Question

我正在为我的作业制作心脏游戏，但我不知道如何将每个元素列入列表列表中：

>>>Cards = [[["QS","5H","AS"],["2H","8H"],["7C"]],[["9H","5C],["JH"]],[["7D"]]]

我想到的是：

for values in cards:
    for value in values:

但我想我的元素有2个列表。如何计算卡中有3个和1个列表的那个？

Answer 1

像这样：

>>> Cards = [[["QS","5H","AS"],["2H","8H"],["7C"]],[["9H","5C"],["JH"]],["7D"]]
>>> from compiler.ast import flatten
>>> flatten(Cards) 
['QS', '5H', 'AS', '2H', '8H', '7C', '9H', '5C', 'JH', '7D']

正如nacholibre指出的那样，compiler包已被弃用。这是flatten：

的来源

def flatten(seq):
    l = []
    for elt in seq:
        t = type(elt)
        if t is tuple or t is list:
            for elt2 in flatten(elt):
                l.append(elt2)
        else:
            l.append(elt)
    return l

Answer 2

稍微晦涩的oneliner：

>>> [a for c in Cards for b in c for a in b]
['QS', '5H', 'AS', '2H', '8H', '7C', '9H', '5C', 'JH', '7', 'D']

您可能想要给出a，b和c更具描述性的名称。

Answer 3

如果您的卡片以笨重的方式嵌套：

>>> Cards = [[["QS","5H","AS"],["2H","8H"],["7C"]],[["9H","5C"],["JH"]],["7D"]]
>>> def getCards(cardList,myCards=[]): #change this to myCards, and pass in a list to mutate this is just for demo
        if isinstance(cardList,list):
            for subList in cardList:
                getCards(subList)
        else:
            myCards.append(cardList)
        return myCards
>>> getCards(Cards)
['QS', '5H', 'AS', '2H', '8H', '7C', '9H', '5C', 'JH', '7D']

将递归遍历列表并查找所有元素。这些是我运行的时间比较所选flattern方法的性能：

>>> print(timeit.timeit(r'getCards([[["QS","5H","AS"],["2H","8H"],["7C"]],[["9H","5C"],["JH"]],["7D"]],[])',setup="from clas import getCards"))
5.24880099297
>>> timeit.timeit(r'flatten([[["QS","5H","AS"],["2H","8H"],["7C"]],[["9H","5C"],["JH"]],["7D"]])',setup="from compiler.ast import flatten")
7.010887145996094

Answer 4

您的列表是一个不完整的嵌套列表，因此您可以先将其设为矩形，using the procedure explained here，然后展平生成的numpy.ndarray。

如果最后一个元素['7D']是[['7D']]，则下面的“ifs”也是不必要的（那么其他答案也可以）。

import numpy as np
collector = np.zeros((3,3,3),dtype='|S20')

for (i,j,k), v in np.ndenumerate( collector ):
    try:
        if not isinstance(cards[i], str):
            if not isinstance(cards[i][j], str):
                collector[i,j,k] = cards[i][j][k]
            else:
                collector[i,j,0] = cards[i][j]
        else:
            collector[i,0,0] = cards[i]
    except IndexError:
        collector[i,j,k] = ''

print collector[collector<>''].flatten()

Answer 5

使用生成器，可以编写更加可读的flatten实现：

def flatten(l):
    if isinstance(l, list):
        for e1 in l:
            for e2 in flatten(e1):
                yield e2
    else:
        yield l

或者，如果您使用的是Python 3.3，它添加了yield from语法：

def flatten(l):
    if isinstance(l, list):
        for e in l:
            yield from flatten(e)
    else:
        yield l

结果：

>>> list(flatten([[["QS","5H","AS"],["2H","8H"],["7C"]],[["9H","5C"],["JH"]],[["7D"]]]))
['QS', '5H', 'AS', '2H', '8H', '7C', '9H', '5C', 'JH', '7D']

Answer 6

使用2个嵌套itertools.chain来展平列表：

In [32]: Cards
Out[32]: [[['QS', '5H', 'AS'], ['2H', '8H'], ['7C']], [['9H', '5C'], ['JH']], ['7D']]

In [33]: from itertools import chain

In [34]: [k for k in chain.from_iterable([i for i in chain.from_iterable(Cards)])]
Out[34]: ['QS', '5H', 'AS', '2H', '8H', '7C', '9H', '5C', 'JH', '7', 'D']

Answer 7

此解决方案对于任何类型的嵌套列表或元组都非常健壮（要添加其他可迭代类型，只需在下面的代码中添加更多or isinstance(...)。

它只是递归地调用一个展开自身的函数：

def unfold(lst):
    output = []
    def _unfold(i):
        if isinstance(i, list) or isinstance(i, tuple):
            [_unfold(j) for j in i]
        else:
            output.append(i)
    _unfold(lst)
    return output

print unfold(cards)
#['QS', '5H', 'AS', '2H', '8H', '7C', '9H', '5C', 'JH', '7D']

Answer 8

使用Rosetta Code中的Flatten a list，您可以这样做：

>>> def flatten(lst):
    return sum( ([x] if not isinstance(x, list) else flatten(x)
             for x in lst), [] )

>>> Cards = [[["QS","5H","AS"],["2H","8H"],["7C"]],[["9H","5C"],["JH"]],["7D"]]
>>> flatten(Cards)
['QS', '5H', 'AS', '2H', '8H', '7C', '9H', '5C', 'JH', '7D']
>>>

该解决方案仅展平嵌套列表 - 而不是元组或字符串。

Answer 9

from itertools import chain, imap

l= [[["QS","5H","AS"],["2H","8H"],["7C"]],[["9H","5C"],["JH"]],[["7D"]]]

k = list(chain.from_iterable(imap(list, l)))
m = list(chain.from_iterable(imap(list, k)))

print m

输出：['QS'，'5H'，'AS'，'2H'，'8H'，'7C'，'9H'，'5C'，'JH'，'7D']

Itertools很棒！

如何获取列表列表中的每个元素？

9 个答案: