如何比较python中列表/集合的列表?-Howtocomparealistoflists/setsinpython?

29  

So you want the difference between two lists of items.

所以你想要两个列表的区别。

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]]

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:

首先，我将列表的每个列表都转换成一个元组列表，这样，tuple可以是可洗的(列表不是)，所以您可以将元组的列表转换为一组元组:

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

Then you can make sets:

然后你可以做集合:

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

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

编辑(sdolan建议):您可以在一行中为每个列表完成最后两个步骤:

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

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).

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(在我们的例子中是列表的列表)。

and find the symmetric difference between the sets:

并求出集合之间的对称差:

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

Note first_set ^ secnd_set is equivalent to symmetric_difference.

注意，first_set secnd_set等价于symmetric_difference。

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:

另外，如果您不想使用集合(例如，使用python 2.2)，那么就很简单了。例如,列表理解:

>>> [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]]

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

或者使用功能过滤器命令和lambda函数。(你必须测试这两种方式并结合)。

>>> 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]]

3  

Not sure if there is a nice function for this, but the "manual" way to do it isn't difficult:

不确定是否有一个很好的函数，但是“手动”的方法并不难:

differences = []

for list in firstList:
    if list not in secondList:
        differences.append(list)

2  

>>> 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', '3c3c3c', '3333'], ['Test4.doc', '4d4d4d', '4444']] 


>>> z = [tuple(y) for y in First_list]
>>> z
[('Test.doc', '1a1a1a', '1111'), ('Test2.doc', '2b2b2b', '2222'), ('Test3.doc', '3c3c3c', '3333')]
>>> x = [tuple(y) for y in Secnd_list]
>>> x
[('Test.doc', '1a1a1a', '1111'), ('Test2.doc', '2b2b2b', '2222'), ('Test3.doc', '3c3c3c', '3333'), ('Test4.doc', '4d4d4d', '4444')]


>>> set(x) - set(z)
set([('Test4.doc', '4d4d4d', '4444')])

1  

i guess you'll have to convert your lists to sets:

我猜你必须把你的列表转换成集合:

>>> a = {('a', 'b'), ('c', 'd'), ('e', 'f')}
>>> b = {('a', 'b'), ('h', 'g')}
>>> a.symmetric_difference(b)
{('e', 'f'), ('h', 'g'), ('c', 'd')}

0  

http://docs.python.org/library/difflib.html is a good starting place for what you are looking for.

http://docs.python.org/library/difflib.html是你寻找的一个好的起点。

If you apply it recursively to the deltas, you should be able to handle nested data structures. But it will take some work.

如果您将它递归地应用到deltas中，您应该能够处理嵌套的数据结构。但这需要一些工作。

0  

By using set comprehensions, you can make it a one-liner. If you want:

通过使用集合理解，您可以使它成为一行程序。如果你想要:

to get a set of tuples, then:

得到一组元组，然后:

Differences = {tuple(i) for i in First_list} ^ {tuple(i) for i in Secnd_list}

Or to get a list of tuples, then:

或者得到一个元组列表，然后:

Differences = list({tuple(i) for i in First_list} ^ {tuple(i) for i in Secnd_list})

Or to get a list of lists (if you really want), then:

或者得到一份清单(如果你真的想要的话)，那么:

Differences = [list(j) for j in {tuple(i) for i in First_list} ^ {tuple(i) for i in Secnd_list}]

PS: I read here: https://stackoverflow.com/a/10973817/4900095 that map() function is not a pythonic way to do things.

PS:我在这里读到:https://stackoverflow.com/a/10973817/4900095那个map()函数不是一个python的做事方法。