Question

我有这样的名单：

a = [('JoN', 12668, 0.0036), ('JeSsIcA', 1268, 0.0536), ('JoN', 1668, 0.00305), ('King', 16810, 0.005)]
b = [('JoN', 12668, 0.0036), ('JON', 16680, 0.00305), ('MeSSi', 115, 0.369)]

我希望结果列表如下：

result = [(('JoN', 12668, 0.0036), ('JoN', 12668, 0.0036)), (('JoN', 1668, 0.00305), ('JON', 16680, 0.00305)), (('King', 16810, 0.005), None), (None, ('MeSSi', 115, 0.369))]

我尝试过嵌套循环，集合，地图，zip但无法实现此输出。请帮助我。

Answer 1

首先使用第一个（使用a）和第三个项目作为键将b和str.lower()转换为词典，然后再循环使用列表中键的并集理解得到所需的输出：

>>> from pprint import pprint
>>> dct_a = {(x[0].lower(), x[2]): x for x in a}
>>> dct_b = {(x[0].lower(), x[2]): x for x in b}
>>> out = [(dct_a.get(k), dct_b.get(k)) for k in set(dct_a).union(dct_b)]
>>> pprint(out)
[(('JoN', 12668, 0.0036), ('JoN', 12668, 0.0036)),
 (('JoN', 1668, 0.00305), ('JON', 16680, 0.00305)),
 (('King', 16810, 0.005), None),
 (('JeSsIcA', 1268, 0.0536), None),
 (None, ('MeSSi', 115, 0.369))]

Answer 2

from string import lower
from itertools import groupby
from operator import itemgetter

def compose(f, g):
    def h(*args, **kwargs):
        return f(*g(*args, **kwargs))
    return h

def lower_first(*args):
    return (lower(args[0]),) + args[1:]

sorting_key = compose(lower_first, itemgetter(0, 2, 1))

grouping_key = compose(lower_first, itemgetter(0, 2))

output = [tuple(v) for k,v in groupby(sorted(a+b, key=sorting_key), 
                                      key=grouping_key)]

将output作为

[(('JeSsIcA', 1268, 0.0536),),
 (('JoN', 1668, 0.00305), ('JON', 16680, 0.00305)),
 (('JoN', 12668, 0.0036), ('JoN', 12668, 0.0036)),
 (('King', 16810, 0.005),),
 (('MeSSi', 115, 0.369),)]

然后添加None值非常简单：

final_output = [ elem if len(elem) >= 2 
    else ((None,)+ elem) if elem[0] not in a else elem + (None,) 
    for elem in output
]

给出：

[(('JeSsIcA', 1268, 0.0536), None),
 (('JoN', 1668, 0.00305), ('JON', 16680, 0.00305)),
 (('JoN', 12668, 0.0036), ('JoN', 12668, 0.0036)),
 (('King', 16810, 0.005), None),
 (None, ('MeSSi', 115, 0.369))]

但是你需要小心，因为用列表来陈述这样的问题通常会掩盖关系连接的问题，这些问题会由具有适当索引的系统来处理，例如pandas.DataFrame似乎更有可能由于其native join and merge capabilities.

，您需要的数据结构类型

Python - 通过检查它们的值来合并2个元组列表

2 个答案: