我在 Python 中有两个具有相同 len 的列表列表(在这个例子中假设为 3)。
A = [['Horse','Duck','Goat'],['Rome','New York'],['Apple','Rome','Goat','Boat']]
B = [['Carrot','Duck'],['Car','Boat','Plane'],['Goat','Apple','Boat']]
我想匹配每一行中的元素并创建一个新的公共(public)元素列表。我需要的结果输出是:
c = [['Duck'],[],['Apple','Goat','Boat']]
和，
d = [1,0,3] ; where d is a list with the count of common elements at each row.
请注意，在列表列表的每一行中，元素可以以任何顺序出现。
请您参考如下方法:
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]
