Python 比较单词并删除两个列表中的重复项

Question

I have the following lists:

list_1_test = [['hi','there','how'],['we','are','one']]
list_2_test = [['hi','you','how'],['we','not','one']]

I wish to compare the words in the list and get the following output:

list_3_test = [['there','you'],['are','not']]

I know how to do this in a simple list, for example:

list_1_test = ['hi','there','how']
list_2_test = ['hi','you','how']

list_3_test=[]
for i in list_1_test:
    if i not in list_2_test:
        list_3_test.append(i)

for i in list_2_test:
    if i not in list_1_test:
        list_3_test.append(i)   

and the result is

['there', 'you']

But when it comes to list of lists, my brain is fried. The order matters. Any help is much appreciated.

Answer 1

If order doesn't matter, you can use set operations + list comprehension:

out = [list(set(l1).union(l2) - set(l1).intersection(l2)) for l1, l2 in zip(list_1_test, list_2_test)]

Output:

[['you', 'there'], ['are', 'not']]

If order matters, you can use dict.fromkeys :

out = []
for l1, l2 in zip(list_1_test, list_2_test):
    one = dict.fromkeys(l1).keys()
    two = dict.fromkeys(l2).keys()
    out.append(list(one - two) + list(two - one))

Output:

[['there', 'you'], ['are', 'not']]

Answer 2

You can use the sets and symetric_difference method (^):

list3 = []
for l1, l2 in zip(list_1_test, list_2_test):
    sym_dif = set(l1)^set(l2)
    list3.append(list(sym_dif))

Previous code in list comprehension format:

list3 = [set(l1)^set(l2) for l1, l2 in zip(list_1_test, list_2_test)]

Answer 3

all_list = []
for i in list_1_test:    
    all_list.append(' '.join(i))

for i in list_2_test:
    all_list.append(' '.join(i))

list_str = ' '.join(all_list)    

result = []
for i in list_str.split():
    if list_str.count(i) == 1:
        result.append(i)
        
print(result)