Python的 - 比較的套

Question

兩個列表我有兩個列表：

list1 = [ 
    set(['3105']), 
    set(['3106', '3107']), 
    set(['3115']), 
    set(['3122']), 
    set(['3123', '3126', '286']) 
] 

和

list2 = [ 
    set(['400']), 
    set(['3115']), 
    set(['3100']), 
    set(['3107']), 
    set(['3123', '3126']) 
] 

如何比較這些列表的交集，因此，舉例來說，如果3126是某處任何這兩個列表的集合，它將追加另一個列表與3126.我的最終目標是追加一個單獨的列表，然後採取列表的長度，所以我知道列表之間有多少匹配。

Answer 1

您可以拼合的套兩個列表成爲集：

l1 = set(s for x in list1 for s in x) 
l2 = set(s for x in list2 for s in x) 

然後你就可以計算交會：

common = l1.intersection(l2) # common will give common elements 
print len(common) # this will give you the number of elements in common. 

結果：

>>> print common 
set(['3123', '3115', '3107', '3126']) 
>>> len(common) 
4 

Answer 2

你不得不合並所有集合;採取在兩個列表集合的聯合，再取這兩個工會的交集：

sets_intersection = reduce(set.union, list1) & reduce(set.union, list2) 

if 3126 in sets_intersection: 
    # .... 

Answer 3

>>> common_items = set().union(*list1) & set().union(*list2) 
>>> common_items 
set(['3123', '3115', '3107', '3126']) 
>>> '3126' in common_items 
True 

時序比較：

>>> %timeit reduce(set.union, list1) & reduce(set.union, list2) 
100000 loops, best of 3: 11.7 us per loop 
>>> %timeit set().union(*list1) & set().union(*list2)  #winner 
100000 loops, best of 3: 4.63 us per loop 
>>> %timeit set(s for x in list1 for s in x) & set(s for x in list2 for s in x) 
10000 loops, best of 3: 11.6 us per loop 
>>> %timeit import itertools;set(itertools.chain.from_iterable(list1)) & set(itertools.chain.from_iterable(list2)) 
100000 loops, best of 3: 9.91 us per loop