将两个列表平均划分到其他3个列表上

Question

我正在努力想出一个有效但简单的解决方案来解决以下问题：

我有两张单子：

list_dicts_1 = [
{"name": "Suarez", "footed": "right-footed", "color": "black"}
{"name": "Suarez2", "footed": "right-footed2", "color": "black2"}
{"name": "Suarez3", "footed": "right-footed3", "color": "black3"}
{"name": "Suarez4", "footed": "right-footed4", "color": "black4"}
{"name": "Suarez5", "footed": "right-footed5", "color": "black5"}
{"name": "Suarez6", "footed": "right-footed6", "color": "black6"}
]


list_dicts_2 = [
{"name": "Coutinho", "footed": "left-footed", "color": "orange"}
{"name": "Coutinho2", "footed": "left-footed1", "color": "orange2"}
{"name": "Coutinho3", "footed": "left-footed2", "color": "orange3"}
{"name": "Coutinho4", "footed": "left-footed4", "color": "orange4"}
{"name": "Coutinho5", "footed": "left-footed5", "color": "orange5"}
{"name": "Coutinho6", "footed": "left-footed6", "color": "orange6"}
]

我想迭代这些dict列表并将它们分配给3个空列表：

list_1 = []
list_2 = []
list_3 = [] 

期望输出：

list_1 = [
{"name": "Suarez", "footed": "right-footed", "color": "black"}, 
{"name": "Suarez4", "footed": "right-footed4", "color": "black4"},
{"name": "Coutinho", "footed": "left-footed", "color": "orange"},
{"name": "Coutinho4", "footed": "left-footed4", "color": "orange4"}
]        

list_2 = [
{"name": "Suarez2", "footed": "right-footed2", "color": "black2"}, 
{"name": "Suarez5", "footed": "right-footed5", "color": "black5"},
{"name": "Coutinho2", "footed": "left-footed2", "color": "orange2"},
{"name": "Coutinho5", "footed": "left-footed5", "color": "orange5"}
]

list_3 = [
{"name": "Suarez3", "footed": "right-footed3", "color": "black3"}, 
{"name": "Suarez6", "footed": "right-footed6", "color": "black6"},
{"name": "Coutinho3", "footed": "left-footed3", "color": "orange3"},
{"name": "Coutinho6", "footed": "left-footed6", "color": "orange6"}
]

我想把这三张单子平分在三张空单子上。字典列表中的每一项只能在空列表中出现一次。所以字典列表的第一行应该放在列表1中。然后第二行的目录应该放在目录2中，以此类推，直到目录里什么都没有了。你知道吗

有没有一个简单的方法来实现这一点？你知道吗

Answer 1

迭代器在这里非常有用：您可以使用^{}获得一个迭代器，该迭代器在list_1、list_2和list_3上无限循环。然后简单地迭代list_dicts_1和list_dicts_2中的dict并将它们附加到从cycle获得的列表中：

import itertools

list_1, list_2, list_3 = lists = [], [], []
# make an iterator that endlessly loops over list_1, list_2 and list_3
itr = itertools.cycle(lists)

# loop over the dicts
for dic in itertools.chain(list_dicts_1, list_dicts_2):
    # and append the dict to the next list
    next(itr).append(dic)

我使用^{}将list_dicts_1和list_dicts_2组合成一个iterable，并使用^{}函数手动推进cycle迭代器。你知道吗

结果：

>>> list_1
[{'name': 'Suarez', 'footed': 'right-footed', 'color': 'black'},
 {'name': 'Suarez4', 'footed': 'right-footed4', 'color': 'black4'},
 {'name': 'Coutinho', 'footed': 'left-footed', 'color': 'orange'},
 {'name': 'Coutinho4', 'footed': 'left-footed4', 'color': 'orange4'}]
>>> list_2
[{'name': 'Suarez2', 'footed': 'right-footed2', 'color': 'black2'},
 {'name': 'Suarez5', 'footed': 'right-footed5', 'color': 'black5'},
 {'name': 'Coutinho2', 'footed': 'left-footed1', 'color': 'orange2'},
 {'name': 'Coutinho5', 'footed': 'left-footed5', 'color': 'orange5'}]
>>> list_3
[{'name': 'Suarez3', 'footed': 'right-footed3', 'color': 'black3'},
 {'name': 'Suarez6', 'footed': 'right-footed6', 'color': 'black6'},
 {'name': 'Coutinho3', 'footed': 'left-footed2', 'color': 'orange3'},
 {'name': 'Coutinho6', 'footed': 'left-footed6', 'color': 'orange6'}]

---

另一种选择是将两个输入列表串联成一个巨大的列表，然后对其进行切片：

list_ = list_dicts_1 + list_dicts_2
list_1 = list_[::3]
list_2 = list_[1::3]
list_3 = list_[2::3]

但是，串联列表的成本有点高，因此如果列表非常大，应该避免这种情况。你知道吗

Answer 2

可以使用while循环：

while True:
    try:
        list_1.append(dicts[0])
        del dicts[0]

        list_2.append(dicts[0])
        del dicts[0]

        list_3.append(dicts[0])
        del dicts[0]

    except IndexError:
        break

编辑：

也可以使用for循环完成！你知道吗

for i in range(len(dicts)//3):
     list_1.append(dicts[0])
     del dicts[0]
     list_1.append(dicts[0])
     del dicts[0]
     list_1.append(dicts[0])
     del dicts[0]

编辑2:

它可以做得更好的方式，仍然很容易理解一个初学者

for i in range(len(dicts)//3):
    list_1.append(dicts[i*3])
    list_1.append(dicts[i*3+1])
    list_1.append(dicts[i*3+2])