從列表的2個列表中明智地提取公共元素

Question

我有兩個列表，列表中有相同的len在Python中（對於這個例子來說是3）。

A = [['Horse','Duck','Goat'],['Rome','New York'],['Apple','Rome','Goat','Boat']] 

B = [['Carrot','Duck'],['Car','Boat','Plane'],['Goat','Apple','Boat']] 

我想匹配每行中的元素並創建一個新的公共元素列表。我需要得到的輸出是：

c = [['Duck'],[],['Apple','Goat','Boat']] 

，並

d = [1,0,3] ; where d is a list with the count of common elements at each row. 

注意列表清單的每一行中的元素可以以任意順序出現。

Answer 1

使用list comprehension和zip：

>>> A = [['Horse','Duck','Goat'],['Rome','New York'], 
     ['Apple','Rome','Goat','Boat']] 
>>> B = [['Carrot','Duck'],['Car','Boat','Plane'], 
     ['Goat','Apple','Boat']] 
>>> c = [[x for x in a if x in b] for a, b in zip(A, map(set, B))] 
>>> d = [len(x) for x in c] 
>>> # or d = list(map(len, c)) # you can omit `list` in python 2.x 
>>> c 
[['Duck'], [], ['Apple', 'Goat', 'Boat']] 
>>> d 
[1, 0, 3] 

Answer 2

替代列表理解：

c = [[x for x in y if x in B[i]] for i, y in enumerate(A)] 
# [['Duck'], [], ['Apple', 'Goat', 'Boat']] 

d = [len(x) for x in c] 
# [1, 0, 3] 

或者，你也可以使用：

res = [set(x) & set(y) for x, y in zip(A, B)] 
# or [set(x).intersection(y) for x, y in zip(A, B)] as @Chris_Rands suggested 
# [{'Duck'}, set(), {'Apple', 'Goat', 'Boat'}] 

通知這最後的格式一個不是你指定的那個，但是它使用set相交爲這些類型的操作構建的離子。