Question

好的，所以我最终会得到很多像这样的元组列表。

l1 = [('A', 1), ('B', 2), ('C', 3), ('D', 4), ('E', 5)]
l2 = [('A', 1), ('C', 2), ('F', 8), ('D', 15), ('X',21)]
l3 = [('C', 1), ('F', 4), ('R', 16), ('D', 19), ('Z', 31)]

每个列表中的位置已按数字排序。

我需要做的是构建一个元组列表，它将所有列表中的信息组合成这样的格式。

final_list = [('A', 1, 1, 0), ('B', 2, 0, 0), ('C', 3, 2, 1), ('D', 4, 15, 19), ('E', 5, 0, 0), ('F', 0, 8, 4), ('X', 0, 21, 0), ('R', 0, 0, 16), ('Z', 0, 0, 31)]

最重要的部分是元组本身按名称，列表1，列表2，列表3等顺序排列，当列表中没有字母/名称时，列表位置为零。

这是否太复杂了，我是否已将自己编入一个洞？

Answer 1

您可以分两步完成此操作：

首先构造一个dict，然后构造final_list

from collections import defaultdict

li = [l1, l2, l3, ...]
temp_dict = defaultdict(tuple)
for l in li:
    for e in l:
        temp_dict[e[0]] += (e[1],)

# Got {'A': (1, 1, 0), 'B': (2, 0, 0), ... }

final_list = [(key, ) + temp_dict[key] for key in temp_dict]

在REPL中尝试过：

>>> l1 = [('A', 1), ('B', 2), ('C', 3), ('D', 4), ('E', 5)]
>>> l2 = [('A', 1), ('C', 2), ('F', 8), ('D', 15), ('X',21)]
>>> l3 = [('C', 1), ('F', 4), ('R', 16), ('D', 19), ('Z', 31)]
>>> from collections import defaultdict
>>> li = [l1, l2, l3,]
>>> temp_dict = defaultdict(tuple)
>>> for l in li:
...     for e in l:
...             temp_dict[e[0]] += (e[1],)
...
>>> temp_dict
defaultdict(<type 'tuple'>, {'A': (1, 1), 'C': (3, 2, 1), 'B': (2,), 'E': (5,), 'D': (4, 15, 19), 'F': (8, 4), 'R': (16,), 'X': (21,), 'Z': (31,)})
>>> final_list = [(key, ) + temp_dict[key] for key in temp_dict]
>>> final_list
[('A', 1, 1), ('C', 3, 2, 1), ('B', 2), ('E', 5), ('D', 4, 15, 19), ('F', 8, 4), ('R', 16), ('X', 21), ('Z', 31)]

Answer 2

不，它并不太复杂，尽管您可能希望考虑使用字母键入的列表（或元组）的字典。 OTOH，你当前的设置比dict使用更少的内存。

#! /usr/bin/env python

l1 = [('A', 1), ('B', 2), ('C', 3), ('D', 4), ('E', 5)]
l2 = [('A', 1), ('C', 2), ('F', 8), ('D', 15), ('X',21)]
l3 = [('C', 1), ('F', 4), ('R', 16), ('D', 19), ('Z', 31)]


def main():
    dlist = [dict(u) for u in (l1, l2, l3)]

    keys = sorted(set(sum([d.keys() for d in dlist], [])))

    final = [(key,) + tuple([d.get(key, 0) for d in dlist]) for key in keys]

    for row in final:
        print row


if __name__ == '__main__':
    main()

Answer 3

再次更新！ 功能更加紧凑。

*

l1 = [('A', 1, 3), ('B', 2, 5), ('C', 3, 16), ('D', 4, 19), ('E', 5, 20)];
l2 = [('A', 1), ('C', 2), ('F', 8), ('D', 15), ('X',21)];
l3 = [('C', 1), ('F', 4), ('R', 16), ('D', 19), ('Z', 31)]

for k in l1,l2,l3:
    # 'dummy to keep track of tuple size'
    k.append(('dummy',len(k[0][1:])))

dict_lx =  [ dict(map(lambda x: (x[0], x[1:]), l)) for l in l1,l2,l3 ]
sorted_keys = sorted(set(reduce(lambda x,y : x + y.keys(), dict_lx, [])))

final = map(lambda key: tuple(
    [key] + reduce(lambda x,y: x + list(y.get(key, y.get('dummy')[0]*[0])), dict_lx, [])), sorted_keys)

final = filter(lambda x: x[0] != 'dummy', final)
print final
#[('A', 1, 3, 1, 0), ('B', 2, 5, 0, 0), ('C', 3, 16, 2, 1), ('D', 4, 19, 15, 19), ('E', 5, 20, 0, 0), ('F', 0, 0, 8, 4), ('R', 0, 0, 0, 16), ('X', 0, 0, 21, 0), ('Z', 0, 0, 0, 31)]

*

Python结合了多个元组列表

3 个答案: