Question

我有一个问题。我使用组合从列表中获得不同的组合。但是，如果我希望得到更具体的升序（3D，4D，5D）组合，而不是（4D，5D，8D），我该怎么做？

例如：

from itertools import combinations

cards = ["3D", "4D", "5D", "6D"]
comb = []
for j in combinations(cards, 3):
    comb.append(list(j))
for j in combinations(cards, 4)):
    comb.append(list(j))

但我得到的输出如下：

["3D", "4D", "5D"], ["3D", "4D", "6D"], ["3D", "5D", "6D"], ["4D", "5D", "6D"], ["3D", "4D", "5D", "6D"]

如何获得这样的输出？

[["3D", "4D", "5D"], ["4D", "5D", "6D"], ["3D", "4D", "5D", "6D"]]

Answer 1

逐步浏览列表并将每个连续3个元素压缩在一起。然后在最后添加完整的卡片列表。

print ([list(t) for t in zip(*(cards[i:] for i in xrange(3)))]+[cards])
[['3D', '4D', '5D'], ['4D', '5D', '6D'], ['3D', '4D', '5D', '6D']]

如果您想在较长的列表和第5组元素上使用它，您可以更改xrange to xrange(5)：

cards = ["3D", "4D", "5D", "6D","7D","8D","9D"]
print ([list(t) for t in zip(*(cards[i:] for i in xrange(5)))]+[cards])
[['3D', '4D', '5D', '6D', '7D'], ['4D', '5D', '6D', '7D', '8D'], ['5D', '6D', '7D', '8D', '9D'], ['3D', '4D', '5D', '6D', '7D', '8D', '9D']]

如果您的卡片列表未按排序顺序排列，您可以在list之前调用sort方法zip：

cards = ["5D", "6D","3D", "4D"]    
cards.sort()
print [list(t) for t in zip(*(cards[i:] for i in xrange(3)))]+[cards]#
[['3D', '4D', '5D'], ['4D', '5D', '6D'], ['3D', '4D', '5D', '6D']]

Answer 2

使用以下代码：

from itertools import combinations

def check(lst):
        for i in range(len(lst)):
            try:
                    if lst[i+1]-lst[i] != 1:
                            return False
            except IndexError:
                    pass
        return True

cards = ["3D", "4D", "5D", "6D"]
comb = []
for j in combinations(cards, 3):
    comb.append(list(j))
for j in combinations(cards, 4):
    comb.append(list(j))

subs = []
for i in comb:
        subs.append([int(item[0]) for item in i])

subs = [item for item in subs if check(item) == True]

comb = []

for i in subs:
        comb.append([str(item)+'D' for item in i])

print comb

此代码使用以下代码将每个子列表转换为整数列表：

>>> item = ['3D', '4D', '6D']
>>> [int(val[0]) for val in item]
[3, 4, 6]
>>>

然后我们确保这些数字直接上升：

>>> def check(lst):
...         for i in range(len(lst)):
...             try:
...                     if lst[i+1]-lst[i] != 1:
...                             return False
...             except IndexError:
...                     pass
...         return True
... 
>>> check([3, 4, 6])
False
>>> check([3, 4, 5])
True
>>> check([3, 4, 5, 6])
True
>>> check([3, 4, 5, 8])
False
>>>

然后使用上面的逻辑，我们过滤掉所有内容，给出以下内容。

运行：

bash-3.2$ python check.py
[['3D', '4D', '5D'], ['4D', '5D', '6D'], ['3D', '4D', '5D', '6D']]
bash-3.2$

Itertools，组合

2 个答案:

运行：