestick Maker'}

然后将两者合并是一块蛋糕.

names_and_jobs = {id: names[id] + (jobs[id],) for id in names}