Python - 在理解中比较两个列表

Question

我试图理解理解是如何运作的.

我想循环遍历两个列表,并比较每个列表以找出差异.如果一个/多个单词不同,我想打印这个单词.

我喜欢这一切的代码,这就是为什么我对理解感兴趣.

Answer 1

在"一个很好的代码行"中执行它是代码高尔夫,并且被误导.让它变得可读.

for a, b in zip(list1, list2):
    if a != b:
       print(a, "is different from", b) 

这在任何重要方面都没有区别:

[print(a, "is different from", b) for a, b in zip(list1, list2) if a!=b]

除了扩展版本比理解更容易阅读和理解.

Answer 2

像kriegar建议使用套装可能是最简单的解决方案.如果你绝对需要使用列表理解,我会使用这样的东西:

list_1 = [1, 2, 3, 4, 5, 6]
list_2 = [1, 2, 3, 0, 5, 6]

# Print all items from list_1 that are not in list_2 ()
print(*[item for item in list_1 if item not in list_2], sep='\n')

# Print all items from list_1 that differ from the item at the same index in list_2
print(*[x for x, y in zip(list_1, list_2) if x != y], sep='\n')

# Print all items from list_2 that differ from the item at the same index in list_1
print(*[y for x, y in zip(list_1, list_2) if x != y], sep='\n')

Answer 3

如果您想比较两个列表的差异，我认为您想使用set.

s.symmetric_difference(t)   s ^ t   new set with elements in either s or t but not both

例子：

>>> L1 = ['a', 'b', 'c', 'd']
>>> L2 = ['b', 'c', 'd', 'e'] 
>>> S1 = set(L1)
>>> S2 = set(L2)
>>> difference = list(S1.symmetric_difference(S2))
>>> print difference
['a', 'e']
>>> 

一行形式？

>>> print list(set(L1).symmetric_difference(set(L2)))
['a', 'e']
>>> 

如果你真的想使用列表理解：

>>> [word for word in L1 if word not in L2] + [word for word in L2 if word not in L1]
['a', 'e']

随着列表大小的增长，效率会大大降低。