如何确定数字组合是否在另一个数字组合中？

Question

有没有办法找出数字组合（存储在列表中）是否在更长的数字组合中（存储在单独的列表中）？

E.g。

mylist = [(1, 4, 7), (3, 6, 9)]

serieslist = list(itertools.combinations((range(1, 50)), 5))
>> [(1, 2, 3, 4, 5), (1, 2, 3, 4, 6), (1, 2, 3, 4, 7)...etc...]

在上面的示例中，我想要在数字(1, 4, 7)的组合中返回数字(1, 2, 3, 4, 7) 的组合。

具体来说，我不想要将(1, 2, 3, 4, 7)分成三个组合的进一步组合。

理想情况下，我想将此内容写入for语句，以便将mylist的每个元素与serieslist的每个元素进行比较。

Answer 1

使用sets来查看你的元组是否是较大元组的一部分：

if set(short_tuple).issubset(longer_tuple):
    # all elements of short_tuple are in longer_tuple

您希望将short_tuple转换为一次：

for short_tuple in mylist:
    short_tuple_set = set(short_tuple)

    for combo in itertools.combinations((range(1, 50)), 5):
        if short_tuple_set.issubset(combo):
            # matched!

生成所有保证匹配的组合会更有效率：

for short_tuple in mylist:
    short_tuple_set = set(short_tuple)

    remainder = (i for i in range(1, 50) if i not in short_tuple_set)
    for combo in itertools.combinations(remainder, 5 - len(short_tuple)):
        combo = sorted(combo + short_tuple)

每个combo是1到49之间的5个数字的有效组合，其中包含short_tuple的所有3个数字，没有必须创建所有可能的组合

如果您将这些作为生成器函数创建，则可以验证它们是否生成相同的输出（除了元组与列表之外; sorted()返回列表）：

>>> def set_test(mylist):
...     for short_tuple in mylist:
...         short_tuple_set = set(short_tuple)
...         for combo in itertools.combinations((range(1, 50)), 5):
...             if short_tuple_set.issubset(combo):
...                 yield combo
... 
>>> def create_combos(mylist):
...     for short_tuple in mylist:
...         short_tuple_set = set(short_tuple)
...         remainder = (i for i in range(1, 50) if i not in short_tuple_set)
...         for combo in itertools.combinations(remainder, 5 - len(short_tuple)):
...             combo = sorted(combo + short_tuple)
...             yield combo
... 
>>> all(a == tuple(b) for a, b in itertools.izip_longest(set_test(mylist), create_combos(mylist)))
True

但第二种方法是所以更快：

>>> timeit('list(f(mylist))', 'from __main__ import set_test as f, mylist', number=10)
14.483195066452026
>>> timeit('list(f(mylist))', 'from __main__ import create_combos as f, mylist', number=10)
0.019912004470825195

是的，这几乎 1000次更快。

Answer 2

如果您的所有序列只包含唯一的数字（没有重复）：

a = (1,4,7)
b = (1,2,3,4,7)
a_in_b = all(x in b for x in a)