Question

我想在两个python_lists之间执行完全外部联接。考虑它们都有相同的键：（名称和宠物）。除此之外，它们具有不同的金额，两者都应显示在result_list中。如何在python中实现？

list_a = [['John','Cat',10], ['Adam','Dog',20]]
list_b = [['John','Cat',40], ['Laura','Fish',60]]
result_list = [['John','Cat',10, 40], ['Adam','Dog',20,0], ['Laura','Fish',0, 60]]]

Answer 1

您可以使用字典来更有效地执行字典操作，该方法可以比较每对列表。

list_a = [['John','Cat',10], ['Adam','Dog',20]]
list_b = [['John','Cat',40], ['Laura','Fish',60]]

d = {}

for person, pet, v in list_a + list_b:
    if (person, pet) in d:
        d[(person, pet)]+=v
    else:
        d[(person, pet)]=v

final_list = [[key[0],key[1],val] for key,val in d.items()]

输出

[['John', 'Cat', 50], ['Adam', 'Dog', 20], ['Laura', 'Fish', 60]]

注意到这不是OP要求的，然后

编辑。

对于预期的输出，代码可以更改为：

list_a = [['John','Cat',10], ['Adam','Dog',20]]
list_b = [['John','Cat',40], ['Laura','Fish',60]]

d = dict(((person,pet),[v,0]) for (person,pet,v) in list_a)

for person, pet, v in list_b :
    if (person, pet) in d:
        d[(person, pet)][1] = v
    else:
        d[(person, pet)] = [0,v]


final_list = [[key[0],key[1],val[0],val[1]] for key,val in d.items()]
#python3.x
final_list = [[*key,*val] for key,val in d.items()]

输出：

[['John', 'Cat', 10, 40], ['Adam', 'Dog', 20, 0], ['Laura', 'Fish', 0, 60]]

Answer 2

有点傻，但是一种直接的方法是从dict逐步构建它：

d = {}
for name, pettype, val in list_a:
    d[name, pettype] = [val, 0]

for name, pettype, val in list_b:
    try:
        d[name, pettype][1] = val
    except KeyError:
        d[name, pettype] = [0, val]

# Use .items() instead of .viewitems() on Python 3.x
result_list = [[name, pettype, vala, valb] for (name, pettype), (vala, valb) in d.viewitems()]

在Python 3.6之前，顺序不一致；在3.6及更高版本中，您将依次看到list_a中的所有名称/宠物，然后依次查看list_b中未出现在list_a中的所有名称/宠物。

如果您真的讨厌重复自己，或者需要加入许多此类列表，则可以将其调整为：

d = {}
all_lists = [list_a, list_b, ..., list_n]
for i, curlist in enumerate(all_lists):
    for name, pettype, val in curlist:
        try:
            vals = d[name, pettype]
        except KeyError:
            vals = d[name, pettype] = [0] * len(all_lists)
        vals[i] = val
result_list = [[name, pettype] + vals for (name, pettype), vals in d.viewitems()]

或使用collections.defaultdict来避免异常处理代码：

from collections import defaultdict

all_lists = [list_a, list_b, ..., list_n]
d = defaultdict(lambda: [0] * len(all_lists))
for i, curlist in enumerate(all_lists):
    for name, pettype, val in curlist:
        d[name, pettype][i] = val
result_list = [[name, pettype] + vals for (name, pettype), vals in d.viewitems()]

在python 2.7中加入两个列表列表

2 个答案: