如何比较python中的列表/集合列表?

2022-01-17 00:00:00 python list set tuples compare

问题描述

比较两个列表/集合并输出差异的最简单方法是什么?是否有任何内置函数可以帮助我比较嵌套列表/集合?

What is the easiest way to compare the 2 lists/sets and output the differences? Are there any built in functions that will help me compare nested lists/sets?

输入:

First_list = [['Test.doc', '1a1a1a', 1111], 
              ['Test2.doc', '2b2b2b', 2222],  
              ['Test3.doc', '3c3c3c', 3333]
             ]  
Secnd_list = [['Test.doc', '1a1a1a', 1111], 
              ['Test2.doc', '2b2b2b', 2222], 
              ['Test3.doc', '8p8p8p', 9999], 
              ['Test4.doc', '4d4d4d', 4444]]

预期输出:

Differences = [['Test3.doc', '3c3c3c', 3333],
               ['Test3.doc', '8p8p8p', 9999], 
               ['Test4.doc', '4d4d4d', 4444]]

解决方案

所以你想要两个项目列表之间的差异.

So you want the difference between two lists of items.

first_list = [['Test.doc', '1a1a1a', 1111], 
              ['Test2.doc', '2b2b2b', 2222], 
              ['Test3.doc', '3c3c3c', 3333]]
secnd_list = [['Test.doc', '1a1a1a', 1111], 
              ['Test2.doc', '2b2b2b', 2222], 
              ['Test3.doc', '8p8p8p', 9999], 
              ['Test4.doc', '4d4d4d', 4444]]

首先，我将每个列表列表转换为元组列表，因为元组是可散列的(列表不是)，因此您可以将元组列表转换为一组元组:

First I'd turn each list of lists into a list of tuples, so as tuples are hashable (lists are not) so you can convert your list of tuples into a set of tuples:

first_tuple_list = [tuple(lst) for lst in first_list]
secnd_tuple_list = [tuple(lst) for lst in secnd_list]

然后你就可以做套路了:

Then you can make sets:

first_set = set(first_tuple_list)
secnd_set = set(secnd_tuple_list)

编辑(由 sdolan 建议):您可以为单行中的每个列表完成最后两个步骤:

EDIT (suggested by sdolan): You could have done the last two steps for each list in a one-liner:

first_set = set(map(tuple, first_list))
secnd_set = set(map(tuple, secnd_list))

注意:map 是一个函数式编程命令，它将第一个参数中的函数(在本例中为 tuple 函数)应用于第二个参数中的每个项目(即在我们的例子中是一个列表的列表).

Note: map is a functional programming command that applies the function in the first argument (in this case the tuple function) to each item in the second argument (which in our case is a list of lists).

并找出集合之间的对称差:

and find the symmetric difference between the sets:

>>> first_set.symmetric_difference(secnd_set) 
set([('Test3.doc', '3c3c3c', 3333),
     ('Test3.doc', '8p8p8p', 9999),
     ('Test4.doc', '4d4d4d', 4444)])

注意 first_set ^ secnd_set 等价于 symmetric_difference.

此外，如果您不想使用集合(例如，使用 python 2.2)，它也很简单.例如，使用列表推导:

Also if you don't want to use sets (e.g., using python 2.2), its quite straightforward to do. E.g., with list comprehensions:

>>> [x for x in first_list if x not in secnd_list] + [x for x in secnd_list if x not in first_list]
[['Test3.doc', '3c3c3c', 3333],
 ['Test3.doc', '8p8p8p', 9999],
 ['Test4.doc', '4d4d4d', 4444]]

或使用功能性 filter 命令和 lambda 函数.(你必须测试两种方式并结合起来).

or with the functional filter command and lambda functions. (You have to test both ways and combine).

>>> filter(lambda x: x not in secnd_list, first_list) + filter(lambda x: x not in first_list, secnd_list)

[['Test3.doc', '3c3c3c', 3333],
 ['Test3.doc', '8p8p8p', 9999],
 ['Test4.doc', '4d4d4d', 4444]]

相关文章